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Table of contents :
Notes
Classical Mechanics
1 Newton basic
2 NM advanced
3 Rigid body basic
4 Central Force
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5 Rigid body advanced
6 Lagrangian mechanics
7 Hamilton Mechanics
8 small oscillations
Electrodynamics
Introduction to EMT
Electrostatic
Dielectrics
Magnetostatic
Electromagnetic Induction
Maxvels equations
Electromagnetic waves
Relativistic ED and Radiation
Mathamatical methode
calculas
Linear algebra
vectors
Infinite Series
DE and special functions
integral transformation
complex analysis
advanced topics
Quantum mechanics
Old Quantum theory
Mathamatical background
Basic Postulates
One Diamentional Potential
Angular Momentum
Identical Particles
Perturbation Variation
WKB And Scattering
Thermodynamics and statistics
Basic thermodynamics
Thermodynamic Potential and Applications
Basic Consepts of Statical Mechanic
Random Walk and Probability
Ensembles in Statical Mechanics
Quantum Statistical Mechanics
Ideal Fermi system
Ideal Bose System
Questions
CM
Basic Newtonian pt 1
Basic Newtonian pt 2
Angular momentum of a rigid body pt1
Angular momentum of a rigid body pt2
advanced Newtonian
central force
Lagrangian
Hamilton and phase space
canonical transformation
small oscilations
ED
Electrostatic 1
Electronics 2
Capacitor Dielectric
Magnetostatic
Magnetostatic 2
Electromagnetic Induction
Maxwell's Equations
ED Advanced
Electromagnetic wave
mathamatical method
vector
Linear Algebra
DE and Special functions
integral transformation
The Complex Analysis
Probability
Numerical methode Group theory Tensor calculas Greens functions
QM
Old Quantum Theory
basic formulation pt1
basic formulation pt2
1D Potential pt1
1D Potential pt2
3D System
Angular Momentum
Approximation methode 1
Approximation methode 2
Thermodynamics and statistics
Kinetic theory
Second law
Maxvells relation
Equation of state
Entropy Calculation
Phase transition
Microcanonical enseble
Canonical Ensemble pt1
Canonical Ensemble pt2
Quantum statistical Mechanics
Two state and Other systems
Ideal Fermi system
Ideal Bose System
Exam
CM
EMT
Citation preview
11 oo : oo : oo
N e-wtonian Mechanics: Basics Sk J ahiruddin*
*Assistant Professor Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay M.Sc Physics 2009-2011 batch He ranked 007 in IIT J AM 2009 and 008 (JRF) in CSIR NET June 2011 He has been teaching CSIR NET aspirants since 2012
1 @Sk J ahiruddin, 2020
Newtonian Mechanics: Basics
11 oo : oo : 03 1 @Sk J ahiruddin , 2020
Newtonian Mechanics: Basics
Contents 1 Basic Concepts
5
1.1
Mot ion in plane polar coordinat e
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1.2
velocity and acceleration in polar coordinat e .
12
1.3
Basic kinematic equations . . . . . . . . . . .
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2 Force and Newton's Laws
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26
2.1
Constant force, free body diagram, friction . .
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2.2
Solving differential equation, Dissipative force
51
2.3
Simple Harmonic Motion . . . . . . . . . . . .
65
Conservation of Momentum and Energy
78
3.1
Center of mass
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3.2
Center of m ass frame
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3.3
conservation of moment um .
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3.4
3.5
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Newtonian Mechanics: Basics
Conservation of Energy in one and two dimension1 16 3.4. 1
Power . . . . . . . . . . . . . .
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3.4.2
Gravitational potential energy:
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3.4.3
Potent ial energy of a spring
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3.5. 1
Collision in one dimension .
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3.5.2
1D collision in COM fr ame .
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3.5.3
Collision in two dimensions
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Collisions
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Variable mass problem 4.1
Rocket P roblem . . . 4.1.1 4.1.2
4. 2
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Derivation of equation of motion
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Rocket motion when no external force is present
4. 1.3
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Rocket in gravitational field
Moment um flux . . . . . . . . . . .
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11 oo : oo : oa 4.1.3 4.2
Rocket in gravitational field
Momentum flux . . . .
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Newtonian Mechanics: Basics
Chain problems . . . . . . . . . . . . . . . . .
157
4.3.1
Equation of motion of a falling chain .
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4.3.2
Chain falling on a weight machine . . .
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Other problems on variable mass . .
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4.4.1
Cart with leaking water tank
4.4.2
Sand on cart . . .
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4.4.3
Growing raindrop
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Newtonian Mechanics: Basics
Basic Concepts
Before we start our discussion on Basic Newtonian mechanics Let's revise our concepts of dimensional analysis first . You know t hat M stand for mass, L for length, and T for t ime. For example, we'll write a speed as [v] = L / T and t he gravitational constant as [G] = £ 3 / (MT 2 ) You can figure t his out by not ing that Gm 1m 2 / r 2 has the dimensions of force, which in t urn has dimensions M L / T 2 , from F =ma. . Example 1.1: A mass m hangs from a massless string of length f and swings back and forth in t he plane of the
paper. The acceleration due to gravity is g. What can we say about t he frequency of oscillations? The only quantit ies having dimension given in the problem 2 are [m] = M [f] = L , and [g] = L / T . But there is one more quantity, t he maximum angle 00 which is dimensionless (and easy to for get) . Our goal is to find t he frequency, which has units of 1/T. The only combination of our given dimensionful quantities that has units of 1/ T is g/f. But we can't rule
11 oo : oo : 13 are [m] = M [£] = L , and [g] = L / T 2 . But there is one more quantity, t he maximum angle 00 which is dimensionless (and easy t o forget ). Our goal is to find t he frequency, which has units of 1/T. The only combination of our given dimensionful quantities t hat has units of 1/ T is g/£. But we can't rule out any 00 dependence, so the most general possible form of
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the frequency is w
g
= f (0o)
£
where f is a dimensionless function of the dimensionless variable 0o More general method would be to t ake the dimensionful quantities raised t o arbit rary powers (ma£bgc in this problem) , and t hen writ e out the units of this product in t erms of a, b, and c. If we want to obtain units of 1/T here, then we need
M aLb
L
c
T2
l T
Matching up the powers of the three kinds of units on each side of this equation gives
M : a = 0,
L : b + c = 0,
T : - 2c = - 1
The solution to this syst em of equations is a = 0, b = - 1/ 2, and c = 1/ 2, so we have reproduced the -Jg/i result. Example 1.2: How does the speed of waves in a fluid
11 oo : oo : 16 The solution to this system of equations is a = 0, b = - 1/ 2, and c = 1/ 2, so we have reproduced the ~ result. Example 1.2: How does the speed of waves in a fluid depend on its density, p, and " bulk modulus," B (which
has units of pressure, which is force per area)? Answer with dimensional analysis
[v] = L / T [email protected]
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[p]=M/L
3
and
Tal vdv = adx > v = v5 + 2ax T hese equations are also written as in your + 2 textbooks
(i) v=u +at 2
(ii) s = ut + 1/2at 2 2 (iii) v = u + 2as (iv)
s=
1 (u + v)t 2
00: 00: 54 2
(ii) s = ut + 1/ 2at 2 2 (iii) v = u + 2as 1 (u + v)t 2
s=
(iv)
Example 1.5: From an elevated p oint A , a st one is project ed vertically upward . When the stone reaches a distance h below A, its velocity is double of what it was at a height h above A. Show that the greatest height obtained by the stone above A is 5h/ 3 [email protected]
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Solution: When t he stone reaches a height h above A
vf = u
2
-
2gh
and when it reaches a dist ance h below A
v~ = u
2
+ 2gh
since the velocity of the stone while crossing A on its return journey is again u vertically down. Also, v2
= 2v1 (by problem) Combining all u
2
=
10 - gh 3
Maximum height u2
H =2g
10 gh 3 2g
Example 1.6: The relation 3t
5h 3
= v'3x + 6 describes the
11 oo :oo : s1 u2
10 gh
5h
2g
3 2g
3
H =-
Example 1.6: The relation 3t = v'3x + 6 describes t he displacem ent of a particle in one direction, where x is in m etres and t in seconds. Find the displacement when t he velocity is zero. Solution: ~=3t-6 Squaring and simplifying x [email protected]
= 3t2 -
12t
+ 12
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@Sk Jahiruddin, 2020
= ~ = 6t - 12 dt v = 0 gives t = 2s
V
Thus we get displacem ent x
=0
projectile motion: Consider a ball thrown t hrough the air, not necessarily vertically. Let x and y be t he horizontal and vertical positions, respectively. The force in the x direction is Fx
=
0, and the
force in t hey direction is Fy = -mg. So F = m a gives
x= O '
an d
••
y
=
-g
(1.11 )
If t he initial position and velocity are (X , Y) and (Vx, Vy) , t hen we can easily integrat e the above equation to obtain (1.12)
11 oo : 01 : oo If t he initial position and velocity are (X , Y) and (Vx, Vy) , t hen we can easily integrat e the above equation to obtain (1.12)
Integrating again we get
and
y(t) = Y
+ Vyt -
1
gt
2
2
(1.13)
These equations for the speeds and positions are all you need to solve a projectile problem.
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Newtonian Mechanics: Basics
Example 1.7: (a) For a given init ial speed, at what inclination angle should a ball be thrown so that it t ravels
the maximum horizontal distance by the time it returns to the ground? Assume that the ground is horizontal, and t hat the ball is released from ground level. (b) What is t he optimal angle if the ground is sloped upward at an angle /3 (or downward , if /3 is negative)?
Solution: (a) Let the inclination angle be 0, and let the initial speed be V. Then t he horizontal speed is always Vx = V cos 0, and t he init ial vertical speed is Vy = V sin 0 The first thing we need to do is find t he time t in the air. We know that the vertical speed is zero at time t / 2, because the ball
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initial speed be V. Then t he horizontal sp eed is always Vx = V cos 0, and t he init ial vertical speed is Vy = V sin 0 The first thing we need to do is find t he time t in the air. We know that the vertical speed is zero at time t / 2, b ecause the ball is moving horizontally at t he highest point. So t he second of equation (1.12) gives Vy= g(t/2). Therefore, t = 2Vy/g. 17 The first of equation (1.13) tells us t hat the horizontal distance t raveled is d = Vxt. Using t = 2Vy / g in this gives The sin 20 factor has a maximum at 0
=
1r /
mum horizontal dist ance traveled is t hen dmax
4. The maxi-
=V
2
/
g
(b) As in part (a), t he first t hing we need to do is find t he t ime t in the air. If the ground is sloped at an angle f3, t hen the equation for the line of the ground is y = (tan /3)x . The path of the ball is given in terms oft by
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Newtonian Mechanics: Basics
x = (V cos 0)t,
and
y
.
1
= (V Sln 0)t - 2gt
2
(1.14)
where 0 is t he angle of the throw, as measured with respect to t he horizontal (not t he ground). We must solve for t he t t hat makes y = (tan f3)x , because t his gives the t ime when t he path of t he ball intersects t he line of the ground. Using equation (1.14) we find t hat y
t=
= (tan /3)x when
2V. -(sin 0 - tan /3 cos 0) g
(1.15)
11 oo : 01 : os equation (1.14) we find that y
t=
= (tan j1)x when
2V. - (sin 0 - tan /1 cos 0)
(1.15)
g
(There is, of course, also the solution t = 0.) Plugging this into the expression for x in equation (1.14) gives
2v x = g
2
2
(sin 0 cos 0 - tan (3 cos 0)
(1.16)
We must now maximize this value for x, which is equivalent to m aximizing t he distance along t he slope. Setting the derivative wit h respect to 0 equal to zero, and using the double-angle formulas, sin 20 = 2 sin 0 cos 0 and cos 20 = cos2 0 - sin2 0, we find t an (3 = - cot 20. This can be rewritt en as t an j1 we have
= -
tan (1r / 2 - 20). Therefore, j1
1 0= 2 jahir@p hysicsguide .in
@Sk Jahir uddin, 2020
7f
/1+ -
2
24
= -(1r / 2
- 20) , so
( 1.17) P hysicsguide
Newtonian Mechanics: Basics
In ot her words, t he throwing angle should bisect t he angle b etween the ground and the vertical.
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Force and Newton's Laws
Det ailed discussion on Newton's Laws can be found in any good elementary mechanics book. All you need to know is the equation (2.1 ) F = ma and its applications.
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good elementary mechanics book. All you need to know is t he equation F = ma (2.1 ) and its applications. T here are many problems associated wit h Newton 's laws. They can be divided into two broad cat egories. First , are t he problems where forces are constant , where you need to draw t he free body diagram. Second, are t he problems where force varies wit h t ime and you need t o solve a differential equation.
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Constant force, free body diagram, friction
Example 2.1: Three freight cars each of mass M are pulled with force F by an engine as shown in figure.
Friction is negligible. Find t he forces on each car .
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tion Example 2.1: Three freight cars each of mass M are pulled with force F by an engine as shown in figure.
Frict ion is negligible. Find t he forces on each car. M
M
M
• a
J_J I LJCJ I LJ!..J I LJ : 3
2
1
Figure 2.1:
Solution: Before drawing the for ce diagram, it is worth thinking about t he system as a whole. The cars are joined and are thus const rained to have the same acceleration. Because t he total mass is 3M, t heir acceleration is F
a = 3M Now we will draw t he free body diagram. The force diagram for t he end car, no 3, is shown, where W is the weight, N is t he upward force exerted by the t rack, and F3 is t he force exerted on car no 3 by car no 2.
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Newtonian Mechanics: Basics
N
-
M
a M
,- --- -----, I I
I ________ .JI 1
w
a
M
----------
-
a
--------- -
'
I
3
-
I
F3
I
2
I
I_ - - - - - - - - .J
F2
II_ _ 1 I _ _ _ _ _ _ _ _J
F
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N
-
M
a M
----------
1 I I
I
I
' - - - - - ____ .J
a
M
----------
I
3
-
F3
Fj
I
2
-
a
---------F2
I
'--------- -'
. I '--------- -'
I
1
F
w
Figure 2.2: The vertical acceleration is zero, so that N horizontal equation of motion is
=M
W . The
F 3M
F 3
Now let us consider t he middle car, no 2. The ver·t ical forces are as before, and we omit t hem . F~ is the force exerted by the last car , and F 2 is the force exerted by car no 1. The equation of motion is
By Newton's third law, F~ = F 3 , so that F~
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•
F / 3. since
P hysicsguide
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Newtonian Mechanics: Basics
a= F / 3M we have /
r,
"\
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@Sk J ahiruddin, 2020
a= F / 3M we have F2=M
F 3M
F
+3
2F 3
The horizontal forces on the first car, no 1, are F, to the right, and , 2F F 2 = F2 = 3 to the left. Each car experiences a net force F / 3 to the right.
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Newtonian Mechanics: Basics
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Newtonian Mechanics: Basics
Example 2.2: In this two examples we will learn how to find the relations between the accelerations in con-
strained motion. Example-I: Wedge and block system : A block slides on a wedge (a planar surface) which in t urns slides on a horizontal table, as shown in t he figure 2.3 (a). The angle of t he wedge is 0 and its height is h. How are the accelerations of the block and the wedge related? Neglect friction. Example-2: Pulley system: The pulley system shown in figure 2.3 (b), is used to hoist a block. How does t he accele1~ation of the end of the rope relate to the acceleration of the block? I
X
X
h
h
. l
X
y
)(
0
x -X - •
(a)
(c)
(b)
Figure 2.3:
Solution: Example-I: Because the wedge is in contact with the t able, we have the trivial constraint that the vertical acceleration of the wedge is zero. To find the less obvious [email protected]
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with t he table, we have the trivial constraint that the vertical acceleration of the wedge is zero. To find the less obvious [email protected]
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Newtonian Mechanics: Basics
constraint t hat t he block slides on t he wedge, let X be t he horizontal coordinate of the end of t he wedge and let x and y be t he horizontal and vertical coordinates of the block, as shown in figure 2.3 (c). From the geometry, we see t hat
h -y
= tan0
x- X h - y = (x - X ) tan 0 Differentiating twice with respect to time and rearranging, we obtain t he constraint equation for the accelerations: ••
jj
= (X - x) tan 0
Example-2: Using the coordinates indicated, the length of t he rope is given by l= X
+ 1rR + (X
- h) + 1rR
+ (x - h)
where R is the radius of each pulley. Hence .. X
=
1 .. - -x 2
The block accelerates half as fast as the hand, and in the opposite direction.
11 oo : 01 : 2s opposite direction.
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Example 2.3: Two masses, M 1 and M 2 , are connected by a string that passes over a pulley. The pulley is accel-
erating upward at rate A , as shown in figure below, and t he gravitational force on each mass is W i = Mig. The problem is to find the rate at which t he masses accelerate and the tension Tin t he string. A
T
T
1
2
,.........___, Xp
1 --
I
I
Figure 2.4:
Solution: From t he force diagrams, we have
TT -
= W2 = W1
M1ii1
(2.2)
M2ii2
We have two equations but three unknowns: y 1 , y2 , and T.
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' T- W1 = M1ii1 T - W2 = M2ii2
(2.2)
\"!ve have two equations but three unknowns: y 1 , y2 , and T. We need a t hird equation, which is the constraint equation that relates the accelerations. The coordinates are shown in 32
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the drawing. If YP is measured to the center of the pulley of radius R , then if l is the length of t he string, we have
Different iating twice wit h respect to time, we find
Using A = jjp, we have the constraint condition
(2.3) Equations (2.2) and (2.3) are easily solved:
T = 2(A+g) M1M2
M1 +M2 .. (2A + g)M2 - M1g Yi= M1 + M2 .. (2A + g)M1 - M 2g 2 y = M1 + M2
Note t hat the result is reasonable. If the masses are identical, t hey accelerate equally and t he tension is M (A + g). If eit her mass vanishes, t he other mass falls freely under gravity. If .
.
..
.
.
.
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Note t hat t he result is reasonable. If t he masses are identical, they accelerate equally and t he tension is M (A + g) . If eit her mass vanishes, t he ot her mass falls freely under gravit y. If A = 0, t he apparatus is known as "Atwood 's machine," a standby in lecture demonst rations and elementary labs. The purpose of Atwood's machine is to '' decrease gravity" - t he larger mass descends more slowly t han if it were in free fall. [email protected]
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Example 2.4 : Mass m is whirled at inst antaneous speed v on t he end of a st ring of length R . T he mo-
t ion is in a vert ical plane in the gravit at ional field of t he Earth. The forces on m are the weight W = mg down and the string force T t oward t he center. T he string makes instantaneous angle 0 wit h the horizontal. Find T and t he tangent ial acceleration at any instant A
A
r
0 /
/
-- --- -.....
V
'' '
/
m
/
I
\
I I
I
0
g
I
I
T
I I
\ \
0 \\ \
I
\
' ' ..... _____ .......
w
/ / /
g
/
0
Figure 2.5:
Solution: T he diagram shows the forces and unit vectors ,
A
A
rn,
,.
, r
•
rr,
T T7
•
/'l
I
1
,. ,
I
•
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Figure 2.5:
Solution: The diagram shows the forces and unit vectors r and 0 The radial force is - T - W sin 0, so the radial equation A
of motion is
-(T
+ W sin 0) = mar 2 = m r - r0
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(2.4)
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The tangent ial force is - W cos 0. Hence
-Wcos0 =
ma0
.
..
(2.5)
= m(r0 + 2r0) 2
0
since r = R = constant, ar = - R equation (2.4) gives
mv
2
2
v T = - - - W sin 0 = m-
R
l
R
-
2
-v / R , and
gR Sln . 0 V2
A string can pull but not push , so that T cannot be negative. This requires that mv 2 / R
> W sin 0. The m aximum value of
W sin 0 occurs when the mass is vertically up; in this case 2 mv / R > W. If t his condition is not satisfied , t he mass does not follow a circular path but starts to fall; f is no longer zero. The tangent ial acceleration is given by equation (2 .5) . since
r=
0 we have
••
a0
= R0
TT T
/"\
11 oo : 01 v ........
: 36
lS con 1 lOn IS no Sa IS e , ., . . . . e mass
oes
not follow a circular path but starts to fall; r is no longer zero. The tangential acceleration is given by equation (2 .5) . since r = 0 we have a 0 = R0 Wcos 0 ••
m
= - gcos0 The whirling block has tangent ial acceleration t hat varies between +g and -g, no matter what t he value of v . On t he downswing t he tangential speed increases, on t he upswing it decreases. If we wanted to swing t he mass with constant velocity, we would need to make a0 = 0 by giving T a tangent ial [email protected]
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component W cos 0. Now we will consider rotational motion, t ranslational motion, and const raints. Example 2.5: A horizontal frictionless table has a small hole in its center . Block A on the table is con-
nected to block B hanging beneath by a string of negligible mass which passes through the hole. Init ially, B is held stationary and A rotates at constant radius r 0 with st eady angular velocity w 0 . If Bis released at t = 0, what is its acceleration immediately afterward?
I
I
\
/
.,.,---z--
' ...... _
----
'A\
/-+--/---._.,,,,, ,,,
z
00 : 01 : 38
T
Me
z
Figure 2.6:
Solution: The force diagrams for A and B after t he moment of release are shown in the sketches.
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For t he block on the table we need consider only horizontal forces. The only such force acting on A is the string force T . The forces on B are t he string force T and its weight W B For A it is natural to use polar coordinat es r, 0, while for B t he linear coordinate z is sufficient, as shown in t he force diagrams. As usual, t he unit vector r is radially outward. For convenience, we have taken z t o b e posit ive in the downward direction. The equations of motion are
-T =
MA ••
0
..r-r 0·2 •
= MA(r0 + 2r0) rn
radial , A t angential , A
(2.6)
(2.7) Inn\
11 - T = MA
00:01 :41
..r-r 0·2
••
0
radial , A
(2.6)
•
= MA(r0 + 2r0 )
t angential , A
(2.7)
WB-T = MBi
vertical, B
(2.8)
Because t he length of the string, l , is constant, we have
(2.9)
r+z=l
Differentiating this twice with respect to time gives the constraint equation r = -z (2 .10) ••
••
The negative sign means that if mass A moves outward, mass B would rise. Combining equations (2.6), (2.8), and (2. 10), we find
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Immediately aft er B is released , r
Newtonian Mechanics: Basics
•
= r 0 and 0 = w 0 . Hence (2 .11 )
i(O) can be positive, negative, or zero depending on the value of the numerator in equation (2. 11 ). If w 0 is large enough , block B will begin to rise after release. Before release, r = 0, but immediately after , the acceleration has a finite value. It is evident t hat because forces can be applied suddenly, acceler-
11 oo : 01
: 44
i(O) can be positive, negative, or zero depending on the value of the numerator in equation (2. 11). If w 0 is large enough , block B will begin to rise after release. Before release, r = 0, but immediately after, the acceleration has a finite value. It is evident t hat because forces can be applied suddenly, acceleration can change abruptly-acceleration can be discontinuous in time. In contrast, position and velocity are time integrals of acceleration and are therefore continuous in time.
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Example 2.6: A block of mass M rests on a fixed plane inclined at an angle 0. You apply a horizontal force of Mg on the block, as shown in figure. Assume t hat the friction force between t he block and t he plane is large enough to keep the block at rest. What are t he normal and friction forces (call them N and Ff ) that t he plane exerts on the block? If the coefficient of static fri ction is µ, for what range of angles 0 will t he block in fact remain at rest?
11 oo : 01
: 46
enoug o eep and fri ction forces (call them N and Fr ) that the plane exerts on the block? If the coefficient of static friction is µ , for what range of angles 0 will t he block in fact remain at rest?
Mg
Figure 2.7:
Solution: Let 's break the forces up into components parallel and perpendicular to the plane. (The horizontal and vert ical comp onents would also work, but t he calculation would b e a little longer.) The forces are N , Fr, the applied Mg, and the weight Mg , as shown in figure below. Balancing the forces parallel and perpendicular to the plane gives, respec-
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tively (wit h upward along t he plane t aken to be positive
Fr= Mgsin0 - Mgcos0 N = Mg cos 0 + Mg sin 0
N
11 oo : 01
: 49
N = Mg cos 0 + Mg sin 0
N
Mg
Figure 2.8: The coefficient µ t ells us that Ff I < µN. Using equation above, t his inequality becomes
Mgl sin0 - cos0 :::; µMg(cos0
+ sin0)
(2. 12)
The absolute value here signifies t hat we must consider two cases: Case-1 : If tan 0 > l , then (2.12) becomes
sin0- cos0 < µ(cos0 [email protected]
+ sin0) =? tan0 < ~ + µ -µ
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We divided by 1 - µ, so this inequality is valid only if µ < l. But if µ > l , we see from t he first inequality here t hat any value of 0 (subject to our assumpt ion, tan 0 > l ) works. Case-2: If tan 0 :::; 1, then (2.12) becomes - ~in A -1--
r'()~
A
...~ - ~ I dz
Iz
z
CT
R
l ~~~~ ,_ , _,
I
I I I I
dz
3 ,: Z =R
R
8
Figure 3.6:
Solution: From symmetry it is apparent t hat the center of mass lies on the z axis. Its height above t he equatorial plane is Z= 1 zdM M
11 oo : 04 : 11 Solution: From symmetry it is apparent that the center of mass lies on the z axis. Its height above the equatorial plane is Z = l zdM M The integral is over three dimensions, but the symmetry of the situation lets us treat it as a one-dimensional integral. We mentally subdivide t he hemisphere into a pile of thin disks. Consider a circular disk of radius r and thickness dz . Its vol2 ume is dV = 1rr dz , and its mass is dM = pdV = (M/V) (dV) ,
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where V = ~1r R 3 . Hence
z=
M
l M
V zdV
l
R
1rr 2 z dz
V z=O To evaluate the integral we need to find r in terms of z . 2
Since r = R
2
-
z
2
,
we have R
z
(R
2
-
z
2
)
dz
0 7r
V
1 z2 R 2 - ~ z4 2 4
1 R4 - ~R4 V 2 4 11rR4 3 _ 1 =- R 7r
R 0
11
00:04:14
0
3.2
Center of mass frame
Center of mass frame (COM frame, or CM frame) is a frame where the total momentum of all t he particles is ZERO. Consider a frame S and another frame S' that moves at constant velocity u wit h respect to S as sho,"ln in the figure below. Given a system of particles, the velocity of t he ith particle in S is related to its velocity in S' by
(3.9) 92
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y' y
S'
L..----• X,
s -----• X Figure 3.7:
Momentum of t he particle in the S and S' frame are t hen
11 oo : 04: 16 Figure 3. 7:
Momentum of the part icle in the S and S' frame are t h en
P =
L m iv i;
(3. 10)
Now we will make S' frame as COM frame (where t he total moment um of a system of part icles is zero), COM frame and CM frame both refer to center of mass frame. If the total momentum is P
I: mivi in fram e S, t hen t h e CM frame is
the frame S' that moves with velocity
P _ I: miv i U =
M =
with respect to S, where M _
(3. 11 )
M
I: m i is t h e total mass.
Then t he momentum in the S' frame is zero. V ·i
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p M
= P- P =O
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(3. 12)
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The CM frame is extremely useful.
Physical processes
are generally much more symmetrical in t his fr ame, and t his makes the results more transparent. The Cl\/1 frame is somet imes called the '' zero moment um'' frame. But t he '' center of mass'' name is commonly used because the center of mass of t he part icles doesn 't move in the CM frame, for t he following reason . You h ave already studied the t h e position of the center of m ass which is defined by
_I: miri R c:rvr = M
(3. 13)
11 oo : 04: 19 ing reason . You have already studied the t he position of the center of mass which is defined by _L miri R CM= M
(3.13)
If we t ake derivative of the above equation, we get M a cM
L m,i~= LFi=
F total
(3. 14)
So as far as the acceleration of t he CM goes, we can treat t he system of particles like a point mass at the CM, and then just apply F = ma to this point mass. since t he internal forces cancel in pairs, we need only consider the external forces when calculating F total Along with the CM frame, you will work in lab frame. There is not hing at all special about lab frame. It is simply t he frame (assumed to be inertial) in which t he condit ions of the problem are given. Any inertial frame can be called the '' lab frame." Solving problems often involves switching back and fort h between the lab and CM frames. For example, if [email protected]
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t he final answer is requested in t he lab frame, t hen you may want to transform the given information from the lab frame to the CM frame where things are more obvious, and then transform back to t he lab frame to give the answer Example 3. 7: A mass m with speed v approaches a stat ionary mass M as shown in t he figure. What is t he velocity of center of mass (COM) frame? What are the velocities of the particles in the COM frame?
11 oo : 04: 21 Example 3. 7: A mass m with sp eed v approaches a stat ionary mass M as shown in the figure. What is t he velocity of center of mass (COM) frame? What are the velocities of the particles in the COM frame? m •
M
V
•
)Ii
Solution: The velocity of the COM frame is the sum over t he moment um of t he lab frame divided by t he total mass.
P
U =-
Af
I: m ·v · --'l
'l
M
mv+M ·O m+M
mv m+M
The velocities of t he two masses in the COM frame is then Vm
= v - u =
Mv m + M'
and
VM
=
0- u = -
mv m +M
We will come back to this example after we learn kinetic energy. [email protected]
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Center of mass frame for two body system: Here we construct the COM frame of two body problem which will be very much useful for our study of tv\ro body problems later. Consider the case of a two-part icle syst em with masses m 1 and m 2 as shown in the part (a) of the figure below. 2
I
m;:,
z
11 oo : 04: 24 very much useful for our study of tv\ro body problems later. Consider t he case of a two-part icle system with masses m 1 and m 2 as shown in the part (a) of the figure below. z
z z'
y X
y
(a)
X
(b)
Figure 3.9:
In t he init ial coordinate system x, y, z, ( part (a) ), the particles are located at r 1 and r 2 and their center of mass is at R == m 1r 1 + m 2 r 2 m 1 +m2
We now set up the center of mass coordinate system, x', y', z', (part (b) ), wit h its origin at t he center of mass. The origins of t he old and new system are displaced by R . T he center of mass coordinates of t he two particles are r ~ == r 1 r ~ == r 2 [email protected]
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Center of mass coordinates are t he natural coordinates for an isolated two-body system. Such a system has no external forces, so t he motion of the center of mass is t rivial-it moves uniformly.
11 oo : 04: 21 Cent er of m ass coordinates are t he natural coordinat es for an isolated two-body syst em. Such a syst em has no external forces, so t he m otion of the cent er of mass is trivial-it moves uniformly. By t he definition of center of m ass we have
Distance from m 1 t o m 2 , as a vector , is
Hence
3.3
conservation of momentum
Total external force F acting on a syst em is related to t he total momentum P of t he system by (3. 15) Consider t he implications of t his for an isolated syst em where there is no ext ernal force. In this case
dP dt [email protected]
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=
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T he t otal moment um of an isolated syst em is constant . T his is the law of conservation of momentum.
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The total momentum of an isolated system is constant. T his is the law of conservation of momentum. The integral form of equation (3.15) is t
Fdt = P (t) - P (O)
(3. 16)
0
We see t hat the change in moment um is the t ime integral of the force. To produce a given change in the momentum in t
F dt have the appropriate
time interval t requires only t hat 0
value; we can use a small force acting for much of the time or a large force acting for only part of the interval. t
F dt is called the impulse.
The integral 0
Now we'll solve some examples involving conservation of momentum and concept of impulse.
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11 oo : 04: 32 98
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Example 3.8: A car of mass M is moving with a uniform velocity v on a horizontal road when a small brick
of mass m drops on it from above and stuck wit h it. W hat will be t he velocity of the car brick combo after the event? m m M
M
V
V
Figure 3.10:
Solution: In the horizontal direct ion, t here is no exter-
nal force. since t he hero has fallen vertically, so his initial horizontal momentum = 0 Initial horizontal moment um of the system = M v towards right T he brick sticks to t he roof of the car, so t hey move wit h equal horizontal velocity V. Final horizontal momentum of the
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system is t hen
(M +m)V Momentum conservation gives
M v = (M +m)V Hence we get
V =
Mv M+m
Example 3.9: A man of mass m is standing on a platform of mass M kept on smooth ice. If t he man starts moving on the platform with a speed v relative to the platform, with what velocity relative to the ice does the platform recoil?
Solution: Consider the situation shown in figure below. Suppose the man moves at a speed w towards right and the platform recoils at a speed V towards left. Both the speed are relative to the ice. Hence, the speed of t he man relative to the platform is V + w . Now as given in the question the man's relative velocity to the ice is v, then
V + w =v
w=v-V
11 oo : 04: 37 w=v- V
V +w=v
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w Platform
V
Ice
Figure 3 .11:
Take the platform and the man together to b e the syst em , there is no external horizontal force on the system . The linear momentum of the syst em remains constant. Init ially, both the man and t he platform were at rest . So the initial momentum 1s zero. •
O= MV -mw Replacing the value of w
MV =m(v- V ) V = mv M+m
11 oo : 04 : 40
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Example 3 .10: T wo ident ical blocks a and b each of mass m slide without friction on a straight track. They are attached by a spring with unstret ched length l and spring constant k; t he mass of the spring is negligible compared to the mass of the blocks. Init ially t he system is at rest . At t = 0, block a is hit sharply, giving it an instantaneous velocity v 0 to t he right. Find the velocity of each block at later times.
Look part (a ) of the figure below. The sit uation is drawn there. since the system slides freely after the collision, t he cent er of mass moves uniformly and therefore defines an inertial frame. Let us t ransform to center of mass coordinates. The center of mass lies at R = mra +mrb m+m 1 = (ra + rb) 2
00: 04: 42
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vb(O)
=0
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t--- - ra - - -, ', ra - rb ' Laboratory
=v 0
..
va(O)
coordinates
--~ : R rb - i I
b
r11lo11
a
a
b Q
I I
i- f1, -
I
~ r'a -
rb 1
(a)
l' 'J
I I
!\
Center of mass coordinates
Figure 3.12:
R always lies halfway between a and b. The center of mass
coordinates of a and b are r~
= Ta =
r~
=
1
2
-
(ra - rb)
Tb -
=-
R
1
2
R
(ra - rb)
= - r'a
In t he oart (b) t hese coordinates are drawn. The instanta-
11
oo : 04: 4s
- ra - rb 2
r~ = rb - R 1 = - (ra - rb)
= - r'a
2
In t he part (b) t hese coordinates are drawn. The instantaneous length of t he spring is r a - rb = r~ - r~ . The instantaneous departure of the spring from its equilibrium length l is
r a - rb - l = r~ - r~ - l. The equations of motion in the center
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of mass system are
mr'a = - k (r'a mr~ = +k (r~ -
r'b - l)
r~ - l)
The form of these equations suggests that we subtract them, obtaining
m (r: - r~) = - 2k (r: - r~ - l) It is n atural to int roduce t h e departure of the spring from its equilibrium length as a variable. Lett ing u
=
r~ - r~ - l , we
have
mu+ 2ku =
0
This is t he equation for simple h armonic motion. The general solution is
u = A sin wt + B cos wt
2k/m. since the spring is unstretched at t = 0, u(O) = 0 which requires B = 0. Then u = A sin wt so that u = Awcoswt. At t = 0 where w =
11
00:04:47
u = A sin wt + B cos wt where w = 2k/m . since the spring is unstretched at t = 0, u(O) = 0 which requires B = 0. Then u = A sin wt so that u = Awcoswt. At t = 0 u(O) and since u
= r~ - r~ -
l
= Aw cos(O)
= r a - rb -
l
= Vo [email protected]
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so that
A= vo/w Therefore
u = (v 0 /w) sin wt and
u = v 0 cos wt . . andVa ' = -vb, ' we h ave since va' - vb' = u, 1 va = -vb = vo cos wt 2 I
I
The laboratory velocities are •
Va = R + v~ vb = R + v~ •
•
since R is constant, it is always equal to its init ial value
11 oo : 04 : so •
Va=
R + v~ •
vb = R + v~ •
since R is constant, it is always equal to its init ial value
.
R
=
=
1
(va(O)
2 1 -Vo 2
+ Vb(O))
Putting these results together gives Va=
Vb =
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Vo 2
Vo 2
(1 +coswt) (1 - cos wt)
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Example 3.11: A loaded spring gun, init ially at rest on a horizontal frictionless surface, fires a marble at angle
of elevation 0 as shown in part (a) of the picture below. The mass of the gun is M , the mass of the marble ism, and the muzzle velocity of the marble (the speed with which the marble is ejected, relative to t he muzzle) is v 0 What is the final motion of the gun? v 0 sin 0
.,,,--.,,. Vo COS ()
X
(a)
v, {b)
-
Vf
11 oo : 04: s3 cos 0 M
v,
X
(a)
(b)
Figure 3.13:
Take the physical system to be the gun and marble. Gravity and the normal force of the table act on t he system. These external forces are both vertical. Because there are no horizontal external forces, the x component of t he vector equation
F = dP / dt is O = dP.x
dt
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So, Px is conserved:
Px, initial = Px, final Let the initial t ime be prior to firing the gun. Because the system is initially at rest, Px, initial = 0, After t he marble has left the muzzle, t he gun recoils to the left with some speed V1, and its final horizontal moment um is - MV1 Finding t he final velocity of the marble involves another important concept. Physically, the marble's acceleration is due to t he force of the gun, and t he gun 's recoil is due to the reaction force of the marble. Look at the part (b) of the figure.
11 oo : 04: 55 Finding t he final velocity of t he marble involves another important concept. Physically, the marble's acceleration is due to t he force of the gun, and the gun's recoil is due to the reaction force of the marble. Look at t he part (b) of the figure. The gun stops accelerating once the marble leaves the barrel, so at t he instant the marble and the gun part company, the gun has its final speed - V1 At t hat same instant t he speed of the marble relative to t he gun is v 0 Hence, t he final horizontal speed of the marble relative to t he table is v 0 cos 0 - V1 . By conservation of horizontal momentum, we t herefore have
or
V _ mv0 cos0 J- M +m
Interesting, isn 't it? What is t he impact the marble give to t he floor? [email protected]
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Example 3.12: A block of mass Mis free to slide on a friction-less horizontal floor. The block has a cylindrical
cavity of radius R in the middle of it. The center of mass (CM) of the block lies on the dashed line passing through t he center of t he cavity (see figure). Init ially t he CM of the block is at a horizontal distance X 1 from the origin, Now a point particle of mass m is released from point A into the cavity. There is negligible friction between t he particle and t he cavity surface. When the particle reaches point B , t he CM of t he block is at a
11 oo : 04 : sa t he CM of t he block is at a horizontal distance X 1 from t he origin, Now a point particle of mass m is released from point A into the cavity. There is negligible friction between t he particle and the cavity surface. When the particle reaches point B , t he CM of the block is at a [JAM distance X 2 from the figure. Find (X 2 - X 1) 2009]
----A
- ~ 8
R
X1
Figure 3.14:
Solution: Center of mass of the total system should be [email protected]
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conserved as there is no external force . The COM of t he combined system initially is MX1
+ m(X1 - R) M+m
After t he ball reaches the other side t he COJVI is
11 oo : os : oo M+m After t he ball reaches the other side t he COM is
Equating the initial and final coordinate of COlVI
Hence we get
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Example 3.13: In t his example we will measure t he speed of a bullet with the help of spring mass system. Consider a block of soft wood on a horizontal frictionless surface. A compression spring with spring constant k and uncompressed length l connects t he block to a wall. The block has mass M , and the spring has negligible
11 oo : os : 03 speed of a bullet wit h the help of spring mass syst em. Consider a block of soft wood on a horizontal frictionless surface. A compression spring with spring const ant k and uncompressed length l connect s t he block to a wall. The block has mass M , and t he spring has negligible mass. At t = 0 a gun fires a bullet of mass m and speed v 0 into the block. The block will absorb the bullet . We don't know v 0 , but we will be able to measure the maximum compression of the wood block, Xmax, after it absorbs the bullet. So we'll find vo in t erms of
k
Xmax •
m M
x~ 0
F igure 3.15:
Our syst em is the bullet and t he block. Let 's say t he block moves back at initial speed ¼ due t o the impulse. By conservation of momentum applied at very short times aft er the [email protected]
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collision
mvo = (M +m)¼
(3.17)
Conservation of moment um is accurate here, because during the very short t ime of t he collision the horizontal force of t he
11 oo : os : 06 collision
m vo = (M +m) ¼
(3.17)
Conservation of momentum is accurate here, because during the very short time of t he collision the horizontal force of the spring has very lit tle t ime to act. During t he ensuing time, however , t he spring force has plenty of t ime to act-it brings t he system momentarily to rest. Measuring how far the syst em moves can t ell us the speed of the bullet. After the initial impulse, the equation of mot ion of the system is (A1
+ m)x =
- kx
The spring length does not appear in the equation of motion because we have taken a coordinate syst em with x = 0 when the spring is uncompressed. We recognize the equation for simple harmonic motion , which has t he general solution x =
A sin wt + B cos wt
(3.18)
where
k
w=
M+m
Using the init ial conditions x(O) = 0, x(O) = ¼ in equation (3.18) we find A = ¼ /w and B = 0. The position and sp eed of the system are then
¼.
x = - s1n wt w
x= [email protected]
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(3.19)
¼cos wt 111
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The syst em first comes t o rest at t = t f when wt f = 1r/ 2 which is the maximum compression of t he spring. Using equa.
.
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The system first comes to rest at t = t f when wt f = 1r/ 2 which is t he maximum compression of t he spring. Using equati . (1. l 7J. ( 1.1~) and 1° .t-0)
= -;=====:::::J:::~ J. (.\/ - m) =
Xmax
so that
Vo=
Jk (M + m) m
X max
Example 3.14: A rubber ball of mass 0.2kg falls to t he floor. The ball hits with a speed of 8m/ s and rebounds with approximately the same speed. High speed photographs show t hat t he ball is in contact with t he floor
for '6.t = 10- 3s. What can we say about t he force exerted on t he ball by t he floor? The situation is drawn in figure 3. 16 (a)
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F
(a)
z
(b) - - - - - T - -:,-,,. I
l l
I I
Fav -----i----
1 I I I I I I
Figure 3.16:
The momentum of t he ball just before it hits the floor is P a = - l.6k kg · m/s and its momentum 10- 3s later is P b = + l.6kkg · m/s. Using ft:, Fdt = P b - P a gives F dt = 1.6k - ( - 1.6k) = 3.2k kg· m/s A
ft:,
A
A
A
Although t he exact variation of F with time is not known, it is easy to find t he average force. If the collision time is f:).t = tb - t a, t he average force F av acting during the collision •
18
ta+~t F av l:).t =
[email protected] ,..-...r't,
.,. , •
, ,.
_,..._,...
F dt
ta
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,
.,,.
.
•
,.
.r
1
11 oo : os : 13
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A
= 3 ·2k
F
kg · m/s = 3200k N 10-3s
av
The average force is directed upward, as we expect. The inst antaneous force on the ball is even larger at t he peak, as shown in part (b) of the figure. If t he ball hits a softer surface t he collision t ime is longer , and the peak and average forces are less. So far we have neglected the impact by gravity. So we calculate it now. F = F ball + F grav A
= F ball - Mgk The impulse equation t hen becomes 10- 3
10- 3 A
F ball 0
A
M gk dt = 3.2k kg· m/s
dt 0
The impulse due to the gravitational force is 10- 3
10- 3 A
Mgk dt 0
A
= - Mgk
dt
= -(0.2) (9.8) ( 10-
3
)
k
0 3
= - 1.96 x 10- k kg• m/s This is small in this problem , but in some problems the impact due to gravity is comparable with the impact due to the velocity change. The next problem deals wit h it
11 oo : os : 16 T his is small in this problem , but in some problems the impact due to gravity is comparable with the impact due to t he velocity change. T he next problem deals wit h it [email protected]
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Example 3.15: A ball weight ing 100 gm, released from a height of 5 m, bounces perfectly elastically off a plate. The collision t ime between t he ball and t he plate is 0.5 s. The average force on the plate is approximately [NET June 2017] (a) 3 N (b ) 2 N (c) 5 N (d ) 4 N Solution: As we see the velocity in which t he ball strike •
lS
2
v = 2gs = 2 x 10 x 5 = 100
V
=
10
We take positive z axis upward so t hat the downward velocity is negative " V = -lOk Init ial moment um is Pi = m v =
0.1 x (- l Ok) = - lk
As it bounces perfectly elastically wit h t he same velocity (opposite direction) t hen t he final moment um is "
"
Pr = m vr = 0.1 x (10k) = lk Change in moment um due to velocity change is
11 oo : os : 1a posite direction) then the final momentum is A
A
Pr= mvr = 0.1 x (10k) = lk Change in momentum due to velocity change is
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So, t he force due to velocity change is F ball
~p 2 " " = = - k = 4kN ~t 0.5
This is the force on the ball by the plate. We are dealing the force on the plate here. According to t he Newton's third law t he ball exert same force on t he plate but in opposite direction. So t he force on t he plate is A
F plate
=
-4kN
Don't forget gravitational force and the impact due to gravity. Ball exert gravitational force on the plate downwards which •
18
F grav =
"
"
-Mgk = - lkN
So the total force on the plate by the ball is 4 + 1 = 5N
We will come across the impact of gravity when we do the chain problem.
3.4
Conservation of Energy in one and two
dimension
11 oo : os : 21 chain problem.
3.4
Conservation of Energy in one and two
dimension Newton's law in ID is
d2 x m dt2
=
dv mdt =F(x)
F(x)
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We can solve t his by integrating mdv /dt = F(x) with respect to x Xb xb dv m -dx= F(x)dx (3.2 1) Xa di Xa we write the differential dx as dx =
dx dt
Then the integration of LHS is xb
m Xa
dv -dx=m di =m =m
So we get
dt = v dt
11 oo : os : 24 1
')
= -mv . .
b
2
So we get Xb
F (x)dx
(3.22)
Xa
The quantity ~ mv 2 is called t he kinetic energy K , and t he lefthand side can be written K b - K a. If at any point a particle [email protected]
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1 2 with mass m has velocity v , we will call the quantity mv as 2 t he kinetic energy of the particle. Xb
F (x) dx is called the work Wba by the
The integral Xa
force F on the particle as t he particle moves from a to b. Our relation now takes the form
(3.23) This is the Work Energy theorem in one dimension. This means t he work done by t he force to move the particle from a to b is the difference between their kinetic energies at these two points. We then define the potential energy difference of the two positions as Xb
½ - Va=
F( x )dx
(3.24)
Xa
So the difference in potential energy is equals to t he difference in kinetic energy.
11 oo : os : 26 positions as Xb
½ - Va =
F( x) dx
(3.24)
Xa
So the difference in potential energy is equals to the difference in kinetic energy. The potential energy at any point x is defines as the integration of force from a reference point x 0 to the point x with a negative sign . X
V(x)
-
F (x') dx'
(3.25)
Xo
Sum of kinetic and potential energy is the total mechanical [email protected]
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energy of the particle. 1
mv 2
2
+ V(x) = E
(3.26)
If potential energy is of a particle is positive at any point we understand t hat we have done some work to take the particle at that position and if we release the particle then it goes to t he Zero potent ial energy level. And if the potential energy is negative then we say that the particle is bounded and ''we" need to perform work to take t he particle to the zero energy level. The potential energy is the integration of force, thus force is the derivative of potential.
F(x) = _dV(x) dx
(3.27)
Given V( x), it is easy to take its derivative to obtain F(x) .
11 oo : os : 29 is the derivative of potential.
F(x) = _dV(x) dx
(3. 27)
Given V (x), it is easy to take its derivative to obtain F(x) . But given F(x), it may be difficult (or impossible) to perform the integration in equation (3.27) and write V (x) in closed form. The function V (x) is well defined (assuming that the f'orce is a function of x only, and if needed it can be computed numerically to any desired accuracy Given any force (it can depend on x, v, t , and/or what ever), the work it does on a particle is defined by W _ J Fdx . If the particle starts at x 1 and ends up at x 2 , then no matter how it get s there (it might speed up or slow down, or reverse [email protected]
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direction a few times), we can calculate t he total work done on it by all t he forces in t he setup, and then equate the result with the change in kinetic energy, via X2
X2
W total -
F total
dx =
m
v dv
dx
dx =
1
2
m v2
2
-
1 2
2
mv1
(3.28) Total mechanical energy of a syst em is conserved if there is no dissipative force. We call it the conservative force t hen.For conservat ive forces, t he work done is independent of how t he particle moves. A force that dep ends only on position (in one dimension) has t his property, because t he integral in equation (3.28) depends only on the endpoints. The W = J Fdx int egral is t he (signed) area under the F vs. x graph, and t his ::i.r P::i. i~ inr1PnPnr1Pnt nf hnw t.h P n ::i.r tirl P O'n P~ fr nm r1 tn
'rn
11 oo : os : 31 part icle moves. A force that dep ends only on position (in one dimension) has t his property, because t he int egral in equation (3.28) dep ends only on the endpoint s. The W = J F dx int egral is t he (signed) area under t he F vs. x graph, and t his area is indep endent of how t he particle goes from x 1 t o x 2 For non conservative forces, t he work done dep ends on how t he particle moves. T his happens when forces depend on t or v, because it then matters when or how quickly t he part icle goes from x 1 to x 2 . Consider frictional forces. If you move a brick sliding across a t able with friction from x 1 to x 2 , t hen the work done by friction equals - µm gl~ x . If you slide t he brick again and take it back to x 1 , t he work done is again - µm g ~ x as friction always opposes the motion , t he contribut ions to t he W = J F dx int egral are always add up.
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Power
Power is t he rate at which work is done. T he work ~ W by force F on a syst em as it moves t hrough a short distance ~r is F · ~r. If the displacement takes place during time ~ t, t hen t he rate of work is ~ W ~ F. ~r ~t ~t In t he limit ~ t ➔ 0, ~r / ~ t ➔ v , so we have
dW = F . V
dt
(3. 29)
Power can b e eit her posit ive or negative depending on whether t .h P ,xr()r k- i ~
() TI
()r
h,,
t .h P
~,,~tPrn
11 oo : os : 34 t
➔
v , so we
dW = F.
V
(3.29)
dt
Power can be either positive or negative depending on whether the work is on or by the system. 3.4.2
Gravitational potential energy:
We will have detailed discussion on gravitation later in a separate chapter. For now it is sufficient to know that t he gravitational force on earth attracts each object with a force Mg towards earth where g is t he gravitational constant. So the gravitational potential energy is (3.30)
mgz
where z is the upward displacement from the ground which is taken as the reference of zero potential energy. The gravitational potential energy is positive as if we release any part icle [email protected]
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at upward displacement z t he gravitational force does work and takes back the particle to t he zero potential energy level. If we release a particle ar rest at some height, it comes down and it's potential energy manifest as kinetic energy.
3.4.3
Potential energy of a spring
The forc e which spring exert when elongated or compressed ,
,
•
1
TT
,
'
,
11 oo : os : 36
3.4.3
Potential energy of a spring
The forc e which spring exert when elongated or compressed by length x is -kx. Here we have taken x to be zero at the equilibrium so that the variable x counts the displacement from the equilibrium position. So the work done is X
W =
}
kxdx = 0
kx
2
2
In two dimension force and displacement may not need to be on the same direction as shown in t he figure below.
F
Figure 3.17:
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The work done is then
W =
~
F · dr=
Fcos 0dr
(3.31)
Example 3.16: This is one of the most basic problem
where you need to apply t he conservation of energy. A small block of mass m starts from rest and slides along a frictionless loop-t he-loop as shown in the figure. What
11 oo : os : 39 Example 3.16: This is one of the most basic problem where you need to apply the conservation of energy. A small block of mass m starts from rest and slides along
a frictionless loop-the-loop as shown in the figure. What should be t he initial height z, so that m completes t he loop without being fallen down.
m a
mv 2 r
rng+N
z
Figure 3.18:
Solution: Initially the whole energy was kinetic energy
Now if the mass has to complete the loop without falling [email protected]
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down, t hen it must go at the top of t he loop which is t he position a as shown in t he figure. It is easy to see that a is at distance 2R above the ground. So the energy at a is U1 = mg(2R) T T
,
TT
1
')
.
11 oo : os : 42 posit ion a as shown in the figure. It is easy to see that a is at distance 2R above the ground. So the energy at a is U1
== mg(2R)
Equating Ei and E f we get 1 2 mgz == mv + mg(2R) 2
At the position a, the loop the total downward force is N + mg , where N is the normal force exerted by the loop. If mv
2
the block is not to fall down then the total upward force - r must balance the downward force. mv
2
N +mg== R
Substitut ing t he value of v N +mg==
mv 2 R
2mgz
R
- 4mg
If the block just touches loop at point a, t hen N == 0, which would be the lowest limit on N and thus the lowest limit on z.
z 0 + m g == 2mg R - 4mg z == 5R/2
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Example 3.17: This is another basic problem where you need to apply the conservation of energy and momentum together. A small cube of mass m slides down a circular oath of radius R cut into a large block of mass
11 oo : os : 44 Example 3.17: This is another basic problem where you need to apply t he conservation of energy and mo-
mentum together. A small cube of mass m slides down a circular path of radius R cut into a large block of mass M , as shown. M rests on a table, and both blocks move wit hout friction. The blocks are initially at rest, and m starts from the top of the path. Find the velocity v of t he cube as it leaves the block. ----
--- ~
R:
h
I I I
~
Ri v~
--V
1h -R Figure 3.19:
Solution: In the initial position (at the left of the figure) t he cube was at rest. No kinetic energy. So total energy was potential energy only
When the cube left t he block (as shown in t he right of the figure) t he velocities of both the block and cube cont ributes to t he K.E, and the potential energy will be mg( h - R) as the [email protected]
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ball will be at height h - R from t he ground. 1
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ball will be at height h - R from t he ground. 1
K = -mv f 2
2
1
')
+ -MV'"'· 2 '
U1 = mg(h- R )
So the final energy
1 2 E1 = -mv 2
1 2 -MV 2
+
+ mg(h - R)
Equating initial and final energy 1
2
mv
2
+
1
2
MV
2
+ mg(h -
R) = mgh
we get 1
mv
2
+
1
2
MV -mgR=O
2 2 Now we apply conservation of momentum Initial momentum was zero ~ = 0 Final momentum should also be zero P1 = mv-MV
=
0
Hence we get V=
3.5
M M +m
2gR
Collisions
There are two basic types of collisions among particles, elastic ones and inelastic. In elastic collisions the total mechanical [email protected]
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energy (kinetic + potential) is conserved) . In inelastic collisions some kinetic energy is lost so the total mechanical energy is not conserved . The lost energy t r ansform into heat. The moment um is conserved both in elastic and inelastic collisions because of absence of external force.
3.5.1
Collision in one dimension
Example 3.18: Consider the same situation as in ample Example 3.7:. A mass m with speed v proaches a stationary mass M as shown in the ure.what are t he final velocit ies of the particle after collision. (in the lab frame)? m
M
V
(1)
apfigthe
..... m
..
mM
•
•-•
ex-
Vf
(3)
(2)
Figure 3. 20:
Solution: Let Vf and ½ be t he final velocities of the mass m and M respectively. Then conservation of momentum and energy give, respectively, mv
1
-mv
2
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2
+ 0 = mvf + M½
+0 =
1 2 -mvf
2
127
1
2
+ -2 M½
Physicsguide
11 oo : 05 : 52 mv
1 2 - mv
2
+ 0 = mvf + M¼ +0=
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1 2 - m vf
2
1
2
+ -2 M½
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We must solve t hese two equations for t he two unknowns Vf and ½. Solving for ½ in the first equation and substituting into t he second gives 2
2
m (v - Vf) m v = m vf + M - -Af2 --2
2
=::::.~►
0 = (m + M)vl - 2mvvf
=::::.~►
0 = ((m + M )vf - (m - M )v) (vf - v)
+ (m -
M )v
2
One solut ion is Vf = v, it satisfies conservation of energy and momentum. But in this solut ion the collision does not happ en. The particles miss each other. The Vf = v (m - M) /( m + M ) root is t he one we want. Substitut ing t his Vf in momentum conservation m v + 0 = mvf + M¼ we get ½.
(m Vf
=
m
M)v
+M
'
and
½ _ f-
2mv m+ M
Now we will consider t he general collision in 1D where both t he particle was moving before t he collision . Let t he collision be as drawn in the figure below
m
•
Vj
M
•
•
•Vi
c> Collisi on
(1)
M
m
•Vf •
•Vr
(2)
Figure 3.21:
•
11 oo : 05 : 55 Vj
Collision
Vf
(1)
(2)
Figure 3.21:
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The masses be m and M was moving wit h velocit ies Vi and ½ before t he collision, and Vf and ½ are the fin al velocities after t he collision. Conservation of momentum and energy
(3 .32) 1
2
2mvi
1
2
+ 2M½ =
1
2
2mvf
+
1 2 2M½
(3.33)
Rearranging
= M (½ - ½) 2 2 m (vf - vl ) = M (½ - ½ )
m (vi -
Vf)
Dividing t he second equation by the first gives vi+v£ Therefore,
(3.34)
= ½+ ½.
(3.35) Here we get one very important result. In a 1- D elastic collision, the relative velocity of the two particles after the collision is the negative of the relative velocity before the collision. Now do some algebra and prove that 'YYl
_
l\ If
') AIf
11 oo : 05 : 57 collision, the relative velocity of the two particles after the collision is the negative of the relative velocity before the collision.
Now do some algebra and prove that VJ
m- M 2M = - - - V i + - - -¼ m+M m+M 129
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V1 =
2m m- M M Vi M Vi
m+
m+
(3 .37)
Perfectly inelastic collision: When perfectly inelastic bodies moving along t he same line collide, they stick to each ot her. Let m and M be t he masses, Vi and ¼ be t heir velocit ies before t he collision and V be the common velocity of the bodies after t he collision. By t he conservation of linear momentum, mvi
+ M¼ = mV + MV V = mvi + M¼ m+M
let us calculate how much kinetic energy is lost in this process. The kinetic energy before the collision is 1
2
2mvi
1
+ 2M¼
2
and t hat aft er t he collision is ½(m + M ) V 2 . Using the expression of final velocity, t he loss in kinetic energy due to the
11 oo : 06 : oo 2
pression of final velocity, t he loss in kinetic energy due to t he
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collision is 1 2 1 2 -mv-2 + -MV2 - - (m + M) V 2 2 2 1 2 1
2
1 mM (v; + ½ - 2vi¼) 2 m+M 2
mM (vi - ¼) 2 (m + M)
2
Coefficient of restitution: For a perfectly elastic collision velocity of separation = velocity of approach and for a perfectly inelastic collision velocity of separation = 0 In general, collisions are neither perfectly elastic nor perfectly inelastic. So we can write velocity of separation = e(velocity of approach) where O < e < 1. The constant e depends on t he materials of t he colliding bodies. This constant is known as coefficient of restitution. If e = 1, the collision is perfectly elastic and if
11 oo : 06 : 03 velocity of separation= e(velocity of approach) where O < e < 1. The constant e depends on the materials of t he colliding bodies. This constant is known as coefficient of restitution. If e = 1, the collision is perfectly elastic and if e = 0, the collision is perfectly inelastic. 3.5.2
1D collision in COM frame
In t he COM frame t he total momentum is zero bot h before and after the collision. As t he energy is conserved then t he 131
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particles just exchanges t he velocit ies in the COM frame. This can be proved in one dimension collisions as follows. We'll discuss the two dimensional case later. Suppose the particle with mass m 1 and m 2 are moving with velocities u~ and u 2 in t he COM before the collision. After t he collision the velocit ies are v~and v;. In the COM frame t he total momentum are always zero. So
from m1u~
+ m2u; =
0 we have
Hence m1v~
+ m2v; = 0 v~/v; = -m2/m1 = U1 I U2 I
I
11 oo : 06 : os Hence m1v~
+ m2v; = 0 v~/v; = -m2/m1 = U 1 I U2 I
I
The kinetic energy before and after t he collision are conserved. So ,
We have already seen V1
v'2 [email protected]
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Now in t he energy conservation equat ion substit ute u~ = ku; and v~ = kv; . Then you get >
So we get two set of solutions. One is the trivial one v~ u~; = u~ . Here the particle missed. The real solution is
v;
So what we have proved is, in COM frame lD collision the particle velocities remains same in magnitude but opposite in direction, t he velocities are just reversed. Example 3.19: Now consider the examples Example 3. 7: and Example 3.18: where we have studied a simple lD collision. in the first example we have calculated the velocity of COM frame, t he initial velocities in COM
11 oo :06 : oa direction, t he velocities are just reversed. Example 3.19: Now consider the examples Example 3. 7: and Example 3.18: where we have studied a simple 1D collision. in t he first example we have calculated the velocity of COM frame, t he initial velocities in COM
fr ame and in t he second example we calculated t he final velocities in t he lab frame using conservation of energy and momentum. Now we know that t he velocities simply reversed in COM frame then we cal easily calculate the final velocities in the lab frame more easily.
mv Solution: The velocity of COM frame was - - - In m+M the COM frame the velocities of mass m and M was u~ = Mv mv . and u~ = M respectively. Now in COM frame
m+ M
m+
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t he velocit ies just reversed, so in t he COM frame t he velocities after collision are simply V
,
-
1 -
-
Mv ----· m+ M '
, V2
mv = m+M
To calculate t he velocities in lab frame after the collision can be calculates adding t he velocity of COM frame with v~ and v~. m-M Vlab = v' + mv i i m+M m+M and lab , mv 2mv V2 = V2 + M m+ m+M This agrees wit h t he lab frame velocities calculates in example Example 3.18: .
00 : 06 : 10 and
2mv = + m+ m+M This agrees with the lab frame velocities calculates in example Example 3.18: . lab V2
mv M
, V2
Here is a schematic of general 1D collision both in Lab and in COM frame Lab after collision
Lab before
collision
0
·
v, . v,' + vcm = -u 1+2vcm
Transform to CM frame
0
·
V2- V2'+ vcm -u 2+ 2vcm
'(} Transform back to the Lab frame CM before collision
m,
CM after collision
Vl I
= -u l
-o
V 2I
I
O
= -U2
I
·
Figure 3. 22:
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Collision in two dimensions
In two dimension one extra moment um equation comes as we need to divide the momentum into the coordinate axes and apply conservation of momentum in each axis separately. So we have two momentum conservation equations and one energy conservation equation. In t hree dimension we have t hree momentum equations and one energy equation. We'll illustrate wit h one example.
Example 3.20: A billiard ball wit h sp eed v approaches an identical stationary one. The balls bounce off each other elastically, in such a way that the incoming one
11 oo : 06 : 13 ree momen um equa ions an illustrate wit h one example.
one energy equa 10n.
Example 3.20: A billiard ball wit h sp eed v approaches an ident ical st ationary one. The balls bounce off each
other elastically, in such a way that the incoming one get s deflected by an angle 0 as shown in the figure. What are the final speeds of the balls? What is t he angle which t he stationary ball is deflected?
m
•
V
at
•m
m
•
,.,
--- -
VJ
•m
Figure 3. 23:
Solution:
Let
Vf
and ½ be the final speeds of the balls. 135
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T hen conservation of momentum in x and y direction gives
mv = mvf cos 0 + m¼ cos 0 = m Vf sin 0 - m ¼ sin
(3.38)
Conservation of energy gives 1
2
1
2
-mv = -mvf 2
2
1
2
+ -m½ 2
(3.39)
We must solve these t hree equations for the t hree unknowns Vf, ¼, and . There are various ways to do t his. You may
11 oo : 06 : 16 1
1
2
2
-mv = -mvr 2 2
1
2
+ -m½ 2
(3.39)
We must solve these t hree equations for the three unknowns V£, ½ , and cp. There are various ways to do t his. You may proceed by this way. From t he moment um equations we eliminate cp by taking cp terms in one side of t he equation , square and add.
= m v + mvrcos0 2 2 2 2 2 2 2 2 m ½ cos cp = m v + m vf cos 0 + 2m vv£ cos 0
m½coscp ~~►
0 = mvrsin0 - m½sin cp Add these and get
v
2
-
2vvr cos 0 + v; =
2
½
(3.40)
Now eliminate ½ by combining t his with the energy conservation equation to get.
vr = vcos0 The energy conservation equation then gives [email protected]
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½ = vsin0
(3.41)
Substitute v f and V1 in momentum conservation equation in y direction. Now it looks
m (v cos 0) sin 0 = m (v sin 0) sin cp This gives
(3.42)
11 oo :06 : 1a y direction. Now it looks
m( v cos 0) sin 0 = m( v sin 0) sin
X
Mgx L
~
X+t.X
Mg (x+t.x) L ~
(1)
(2)
(3)
The momentum of the chain at situation (2) is A.nitial
= MV
The mass of the hanging portion is M
11 oo : 07 : 1 a e momen um o ..A.nitial
= MV
The mass of the hanging portion is
M
LX
External force at situation (2) is Mg
F external ( [email protected]
t) = L
X
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The momentum at situation (3) is ..A.nitial
+ !::l.P = M (V + f::l.v)
External force at situation (3) is F external ( t
Mg
+ f::l. t) = L (X + f::l. X)
Now you see that the force is not constant. But you can make f::l. x as small as possible and effectively the force becomes
constant Fexterna1(t) =
:g x in the motion from x to x + !::,.x
So the change in moment um is the integration over force.
f::l. p
=
M f::l. V
t+D.t
=
t
M gx f::l.t F external dt ~ l
Divide by f::l.t and make the limit of f::l.t equation of motion
➔
0, we get the
11 oo : 07 : 20 t+~ t
I:),. p = M I:),. V =
t
M gx /:),.t F external dt ~ l
Divide by /:),.t and m ake the limit of equation of motion dv
2
d x
gx
dt
dt 2
L
/:),. t ➔
0, we get t he
The general solut ion of t he different ial equation is x(t)
= Aelt + B e- lt
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We will apply t he init ial condit ions to t he solut ion. At t ime t = 0 (sit uation (1) )
x(O) = A+ B = Lo At t = 0 t he chain is at rest
Hence
A = B = Lo/2 and t he final solut ion X
4.3.2
Lo
(t) = 2
9-t
eL
- 9-t
+e
L
= L 0 cosh ft L
Chain falling on a weight machine
11 oo : 07 : 23 x (t ) == -
4.3.2
eft
+ e - ft == L 0 cosh
-t L
Chain falling on a weight machine
A chain of mass M and length L is suspended vertically with its lower end touching a scale. The chain is 1~e1eased and falls onto t he weight machine. What is t he reading of the scale when a length x of the chain has fallen? The reading of the weight machine will come from two parts. The first is obviously the weight of t he chain accumulated on the machine. As the chain has fallen t he distance x [email protected]
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on t he weight machine, t he weight due to t his part is
W1 = Mgx/L The second component contributing to the weight is due to the momentum delivered by the chain to the weight machine. As t he chain is falling freely the velocity of t he chain parts at t he instant of hitting t he scale is simply found from v
2
= 2gx
So t he momentum dp delivered in time dt dp = d(m v) = vdm
where the v has t aken to b e constant as t he momentum is delivered in very small time dt .
11 oo : 07 : 26 dp = d(mv) = vdm
where the v has t aken to be constant as t he momentum is delivered in very small time dt. The chain is moving in velocity v, t he length of the chain falls down to the machine in time dt is dx = v dt. The mass per unit lengt h is M / L. So the mass dm which falls on the weight machine in time dt is nothing but dm
M - dx
=
L
=
M - vdt
L
So the momentum dp delivered in time dt dp = d(mv)
=
M M 2 v dm = v L vdt = L v dt 161
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So the force which will be exerted by t he chain to the weight machine is the rate of change of this momentum delivered dp dt
As v 2
= lim '6.p = M v 2 ~t➔O
L
'6.t
= 2gx, so the force due to momentum delivered becomes dp
M 2 - v
W2= - = dt L
=
X
2Mg-
L
The t otal force on the scale during the fall of t he chain is therefore X X X Mg-+ 2Mg- = 3MgL L L r"i1
1
Jl
I
•
,
•
J
,
•
l
l
rr,1
11 oo :07 : 2a The total force on the scale during the fall of t he chain is therefore X X X Mg-+ 2Mg- = 3MgL
L
L
Check that energy is not conserved in this problem. The chain falling on a weight machine problem can be solved in other ways. Think on the other methods.
4.4
Other problems on variable mass
4.4.1
Cart with leaking water tank
A municipality car of empty mass M 0 contains water tank of m. At t = 0 a constant horizontal force F is applied in the direction of rolling and at the same water tank gets a leakage [email protected]
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and water starts flow out at constant rate dm/ dt . Find the speed of the car when all t he water is gone. Assume the car is at rest at t = 0. Ignore friction. I think this is a very common situation and most of you have seen a gaddi with water leaking from it. Our system consists of t he freight car and water initial mass Mo at t = 0. The bottom is opened at t = 0,. We see that the water starts leaking at steady rate. We call the rate of water leakage is , , which means "/ amount of water falls per second. So that
11 oo : 07 : 31 Our system consists of t he freight car and water initial mass Mo at t = 0. The bottom is opened at t = 0,. We see t hat t he water starts leaking at steady rate. We call t he rate of water leakage is ry, which means ry amount of water falls per second. So t hat
dm/dt = ry Total mass of the car plus water system is
M(t) = M0 + m(t) Total mass M also decreases with t ime as t he water gets out . dM/ dt = - ry. Then
M (t) = Mo - ryt Now we will apply conservation of moment um to solve the problem.
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Say at time t t he cart has mass M (t ) and velocity v(t) which is sit uation (1) in t he figure below. T he moment um is
P (t) = M v
-v
F
F
11 oo : 07 : 33 = Mv
P (t)
F
V
) v+~v
F
~ B v~v (1)
(2)
Figure 4.4:
At time t + fl.t t he cart has mass M - fl.m and velocity v + C:l. V which is situation (2) in the figure below. The small amount of water dm which has just come out from the cart is also moving horizontal! with velocity v + fl.v . Its moment um in vertical direction does not count as out force and motion are in horizontal direction. The momentum is now P(t + fl.t )
= (M - fl.m)(v + fl.v) + fl.m(v + fl.v)
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The change in moment um fl.P
= P (t + fl.t ) - P (t)
~
M fl.v
(ignoring second order term fl. mfl.v, we always do that in variable mass problems ) So the equation of motion
11 oo : 07 : 36 (ignoring second order term ~ m ~ v, we always do that in variable mass problems ) So the equation of mot ion dv dt
F = dP = Mdv dt dt
F
F
M
M o-,t
Water t ank is emptied at time t 1, so that t1
= m /,
Then we integrate the equation of motion noting t hat at t = 0 : v = 0 and at t = t 1 : v = v f v(t1)
t1
dv = 0
0
v (t1) = -
Pdt' Mo - , t' F Mo - , t1 ln Mo
'
4 .4 .2
F
'
In
Mo M 0 -m
Sand on cart
A car of empty mass M 0 start s moving due to a const ant horizontal force F as shown in the figure below. Sand spills on the car from a stationary hopper. The velocity of sand loading [email protected]
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is constant and equal to , kg/s. Find the t ime dependence of the velocity and the acceleration of the flatcar in the process of loading. Ignore friction . The situation is drawn in the figure below. Suppose at
11 oo : 07 : 39 the velocity and the acceleration of the flatcar in the process of loading. Ignore f1·iction. The situation is drawn in the figure below. Suppose at t ime t, the velocity of the sand plus cart syst em is v . At t ime mass of sand has already been deposited , so the total t ime mass 1s (situation(l)) M = Mo +,t
,t •
After time ~ t the mass ~ m = dm has been deposited to the cart and velocity has become v + ~ v (situation (2) ).
F
F
-
-
> v
v+ t. v
B
(1)
(2)
Figure 4.5:
There is one fundamental difference in this problem. The [email protected]
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mass ~ m which will be deposited in t ime ~ t do not have an initial horizontal velocity with respect to the lab frame in which we are observing. This is the fundamental mist ake .
.
11
00:07:41
mass l:lm which will be deposited in t ime l:lt do not have an initial horizontal velocity with respect to the lab frame in which we are observing. This is the fundamental mistake students do here. Although the source of t he sand is moving, but from our viewpoint t he sand is just being deposited to t he cart, the velocity of the sand source has nothing to do with our system and force. The mass may also be raindrops or ice flowing from the sky. In all these cases t here is no initial horizontal velocity of the mass deposited. Initial momentum is
P (t ) = M v Final momentum is
P (t+ l:lt) = (M + l:lm)(v+ l:lv) So the change in moment um is
l:lP
=
P (t
+ l:lt) -
P (t)
dP dt
Mdv dt
=
M l:lv
+ l:lmv
So the force
F
=
=
+ v dm dt
We know that
M(t ) = Mo + r-.; t [email protected]
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and
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dm
dM
11 oo : 07 : 44 [email protected]
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and
dm dM dt = dt = 1
So the equation of motion becomes
dv F = (Mo + ryt) dt
+ vry
dv (F - 1 v) = (Mo+ ryt ) dt v dv t dt 0 ( F - ryv ) 0 (MO + ryt) Perform t he integration, it is not tough _ _!_
,
_ _!_
ln[F - ,v]~
=
_!_
'Y
[ln (m0 + ryt)]~
I In mo + ryt mo F 'Y F mo + ryt = ln F - 1v mo F mo + ryt F - 1v mo
ln F - 1 v
'Y
ln
Simplifying Ft
v = --m0
+ 1t
t he velocity is along the direction of t he force. So writing vectorically Ft v = --m0 + 1 t
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11 oo : 07 : 46 v = ---
m0+,t
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The acceleration
dv a =dt
4.4.3
F
Growing raindrop
A raindrop of initial mass M 0 starts falling from rest under the influence of gravity. Assume that the drop gains mass from the cloud at a rate proportional to the product of its inst antaneous mass and its instantaneous velocity: dM
dt
= kMV
where k is a constant . We'll calculate t he equation of motion and t hen find t he t erminal velocity. Weill proceed like previous problems. Consider the change in moment um of the drop as it gains mass during the time interval from t to t + ~ t Initial moment um at time t P(t) = MV
Final momentum at t ime t + ~t P (t [email protected]
+ ~t) = (M + ~M)(V + ~V) 169
P hysicsguide
11 oo : 07 : 49 Final momentum at t ime t P (t
+ ~ t) =
+~t
(M
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Change in momentum
The force applied is the gravitational force F
=
M
=
g
dP dt
=
MdV dt
VdM + dt
As it is given that d:: = kMV, the equation of motion becomes MdV kMV 2 = M dt + g or dV dt
= - kV2 g
I'm sure you have solved this differential equation while you studied the disspative force problem in section 2.2. So solve t he problem completely. From t he equation of motion you see t hat the acceleration decreases as the falling drop gains speed. The t erminal is the velocity when the acceleration becomes ZERO and t hus when v = Vterminal, ~i; = 0, hence V'terminal
=
g/ k
Here we end our discussion on Basic Newtonian Mechanics. Next is Advanced Topics in Mechanics
00: 00: 00
N e-wtonian Mechanics: Advanced
® +
Sk J ahiruddin *
*Assistant Professor
" " Sister Nibedita Govt. College, Kolkata " Author was the topper of IIT Bombay M.Sc Physics 2009-2011 bat]" • He ranked 007 in IIT JAM 2009 and 008 (JRF) in CSIR NET June 2~ 0 _ He has been teaching CSIR NET aspirants since 2012
(!>
X
1 @Sk Jahiruddin, 2020
Newtonian Mechanics: Advanced
11 oo : oo : oo
N e-wtonian Mechanics: Advanced Sk J ahiruddin *
*Assistant Professor Sister Nibedita Govt. College, Kolkata Author was t he topper of IIT Bombay M.Sc Physics 2009-2011 batch He ranked 007 in IIT JAM 2009 and 008 (JRF) in CSIR NET June 2011 He has been teaching CSIR NET aspirants since 2012
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Newtonian Mechanics: Advanced
11 oo : oo : 03 1 Newtonian Mechanics: Advanced
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Contents 1
2
Conservation of energy in three dimension
3
1.1
4
Conservative force: . . . . . . . . . . . . . . .
Collision in 2D in Center of Mass frame
18
2.1
Formulation of t he problem
2.2
Elastic scattering when target particle at rest 2.2.1
•
•
•
•
•
•
•
•
•
•
•
2
•
•
•
•
•
•
•
•
•
•
18 23
Kinetic energies in the Lab and COM frame • • • • •
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•
•
29
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,
11 oo : oo : os
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Newtonian l\/[echanics: Advanced
Conservation of energy in three dimension
T he conservation of energy can be easily extended t o three dimension. Potent ial energy in t hree dimension becomes dependent on t he distance from t he origin (radius vector) and it is t he integration of forces wit h t he radius vector from a reference point ro . r
U(r)
F (r') · dr'
-
(1.1)
ro
Force and displacement may not be in t he same direction as shown in t he figure below. In two dimension case t his becomes F r cos 0. In t hree dimension we need to perform line integral. T he integrat ion of force give t he difference of kinet ic en-
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ergy as before.
F =mdv dt b
F · dr a
=
dv m--dr dt a b dv m dt · vdt b
(1.2)
a
bm
a
1.1
d
2 dt (v2) dt
Conservative force:
A force F acting on a particle is conservative if and only if it satisfies two conditions: (i) F dep ends only on the particle's posit ion r ( and not on t he velocity v , or t he t ime t , or any other variable; that is, F = F (r) (ii) For any two points 1 and 2 , the work W (l ➔ 2) done by F is the same for all paths b etween 1 and 2 A force F (r) to be conservative a necessary and sufficient condition is to h ave the curl of the force to be zero.
(1.3) Conservative force means the work done is path independent . \"!ve will discuss t hese ooint in details. Let us first write the
11 oo : oo : 1o and su . . . . . cient condition is to h ave the curl of the force to be zero.
(1.3) Conservative force means the work done is path independent. We will discuss these point in details. Let us first write the force and potential energy relation in three dimension. Let us [email protected]
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consider a particle acted on by a conservat ive force F (r), wit h corresponding potent ial energy U (r), and examine the work done by F (r) in a small displacement from r to r + dr. We can evaluate this work in two ways. On t he one hand, it is, by definition ,
W(r
➔
r + dr) = F (r) · dr
(1.4)
= Fxdx + Fydy + Fzdz
for any small displacement dr with components (dx , dy , dz) . On t he other hand, we have seen that the work W (r ➔ r + dr) is t he same as (minus) the change in PE in the displacement:
W(r
➔ r
+ dr) =
+ dr) - U(r )] = - [U (x + dx, y + dy , z + dz) -dU = -[U(r
U (x, y , z) ]
(1.5) We know from t he definition of derivative dU = U(x + dx, y + dy, z + dz ) - U(x, y, z) au au au ax dx + ay dy + a z dz
(1.6)
So the work done is
W(r
➔
r + dr) = -
au ax dx
+
au ay dy
+
au a z dz
(1. 7)
11 oo : oo : 13 -
-
ax
y
az
So t he work done is
au au au axdx + ay dy+ az dz
W (r -+ r + dr ) = -
(1. 7)
The two expression of work in equat ion (1.4) and (1.7) are equivalent . So we get
au Fx = - ax'
au Fy = - ay'
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Fz = -
au az
(1.8) P hysicsguide
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Newtonian Mechanics: Advanced
T hat is, F is t he vector whose t hree components are minus t he three partial derivatives of U wit h resp ect t o x , y , and z . A slightly more compact way to writ e this result is t his:
F =
-x au _ -s, au _ zau ax ay az
(1.9)
From vect or calculus we know t hat t he gradient of a scalar function is defined as
nf _ V
-
af af af X ax + y ay + z az A
A
A
(1.10)
So the force is the gradient of t he potential with a negative sign F =-V U (1.11 ) •
This important relation gives us the force F in t erms of derivat ives of U, just as the definit ion (1.1) gave U as an integral of F. When a force F can b e expressed in the form (4.33), we say t hat F is derivable from a potential energy. Thus, we have sho"'rn that any conservative force is derivable from a potential energy.
11 oo : oo : 16 of F. When a force F can be expressed in the form (4.33) , we say t hat F is derivable from a potential energy. Thus, we have sho~rn that any conservative force is derivable from a potential energy. Example 1.1: The potential energy of a certain particle
is U = A xy 2 + B sin C z , where A, B and Care constants. What is the corresponding force?
Solution: To find F we have only to evaluate the three partial derivatives. In doing this, you must remember that [email protected]
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auI ax is found by different iating wit h respect to X' treating y and Z as constant, and so on. Thus au j ax = Ay 2 , and so on, and the final result is F = - (xAy
2
+ y2Axy + zBC cosCz )
Now we prove the statements about the conservative force stated before. Suppose a force is conservative which essentially means that the work done to move one point to another point is path independent. Look at the figure below. We do work to move particle from point A to B. Path I is the path AC B , Path II is the path AD F B. The work done is same for both paths.
F · dr
WAcB = ACB
F · dr
= WAD FB = ADFB
Path I C ---------.
s
11 oo : oo : 1s ACB
ADFB
Path I C -----------
s
D
A
Path II Figu1·e 1.1 :
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So
F · dr ACB
F-dr=O ADFB
Now t he sign of line integral changes when we reverses the direction of integration.
F · dr + ACB
F · dr = 0 BFDA
hence
F · dr = 0 So the work done in traveling the whole path once in a closed cycle is ZERO. From the stokes theorem we know
F · dr = ~() if thP lin P intPP'r::l.l nf
::i.
(V x F ) · n ds
,rPrtnr ::l.r n,,nn
:=i.
rln~Pn r .,,r,rP i~ 7.Prn
11 oo : oo : 21 From the stokes theorem we know
F · dr =
(V x F ) · nds
So if t he line integral of a vector around a closed curve is zero then the curl of t he vector must be zero. V x F = O
Finally we summaries our discussion of conservative force If a force F is conservative in a region then all these following condition satisfies. Any one of the following five conditions implies all the others. • V x F = 0 at every point of the region. [email protected]
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• § F ·dr = 0 around every simple closed curve in t he region. • F is conservative, that is
JJ F · dr is independent of the
path of integration from A to B. course, lie entirely in t he region.)
(The path must, of
• F · dr is an exact differential of a single valued function.
• F = - V · U , where U is a scalar potential.
11 oo : oo : 23
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Newtonian Mechanics: Advanced
Example 1.2: A particle of mass m moves in the xy plane so t hat its position vector is
r
= acoswti + bsinwtj
where a, b and w are positive constants and a > b. (a) Show t hat t he particle moves in an ellipse. (b) Show t hat t he force acting on t he particle is always directed toward t he origin. (c) Find t he kinetic energy at points A and B of t he figu1~e as shown below. (d) Find t he work done by t he force field in moving the
11 oo : oo : 26 directed toward the origin. (c) Find the kinetic energy at points A and B of t he figure as shown below. (d) Find t he work done by t he force field in moving the particle from A to B. (e) Show that the total work done by the field in moving the particle once around t he ellipse is zero. (f) Show that the force field is conservative. (g) Find the potential energy at points A and B. (h) Find t he total energy of the particle and see t hat it is constant.
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y 8
m b - i - - - - - - ~-
wt A --L-.,;;.~-----.i-,::...::....-a
X
11 oo : oo : 29
Figure 1.2:
Solution: (a) The position vector is r = xi
+ yj
=
acoswti + bsin wtj
So
x
a cos wt, y = bsin wt
=
Then
+ (y/b) = cos wt+ sin wt= 1 t he ellipse is so given by x 2 / a 2 + y 2 /62 = 1 (b) Assuming t he (x/a)
2
2
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2
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particle has constant mass m, the force acting on it is 2
2
dv d r d F = m dt = m dt 2 = m dt 2 [(acoswt)i
=m
2
+ (bsin wt)j ]
2
[- w a cos wti - w b sin wtj] 2
= -mw [a cos wti
+ bsin wtj ] =
2
-mw r
which shows that t h e force is always directed toward the ori•
gin.
11 oo : oo : 31 = =
2 2 m [-w acoswti - w bsinwtj ] 2 2 -mw [a cos wti + b sin wtj ] = -mw r
which shows that t he force is always directed toward t he origin. •
(c) Velocity v =
dr /dt
=
-wasin wti + wbcoswtj
Kinetic energy
~mv
2
2 2
= ~m (w a
sin
2
2 2
2
wt+ w b cos wt)
Kinetic energy at A[ where cos wt = 1, sin wt = O] = Kinetic energy at B [ where cos wt= 0, sin wt= 1] =
1
2 2 mw b
2
1 2 2 mw a 2
(d) Work done B
B
F · dr =
2 2 ( -mw r) · dr = -mw
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Work done
=
r · dr
12
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1
2mw2 (a2 - b2)
=
1
1
2mw2a2 - 2mw2b2
= kinet ic energy at A - kinetic energy at B (e) We can find the work done by directly integrating the force also. See that at A and B , t = 0 and t = 1r /2w respec-
11 oo : oo : 34 Work done = 2mw2 (a2 - b2) = 2mw2a2 - 2 mw2b2 = kinetic energy at A - kinetic energy at B (e) We can find the work done by directly integrating the force also. See that at A and B , t = 0 and t = 1r /2w respect ively. Then t he Work done in moving the particle from A to
B B
F · dr A 1r/2w
[-mw ( a cos wti + b sin wtj )] • [-wa sin wti
+ wb cos wtj ]
2
0 1r/2w
mw
3
(a
2
-
2
b ) sinwtcoswtdt
0
1 2 2 2 2 = -mw ( a - b ) sin wt 2
1r/2w
0
1 2 2 = -mw ( a 2
2
b
-
)
Similarly for a complete circulation around t he ellipse, t goes from O tot= 21r /w. 21r / w
2
3
2
mw ( a - b ) sin wt cos wtdt
Work done 0
l 2 2 = - mw ( a
2
-
2 b ) sin wt
21r / w
2
= 0 0
You can do as you did in part (d ) expect the integral goes from one point to t he same point, say A to A and you get [email protected]
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zero. A
A
F · dr = A
2 2 ( -mw r) · dr = -mw
A
1
n
(A
A
r · dr A
1
11 oo : oo : 36 zero. A
A
F · dr =
A ( -mw
2
r ) · dr = -mw
2
r · dr A
(f) F
= -mw
2
r = -mw 2 (xi + yj )
We calculate t he curl of force •
•
I
V x F =
=
• I
+k
k
J
a;ax 8/8y a;az - mw 2 x -mw2 y 0
a a ( ?. ) 8y (0) - 8z -mw~y i)
ax
(- mw 2y) -
i)
8y
. a ( 2) + J 8z -mW X
-
a 8x (0)
(- mw2 x)
=0 As t he curl of t he force is zero, t he force is conservative.
(g) Since the field is conservative t here exists a pot ential V such t h at
F = -mw
2 •
x1 -
2 •
mw YJ =
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nv - v =
8V. 8V . 8V k 1 Jax 8y az P hysicsguide
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T hen l'"\ T T
I I'"\
?
l'"\ T T I I'"\
?
l'"\ T T / I'"\
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Then
av;ax = mw x, av/ay = mw y, av;az = 0 2
2
from which, omitting the constant, we have V =
1 2 2 -mw x
2
1 2 ? + -mw y~
2
=
1 2 ( ? -mw x~
2
?) + y- =
1 ? ? -mw~r~
2
which is t he required potential. (h) Kinetic energy at any point
Potent ial energy at any point
1 ? 2 V = -mw~r 2
1
.
= mw ( a cos wt + b s1n wt) 2
2
2
2
2
2
Add and get
The energy depends only on the constants a and b so the energy is constant.
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11 oo : oo : 42 15
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Example 1.3: (a) Show t hat the force field F defined by
is a conservative force field . (b) Find a pot ential corresponding to t he force.
Solution: (a) T he force field F is conservative if and only if curlF = V x F = O •
I
a;ax
V x F =
=
•
J
k
8/ 8y
a/az
•
I
=0 (b) As we know the force field F is conservat ive if and only if there exist s a scalar function or potent ial V (x, y, z) such
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that F
= -
Newtonian Mechanics: Advanced
grad V
- VV
=
.
F = _ vv = _ av i _ av j _ av k
ax 2 3 = (y z
-
8y az 2 3 6xz ) i + 2xyz j
2 2
+ (3xy z
-
2
6x z ) k
Hence if F is conservative we must be able to find V such that
av/ ax = 6xz
2
-
8V/8y = -2xyz
2 3
y z 3
2
2 2
av/ az = 6x z - 3xy z
To find the potential we need to integrate each component of the force and omit the common terns. Then 2 2
V = 3x z
-
2 3
xy z
+ 91 (y, z )
where 91 (y , z) is a function of y and z
constant), we have
stant)
11 oo:oo:47
stant) jahir@p hysicsguide .in
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The trick is t o add all t hree V' s and t ake one term only once. 3x 2 z 2 and -xy 2 z 3 both t he terms have appeared twice. We take both the t erm only one t ime. So the required potent ial is V
2
2 2
= 3x z
-
2 3
xy z
+c
Collision in 2D in Center of Mass frame
2.1
Formulation of the problem
Consider two part icles of masses m 1 and m2 with velocities v 1 and v 2 respect ively. The center of mass velocity V is V =
+ m 2v 2 m1 + m2
m 1V 1
As shown in t he part (a) of t he figure below, V lies on t he line joining v 1 and v 2 The velocit ies in t he COM system are V 1c
= V1
-
V
m2
- - - ( V1 -
V 9)
11 oo : oo : 49 As shown in the part (a) of t he figure below, V lies on the line joining v 1 and v 2 The velocit ies in t he COM syst em are V1c
= V1 =
-
V
m2 - - - (v1 - v 2) m1 m2
+
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(a)
(b)
Figure 2.1:
and V2c
= V2 - V =
-m1 - - - (v1 - v 2) m1 m2
+
v 1c and v 2c lie back to back along the relative velocity vector v = v 1 - v 2 as shown in the part (b) of the figure.
The momenta in t he COM system are P lc
= m1 V1c m1m2 = - - - (v1 m1
+ m2
= µv P 2c
=
m2 V 2c
- v 2)
11 oo : oo : s2 P lc
= m1 V1c m1m2 = - - - (v1 - v 2) m 1 +m2 = µv
P 2c
=
= = [email protected]
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m2 V 2c
-m1m2 - - - (v 1 - v 2) m1 +m2 -µv 19
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Hereµ= m 1m2/ (m1 + m2) is t he reduced mass of the system, t he natural unit of mass in a two-particle syst em. The total momentum in the COM system is zero. The total momentum in t he Lab frame is
and since total momentum is conserved in any collision, V is constant. We can use this result to help visualize the velocity vectors before and after t he collision. Now look at the figure below. Part (a) visualizes the path of two colliding particles before and after the collision in the lab frame. Part ( b ) shows the init ial velocities in the Lab and COM systems. All the vectors lie in the same plane, and v 1c and v 2c must be in opposite direction as t he total momentum in the C system is zero. After t he collision, as shown in part ( c ) , t he velocities in t he COM system are again in opposite direction. Part (c) also shows t he final velocities in the lab system. Note that the plane of part (c) is not necessarily the
11 oo : oo : 55 in the C system is zero. After t he collision, as shown in part ( c ) , t he velocities in the COM syst em are again in opposite direction. Part (c) also shows t he final velocities in the lab system. Note that the plane of part (c) is not necessarily the plane of part (a)
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~ - __ __ ,.,.. ,,, ,,/4 V2
' , ', v' I V1 c
-
- -----~
~
V2
V
,
V
JI V1 c
Y2
(a)
2c
v'2
(b)
(c)
Figure 2.2:
We will derive the mathematical formulation for the elastic collision and leave t he inelastic collision as the treatment will be complicated. Conservation of energy applied to the COM system gives, for elastic collisions, 1
2 2mi V i c
+
1
2 2 m2v2c
=
1
,2 2mi Vi c
+
1
,2 2 m2V2c
Total momentum is zero in t he C system. We therefore have
11 oo : oo : s1 2 2 m1 Vi c
+
2 2m2v 2c
=
,2 2 m1 V i c
,2 2 m2v2c
+
Total moment um is zero in t he C syst em. We therefore have
Using momentum conservation to eliminate v2c and t he energy equation gives 1 2
2 Vlc
1
= 2
m1
+
v;cfrom
m2 1 m2
or
(2.1) [email protected]
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Similarly, (2.2)
In an elastic collision, the speed of each particle in the COM syst em is t he same before and after the collision; the velocity vectors simply rotate in the scattering plane. Now consider one of t he particles, m 2 is init ially at rest in the laboratory. So v 2 = 0. This case is important as it happens in many real experiment. In our earlier discussion of velocities before the collision, substitute v 2 = 0. V
m1 = ---V1 m 1 +m2
V1 c = V 1 -
V
m2 = - - - -V 1
m,
+ m')
11 oo : 01 : oo substitute v 2 = 0. V
=
m1
---V1
m 1 Vi c
=
V 2c =
V1 -
- V
+ m2 V
=
m2
- - - -V i
+ m2
m 1 m1
= - - - - -V 1 m1
+ m2
The situation is explained in t he figure below. Here the trajectories after t he collision in the COM and Lab systems are shown. You see that v ~ makes angle 01 , and v ~ makes angle 02 in Lab frame. In the COM frame the velocity vectors rotates with angle 8 which is called the scattering angle. Because t hese angles are in L , t hey are in principle [email protected]
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surable in the lab. The velocity diagrams can be used to relate 01 and 02 to the scattering angle 8 , as we shall do just now Lab V
---►
COM V1
~--·------► Y2c Y 1c
m1 ..._ _ __,.,■ m 2
v 1'
_____ , -----.,,___......,..,~-:= -
0
- - ~-7-: ?~ 0
---- -
Y 1c
Figure 2.3:
Y2c
11 oo : 01
: 02
Figu1·e 2.3:
2.2
Elastic scattering when target particle at rest
Now consider the elastic scattering of a particle of mass m 1 and velocity v 1 from a second part icle of mass m 2 at rest . The scattering angle 8 in t he COM syst em is unrestrict ed, but t he conservation laws impose limitations on t he laboratory angles, as we now show. The center of mass velocity has 23
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magnit ude V =
m1v1 m1 + m2
(2 _3)
The velocity of COM is parallel t o v 1 . The initial velocities in the COM syst em are V 1c
V 2c
=
=
m2
- - - - V1
m1 +m2 m1
(2.4)
- - - - - v1
m 1 +m2
The mass m 1 is scattered t hrough angle 8 in the COM syst em as shown in figure above. This essentially means that the mass m 1 deviates an angle 8 from its original direction in the COM
11 oo : 01 : os V 2c
=
- - - - - V1
m1 +m2
The mass m 1 is scattered t hrough angle 8 in t he COM system as shown in figure above. This essentially means that the mass m 1 deviates an angle 8 from its original direction in t he COM frame. Now look at the diagram below. This is velocity diagram of the collision. Look at the figure 2.3 first . You must see t hat the initial direction of m 1 is same both in Lab and COM frame (although the magnitude of velocities are different) as well as t he velocity of COM frame which is V. 01 is t he angle between t he initial and final direction of mass m 1 in t he Lab frame. 8 is t he angle between the initial and final direction of mass m 1 in t he COM frame. As initial direction of m 1 is same both in Lab and COM frame we draw t he vector direction of V , v1, V1c along same line.
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;
v'1
•
V1
01
V
,
; ; V1 c ; ;
;
8
-- --- • V1 c
Figure 2.4:
Now it should be clear from t he velocity addition rule that
11 oo : 01 : os Figure 2.4:
Now it should be clear from t he velocity addition rule that I
.
v 1 Sln
0=
I
.
e
V i c Sln -
and v~ cos 0
= V + v~c
Hence we write
e V1c Sln tan 01 = - - - - V + v~c cos 8 I
•
since the scattering is elastic, v~c =
V1c •
Hence
sin 8 tan 01 = - - - - V + V1c cos 8 sin8 (V /v1c) + cos 8 V1c
From (2.3) and 2.4 we get V/v1c = m1 / m2. So, [email protected]
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sin8 tan0 1 = - - - - - - (m1/ m2) + cos 8
(2.5)
You can easily check few important results
(2.6) and
A
11 oo : 01
: 1o
You can easily check few important results
(2.6) and
(2.7)
(ii)
(iii) If m 1 < m 2 , all scatt ering angles for m 1 in LS are possible. (iv) If m 1 = m 2 , scattering only in t he forward hemisphere (0 :::; 90°) is possible. (v) If m 1 > m 2 , t he maximum scattering angle is possible,
01max, being given by
We'll prove some important relations relating t he scattering angle of target mass m 2 in Lab and COM frame. Look at t he figure below. The scatt ering angles in Lab and COM frame has been drawn together in upper left figure (a), the COM frame alone in upper right (b) and t he vect or diagram 26
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of fin al state of target mass both in Lab and COM m 2 below
(C). COM
Lab
v;
00 : 01 : 13 (C). COM
Lab
v; 1
,:,___-==-:::-::--=::,:: :=-v.,_)_-r ~ _ 0
02
0_ 7r-0
COM fl'lmo
(a)
(b) I
/ v'
le
0
V
(c)
Figure 2.5:
We get by balancing the velocity vectors along y axis
. 02 = V2c ' Sln . 8V2' Sln And by balancing the velocity vectors along x axis
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Divide first equation by second tan 02 = v I
V 2c
sin8 -
cos 8
11 oo : 01
: 1s
Divide first equation by second tan02
=
sin8 _v____ , - cos 8 V 2c
From equations (2.1 ), (2.2) , (2.3) and (2.4) we easily see t hat the magnitude of velocity of COM frame and t he velocity of m 2 after the collision in COM frame are equal, i.e
(2.8) Hence
sin8 8 tan 02 = = cot 1 - cos 82 We may write this as tan0 2 = tan
7r
8
2
2
(2.9)
(2. 10)
Thus 02 =
1
- (1r - 8 )
(2.11)
2
This is an important result. In t he special case of m 1
=
m2, we already know m1
= m2
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Combining wit h (2.11) we get
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: 1s
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Combining with (2.11) we get pi
01
•
+ 02 = 2 ;
when
m1 = m 2
(2.12)
This result we have proved before.
2.2.1
Kinetic energies in the Lab and COM frame
KE in lab frame
KE in COM frame
Using equation (2.4) we have
Ti c
1
m 1m 2
2
m 2
= - - - - -Vi = - - - - T i
2 m 1 + m2 m 1 + m2 This result shows t hat the init ial kinetic energy in the COM system Tc is always a fraction m 2 / ( m 1 + m 2 ) < 1 of the init ial LAB energy. For the final COM energies, we find 2
T1
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00 : 01 : 21
29
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and
T he ratio of KE of m 1 before and aft er collision
Using cosine law in figure 2.4
2 V 'l c
=
2 VI'
2
+V -
2VI'V COS 01
Hence
We already know and
V
We also know v~c sin 8 = v~ sin 01 Using t his we get , sin8 V1 . cSln 01
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: 23
, sin8 Vlcsin 01
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and using (2.5) sine cos 01 sin 8 m1 = - - = cos 8 + tan 01 m2 sin01 So t hat cos 8 + _m_i m2
Doing the substitut ions and some algebra (do yourself) we get
2
m1
cos 8 + -
m2
T his simplifies to
T{ = 1 T1
2m1m2 (l -cos e- ) 2 (m1 + m 2)
(2.13)
This is t he ratio of the KE of t he projectile part icle m 1 before and after collision. Do each step in this formulation carefully. There can be many problems which can be formed based on t his discussion. If you are thinking why did we calculate t he ration of KE of the project ile particle, I would like to mention that in many experiment we have the idea about t he ratio of the KE of t he projectile and we need t o calculat e at which
11 oo : 01
: 26
this discussion. If you are thinking why did we calculate the ration of KE of the projectile particle, I would like to mention that in many experiment we have the idea about the ratio of the KE of t he projectile and we need to calculate at which 31
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angle t he particle detector should b e placed, i.e, calculate 01 . We will see in an example shortly. Prove yourself t hat (2. 14)
Example 2.1: What is the maximum angle that 01 can attain for the case V > vie? What is 01 max for m1 and m1
>> m2
= m2?
Solution: The scattering angle 8 depends on the details of t he interaction , but in general it can assume any value. If m 1 < m 2 , it follows from equation (2.5) or t he geometric construction in part (a) of the figure below that 01 is unrestricted. However, the sit uation is quite different if m 1 > m 2 . In this case 01 is never greater than a certain angle 01,max · As figure (b) shows, the maximum value of 01 occurs when v~ is perpendicular to v ~c· Now look at the geometry of part (b) of the figure. You see
(don't forget v~r.
= V1c) If
00 : 01 : 28 c·
Now look at the geometry of part (b) of t he figure. You see
(don 't forget v~c
= V1c)
If
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and the maximum scattering angle in Lab frame approaches zero. Increasing
e
I
----- .... ....
/
/
1/
\
I
'''
\
\
I
\
\
01,max /
E)
\
D ____ ':
V
I
/
\
I \
I \
(a)
\
''
I
....
I
/
'
-- ----
/
....
/
/
(b)
Figure 2.6:
Think on your fam iliar example, if m 1 /m2 < 1 as in (a), t his is like a flow of sm all m arbles hitting a big ball; t he marbles scatter in all directions.
On t he other h and , if a
moving big ball hits small m arble, then m 1 /m2 big ball do not deflect much. For m1 = m2
>>
1 and t he
11 oo : 01
: 31
marbles scatter in all directions.
On the other hand, if a
moving big ball hits small marble, then m 1 /m2 >> 1 and the big ball do not deflect much. For
m1
=
m2
tan 01 =
sin8 1 + cos 8-
=
tan (8 / 2)
so that
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Example 2.2: If a particle of mass m 1 collides elastically wit h one of mass m 2 at rest, and if m 1 mass is
scattered at an angle 01 and t he mass m 2 recoils at an angle 02 with respect to t he line of motion of the incident particle, (see figure 2.3) t hen show that sin(202 + 01 ) sin01
m1 m2
Solution: This can be easily done with the help of relation between angles in COM and Lab frame. As we know from equation (2.5) tan01 =
sin8 ------(m1/m2) + cos 8
Again we know from (2 .11) 8 =
7r -
202
or sin 8 = sin(1r and
-
202) = sin 202
11 oo : 01
: 34
or sin 8 = sin(1r
-
202) = sin 202
and Using t hese tan 01 =
sin01 cos 01
sin8 -----(m1/m2) +cos 8 sin 202 (m1/m2) - cos 202
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Doing cross mult iplication and rearrangement 1
m sin 01 = sin 01 cos 2 + cos 01 sin 202
= sin (01 + 202)
m 2
hence
m1 m2
sin(202 + 01) sin 01
Example 2.3: P art icles of m ass m 1 elastically scat ter from part icles of m ass m 2 at rest . (a) At what LAB angle should a part icle det ector be set to detect part icles t hat lose one-third of t heir m oment um? (b) Over what range m 1/ m 2 is this possible? (c) Calculate t he scattering angle for m 1/m2 = 1
Solution: (a) In t he Lab frame given t hat , 2 m1V1 = ~m1V1
==> ~>
2 V1 = ~V1 ,
11 oo : 01
: 36 =1
(c) Calculate the scat tering angle for m 1 / m 2
Solution: (a) In t he Lab fr ame given t hat ~
, 2 m 1Vi = 3 m1 V1
Ratio of final and initial frame is 2 lm v' T'1 ? 1 1 v~2 2 lm v T1 2 1 1 ~
vr
,
Vi =
~>
2
3 V1
kinetic energy of particle m 1 in Lab 2 3
2
= 1-
2m 1m2 (m1 + m2)
2
(1
- cos e )
This equation can b e solved for cos 0, giving 2 5 (m1 + m2) cos 8 = 1 - - - - - - = 1 -y l 8m1m2 35
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2
5 m1 + m2 1 m 1m 2 We need 01 here so use equation (2.5) .
tan 01
=
✓2y - y2
sin8 -----cos 8 + m1/m2
This is the angle where the det ector should be placed. We know the values of m 1 and m 2 before the experiment , so the calculation of 01 can be done from t he knowledge of the masses. (b) Because tan 01 must be a real number , only values for m 1 /m2 where 2 - y > 0 are possible. Therefore, 2
_ 5 (m1 + m2) > 2 0 18m1m 2 -
we write t his as
11 oo : 01
: 39
m 2 where 2 - y > 0 are possible. Therefore, 2
_ 5 (m 1 + m2) > 2 0 18m1m 2
we write t his as 2
-5 Now t aking x
-5> 0
+ 26
= m 1 /m2 we write this equation as -5x
2
+ 26x -
5>0
If we make the inequality as equality, i.e, -5x 2 + 26x - 5 = 0, 1 we get x = , 5 Hence 5
1
m1 - 0
;
Hint and discussion: Put t he force in t he equation of orbit - Equation-1 .2. Get the equation
µA
l --
z2
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Central Force
Substitute uk
l
l
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Central Force
Substitute V =
µk l u----l2 1 - ~ z2
To get 1-
µ>..
l2
V
=0
. Then examine for three cases - 1) >..
.. >
2
l
µ
and 3)
>.. = ~. Convince yourself that only in 1st case you get some kind of orbit and in case two and three the particle spirals in toward the force center.
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Central Force
Substitute V =
µk l u----l2 1 - ~ z2
To get 1-
µ>..
l2
V
=0
. Then examine for three cases - 1) >..
.. >
2
l
µ
and 3)
>.. = ~. Convince yourself that only in 1st case you get some kind of orbit and in case two and three the particle spirals in toward the force center.
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Central Force
Substitute V =
µk l u----l2 1 - ~ z2
To get 1-
µ>..
l2
V
=0
. Then examine for three cases - 1) >..
.. >
2
l
µ
and 3)
>.. = ~. Convince yourself that only in 1st case you get some kind of orbit and in case two and three the particle spirals in toward the force center.
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43
Physicsguide
11 oo : 02 : os [email protected]
42
Physicsguide
@Sk Jahiruddin, 2020
Central Force
Substitute V =
µk l u----l2 1 - ~ z2
To get 1-
µ>..
l2
V
=0
. Then examine for three cases - 1) >..
.. >
2
l
µ
and 3)
>.. = ~. Convince yourself that only in 1st case you get some kind of orbit and in case two and three the particle spirals in toward the force center.
[email protected]
43
Physicsguide
11 oo : 02 : 01 [email protected]
42
Physicsguide
@Sk Jahiruddin, 2020
Central Force
Substitute V =
µk l u----l2 1 - ~ z2
To get 1-
µ>..
l2
V
=0
. Then examine for three cases - 1) >..
.. >
2
l
µ
and 3)
>.. = ~. Convince yourself that only in 1st case you get some kind of orbit and in case two and three the particle spirals in toward the force center.
[email protected]
43
Physicsguide
11 oo : 02 : 1o [email protected]
42
Physicsguide
@Sk Jahiruddin, 2020
Central Force
Substitute V =
µk l u----l2 1 - ~ z2
To get 1-
µ>..
l2
V
=0
. Then examine for three cases - 1) >..
.. >
2
l
µ
and 3)
>.. = ~. Convince yourself that only in 1st case you get some kind of orbit and in case two and three the particle spirals in toward the force center.
[email protected]
43
Physicsguide
11 oo : 02 : 12 [email protected]
42
Physicsguide
@Sk Jahiruddin, 2020
Central Force
Substitute V =
µk l u----l2 1 - ~ z2
To get 1-
µ>..
l2
V
=0
. Then examine for three cases - 1) >..
.. >
2
l
µ
and 3)
>.. = ~. Convince yourself that only in 1st case you get some kind of orbit and in case two and three the particle spirals in toward the force center.
[email protected]
43
Physicsguide
11 oo : 02 : 1 s [email protected]
42
Physicsguide
@Sk Jahiruddin, 2020
Central Force
Substitute V =
µk l u----l2 1 - ~ z2
To get 1-
µ>..
l2
V
=0
. Then examine for three cases - 1) >..
.. >
2
l
µ
and 3)
>.. = ~. Convince yourself that only in 1st case you get some kind of orbit and in case two and three the particle spirals in toward the force center.
[email protected]
43
Physicsguide
11 oo : 02 : 1 a [email protected]
42
Physicsguide
@Sk Jahiruddin, 2020
Central Force
Substitute V =
µk l u----l2 1 - ~ z2
To get 1-
µ>..
l2
V
=0
. Then examine for three cases - 1) >..
.. >
2
l
µ
and 3)
>.. = ~. Convince yourself that only in 1st case you get some kind of orbit and in case two and three the particle spirals in toward the force center.
[email protected]
43
Physicsguide
11 oo : 02 : 20 [email protected]
42
Physicsguide
@Sk Jahiruddin, 2020
Central Force
Substitute V =
µk l u----l2 1 - ~ z2
To get 1-
µ>..
l2
V
=0
. Then examine for three cases - 1) >..
.. >
2
l
µ
and 3)
>.. = ~. Convince yourself that only in 1st case you get some kind of orbit and in case two and three the particle spirals in toward the force center.
[email protected]
43
Physicsguide
11 oo : 02 : 23 [email protected]
42
Physicsguide
@Sk Jahiruddin, 2020
Central Force
Substitute V =
µk l u----l2 1 - ~ z2
To get 1-
µ>..
l2
V
=0
. Then examine for three cases - 1) >..
.. >
2
l
µ
and 3)
>.. = ~. Convince yourself that only in 1st case you get some kind of orbit and in case two and three the particle spirals in toward the force center.
[email protected]
43
Physicsguide
11 oo : 02 : 2s [email protected]
42
Physicsguide
@Sk Jahiruddin, 2020
Central Force
Substitute V =
µk l u----l2 1 - ~ z2
To get 1-
µ>..
l2
V
=0
. Then examine for three cases - 1) >..
.. >
2
l
µ
and 3)
>.. = ~. Convince yourself that only in 1st case you get some kind of orbit and in case two and three the particle spirals in toward the force center.
[email protected]
43
Physicsguide
11 oo : 02 : 2a [email protected]
42
Physicsguide
@Sk Jahiruddin, 2020
Central Force
Substitute V =
µk l u----l2 1 - ~ z2
To get 1-
µ>..
l2
V
=0
. Then examine for three cases - 1) >..
.. >
2
l
µ
and 3)
>.. = ~. Convince yourself that only in 1st case you get some kind of orbit and in case two and three the particle spirals in toward the force center.
[email protected]
43
Physicsguide
11 oo : 02 : 30 [email protected]
42
Physicsguide
@Sk Jahiruddin, 2020
Central Force
Substitute V =
µk l u----l2 1 - ~ z2
To get 1-
µ>..
l2
V
=0
. Then examine for three cases - 1) >..
.. >
2
l
µ
and 3)
>.. = ~. Convince yourself that only in 1st case you get some kind of orbit and in case two and three the particle spirals in toward the force center.
[email protected]
43
Physicsguide
11 oo : 02 : 33 [email protected]
42
Physicsguide
@Sk Jahiruddin, 2020
Central Force
Substitute V =
µk l u----l2 1 - ~ z2
To get 1-
µ>..
l2
V
=0
. Then examine for three cases - 1) >..
.. >
2
l
µ
and 3)
>.. = ~. Convince yourself that only in 1st case you get some kind of orbit and in case two and three the particle spirals in toward the force center.
[email protected]
43
Physicsguide
11 oo : 02 : 3s [email protected]
42
Physicsguide
@Sk Jahiruddin, 2020
Central Force
Substitute V =
µk l u----l2 1 - ~ z2
To get 1-
µ>..
l2
V
=0
. Then examine for three cases - 1) >..
.. >
2
l
µ
and 3)
>.. = ~. Convince yourself that only in 1st case you get some kind of orbit and in case two and three the particle spirals in toward the force center.
[email protected]
43
Physicsguide
11 oo : 02 : 38 [email protected]
42
Physicsguide
@Sk Jahiruddin, 2020
Central Force
Substitute V =
µk l u----l2 1 - ~ z2
To get 1-
µ>..
l2
V
=0
. Then examine for three cases - 1) >..
.. >
2
l
µ
and 3)
>.. = ~. Convince yourself that only in 1st case you get some kind of orbit and in case two and three the particle spirals in toward the force center.
[email protected]
43
Physicsguide
11
00:02:41
[email protected]
42
Physicsguide
@Sk Jahiruddin, 2020
Central Force
Substitute V =
µk l u----l2 1 - ~ z2
To get 1-
µ>..
l2
V
=0
. Then examine for three cases - 1) >..
.. >
2
l
µ
and 3)
>.. = ~. Convince yourself that only in 1st case you get some kind of orbit and in case two and three the particle spirals in toward the force center.
[email protected]
43
Physicsguide
11 oo : oo : oo
Rigid Body: Advanced Sk J ahiruddin*
*Assistant Professor Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay M.Sc Physics 2009-2011 batch He ranked 007 in IIT J AM 2009 and 008 (JRF) in CSIR NET June 2011 He has been teaching CSIR NET aspirants since 2012
1 @Sk J ahiruddin, 2020
Rigid Body: Advanced
11 oo : oo : 03 1 Rigid Body: Advanced
@Sk Jahiruddin, 2020
Contents 1 Introduction
2
1.1
Velocity and Acceleration in Rotating Frame .
5
1.2
Coriolis and Centrifugal forces . . . . . . . . .
9
Moment of Inertia Tensor 2.1
3
4
MOI Tensor and Principle axis .
•
•
•
•
•
•
•
•
Angular Momentum of Rigid Body
3.2
21
26 •
26
Different expressions of Rotational Kinet ic Energy of Rigid body . . . . . . . . . . . . . . .
30
3.1 Angular Momentum
4
16
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•
G eneral motion of Rig id body
32
4.1
Rotation . . .
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32
4.2
Euler 's angle .
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36
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11 oo : oo : 06 •
jahir@physicsguide .in
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@Sk Jahir uddin, 2020
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P hysicsguide
Rigid Body: Advanced
4.3
Angular velocity in terms of Euler 's angles . .
37
4.4
Euler Equations . . . . . . . . . . .
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39
4. 5
Force free motion of symmetric top
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40
4.6
MOI ellipsoid . . . . . . . . . . . .
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44
11 oo :oo :oa
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1
Physicsguide
Rigid Body: Advanced
Introduction
Rigid bodies, defined to b e a collection of N points const rained so that t he distance b etween the points is fixed. i.e ri -
rj
= Constant
(1.1)
for all i, j = 1, ... , N . A simple example is a dumbbell (two masses connected by a light rod) , or t he pyramid drawn in t he figure. In both cases, t he distances between t he masses is fixed .
Often we will work wit h continuous, rather than discrete, bodies simply by replacing Lti mi ➔ J drp(r ) where p(r ) is
11 oo : oo : 11
Often we will work with continuous, rather than discrete, bodies simply by replacing Lti mi ➔ J drp(r) where p(r) is the density of t he object. A rigid body has six degrees of [email protected]
Physicsguide
4
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Rigid Body: Advanced
freedom 3 Translation
1.1
+ 3 Rotation
Velocity and Acceleration in Rotating Frame
Any external coordinate syst em which is fixed in space is called space fixed axis as shown in OXYZ axis in the figure. It is convenient to define a second set of axes which is fixed with the body, i.e. it moves or rotates with the body. We denote t his set of axes by Cxyz where C is t he origin of this body-fixed'' system of coordinates. Frequently (but not necessarily) C is chosen to be the centre of mass of the body z
y
C
X
O·) : < - - - - - - - Y
X
11 oo : oo : 13
o~
______ v
X
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5
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Rigid Body: Advanced
Let us consider t he situation where Cxyz axes rotate about t he OZ axis where t he points O and C coincide. Our interest will be to find t he relationship between dynamical quantities in the rotating frame of reference with those in the fixed frame. Let us consider what happens to a vector when it is rotated about a fixed axis by an angle 6
~►
p
=-
q
P
-- p -- oq q
Integrating we get
F2 = Plogq Now different iate F 2 with respect to P to get
Q = log q ⇒ q =
eQ
Now the new Hamiltonian is p2 1 H = 2 + 2q2
p 2 + 1 e-2Q = E 2
Constant
11 oo : 01
: 36
Now the new Hamiltonian is
p2 1 H - -2 +2q-2
p 2
+ 1 e-2Q =
E
Constant
2
Now
8H = ( P 2 + l ) e- 2Q = 2E P· = - --
8Q
Integrating
P = 2Et + C We also get p 2+ 1
2
2Et + 2Ct +
2E 35
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c2 + 1 E 2
P hysicsguide
@Sk Jahiruddin, 2020
Hamiltonian .Nlechanics
Q = log
2 c + 1 2 2Et + 2Ct + - -
2E
Example 4 .20: If a generating function of typ e 1 is given by
Find t he new canonical variables Q and P
Solution: We see Q p=- =q2 8F1
an d
oq
Writing t he new coordinates as function of t h e old ones gives
Q = pq
2
and
P =
~
q
11 oo : 01
: 39
Writing the new coordinates as function of t he old ones gives Q
= pq
2
P =
and
!q
Example 4.21: Prove t hat the following transformation is canonical and find a suitable generating function for it. Q
p
= ln
q
2
P =-
and
q
2
+l
p_ q
Solution: We know that for type 1 generating function
_8F_1
oq
[email protected]
= P = qeQ
Physicsguide
36
@Sk J ahiruddin, 2020
Hamiltonian Nlechanics
Which gives us
And p = _
8F1
2
= _ q eQ
2
8Q ⇒ g(Q)
=
2
dg
q 2
dQ
+l
eQ
So the generating function is given by 2
q
2
+1
eQ
2
p
q
q
2
+1
eQ
11 oo : 01
: 42
So the generating function is given by 2
q
2
4.4
+1
eQ
Canonical transformation and cyclic co-
ordinates Let us imagine t hat we find coordinates qi t hat are all cyclic. Then
Pi= o, so
are all constant. If H is conserved , then:
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37
Physicsguide
Hamiltonian Niechanics
is also constant in t ime. In such a case the remaining equations of motion:
All coordinates are linear in t ime and the motion becomes very simple. We might imagine searching for a variable transformation to make as many coordinates as p ossible cyclic. The search of such types of coordinat es is guided by t he canonical transformation as we can not t ransform the coordinates and momentums arbitrarily and make the new coordinates cyclic.
11 oo : 01
: 44
mation to make as many coordinates as possible cyclic. The search of such types of coordinates is guided by t he canonical transformation as we can not t ransform the coordinates and momentums arbitrarily and make t he new coordinates cyclic.
5
Poission Bracket
Poisson bracket of two dynamical variables u(p , q, t) and v(p.q.t) is defined by
(5 .1) It is easy to check t hat the following fundamental Poisson bracket relations are satisfied:
There are a few other properties of note. These include: [email protected]
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Hamiltonian lVIechanics
{u, u} = 0
{u ,v} =-{v,u} {au + bv, w}
= a{ u, w} + b{ v, w}
+ {u, w }v
{uv, w} = u{v, w} { U, { V,
W}} + {V, { W, U}} + {W, { U, V}} = 0 df
{f(u) ,v} = du {u ,v } r
,. /
, -....
df
r
"
11 oo : 01
: 47 {uv, w} = u{v, w} + {u, w }v
{ U, { V,
W}}
+ {V, { W, U}} + {W, { U , V}} =
Q
df
{J (u),v } = du {u ,v } df
{u, f (v)} = dv {u, v}
(5.3) Now we need to learn one very important Theorem: A transformation Qj canonical if and only if:
=
Qj(q,p, t) and Pj
=
Pj (q ,p, t ) is
Poission bracket of any two dynamical variable is invarian under a canonical transformation , i.e
[A, B ]q,p = [A , B ]Q,P
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39
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(5.5)
Physicsguide
Hamiltonian Nlechanics
Example 5.22: Consider t he following transformation
(a) Is the transformation canonical? (b) Determine the new Hamiltonian. (c) Write down the Hamilton's equations for both H ~rn ilt"ni~nc:: ~nrl c::"l,rP f()r th P , r~ri~ hlPc::
11 oo : 01
: 49
(a) Is the transformation canonical? (b) Determine the new Hamiltonian. (c) Write down the Hamilton's equations for both Hamiltonians and solve for the variables. (d) Check that t he solutions set of both old and new Hamiltonian are same.
Solution: Calculate Poisson Bracket for t he new coordinates with respect to old coordinates
QP
= oQoP _oQoP = I [ ' ]q,p oq op op oq So t he transformation is canonical. (b) q = vf-P and p = 2Qvf-? Substituting the values of p and q in the old Hamiltonian to get t he new Hamiltonian as function of new coordinates.
~ (1 + 4Q
H'(Q, P ) = -
2
)
(c) Hamilton's equations in old coordinates •
•
q = p;
p = - q
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40
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Physicsguide
Hamiltonian Nlechanics
Hamilton's equation in new coordinates
. oH' Q= - = oP
1 ( 2) - - 1 + 4Q 2
p = _oH' = 4PQ oQ
11 oo : 01 : s2 . Q
=
8H' -= 8P
1 ( 2) - - 1 + 4Q 2
p == _8H' == 4PQ 8Q
We solve for old coordinates
q == A sin t + B cos t
q == p == - q;
➔
p == - q == p ;
➔ p
== C sin t + D cos t
At t == O; q == q0 and p == p0 . Hence the solution q == qo cost + Po sin t
and p(t) == - qo sin t + Po cost
For new coordinates the equations are not very easy to solve. However we can just substitute the solut ions in old coordinates to new coordinat es with t he t ransformation given in the problem. Q t = p == - qo sin t + Po cos t () 2q 2 (q0 cos t + Po sin t ) 2
P (t ) = -q == - (qocos t+ Po sin t)
.
2
Example 5.23: Show that t he time indep endent transformation given belovv is canonical and obtain a suitable generating function of type F3 (p, Q) p == p + q2 jahir@p hysicsguide .in
@Sk Jahir uddin, 2020
+ pq2;
Q == tan- 1 q
41
P hysicsguide
Hamiltonian :Niechanics
Solution: 2
Q p = 8Q8P _ 8Q8P == l + q == l [ ' ]q,p 8q 8p 8p 8q l + q2
11 oo : 01
: 55
Solution: Q p
[ '
= 8Q8P _ 8Q8P = l
]q,p
8q 8p
l
8p 8q
2
+q + q2
= l
So the transform ation is canonical. It is given that we need to find the generating function of the type F3 (p, Q). Hence we need to write q and Pin terms of Q and p. q=tanQ;
P = p (l
+q ) +q 2
2
So
p sec Q + tan Q 2
=
2
8F3
q = tan Q = - -
8p
2
P = p sec Q
+ tan
2
8F3
Q = - DQ
From first equation F3 = -p tan Q + f 1 ( Q) and from 2nd equation F3 = -p tanQ - tanQ + Q + f 2 (p) . So t he generating function is F3
=
-p tan Q - tan Q
+Q
otherwise we can write at first
dF = 8 F3d 8 F3dQ 3 8p p + 8Q
= - tan Qdp F3 [email protected]
@Sk Jahiruddin, 2020
=
2
(p sec Q
+ tan
-p tan Q - tan Q 42
2
Q )dQ
+Q Physicsguide
Hamiltonian l\/Iechanics
Example 5.24: F ind the conditions that need to be sat-
11 oo :01 : s1 @Sk Jahiruddin , 2020
Hamiltonian Nlechanics
Example 5.24: Find t he conditions that need to be satisfied by the real constants a , /3, r', 8 so t hat t he t rans-
formation is canonical.
Solution: -y- l 'Y fJ - l _ 8Q8P 8Q8P _ [Q, P ]qp- 8 8 - 8 8 - apryq xO -aq /38q
'
q
p
p
q
= a/38q'Y+fJ - I If the Poisson Bracket is t o be 1 t hen a/38 = - 1 and 1+8- l 0
jahir @physicsguide.in
@Sk Jahiruddin, 2020
43
=
P hysicsguide
Hamiltonian Nlechanics
11 oo : 02 : oo 43
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Hamiltonian Nlechanics
Example 5.25: A particle t hat has coordinatedependent mass mq 2 is attached to a spring t hat has spring constant k = mg. The particle's Hamiltonian is
H =
1 2 2mq--? + -2 mgq p2
(a) Derive t he HAmilton's equation of motion. (b) Show that t he coordinat e transfo1~mation z
=
1 - q2 and Pz 2
p
=-
q
is a canonical transformation. Derive t he Hamilton equat ions of motion for z and Pz and find t he general solutions for z(t) and Pz(t ) in terms of int ial values z (O) and Pz(O). (c) Let us define a new coordinate P which is equals to new Hamiltonian. p2
P = H =
z
2m
+mgz
Find another coordinate Q which is canonical to P. (Q and P that make up a pair of canonical coordinates.) Derive t he Hamilton 's equation for Q and P, and solve Q (t) and P (t) in terms of init ial values of Q and P.
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11 oo : 02 : 02
44
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Hamiltonian :Niechanics
Solution: (a) 8H q= 8p •
.
p
P= -
- -·
mq2'
8H 8q
=
p2
mq 3
-mgq
(b) 8 z 8pz 8 z 8pz l [z, pz] qp= 8 8 - 8 8 =q--0= 1 ' q p p q q
So the t ransformation is canonical. Let us try to find a second type generating function F2(q, Pz) . We have studied t he convent ion of Q and P as the transformed coordinates. Here the new variables are Q = z and P = Pz· >
and
8F2
Z
, 1 2 = Bpz = 2q + f (pz)
iF
As t he partial derivative Pz2 matches with the definition of z (z = ½q2 ), we can take f'(P z) = 0, then. So
Well, now you see that the Hamiltonian becomes in t erms of new variables p2 H= + mgz
:n 2
11 oo : 02 : os Well, now you see that t he Hamiltonian becomes in terms of new variables p2 H = + mgz 2
:n
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45
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Hamilt onian l\/Iechanics
This hamiltonian is exactly same hamiltonian as a particle in a gravitational field. So we get .z = Pz m
d an
. Pz = -mg
So the solution Z
1 2 = - gt 2 Pz
+
Pzo
- t m
+ Zo
= -mgt + Pz
0
(c) Given P = H =
p2 z
+mgz
2m We need to find a Q, such t hat the transformation is canonical. Lets t ry to find a generating function F2 (z, P ).
EJF2 ✓ oz = Pz =2m(P - mgz ) Hence 2~ /2 3 F2( z , P ) = - - - - (P - mgz ) + f (P) 3 mg 2 v'2 = - - - (P - mgz )312 + f (P ) 3g m 1
/2 .
,
1
In
..
.
11 oo : 02 : oa P)
F 2(z,
3 2
= - - - - (P -
mgz) 1
3 mg
+ f (P)
2~ 3 2 = - - - (P - m gz ) 1 + f (P ) 3g m 2 1 2 - (P - mgz) l m
+ J'(P )
46
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Now use P =
Hamiltonian .Nlechanics
2
J~ + mgz to get 2
Q
= _ Pz + J'( Pz + mgz) 2m
mg
- !-i~ - We could not have
If we choose f' (P) = 0.We get Q = chosen f' (P) = 0. But as we see
{Q, P} = {-
p
p2 z ,
z
mg 2m
+mgz } = l
So P and Qare canonical variables. If we have chosen J' (P) i= 0, then the Q would have been more complex. And as our task is t o choose just another Q which is canonical to P , why should we go for more complex Q when we have already have got simpler Q. So our Q is just Q
= _ Pz mg
Now we have H = P so
P=
-
oH = oQ
0
and
.
oH
Q =-= 1
oP
Hence the solut ion is p -
p ,..
n -
+ -L
n ,.
11 oo : 02 : 1o P= -
f)H
aQ
= 0 and
.Q =oH -= 1 oP
Hence t he solut ion is
P = Po
Q=t
and
+ Qo
Easy, isn't it ?? In t his problem you have seen t hat an otherwise complex problem can be made simple by use of canonical t ransformation. 47
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6
Hamiltonian Nlechanics
Equation of motion of a dynami-
cal variable let us consider a dynamical variable (function of Q, p , t ) u u(Q, p , t) Then we have by application of partial derivative . u
=
au . OQi Qi
+
au . api Pi
+
=
au at
Using Hamilton 's equation of mot ion we get .
au a H
au aH
OU
aQi api
api oq,i
at
u= - - - -
-+-
In t erms of Poission Bracket we writ e t he above equation more concisely as
(6. 1) • If u
=
• If u
= Pi, t hen
Qi t hen
11 oo : 02 : 13 • If u
= qi then
• If u
= Pi, then . api aH Pi = {Pi, H} + at = - aqi
These two equations are Hamilton 's equation of motion. • Now if u
=
H then
if = {H H}
'
aH + at
=
aH at
48
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• Next consider what we can say if a quantity u is conserved. Here:
u= 0=
au {u H } + ' at
So we can say if
au at
=
0
and {u, H}
=
0⇒u
is conserved
Another fact, is that if u and v are conserved t hen so is {u , v}. This could potentially provide a way to compute a new constant of motion. To prove it, first consider the sp ecial case where:
8u =av= O at at
~~>
{u, H} = {v, H} = O
then using the J acobi identity we have:
{H, {u,v}}
=
- {u,{v, H }} - {v, {H ,u}}
=
- {u, O} - {v, O}
=
0
11 oo : 02 : 1 s then using t he Jacobi identity we have:
{H, {u,v} } = -{u, {v, H }}-{v,{ H ,u} } = - {u,O}- {v, O}
=
For t he most general case we have
du au dt = {u, H} + at = O
au {u, H} = - at
And dv
dt
= { v , H}
av
+ at
av
{v, H} =-at
O
=
49
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So we get
+ {v, { H , u} }
{{u, v }, H} = {u, {v, H}}
==-
av
u, a
t
+
au v, at
a
at {u , V}
Now the equation 6.1 for tha variable {u, v} is
a
d
dt {u, v} = {{u, v}, H} + at {u, v} Hence we get d
dt {u, v} = 0
7
Infinitesimal Canonical Transfor-
m rition
0
11 oo :02 : 1a dt {u, v} = 0
7 Infinitesimal Canonical Transformation Let us now consider t he generat ing function:
F2( q, P , t) = qi~
+ EG( q, P , t)
where F2 = qi~ is an identity transformation, and El t+ /3
11 oo : 02 : s4 as /3 = - = - t±m aa
dq J2ma - (mwq) 2
⇒ t+/3
mw 2 2a q
1 = ±-arcsin w
integrating we get
q=
±
-
2a
mw
. s1n(w(t + /3) ) 2
so /3 is relat ed to t he phase. Next we consider p and use t his result t o obtain: P=
as aq =
±J2ma - (mwq)
2
=
±-hma cos(w(t + fJ ))
63
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So we get our familiar solut ion of p and q. To complete t he story, t he constants a and /3 must be connected wit h t he init ial conditions q0 and p0 at t ime t = 0. By squaring p and q and using t = 0 we get
Now dividing p and q at t
= 0 we get
tan /3
= mw qo Po
The choice q0 = 0 and hence /3 = 0 corresponds t o st art ing t he mot ion with t he oscillator at its equilibrium posit ion q = 0 .
9.2
Separation of variable and H-J method
11 oo :02 : s1 The choice q0 = 0 and hence /3 = 0 corresponds to starting the motion with the oscillator at its equilibrium position q = 0 .
9.2
Separation of variable and H-J method
Separation of variables is t he main technique to solve t he H-J equation. In particular, for a time independent H where
if = 8H = 0
(9. 13)
at
we can always separate t ime by taking: (9. 14)
where a 1 has been chosen as the separation constant, then plugging t his into the t ime dependent H-J equation yields 64
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(just as in our Harmonic Oscillator example) :
This result is referred to as the time independent HamiltonJacobi equation. since H = 0 , His conserved, and equal to a constant a 1 . If H = E then this constant is energy, a 1 = E. The function W is called Hamilton's characteristic function. Now let's try to solve the central force Kepler problem by H-J method. •
Example 9.28: As an extended example, let us consider the Kepler problem of two masses m 1 and m 2 in a
11 oo : 02 : sg method.
Example 9 .28: As an extended example, let us consider t he Kepler problem of two masses m 1 and m 2 in a central potential (with t he CM coordinate R Lagrangian is:
L =
!mr 2
2
-
V (r)
1
where
m
=
0). The
1
1
m1
m2
- +-
and here m is t he reduced m ass. Any V (r) conserves L = r x p , so the mot ion of r and p is in a plane perpendicular to L. The coordinates in t he plane can be taken as (r, 'ljJ), so:
L= m 2
r2 + r 2 ~ 2
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V(r )
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'lj; is a cyclic coordinate 2 .
P 'ljJ
= mr 'ljJ
is a constant
Now check that we can write the hamiltonian as H =
+ V (r)
1 2m
As 'lj; is cyclic, then
P 'ljJ
O'.'ljJ
= a1 =
E
is constant. Using:
then the t ime independent H-J equation is:
(9 .15)
11 oo : 03 : 02 As 'ljJ is cyclic, then Pv; _ av; is constant. Using:
then the t ime independent H-J equat ion is: 1 2m
After simplification a,2
2m (a.1 - V(r)) -
t r
Integration gives
Transformation equations are dr
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and
9.3
Action-Angle Variables
For many problems, we may not be able to solve analytically for the exact motion or for orbital equations, but we can still characterize the motion. For periodic systems we can
11 oo : 03 : os 9.3
Action-Angle Variables
For many problems, we may not be able to solve analytically for t he exact motion or for orbital equations, but we can still characterize t he motion. For periodic systems we can find t he frequency and t ime period by exploit ing action-angle variables. For periodic motion, we replace P = a 1 by t he action variable J = pdq T he coordinate conjugate to J is t he angle variable
aw w=-8J
After some derivatons and arguments we get v( J) is the frequecy of the periodic motion.
8H (J ) V=--8J
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Example 9.29: Consider a general Hamiltonian
H =
p2 2m
+ a xiv
Find the time period of a particle in t his Hamiltonian using action angle variable.
11 oo : 03 : 01 =
H
+a
2m
x iv
Find t he t ime period of a par t icle in t his Hamiltonian using action angle variable.
Solution: As the potent ial is symmetric then the m agnitudes of t he turning points are same
E l/v Xturning
=± a ✓2m (E - axv)dx 0
a
= 2 2m E Now we h ave E
= ax2.
1 - -xvdx
E
o Substituite Q
z = -x
V
-E
so
x=
zE
l/ v
'
Then
J
= 2J2m E
E a
dx =
•
1/ v
E a
l
1
V
1/ v
l
- zv1 _ 1d z V
zt- 1J1 -
z dz
0
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E l/v l I'(l /v)f (3/2)
a
v
E l/v I'
a
r (t + ~) (t + 1) ½v0f r (~+ ~)
11 oo : 03 : 1o 1 I'(l / v)I'(3/ 2)
=2 2mE
(e + ~) l / v I' (e + 1) ½J7r I'
V
= 2 2mE
E
r (~ + ~) E l / v r (e + 1) = ✓21rmE a r (e + ~) E t+~ = J al f v r (e + ~) a
~ r (t+ 1) P ut t he constants inside anot her constant
and writ e Taking derivative with resp ect to J
=
E t +~ - l 8E 8J
k
T hen the frequency
BE 1 1 V=-=k/E v-2 8J Then the T ime period
T
1
= - =
1
1
E v- 2/k
V
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Hence the time period proportional to 1
T ex E v-
1 2
11 oo : 03 : 12 Hence t he time p eriod proportional to 1
T ex E v-
1 2
Example 9.30: Let us consider the Harmonic oscillator problem again
1 2 2 2 2 H = m (p + m w q ) = E = a = constant 2
Now J is " ._,
._,
where a is the constant total energy and w If we substitute q =
,
to
2
= k/m.
/2
a 2 sin 0 Then t he integral reduces mw 27T
J= 2a cos2 0d0 w 0 Limits should b e over one complet e cycle of q.
The integral after integration J= 21ra w
Solving for a
a
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The frequency of oscillation is t hen
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The frequency of oscillation is then 8H w 1 -=v=-=8J 2n 2n
k m
Example 9.31: A particle of mass m moves in one dimension subject t o t he potent ial
k U(x) = -sin-2-(x-/ a- ) (a) Obtain an integral expression for Hamilton's characteristic function. (b) Under what conditions may action-angle variables be used? (c) Assuming that action-angle variables are perm1ssible, det ermine the frequency of oscillation by the action-angle method. (d) Check your result for the oscillation frequency in the limit of small oscillations. •
Solution: ( a) As the total energy is conserved. H
P2
=-
2m
k
+ - -2 - - = E = sin (x/ a)
constant
The motion is therefore between the t urning points. [email protected]
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11 oo : 03 : 11 The motion is therefore between t he t urning points. [email protected]
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ing points are t hose where total energy becomes equal with potential energy x_(E) = n1ra + a sinx+(E) = (n
+ l )1ra -
1
k/ E
asin-
1
k/ E
where n is any integer. The Hamilton J acobi equation is then 1 2m
dW
dx
2
k
+ - 2- - =Q sin (x/ a)
Which We may t hen write X
W(x, E ) = v2m,
dx'
E-
2
k
sin (x' / a)
X- (E)
The lower limit may be left as unspecified ; this only changes t he result by a constant. (b) We need t hat the motion is bounded. In our case, x _(E) < x < x+(E ) (c) The action is
E-
dx'
2
k
sin (x' / a)
This integration is tough. However if we substit uite
cos(x/a) = [email protected]
k l 1 - -cos -u
E
72
2
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11 oo : 03 : 20 This integration is tough. However if we substituite
k 1 1 - -cos -u
cos(x/a) =
E
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Then J becomes 21r
J = aJ2mE
1 - cos u du----
E+k - COSU E -k
0
This integral can be solved by method of contour integration by using techniques of complex analysis. However t he standard integral table may be used 21r
I = 0
du 1 21r b - cosu
1
Jb 2 -
Using t his integral we got, with b = (E J
1
+ k)/(E -
k), 2
J
= ~a( vfE - v'k) , E = ~ +v'k a
Note t hat the minimum energy is frequency is given by
v( J) =
8E J -= -+ 8J ma 2
Emin
= k. The oscillation
2k ma2
2E ma 2
2
(d) With U(x) = k/ sin (x/a) we have
U'(x) = _ 2k . cos(x/a) a
3
sin (x/a) 4
U"(x) = 2k . sin (x/a) a2
+ 3sin
2
2
(x/a) cos (x/a) sin6 (x/ a)
11 oo : 03 : 23 U'(x) = _ 2k . cos(x/a) a
sin 3 (x/a) 4
U"(x) = 2k . sin (x/ a)
+ 3sin
a2
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2
2
(x/a) cos (x/a) 6 sin (x/ a)
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Setting U' (x*) = 0 we obtain x* = (n + ~) 1ra, where n is an integer . At any of these equilibria, U'' (x*) = 2k/ a 2 . Therefore, the frequency of small oscillations is
Ws.o.
=
U'' (x*)
which agrees t he result of (c)
m
2k ma2
00: 03 : 25
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Setting U' (x*) = 0 we obtain x* = (n + ½ ) 1ra, where n is an integer . At any of these equilibria, U'' (x*) = 2k/a2 . Therefore, t he frequency of small oscillations is
W s.o.
=
U'' (x*) m
2k ma2
which agrees the result of (c)
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Setting U' (x*) = 0 we obtain x* = (n + ½ ) 1ra, where n is an integer . At any of these equilibria, U'' (x*) = 2k/a2 . Therefore, t he frequency of small oscillations is
W s.o.
=
U'' (x*) m
2k ma2
which agrees the result of (c)
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Setting U' (x*) = 0 we obtain x* = (n + ½ ) 1ra, where n is an integer . At any of these equilibria, U'' (x*) = 2k/a2 . Therefore, t he frequency of small oscillations is
W s.o.
=
U'' (x*) m
2k ma2
which agrees the result of (c)
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Small Oscillation Sk J ahiruddin*
*Assistant Professor Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay M.Sc Physics 2009-2011 batch He ranked 007 in IIT J AM 2009 and 008 (JRF) in CSIR NET June 2011 He has been teaching CSIR NET aspirants since 2012
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Small Oscillat ion
11 oo : oo : 03 1 @Sk Jahir uddin , 2020
Small Oscillation
Contents 1 Stability and equilibrium
3
1.1
Bounded and Unbounded motion
•
•
•
•
4
1.2
Integrat ion of equation of mot ion in lD .
•
•
•
5
•
•
•
2 Small oscillation 2.1
8
Small oscillation in more t han one dimension .
10
3 General formulation of Small Oscillation
13
3.1
Matrix formulation of small oscillation problem
17
3.2
Rigorous solut ion of Linea1· t riatomic molecule
25
3. 3
Making t he non quadratic potent ial quadrat ic
32
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n
11
r--,.
0
11
, •
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Small Oscillat ion
Stability and equilibrium
The result F = - dU/ dx is useful not only for computing the force but also for visualizing the st ability of a system from the potent ial energy plot. Suppose t here is a force on the part icle is F = - dU/ dx , and the system is in equilibrium where there is no force i.e. dU / dx = 0. If t his occurs at a minimum of U it is a st able equilibrium whereas if it is at a maximum of U, the equilibrium is unstable. Say, dU/ dx = 0 occurs at some point x 0 . To t est for stability we must determine whether U has a minimum or a maximum at x 0 . One needs to examine d2 U/ dx 2 at x 0 . Potential Energy, U
Unstable E ulllbrium Neutral equilibrium Position, x
Stable Equilibrium
11 oo :oo :oa
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-+ Stable;
-+ Unstable; -+ Neutral
If t he second derivative is posit ive, t he equilibrium is stable; if it is negative, t he system is unstable. If d2 U/ dx 2 = 0, one must look at higher derivatives. If all derivatives vanish so t hat U is const ant in a region about x 0 , t he system is said to be in a condition of neutral.
1.1
Bounded and Unbounded motion
In Mechanics we generally divides the motion in two broad categories. The Bounded system, where energy is negative and t he Unbounded system where the energy is positive. Bounded system may be periodic or non periodic. In this section we will discuss the periodic bounded system t heir oscillation and frequencies.
since kinet ic energy K = E - U can never be negative, the
11 oo : oo : 1o cillation and frequencies.
since kinetic energy K = E - U can never be negative, the motion of the syst em is constrained to regions where U < E.
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Small Oscillation
The points where E = V (K.E = 0) are called t urning points. The following figure is an illustration of bounded motion of one potent ial wit h two different energy and turning points. Potential energy
Total energy E1
E1 1 - - - - - - + - - - - r , - - - - - - - L - - - - - - - + -
Total energy E2 E2
1 -- - - . - -- \.----- -r t --
--'-- - - --,tL-- - . -
Kinetic energy at xA
Potential energy at xA X
1.2
"'
Integration of equation of motion in 1D
1 E=-m 2
dx dt
2
dx
+ V(x)
dt
2
- [E - V(x) ] m
(1.1)
Hence
dx
11 oo : oo : 13 1
E = -m 2
dx dt
2
dx
+ V(x)
2
- [E - V( x) ]
dt
m
(1.1)
Hence dt
=
dx
(1.2)
-----;:=================-
~ [E - V (x)]]
So the t irne p eriod as a function of energy is twice the tirne [email protected]
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Small Oscillation
t he particle take to go from T(E)
=
x1
to
x2
dx
m
2
2
x 1 (E)
✓[E - V (x)]
(1.3)
Example 1.1: Example: Consider t he potential shown in the figure. Find the p eriod of oscillation
E-- 3U0/2
- - - - U=U0 - - - - - - - U=O +(-----+) ....
b b Time period is twice t he tirne taken to reach from starting point to end point
T= 2
b
O
dx
2b
dx
--+2 1
/2E
b
1
j2(E- Uo)
11 oo : oo : 16 Time period is twice t he t ime taken to reach from start ing point t o end point b
2b
dx
T = 2
2E
0
+ 2
m 26 = 2b 3Uo +
m
m Uo m Uo
1
v'3
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2(E-Uo)
b
m
= 2b 1 +
dx
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Example 1. 2 : Determine t he period of oscillation , as a
function of the energy, when a part icle of m ass m moves in fields for which t he potent ial energy is (b) U
2
= - Uo / cosh ax,
- Uo < E < 0,
(a) is already d one in the action angle variable. Here we will get sam e ans by doing similar integration. (E / A)l /n
T = 2
(2rri)
J (E - A xn)
0
= 2
2m E
•
dx
1/n
E A
1 0
dy
J(l - yn)
By t he substit ut ion yn = u t he integral is reduced to a beta function, which can be expressed in t erms of gamma funct ions:
E -- ·
E
-
A
l /n
f (l / n)
r (~ + 1/n)
(
11 oo :oo :1a By t he substitution yn = u t he integral is reduced to a beta function , which can be expressed in terms of gamma functions:
E
21rm
-- ·
E
1/ n
f (l /n)
-
r (~ + 1/n)
A
Do (b) and (c) similarly. The ans are
(b) T = (1r/a) ✓(2m/ E l), (c) T= (1r/a) ✓[2m/(E+Uo)]
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Small Oscillation
Small oscillation
Nearly every bound system oscillates like a harmonic oscillator if it is slightly perturbed from its equilibrium position. >.
0) ~
a>
C
w
r
--------r-----I
\ \
1
/
Parabola
1
I
\ I
I I
/ I
I
//
I I
. -.: .
/..
Look at the figure. Expand U (r) about r 0 , the position of the potential minimum.
dU
U (r) = U (r n) + (r - rn) .
+
l
-=- (r - rn)
2
2
dU . "
+ ···
11 oo : oo : 21 Look at the figure. Expand U(r) about r 0 , t he position of the potential minimum.
dU U(r) = U (ro) + (r - ro) dr
+
2 d U 2 2 (r - ro) dr2
1
ro
+ ... ro
However, since U is a minimum at r0 , (dU/dr ) ro = 0. Furthermore, for sufficiently small displacements. we can neglect the terms beyond the third in the power series. In this case, kx2 U(x) = constant + 2 This is the potential energy of a harn1onic oscillator, where
d2 U k = dr 2
the effective spring constant
;
ro
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Frequency of small oscillation is
W =
k
1 d2 U
m
m dr 2
Example 2.3: The time period of a particle of mass m , undergoing small oscillations around x = 0 , in the
potential V =
(a)
1r ,
(C) 21r i\
mL 2
-
mL 2 (b) 21r , 2Vo
Vo mL l /,..
Vo cosh (f) , is
2
[NET Dec 2018]
11 oo : oo : 23 (a)
m L2 7r
(b) 21r
Vo
(c) 21r
mL
m L2 2V0
2
(d) 21r
Vo
Solution: F irst make
2mL
2
Vo
!~ = 0 to find the equlibrium point . dV
Vo I I Vo cosh(L) = 2 ex L + e- x L X
Vo ex/L - e- x /L
➔-
dx
2£
So dV = O
x= O
dx
Now the double derivat ive
Vr _O
2£2 e
x/L
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+
e
- x/L
=
1 £ 2V ;
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Small Oscillat ion
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Vo dx2 x=O £ 2 So t he frequency of sm all oscillation is
w=
l d2 V
m dx2
Vo x=O
m£ 2
Hence the time p eriod
T = 21r/w = 21r
2.1
m L2 Vo
Small oscillation in more than one di-
11 oo : oo : 26 T
2.1
21r/w
=
=
21r
m L2
Vo
Small oscillation in more than one di•
mens1on You have already know the theory in small oscillation in 2 and 3 dimension, I'll not repeat t hat . You can proceed in both way. Either calculate the equation of motion and solve directly or just calculate the V and T matrix and need to find t he frequencies of small oscillation by getting
we will proceed wit h interesting examples. As the first and most simple system of vibrating mass points, we consider the free vibration of two mass points, fixed [email protected]
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Small Oscillation
to two walls by springs of equal spring constant, as is shown in t he figure
I I
-
•, X
.
I
I
,+I
I
- •t x 2
:-
' I I I
-kx l
11 oo : oo : 29 I
. X I , I
-· I
I I I
'
-kx l
The two mass points shall have equal masses. The displacements from the rest positions are denoted by x 1 and x 2 , respectively. We consider only vibrations along t he line connecting the mass points. When displacing the mass 1 from the rest posit ion, there acts the force -kx 1 by t he spring fixed to t he wall, and the force +k (x 2 - x 1 ) by the spring connecting the two mass points. Thus, the mass 1 obeys t he equation of motion
For the second mass we get
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\"Ne first determine the possible frequencies of common vibrat ion of t he two particles. The frequencies that are equal for all particles are called eigenfrequencies. The related vibrational states are called eigen- or normal vibrations. These definitions are generalized for a N -particle system. We take the solution i
P
h"t h n ~rt.ir 1P~ ~ h ~ll ,,ihr~tP u rit.h t hP ~~.mP frPn 11 Pnr,, r,,
11 oo : oo : 31 st at es are called eigen- or normal vibrations. These definit ions are generalized for a N -part icle system. We take the solut ion
i.e., both particles shall vibrate with the same frequency w . The specific typ e of the solution, be it a sine or cosine function or a superposition of both, is not essential. We would always get the same condition for t he frequenty, as can be seen from the following calculation. Put t he solut ion in the equation of motion 2
+ 2k) -
=0 2 - A1k + A2 ( -mw + 2k) = 0 A1 ( -mw
A2k
(2.1)
The system of equations has nontrivial solut ions for the amplit udes only if the det erminant of coefficients D vanishes: -mw 2
+ 2k
- k
-k -mw
2
+ 2k
= ( -mw
2
+ 2k)
2 -
k
2
=
0
We t hus obtain an equation for det ermining the frequencies: 4
w
-
k 2 k2 4-w + 3- 2 = 0
m
m
12
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The posit ive solut ions of t he equation are t he frequencies 3k
k
m
m
These frequencies are called eigenfrequencies of t he system; t he corresponding vibrations are called eigenvibrations or nor-
11 oo : oo : 34 3k
k
m
m
These frequencies are called eigenfrequencies of t he syst em; t he corresponding vibrat ions are called eigenvibrations or normal vibrations. To get an idea about the t ype of t he normal vibrations, we insert the eigenfrequency into t he equat ion (2.1) For the amplitudes, we find
A1 =-A2
for
w1
=
3k m
and
k m
3 General formulation of Small Oscillation Consider a syst em wit h { qi} as the generalized coordinat es. Since the system is conservative, V (q1 , q2 , ... , qN) . Lagrange defined equilibrium as a configuration in which all generalized . h , 1.e. . 8V 0 forces van1s Bqi = .
a2v
- - > 0 Signifies Stable equilibrium 8qi8qj
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and
Small Oscillation
a2 v
- - - < 0 Signifies unstable equilibrium 8qi8qj
Without loss of generality, let us shift t he equilibrium posiJ
•
J
j
,
•
•
I
A\
Tf'
, 1
•
11 oo : oo : 36 and
a2 v
- - - < 0 Signifies unstable equilibrium 8qi8qj
Without loss of generality, let us shift the equilibrium position to the origin (q 1 = q2 = ... , = qN = 0). If the syst em is disturbed to a configuration {qi}, we can write,
V (q1 ' q2 '
. . .)
= V (0' 0' . . .) +
L 1,
+
Higher order terms
where t he partial derivatives are evaluated at the p osition of equilibrium and all higher order terms'' which involve third order or higher corrections are neglected. If t he potential energy is measured from its minimum value, we choose V (O, 0, ... ) = 0. Along with
av
= O
8qi
0
The leading term in the change in potential energy is then
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for stable equilibrium. Let us write
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Small Oscillation
for stable equilibrium. Let us write
so that 'l ,J
Now, the kinetic energy in general a quadratic in generalised velocities, and can be written as
The coefficients m i j are in general functions of the coordinates Qk, but may be expanded in Taylor series about the equilibrium position ffiij ( Ql ' Q2' .. . )
=
ffiij ( 0'
0' ... ) +
L k
0
It t urns out t hat t he quantities ;"i.i and t he higher order qk 0 derivatives are negligibly small so that t he coefficients m i j s can be essentially treated as constants having t he same values as they would have in t he equilibrium position . Thus around the equilibrium posit ion, 8
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11 oo : oo : 42 15
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Then the Lagrangian has the following structure: £ = T - V =
1
2 I: (T;jl/;C/j -
V;jq;qj)
i,J
Now if you solve the Lagrange's equation of motion
for the given Lagrangian you will get (you need not to do the general case by hand, you will do in the examples)
.
1,
The equation is a homogeneous equation in Ai s and t he condition for existence of t he solution is
This is equation is called t he secular equation. We are only interested in real roots of the above equation. w~s det ermined from t his equation are known as characteristic frequencies or eigenfrequencies.
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Small Oscillation
Matrix formulation of small oscillation
problem Using symmetry properties of V and T (symmetric matrix) we can write the secular equation as
.
1,
where A = w 2 . Define a column vector
A= • • •
The matrices V and T are given by
V=
½1
V12
• • •
Vi N
T11
T12
• • •
T 1N
½1
V22
•
• •
V2N
T21
T22
•
• •
T 2N
•
•
• •
• • •
• • •
V N1
•
•
•
V N2
•
• • •
• •
•
VNN
andT=
•
T N 1 T N2
• • •
• • •
• • •
TNN
So t he matrix equation can b e written as a m atrix equation
(3.1) Not e that this equation is not in the form of an eigenvalue equation as VA = AA but VA = ATA. If T is invertible, (in
11 oo:oo:47 (3.1) Note that this equation is not in the form of an eigenvalue equation as VA = AA but VA = ATA. If T is invertible, (in [email protected]
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general it is) one can get an eigenvalue equation w 2 I A.
r - VA 1
=
since we have N homogeneous equations, we have N modes, 2 i.e. N solutions for w . Let us denote the k -th mode frequency by = Ak. Let the vector A corresponding to this mode be written as
wi
•
•
•
We have
Taking conjugate of this equation and changing the index k to i, we get ~
where we have used A to denote t he t ranspose of t he matrix ~ A .. From V A k = AkT A k we get by multiplying with A k
~
We have V A k = AkT A k. Multiply by Ai in LHS. You get ~
~
A ,iV A k = AkAiT A k
11 oo : oo : 49 ~
\'lve have V Ak = AkT Ak. 1\1:ultiply by Ai in LHS. You get ~
~
AiV Ak = AkAiT Ak
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~
We also have AiV = AiAiT. Multiply by Ak in RHS . You get ~
~
AiV A k = AiAiT A k From the last two equations ~
(Ak - Ai) AiT Ak = 0 Thus, if the eigenvalues are non-degenerate, i.e. if Ai we get the orthogonality condition
i=
Ak,
Note that this is different from the orthogonality condition on eigen vectors for a regular eigenvalue equation. Since (3. 1) does not uniquely determine A , we define normalization condition as
(3.2) The normalization of the eigenvectors are not necessary unless specifically stated in the problem. We can get the idea of normal coordinates without normalization as we see in the examples. Fundamentally speaking vve have a set of coordinate 1J. We are trying to achieve another set of coordinates
11 oo : oo : s2 e e1genvec ors are no necessary unless specifically stated in the problem . We can get t he idea of normal coordinates without normalization as we see in the examples. Fundamentally speaking we have a set of coordinate rJ . We are trying to achieve another set of coordinates ( , which are related by ~>
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r~ _-
A-1,n •I
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Small Oscillation
In the form of ( t he equations look like ..
(i
2
+w ( = 0
The set of ( coordinat es are called normal coordinates and these vibrations in the coordinate of ( are called normal mode. Example 3.4: Consider two masses m 1 = 2m and m 2 = m connected by t hree springs, as shown in the following figure. Find t he frequency of oscillation and normal
modes. k = 2k 3
Solution: The Lagrangian of t he system is £
=
1 ·2 1 ·2 1 2 1 2 1 2 m1x1 + m2x 2 - k1 x 1 - k2 (x2 - x 1) - k3x2 2 2 2 2 2
Define T and V matrix
11 oo : oo : 55 Solut ion: T he Lagrangian of t he system is £
=
1 ·2 1 ·2 m1x 1 + m2x 2 2 2
-
1 2 1 2 1 2 k1x1 - k2 (x2 - x1) - k3x 2 2 2 2
Define T and V m atrix
m1 0 0 m2
5k -k -k 3k
•
'
Non trivial solution exist if
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which gives
5k - 2mw 2 -k 3k - mw 2 -k
=0
The secular equation has t he solution w2 = 7 k /2m and 2k / m. The normal modes are now found by substit ut ing t he values of w's in to the secular equation and evaluation of eigenvectors
Writing A =
7k 2m
'
we get ' for w2
=
7 k 2m
5k -k -k 3k
2m 0 O m
For t his normal mode we have A 2 unnormalized
1
-2
= - 2A 1 so t hat we get
11 oo :oo : s1 - 2A 1 so t hat we get
For t his normal mode we have A 2 unnormalized
1
-2 For w2 =
';:,,
a similar calculation gives
1
A=
1
So we get the unnormalised transformation matrix 1
A=
1
-2 1 21
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We have taken the coordinates as x 1 and x 2 . The normal coordinates ;- -_ A-1-n ~
~>
•I
1 -1
2 If we write w1 =
?k 2m
and w 2 =
1 ~, the general solution is
where a 1, a 2 , b1, b2 are the constants which to b e determined by the initial conditions. The solution can also be written by in t erms of old (x 1, x 2) coordinates 1 1
cosw2t + c2
1 1
sin w2t
11 oo : 01 : oo by the initial conditions. The solution can also be written by in t erms of old (x 1,x 2) coordinates 1 1
1
1
sin w2t
1
1
-2
-2
sinw1t
where c1, c2, c3 , c4 are the constants which to be determined by t he initial conditions x1(0), x2(0), i1(0), i2(0).
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Example 3 .5 : Three masses m each , init ially located equidistant from one another on a horizontal circle of radius R. They are connect ed in pairs by three springs of force constant k each and of unstretched length 21r R/ 3. The spring threads the circular tract so t hat the mass is constrained to move on t he circle. Find the normal modes.
11 oo : 01
: 02
Solution: Let 01 , 02 and 03 be t he angular displacements of t he three masses from their equilibrium positions. The Lagrangian is given by £ = T - V, where,
021 + 02 + 02 2 3 and
23
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T and V matrices are 2
V
= kR
2
1 0 0
- 1 - 1
- 1
T = mR
2
- 1 -1 -1 2
2
0 l 0
0 0 1
You can write the secular equation as before and easily show that the normal modes are given by w = 0 and w = (doubly degenerate). For w = 0, the secular equation is -k -k -k 2k -k 2k
R
2
_.,,.
_.,,.
')0
=0
~
11 oo : 01 : os egenerate . For w
= 0, t he secular equation is R
2k -k -k -k 2k -k -k -k 2k
2
=0
which gives A 1 = A2 = A 3 . T he normalized normal coordinate is given by from ATTA = I. We get 1
J3mR 2 For w
2
1 1 1
= 3k/m, we have -k -k -k -k - k - k -k -k -k
A1 A2 A3
=0
which gives A1 + A2 + A3 = 0. The eigenvectors cannot be uniquely fixed. However , if we use the ort hogon ality with [email protected]
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t he zero mode, we can t ake t hese two choices (2 , - 1, 1) and (0, 1, - 1) which are ort hogonal to t he first vector and t hemselves ort hogonal. Normalization gives 1
J6mR 2
3.2
2 - 1 - 1
• )
1
J2mR 2
0 1 - 1
Rigorous solution of Linear triatomic
molecule
11 oo :01 : oa ✓6mR2
3.2
✓2mR2
- 1
- 1
Rigorous solution of Linear triatomic
molecule We now consider t he model of linear t riatomic molecule. In t he equilibrium configuration of the molecule, two atoms of mass m are symmetrically located on each side of an atom of mass M as shown in the figure.
All three atoms are on one straight line, t he equilibrium distances apart are denoted by b . For simplicity, we shall first consider only vibrations along t he line of the molecule, and the actual complicated interatomic potential will be approximated by two springs of force constant k joining the t hree 25
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atoms. The positions of three atoms are denoted by x 1 , x 2 , and x 3 In these coordinat es, t he potential energy is
V
k
= 2 (X2 -
X1 -
b)
2
k
+ 2 (X3 -
X2 -
b)
2
We now int roduce coordinates relative to the equilibrium posit ions:
11 oo : 01 V
=-
: 1o
2
b) + - (x3 -
(x2 -
X1 -
Xo2 -
X o1 =
X2 -
b)
2
2 2 We now int roduce coordinates relative to the equilibrium posit ions:
where
b=
Xo3 -
Xo2
The potential energy is now
or V =
k ('r/12 + 2rJ22 + 'r/32 - 2rJ1 'r/2 - 2rJ2rJ3 )
2
Hence the V matrix is
k -k
0 2k - k
V =
-k 0 -k
k
The kinetic energy is
m ( ·2
-2
T = 2 'r/1 + 'r/3)
+
26
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M 2
-2
'r/2
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Small Oscillation
So t he T matrix is
O 0 0 JVJ 0 0 0 m
m
T =
The secular equation is
11 oo : 01
: 13
O 0 0 M 0 0 0 m
m
T = The secular equation is
-k
0 -k
2
2k-w M -k k -w 2 m
= 0
Evaluate the det erminant. You will get the equat ion
The solutions are
k m
'
w3
=
k m
2m 1+ M
Now insert the values of t he frequencies into t he equation of motion
(V -
w
2
T) A= 0
2 (mw - k) a1 +
ka1 Oa1
ka2 2 + (Mw - 2k) a2 + ka2
+Oa3 +ka3 2 + (mw - k) a3
=0 =0 =0 (3.3)
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one by one Insert ion of w = w1 = 0 into (3.3) yields a1 = a2 = a3. The eigenfrequency w 1 = 0 does not correspond to a vibrational motion. but reoresents onlv a uni- form translation of the
11 oo : 01
: 1s
one by one Insert ion of w = w1 = 0 into (3.3) yields a1 = a2 = a3. The eigenfrequency w 1 = 0 does not correspond t o a vibrational motion, but represents only a uni- form translation of the ent ire molecule: • ➔ o ➔ • 1 2
Inserting w = w2 = (k/m) ! into (3.3) yields a 1 = - a3 , a 2 = O; i.e., the cent ral atom is at rest , while t he outer atoms vibrate against each other: +-- • o •
➔
Inserting w = w 3 = {k/m(l + 2m/ M )} 1 into (3.3) yields a 1 = a 3 , a 2 = -(2m/M)a 1 , i.e. , t he two outer atoms vibrate in phase, while the central atom vibrates with opposite phase and with another amplitude: • ➔ +-- o • ➔ 1 2
So without normalization the normal coordinates are (1
= 'r/1 + 'r/2 + 'r/3
(2
=
(3
'r/1 -
= 'r/1
-
(3.4)
'r/3
2m M 'r/2 + 'r/3
With the equations
k
• ••
m
+
2k M
(3 = 0
(3.5) Now we will evaluate t he normalized eigenvectors seriously. • For w = 0, t he equation VA - w 2T A = 0 can be written 28
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as
(
k
-k
O \
( A1 \
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: 1a
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Small Oscillation
as
k -k 0 - k 2k - k 0 -k k
A1 A2 A3
= 0
which gives A 1 = A2 = A3. Taking each of t hem to b e equal to 1 (in some unit) we get
A=
l
~
1 1 1
where N is to be fixed by normalization. Normalization condition gives
1 N •
1.e. M
0 0 0 M 0 0 0 m
m
1 1 1
+ 2m =
J2m+M • Consider w =
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= 1
N . Thus t he normalized vector is 1
2
1 1 1
1 1 1
! . For t his case, we have 0
-k
0
-k
k
-k
0
-k
0 29
= 0
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which gives A2 = 0 and A 1 = A 3 . Using the orthogonality with t he mode w 1 , we take t his mode to be 1
1
0
vR
- 1
Normalization condition 1 N
O 0 0 M 0
0
0
-1
1
m
1 0 -1
0
m
= 1
fixes N = 2m and the displacement vector is 1
1
0
v'2m
- 1
which shows the central mass to be at rest while the end masses displace by the same amount in the opposite direction. 2
• For w =
! + ~, t he equation is -2km M
-k 0
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-k
0
kM
-k
m
-k
-2km M
30
A1 A2 A3
= 0
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: 23
m
-k
0
jahir@physicsguide .in
- 2 km M
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Small Oscillation
which gives A 1 = A3 and A 2 = -2;/ A 1. The normalization can be done as in the previous cases. The displacement vector is given by 1
M 2m M + 4m 2
- 2m/M 1
We can now get t he new set of coordinates. As we have started with the set of coordinates 'T/ and found t he normalized eigenvectors t hen the relation between the old set of coordinate and new set of coordinate is rJ
= A(
The A matrix is A=
1
1
1
✓2m+M 1 ✓2m+M 1
J2m+M
1
1 ✓2m
1
1
V2rn
1
JVI
0
2mM
- 1
+4m2
-2m/M 1
✓2mM~4m2 2m M2+2m!Vf
0
1
1
J21n+M
✓2m
M 2mM+4m2
We get t he new coordinate set after some tedious calculation which you need not to do, just see t he results .
A-1
../m
../M
../m
J 2rri+M
J 2m+M
J 2m+M
=
1 2
O
1 2
M
2m
M
J4m+2NI
J 2m+M
J4m+2lllf
11 oo : 01
: 26
which you need not to do, just see the results.
rm
m
rm
J2m+M
J2m+M
J2m+M
A-1=
1
1 2
0
2
M J4m+2M
2m
J 2m+M
M J4rri+2M
31
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Verify that AA- 1 = I Hence (1
1 m+M
= ✓2
(2 =
Vffi'T/1
+ ✓-ivf '//2 + Vffi'f/3
1 - (rJ1 - rJ3) 2
The equation of motions are in normalized coordinates
3.3
Making the non quadratic potential quadrc
Example 3.6: Consider a particle of mass m moving in two dimensions in a potential
V (x,y)
1
2
1
= - kx + Aox 2
2
22
y
1
4
+ 4 A1X,
k , Ao , A1 > 0
(a) At what point (x 0 , y0 ) is the particle in stable equilibrium? (b) Give the Lagrangian appropriate for small oscillations about t his equilibrium position. (c) What are the normal frequencies of vibration in (b)?
11 oo :01 : 2a at point x 0 , y 0 is t e partic e in sta e equilibrium? (b) Give the Lagrangian appropriate for small oscillat ions about t his equilibrium position. (c) What are the normal frequencies of vibration in (b)?
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Solution: (a) A point where
and
a2 v
ax2 (dx)2
a2 v
+ 2 8x8y dxdy +
a2 v
ay2 (dy)2 > 0
is a point of stable equilibrium. For t he given potential we find two such points, k/ Ai , 0 and - k/ Ai , 0 (b) V is a minimum at a point of stable equilibrium (xo, Yo) . At a neighboring point (x, y), we have, to second order of the small qualit ies x - xo, y - Yo,
V (x, y) = V (xo, Yo) 1
+2
a2 v 8x2
2
x· Y Ql
(x - xo)
(x - xo) (y - Yo)
Q
a2v
+ 8y2
+2
(y - Yo)
2
xo ,Yo
Evaluate the derivatives for t he equilibrium point ;1, 0 . You will get upto second order approximation of x - x 0 and
11 oo : 01
: 31
fi,
Evaluate the derivatives for the equilibrium point 0 . You will get upto second order approximation of x - x 0 and
y-yo
2k
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Shift the coordinate system to the new origin
X
I
=
k X-
A'
f;, 0
.
I
y =y
and take t he new origin as the reference level for potential energy. Then
And the Lagrangian is l ( X. i2 +y. i2) - -1 k L = T - V = -m 2 2
Similarly for t he other point of equilibrium, we set X
II
=x+
k
y
II
=
A' II II l . I I x , y rep ac1ng x , y
(c) The secular equation
y
11 oo : 01
: 33 II
y = y 1 . / I x , y rep ac1ng x , y II
II
(c) The secular equation
2k - mw 2 0
0 k>-.o _ >-.1
=0 mw2
The solutions are
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These are the normal angular frequencies for small oscillations of the system , about both of the points of equilibrium.
Example 3.7: Two mass points of equal mass m are connect ed t o each other and to fixed points by three equal springs of force constant k, as shown in t he diagram. The equilibrium length of each sprint is a . Each mass point has charge +q, and t hey repel each other according to Coulomb law. Set up the secular equation for the eigenfrequencies. ,n
k
m
+q
a
+q
k a
Solution: This problem also has two generalized coordinates x 1 and x 2 , t he coordinates of the two masses start ing from left . x 10 = a and x 20 = 2a are the init ial posit ions of the
11 oo : 01
: 36
Solution: This problem also has two generalized coordinates x 1 and x 2 , t he coordinates of t he two masses start ing from left . x 10 = a and x 20 = 2a are t he init ial posit ions of t he masses. Let rJi and T}2 be t he displacements from the equilibrium posit ions. (rJ1 = x 1 - x 10, T}2 = x 2 - x 2o) So kinetic energy operator will be
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The T matrix will be
T
0 0 m
m
As far as potential energy is concerned, besides t he contribut ion from the t hree springs, we will have one extra contribut ion due to t he Coulomb repulsion between t he two charges V
=
1 2 1 2 krJ1 + k (rJ2 - rJ1 ) 2 2
=
1( 2 ') ) 2 2krJ1 - 2kr}1T/2 + 2krJ2
+
1 2 krJ2 + 2 41rEo
+
q2 X1 -
X2
q2 4 I JrEo T/2 - T/1 - a
Because rJ2 - rJi / a / 2
= Q
2co
Problem: A sphere of radius R 1 has charge density p uniform within its volume, except for a small spherical hollow region of radius R 2 located a distance a from the center. [email protected]
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Electrostatics
Find t he field E at the center· of the hollow sphere.
Solution: Consider an arbitrary point P of the hollow region as shown in figure 3. 7 and let
OP = r,
Q'P = r',
00' = a
'
r' = r - a
11 oo : 01
: 1o
. ons1 er an ar 1trary point region as shown in figure 3. 7 and let
OP=r,
00' = a
Q'P=r',
p
, Rt
'
''
'
r' = r - a
\~ ""R2 ~ ., ' '' ;'
\
'('a
/,
Figure 3. 7: Electric field inside hole
If t here were no hollow region inside the sphere, t he elect ric field at the point P would be
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Electrost atics
If only t he spherical hollow region has charge density p t he electric field at P would be p I E 2 = -r 3so
11 oo : 01
: 13
If only t he spherical hollow region has charge density p t he electric field at P would be
E 2 =p-r' 3Eo
Hence t he superposit ion t heorem gives t he electric field at P as p E = E1 - E2 = - a 3Eo
Thus the field inside the hollow region is uniform. This of course includes t he center of t he hollow. There are many examples in t he standard books like Griffiths. Please do some exercises from Griffiths also.
4
Divergence and Curl of Electric Field
Divergence and curl of Electric field is given by (Remember, here we are dealing with t he electrostatic case, when t he charges of the syst em are not moving. In the dynamic case
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t he cur1 will be different) p
V·E = -
EQ
(4.1)
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: 1s
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Electrostatics
t he cur 1 will be different) p
V·E = -
(4.1)
V xE = O
(4. 2)
Eo
We write E (x ) in terms of a scalar function V(x ) as
E (x ) = - VV (x )
(4.3)
The very important function V (x) is called the electric potent ial. 1 p (x') d3 x' V(x) = --(4.4) 41rEo Ix - x' Applying it with t he divergence of electric field we get
This is called Poisson's equation. In regions of space where t he charge density vanishes, we're left solving the Laplace equation
(4.6)
5
Electric Potential
It is instructive to construct V(x ) in another way. Let - V (x) be the line integral of E along a curve r from a [email protected]
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reference point x 0 to x
28
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Electrostatics
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: 1a
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Electrostatics
reference point x 0 to x
V(x) = -
r
E · df
(5.1)
This equation says that V (x) is equal to the work that must be done by an external force, acting against the electric force, to move unit charge from x 0 to x along t he path r , because - E is t he force per unit charge that t he external agent must exert. Example: T he electric potential of some configuration is given by the expression e->-.r
V(r) = A -
r
where A and ,\ are constants. Find t he electric field E (r), t he charge density p(r), and the total charge Q .
Solution:
0 E =- v'V= -A-
or
=-A
e- >-.r
r
r ( - A) e->-.r - e->-.r
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r2
29
A
r A
=
Ae- >-.r (1
+ Ar)
r
r2
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Electrostatics
11 oo : 01 [email protected]
: 21
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Electrost atics
So the potential p=EoV·E
= EoA e-,\r(l + Ar) V ·
r"' r2
But as you know r"' r2
And from t he properties of the delta function
The gradient
a
v7 (e-,\r (l + Ar)) = f Br (e- ,\r( l + Ar)) =
f { - Ae- ,\r(l +Ar) + e- ,\r A} = f (-A 2re- ,\r)
So we get
And hence the charge density p [email protected]
= EoA 30
Physicsguide
00 : 01 : 23
p
= EoA
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Hence t he total charge
- Ar _e _ 41rr 2 dr r 00
2
= EoA 41r - .\ 41r
r e-Ar dr 0
6
Surface charge and discontinuity of electric field
Now consider an infinite plane, which we take to be z = 0 , carrying uni£orm charge per unit area, a . We again take our Gaussian surface to be a cylinder, this t ime with its axis perpendicular to the plane as shown in the figure. In t his context, the cylinder is sometimes referred to as a Gaussian '' pillbox'' (on account of Gauss' well known fondness for aspirin).
11 oo : 01
: 26
31
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E
-- --
--
z=O
- - .... E
Figure 6.1: Discontinuity of electric field when surface charges are present
On symmetry grounds, we have
E = E (z) z Moreover, the electric field in t he upper plane, z > 0, must point in the opposite direction from t he lower plane, z < 0, so t hat E(z) = -E(-z). The surface integral now vanishes over the curved side of the cylinder and we only get contributions from t he end caps, which we take to have area A . This gives
o-A E · dS = E (z) A - E (-z) A = 2E(z) A = S Eo
11 oo : 01 : 2a of the cylinder and we only get contributions from the end caps, which we take to have area A . This gives
aA
E · dS = E (z) A - E (-z) A = 2E(z) A = -
Eo
S
The electric field above an infinite plane of charge is t here32
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Elect rostatics
fore (J"
E (z) = -
2Eo
Note that the electric field is independent of t he distance from the plane! This is because the plane is infinite in extent: t he further you move from it, the more comes into • view. There is another important point to take away from t his analysis. The electric field is not continuous on either side of a surface of constant charge density. We have ' I -
-
I I
__ /!-
✓ L-_1
-
-.
--f
-
+
+ +
.,...---7
-
> ~
+
Figure 8. 1: Conducting sphere in an Electric field
We'll learn how to compute t he electric field in this, and related, situations later.
9
Image Problem
11 oo : 02 : 1o
9
Image Problem
For particularly simple situations, there is a rather cute method t hat we can use to solve problems involving conductors. Although t his technique is somewhat limited, it [email protected]
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does give us some good int uition for what's going on. It's called t he method of images. Example: A charged particle near a conducting plane: Suppose a point charge q is held a distance d above an infinite grounded conducting plane. What is t he potential in t he region above the plane? It's not just (1/ 41rt 0 ) q/n, for q will induce a certain amount of negative charge on t he nearby surface of t he conductor; the total potential is due in part to q directly, and in part to this induced charge. But how can we possibly calculate t he potential, when we don't know how much charge is induced or how it is distributed?
We're looking for a solution to t he Poisson equation with a delta-function source at z = d = (0, 0, d), together with t he requirement that
R , the contribution from both the surface and volume bound charges need to be calculated
Q tot = (kR) (41r R
2
)
+ (-3k) : 7rR
3
=0 '
so
One important result you need to remember; t he electric field produced by a uniformly polarized sphere of radius R •
IS
Inside: E
=-
1 P, 3Eo
for r < R
Outside: the potential and field is just like a perfect dipole at the origin
V =
1 p ·f 41rEo r 2 '
for
r> R
11 oo : oo : 16 dipole at t he origin 1 V = p ·f 41rEo r 2 '
3
for
r > R
Electric Displacement Vector
We learned that the electric field obeys Gauss' law
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Dielectrics
This is a fundamental law of Nature. It doesn't change just because we're inside a material. But, from our discussion above, we see that there's a natural way to separate the electric charge into two different types. There is the bound charge Pb that arises due to polarisation. And then t here is anything else. This could be some electric impurit ies t hat are stuck in the dielectric, or it could be charge t hat is free to move because our insulator wasn't quite as good an insulator as we originally assumed. The only important thing is that t his other ch arge does not arise due t o polarisation. We call t his extra charge free charge, pf . Within the dielectric, t he total charge density is P = Pb+ PJ ~
and we know Pb
~
= - V .P , hence
(3.1)
11 oo : oo : 19 '
(3.1) ......
-+
and we know Pb= -V.P , hence
Eo V · E
= p=
Pb+ PJ = - V · P
+ PJ
(3.2)
we write t he equation as
(3.3) Now define D [email protected]
EoE
+P
(3.4) Physicsguide
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Dielectrics
In integral form
D · da =
Qfenc
(3.5)
where Q ! enc denotes the total free charge enclosed in the volume. That 's quite nice. Gauss' law for the displacement involves only the free charge; any bound charge arising from polarisation has been absorbed into the definition of D
3.1
Linear Dielectrics
In general, the polarisation P can be a complicated function of the electric field E . However , most materials it t urns out
11 oo : oo : 21 3.1
Linear Dielectrics
In general, the polarisation P can be a complicated function of the electric field E . However, most materials it t urns out t hat P is proport ional to E. Such m aterials are called linear dielectrics. They have
(3.6) where Xe is called t he electric susceptibility. It is always posit ive: Xe > 0 Any function that has P (E = 0) = 0 can be Taylor expanded as a linear term + quadratic + cubic and so on. For j [email protected]
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suitably small electric fields, t he linear term always dominates. To determine when t he quadratic and higher order terms become important, we need to know the relevant scale in the problem. For us, t his is the scale of electric fields inside the atom. In most situations, t he applied electric field leading to the polarisation is a t iny perturbation and the linear t erm dominates. The linearity fails for suitably high electric fields. There are exceptions to linear dielectrics. Perhaps the most striking exception are materials for which P # 0 even in the absence of an electric field. Such materials - which are not particularly common - are called ferroelectric. For
11 oo : oo : 24 There are exceptions to linear dielectrics. Perhaps the most striking exception are materials for which P # 0 even in t he absence of an electric field. Such materials - which are not particularly common - are called ferroelectric. For what it's worth, an example is B aTi0 3 . Well, for linear dielectrics D = EoE
+P
=
EoE + EoXeE
Eo (1 + Xe) E
=
(3.7)
So we define
(3.8)
D =EE where
E
Eo (1 + Xe)
(3.9)
This constant E is called t he permittivity of t he material. (In vacuum, where t here is no matter to polarize, the [email protected]
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Dielectrics
ceptibility is zero, and t he permittivity is Eo . That's why Eo is called the permittivity of free space. The relative permittivity, or dielectric constant, of the material, is defined as Er -
1 + Xe
E
=-
Eo
(3.10)
Example: A metal sphere of radius R 1 carries a charge
Q . It is surrounded, out to radius R 2 by linear dielectric material of permittivity E. Find the potential at the center
11 oo : oo : 26 Eo Example: A metal sphere of radius R 1 carries a charge
Q . It is surrounded, out to radius R 2 by linear dielectric material of permittivity E. Find the potential at the center (relative to infinity) .
Solution: To compute V, we need to know E ; to find E , we might first try to locate t he bound charge; we could get t he bound charge from P , but we can't calculate P unless we already know E (from the relation P = EoXeE ) We can calculate D , as § D · da = Q ! en c
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Dielectrics
Q R1
11 oo : oo : 29
Figure 3.1: Conductor surrounded by dielectrics
for all points r > R 1 (Inside the metal sphere, of course, E = P = D = D = 0. Once we know D , it is a trivial matter to obtain E , using D =EE
for R 1 < r < R 2
Q 2f , 41rEr
Q 2 r, for r > R2 41rEor
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Dielectrics
The potential is t hen R2
0
V =-
E-dl =00
R2
41r
41rE0 r 2
00
R1
Q
Q
1
0
Q 41rEr 1
dr
2
+ Eo R 2 ER1
dr-
(O)dr R1
1 ER 2
11 oo : oo : 32 00
00
R1 R2
Q 41r
41rE0 r 2 0
Q
dr-
47rEr 2
1
(O)dr R1
1
1
+ EoR 2 ER1
ER 2
The polarization is present inside the dielectric material EoXeQ " P = EoXeE = r; 2 47rEr The bound volume charge
R1 < r < R2
Pb= - V · P
=0
The volume surface charge EoXeQ
at t he outer surface,
41rER~'
-EoXeQ 41rER i '
4
at the inner surface.
Boundary Conditions in dielectric (4.1)
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Dielectrics
-+
we write t he same equation in terms of E
(4.2) The boundary condit ions in terms of Electric field ...
11 oo : oo : 34 we write t e same equation in terms o
(4.2) The boundary conditions in terms of Electric field ... ~
~
E above -
E below
II
II
E above -
1
(4.3)
= -a Eo
_
E below -
(4.4)
O
Boundary conditions in terms of potentials OV above E above
On
OVbelow E below
-
Vabove
=
On
=
V below
- a1
(4.5)
(4.6)
We use the boundary conditions depending on the problems given.
5
Energy in Dielectrics
Energy stored in a dielectric system is
w=!
D -EdT
2
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Dielectrics
Example: A sphere of radius R is filled with material of dielectric constant Er and uniform embedded free charge p f.
What is t he energy of this configuration?
00: 00: 37
Example: A sphere of radius R is filled with material of dielectric constant Er and uniform embedded free charge pf. What is t he energy of this configuration?
Solution: The displacement vector
D(r) = The elect ric field
E (r) =
Then t he electrostatic energy without considering the 2 dielectrics (W = ~ J E dT)
Pi 3Eo
0
21r
2
RG
1
00
-
R
r
4
2
41rr dr
2 R5
9Eo PJ
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The total energy (W =
Physicsguide
Dielectrics
1f D · EdT)
11 oo : oo : 39 @Sk J ahiruddin , 2020
Dielectrics
The total energy (W =
if D · EdT ) R
Pi Pi 2 2 r 41rr dr 3 3EoEr 0 00 1 ? PiR3 PiR3 41rr~dr + 3 4 3Eo R r 21r 2 Rs - 1 + 1 9EoPi 5Er
Let's check that W2 is the work done on t he free charge in assembling t he system. We start with the (uncharged, unpolarized) dielectric sphere, and bring in the free charge in infinitesimal inst allments (dq) , filling out t he sphere layer by layer. When we have reached radius r', t he electric field •
IS
E(r) =
Pi r (r < r') 3EoEr p r'3 i r (r' < r < R) 2 3EoEr r Pi r'3 " 3Eo r2 r (r > R)
The work required to bring the next dq in from infinity to
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Dielectrics
11 oo : oo : 42 15
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Dielectrics
r' is R
dW = - dq
r'
E · di +
E · di R
00
R
= - dq
00
1
1
R
Er
-+-
1 -dr + r2 1
1
r'
R
dq
---
R
dW = -dq
p1r'3 r' 1 -dr 3EoEr R r 2
r'
E · di +
E · di R
00
R
= -dq
00
1
1
R
Er
-+-
r'
1 p 1r'3 -dr+2
r
-
3EoEr
1 1 --r' R
R
1
r
dr 2
dq
This increases t he radius (r') :
so t he total work done, in going from r' = 0 tor' = R , is
W =
41rp
2
f
R
3Eo
21r
1
2
9EoPJ j [email protected]
R
1
1- -
5
r' dr'
Er
0
1
+-
Er
0
Rs -1+ l 5Er
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Dielectrics
Example: A spherical conductor, of radius a, carries a charge Q. It is surrounded by linear dielectric material of
susceptibility Xe, out to radius b. Find the energy of this configuration.
Rz
Q R1
Figure 5.1: Conductor surrounded by dielectrics: Energy
Solution: The displacement vector is D =
0, _Q......,,. f
41rr2
'
(r < a) (r > a)
So the electric field
E=
(r < a) (a< r < b) (r > b)
11 oo:oo:47 So t he electric field (r < a)
E=
(a< r < b) (r > b)
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Dielectrics
The work done
w = !2
D. E dT
1 Q2 - - -47r 2 (41r) 2
Q2
1
E b
1
- 1 -81r E r
Q2
b
l 1 -2 2r2dr
a r
r
1
- 1
Eo
r
1
1
+a
1
- -
-
(1 + Xe) a b Q2 1 Xe -+81rEo (1 + Xe) a b 81rEo
5.1
1
00
+-
Eo
b
1
r
2
dr
00
b
1 +-b
Forces on Dielectrics
Force is the derivative of energy F = dW
dx
(5.2)
If capacitance of a system is changing wit h respect to x then t he force (5.3) ..
11 oo : oo : so If capacitance of a system is changing with respect to x then the force (5.3)
Example: Capacitor partially filled with dielectrics
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Dielectrics
Figure below shows parallel plat es of size f x w and separation d , partially filled with a dielectric slab. The dimensions of the slab are x x w x d , with x < f . What is the capacitance? Assume the separation d is small compared to x and f - x, so that end effects are negligible. -Q d
+Q X
e-x
Figure 5.2: A dielectric slab inserted part-way between charged plates. A force is pulling the slab into the space between t he plates
Let the potential difference between the plates be V . The key to this example is t hat the electric field E is the same in the two regions (dielectric and vacumm) between t he plates. By planar symmetry, the field is uniform in both regions and normal to the plates. But E , being normal to the
11 oo : oo : s2 Let the potential difference between the plates be V . The key to this example is t hat the electric field E is the same in the two regions (dielectric and vacumm) between t he plates. By planar symmetry, the field is uniform in both regions and normal to the plates. But E , being normal to the plates, is tangent to t he boundary surface of t he dielectric. Since the tangent ial field must be continuous, t he field is t he same on either side of t he boundary. Thus the electric j [email protected]
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Dielectrics
field is E = 11 V / d in both regions, where 11 is normal to the plates. To determine t he capacitance we must relate V to the charge ±Q on the plates. Apply Gauss's Law to a surface surrounding one plate, e.g. t he lower plate: The flux of D is equal to Q . That is, since D is EE in t he dielectric (of area xw) and EoE in t he vacuum region [ of area (£ - x )w] EExw
+ EoE(R -
x)w
=
Q
Substit uting E = V / d we find that the capacitance is
We might have anticipated this result by noting t hat t he system is equivalent to two capacitors in parallel, with C 1 = EXW / d and C 2 = Eo (£ - x )w / d. Then t he combined capacitance is C1 + C2.
11 oo : oo : 55 We might have anticipat ed t his result by noting t hat t he system is equivalent to two capacitors in parallel, with C 1 = EXW / d and C2 = Eo(R - x )w / d. Then t he combined capacitance is C1 + C2 . Now what is t he force on t he dielectric slab? We can have two cases. The plates are with fixed charge: Plat es are isolated so t hat t heir charges ± Q are constant. The work done by t he electrostatic force if t he slab moves a dist ance dx into [email protected]
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Dielectrics
t he capacitor is Fdx Because the syst em is isolated , conservation of energy says t hat t he work is equal to - dU where U is the tot al energy of t he capacitor. T herefore t he inward force on t he slab is F = - dU / dx To calculate U for this example it is natural to use t he equation U = Q 2 / 2C because Q is constant . The capacitance we calculated. Thus the energy is
U(x) =
Q2d 2w [Ex+ Eo(f - x) ]
So the force on the slab is
F
=
2
Q d (c - to) 2 2w [EX + Eo (R - x)]
For fixed charge the force decreases as x increases. We may also express F in terms of the potent ial difference be-
11 oo :oo : s1 F=
E -
Eo
+ Eo ( .e -
2w [EX
x)]
2
For fixed charge the force decreases as x increases. We may also express F in terms of the potential difference between t he plates, V = Q / C, as F
=V
2
w
(E - Eo) 2d
Plates with fixed potential difference: Now suppose t he plates in Fig. 5.2 are connected to a battery so that the potential difference V is fixed. In this case the work done by the electrostatic force F is not equal to -dU, because j [email protected]
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Dielectrics
additional energy is supplied by the battery. If the slab moves by distance dx , then charge dQ is transferred to the plates from the battery, and so the batt ery supplies energy (d) V. The conservation of energy in this case is dU = (dQ )V - Fdx
Here it is natural to use the equation U = CV 2 / 2, because V is const ant. Also, Q = CV so dQ = (dC )V. Inserting t hese relations into the conservation law gives
Thus the for ce on t he slab is
11 oo : 01 : oo by the electrostatic force F is not equal to -dU, because j [email protected]
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Dielectrics
additional energy is supplied by the battery. If the slab moves by distance dx, then charge dQ is transferred to the plates f1"om the battery, and so the battery supplies energy (d) V. The conservation of energy in this case is dU = (dQ) V - Fdx
Here it is natural to use the equation U = CV 2 / 2, because V is constant. Also, Q = CV so dQ = (dC )V. Inserting t hese relations into the conservation law gives
Thus the force on t he slab is F =
! dC v2= V 2 dx
2
w
(E - Eo) 2d
Again, a reminder, do all t he exercises which are given inside t he chapters of Griffit hs.
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11 oo : oo : oo
Magnetostatics Sk J ahiruddin Assistant Professor Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay M.Sc Physics 2009-2011 batch He ranked 007 in IIT JAM 2009 and 008 (JRF) in CSIR NET J t1ne 2011 He has been teaching CSIR NET aspirants since 2012
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Magnetostatics
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Magnetostatics
Contents 1
Magnetostatics
4
1.1
Steady current .
1.2
Force on a moving charge
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6
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8
Electric current as a source of magnetic field
9
1.3. 1
Biot-Savart Law .
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9
1.3.2
Ampere's Law . .
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13
1.2.1 1. 3
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Force on current carrying wire
1.4
Surface Currents and Discontinuit ies
1.5
Magnetic Vector Potential
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17
1.6
Magnetic Dipole . . . . . .
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20
1.6.1 1.7
Torques and Forces on Magnetic Dipoles 28
Electric Current . . . . .
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28
1. 7.1
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Current Density . 2
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11 oo : oo : 06 1.7. 1
Current Density .
•
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2
j [email protected]
•
•
30
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l\lfagnetostatics
1.7.2
Conservation of Charge .
•
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30
1. 7.3
Ohm's Law . . . . . . .
•
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31
2 Magnetic Field in a m edium
35
2. 1
Bound Currents ..
•
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37
2.2
Magnetic intensity
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39
Linear m agnetic material . . .
•
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40
Electromagnet ic Boundary condit ions
•
•
•
•
42
2.2. 1 2.3
2.3.1
2.3.2
2.3.3
Magnetostatic Boundary condition inside a medium . . . . . . . . . . . . .
42
Boundary condition in both elect ric and magnetic field . . . . . . . . . .
42
Example: Field inside the gap
45
•
•
•
•
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Magnetostatics
Charges give rise to electric fields. Current give rise to magnetic fields. In t his section, we will study the magnetic fields induced by steady currents. This means t hat we are again looking for t ime independent solutions to the Maxwell equat ions. We will also restrict to situations in which the charge density vanishes, so p = 0. We can t hen set E = 0 and fo cus our attention only on the magnetic field. We're left with two Maxwell equations to solve:
(1.1) and
(1.2) If you fix the current density J , these equations have a unique solution. Our goal in this section is to find it.
1.1
Steady current
Before we solve (1.1) and (1.2) let 's pause to think about the kind of currents t hat we're considerine: in this section. Be-
11 oo : oo : 11 1.1
Steady current
Before we solve (1.1) and (1.2) let 's pause to think about the kind of currents t hat we're considering in t his section. Because p = 0 , there can't be any net charge. But, of course, [email protected]
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we still want charge to be moving! This means that we necessarily have both positive and negative charges which balance out at all points in space. Nonetheless, t hese charges can move so t here is a current even t hough t here is no net charge t ransport. This may sound artificial, but in fact it's exactly what happens in a typical wire. In that case, there is background of positive charge due to the lattice of ions in the metal. Meanwhile, the electrons are free to move. But they all move together so t hat at each point we still have p = 0. The continuity equation, which captures the conservation of electric charge, is
(1.3) Since t he charge density is unchanging (and, indeed, vanishing) , we have
(1.4) Mathematically, this is just saying t hat if a current flows
11 oo : oo : 13 Since t he charge density is unchanging (and, indeed , vanishing) , we have (1.4) v' · J = O Mathem atically, this is just saying that if a current flows into some region of space, an equal current must flow out to avoid the build up of charge. Note that this is consistent with (1.1) since, for any vector field, v' · (v' x B ) = 0 j [email protected]
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Force on a moving charge
The most basic form of the magnetic force is the force on a charge q moving with velocity v , exerted by a magnetic field B , which is F=qvxB (1.5) The SI unit of magnetic field is the t esla (T ). One t esla is defined as one newton per ampere-met er. By (1.5) the force on 1 C moving 1 m/ s perpendicular to a field of 1 T is 1 N . A magnetic field of 1 T is a rather strong field. The largest fields t hat can b e produced by conventional electromagnets are about 2 T . For high-field research in laboratories, fields of 10 - 12 T , produced with superconducting mag- nets, are used. The record for steady-state fields is about 30 T . At the surface of the Earth the magnetic field is about 0.5 x 10- 4T . Very large and very small fields are important in astronomy; e. g.10 8 T near pulsars, and 10-10 T in interst ellar space in the galaxy.
11 oo : oo : 16 . At t he surface of t he Eart h the magnet ic field is about 0.5 x 10- 4T . Very large and very small fields are important in ast ronomy; e. g.10 8 T near pulsars, and 10- 10 T in interst ellar space in the galaxy. The direction of t he magnet ic force on q is sideways, i.e., perpendicular to v and to B , as shown in Fig. 8.1 for a posit ive charge. Therefore, the magnet ic force does no work
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on q: dW = F · dx = F · v dt = 0
(1.6)
The magnetic force affects t he direct ion of motion of q, but not its kinetic energy. Because the magnetic force does no work, it is not possible to define a pot ent ial energy funct ion for t he magnet ic force. The magnet ic force is velocity dependent , which is quit e different from t he ot her fundamental forces we encounter in physics. B
V
q
F
11 oo :oo :1a
Figure 1.1: The magnetic force. The force on a moving charge is sideways, perpendicular to both v and B. The figure shows F and the circular trajectory for a positive charge in a uniform field.
Equation (1.5) may be taken as the definit ion of t he magnetic field. From the measurement of t he force on a [email protected]
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Magnetost atics
test charge q, e.g. by observing its deflection for a known velocity, t he defined quantity B could in principle be deduced from (1.5) When both electric and magnetic field is present t hen t he force equation is known as lorentz force equation
F = q(E
1. 2 .1
+V
X
B)
(1.7)
Force on current carrying wire
A current-carrying wire in a magnetic field experiences a force, from the magnetic force on the individual moving charges t hat make the current . Electric current is motion of r.h::-i.r lfP. . ::-i.nrl ( 1 ..~l iR t.hP. forr.P. on
FI.
movin lf r.h ::-i.r lfP. . ~11nnoRP.
11 oo : oo : 21 A current-carrying wire in a magnetic field experiences a force, from the magnetic force on the individual moving charges t hat make the current . Electric current is motion of charge, and (1.5) is t he force on a moving charge. Suppose t he current consists of part icles wit h charge q and linear density n 1 , moving with mean velocity v . T hen the net force dF on a small segment df of t he wire is
(1.8) because n 1df is t he nt1mber of moving cha1~ges in df . T he current I in t he wire is qn 1 v, and v is parallel to df, so the total force on the wire is j [email protected]
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F=
1.3
. wrre
I d£
X
B
(1.9)
Electric current as a source of magnetic field
Where does a magnetic field come from? W hat is its source? The most familiar source is a permanent magnet , i.e., a piece of magnet ized iron or ot her ferromagnet ic material. But at a more basic level t he source of B is to be found in electric current. We will study magnet ic materials, and ,.., ,.... ,.... + 1-- ,... + + 1-- ,.... .,,..,,... ,... ,..,..,,.., ,.... + ~ ,... ,... .,,.., ,..1 .,::: ,.... 1 ,..1 ,..,. .C ,... .C,.....,.,,. ,..,. .,,..,,... ,... ,..,..,,.., ,.... + ,... ,..,. .,,..,,... ,.... ,.., .C..,, ,..,. .,,..,,...
11 oo : oo : 24 The most familiar source is a permanent magnet , i.e., a piece of magnetized iron or ot her ferromagnetic material. But at a more basic level t he source of B is to be found in electric current. We will study magnetic materials, and see t hat the magnetic and field of a ferromagnet comes from properties of atomic electrons - t heir spin and orbital motion - which, although not classical currents, do involve dynamics of charged particles. However, now we are concerned with B from macroscopic steady currents.
1.3.1
Biot-Savart Law
The magnetic field of a steady line current is given by the Biot-Savart law: The field dB at a point P, due to an in-
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finitesimal current element I df at a point P ' . Figure below shows the geometry. For this elemental case the Biot-Savart law is A
Biot-Savart law
I dC
p
r
11 oo : oo : 21
I df
Figure 1.2: Elemental form of the Biot-Savart law. Id/!, is t he source of magnetic field; and dB is the resulting field at P.
dB
=
µo Id/!, x
r
(1.10)
41r r 2 Here r is the distance between P' (the source point) and P (the field point), r is the unit vector in the direction from P' to P, and r = rr. Note that t he Biot-Savart law is an inverse-square law, like Coulomb's law, but the direction of t he magnetic field is azimuthal, around the axis of I d/!,. Do few examples from Griffiths. j [email protected]
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Ampere's Law
The integral of the current density over t he surface S is the same thing as t he total current I that passes through S. Ampere's law in integral form then reads
B · dr = µ of
(1.11)
C
For t he problems where cylindrical symmetry is there, it is
11 oo : oo : 29 Ampere's law in integral form then reads
B · dr = µof
(1.11)
C
For t he problems where cylindrical symmetry is there, it is easy to apply Ampere's law to calculate Magnetic field. Consider an infinite, straight wire carrying current I . We'll take it to point in t he z direction. The symmetry of the problem is jumping up and down telling us t hat we need to use cylindrical polar coordinates, (r, rp , z ), where
r=
J x 2 + y 2 is t he radial distance away from t he wire. We take the op en surface S to lie in the x - y plane,
centered on the wire. For t he line integral in (1.1 1) to give something t hat doesn't vanish , it's clear t hat the magnetic field has to have some component t hat lies along t he circumference of the disc.
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s C
11 oo : oo : 32 s C z ~
r
Figure 1.3: Magnetic field of long straight vvire
But, by t he symmetry of t he problem, t hat's actually t he only component that B can have: it must be of the form B = B (r )cp . (If this was a bit too quick, we'll derive t his more carefully below). Any magnetic field of t his form automatically satisfies the second Maxwell equation V · B = 0. We need only worry about Ampere's law which tells us 21r
B · dr C
=
B (r)
rd(f)
=
21rrB (r)
=
µof
0
We see that the strength of the magnetic field is
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(1.12) P hysicsguide
Magnetostatics
The magnetic field circles the wire using the ''right-hand rule" : stick the t humb of your right hand in the direction of the current and your fingers cur1 in the direction of the magnetic field.
11 oo : oo : 34 e magnetic e circ es t e wire using t e rig t- an rule" : stick the t humb of your right hand in the direction of the current and your fingers cur1 in the direction of the magnetic field.
1.4
Surface Currents and Discontinuities
Consider the flat plane lying at z = 0 with a surface current density that we'll call K . Note that K is t he current per unit length, as opposed to J which is the current per unit area. You can think of the surface current as a bunch of wires, all lying parallel t o each ot her. We'll take t he current to lie in t he x-direction: K = K x as shown below. z y
K
Figure 1.4: Surface Current
From our previous result, we know that the B field j [email protected]
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should curl around t he current in the right-handed sense. But, with an infinite number of wires, t his can only mean t hat B is oriented along t he y direction. In fact , from the
00: 00: 36 l\/fagnetostatics
should curl around t he current in the right-handed sense. But, with an infinite number of wires, t his can only mean t hat B is oriented along t he y direction. In fact , from the symmetry of t he problem, it must look like z y
I
I I I I
B
Figure 1.5: Magnetic field due t o a Surface Current
wit h B pointing in the -y direction when z > 0 and in t he +y direction when z < 0 we write
B = - B (z) y wit h B (z) = -B(- z). We invoke Ampere's law using the following open surface:
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11 oo : oo : 39 @Sk J ahiruddin, 2020
1\/Jagnetostatics
z y
I I I
I
I
I
I I
)
I
I
I I
I I
Figure 1.6:
wit h length L in t he y direction and ext ending to We have B · dr
±z.
= LB (z ) - LB(-z) = 2LB(z) = µ 0 KL
C
so we find that the magnetic field is constant above an infinite plane of surface current B (z)
= µo K 2
z >0
This is rather similar to the case of the electric field in t he presence of an infinite plane of surface charge. The analogy with electrostatics continues . The magnetic field is not continuous across a plane of surface current . We have
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11 oo : oo : 42 15
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In fact, t his is a general result that holds for any surface current K . We can prove this statement by using the same curve that we used in the Figure above and shrinking it until it barely touches t he surface on both sides. If the normal to t he surface is n and B ± denotes the magnetic field on either side of the surface, then (1.13)
Meanwhile, the magnetic field normal to the surface is continuous. We can write t he boundary conditions as (1.14)
and E above ll
-
E below ll _ K - µo
(1.15)
When we looked at electric fields, we saw that the normal component was discontinuous in the presence of surface charge while the tangential component is continuous. For magnetic fields, it's t he other way around: t he tangential component is discontinuous in the presence of surface currents.
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11 oo : oo : 44 currents.
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1.5
l\tlagnetostatics
Magnetic Vector Potential
At this point it is useful to compare the principles of magnetostatics and electrostatics, given in t he table below. Electrostatics
Magnetostatics
V x B = µ 0J V-B = 0 B cur ls around I
V · E = p/ Eo
V x E=O E di verges from q
Note this difference: The electric field has scalar sources, but the magnetic field has vector sources. Electric charge, i.e., a point source of E from which E diverges, is a common property of elementary particles. Motion of charged particles-current -is the source of B , around which B curls. Magnetic charge, i.e., a point source from which B diverges in t he rest frame of the particle, apparently does not exist in nature, or at least it has not been observed. Such a hypothetical charge is called a magnetic monopole. Dirac showed t hat it is possible to construct a consistent quantum theory wit h both electric charges and magnetic monopoles . However , t he fundamental magnetic charge g
11 oo:oo:47 Such a hypothetical charge is called a magnetic monopole. Dirac showed t hat it is possible to construct a consistent quant um theory with both electric charges and magnetic monopoles. However, t he fundamental magnetic charge g [email protected]
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and electric charge e would necessarily be quantized , and satisfy t he relation eg = n/2, where n is an integer. Many experimenters have searched for magnetic monopoles, but so far t he results are negative. Some speCUlative t heories of high-energy physics, such as grand unified field theories, predict t he existence of very massive magnetic monopoles, too massive to be produced at current high-energy accelerators, but which might have been produced in t he big bang. Searching for magnetic monopoles continues to be an interesting experimental challenge If magnetic monopoles do not exist, t hen t he equation V · B = 0 is a universal equation of magnetism. Whether magnetic monopoles exist or not, t he source of the magnetic fields we encounter in physics are not point magnetic charges but rather currents of electric charge, corresponding to the source equation V x B = µoJ. We found in electrostatics t hat it is useful to introduce a scalar potent ial V (x) for t he electrostatic field, such t hat E = - VV. This guarantees that V x E = 0. In an analo-
11 oo : oo : so We found in electrostatics t hat it is useful to introduce a scalar potential V (x ) for t he electrostatic field, such t hat E = - VV. This guarantees that V x E = 0. In an analogous way we may int roduce a vector potent ial A (x) for the
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magnetic field, such t hat
(1.16) This guarantees that V · B = 0 However , (1.16) does not uniquely determine A for a given magnetic field B : If f (x ) is an arbitrary scalar funct ion, t hen A + V f has t he same curl as A (namely B) because V x V f is identically 0. Therefore we may impose a condition on A called a gauge condition, to remove t his ambiguity. The Coulomb gauge condit ion, which we will use in magnetostatics , is
(1.17) Taking (1.16) and (1. 17) together is still not enough to make A (x) unique, because adding a constant does not change eit her V x A or V · A. But imposing an appropriate
11 oo : oo : s2 Taking (1.16) and (1.17) together is still not enough to make A (x ) unique, because adding a constant does not change either v7 x A or v7 •A. But imposing an appropriate boundary condition, such as requiring A -+ 0 at infinity, makes A unique.
Example: Consider the uniform field B j [email protected]
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vector potent ial function , satisfying (1.16) for the uniform field and the Coulomb gauge condition (1.17) is
A (x ) =
1 B x x 2
=
1
2
" " B0 (-yi + xj )
An example of uniform B is t he field inside a solenoid, so A
for B = Bok we may picture a long, t ightly and uniformly wound solenoid, whose axis is the z axis. Note for t his case t hat A (x ) is parallel to t he surface currents (azimut hal) and t hat A (x ) curls around the B field. The analytic expression of vector potential in vector notation is A (x ) = µo d3 x' J (x') (1.18) 41r v x - x'I
1.6
Magnetic Dipole
Vector potential, while in the general coordinate as shown
11 oo : oo : 55 1.6
Magnetic Dipole
Vector potential, while in t he general coordinate as shown below figure , can be expanded as
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Figure 1.7: Multipole expansion of vector potential
As in t he mult ipole expansion of V, we call the first term (which goes like 1/ r) t he monopole term , the second (which goes like 1/ r 2 ) dipole, t he third quadrupole, and so on .
A (r) = µof
l
41r
r
1 di ' + r2
r' cos adI' ( 1 1 () \
11 oo : oo : sa goes like 1/r 2 ) dipole, t he third quadrupole, and so on.
!
A (r) = µof
r
41r
1
+r 3
1
di ' + -
r' cos adI'
r2
(1.19)
')2 r (
-3 cos2a - -1
2
2
dI / + · •·
The magnetic monopole term is always zero The dipole t erm A dip(r )
=
µo f 41rr 2
r' cos adI'
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=
µof
(r · r') di'
41rr 2
(1.20)
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We define the dipole moment
m
f
da = fa
(1.21)
Then t he vector potent ial for the t he dipole term A
0 m x r · ()µ A dip r 41r r 2
(1.22)
Magnetic field of a magnetic dipole can be written as
(1.23)
The m agnetic field of a (perfect) dipole is easiest to calculate if we put m at t he origin and let it point in the z
11 oo : 01 : oo (1.23)
The magnetic field of a (perfect) dipole is easiest to calculate if we put m at t he origin and let it point in t he z -direction see figure below.
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z
e
m
I I I I I I ............... I .......... ', I cl> ' ,.......I
y
X
Figure 1.8: Calculation of magnetic field due to a magnetic dipole . . According to Eauat1on Ar1,n (r )
A
µ0 m x r
=-
~
. t he oot ent1al
11 oo : 01
: 03
Figure 1.8: Calculat ion of m agnetic field due to a m agnetic dipole . . µ 0 m x r" . According to Equation A dip ( r) = t he potent ial 2
41r
at point (r, 0, ) is
r
A . ( ) = µ 0 m sin 0). dip r 41f r 2 'fJ
(1.24)
Hence the magnetic field B dip(r )
= V x A = µo~ (2 cos 0f + sin 00)
(1. 25)
41rr
Example:
Find the m agnetic dipole moment of the
book type loop shown in Figure below. All sides have length w, and it carries a current I. 23
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z
1 i
Magnetostatics
/2
w
~ Jl>?w
..
I
y
tw
i
w B
+
A
.
I
A
Figure 1.9:
Solution: This wire could be considered t he superposi-
11 oo : 01 : os Figure 1.9:
Solution: This wire could b e considered t he superposit ion of two plane square loops The '' extra'' sides ( AB) cancel when t h e two are put together , since t he currents flow in opposite directions. The net magnetic dipole moment is
v'2Iw 2 ,
its magnitude is
and it points along the 45° line
z =y Example: A circular loop of wire, with radius R , lies in t he xy plane (centered at the origin) and carries a current
I running counterclockwise as viewed from the positive z
(a) What is its m agnetic dipole moment? j [email protected]
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(b ) What is the (approximate) m agnetic field at p oints far from t he origin ? (c) Show t hat, for points on t h e z axis, your answer is con-
sistent with the exact field B = µ~I (R2 : :2 ) 312 when z
(a) m =Ja= I 1rR 2 z
Solution: (b) B ~ I
\
r-..
,,
>> R.
;f; ~f 1
2
(2 cos 0f + sin 00) .
/)
A
/
('
.
r. \
r,
11 oo :01 : oa (a) m =Ja= I 1rR
Solution: (b) B ~
f ~f 1
2
2
z
(2 cos 0f + sin 00)
(c) On the z axis, 0 = O,r = z, f = z( for z > 0). So
B ~ B ~ _µo_I_R_2 z 2z 3 for z < 0, 0 = in place of z 3
1r,
f =
The original ans
R ) to B ~ µ 0 I R 2/ 2 z
-z,
so the field is the sam e, with z
µo f
B = 3.
R2
) ( 2 2 312 2 R +z
reduces (for z
3
>>
So t he ans are same.
Example: (a) A disk of radius R , carrying a uniform surface charge a , is rotating at constant angular velocity w . Find its magnetic dipole moment. (b) A spherical shell of radius R , carrying a uniform surface charge a , is set spinning at angular velocity w. Find t he m agnetic dipole moment of the spinning spherical shell. [email protected]
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Also find t he m agnetic vector potential of the spinning shell. 2
Solution: (a) For a ring, m = I 1rr . So t he current due to a ring wit h radius extended to small amount r ➔ r + dr is (see figure (a) below )
11 oo : 01
: 1o 2
Solut ion: (a) For a ring, m = I 1rr . So the current due to a ring with radius extended to small amount r-+ r + dr is (see figure (a) b elow ) z
R in0 Rd0
r r+dr
(b)
(a)
F igure 1.10:
I -+ O"Vdr = O"wrdr So t he magnetic dipole moment R
m
2
1rr 0"wrdr
=
=
4
7rO"wR /4
0
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Magnetost atics
(b) See figure (b) above. T he total charge on t he shaded ring is dq = 0"(21rR sin 0) R d0. The time for one revolut ion is dt
= 21r /w .
So the current in the ring is I =
%; =
O"wR 2 sin 0d0 . The
11 oo : 01
: 13
(b) See figure (b) above. The total ch arge on the shaded
= a(2nR sin0)Rd0 . The
ring is dq is dt
=
21r /w
time for one revolution
.
So the current in the ring is I = ~~ = aw R 2 sin 0d0. The area of t he ring is 1r(R sin 0) 2 , so the magnetic moment of
= (awR 2 sin 0d0) 1r R 2 sin2 0,
t he ring is dm
and the total
dipole moment of the shell is 7r
4 m = awn R
4 sin 0d0 = ( 4/ 3)aw1r R 3
0
Hence ill =
41r CJW R 4"Z
3
Dipole term of the vector potential is . ( ) _ µ 0 m x r _ µ 0 m sin 0 ;;._ A dip r - - - -2 - '-P 2 41r r 41r r In this problem 4
A . _ µo 41r
R 4sin 0 J., _ µoaw R sin 0). 41r 3 aw r2 \f' 3 r2 '-P
dip -
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Torques and Forces on Magnetic Dipoles
Torn11 P. on
;::i.
mFJJ!nP.tir rlinolP. of rlinol P. mom P.nt. m. w hil P.
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l\tlagnetostatics
Torques and Forces on Magnetic Dipoles
Torque on a magnetic dipole of dipole moment m while placed in magnetic field B
N = m xB
(1. 26)
Force on a magnetic dipole is zero when placed in a uniform field. But when placed in a non-uniform field t hen there is a net force acts F = V (m · B ) (1.27) Potential energy of a magnetic dipole while placed in a magnetic field is U = - m ·B (1.28)
1. 7
Electric Current
The conceptually simplest example of an electric current is t he current in a thin conducting wire. In an ideal onedimensional wire the current I is defined as t he net charge passing a point P per unit time. In a real wire I is defined as t he charge per unit time passing through a cross section
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of the wire at P.
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of the wire at P . (1.29)
The unit of current is the ampere (A), which is the basic electric unit in t he SI system. If t he wire carries a current of 1 A at point P, t hen 1 C of net charge passes P each second. If t he current in a wire is due to charges q moving with mean velocity v, and t he charges have linear density nL ( = number of charge carriers per unit length) , then
(1.30)
In n is the charge carrier density per unit volume and A is t he surface area of t he current carrying wire, q is t he amount of charge each particle carries, t hen t he current is
I = qnAv
(1.31)
What is the current for an orbit ing electron? Well t he charge is e, velocity is v and the radius of atom is taken to be R . Then t he electron crosses a point once in one t ime period. So t he current is
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11 oo : 01
: 21 T
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As T is 21r R / v , t he current I =
ev 21rR
(1.32)
You will need t his expression many t imes
1. 7 .1
Current Density
If the current in a volume of space is due t o charges q with volume number density n ( = number of charge carriers per unit volume) moving wit h mean velocity v , then t he current density is J = qnv (1.33) Note t hat the unit s of qnv are A/ m 2 . Current density is t he flux of electric charge. In general flux is equal to density t imes velocity.
1. 7.2
Conservation of Charge
The net charge of an isolat ed syst em is constant. But charge is not only conserved overall in a syst em , it is also conserved point by point t hroughout t he syst em . This local conservat ion of charge is described mathematically by t he continuity [email protected]
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point by point throughout the system. This local conservat ion of charge is described mathematically by the continuity j [email protected]
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J\/fagnetostatics
equation
V. J = - fJp fJt
(1.34)
Here p(x , t) is the volume charge density (=charge per unit volume), and J (x , t) is the volume current density (= current per unit area). Equation (1 .34) is universally true, for arbitrary time dependence. Integral form of the continuity equation,
d
s
J · dA = - -
dt
(1.35)
states that the rate of charge passing outward through the closed surface S is equal to the rate of decrease of charge in the enclosed volume V. Equation (1.34), or equivalently (1.35) , is a basic equation of electrodynamics, expressing local conservation of charge.
1.7.3
Ohm's Law
How is the current related to t he potential gradient in a conductor? If two terminal points on a conductor are held at a constant potential difference V , e.g., by connecting them to the electrodes of a bat tery, then in equilibrium a steady
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How is the current related to the potential gradient in a conductor? If two terminal points on a conductor are held at a constant potential difference V , e.g. , by connecting them to the electrodes of a battery, then in equilibrium a steady 31
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current flows through the conductor. Let I be the total current at either point, i.e., the integrated flux J J · dA t hrough a surface inside the conductor surrounding the point . It is found empirically t hat formany cases t he current and potent ial difference are proportional,
V = IR
(1.36)
The constant of proportionality R is called the resist ance of t he conductor. The SI unit of resistance is the ohm (D), defined by ID = 1V / A . The reciprocal 1/ R is called the 1 conductance, and its unit is n- , or siemens (S) . Ohm's law holds to a very good approximation for many conductors. However, it is not a universal principle, there are examples where it does not hold. The resistance R of a sample of matter is a function of t he geometry (size and shape) of the sample, and of the material composition . For example, the resistance of a uniform cylinder, of length L and cross section A , is proportional to L and inversely proport ional to A ,
R -
nT I Ll
/ 1 ~7 \
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cylinder, of lengt h L and cross section A , is proport ional to L and inversely proport ional t o A ,
R = pL / A
(1.37)
The parameter p (not to be confused with charge density 32
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p(x ) ! is an int rinsic property of t he material called t he resistivity. We may also write a local form of Ohm's law, which is a more basic equation.
J (x ) = aE (x )
(1.38)
Again, a is the conductivity.
Example Two long coaxial metal cylinders (radii a and b) are separat ed by material of conduct ivity a as shown in figure below . If they are maintained at a potent ial difference V , what current flows from one to the other , in a length L? E
a
----------- ---------------
\
I I I
I I
b------ - - - -- -- - - - ----- --- - - J
L
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b--------------------------J
L
Figure 1.11:
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Solution: The field between the cylinders is
where A is t he charge per unit length on the inner cylinder. The current is
I=
J- da = a
E · da =
a -AL
Eo
The integral is over any surface enclosing t he inner cylinder. The potential difference between t he cylinders is a
V= b
A E· dl = - ln 27rEO
21raL
I = ln(b/a) V So the resistance of the syst em is
R = ln(b/ a) 21raL
b a
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I = ln(b/ a ) V So the resist ance of the syst em is
R = ln(b/ a) 21raL
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Magnetostatics
Magnetic Field in a medium
Electric fields are creat ed by charges; magnetic fields are created by currents. We learned in our first course that the simplest way t o characterise any localised current distribut ion is through a magnetic dipole moment m . For example, a current I moving in a planar loop of area A with normal ft has magnetic dipole moment, m
= ! Aft
Magnet ic vector potent ial and magnetic field due to a magentic dipole are
A (r) = µo m x r => B (r) = µo 41r r 3 41r
3(m • f )f - m r3
(2.1)
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gentic dipole are
A (r)
=
µo m x r =;- B (r) 4n r 3
=
µo 4n
3(m • f )f - m r3
(2.1)
Current loops, and their associated dipole moments, already exist inside materials. They arise through two mechan1sms: •
• Electrons orbiting the nucleus carry angular momentum and act as magnetic dipole moments. • Electrons carry an intrinsic spin. This is purely a quanj [email protected]
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t um mechanical effect. This too contributes to t he magnetic dipole moment. We define t he magnetisation M to be t he average magnetic dipole moment p er unit volume. In most (but not all) materials, if there is no applied magnetic field then the different atomic dipoles all point in random direct ions. This means t hat, after averaging, (m) = 0 when B = 0. However , when a magnetic field is applied , t he dipoles line up. The magnetisation typically takes t he form M ex B. We' re going to use a slightly strange notation for the proport ionality constant . (It's historical but , as we'll see, it turns out to simplify a later equation)
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applied, the dipoles line up. The magnetisation typically takes t he form M ex B. We're going t o use a slight ly strange notation for t he proport ionality constant . (It's historical but, as we'll see, it t urns out to simplify a later equation)
M = 1 Xm B µo l + Xm
(2.2)
where Xm is the magnetic susceptibility. The magnetic properties of materials fall into t hree different categories. The first two are dictated by t he sign of Xrri : • Diamagnetism: - 1 < Xm < 0. The magnetisation of diamagnetic materials points in t he opposite direction t o the applied magnetic field. Most metals are [email protected]
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netic, including copper and gold. Most non-metallic materials are also diamagnetic, including, import antly, water with Xm ~ -10- 5 . This means, famously, that frogs are also diamagnetic. Superconductors can be t hought of as " perfect" diamagnets wit h Xm = -1 . • Paramagnetism: Xm > 0. In paramagnets, t he magnetisation points in the same direction as t he field. There are a number of paramagnetic met als, including Tungst en, Cesium and Aluminium. • Ferromagnetism: M i= 0 when B = 0. Materials with t his property are what you usually call '' magnets'' . They're
11
00:01 :41
are a number of paramagnet ic metals, including Tungst en, Cesium and Aluminium. • Ferromagnetism: M # 0 when B = 0. Materials with t his property are what you usually call '' magnets'' . They're t he t hings stuck to your fridge. The direction of B is from t he sout h pole to the nort h. Only a few elements are ferromagnetic. The most familiar is Iron. Nickel and Cobalt are other examples.
2.1
Bound Currents
When a material becomes magnetised (at least in an anisotropic way), t here will necessarily be regions in which there is a current. This is called the bound current. [email protected]
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Magnetost atics
Let 's first give an intuitive picture for where t hese bound currents appear from. Consider a bunch of equal magnet ic dipoles arranged uniformly on a plane like t he left pict ure : The currents in t he interior region cancel out and we're left only wit h a surface current around t he edge. We know t hat t his is surface current a K. We'll follow this notation and call the surface current arising from a constant, internal magnet isation K bound· M
M
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and call the surface current arising from a constant, internal magnetisation K bound . M
M
Bound surface current
Bound surface and volume current
Figure 2.1: Origin of bound surface and volume current
Now consider instead a sit uation where the dipoles are arranged on a plane, but have different sizes. We'll put t he big ones to t he left and the small ones to t he right, like In t his case, the currents in the interior no longer cancel. As we can see from the right side of the picture pict ure, t hey go into the page. since M is out of the page, and we've arranged t hings so that M varies from left to right , this suggests that j [email protected]
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J
bound
f"..J
v7
X
J\/fagnetostatics
M.
Bound surface and volume currents are K b ound
=M
ft
(2.3)
M
(2.4)
X
and J bound
= v7
X
Note that t he bound current is a steady current, in the
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and J bound
=V
X
M
(2.4)
Note that t he bound current is a steady current, in the sense that it obeys V · J bou11d = 0 .
2.2
Magnetic intensity
Recall t hat Amp'ere's law describes the magnetic field generated by static currents. We've now learned t hat, in a material, there can be two contribut ions to a current: the bound current J bound that we've discussed above, and t he current J free from freely flowing electrons that we were implicit ly talking. Amp'ere's law does not distinguish between t hese two currents; the magnetic field receives cont ributions from both. V
X
B
[email protected]
= µo (J free + J bound ) = µoJ free + µo V x M 39
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(2. 5) P hysicsguide
Magnetostatics
We define the Magnetic Intensity or magnetising field, H as
(2.6) This obeys
V x H=J free and in integral form
(2.7)
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H = -B-M µo
(2.6)
V x H = J rree
(2.7)
This obeys and in integral form
H ·di = Irenc
(2. 8)
We see that the field H plays a similar role to the electric displacement D ; the effect of the bound currents have been absorbed into H , so that only the free currents cont ribute. Note, however, that we can't quite forget about B entirely, since it obeys V · B = 0 . In contrast, we don 't necessarily have '' V · H = O'' . Rather annoyingly, in a number of books H is called t he magnetic field and B is called t he magnetic induction. But this is stupid terminology so we won't use it.
2.2.1
Linear magnetic material
For most of t he magnetic material the magnetization vector and magnetic intensity are proportional. They are called j [email protected]
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linear magnetic material
(2.9) The constant of proport ionality Xm is called the magnetic
11 oo : 01 : s2 linear magnetic material
(2.9) The constant of proportionality Xm is called the magnetic susceptibility; it is a dimensionless quantity t hat varies from one substance to another - positive for paramagnets and negative for diamagnets. From the definition of H
or B =µH
(2.10)
where (2.11)
µ is called the permeability of the mat erial.
Do some example from Griffiths.
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2.3
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Electromagnetic Boundary conditions
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2.3
Electromagnetic Boundary conditions
2.3.1
Magnetostatic Boundary condition inside a medium
Boundary conditions inside the magnetic medium. (2.12)
and II H above
-
- K
II
H below -
A
f X n
(2.13)
t his can also be written as in terms of B
11 B above
2.3.2
-
11 B below -- µo (K
X
n")
(2. 14)
Boundary condition in both electric and magnetic field
In the presence of surface charge, the electric field normal to t he surface is discontinuous, while the electric field tangent to the surface is continuous. For magnetic fields, it's t he other way around: in the presence of a surface current, the magnetic field normal to t he surface is continuous while t he magnetic field tangent to the surface is discontinuous. [email protected]
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Magnetostatics
What happens with dielectrics? Now we have two opt ions of the electric field, E and D , and two options for the magnetic field, B and H . They can 't both b e continuous because they're related by D = EE and B = µH and we'll be interested in sit uation where E( and possibly µ) are different on either side. Nonetheless, we can use the same kind of computations t hat we saw previously to derive the boundary conditions. Roughly, we get one boundary condit ion from each of the Maxwell equations.
The tangential component of the electric field Is continuous.
The normal component of the electric field is discontinuous
Figure 2.2: Boundary conditions of electric field
For example, consider t he Gaussian pillbox shown in the left-hand figure above. Integrating t he Maxwell equation V · D = p free t ells us that the normal component of D is discontinuous in the presence of surface charge, (2.15) j [email protected]
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j [email protected]
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where n is the normal component pointing from 1 into 2. Here CJ refers only to the free surface charge. It does not include any bound charges. Similarly, integrating V · B = 0 over t he same Gaussian pillbox tells us that the normal component of the magnetic field is cont inuous. (2.16)
To det ermine the tangential components, we integrate t he appropriate field around the loop shown in the righthand figure above. By Stoke's theorem, this is going to be equal to t he integral of the cur1 of t he field over the bounding surface. This tells us what the appropriate field is: it 's whatever appears in the Maxwell equations with a curl. So if we integrate E around the loop, we get the result
nx
(E 2 -
E 1)
=
o
(2.17)
Meanwhile, integrating H around t he loop tells us t he discontinuity condit ion for t he magnetic field (2.18)
where K is the surface current.
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2.3.3
1\/Jagnetostatics
Example: Field inside the gap
Here is a very interesting problem. AC-magnet is shown in Figure . All dimensions are in cm. The relative permeability of t he soft Fe yoke is 3000 . If a current I = 1 amp is to produce a field of about 100 gauss in t he gap, how many t urns of wire are required? Fe
T9 1
20 1
2
1 7~
1--7
T 9 1
Figure 2.3: Magnetic field in the gap
Solution: You must know t he boundary condit ion of t he magnetic field. Consider t he surfaces which act like the cross section of t he magnet (red colored in t he image) . The
11 oo : 02 : os Solution: You must know t he boundary condit ion of t he magnetic field . Consider t he surfaces which act like t he cross section of t he magnet (red colored in t he image) . The
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normal component of B is continuous
And there are no parallel component of Magnetic field (remember t hat in a solenoid t he field is along the axis only) . So, t he magnetic intensity in the gap is H = J}_, while that µo inside t he magnet is H = µoµ B , where µr is t he relative perr meability of the iron. Ampere's circuital law
H · dl = N I applied to the closed loop passing through inside the magnet and covering the whole magnet
hence 1 d + -(4l - d) µr now apply the given data - l = 20 cm = 0.2 m , d = 2 cm = 0.02 m, and B = 100 gauss = 100 x 10- 4 Tesla.
N
= 100
X
10-
4
Ll.'71" >< 1 n-1 >< 1
0.0
2+
0.2
X
4 - 0.02
~nnn
11 oo : 02 : 01 j [email protected]
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normal component of B is continuous
And there are no parallel component of Magnetic field (remember that in a solenoid t he field is along the axis only) . So, the magnetic intensity in the gap is H = J}_ , while that µo inside t he magnet is H = µoµr B , where µr is t he relative permeability of the iron. Ampere's circuital law
H · dl = N I applied to the closed loop passing through inside the magnet and covering the whole magnet
hence 1 d + -(4l - d) µr now apply the given data - l = 20 cm = 0.2 m , d = 2 cm = 0.02 m, and B = 100 gauss = 100 x 10- 4 Tesla.
N
= 100 41r
X
4
1010- 7 X 1 X
0.0
2
+
0.2
4 - 0.02 3000
X
= 161 t urns j [email protected]
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Electromagnetic Induction Sk J ahiruddin Assistant Professor Sister Nibedita Govt. College, Kolkata Author was t he topper of IIT Bombay M.Sc Physics 2009-2011 batch He ranked 007 in IIT JAM 2009 and 008 (JRF) in CSIR NET J t1ne 2011 He has been teaching CSIR NET aspirants since 2012
Contents 1 Changing Electric and Magnetic Field
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Motional emf .
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1.2 Faraday 's La~r .
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1.3 Induct ance . . .
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1.3.1
Self Inductance . .
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1.3.2
Mutual Inductance
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1.1
1.4
Magnetostatic Energy
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1.1
P hysicsguide
Electromag11etic Induction
Changing Electric and Magnetic Field Motional emf
When charge moves there acts the Lorenz force on t hem F
= qE + qv x B
(1.1)
The charges move according t he force experienced by t hem. The motion of t he charges generates a current and t hus an emf on t he medium in which t he charges are moving. This is called motional emf. Lorentz force law is very useful to determine t he direct ion of t he induced current . To calculate t he magnitude we need to know t he Faraday's law of induction.
1.2
Faraday's Law
When the magnetic flux ( = J B · dS) somehow changes in a closed conducting system (be it a closed loop of wire or
11 oo :oo :oa When the magnetic flux ( q> = J B · dS) somehow changes in a closed conducting syst em (be it a closed loop of wire or [email protected]
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somet hing a bend and closed rod) t here creat es an emf
(1.2) The significance of t he negative sign is, t he induced emf t ries t o oppose t he change of the magnetic flux . The basic principle t he Faraday's law states is: A changing magnetic field induces an electric field.
E · dl = - dq> = -
dt
aB. da
at
(1.3)
This is Faraday's law, in int egral form. We can convert it to differential form by applying Stokes; theorem: \7
X
E= -
aB
at
(1.4)
This equation t ells us t hat if you change a magnetic field , you'll creat e an electric field. In t urn, this electric field can be used to accelerate charges which , in this context , is usually thought of as creating a current in wire. The process of creat ing a current t hrough changing magnet ic fields is called induct ion. Faradav's law tells us t hat if vou chan e:e the mae:netic
11 oo : oo : 11 usually thought of as creating a current in wire. The process of creating a current t hrough changing magnetic fields is called induction. Faraday's law tells us that if you change the magnetic flux t hrough S then a current will flow. There are a number [email protected]
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Electromagnetic Induction
of ways to change t he magnetic field. You could simply move a bar magnet in the presence of circuit, passing it through the surface S; or you could replace t he bar magnet with some other current density, restricted to a second wire C', and move t hat; or you could keep the second wire C' fixed and vary the current in it, perhaps turning it on and off. All of these will induce a current in C. However, there is then a secondary effect . When a current flows in C, it will create its own magnetic field. This induced magnetic field will always be in the direction that opposes the change. This is called Lenz's law. If you like, " Lenz's law'' is really just the minus sign in Faraday's law. Example A metal bar of mass m slides frictionlessly on two parallel conducting rails a distance l apart (see figure below) . A resistor R is connected across t he rails, and a uniform magnetic field B , pointing into the page, fills the entire region.
11 oo : oo : 14 uniform magnetic field B , pointing int o the page, fills the entire region.
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V
m
Figure 1.1: Current creat ed by moving bar in a magnetic field
(a) If the bar moves to the right at speed v, what is the current in the resistor? In what direction does it flow? (b) What is the magnet ic force on the bar? In what direction? (c) If the bar starts out wit h speed v 0 at time t = 0, and is left to slide, what is its speed at a later t ime t ? (d) The initial kinetic energy of t he bar was, of course,
11 oo : oo : 16 (c) If the bar starts out with speed v 0 at time t = 0, and is left to slide, what is its speed at a later time t ?
~
(d) The initial kinetic energy of the bar was, of course, Check that the energy delivered to t he resistor is
mv5.
exactly
~mv5.
Solution: (a) E = -~if!t = - B l9lf = B lv·E = IR dt ' ⇒ I = BAv. The minus sign tells you t he direction of flow [email protected]
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Electromagnetic Induction
of the current. (v x B) is upward, in the bar, so t he current is downward through the resistor . (b) F
= IlB =
2
to the left.
B~ v,
v . Hence V
= Voe -
B2l2
mR
t
(d) The energy goes into heat in the resistor. The power 2
delivered to resistor is 1 R , so dW - I 2R - B 2z2v2 R - B 2z2
dt -
we have taken a
R2
-
2 2
B l • rnR'
So
dW
dt
-
2 - 20.t
R Vo e
= amv02 e- 2at
The total energy delivered to the resistor is oo
e- 2atdt =
e-2at oo amv02 - -
- 2a
n
= a.mv 2 -
1
°2a
1
= -mv 2 2
°
11 oo : oo : 19 we have taken a = -
2 2
B l
•
mR '
So
dW
dt
= amv02e- 2at
The total energy delivered to the resistor is W
oo
2 =am v0
e- 20.tdt
1
e - 20.t oo
=
0
amv02 - - 2a
= 0
amv 02 2a
1
=
-mv 2 2
°
Example: Faraday's disk generator: A metal disk of radius a rotates with angular velocity w about a vertical axis, t hrough a uniform field B , pointing up. A circuit is made by connecting one end of a resistor to the axle and the other end to a sliding contact , which touches the outer edge of the disk (See figure) . Find the current in the resistor. [email protected]
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B
B (Sliding contact)
I R
Figure 1.2: Current created by rotating metal disk
Solution: The speed of a point on the disk at a distance s from t he axis is v = ws , so t he force per unit charge is f mag = v x B = wsB§. The emf is therefore ra
ra
, ,D ,.,,2
11 oo : oo : 22 Solution: The speed of a point on the disk at a distance s from the axis is v = ws , so t he force per unit charge is f mag = v x B = wsB§. The emf is t herefore a
a
fmagds = wB 0
0
So t he current is
wBa 2 sds = - 2 2
I = ~ = wB a R 2R The flux law or Faraday-Letz rule can also be written as in terms of electric field E · dl
j [email protected]
= - d dt
(1.5)
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Example: A uniform magnetic field B (t), pointing straight up , fills t he shaded circular region, made by conducting material, as shown in t he figure below. If B is changing with t ime, what is the induced electric field?
B(t)
I
\
-- -- ..... ' ..... _-- S
\
I
11 oo : oo : 24 I
\
'
.....
_--
S
\
"-"" I ......
Amperian loop Figure 1.3: Induced Current by changing magnetic field
Solution: E points in the circumferential direction, just like t he magnetic field inside a long straight wire carrying a uniform current density. Draw an Amperian loop of radius s, and apply Faraday's law: d1> d 2 2 dB E · dl = E(21rs) = - dt = - dt (1rs B (t)) = -1rs dt [email protected]
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Hence
E = -~dB 1J 2 dt If B is increasing, E runs clockwise, as viewed from above. Example: A line charge A is glued onto t he rim of a
wheel of radius b,. T he spokes are made of some nonconducting material. The wheel is then suspended horizontally, as shown in figure below so t hat it is free to rotate. In the central region, which is made by conducting material, out to radius a, there is a uniform magnetic field B 0 , pointing up. Now someone turns the field off. What happens?
11 oo : oo : 26 ducting material. The wheel is then suspended horizontally, as shown in figure below so that it is free to rotate. In the central region, which is made by conducting material, out to radius a, there is a uniform magnetic field B o, pointing up. Now someone turns the field off. What happens?
j [email protected]
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Electromag11etic Induction
E l
11 oo : oo : 29
E Rotation direction
di
Figure 1.4: Charged disk rotates because of changing B
solution: The changing magnetic field will induce an electric field, curling around the axis of the wheel. This
electric field exerts a force on the charges at the rim, and the wheel st arts t o t urn. According t o Lenz's law, it will rotat e in such a direction t hat its field tends to restore t he upward flux. The mot ion, then, is counterclockwise, as viewed from above. Faraday's law, applied to the loop at radius b , says
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dif> E-dl = E (21rb) = - - = dt
Physicsguide
11
Electromagnetic Induction
-
2 dB 1ra dt '
2
or
E = _a dB 4J 2b dt
The torque on a segment of length dl is (r x F) , or b)..E dl. The total torque on t he wheel is therefore
11 oo : oo : 32 2b dt
T he torque on a segment of length di is (r x F ), or b>..E dl. The total torque on t he wheel is t herefore
N
a 2 dB 2b dt
= b>..
dl = -b>..1ra 2 dB dt
T he angular moment um imparted to t he wheel is 0
N dt
=
->..1ra2 b
dB
=
>..1ra 2 bB0
Bo
It doesn 't matter how quickly or slowly you turn off the field; t he resulting angular velocity of t he wheel is the same. Now the quest ion is - where is the angular momentum coming from? wait for the next section.
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1.3 1.3.1
12
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Electromagnetic Induction
Inductance Self Inductance
Suooose that a constant current I flows along some curve
11 oo : oo : 34 ance 1.3.1
Self Inductance
Suppose that a constant current I flows along some curve C. T his gives rise to a magnetic field and hence a flux = J8 B · dS t hrough t he surface S b ounded by C . Now increase t he current I . This will increase t he flux . But we've just learned t hat t he increase in flux will, in t urn, induce an emf around t he curve C. T he minus sign of Lenz's law ensures t hat this acts to resist t he change of current. T he work needed to build up a current is what 's needed to overcome t his emf. If a current I flowing around a curve C gives rise to a flux = J8 B · dS t hen t he inductance L of t he circuit is defined to be
(1.6) The inductance is a property only of our choice of curve C . Example: Self inductance of a Solenoid: A solenoid consists of a cylinder of length l and cross-sectional area A . We take l >> -JA so t hat any end-effects can be neglected. A wire wrapped around t he cylinder carries current I and j [email protected]
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Electromagnetic Induction
winds N t imes per unit length. The magnetic field t hrough t he cent re of t he solenoid t o be T 1\ T
I
-t
,-, \
00: 00: 36 Electromagnetic Induction
winds N t imes per unit length. The magnetic field t hrough t he cent re of the solenoid to be
B = µoIN
(1. 7)
Figure 1.5:
This means that a flux through a single turn is 1> 0 = µ 01 NA. The solenoid consists of Nl t urns of wire, so the total flux is with V = Al the volume inside t he solenoid. The inductance of the solenoid is there£ore
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1.3.2
Mutual Inductance
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11 oo : oo : 39 @Sk J ahiruddin, 2020
1.3.2
Electromagnetic Induction
Mutual Inductance
Say there are two loops placed in reasonably closed distance and current 11 passes through loop 1. The current will create magnetic field. T hat magnetic field will create magnetic flux around the second loop (loop 2) . It can be proved t hat the flux flown in the second loop due to current in 1st loop is proportional to I 1 .
~Loop2
~---t-r-:::::
~Loopl
Figure 1.6: Mutual inductance
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Electromagnetic Induction
11 oo : oo : 42 [email protected]
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Electromagnetic Induction
(1.8) where M 21 is the constant of proportionality; it is known as the mutual inductance of t he two loops.
M 21 is a purely geometrical quantity, having to do wit h t he sizes, shapes, and relative positions of the two loops. If we reverse the situation: current 12 passes t hrough loop 2. The current will create magnetic field. That magnetic field will create magnetic flux around loop 1. It can be proved that the flux flown in the loop 1 due to current in loop 2 is proportional to 12 with the same proportionality constant . i.e
(1.9) and
(1.10) Whatever the shapes and positions of the loops, the flux t hrough loop 2 when we run a current I around loop 1 is ident ical to t he flux through loop 1 when we send t he same current I around loop 2. We drop the subscripts and call t hem both M .
j ahir@physicsguide. in
16
Physicsguide
11 oo : oo : 44 t hem both M .
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Electromagnetic Induction
Example: A short solenoid (length l and radius a, with n 1 turns per unit length) lies on t he axis of a very long
solenoid (radius b, n 2 t urns per unit length ) as shown in Figure below. Current I flows in the short solenoid. What is the flux t hrough t he long solenoid? What is t he mutual inductance of the system?
Solution: Since the inner solenoid is short , it has a very complicated field; moreover , it puts a different flux t hrough each turn of the outer solenoid. It would be a very tough task to compute the total flux this way. However, if we use t he equality of the mutual inductances, the problem becomes very easy. Just look at the reverse sit uation: run the current / t hrough the outer solenoid, and calculate the flux through t he inner one. The field inside t he long solenoid is constant:
B = µ 0n 2 I so t he flux through a single loop of the short solenoid is 2
B1ra = µ 0n 2 I 1ra
2
There are n 1 l t urns in all, so the total flux t hrough the inner solenoid is
11 oo:oo:47 There are n 1 l t urns in all, so the total flux through the inner solenoid is
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Electromagnetic Induction
This is also the flux a current J in the short solenoid would put through the long one, which is what we set out to find. Incidentally, the mutual inductance, in this case is
1.4
Magnetostatic Energy
The definition of inductance is useful to derive the energy stored in the magnetic field. Let's take our circuit C with current I. We'll t ry to increase the current. The induced emf is E == _ dct> == - L dI dt dt As we mentioned above, the induced emf can be thought of as the work done in moving a unit charge around the circuit. But we have current I flowing which means that, in time bt, a charge I bt moves around the circuit and the amount of work done is
dI bW == Eibt == - LI dt bt dW == - LidI == _ LdI J.L
J.L
C")
2
J.L
11 oo : oo : 49 of work done is dI et
where k is a constant, and e = 1/ ,FoJio
Solution:
Well' t h is can be taken as a reverse type
of problem where we will calculate t he charge and current distribut ions from t he given potent ials. In generally we cal,
'
. ,
.
.
11 oo : 01 : os Solution: Well' this can be taken as a reverse type of problem where we will calculate the charge and current distributions from the given potentials. In generally we calculate t he opposite. First we'll det ermine t he electric and m agnet ic fields, using B = v' x A and E = - v'V -
E=-
8A
at
~!
µok ,, = - 2 (et - xl)z
and
B = "v
X
A = -µo -k 8 (et4e 8 x
X
) 2y,, =±-etµok ( 2e
X
)y,,
(plus, for x > O; minus, for x < 0). These are for x < et; when lxl > et E = B = 0. see figure below. Calculating every derivative in sight, We find j [email protected]
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JVIaxwell's equations
-et
et X
~ket 2
II
1,-~
11 oo : 01 : oa X
~kct 2
~kt 2
-et X
Figure 3 .1: charge and current from given potent ials
and
aE
at
µokc,_ = - 2 z;
_aB _ _ ± -µo_ky"
at
2
As you can easily check, Maxwell's equations are all satisfied , with p and J both zero. Notice, however , that B has [email protected]
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Maxwell's equations
a discont inuity at x = 0 , and this signals the presence of a surface current K in the yz plane; boundary condition 1...B I 1...B I = K 1 x n gives µ1 1 µ2 2
kt y = K
X
and hence
K
= ktz
X
11 oo : 01 lµ1 B 1II - lµ2 B 2II
: 1o
= K J x n gives kty = K
X
X
and hence
K
3.2
= ktz
Gauge Transformation
From the relations of fields and potentials you can easily verify t hat if you change the electric and magnetic vector potentials according to the following relation t hen t he elect ric and magnetic field remains unchanged.
A'= A + v'A V' = V- 8A
(3.8)
at
A is a scalar function of position and time A = A(x .y.z, t) Gauge transformations are broadly classified into Coulomb's Gauge and Lorentz 's gauge. The Gauge depends on t he choice of A Coulomb's Gauge:
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P hysicsguide
Maxwell's equations
In which case we apply Coulomb's Gauge? Of course in static configuration. Where the charges, currents and fields are not changing with respect to time. What is t he condit ion of A? Well the gauge choice tells us
11 oo : 01
: 13
In which case we apply Coulomb's Gauge? Of course in static configuration. Where the ch arges, currents and fields are not changing with respect to time. What is t he condition of .X? Well the gauge choice tells us t hat v7 · A = 0 for both b efore and after t h e transformation of the potentials. Take divergence both side of the relation A' = A
+ v7 .X.
You get
v7 · A' = v7 · A
+ v7 · v7 .X
hence as both v7 · A = 0 and v7 · A' = 0. We get 2
v7 .X = 0 That is the choice of .X of Coulomb's gauge
Lorenz gauge:
av v7 ·A = -µoEo- -t 8
(3.10)
Where will we apply Lorentz gauge? Definitely in dynamic situation where t h e charge currents and fields changes with time. j [email protected]
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Physicsguide
Maxwell's equations
And what is t he choice of .X? . Well t he relation v7 • A =
-µ 0 c0 ~~ is valid both before and after t he t ransformation . •
1.e
11 oo : 01
: 1s
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Maxwell's equations
And what is t he choice of A? . Well t he relation V · A = -µ 0 c0 ~1; is valid both before and after t he t ransformation . •
1.e
V . A = - µoEo
av
av' V ·A'= -µoEo -
at ;
at
Now you again take divergence both side of t he relation A'= A + V A. You get
V · A' = V · A
+ V · VA
Use Lorentz gauge conditions to get
t hen apply gauge transformation basic criteria V'
=V-
-a>-. at
So we get
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or
28
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lVIaxwell's equations
11 oo : 01
: 19
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Maxwell's equations
or
av -µoEo at
2
a A av + µoEo at2 = -µ oEo at
2
+V A
hence the choice of A in Lorentz gauge is
Now we explore more with some examples Does the following set of potent ials represents a physical electromagnetic field? Example:
A = (2x -wt)i+(y- 2z)]+ (z -
cp = 3xyz-4t;
2eiwt ) k
Solution: The most obvious way is to calculate t he Elect ric and magnetic field from the given potentials and check
whether they satisfy the Maxwell's equations. But the shortcut in t hese type of problems is to check the gauge condit ion. As t he potentials are t ime dependent then you check whether t he the Lorentz gauge V · A = -µ 0Eo ~~ condition is valid here. Which of the following t ransformations
Example:
(V, 1 ) >
>
(V' ,
A') of the electrostatic potential V and t he
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Maxwell's equations
11 oo : 01
: 21
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vector potential June 2015]
Maxwell's equations
A is a gauge transformation?
/
➔,
➔
/
➔,
➔
[NET
A
(a) (V = V +ax,A = A +atk) (b) (V
=
V
+ ax, A
=
➔,
/
A
A - atk)
➔
A
(c) (V =V +ax, A = A+ati) (d) (V'
=
V
+ ax , 1 =
A- ati)
Solution: As t he transformation condit ion of t he potent ial is the same so you definitely get the A from there. As V' = V - 9ft, then A= -axt. Hence the vector potential
A' = A + VA; Example: The scaler and vector potent ials cp(;i:, t) and (;i:, t) are determined up to a gauge transformation
A
qJ , , qJ'
= qJ - ~ and
A , , A' = A + V (
where ( is an
arbitrary continuous and differentiable function of If we further impose the Lorenz gauge condit ion
;i: and t.
~.A+! it = 0 t hen a E;ossible choice for t he gauge function ~(;i:, t) is (where w, k are nonzero constants wit h w = c
Ji )
(a) cos wt cosh (c)cosh wt cos
[NET Dec 2014]
1.;i: 1. ;i:
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1. ;i: 1. -t
(b) sinh wt cos (d )cosh wt cosh 30
Physicsguide
11 oo : 01
: 24
C
➔
(a)cos wt cosh k .? (c)cosh wt cos
1.?
j [email protected]
➔
(b )sinh wt cos k.? (d )coshwt cosh
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Maxwell's equations
Solution: Check t he Lorentz gauge condition over >. (the t here )
2 a2t V t = µoEo at2
for all the opt ions. The ans is (d) . Don't for get to use
1 2 C =-µoEo
4
w
2
k2
Momentum in EM Field
4.1
Poynting's theorem
We know that ELECTROMAGNETIC ENERGY DENSITY, i.e, the energy stored in electromagnetic fields, per unit volume, 1s 1 U=2 •
Now, Suppose we have some charge and current configurat ion which , at t ime t, produces fields E and B . In t he next instant , dt , the charges move around a bit. Now the question is -- How much work, dW, is done by the electromagnetic forces acting on these charges, in
11 oo : 01
: 26
instant, dt , the charges move around a bit. Now the question is -- How much work, dW, is done by the electromagnetic forces acting on these charges, in [email protected]
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31
Physicsguide
lVIaxwell's equations
t he interval dt? After some calculation we can find that the rate at which work is done on all the charges in a volume Vis
dvV dt
1 d dt v 2
s
(E x B ) · da (4.1)
where S is the surface bounding V. This is Poynt ing's theorem; it is the '' work- energy t heorem'' of electrodynamics. The first integral on the right is the total energy stored in the fields , J ud3r The second term evidently represents t he rate at which energy is t ransported out of V, across its boundary surface, by the electromagnetic fields. Poynting's theorem says, then, that the work done on the charges by the electromagnetic force is equal to the decrease in energy remaining in t he fields, less the energy that flowed out through the surface. The energy per unit t ime, per unit area, transported by t he fields is called the Poynting vector: 1 S - (Ex B ) (4.2) µo
11 oo : 01
: 29
The energy pe1~unit time, per unit area, transported by the fields is called the Poynting vector: 1 S _ - (E x B ) (4.2) µo S · da is the energy per unit time crossing the infinitesimal surface da - the energy flux. Generally, S is called the 32
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Maxwell's equations
energy flux density). we can write Poynt ing theorem more compactly d
dW
S · da
dt
(4.3)
s
If there is no charge and current - empty space - no work is done on the charges -
au dT = at
~~ = 0 S-da= -
(V. S)dT
so we get
au =-V. S at
(4.4)
This is the '' cont inuity equation'' for energy - u (energy density) plays t he role of p ( charge density), and S takes the part of J (current density). It expresses local conservation of electromagnetic energy. Example: A long coaxial cable carries current I. The current flows down the surface of t he inner cylinder , radius
11 oo : 01
: 31
of elect romagnetic energy. Example : A long coaxial cable carries current I . The
current flows down t he surface of t he inner cylinder, radius a, and back along t he outer cylinder , radius b as shown in figure below.
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Maxwell's equations
•
b a
I
I -
Figure 4.1:
(a) Find t he magnet ic energy stored in a sect ion of length l.
(b) Now, assume that the two conductors are held at potential difference V, and carry current I . Calculate the power t ransported down the cables, Solution: (a) According to Ampere's law, the field between t he cylinders is
11 oo : 01
: 34
Solution: (a) According to Ampere's law, the field between the cylinders is B = µof
cp
21rs
outside the field is ZERO. Energy per unit volume is I
µ 0f
2µo
21rs
2
µ 0f
2
81r2s2
The energy in a cylindrical shell of length l , radius s, and [email protected]
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:Nlaxwell's equations
t hickness ds, then , is µof 2
ds s
81r2s2
Integrating from a to b, we have: W = µof 2z ln
b
a
41r
This energy calculation is also a very simple way to calculate t he self-inductance of t he cable. The energy can also be 2 written as ½Lf . Comparing the two expressions, you see
L = µol ln 21r
b a
(b) Take t he charge density of the inner cylinder to be A. Then t he electric field is (Apply Gauss law) A l
11 oo : 01
: 37
(b) Take t he charge density of the inner cylinder to be A. Then t he electric field is (Apply Gauss law)
E=
A
~§
27rEQ S
We have already calculated the magnetic field
B = µo l!;p 21r s
So the Poynting's vector is
S
l (E x B ) =
=
µo
z
Al l 41r2Eo s2
35
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Poynting's vector represents the energy transported p er unit t ime p er unit area. So the Power (energy flow per unit time) is t he surface integral over the Poynting's vector b
S · da = a
Al S21rsds = 27rEo
b
1 Al -ds = ln(b/ a)
27rEo
a S
Now t he potential is b E.
V=
A
di=
27rEO
a
b
1 -ds
a S
=-
A
ln(b/a)
27rEO
Hence
P = IV ·
'
4.2
as it should be
Force on char!!es - Stress Tensor
11 oo : 01
: 39
Hence
P = IV ·
'
4.2
as it should be
Force on charges - Stress Tensor
Tot al electromagnet ic force on t he charges in volume V is
(E + v x B )pd r = 3
F= V
This can be calculated as (you don't need t he derivation. If you st ill want , t hen see Griffit hs.)
F=
d T -da - Eoµ os dt ++
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3
Sd r
(4. 5)
V
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Maxwell's equations
++
where T is t he Maxwell's st ress t ensor defined component wise as •
1
(4.6)
T he indices i and j refer to t he coordinates x, y, and z, so t he stress t ensor has a total of nine component s
(Txx, Tyy, Txz, Tyx, and so on) The Kronecker delta,
Oij,
is 1 if t he indices are t he same
11 oo : 01
: 42 x,
The Kronecker delta,
Oij,
is 1 if the indices are the same
and zero otherwise
Thus components can be calculated as
Txx = -l Eo ( E x2 - Ey2 - E z2) + - l ( B x2 - B y2 - B z2) 2 2µo 1 Txy = Eo (E xEy) + - (B xB y) µo Dot products of T with a vector a can be defined as (a · T )J· -.
~ a-~-· L....t 1., 1.,J '
.
i=x,y,z
i=x,y,z
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Maxwell's equations
In the static case t he second term drops out, and the electromagnetic force on the charge configuration can be expressed entirely in te1~ms of t he stress t ensor at the boundary: F =
H
H
T -da
( static )
(4.7)
s
Physically, T is the force per unit area (or stress) acting on the surface. More precisely, ~ j is the force (per unit area) in the i t h direction acting on an element of surface oriented in the 1 th direction - '' diae:onal' elements
11 oo : 01
: 44
f-+
Physically, T is t he force per unit area (or stress) acting on the surface. More precisely, ~ j is the force (per unit area) in t he i t h direction acting on an element of surface oriented in t he j t h direction - '' diagonal' elements (Txx, Tyy, T22 ) represent pressures, and '' off-diagonal'' elements (Txy , Txz , etc.) are shears. To be more illustrative, suppose you are asked to calculate t he force in t he z direction. If you know t he stress tensor then the force can be directly calculated as f-+
(T ·da )z = Tzxdax + Tzyday + Tzzdaz To calculate t he force using stress tensor is a very lengthy task, so we avoid t hat calculation altogether.
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4.3
Maxwell's equations
Momentum density
Momentum stored in the field can be represented by
(4.8) V
Momentum stored per unit volume is called the moment um density (A O \
11 oo : 01
: 47 (4.8) V
Momentum stored per unit volume is called t he moment um density (4.9) g = µocoS = Eo(E x B ) Momentum per unit t ime flowing in through the surface is represented by the integral of Maxwell's stress tensor . f-+
T -da s
In empty space 8g
f-+
- = v'· T
at
(4.10)
This is the '' continuity equation'' for electromagnetic momentum, with g (momentum density in the role of p ( charge "' density) and - T playing the part of J; it expr·esses t he local conservation of field momentum. Angular moment um stored in the field per unit volume •
IS
£ = r x g = Eo [r x (E x B)] [email protected]
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(4.11) Physicsguide
lVIaxwell's equations
Do some examples and exercises on momentum and angular momentum from Griffiths.
00 : 01 : 49 l j [email protected]
= r x g = Eo [r x 39
@Sk J ahiruddin , 2020
(E x B)]
(4.11) P hysicsguide
:Nlaxwell's equations
Do some examples and exercises on moment um and angular moment um from Griffit hs.
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P hysicsguide
11 oo : oo : oo
Electromagnetic Wave Sk J ahiruddin Assistant Professor Sister Nibedita Govt. College, Kolkata Author was t he topper of IIT Bombay M.Sc Physics 2009-2011 batch He ranked 007 in IIT J AM 2009 and 008 (J RF) in CSIR NET J t1ne 2011 He has been teaching CSIR NET aspirants since 2012
1
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EM wave
11 oo : oo : 03 1
EM wave
@Sk J ahiruddin, 2020
Contents 1
M axwe ll's equations in vacuum - EM wave 1.1
Solution of EM wave . . . . . . .
•
•
•
•
•
•
6
1.2
General solut ion of wave equation
•
•
•
•
•
•
11
1.3
Energy and Momentum in E lectromagnetic Waves
2
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Maxwell's equation in linear medium 2. 1
2.2
3
4
21
Reflection and Transmission at Normal Incidence . . . . . . . . . . . . . . . . . . . . . .
23
Reflection and Transmission at Oblique Incidence . . . . . . . . . . . . . . . . . . . . . .
24
M axwe ll's equations in C onductors 3. 1
16
Attenuation and skin depth 3.1 .1
j [email protected]
. . . . . .
Skin depth in a poor conductor:
2
25 •
•
•
27
•
•
28
Physicsguide
11 oo : oo : os 3.1.1
Skin depth in a poor conductor:
j [email protected]
•
2
EM wave
•
29
Reflection and transmission of EM from a conducting surface . . . . . • • • •
30
3.1.2
Skin depth in a good conductor:
•
4
5
28
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3. 2
•
•
•
•
•
•
Wave guide
33
4.1
TE Waves in a Rectangular Wave Guide . .
35
4.2
TM Waves in a Rectangular Wave Guide . .
38
4.3
Resonance ca,,ity . . . . . . . . .
•
•
•
•
•
•
40
4.4
Dielectric inserted into waveguide
•
•
•
•
•
•
41
Retarded Potential
42
11 oo :oo :oa
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EM wave
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1
Maxwell's equations in vacuum EM wave
Maxwell's equations in vacuum are
(ii) v' · B
=
0,
(iv) v' x B
=
(1.1)
µoEo ~~
They constitute a set of coupled , first-order , partial differential equations for E and B. They can be decoupled by applying the curl to (iii) and (iv):
v7
X
(v7
X
2
E ) = v7 (v7 · E ) - v7 E = v7
aB
at
X
a a2 E = - at (v7 X B) = - µ oEo at2 Similarly for magnetic field V
X
(V
X
2
B ) = V (V · B ) - v7 B = V X =
a
µoEo at (V
X
E)
=
aE µo Eo at
a2 B
-µoEo at2
Or, since v7 · E = 0 and v7 · B = 0 " ' ) T""'O
r,') -r,.
11 oo : oo : 11 µoEo at
X
aB
a
2
= µoEo at (V
X
E) = -µoEo at2
Or, since V · E = 0 and V · B = 0
a2E
2
(1.2)
V E = µoEo 8t2 ' [email protected]
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In vacuum, t hen, each Cartesian component of E and B satisfies the three dimensional wave equation,
v2f
a2 f 2
= I
(1 .3)
v 8t2
So Maxwell's equations imply that empty space supports the propagation of electromagnetic waves, traveling at a speed v=
1
~
=
8
3.00 x 10 m /s
(1.4)
But this is something that we've seen before. It's the speed of light c ! This, of course, is because these electromagnetic waves are light. The emergence of light comes from looking for solut ions of Maxwell's equations in which t he electric and magnetic fields change with time, even in the absence of any external charges or currents. The essence of the physics lies in the two Maxwell equations on the right: if the electric field shakes, it causes the magnetic field to shake which, in t urn, causes the electric field to shake, and so on. The simple calculation that we have just seen represents one of t he most important moments in physics. Not onlv
11 oo : oo : 14 field shakes, it causes the magnetic field to shake which, in t urn, causes the electric field to shake, and so on. The simple calculation that we have just seen represents one of t he most important moments in physics. Not only are electric and magnetic phenomena unified in the Maxwell equations, but now optics - one of the oldest fields in science - is seen to be captured by t hese equations as well. j [email protected]
5
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1.1
Physicsguide
ElVI wave
Solution of EM wave
We've derived two wave equations, one for E and one for B . We can solve these independently, but it's important to keep in our mind that the solutions must also obey t he original Maxwell equations. This will then give rise to a relationship between E and B. Let's see how this works. We'll start by looking for a special class of solutions in which waves propagate in the x -direction and do not depend on y and z. These are called plane-waves because, by construction, t he fields E and B will be constant in t he (y , z) plane for fixed x and t The Maxwell equation V · E = 0 tells us that we must have Ex constant in this case. Any constant electric field can always be added as a solution to the Maxwell equations so, without loss of generality, we'll choose this constant to vanish. We look for solut ions of t he form
11 oo : oo : 16 have Ex constant in this case. Any constant electric field can always be added as a solution to the Maxwell equations so, without loss of generality, we'll choose this constant to vanish. We look for solutions of the form
E = (O, E (x, t), O)
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where E satisfies the wave equation
2-2 a2 E2 - v2E = o c 8t
The most general solut ion to the wave equation takes t he form
E (x, t ) = f (x - et)
+ g(x + et)
Here f (x - et) describes a wave profile which moves to the right with speed e. (Because, as t increases, x also has to increase to keep f const ant). Meanwhile, g(x + ct ) describes a wave profile moving to the left wit h the speed c. The most important class of solutions of this kind are t hose which oscillate wit h a single frequency w . Such waves are called monochromatic. For now, we'll focus on t he rightmoving waves and t ake t he profile to be the sine function.
11 oo : oo : 19 The most important class of solutions of t his kind are those which oscillate wit h a single frequency w . Such waves are called monochromatic. For now, we '11 focus on t he right moving waves and take t he profile t o be the sine function. (We'll look at t he opt ion to take cosine waves or ot her shifts of phase in a moment when we discuss polarisation). We have
E = E osin
w
X
- -t
(1.5)
C
which in generally writ t en as
E = Eo sin(kx - wt ) [email protected]
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7
(1.6) P hysicsguide
EM wave
where k is t he wavenumber . The wave equation requires t hat it is relat ed to the frequency by
(1.7) Equations of t his kind, expressing frequency in terms of wavenumber, are called dispersion relations. Because waves are so important in physics, there's a whole bunch of associated quantit ies which we can define. They are: • The quantity w is more properly called the angular frequency and is t aken t o be posit ive. The actual frequency f = w / 21r measures how often a wave peak passes you by. But because we will only talk about w, we will be lazy and just refer to t his as frequency.
11 oo : oo : 21 quency and is taken to be positive. The actual frequency f = w / 21r measures how often a wave peak passes you by. But because we will only talk about w, we will be lazy and just refer to t his as frequency. • The period of oscillation is T
= 21r / w
• The wavelength of t he wave is ,\ = 21r / k. This is the property of waves t hat you first learn about in school. The wavelength of visible light is between ,\ rv 3.9 x 10-7 m and 7 x 10-7 m . At one end of the spectrum, gamma rays have wavelength ,\ rv 10- 12 m and X -rays around ,\ rv 10- 10 to 10- 8 m. At t he other end, radio waves have ,\ rv 1cm to 10 km. Of course, the elecj [email protected]
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t romagnetic spectrum doesn't stop at t hese two ends. Solutions exist for all A. Although we grow up thinking about wavelength, moving forward the wavenumber k will turn out to be a more useful descript ion of the wave
• E 0 is the amplitude of the wave. So far we have only solved for the electric field. To determine the magnetic field, we use V · B = 0 to tell us that B x is constant and we again set B x = 0. We know that t he other components B y and B z must obey t he wave equat ion ~\ 0:~~ - V 2 B = 0. But their behaviour is dictated by
11 oo : oo : 24 So far we have only solved for the electric field. To determine the magnetic field, we use V · B = 0 to tell us that B x is constant and we again set B x = 0. We know that t he other components B y and B z must obey t he wave equat ion c\ ~t~ - V 2 B = 0. But their behaviour is dictated by what t he electric field is doing through the Maxwell equat ion V x E = -8B / at. This tells us t hat
B = (0,0, B ) with
8B - -t 8
=
8E - --x 8
we find B
=
Eo
=-
-kEo cos(kx - wt) sin(kx - wt )
C
We see that the electric E and magnetic B fields oscillate in phase, but in orthogonal directions. And both oscillate [email protected]
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in directions which are orthogonal to t he direction in which t he wave travels. We can visualize t he wave propagation like t his
k
11 oo : oo : 26 E
k
Figure 1.1: Electromagnetic wave propagation
We can often write t he solut ions of E and Bin complex notation
E ==
Eo-S, ei(kx- wt)
'
B ==
E a z ei(kx- wt)
(1.8)
C
This is strange because t he physical electric and magnetic fields should certainly be real objects. You should t hink of them as simply the real parts of t he expressions above. But the linearity of t he Maxwell equations means both real and imaginary parts of E and B solve the Maxwell j [email protected]
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EM wave
equations. And, more importantly, if we start adding complex E and B solutions, then the resulting real and imaginary pieces will also solve the Maxwell equations . The advantage of this notation is simply t hat it's typically easier to manipulate complex numbers t han lots of cos and sin formulae. However, you should b e aware t hat this notation comes with some danger: whenever you compute something which . ..
.
.
11 oo : oo : 29 ier to manipulate complex numbers t han lots of cos and sin formulae. However, you should be aware t hat this notation comes with some danger: whenever you compute something which isn 't linear in E and B - for example, the energy stored in t he fields , which is a quadratic quantity - you can 't use the complex notation above; you need to take t he real part first .
1.2
General solution of wave equation
Above we have presented a particular solution to t he wave equation. Let's now look at the most general solut ion with a fixed frequency w. This means that we look for solut ions within t he ansatz,
E
= E oei(k ·x-wt)
and
B
(1.9)
= B oei(k ·x-wt)
where, for now, both Ea and Bo could be complex-valued vectors. (Again, we only get t he physical electric and [email protected]
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netic fields by taking t he real part of these equations). The vector k is called the wavevector. Its magnitude, k = k, is the wavenumber and the direction of k points in t he direction of propagation of the wave. The expressions in ( 1.9) already satisfy the wave equations of fields, if w and k obey t he dispersion relation w 2 = c2 k 2 '
,..
.
,
,
..
,.
. ,
11 oo : oo : 31 is the wavenumber and the direction of k points in the direction of propagation of the wave. The expressions in (1.9) already satisfy the wave equations of fields, if w and k obey t he dispersion relation w 2 = c2 k 2 We get further constraints on E 0 , B o and k from the original Maxwell equations. These are
V ·E
=
0 ⇒ ik · E 0 = 0
V ·B
=
0 ⇒ ik · B 0 = 0
'v x E =
~ ~~ ⇒ ik x E 0 =
iwB 0
Linear Polarisation: Suppose that we take E 0 and B 0 to be real. The first two equations above say that both E 0 and B o are orthogonal to the direction of propagation. The last of t he equations above sa)'S that E 0 and B 0 are also orthogonal to each other. You can check that t he t hird Maxwell equation doesn't lead to any further constraints. Using t he dispersion relation w = ck, the last constraint above can be written as
k x (Eo/c) = Bo [email protected]
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@Sk J ahiruddin, 2020 A
This means that t he t hree vectors k , E 0 / c and B o form a right-handed orthogonal triad. Waves of t his form are said to be linearly polarised. The electric and magnetic fields oscillate in fixed directions, bot h of which are transverse to t he direction of propagation.
11 oo : oo : 34 1s means t at t e t ree vectors O orm a right-handed ort hogonal t riad. Waves of t his form are said to be linearly polarised . T he elect ric and magnetic fields oscillate in fixed directions, bot h of which are transverse to t he direction of propagat ion. Evidently, E and B are in phase and mut ually perpendicular; t heir (real) amplit udes are related by k I Bo= - Eo = - Eo W
C
Circular and Elliptic Polarisation: Suppose t hat we now take E 0 and B 0 to be complex . The actual elect ric and magnetic fields are just t he real parts of (1.9) but now t he polarisation does not point in a fixed direct ion. To see t his, write E a = a - i f3 The real part of t he electric field is t hen
E = a cos(k · x -wt ) + f3 sin(k • x - wt ) with Maxwell equations ensuring t hat a · k = ,8 · k = 0. If we look at t he direction of E at some fixed point in space, say t he origin x = 0 , we see that it doesn 't point in a fixed direcj [email protected]
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@Sk J a hiruddin , 2020
t ion. Instead, it rotates over t ime wit hin t he plane spanned by a and ,B (which is t he plane perpendicular to k). A
•
1
•
1
I
1
1
I
1
I
11 oo : oo : 37 t ion. Instead, it rotates over time wit hin t he plane spanned by a and /3 (which is the plane perpendicular to k). A special case arises when the phase of E 0 is ei1r:/ 4 , so that a = 1/31 , with the further restriction t hat a · /3 = 0. Then t he direct ion of E traces out a circle over time in t he plane perpendicular to k. This is called circular polarisation. The polarisation is said to be right-handed if (3 = k x a and left-handed if (3 = - k x a A
A
In general, t he direction of E at some point in space will t race out an ellipse in t he plane perpendicular to t he direction of propagation k. Unsurprisingly, such light is said t o have elliptic polarisation. If t he wave propagate in t he z direction then you take E points in the x direction, and B points in t he y direction
E (z t ) = E '
ei(kz- wt) x 0
'
taking t he real part
E (z, t) = Ea cos(kz-wt+b) x ,
B (z, t) =
! Ea cos(kz-wt+b) y C
We geralize to monochromatic plane waves traveling in an arbitrary direction. Wave vector , k , pointing in the [email protected]
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P hysicsguide
E11 wave
t ion of propagat ion, whose magnitude is t he wave number
11 oo : oo : 40 @Sk Jahiruddin , 2020
EM wave
t ion of propagation, whose magnit ude is t he wave number k. The scalar product k · r is t he appropriate generalization of k z so
E (r , t ) =
B (r , t) =
E oei(k •r - wt) n
1
'(k
-Eoe i
·r-wt
)
(k A
X
n)
C
where
=
1
-k A
X
E
(1.11)
C
n is t he polarization vector .
Because Eis transverse,
A
n· k = 0 The elect ric and magnetic fields (the real part of the complex fields) in a monochromat ic plane wave wit h propagat ion vector k and polarization ft are E (r , t ) = E 0 cos(k · r - wt + c5)n 1 B (r , t) = - E0 cos(k · r - wt+ c5)(k x n) A
(1.12)
C
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P hysicsguide
E M wave
11 oo : oo : 42 15
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1.3
Energy and Momentum in Electromagnetic Waves
Energy per unit volume in electromagnetic field is 1 u =2
(1.13)
for monochromatic plane wave
B
2
=
1
-c2 E
2
=
µoEoE
2
(1.14)
electric and magnetic contributions are equal. t hen the energy density (wave propagating in z direct ion) (1.15) As the wave t ravels, it carries this energy along with it. The energy flux den- sity (energy per unit area, per unit t ime) t ransported by the fields is given by the Poynting vector 1 S =-(Ex B) µo For monochromatic plane waves propagating in t he z direct ion, 2 S = cE 0 cos (kz - wt+ 8)z = cuz
E5
j [email protected]
16
Physicsguide
11 oo : oo : 4s t ion,
j [email protected]
16
Physicsguide
@Sk J a hiruddin , 2020
ElVI wave
moment um density stored in t h e field is 1
g =- S c2
so for plane monochromatic wave 1 E 2 cos 2(k z - wt g = -Eo O
+ u~) z = A
C
1 -uz A
C
7
In the case of light , the wavelength is so short ( rv 5 x 10- m) , 15
and the period so brief ( rv 10- s) , t h at any macroscopic measurement will encompass m any cycles. That 's why we are interested only in time average value. The average of cosine-squared over a complet e cycle is 1 2'
so (1.16)
(1.17)
(1.18) The average power p er unit area t ransported by an elect romagnetic wave is called t h e inten sity :
I
(S)
1
2
= cEo E 0 2
(1.19)
11 oo:oo:47 The average power per unit area t ransported by an elect romagnet ic wave is called t he intensity :
I _ (S) = [email protected]
1
2
2
cEoE 0
(1.19) P hysicsguide
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@Sk J ahiruddin, 2020
E11 wave
When light falls (at normal incidence) on a per£ect absorber , it delivers its momentum t o t he surface. In a t ime
~t, t he momentum transfer is ~p = (g )Ac~ t, so the radiation pressure (average force per unit area) is
(1.20) If the radiation is totally reflected back along its original path, the force per unit area is
2I
P =c
Example: A radio station on t he surface of the Earth t ransmits a sinusoidal wave with an average total power of 50 kW. Assuming the signal is radiated uniformly in all directions above t he ground, what are t he amplitudes, E 0 and B 0 of the respective electric and magnetic fields detected by a satellite at a distance of 100 km from t he transmitt ing antenna at the station?
Solution:
As the station is on t he surface of the earth
t hen all the radiation must have gone above, not inside the
11 oo : oo : so by a satellite at a distance of 100 km from t he transmitting antenna at t he station?
Solution:
As the station is on t he surface of the earth t hen all t he radiat ion must have gone above, not inside the ground. So we t ake a hemisphere centered on t he radio [email protected]
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station wit h a radius r = 100km. (we would t ake full sphere when the radiation will be in the space, like sun) . The surface area of the hemisphere is 2 2 5 A =21rr = 21r (1 x 10 m)
= 6.28
10
x 10 m
2
Since all the radiated power passes through t his surface, the average power per unit area (i.e., t he intensity) is I = p = 5.00 x 104W = 7.96 x 10- 7W / m 2 A 6.28 x 1010 m 2
t he intensity is t he average value of t he P oynt ing vector .
E0 =
2µ 0 cl
== ✓2 (41r
= 2.45
X
X
10- 7 ) (3
X
108 m/s) (7.96
2
10- V / m
Rn = En / ('. = 8.17
X
10- 11 T
X
10- 7W / m 2 )
11 oo : oo : s2 E0
=
2µ 0 cI
= ✓2 (41r
= 2.45
X
X
10- 7 ) (3
X
108m/s) (7.96
X
10- 7W / m 2 )
10- 2V /m 11
Bo = Eo/c = 8.17 x 10- T
Example: You have kept your hand at a distance of 2 m from a 60 W light bulb . find j [email protected]
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EM wave
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(a) t he intensity (b) t he radiation pressure (c) the force acting on your hand (d) t he p eak electric and magnetic fields
Solution : directions.
Assume t he bulb radiates uniformly in all
(a) If t he average power is (P ) = 60W, the intensity at
R = 2m is
2
2
I = (P )/ Area = (P )/41rR = 60W/ 41r(2m) = l .19W/ m (b) The radiation pressure at R = 2m is 2
8
2
9
P rad = I/c = 1.19W / m / 3 x 10 m /s = 3.98 x 10- P a (c) The force on your hand is
2
11 oo : oo : 55 (b) The radiation pressure at R = 2m is Prad
I/c
=
=
2
8
1.19W / m / 3 x 10 m/s
2
=
9
3.98 x 10- Pa
(c) The force on your hand is F = Prad Ahand = (3.98
X
9
10- Pa) (0.10m
X
0.18m)
11
= 7.2 x 10- N (d) We knovv From earlier we have P rad [email protected]
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and Physicsguide
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Jahiruddin, 2020
Ea
Bo = -
EM wave
and so Ea = c✓2Pradµo So
C
E 0 = (3 x 10 m/s) ✓2 (3.98 x 10- 9 Pa) (41r x 10- 7 ) 8
= 29.7 V /m B0
2
=
29. 7V / m 3 x 108 m/s
9 89 10- sT · x
=
Maxwell's equation in linear mediuIJ
Inside matter , but in regions where t here is no free charge or free current, Maxwell's equations become (i) (ii)
V-D V ·B
= =
0
(iii)
'
0
)
(iv)
v x E = _ aB
at
V x H=
aD
at
(2.1)
11 oo : oo : s1 (i)
v7·D == O
(iii)
(ii)
v7·B==O
(iv)
'
'
aB at v7 X H == ao at
(2.1)
aB at aE X B == µE at
(2.2)
v7
X
E == -
In terms of E and B
(ii)
v7 x E == -
(iii)
(i) v7 · E == 0, v7·B == O j
(iv)
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V
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velocity of propagation of E NI wave inside a medium is 1
C
yif]j,
n
(2.3)
V - --- -
(2.4)
n ==
n is the index of refraction of the substance. For most materials, µ is very close to µ 0 , so
(2.5) where
Er
is the dielectric constant. Since
Er
is almost
always greater than 1, light travels more slowly through matter-a fact that is well known from optics.
11 oo : 01 : oo where
Er
is the dielectric constant. Since
Er
is almost
always greater than 1, light travels more slowly t hrough matter-a fact that is well known from optics. energy density would be 1 U=2
(2.6)
Poynting vector 1 s =- (E X B) µ
(2.7)
Intensity
(2.8)
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Boundary conditions
(i) E1Er = E2Et, (ii) B/- = Bf ,
2.1
(2.9)
Reflection and Transmission at Normal Incidence
You have done the derivation before. I just state the results If f3 _ µ1v1 µ 2v2 '
,
=
µ1n2
µ2n1
11 oo : 01
: 03
You have done the derivation before. I just st at e the results If /3
_
µ1 v 1
=
µ 2v2
µ 1n2 µ2n1
t hen the reflected and transmitted component of wave
E
2
EoR =
l
+ (3
Ot
For maximum medium t he p ermeabilities µ are close to their values in vacuum, µ ~ µo then f3
=
v1/v2,
and we h ave (2.10)
Eo1, In t erms of indices of refraction
(2.11)
Eo1, 23
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R eflection and transmission coefficients
IR
R
11
=
EoR Eo1
2
2
n 1 - n2 n1
(2.12)
+n2
and
Jy T == 11 2.2
--
E2V2 E1V1
Eor Eo1
2
4 n 1 n2 (n1
+ n2)
2
(2.13)
Reflection and Transmission at Oblique Incidence
11 oo : 01
2 .2
: 06
R e fle ction and Transmission at Oblique Incide nce cos0r --; cos01 a- /3
a EoR =
(2.14) (2.15)
a + /3
These are known as Fresnel 's equations The reflection and t ransmission coefficients for waves polarized parallel to the plane of incidence are
R
T
Ir 11
--
E2V2 E1V1
IR 11
=
EoT Eo1
EoR Eo1 2
2
cos0r cos01
j a [email protected]
a - /3 a+ /3 =
24
(2.16) 2
a+ /3
2
(2.17) P hysicsguide
ElVI wave
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3
a/3
2
Maxwell's equations in Conductors
Inside conductor charges are free fo move. And free current density follows Ohm's law JJ
= o-E
C- - .... .... . 1\ff _ __......,,....,._11 ,_ --.. ... , ..... ~ : -,...._ ~ - - - ,,..,. . . . --
(3.1)
11 oo : 01 : oa Inside conductor charges are free fo move. And free current density follows Ohm's law
J J = o-E
(3.1)
So our Maxwell's equations becomes .
1
(1) V · E = - p1 , E (ii) V · B = 0,
...)
(
111
'7
v X
.
(1v ) V X
E
8B = - Ot
(3.2)
BE
B = µo-E + µ E
ot
We apply now, t he continuity equation for free charge,
(3.3) Using Ohm's and Gauss Law t his becomes
(3.4) T he solution becomes
(3.5) [email protected]
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P hysicsguide
E11 wave
Thus any initial free charge p(O) dissipates in a characterist ic t ime T Ej a- . This reflects t he familiar fact t hat if you put some free charge on a conductor, it will flow out to t he edges. The time constant T affords a measure of how "good ' a con ductor is: For a "perfect'' conductor, o- = oo and, = O; for a '' good' conductor, , is much less than t he other relevant t imes in the problem.
11 oo : 01
: 11
some free charge on a conductor , it will flow out t o the edges. The time constant T affords a measure of how "good' a con ductor is: For a " p erfect '' conductor, a= oo and a "good' conductor ,
T
T
= O; for
is much less than the other relevant
t imes in the problem. After all free charges disappear Maxwell's equation b ecomes
(i) V · E = 0, (iii) V x E = - ~~ (ii) V · B = 0, (iv) V x B =µ Et~ + µaE
(3.6)
Applying the curl t o (iii) and (iv) , we get modified wave equations for E and B :
a2E
2
V E = µE at2
+ µa
aE
at '
2
a2 B
V B = µE at2
+ µa
aB
at (3. 7)
These equations have plane wave solutions
E(z, t ) =
-
E oei(kz-wt) '
B (z, t ) =
-
B oei(kz-wt )
(3.8)
But with one difference. Now the wave vector has b ecome complex
(3.9) 26
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You can break the wave vector in real and complex part ~
k and get
= k + i~
(3.10)
11 oo : 01
: 13
You can break the wave vector in real and complex part ~
k
= k + i~
(3.10)
and get 1/2
2
Eµ
1+
2
(3.11)
+ 1
EW
and
1/ 2 (J'
- 1
1+
3.1
2
(3.12)
EW
Attenuation and skin depth ~
The imaginary part of k results in an attenuation of t he wave (decreasing amplit ude wit h increasing z) :
E (z, t) =
B (z, t)
Eo e - r;,zei(kz- wt)'
= B oe- r;,z ei(kz - wt)
(3.13) The distance it takes to reduce t he amplit ude by a factor of
1/e (about a t hird) is called t he skin depth: 1 -
d
(3.14)
~
it is a measure of how far t he wave penetrates into the con~ ductor. Meanwhile, t he real part of k determines the wavej [email protected]
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27
P hysicsguide
EM wave
length , the propagation speed, and t he index of refraction, in t he usual way:
11 oo : 01
: 16
@Sk J ahiruddin , 2020
ElVI wave
lengt h , the propagation speed , and t he index of refraction, in t he usual way: .\ ==
ck n== -
2n k '
3.1.1
(3.15)
w
Skin depth in a poor conductor:
For poor conductor a
w
K,
~w
Eµ 2
(3.16)
WE
w
K,
~
w
Eµ
(3.20)
a
l +
2
1/ 2
2
- 1
EW
Eµ
a
2
EW
(3.21)
µaw
1/2 ,...._, ,...._,
2
So for a good conductor
2
(3.22)
µaw For good conductor as a
>> WE so,
you can see (3.23)
or
21r
A= k
A = 21rd or d= 21r "" '
21r ('..J
-
(3.24)
Example : (a) Find t he skin depth of pure water. You are given that for pure water conductivity a= dielectric constant k = 80 .1 and µ ~ µ 0 . j [email protected]
@Sk J ahiruddin, 2020
29
l/ (2.5 x
5
10
),
P hysicsguide
EM wave
11 oo : 01
: 21
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29
EM wave
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(b) Find the skin depth (in nanomet ers) for a typical metal whose conductivity a ~ 107 , the frequency w ~ 1015 with E ~ Eo and µ ~ µo.
Solution: (a) Conductivity shows that the water is a poor conductor. Hence we will take approximation d = 1. rv K,
2 •
(7
d = (2) (2.5 x 10
5
(80.1) (8.85 X l0-12) 41T X 10- 7
)
= 1.19 X 104 m
(b) We have proved that for good conductor d ~
2 µaw
•
Put t he values to get
d
=
8
l X
107
=
1. 3 x 1o-8
=
13nm
So the fields do not penetrate far into a metal
3.2
Reflection and transmission of EM from a conducting surface
Suppose a EM wave is traveling from a medium to a conductor. What will be t he fraction of t he wave reflected and [email protected]
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11 oo : 01
: 23
Suppose a EM wave is traveling from a medium to a conductor. What will be t he fi·action of the wave reflected and j [email protected]
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30
EM wave
@Sk J ahiruddin, 2020
t ransmitted in the conducting surface. Well you can prove ~ t hat if k 2 is the complex wave vector of the conductor, µ 1 , µ 2 are t he magnetic suscept ibilities and v 1 is the velocity of the wave in the first medium then define µ1 v1 ~
~
/3 = - k 2
(3.25)
µ2w
The amplitude of t he reflected and transmitted wave is ~
EoR =
(3.26)
~
1 + /3
~
For a perfect conductor (a = oo), k 2 = oo, so f3 infinite, and so ~
~
Eo1z = - Ea1,
~
EoT = 0
(3.27)
In this case the wave is totally reflected, wit h a 180° phase shift. That's why excellent conductors make good mirrors. You seen t hat mirror is designed by a thin silver coating onto t he back of a pane of glass. The glass has nothing to do with the reflection; it's just there to hold t he silver in t he 1·ight place and transmit light without absorption. Calculate the reflection coefficient for light at an air-to-silver interface Example:
11 oo : 01
: 26
Example: Calculate the reflection coefficient for light at an air-to-silver interface
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at opt ical frequencies (w = 4 x 10
15
/
s)
Solution: 2
R=
~ 2
~
1 - ;3
1 - ;3
~
~
1 + ;3
1 + ;3
where ~
JJ
rv
=
1 - ;3* ~
1 + ;3*
_ µ1V1k~ _ µ 1V1 (k + . ) - -- 2 - -2 'lK,2 µ 2w µ2 w
since silver is a good conductor (a K,2
~
>> EW)
k 2 rv = W
so
Let
=c
(6
X
107 ) ( 47f X 10- 7) = 29 (2) (4 X 1015 )
Then n
( l - 1' - i,1 \ ( l - 1' + i ,1 \
( 1 _ 1') 2 + 1'2
{\
{\')
11 oo :01 : 2a (6
=c
X
10 7 ) ( 41r X 10- 7 ) = 29 (2) (4 X 1015 )
Then
R= 32
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Wave guide
G uided waves are waves which are not free to move everywhere in the space. Dimension of t he medium t hrough which t he wave passe is restricted. The boundary of t he medium must be conductor. An d as the boundary is a p erfect conductor then E
=0
and B = 0 inside the m aterial
itself, and hen ce t he boun dary conditions at t he inner wall are.
(i) (ii)
=O B J_ = 0
E ll
(4. 1)
Let us take the medium is rest ricted by x and y and the wave is propagating along x direction.
11 oo : 01
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EM wave
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X
Conductors
a
------------------
--+---►
z
y
Figure 4. 1: Guided wave
Boundary conditions and the Maxwell's equations gives
(i) and
a2 8x2
a2
+ 8y2 + (w/c)2 -
k2 Ez = 0
(4.2)
11 oo : 01 a2
a2
(i)
Ox2
and
a2
(ii)
: 34
+ [)y2 + (w/c)2 -
a2
Ox2
+ [)y2 + (w/c)2 -
k2 Ez = 0
(4.2)
k2 B z = 0
(4.3)
If Ez = 0, we call t hese T E (t ransverse electric'') waves; if B z = 0, t hey are called T M (''transverse magnetic") waves; if both Ez = 0 and Bz = 0 , we call them TEM waves. You can prove t hat TEM waves can 't occur in hollow rectangular wave guide. [email protected]
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EM wave
TE, mode
⇒
Fz = 0 Bz = exist
T M mode
⇒
Ez = exist B z = 0
TEM mode
4.1
P hysicsguide
⇒
Ez = Bz = 0
TE Waves in a Rectangular Wave Guide
We need solution of (4.3) . t ake
Bz(x, y) = X (x) Y (y) and do t he separation of variable. You get t he solut ion
Bz = Bo cos(m1rx/a) cos(n1ry/b) wit h t he disp ersion relation
(4.4)
11 oo : 01
: 36
Bz = Bo cos(m1rx/a) cos(n1ry/b)
(4.4)
wit h the dispersion relation
k = J(w/c)
2
-
1r
2
[(m/ a)
2
+ (n/b)
2
]
(4.5)
Define
(4.6) Now you see if W
7.5 GHz.
4.2
TM Waves in a Rectangular Wave Guide
The solution is
Ez = F 0 sin
j [email protected]
m1rx
a
•
Sln
n1ry
b
(4.12)
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@Sk J ahiruddin, 2020
if either of t he m and n is zero the solution itself is zero. So the lowest frequency mode is for TM wave in rectangular wave guide is T E11.
Example: A waveguide has a square cross section of side 2a. For Tm modes of wavevector k, the transverse electromagnetic modes are obtained in terms of a function ~ (x, y) which obeys the equation r-
,...,.,.,
,....J)
,
11 oo : 01
: 47
Example : A waveguide has a square cross section of side 2a . For T m modes of wavevector k, t he transverse elect romagnet ic modes are obtained in terms of a funct ion 'lf;(x, y) which obeys t he equat ion 32 8 2 X
+
32 8 2
y
w2 2 -
+ (C
2
k ) 'lf; (x,y) = 0
with t he boundary condit ion 'lf; (±a, y) = 'lf; (x, ±a) = 0. T he frequency w of the lowest mode is given by [NET June 2016] 2 2 2 (a)w2 = c2(k2 + 4;2) (b)w = c (k + ~) 2 2 2 2 2 2 (c)w = c (k + (d)w = c (k +
;:2)
;:2)
Solution: As t he mode is T M , the lowest frequency mode is T M 11 . So t he frequency w of the lowest mode is W 11 = C7r
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2
+
39
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4.3
1 2a
2 =
v'2 C7r 2a
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Resonance cavity
Resonant cavity produced by closing off t he two ends of a rect angular wave guide, at z = Oand at z = d, making a perfect ly conducting empty box. z
11 oo : 01
: 49
Resonant cavity produced by closing off the two ends of a rectangular wave guide, at z = Oand at z = d, making a perfectly conducting empty box. z
►
d
o " r - - - - - - -- t - -
y
X +----------+
b
Figure 4.2: Resonance Cavity
We can prove t hat t he cut off frequencies for both TE and TM modes in a resonance cavity are given by Wzmn
= c1r ✓(l/d) 2 + (m/a)2 + (n / b) 2
Vtmn
=
2 ~ ✓(l/d)
+ (m/a) + (n/ b) 2
40
j [email protected]
(4.13) 2
(4.14) Physicsguide
@Sk J ahiruddin, 2020
EM wave
Example : What should be t he 3rd dimension of a cavity of cross section 1 cm x 1 cm which operates at T E 103 mode at 24 GHz??
Solution: l
=
1, m
=
0, n
=
3. use
11 oo : 01 : s2 Example: What should be t he 3rd dimension of a cavity of cross section 1 cm x 1 cm which operates at T E 103 mode at 24 GHz??
l = l , m = 0, n = 3. use
Solution:
a= 0.01 , b = 0.01 , c = ? use t he formula of 24
2
X
X 10 9
to get 2
2
1 0.01
X 10 8
3
4.4
V zmn
3
+ o+
2
C
Dielectric inserted into waveguide
When t here is a dielectric / magnetic medium inside of a 1 dielectric inst ead of vacuum, just replace c by - -
ftµ
Vmn
=
1
m
2ftµ
a
For most of t he material µ Vmn
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Example:
frequency T E 0
=
~
n
2
2
(4.15)
+ b
µ 0 , so t hen
1
m a
2 ✓ErEOµO C
m
2~
a 41
2
+
2
n b
+ n
2
(4.16)
2
b Physicsguide
EM wave
For a air filled wave guide. t he cut off = 1.8756 GHz. What will be the cut off
00 : 01 : 54 EM w ave
@Sk J ahiruddin, 2020
Example:
frequency T E 0
For a air filled wave guide. t he cut off = 1.8756 GHz. What will be the cut off
frequency if one dielectric of relative permeability E = 9Eo is inserted in t he dielectric? Solution: you have already derived the formula for cut off frequency for t he dielectric inserted wave guide. The ans •
IS
V dielectric = V vacuum / mn
5
mn
;z- = 1· 8756/3 GHz
y cr
Retarded Potential
Suppose you are measuring E and B at a position at a part icular t ime t. If t he system is static then it easy. What if t he source charge and current are changing with time? Electromagnetic signals travels with t he speed of light. So t he charge distribution of little backward time actually contributes in the field calculation. The time we need to calculate the charge and current distribution is called t he retarded time (5.1) [email protected]
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11 oo : 01 : s1
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EM wave
The formula to calculate the potentials t hen
V (r ,t) = -
1
P (r' ' tr) d3 ,
r,
47rEQ
1/,,
A (r , t )
=
µo 1r 4
(5.2) Here p ( r', tr) is t he charge density that prevailed at point r' at the retarded time tr. Because t he integrands are evaluated at the retarded time, t hese are called retarded potential. We will explore through examples. Example:
An infinite straight wire carries the current
l (t) =
0,
for t< 0
l o,
for t > 0
That is, a constant cu1~rent 10 is t urned on abrupt ly at t Find the result ing electric and magnetic fields.
= 0.
I dz
Figure 5.1: Retarded potent ial calculation for a current carrying wire which have started at t = 0 j [email protected]
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Physicsguide
ElVI wa.ve
11 oo : 02 : oo 43
j [email protected]
P hysicsguide
@Sk J ahiruddin , 2020
E M wave
Solution: The wire is electrically neutral, so t he elect ric scalar potential is zero. Let the wire lie along the z axis. The ret arded vector potent ial at point P is
For t < s / c, the '' news'' has not yet reached P , and the potential is zero. For t > s / c, only the segment z < J (ct )2
-
s2
cont ributes. Outside t his range tr is negative, so I (tr) = O; t hus
The electric field is
E (s t ) = ' [email protected]
- fJ A = -
fJt 44
µol oc z 21T" (et )2 - s2
J
P hysicsguide
11 oo : 02 : 03 E (s t ) =
'
-
8A
at
= -
µo l oc z 21r ✓ (et)2 - s 2
44
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and t he magnetic field
A
Ast
➔
oo we recover the static case: E = 0, B = (µ 0 I 0 / 21r s) 0. Find t he electric and magnet ic fields generated. (b) Do t he same problem for delta funct ion current
I (t) = qo8(t) Solution: (a) We proceed exactly as the previous example. We know for t < s/c, A = O; for t > s/c. Now when calculating A we replace the time dependent current, we must replace t he t ime in the expression of current by t he //,, retarded t ime t - - Then you replace //,, = ✓s 2 + z 2 C
A (s, t) =
✓(ct)2 -s2 k(t - ✓s2
2 o
+ z2 / c) ---;::::=:====--dZ ✓s 2 + z 2
11 oo : 02 : 04 we mus rep ace e 1me 1n v~~e expression o curren /1,, ret arded time t - - Then you replace /1,, = ✓ s 2 + z 2 C
A (s, t ) =
✓(ct)2 -s2 k(t - ✓s2
2
+ z2 / c) ---;::::::::======--dZ ✓s 2 + z 2
o
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✓(ct) 2 -s 2
dz
✓(ct) 2 - s 2
1
dz
---;::::========== - ✓s2 + z2 C 0
0
t ln
et+ J(ct) 2 - s 2 s
1
- -J(ct) 2 - s2 C
The electric field E (s t)
'
t
= _
8A
at
s et+ J(et) 2
=
kz ln et+ ✓(et) 2~ s
_µ 0
1 -
s2
s
1
2
-
s
2c2 t
2
e+ -----;:::::==== 2 2 2 J(ct)
s
-
+ 2e2 t - - ----;:::::==== 2e J(et) 2 - s 2 1
et
et
+ -----;::=== -----;::=== 2 2 2 2 ✓(et)
s
-
s
✓(et)
( or zero, fort< s/e)
-
s
11 oo : 02 : oa
46
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Physicsguide
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The Magnetic field B (s, t) = -
8AZA 08
..a /3 on the right-hand side. But this means t hat E1µvpa must be proportional to Eµvpa We only need to determine the constant of proport ionality. To do this, we can look at
11 oo : 01 But t his means that
1
E µvp 0. To t he right of S, however, the curve is concave downwards 2 2 so t hat df / dx is decreasing with c and hence d f / dx < 0 In summary, at a stationary point df / dx
==
0 and
(i) for a minimum, d2 f /dx 2 > 0 (ii) for a maximum, d2 f / dx 2 < 0 (iii) for a stationary point of inflection, d2 f / dx 2 d2 f / dx 2 changes sign t hrough t he point.
==
0 and
Example : Find the positions and natures of t he stationary points of t he function
f (x)
==
2x
3
-
3x
2
-
36x
+2
Solution: The first criterion for a stationary point is t hat df / dx == 0, and hence we set
df == 6x 2 dx from which we obtain
-
6x - 36 == 0
(x - 3)(x + 2) j [email protected]
17
==
0
physicsguide CSIR NET, GATE
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Hence t he stationary points are at x
Basic Calculus
3 and x
- 2.
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Basic Calculus
Hence the stationary points are at x = 3 and x = - 2. To determine the nature of t he stationary point we must 2 2 evaluate d f / dx :
d2f dx 2
= 12x - 6
Now, we examine each stationary point in turn. For 2 2 x = 3, d f / dx = 30. since t his is posit ive, we conclude t hat x = 3 is a minimum. Similarly, for x = - 2, d2 f /dx 2 = -30 and so x = - 2 is a maximum.
1.5
Curvature of a function
Look at the figure 1.3 below. Let 0 be the angle made with the x -axis by the tangent at a point P on the curve f = f (x) , with tan 0 = df /dx evaluated at P. Now consider also t he tangent at a neighbouring point Q on the curve, and suppose that it makes an angle 0 + ~0 with the x -axis, as illustrated in t he figure.
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f(x) C
p
0
,, , -- , , ,, ,, ,- , ,, , ,, , ,, , 0 + 11.0 -- 0 ,, , -,
,
X
Figure 1.3: T wo neighbouring t angents t o t he curve f (x) whose slopes differ by !:l.0. The angular separation of the corresponding radii of t he circle of curvature is also !:l.0
It follows t hat t he corresponding normals at P and Q, which are perpendicular to the respective tangents, also intersect at an angle !:l.0. Furt hermore, t heir point of intersect ion, C in t he figure, will be t he posit ion of the cent re of a circle that approximates the arc PQ , at least to t he extent of having t he same tangents at t he extremit ies of the arc. This circle is called t he circle of curvature. For a finite arc PQ , t he lengths of GP and CQ will not, j [email protected]
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in general, be equal, as t hey would be if f = f (x) were in fact the equation of a circle. But, as Q is allowed to t end to P, i.e. as ~0 ➔ 0, they do become equal, their common value being p, the radius of the circle, known as the radius of curvature. It follows immediately t hat the curve and the circle of curvature have a common tangent at P and lie on the same side of it. The reciprocal of the radius of curvature, p- 1 defines the curvature of the function f (x) at the point p The radius of curvature can be defined more mathematically as follows. The length ~ s of arc PQ is approximately equal to p~0 and, in the limit ~0 ➔ 0, t his relationship defines pas ds d0
(1.12)
It should be noted that, ass increases, 0 may increase or decrease according to whether t he curve is locally concave upwards (i.e. shaped as if it were near a minimum in f (x) ) or concave downwards. This is reflected in the sign of p, which therefore also indicates the position of t he curve (and of the circle of curvature) [email protected]
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relative to the common tangent , above or below. Thus a negative value of p indicat es t hat t he curve is locally concave downwards and that t he t angent lies above the curve. We next obtain an expression for p, not in terms of s and 0 but in t erms of x and f (x) . The expression , t hough somewhat cumbersome, follows from the defining equation (1.12) , the defining property of 0 t hat tan 0 = df / dx f' and t he fact that the rate of change of arc length with x is given by
ds dx
df dx
1+
2 1/ 2
(1.13)
From the chain rule it follows that
ds d0
P=
=
ds dx dx d0
Differentiating both sides of tan 0 = df / dx with resp ect to x gives 2 d0 d f 2 sec 0- = - 2 _ J'' dx dx •
2
2
from which , using sec 0 = 1 +tan 0 = 1 + (! ')
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21
2
,
we can
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@Sk Jahiruddin, 2020
Basic Calculus
obtain dx/d0 as dx d0
1 + tan 0
1 + (!')2
!''
!''
2
Subst it uting t hese result s into the expression of p, we get t he final expression for p 1 + (!')2 p=
3/2
(1.14)
! ''
Example: Show t hat t he radius of curvature at t he
p oint (x, y ) on t he ellipse
x2
y2
a2
+ b2 =
4 2
4
1
2 312 4 4 / (a b ) x )
has magnit ude ( a y + b and the opposit e sign to y . Check t he sp ecial case b = a , for which the ellipse becomes a circle.
Solution: Differentiating t he equat ion of t he ellipse with respect to x gives
2x a2 [email protected]
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+
2y dy _ O b2 dx -
22
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Basic Calculus
and so
dy dx
Taking derivative again b2 a2
b4 a2y3
y - xy' y2
y2
x2
b2
+ a2
where we have used the fact that (x, y) lies on t he ellipse. We note that d2 y / dx 2 , and hence p, has the opposite sign to y 3 and hence to y . Substituting in the final expression of radius of curvature (1.14), we get
PI =
[1 + b4x2 / (a4y2)] 3/ 2 - b4/ (a2y3)
( a4y2
+ b4x2 ) 3/ 2 a4b4 2 312 x )
For the special case b = a, IP reduces to a- (y + and, since x 2 + y 2 = a 2 , this in turn gives PI = a, as expected. 2
2
2
Calculus of Two variables
This section will be updated soon. For t he time being please have some examples to learn t he basics. [email protected]
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Basic Calculus
@Sk J ahiruddin 1 2020
Example: Find the maximum and minimum points of t he function x3
-
y3
-
2xy
+2
Solution: We know for maximum or minimum points of functions of two variables. If f x
=
i! =
0 and
fy =
~[
= 0 Then at point (a, b)
(a, b) is a minimum point if at (a, b), f xx > 0, f yy > 0, and f xx f yy > J;y (a, b) is a maximum point if at (a, b), fxx < 0, f yy < 0, and f xx f yy > J;y (a, b) is neither a maximum nor a minimum point if f xx fy y < J;y Now apply t his to find the answer which is (0, 0) is a saddle point, (-2/ 3, 2/3) is a maxima
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Basic Calculus
Example: Find t he volume of the largest rectangular
parallelepiped wit h edges parallel to t he axes, inscribed x2 y2 z2 in the ellipsoid 2 + 2 + 2 = 1 with the help of a b c Lagrange's multiplier.
Solution: We take a point (x, y , z) in the corner in t he first octant where the box touches t he ellipsoid. Then (x , y, z) sat isfies t he ellipsoid equation and the volume of t he box in the first octant is xyz. There are 8 oct ants, so t he volume will be 8xyz. Our problem now becomes is to maximize f (x, y, z) = 8xyz, where the relation between (x, y, z) is the ellipsoid equation
cp ( X' y' z)
=
x2
y2
+ a
b2
2
z2
+
2 =
C
1
We know in the method of Lagrange's unknown multipliers if f is t he function to be maximized and t he function cp describes a relation between t he variables we write,
F (x, y, z ) =
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f + >..cp = 8xyz + ).. 25
x2
y2
z2
a2 + b2 + c2
physicsg11ide CSIR NET, GATE
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Then we set the t hree part ial derivatives of F equal to
Then we set the three partial derivatives of F equal to
0: oF 2x = 8yz + .\ · - 2 = 0 ox a oF 2y oy = 8xz + .\ · b2 = 0 oF oz
2z
= 8xy + .\ · - 2 = 0 c
Mult iply t he first equation by x, the second by y , and t he third by z, and add to get
3 · 8xyz
+ 2.\
y2
z2
+ b2 +
c2
x2
a2
=0
Using the equation of t he ellipsoid, we can simplify t his to
24xyz + 2,\ = 0
~~>
,\
= - l2xyz
Substituting ,\ into t he oF/ o x equation, we find that
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8y z - 12xy z · x2
2x
=0
a2 1 = -a2 3
The other two equations could be solved in t he same way. We can also easily guess the solution from symmetry 2 2 2 2 t hat the solutions will be y = b and z = c . Hence the maximum Volume is
1
8xyz =
1
8abc
3vl3
Example: Make t he variable t ransformation r
=x+
vt, s = x - vt in t he wave equation 2
2
0y
10 y
ox 2
v 2 ot2
write the transformed form and solve the equation. Solution: The first step is t he most important step
oF oFor oFos oF oF o a -ox = + = + = or + OS or ox OS ox or OS j [email protected]
@Sk Jahiruddin, 2020
and
27
F
physicsguide CSIR NET I GATE
Basic Calculus
@Sk Jahiruddin, 2020
Basic Calculus
and
8F 8F8r 8F8s 8F 8F - = --+-- = v--v- = v at ar at as at ar as
8 ar
a
F
as
Now we can easily derive
a aF a a ax ax ar + as 32p 32p 32p 2 8r 2 + 8r8s + 8s 2
8F ar
8F + -a-s
And
aF
8F a {) = v at 2 at at ar as 2 2 2 8 F 8 F 8 F = v2 - - - 2 - - + - 8r2 aras 8s 2 2
a
8F V ar -
8F VOS
Just put t hese relation in the original differential equation to get 1 8 2F 8 2F -2- -2 = 4- - = 0 v 8t
8r8s
28
j [email protected]
physicsguide CSIR NET, GATE
@Sk Jahiruddin, 2020
l\ If" _ _ 1.L ! _
Basic Calculus
1 -
T __ .L -
-
-- -
1
@Sk J ahiruddin, 2020
3
Basic Calculus
Multiple Integral
We discuss t he multiple integral calculations here Example : Evaluate t he double integral ,.. ,..
(2x - 3y )dxdy ., ., A
A is t he t riangle wit h vertices (0, 0), (2, 1), (2, 0) Solut ion: (2, 1)
,0)
Here, the area is included by the curves y =
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29
1, x =
2.
physicsguide CSIR NET, GAT E
Basic Calculus
@Sk Jahiruddin, 2020
Basic Calculus
So, (2x - 3y )dxdy = 2
2
dy-3
xdx 0
A
2
X
2
0 2
2
X
X
2
ydy
dx 0 x2
0
xdx- - 3 dx 2 8 0 2 3 2 x dx - 0 8 0 3 2 3 2 x 3x 8 5 - --- - 1--3 o 83 0 3 3
=2
Example: " "
2
12y cos xdxdy
V VA
Where A is t he area included by the curves y the x axis and the line x = 1r / 2.
j [email protected]
@Sk J ahiruddin, 2020
n
1
, •
30
= sin x,
physicsguide CSIR NET, GATE
Basic Calculus
.
7r
s1n x
2
2
12y cos xdxdy = 12
y2dy
cosxdx
A
0
0 . 3 sin x
zr. 2
= 12
cosxdx~ 0
0
! .
3
s1n xcosxdx
=4
=
0 7r 4
.
Sln -
2
sin4 x zr. 4x ~ 4
=l
Example: Find the volume above the triangle with vertices (0, 0), (2, 0) , and (2, 1), and below the paraboloid z = 24 - x 2 - y 2
Solution: Volume of the region is given by,
V=
dxdydz X
2
2
dy
dx 0
0
0
[·. · talking idea from t he first problem] X
2
2
(24 - x 0
2
-
2 y )dxdy
0
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31
physicsguide CSIR NET, GATE
@ Sk J a hiruddin, 2020
-X
2 0
dy0
2
= 12 n
2
2
dx
=24
Basic Calculus
-X
2
2 x dx
dy -
0
2
1 x dx - ').
n
3
0 2
3
x dx n
2
dx 0
1
x dx - ?. 4
-X
2 0
y2dy
Basic Calculus
@ Sk J a hiruddin , 2020
2
2
= 12
xdx -
0
x2
2
2
°
X -
2
dy -
0
2
X
2
dy-
0
=6
X
dx
= 24
= 12
X
-
0
1 2 1
-
2
2
3
0 X -
2
-
°
4 4 - 2 - 1 = 131 6 6
0
0
x dx -
x4
dx
-
1
24
1 24 X
0 4
x
4
2
°
Example: A dielectric lamina with charge density p = ky covers t he area between t he parabola y = 16 - x 2
and the x axis. Find t he total charge.
Solution:
Area
-4
+4
We need to det ermine the area ment ioned in t he figure, j a [email protected]
32
physicsguide CSIR NET , GATE
Basic Calculus
@ Sk J a-hiruddin , 2020
.
r
.
r
A
t
-1
l'""'i
')
So t he total charge will be given by, 16-x2
4
kydxdy = k
pdxdy =
Q=
dx -4
2 16-x2
4
=k -4
k 2 k 2
y . dx 2
ydy 0
k
0
4
[256 + x
4
-
2
32x ) dx
-4
256x
=
+
x5
5
x3
- 32
4
3
-4
k 2 X 45 256 X 8 + - 32 2 5 2048 _ 4096 k 2048 2 + 5 3 k 16384 8192k -X-2 15 15
2 X
X
43
3
Example: Evaluate the triple integral
ydxdzdy y= - 2 ..- z= l
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x =y+z
33
physicsguide CSIR NET, GATE
@Sk Jahiruddin, 2020
Basic Calculus
Solution: 2
3
2y+z
ydxdzdy y= - 2
r3
z=l
x=y+z r2
@Sk J ahiruddin, 2020
Basic Calculus
Solut ion: 3
2
2y+z
ydxdzdy y=-2 z=l 3
x=y+z 2
ydy
dz [2y z=l
y=-2 3
2
3
dz y
y=-2 3 3
3
- 2
+ z - (y + z)]
=
z =l
33 + 23
35
3
3
Example : Find the volume between t he planes z = 2x + 3y + 6 and z = 2x + 7y + 8 and over the square in the (x, y) plane wit h vertices (0, 0), (1, 0), (0, 1) , (1 ,
1). Solution:
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34
@ Sk Jahiruddin , 2020
Basic Calculus
(0, 1)
(1, 1)
(0,0)
(1,0)
The volume will be given by, 1
dxdydz ==
V ==
dx 0
1
dy
dz 2x+3y+6
0
1
dy[(2x + 7y + 8) - (2x + 3y + 6)]
dx 0
0 1
1
dy[4y + 2]
dx 0
0 1
1
dx
==4 0
==2
2x +7y+8
1
1
ydy
+2
0
1
dy
dx 0
0
+ 2 == 4
j ahir@physicsguide. in
35
physicsguide CSIR NET, GATE
@Sk J ahiruddin 1 2020
Basic Calculus
Example: Find the volume in the first octant bounded by the coordinate planes and the plane x + 2y
+z = 4
Solution: 1z X
+ 2y + Z
=4
y=2 Y-->
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36
physicsguide CSIR NET, GATE
Basic Calculus
36
j ahir@physi csgui de.in
physicsguide CSIR NET, GATE
@ Sk Jahiruddin, 2020
Basic Calculus
The volume will be given by, 2
V=
dxdydz
=
dy
dx 0
0 2
4-x-2y
4- 2y
0
4- 2y
dy 0
dx{4-x-2y } 0
2
= 4
4-2y
dy 0
dx-
dy
0
0
2
=4
2
0
xdx- 2
M- s -
0 2
0
2y
0
{4 - 2y }ydy
{4 - 2y} dy- 2
2
dx
2
0
[4{4 - 2y} - -{ 16 + 4y 2 [16 -
ydy
2
1
2
4-2y
2
0
1
{4 - 2y }dy - 2
0
4- 2y
2
2
-
2
16y} - 2{4y - 2y }]dy
+ M- sy + 4y
2
J
0 2
2
[8 - 8y + 2y ]dy = 8y -4y + ; 2
2
. 0
=
}6 -¼ + 16 3
j [email protected]
3
2 0
=
16 3
37
physicsguide CSIR NET, GATE
Linear Algebra Sk J ahiruddin Assistant Professor Sister Nibedita Govt. College, Kolkata Aut hor was the topper of IIT Bombay M.Sc Physics 2009-2011 batch He ranked 007 in IIT JAM 2009 and 008 (JRF) in CSIR NET Jt1ne 2011 He has been teaching CSIR NET aspirants since 2012
1
@Sk J ahiruddin , 2020
Linear Algebra
Linear Algebra
@Sk J ahiruddin , 2020
Contents 1
Linear Vector Space
7
1.1
Definit ion . . . .
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7
1.2
Linear dep endence of functions: Wronskian .
10
1.3
Basis . . . . .
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1.4
Inner Product
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1.4.1
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Hilbert Space . .
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1.5
Some useful inequalit ies
1.6
Gram-Schmidt orthogonalisation
. .
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25
1.7
Linear function and Linear Operators
•
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27
1. 7.1
Linear function of vect or
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1.7.2
Linear Operator . . . . .
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28
1. 7.3
Properties of linear operators
•
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30
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Linear Algebra
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4
3. 2 Lower and upper triangular matrices .
•
•
•
64
3.3
Symmetric and ant isymmetric matrices
•
•
•
65
3.4
Orthogonal matrices
•
•
•
66
3.5
Hermitian and anti-Hermitian matrices
•
•
•
68
3.6
Unitary matrices
•
•
•
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•
•
•
•
•
•
•
•
•
•
•
69
3.7
Normal matrices
•
•
•
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•
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•
•
71
. . . . . . . . . .
Eigenvalues and Eigenvectors
4. 1 eigenvalues .
74
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74
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75
4.2
eigenvectors
4.3
Eigenvalue and eigenvectors of normal matrices 77
4.4
Eigenvectors and eigenvalues of Hermitian and anti Hermit ian matrices •
82
Eigenvectors and eigenvalues of a unitary matrix . . . . . . . . . . . . . . . . . . . . . . .
85
•
4.5
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4
•
•
•
•
•
•
•
•
•
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Linear Algebra
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4.6
4. 7
Linear Algebra
Simultaneous eigenvectors and commutation of matrices . . . . . . . . . . . . . . . . . . .
86
Cayley Hamilton and General properties of eigenvalues . . . . . . . . . . . . . . . 4.7.1
Some properties of eigenvalues
4.7.2
Some examples to evaluate eigenvalues 91
•
•
•
•
89
•
•
•
•
89
5 Similarity transformations and diagonalization 99 5.1
Change of basis . . . . . .
•
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•
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•
•
•
•
•
99
5.2
Similarity transformation .
•
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•
•
•
•
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•
•
101
•
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•
•
•
103
5.3
5.2 .1
Unitary transformation .
5.2.2
Physical Significance of Similarity Trans£ormation . . . . . . . . . . . . . . . 105
Diagonalisation of matrices .
6 Function of matrices
j [email protected]
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•
•
•
•
•
•
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•
•
106
111
5
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Linear Algebra
Linear Algebra
@Sk J ahiruddin , 2020
6. 1
6.2
7
8
9
Function of matrices by similarity transformation . . . . . . . . 6.1.1
Trace formula
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112
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•
117
Function of matrix by using power series expansion and Cayley Hamilton t heorem . . . 118
130
Quadratic form 7.1
Basis invariance of Quadratic form
•
•
•
•
•
131
7.2
More examples on Quadratic form .
•
•
•
•
•
136
141
Exercises 8. 1
Ans Keys
8. 2
Solutions .
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♦
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Ans Key of More Exercises . .
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6
150 180
More Exercises with Ans Keys 9. 1
148
•
•
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•
182
physicsguide CSIR NET, GATE
Linear Algebra
(ii) t he set is closed under mult iplication by a scalar (any complex number) to form a new vector Aa,. So if a is an element of t he set, and A is a scalar t hen Aa belongs to the set. The operation being bot h distribut ive and associative
A· (a + b)= A·a +A ·b
(1.3)
(A+µ) · a = A ·a + µ · a
(1.4)
A· (µ. a)=(A ·µ )· a
(1.5)
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7
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Linear Algebra
where A and µ are arbit rary scala1~s; You need to know specifically t hat the set not needed to be closed under multiplication. Means if a and b are two elements of the set then a· b need not necessarily belong to t he set. (iii) t here exists a null vector O such t hat a + 0 = a for all a; (iv) mult iplication by unity leaves any vector unchanged, i.e. 1 · a = a ; (v) all vectors have a corresponding negative vector - a such that a + (-a) = 0 . We note t hat if we restrict all scalars to be real t hen we obtain a real vector space (an example of which is our ,..
.,.
_,
,
.
,
\
.
,
,
Linear Algebra
@Sk J ahiruddin, 2020
where A and µ are arbitrary scalars; You need to know specifically that the set not needed to be closed under multiplication. Means if a and b are two elements of the set then a· b need not necessarily belong to t he set. (iii) t here exists a null vector O such that a + 0
= a for
all a; (iv) multiplication by unity leaves any vector unchanged , i.e. 1 · a = a ; (v) all vectors have a corresponding negative vector -a such that a + (-a) = 0 . We note t hat if we restrict all scalars to be real t hen we obtain a real vector space (an example of which is our familiar t hree-dimensional space); otherwise, in general, we obtain a complex vector space. We note that it is common t o use t he terms 'vector space' and 'space', instead of t he more form al 'linear vector space'. The span of a set of vectors a , b , ... , s is defined as the set of all vectors that may be written as a linear sum of the
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Linear Algebra
Linear Algebra
@Sk J ahiruddi11 , 2020
original set, i.e. all vectors x
= aa + ,Bb + · · · + o-s
(1.6)
t hat result from t he infinite number of possible values of t he (in general complex) scalars a , ,8, . . . , a-. If x in (x = aa + ,Bb+ · · ·+a-s) is equal to O for some choice of a, ,8, ... , o(not all zero), i.e. if
aa + ,Bb + · · · + o-s =
0
(1. 7)
t hen the set of vectors a , b , ... , s , is said to be linearly dependent. In such a set at least one vector is redundant, since it can b e expressed as a linear sum of t he others. If, however, (aa + ,Bb + · · · + a s = 0) is not satisfied by any set of coefficients (other than the t rivial case in which all the coefficients are zero) t hen t he vectors are linearly independent, and no vector in the set can be expressed as a linear sum of t he others. If, in a given vector space, t here exist set s of N linearly indep endent vectors, ( but no set of N + l linearly independent vectors exist ) then t he vector space is said to be N -dimensional. [email protected]
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Linear Algebra
Example: Find whether these following sets form vec-
@Sk J ahiruddin , 2020
Linear Algebra
Example: Find whether these following sets form vec-
tor space or not with t he given dimensions in the bracket? (a) Non singular N x N matrices (dimension = N 2 ) (b) Polynomial functions of x ( dimension = infinite) (c) Absolutely convergent series (dimension = infinite) Ans: (a) No, (b) Yes, (c) Yes Now non singular N x N matrices are not closed under addit ion. We can add A + (-A ) = 0 where O then a singular matrix. Thus the set is not closed and it is not a vector space. Check that for (b) Polynomial functions of x and (c) Absolutely convergent series all the properties of vector space remains valid.
1.2
Linear dependence of functions: Wronskian
Before we proceed we need to know about Linear dependence of functions and the Wronskian. The functions [email protected]
@Sk J ahiruddin, 2020
f 1 ( x), f 2 ( x), · · · , f n ( x) 10
are linearly depen-
physicsguide CSIR NET, GATE
Linear Algebra
Linear Algebra
@Sk J ahiruddin , 2020
dent if some linear combination of them is ident ically zero, t hat is, if there are constants k1 , k2 · · · , kn, not all zero, such t hat For example, sin 2 x and ( 1 - cos 2 x) are linearly dependent since sin2 x - (1 - cos 2 x) = 0 •
But sin x and cos x are linearly independent since there are no numbers k1 and k2 not both zero, such t hat
k1 sin x
+ k2 cos x
is zero for all x We shall be particularly interested in knowing t hat a given set of functions is linearly independent. For this purpose t he following t heorem is useful If f1(x), f2( x), · · · , fn(x) have derivatives of order n- 1, and if the determinant
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Linear Algebra
@Sk J ahiruddin , 2020
Linear Algebra
So, t he functions are independent of each other. (b) Here Wronkskian of the functions is
+ l )ex 0 ex (x + 2)ex l ex (x
= (x + 2)ex{xex - ex} - ex{x(x + l )ex - xex} = (x + 2)(x - l) e2x - x 2 e2x = (x - 2)e2x =/= 0 So, t he functions are independent of each other.
1.3
Basis
If V is an N -dimensional vector space then any set of N linearly independent vectors e 1 , e 2 , . .. , eN forms a basis for V. If x is an arbitrary vector lying in V then the set of N + l vectors x , e 1 , e 2 , . . . , eN , must be linearly dependent and t herefore such t hat
where t he coefficients a, /3, .. . , x are not all equal t o 0, and in particular x =I= 0 . Rearranging previous equation we may [email protected]
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Linear Algebra
Linear Algebra
@Sk J ahiruddin , 2020
write x as a linear sum of the vectors ei as follows: N
x
= x1e1 + x2e 2 + · · · + xNe N = L xiei
(1.10)
i =l
for some set of coefficients x i that are simply related to t he original coefficients, e.g. x 1 = -a/x, x 2 = -/3 /x, etc. Since any x lying in the span of V can be expressed in terms of t he base vectors ei, the latter are said to form a complete set. The coefficients Xi are the components of x wit h respect to t he ei -basis. These components are unique, since if both N
x
N
= L xiei
and
x
= L yiei
i=l
t hen
i =l
N
L (xi - Yi) ei = 0
(1.11)
i =l
which, since the ei are linearly independent , has only t he solution Xi = Yi for all i = 1, 2, ... , N From t he above discussion we see t hat any set of N linearly independent vectors can form a basis for an N -dimensional space. If we choose a different set e~, i 1, ... , N then we can write x as j [email protected]
@Sk J ahiruddin , 2020
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physicsguide CSIR NET, GATE
Linear Algebra
@SIle of an inner product. In effect t he notion of an inneI 111 u I (a b) is a generalisation of the dot product to more a h •t 1·ac · vector spaces. Alternative notations for (alb) are . . ~ simply a •b .
(' I
l. I
'
Linear Algebra
@Sk J ahiruddin, 2020
2 x 3 m atrix as a linear combination of these six m atrices) . Since there are 6 basis vectors, t he dimension of t his space is 6.
1.4
Inner Product
The inner product of two vectors, denoted in general by
(a b), which is a scalar function of a and b. The scalar or dot product, a · b _ a b cos 0, of vectors in real t hreedimensional space (where 0 is the angle between t he vect ors), was introduced in t he last chapter and is an example of an inner product. In effect the notion of an inner product (a b) is a generalisation of the dot product to more abstract vector spaces. Alternative notations for (alb) are (a, b), or simply a ·b. The inner product has the following propert ies:
(i) (a b) = (b a)* (ii) (a Ah + µc) = A(a b) + µ(a c) We note that in general, for a complex vector space, (i)
j [email protected]
@Sl 0 for all A, so we may choose A = re- ia and require that , for all r
This means that t he quadratic equation in r formed by setting the RHS equal to zero must have no real roots. This, in t urn, implies that
which, on taking t he square root (all factors are necessarily posit ive) of both sides, gives Schwarz's inequality. Triangle inequality:
la+ bi
Linear Algebra
e1
e1 = A
e1 A
X2 -
(e1)t · X2 e1 A
X3 -
(e2) t · X3 e2 -
;►
A
e2
e2
=
e2
(e1) t · X3 e1 A
;>
A
e3
=
e3
le3
• •
•
Example: For the given sets of vectors, use the GramSchmidt method to find an orthonormal set. A = (0, 2, 0, 0), B = (3, -4, 0, 0), C = (1, 2, 3, 4)
Solution: e1
j [email protected]
= A / A = (0, 1, 0, 0)
26
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@Sk J ahiruddin , 2020
Linear Algebra
e3= e3/ 1. 7
e 3 = (0, 0, 3, 4)/5
Linear function and Linear Operators
1. 7 .1
Linear function of ve ctor
A function of a vector, say
f (r ),
is called linear if
where a is a scalar.
Example: If F (r) = A x r for any given vector A, t hen is F (r) a linear vector function? Solution:
j [email protected]
~ n
1
T
1
•
1 1 •
nn n A
We know t hat F (r) is linear function if
27
physicsguide CSIR NET, GAT E
T
•
@Sk J ahiruddin , 2020
Linear Algebra
Hence
F(r1 + r2) = A x (r1 + r2) = A x r1 + A x r2 = F(r1) + F (r2) Again
F(ar) = A x (ar) = a(A x r) = aF(r) As both t he condit ions are satisfied then F (r) is a linear vector function.
1. 7.2
Linear Operator
A linear operator operat ing on linear combination of two vectors gives
A (Aa + µb) = AAa + µAb
(1.18)
where A, µ are scalars. Another equivalent d efinition is: 0 is a linear operator if
O(A + B )
=
O(A) + O(B )
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28
and
O(kA)
=
kO(A)
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We say that A 'operates' on x to give the vector y. We note that the action of A is independent of any basis or coordinate system and may be thought of as 'transforming' one geometrical entity (i.e. a vector) into another. Example: Is square root a linear operator? We are asking, is ✓A + B the same as ✓A + JB? The answer is no; taking the square root is not a linear operation. Example: Is taking the complex conjugate a linear operation? We want to know whether A + B = A + B and kA = kA. The first equation is t rue; t he second equation is t rue if we restrict k to real numbers.
2xD + 7 a linear operator? Where 2 d d 2 D and D stands for - and respectively. 2 2
Example: Is x D
2
-
dx
dx
Solution: We see that O is a linear operator if
O (A
+ B) =
O (A)
0 = x 2 D 2 - 2x D A(x) and B = B (x) j [email protected]
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+ O (B)
and
O( kA)
= kO(A)
+ 7 so we take two function of x, A 29
=
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O(A
+B)=
2
(x D
2
-
2x D
+ 7)(A + B )
dA - 2 dA - x dx2 x dx -
2
+
7A
2 d2 B
dB + x dx2 - 2x dx
+7
B
= O(A) + O(B ) Similarly 2
O(kA ) = (x D
2
-
2xD + 7) (kA) = k(x D 2
2
-
2x D
+ 7)A
= kO(A ) As both t he condit ions are satisfied t hen O is a linear operator.
1. 7 .3
Properties of linear operators
If x is a vector and A and B are two linear operators t hen it follows that (A + B)x = Ax + Bx
(AA)x = A(Ax) (AB)x = A(Bx) where in t he last equality we see t hat t he action of two linear operators in succession is associative. The product of two j [email protected]
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linear operators is not in general commutative, however, so t hat in general ABx i= BAx. In an obvious way we define t he null (or zero) and identity operators by
Ox = 0
and
Ix
= x
for any vector x in our vector space. T wo operators A and B are equal if Ax = Bx for all vectors x. Finally, if t here exists an operator A- 1 such t hat
t hen A- 1 is the inverse of A . Some linear operators do not possess an inverse and are called singular, whilst those operators that do have an inverse are t ermed non-singular.
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2
Linear Algebra
Matrix: D efinition and Basic Properties
A M x N matrix is written as
A=
A11
A12
• • •
A 1N
A 21
A 22
• • •
A 2N
• • •
• • •
AMI
AM2
•
•
•
• • •
• • •
A 1111 N
Significance of matrix is deep. A matrix may be a vector, tensor , operator etc. We slowly dig into details. Whe never a m atrix is w ritten , it is w ritten with respect to some basis
Vectors are represented by column matrix
x=
• • •
We note t hat in a different basis e~ the vector x would be represented by a different column matrix containing t he j ahir@physi csgui de.in
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components x~ in the new basis, i.e.
x' =
2.1
• • •
Basic Matrix algebra (A + B)ij = Aij + Bij (AA)ij = AAij
(AB)ij =
L AikB kj k
Example: The matrices A, B and C are given by
A=
2 - 1 3 1
'
Find the matrix D
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B=
1 0 0 -2
- 2 1
'
-1 1
= A + 2B - C
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Solution:
D=
2 -1 3 1
+2
1 0 0 -2
-2 1 -1 1
2 +2x l -(-2) -1+ 2 x 0 - 1 3+2x0-(-1) 1 + 2 x(- 2)-1
2.2
6 -2 4 -4
Multiplication:
The product AB of an M x N m atrix A with an N x R matrix B is itself an M x R matrix P, where N
Pij =
L AikB kj
for i = 1, 2, ... , M ,
j
= l , 2,
... , R
k =l
For example, P
= AB may be written in matrix form
I
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A11
A12
A13
A21
A 22
A23
34
I
B 11
B 12
B 21
B 22
B 31
B 32
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W here
= P 21 = P12 = P 22 = P 11
+ A12 B 21 + A13 B 31 A21B11 + A22B21 + A23B31 A1 1B12 + A 12 B 22 + A 13 B 32 A11 B 11
A21B12
+ A22B22 + A23B32
Mult iplication of more t han two matrices follows naturally and is associative. A(BC) _ (AB)C Mult iplication of matrices is not , in general, commutat ive. AB i= BA in general. T here are sp ecial cases when AB = BA
Example Evaluate P = AB and Q = BA where
A=
3 2 - 1 0 3 2 1 - 3 4
2 -2 3
'
B=
1
3
0 2 1 1
Solution: T he element ~ j of the matrix P = AB is found by taking t he 'scalar product of t he i th row of A with t he j th column of B. For example, P11 = 3 x 2 + 2 x 1 + (- 1) X 3 = 5, P 12 = 3 X (-2) + 2 X 1 + (- 1) X 2 = -6, etc. [email protected]
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So we get
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So we get
3
P = AB =
2
0 3 1 - 3
- 1
2 - 2 3 1 1 0 3 2 1
2 4
5 - 6 8 9 7 2 11 3 7
Similarly we get
Q = BA =
2 - 2 3 l 1 0 3 2 1
3
2
- 1
0
3
2
9 - 11 6 3 5 1
4
10
1 -3
9
5
These results illustrate t hat, in general, two matrices do not commute.
The property t hat matrix multiplication is distribut ive over addit ion, i.e. t hat
(A+ B )C = AC + B C;
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C(A + B) = CA + CB
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t he transpose of its complex conjugate, or equivalently, the complex conjugate of its t ranspose, i.e. Example: Find the complex conjugate of t he matrix 1 2 3i
A=
1+i 1 0
Solution: By taking the complex conjugate of each element we obtain immediately A* =
1 2 - 3i 1- i 1 0
We can prove that
Example: Find the Hermitian conjugate of the matrix 1 2 3i
A=
1+i 1 0
Solution: Taking t he complex conjugate of A and t hen forming the transpose we find
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1
1- i
2 -3i
1
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t he transpose of its complex conjugate, or equivalently, the complex conjugate of its transpose, i.e. Example: Find the complex conjugate of t he matrix 1 2 3i 1+i 1 0
Solution: By taking the complex conjugate of each element we obtain immediately A* =
1 2 -3i 1- i 1 0
We can prove that
Example: Find the Hermit ian conjugate of the matrix 1 2 3i A= 1+i 1 0
Solution: Taking t he complex conjugate of A and t hen forming the t ranspose we find
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1
1- i
2
1
-3i
0
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Check th at you obtain the same result, of course, if you first t ake the transpose of A and then t ake the complex conjugate.
2 .5
Trace
Trance of a matrix is defined by N
Tr A =
A11
+ A 22 + · · · + A NN
=
L Aii i=l
Trace is linear Tr(A ± B)
= Tr A ± TrB
Trace of the product of two matrices is indep endent of t he order of their multiplication
N N N N N N TrAB = L(AB)ii = L L Aij B ji = L L Bjt Aij = L(BA) i =l
i= l j = l
i =l j =l
j=l
Trance is same in cyclic permutation. Tr ABC = Tr BCA = Tr CAB j [email protected]
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2. 6
Linear Algebra
Determinant
You know how to evaluate t he determinant. For example
det A = A
A
A11
A 12
A13
A21
A 22
A 23
A31
A 32
A 33
= - A21 (A12 A 33 - A13A32)
+ A22 (A11A 33 -
A 13A31)
- A23 (A11 A 32 - A12A31)
= A1 1 (A22A33 - A23A32)
+ A13 (A21A32 -
+ A12 (A23A31 -
A21A33)
A22A31)
Properties of determinants
(i) Determinant of t he transpose: The transpose matrix AT has t he same determinant as A itself, i.e.
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A
(ii) Determinant of the complex and Hermitian conjugate:
(iii) Interchanging two rows or two columns: If two rows (columns) of A are interchanged, its determinant changes sign but is unaltered in magnitude. (iv) Common factor: If all t he elements of a single row (column) of A have a common factor, A , then this factor may be removed; the value of the determinant is given by t he product of the remaining determinant and A Clearly this implies that if all the elements of any row (column) are zero t hen A = 0.
It also follows t hat if every element of t he N x N matrix A is multiplied by a constant factor A t hen
AAI = AN A
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(v) Identical rows or columns: If any two rows (columns) of A are identical or are multiples of one another, t hen it can be shown that A = 0 . (vi) Adding a constant multiple of one 1~ow (column) to anot her: The determinant of a matrix is unchanged in value by adding to t he elements of one row (column) any fixed multiple of t he elements of another row (column). (vii) Determinant of a product: If A and B are square matrices of the same order t hen
AB l = IA B
BA
and in the extended version AB . .. G = A B I . . . G = IA IG . . . B I = IA .. . GB
It shows that the determinant is invariant under permutation of the matrices in a multiple product.
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Example: Evaluate the determinant
2 -2
0
1
3
1 0 1 3 -3 4 -2 -2 1 - 2 - 1
A
Solution: Taking a factor 2 out of t he third column and t hen adding t he second column to t he third gives
1
A = 2
0
0 1
1 -1
3
1
0
1
3
1
0
1
0
1
= 2
3 -3 2 -2 -2 1 - 1 - 1
3 -3 - 1 -2 -2 1 0 - 1
Subtracting the second column from the fourt h gives
A =2
1
0
1
3
0
1
0
0 1
3
-3 - 1
-2 1
0
-2
We see t hat the second row has only one non-zero elej [email protected]
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You can find t he inverse of a matrix by (i) Finding cofac44
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tors (ii) Row and column operations. We physics students generally find the inverse by constructing first t he matrix C containing the cofactorsof t he elements of A. Then take the t ranspose of C and divide er the determinant of A.
For a 2 x 2 matrix, the inverse has a particularly simple form. If the matrix is
A= t hen it s determinant - A- is given by A = A11 A22 A12 A21 , and t he matrix of cofactors is
Thus t he inverse is
-
Linear Algebra
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tors (ii) Row and column operations. We physics students generally find t he inverse by constructing first the matrix C containing t he cofactorsof t he elements of A. Then take the t ranspose of C and divide er the determinant of A.
A-1) (
= ( C)i = cki
Al
ik
Al
For a 2 x 2 matrix, the inverse has a part icularly simple form. If the matrix is
A= t hen its determinant -A- is given by A = A11 A22 A 12 A21 , and the matrix of cofactors is
Thus the inverse is
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A-1
=
1
cT
A
A11A22 -
A12A21
A 22
-A12
-A21
A11
Example: Find the inverse of the matrix 2
A=
4
3
l - 2 - 2 -3 3 2
Solution: We first determine A : A
= 2[-2(2) - (-2)3] + 4[(-2)(-3) - (1)(2)] + 3[(1)(3) - (- 2)(-3)] = 11
This is non-zero and so an inverse matrix can be constructed . To do this we need t he matrix of t he cofactors, C, and hence cT. We find
2 C
=
4 -3 1 13 - 18
- 2 7
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and
-8
46
cT
=
2 4
1
- 2 7
13 -3 - 18 -8
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Let t he columns of an M x N matrix b e interpret ed as t he components in a given basis of N (M-component ) vectors v 1, v 2, ... , v N, as follows:
t
t
t
A=
Then the rank of A , denoted by rank A or by R( A), is defined as t he number of linearly independent vectors in t he set v 1, v 2, ... , v N, and equals the dimension of t he vector space spanned by t hose vectors. Alternatively, we may consider the rows of A to contain the components in a given basis of the M (N -component vectors w 1 , w 2 , . . . , w M as follows:
A=
• • •
The rank of A is also equal to the number of linearly indep endent vectors in the set w 1 , w2 , ... , w M . Rank of a matrix in generally determined by row reduction. It can also be det ermined by finding how much j [email protected]
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Let t he columns of an M x N m atrix b e interpret ed as t he com ponents in a given basis of N (M-component) vectors v 1, v 2, . .. , VN, as follows:
t
t
t
Then t h e rank of A , d enoted by rank A or by R(A), is defined as t h e number of linearly indep en dent vectors in t he set v 1, v 2, ... , v N, and equals t he dimension of t he vect or sp ace spanned by t hose vect ors . Alt ernatively, we m ay consider t he rows of A t o contain t he compon ent s in a given basis of t h e M( N -component vectors w 1 , w 2 , ...
, WM
as fol-
lows:
• • •
The rank of A is also equal to the numb er of linearly in dep endent vectors in t he set w 1 , w 2 , ..
. , wM .
R ank of a m atrix in generally det ermined by row redu ct ion. It can also b e det ermined by finding h ow much [email protected]
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submatrices wit h non zero determinant can be found. YOll need to know t hat a su bmatrix of matrix A is any matrix t hat can be formed from t he elements of A by ignoring one, or more t han one, row or column. Now what is row reduction? Row reduction are done by using t hese st eps. 1) Interchange two rows 2) Mult iply (or divide) a row by a (nonzero) constant 3) Add a multiple of one row to another; t his includes subtracting, t hat is, using a negative multiple. Example : Make row reduction of the matrix
2 0 - 1 2 6 5 3 7 2 - 1 0 4 Solution: Step 1: Subtract 3 times t he first row from the second row and subtract t he first row from the third 2 0 - 1 2 row. T his gives: 0 5 6 1 0 - 1 1 2 [email protected]
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Solution: 1 0 1 0 -1 -2 -1 0
The determinant of the matrix is =
2
2
5
3
2
4
8
6
0 1 0 - 1 -2 -1 0 2 2 2 3 1
2
4
2
0
1
6
0 0
- 1 -2 0 0 2 2
1
2 4 0
2
2
3 0
2
4
6 0
1
R3 - R1 R'4 -_ R42 =
=
-2
51
0
0 0
- 1 - 2 0 0 1 1
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0 0
- 1 -2 0 0
⇒
R;
0 3 0 6
2 2
3 0 3 0
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R~ = R4 - R3
=
1 0 0 0 - 1 - 2 0 0
- 2
1
2
0
0
3 0 0 0
Now,
1 0 0 - 2 - 1 - 2 0 = - 2[- 2
1
2
X
3] = 12
3
So this 3 x 3 sub-matrix of t he main matrix have Non-Zero determinant. So, Rank of t he matrix is 3.
Example: Find t he rank of the matrix
1 2
2 - 1 O - 1
1 -2
4 1
0 -3
Solution: Row reduced echelon form will b e 1 2
2 0
1 -2
- 1 - 1
4 1
0
-3
1 0 - 1/ 2 1/ 2 0 1 -1/ 4 7/ 4 0 0 0 0
As the matrix is now have two non zero rows. It will have rank TWO .
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Example: Find the rank of
1 0 2 1
A=
0 2 4 2
0 2 2 1 Solution: After some elementary row operations we get
A=
1 0 2 1
1 0
2
1
0 2 4 2
0 2
4
2
0 2 2 1
0 0 -2 -1
We can't make any more reduction to make any of the more rows and columns to be zero. So the matrix have rank THREE.
Example: Find the rank of the matrix by checking the non singular submatrices.
1 1 0 -2 2 0 2
2
4 1 3
1
Solution: The largest possible square submatrices of A must be of dimension 3 x 3. Clearly, A possesses four such j [email protected]
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2.9
Linear Algebra
Consistency of solutions of linear equations
Suppose you are given t his set of equations
=2 6x + 5y + 3z = 7 2x -y = 4 2x
z
We write the linear equations in the form
Mr = k where
M=
2
0 -1
6
5
3
2 -1
0
X
r
=
y z
'
k=
2 7 4
Now we form t he augmented matrix which we call A . Note t hat t he fi1~st t hree columns of A are just t he columns of M , and t he fourth column is the column of constants on the right hand sides of the equations, i.e the k matrix. So
A= j ahir@physi csgui de.in
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Now let 's consider the general problem of solving m equations in n unknowns. Then M has m rows (corresponding to m equations) and n columns (corresponding to n unknowns) and A has one more column (t he constants). The following summary . outlines t he possible cases. 1) If (rank M) < (rank A), the equations are inconsist ent and t here is no solution. 2) If (rank M ) = (rank A) t here is one solut ion.
=
n (number of unknowns),
3) If (rank M) = ( rank A ) = R < n, t hen R unknowns can be found in terms of the remaining n - R unknowns. .
.
Example: Write and row reduce t he augmented ma-
trix for the given set of equations to find out whether the given set of equations has exactly one solut ion, no solutions, or an infinite set of solutions.
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(a)
X
=4 -y + 2z = 3
X
-y + 2z = 3
- x +y- z
Solution:
The characteristics inhomogeneous matrix is - 1
1
-1
4
1 1
- 1 -1
2 2
3 3
=
[R;
=
R3-R2]
- 1
1
- 1
4
1
-1
2
0
0
0
3 0
So, we get equations
-x + y - z = 4 X
0. x
. .. (i)
-y + 2z = 3
... (ii)
+ 0.y + 0. z = 0
··· (iii)
From equation (iii), we can say t he equation will have infinite number of solutions. (i)
+
(ii) ⇒
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z = 7 and x = y - 11 .
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(b) X
-y + 2z = 5
+ 3y - z = 4 X -y + 2z = 3
2x
Solution:
The characteristics inhomogeneous matrix is 1 - 1
2
2 3 - 1 1 - 1 2
5
4 3
1 - 1 =
[R;
=
R3 - R1] 2 0
3 0
2
5
- 1 0
4
-2
From the last line we are getting that, O.x
+ O.y + O.z =
- 2
This is an inconsistent equation. So there will be No Solution.
2.9.1
Homogeneous Equations
When the constants on the right hand sides are all zero ( k matrix is zero) then t he set of equation are called homogej [email protected]
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x+y = O neous equations. Examples are , and x - y= O X -
2y + 3z = 0
+ 4y 2x + 2y X
6z
=
0
3z = 0
Homogeneous equations are never inconsistent ; t hey always have the solut ion ''all unknowns = O'' (often called the ''trivial solution'' ). If the number of independent equations (that is, t he rank of t he matrix) is t he same as t he number of unknowns, this is the only solution. If the rank of t he matrix is less t han the number of unknowns, there are infinitely many solut ions. Important t heorem to know: A system of n homogeneous equations in n unknowns has solutions other than the trivial solution if and only if t he det erminant of the co efficients is zero. Example: For what values of A does t he following set of equations have nontrivial solut ions for x and y? ( 1 - A)x
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+ (4 - -A)y = 0 59
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@SI..
a 12 a22 -
• • •
• • •
a n1
a n2
>..
• • •
a1n
• • •
a2n
• • •
• • •
• • •
a nn -
=0 A
is called the characteristic equation of A. On expanding the determinant, the characteristic equation t akes t he form
where k' 's are expressible in terms of the elements a i j . The roots of this equation are called the eigenvalues or latent roots or characteristic roots of t he matrix A .
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4.2
Linear Algebra
eigenvectors
Let A is an N x N matrix. Then t he column matrix x which satisfies A x = ,,\x is called the eigenvector of the matrix A. Let us illustrate with an example. Example: Find t he eigenvalues and eigenvectors of t he 5 4 matrix
1 2
Solution: The characteristic equation is [A - .X] 5-,,\ 4 1 2 - ,,\ or
=0
or
=0
(.X-6)(.X -1 ) = 0
• • •
.X = 6 1 '
Thus the eigenvalues are 6 and 1 If x, y be the components of an eigenvector corresponding to t he eigen value ,,\ , then
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[A - ,\J]X
for
=
5- ;\ 4 1 2- ,\
- 1
A= 6, we have
X
=0
y
4
X
1 -4
y
=0
This gives only one indep endent equation -x + 4y
=
0.
Now we choice. We can choose any of the value of either x or y . It is a general convent ion t o take he sm aller value to b e 1. We choose y
=
1 here . So we get the eigenvector
4
1
for
A = 1, ,ve h ave
4 4
X
1 1
y
=0
This gives only one indep endent equation x we choose x
= 1, then
+y =
0. If
we get the eigenvector
1 - 1 [email protected]
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. We must keep into mind that the eigenvectors obtained are not normalized. To normalize we need to divide by t heir modulus. The normalized eigenvectors are
1
v'17
1
4 , 1 •
~
4 1
We will discuss a lot about the physical significance of t he eigenvalue and eigenvectors as we proceed furt her. Let us first start wit h the properties of eigenvalues and eigenvectors of the special matrices.
4.3
Eigenvalue and eigenvectors of normal matrices
We have already told the t he normal matrices are the mat rices which follows At A = AA t . We also showed that both Hermitian and unitary matrices (or symmetric and orthogonal matrices in t he real case) are examples of normal matrices. We now discuss t he properties of the eigenvectors and eigenvalues of a normal matrix. 77
If x is an eigenvector of a normal matrix A with corresponding eigenvalue A then A x = Ax, or equivalently,
(A - A]_)x = 0 Taking B = A - A]_ . Then the previous eq becomes B x= 0 and, taking the Hermitian conjugate, we then have
However, the product BtB is given by
B tB = (A - A]_)t(A - A]_)= (At - A*) (A - A]_)
= At A - A* A - AA t + AA* Now since A is normal, AAt = At A and so
and hence B is also normal. We then find
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hence
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hence
Therefore, for a normal matrix A, t he eigenvalues of At are the complex conj ugates of t he eigenvalues of A. Let us . . now consider two eigenvectors xi and x 1 of a normal matrix A corresponding to two different eigenvalues >..i and Aj . We t hen have
. Ax1
.
== Aj X1
Mult iplying the second equation on the left by (xi) t we obtain
(xi) t AxJ == Aj (xi ) t xj On t he LHS use (AB • • • G)t == e t ••• Bt At and the property just proved for a normal matrix to writ e At xi == >..;xi
From the previous two equations we get
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Example: Prove t h at a normal m atrix A can be written 80
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in t erms of its eigenvalues Ai and orthonormal eigenvectors . xi as N
A= L ..\ixi (xi)t i =l
Solution: You can show t h at bot h sides of the expression give t he same result when acting on an arbitrary vector
y. Since A is normal, we m ay expand y in t erms of the eigen. vectors x i , as shown in previous equation. Thus, we h ave
Ay
=
N
N
i= l
i =l
AL a,ixi = L ai..\ix i
Alternatively, t he action of LHS on arbit rary vector y is
i= l
since ai
= (x i) t
i= l
y.
We see t h at the two expressions
for t h e action of each side of the given equation on y are ident ical; which confirms that t h e equation is correct .
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But (xi ) t xi is t he modulus squared of t he non-zero vec. 2 tor x and is t hus non-zero. Hence Al must equal Ai and t hus be real. The same argument can be used to show t hat t he eigenvalues of a real symmet ric matrix are t hemselves real. Since an Hermitian matrix is also a normal matrix, its e ige nvectors are orthogonal ( or can b e made so using the Gram- Schmidt orthogonalisation procedure ). Alternatively we can prove the orthogonality of the eigenvectors directly. Proof: Consider two unequal eigenvalues t heir corresponding eigenvectors satisfying
Ai
and
Aj
and
Axi = Aixi Axj
=
A j Xj
Taking t he Hermit ian conjugate of t he first equation we find (xi) t At= Ai (xi) t . Mult iplying t his on t he right by x J we obtain
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But ( xi ) t xi is t he modulus squared of t he non-zero vec.
2
t or x and is t hus non-zero. Hence Al must equal Ai and t hus be real. The same argument can be used to show t hat t he eigenvalues of a real symmetric mat rix are t hemselves real.
Since an H ermitian matrix is also a normal matrix, its e ige nvectors are orthogonal ( or can b e made so using the Gram- Schmidt orthogonalisation procedure ). Alternatively we can prove the orthogonality of the eigenvectors directly. Proof: Consider two unequal eigenvalues Ai and AJ and t heir corresponding eigenvectors satisfying
= Aixi Axj = A j X j Axi
Taking t he Hermit ian conjugat e of t he first equation we find (xi) t At = Al (xi) t . Mult iplying t his on t he right by x J we obtain
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Therefore mat rices that are anti-Hermit ian are also normal and so have mut ually orthogonal eigenvectors. T he propert ies of t he eigenvalues are also simply deduced , since if Ax = AX t hen
Hence A* zero).
= - A and so A must be pure imaginary ( or
In a similar manner to t hat used for Hermit ian matrices, t hese propert ies may be proved directly.
4.5
Eigenvectors and eige nvalues of a unitary matrix
A unitary matrix satisfies At = A- 1 and is also a normal matrix, wit h mutually ort hogonal eigenvect ors. To invest igate t he eigenvalues of a unitary mat rix, we note that if Ax = AX then
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and we deduce that AA* = A 2 = 1.
Thus, the eigenvalues of a unitary matrix have unit modulus.
4.6
Simultaneous eigenvectors and commutation of matrices
Two different normal matrices can have a common set of eigenvectors if, and only if, they commute Proof: Let A and B be two N x N normal matrices and . xi be t he i th eigenvector of A corresponding to eigenvalue Ai, i.e. .
.
Ax2 = Aixi
for
i = 1, 2, .. . , N
Init ially we assume that the eigenvalues are all different (i) F irst suppose that A and B commute. Now consider .
ABxi
.
=
BAxi
.
=
B Aixi
=
AiB x
.
2
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combination of the eigenvectors, N
x= L
cixi
i= l
Now consider the following N
ABx = AB L
N
cixi = A L
N
ciµixi =
L ci>..iµixi
i =l
i= l
i= l
N
N
N
and BAx = BAL Cix'i = B L i= l
CiAiXi = L
i =l
ciµi>..ixi
i =l
It follows t hat ABx and BAx are t he same for any arbit rary x and hence t hat (AB - BA)x
=0
for all x . That is, A and B commute. It should be not ed t hat if an eigenvalue of A, say, is degenerate then not all of its possible sets of eigenvect ors will also const it ut e a set of eigenvectors of B. However, provided t hat by taking linear combinations one set of joint eigenvect ors can be found , the proof is still valid and t he result still holds. j [email protected]
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combination of the eigenvectors, N
x
= L cix i i =l
Now consider the following N
ABx = AB L
N
N
cixi = AL ciµixi =
L ci>..iµixi
i= l
i =l
i =l
N
N
N
and BAx =
BAL i= l
Ci Xi
=
BL
CiAiXi =
i= l
L CiµiAiXi i =l
It follows t hat ABx and BAx are t he same for any arbit rary x and hence that
(AB - BA)x = 0 for all x. That is, A and B commute. It should be noted t hat if an eigenvalue of A , say, is degenerate then not all of its possible sets of eigenvectors will also constit ute a set of eigenvectors of B. However , provided t hat by taking linear combinations one set of joint eigenvect ors can b e found, the proof is still valid and the result still holds. jahir@physicsguide. in
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4. 7
Linear Algebra
Cayley Hamilton and General properties of eigenvalues
We know that a scalar,\ is an eigenvalue of the matrix A if and only if ,\ is a solution to t he characteristic equation det(A - -\n) = 0
This is called characteristic equation and will give a polynomial of n degrees for n x n matrix. Now Cayley Hamilton t heorem says every matrix follows its characterist ic polynomial. For illustration I say that if characteristic equation of matrix A is
t hen t he matrix follows aA + bA + cA + dn = 0 2
3
4 . 7 .1
Some properties of eigenvalues
(i) eigenvalues and eigenvectors of a shift ed matrix A - an : A x = -\x j [email protected]
.,..-...,..... ...
- . . . ..
,_,..,...,
⇒
(A - an)x = (-\ - a)x 89
....
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.
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(ii) eigenvalues and eigenvectors of A- 1
:
Ax = AX (iii) The eigenvalues of A and AT are the same (as their characteristic polynomials are the same), but there is no simple relationship between t heir eigenvectors. (iv) eigenvalues and eigenvectors of powers of A :
In general, the eigenvalues of Ak are Ak, and t he eigenvectors are t he same as t hose of A . (v) The eigenvalues of a triangular matrix are clearly its diagonal entries. The eigenvalues of a block triangular matrix (vi) If A is a real matrix with a complex eigenvalue A = µ+iv and corresponding complex eigenvector v = x+ iy, t hen the complex conjugate A* = µ- iv is also an eigenvalue with complex conjugate eigenvector v = x - iy (vii) The sum of the eigenvalues of a matrix equals its t race j [email protected]
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(viii) The product of t he eigenvalues is t he determinant of the matrix.
4 . 7 .2
Some examples to evaluate eigenvalues
Example: Find the eigenvalues and normalised eigenvectors of the real symmetric matrix
3 l 1 -3 3 -3 - 3 1
A=
1
Solution: The characteristic equation is 1- A
l
1
1- A
3 -3
3
-3
-3 - ,\
=0
Expanding out t his determinant gives
(1 - A) [( l - A)(-3 - A) - (-3)(-3)]
+ 1[(-3)(3) - 1(-3 - A)] + 3[1(-3) - (1 - A)(3)] = 0 Simplify to get
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Xl
=
J2' J2'
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v'2
1 1 0
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Repeating the last paragraph, but with the factor 2 on t he RHS replaced successively by ,\ 2 = 3 and ,\ 3 = - 6, gives two further normalised eigenvectors x
2
1
= - (1
x
v'3
3
1
= - (1
v'6
- 1
In t he above example, t he t hree values of ,\ are all different and A is a real symmetr·ic matrix. Thus we expect, and it is easily checked, t hat t he t hree eigenvectors are mut ually ort hogonal, i.e.
Example: (Degenerate eigenvalue) Find t he eigenvalues of the matrix and then construct an orthonormal set of eigenvectors for the matrix
A=
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Repeating the last paragraph, but with the factor 2 on t he RHS replaced successively by ,\ 2 = 3 and ,\ 3 = -6, gives two furt her normalised eigenvectors x
2
1
= - (1
x
v'3
3
1
= - (1
v'6
- 1
In t he above example, t he t hree values of,\ are all different and A is a real symmetr·ic matrix. Thus we expect, and it is easily checked, t hat the t hree eigenvectors are mutually orthogonal, i.e.
Example: (Degenerate eigenvalue) Find the eigenvalues of the matrix and t hen construct an orthonormal set of eigenvectors for the matrix
A=
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Characteristic equation
0
3
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Characteristic equation
0
0 - 2 - ,\
3 0
3
0
1- ,\
1 - ,\
0=
2
= -(1 - -\) (2 + ,\) + 3(3)(2 + ,\) 2 = (4 - -\)(,\ + 2) Thus ,\ 1 = 4, ,\ 2 = - 2 = -\ 3 . The eigenvector for ,\ 1 = 4 1
is found from x =
T
X1
x2
X3
1 0 3 0 -2 0 3 0 1 This matrix equation gives three linear equations which say t hat x = z and y = 0. Hence we choose an engenvector of the form (x, 0, x) and get by normalization
X
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1
1
=-
y12
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1 0 3 0 -2 0 3 0 1 This matrix equation gives three linear equations which says x- = -z and y is arbitrary. So we need to construct two vectors with t he general form (x, y, -x) which will be 1, 0, - 1). orthogonal to the first eigenvector
( 12
Now we see that a general column vector that is orthogonal to x 1 is T x = a b -a This is well fitted with our general form of next two eigenvectors which is (x, y, -x) . So we choose t he first one to be x2
=
T
0 1 0
and t he second one X
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3
1 =~
T
- 1 0 1
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We have taken these two eigenvectors in the form that they are mutually orthogonal and both are orthogonal to t he first eigenvector also as asked in the question.
We don't always get n eigenvectors for a n x n matrix. For example try to find out the eigenvectors of the matrix.
1 2 1 A=
l - 1 1
2
0
1
Solution: The characteristic equation of the matrix A
=
1 2 1 1 - 1 1
2
0
•
IS
1 3
2
2
0 = det (A - Al) = -A + A + 5A + 3 = -(A + 1) (A - 3) There is a double eigenvalue A1 = - 1 and a simple eigenvalue A2 = 3. In case of A1 = 2 we get
A-Ail= A+I =
2 2 1
2 2 1
X
1 0 1 2 0 2
1 0 1
y
2 0 2
z
So we have the only choice
(-2y, y, 2y)T
=0
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eigenvectors are respectively
l T )2 (1, 1, 0) l T ~(1 , -1, 1) and
l T v'6(1 , - 1, -2)
(b) Eigenvalues are 8, - 1, - 1. Eigenvector of corresponding eigenvalue 8 is ½(2, 1, 2)r. Any two vectors orthogonal to (2, 1, 2)T and which are themselves orthogonal would be t he eigenvectors of eigenvalue - 1, for example
1
l T )2(1, 0, -1) and
l T v'18(1, - 4, 1) 8
(c) Eigenvalues are 4, -2 , -2 Eigenvector of corresponding eigenvalue 4 is }z(l , 0, l )T Any two vectors orthogonal to }z(l , 0, l )T and which are themselves orthogonal would be t he eigenvectors of eigenvalue -2, for example }2(-1, 0, l)T and (0, 1, o)T [email protected]
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5
Similarity transformations and diagonalization
5.1
Change of basis
Suppose we take a basis e i : i = 1, 2, ... , N , into our N -dimensional vector space then we may write a vector like
We represent x in this basis by the column matrix as T
We now consider how these components change as a result of a prescribed change of basis. Let us introduce a new basis e ~ : = 1, 2, . . . , N which is related to t he old basis by N
eJ= L S-ijei i=l
Sij
eJ
t he coefficient is the i th component of with respect to t he old (unprimed) basis. For an arbitrary vector x it follows t hat N
x
=
N
J\T
N
L xiei = L xJeJ= L xJ L sijei i =l
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From this we derive t he relationship between the compo-
From t his we derive t he relationship between t he components of x in the two coordinate systems as N
xi
=
L sijXJ j =l
which we write in t he matrix form as
x = Sx' where S is t he transformat ion matrix associated wit h the change of basis. Furthermore, since t he vectors e3 are linearly independent , the matrix S is non-singular and so possesses an inverse s- 1 . Multiplying x = Sx' on the left by s- 1 we find X
' = s-1X
which relates the components of x in t he new basis to t hose in the old basis Comparing x' = s- 1x and ej = L :1 s ij e i we note that the components of x transform inversely to the way in which t he basis vectors e i themselves t ransform. This has to be so as the vector x itself must remain unchanged.
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5.2
Similarity transformation
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5.2
Similarity transformation
We may also find the transformation law for the components of a linear operator under t he same change of basis. Now, t he operator equation y = Ax (which is basis independent) can be written as a mat rix equation in each of the two bases as y
= AX,
y'
= A' X'
But, using x = Sx' we may rewrite t he first equation as
y' =
Sy'= ASx'
s- ASx' 1
Comparing this with y' = A'x' we find that t he components of the linear operator A transform as (5.1) Equation 5. 1 is an example of a similarity transformat ion - a transformation t hat can be particularly useful in converting matrices into convenient forms for computation. Given a square matrix A , we may interpret it as representing a linear operator A in a given basis e i . From equat ion 5 .1 however , we may also consider the matrix A' = j [email protected]
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e same as Hence The eigenvalues are same v ........
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s- 1As -
A' - ,\ ].
,\]. = s - 1 (A - Ali )S
s- 1 s1A -
,\]. = A - ,\].
(iv) The value of t he trace is unchanged: as we know t hat the sum of t he eigenvalues is the t race
TrA' =
1
LA:i = LLL (s- )ij Ajkski i
i
j
k 1
= LLL ski (s- )i_j Ajk i
=
j
k
LL 6kj Ajk = LAjj = j
k
TrA
j
(v) You must know t hat A and A' have different sets of eigenvectors in general.
5.2.1
Unitary transformation
An important class of similarity transformations is that for which S is a unitary matrix; in this case A' = s- 1AS = st AS. Unitary transformation matrices are particularly important , for the following reason. If the original basis e i
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s- 1AS- ..\li
A' - ..\ li
s- 1 s1A -
=
s-1 (A -
..\li)S
..\]_ = A - ..\]_
(iv) The value of t he trace is unchanged: as we know t hat t he sum of t he eigenvalues is t he t race
TrA'
=
LA~i = LLL (s- )ij Ajkski 1
i
=
k
LLL ski (s- ) ijAjk 1
i
=
j
i
j
k
L L 6kjAjk = L Ajj = Tr A j
j
k
(v) You must know that A and A' have different sets of eigenvectors in general.
5.2.1
Unitary transformation
An important class of similarity t ransformations is that for which S is a unitary matrix; in this case A' = s- 1 AS =
st AS. Unitary transformation matrices are particularly important, for the following reason . If t he original basis ei j [email protected]
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is orthonormal and t he transformation matrix S is unitary t hen t he new basis is also ort honormal. This can b e proved as follows.
(e~e3) = L skiekl L srjer r
k
= L ski L Srj (ek er) r
k
=
L szi L SrjOkr = L sziskj = (sts)ij = Oij r
k
k
Furt hermore, in addition to the properties of general similarity t ransformations, for unitary transformations the following hold. (i) If A is Hermit ian or ant i-Hermitian then A' is also Hermit ian or anti-Hermitian respectively, i.e. if At = ±A t hen
(ii) If A is unitary (so t hat At
= A- 1 ) then A' is unitary,
•
since
(A')t A' = (stAs) t (stAs ) = = jahir@physicsguide. in
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stAtAs = s- 1 ].s = 104
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tained in the matrix M. We say that this matrix is an operator which maps the plane into itself. In three dimension this argument can be easily extended, t he mapping then becomes from one point in three dimensional space to another point in t he three dimensional space. j [email protected]
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This transformation from one set of points to another set of points which are caused by the matrix M can be a rotation, reflection or any other kind of transformation. The special case of a linear transformation which preserves the length of a vector is called Orthogonal Transformations. We can easily p1·ove t hat if the matrix M is orthogonal t hen it preserves the length of a vector. det M = l corresponds geometrically to a rotation, and det M = - 1 means that a reflection is involved.
All matrices need not to be orthogonal. But every matrix cause some deformation on t he set of points upon which it acts. Now A' = s- 1 AS is the matrix which describes in the (x', y') system the same deformation that A describes in the (x, y) system.
5 .3
Diagonalisation of matrices
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This transformation from one set of points to another set of points which are caused by the matrix M can be a rotation, reflection or any other kind of transformation. The special case of a linear t ransformation which preserves the length of a vector is called Orthogonal Transformations. We can easily prove t hat if the matrix M is orthogonal then it preserves the length of a vector.
det M = l corresponds geometrically to a rotation, and det M = - 1 means that a reflection is involved. All matrices need not to be orthogonal. But every mat rix cause some deformation on t he set of point s upon which it acts. Now A' = s- 1 AS is the matrix which describes in the (x' , y') system t he same deformation t hat A describes in t he (x, y) system.
5.3
Diagonalisation of matrices
Suppose t hat a linear operator A is represented in some basis e i : i = 1, 2, . . . , N by t he matrix A. Consider a new
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basis
.
xJ
given by N
x j
L
=
s ij e i
i =l
where the x J are chosen to be the eigenvectors of t he linear operator A , i.e. .
AxJ
=
. Aj XJ
In the new basis, A is represented by the matrix A' = s-1AS, which has a part icularly simple form, as we shall see shortly. The element S ij of S is t he i th component, ( in t he old (unprimed) basis,) of t he j t h eigenvector xi of A , i.e. the columns of S are the eigenvectors of t he matrix A :
t t S=
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x2
t ...
XN
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.
.
.
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t hat is, Sij = (xj )
i .
Therefore A' is given by
L L (s- ) ik A kz Szj = L L (s-l ) ik A kl (xj ) = L (s-1 )ik Aj (xj ) = L Aj ( s- ik s kj = Aj bij
1
1
(s- As)ij =
k
l
l
k
l
k
k
1
)
k
So the matrix A' is diagonal with the eigenvalues of A as t he diagonal elements,
A' =
A1 0 0 A2 •
• • •
0 • • •
•
• •
0
• • •
•
0
•
0
AN
So, given a matrix A, if we construct the matrix S that has t he eigenvectors of A as its columns t hen the matrix A' = s- 1 AS is diagonal and has t he eigenvalues of A as its diagonal elements. We require S to be non-singular ( S # 0), t he N eigenvectors of A must be linearly independent and form a basis for the N -dimensional vector space. [email protected]
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It may be shown that any matrix with distinct eigenvalues can be diagonalised by this procedure. If, a general N x N square matrix has degenerate eigenvalues then it may, or may not, have N linearly independent eigenvectors. If it does not then it cannot b e diagonalised.
For normal matrices (which include Hermitian, anti-Hermitian and unitary matrices) the N eigenvectors are indeed linearly independent. Moreover, when normalised, these eigenvectors form an orthonormal set (or can be made to do so). Therefore the matrix S with these normalised • eigenvectors as columns, 1.e. whose elements are S i.1 = (xJ) i , has t he property
(sts) ij =
L
(st)ik (S)kj
=
L
k
=
t j ' i ) X ( X
k
ski skj
=
L
(xi): (xj)k
k
~ = Uij
Hence S is unitary (s- 1 = can be diagonalised by
st)
and the original matrix A
Therefore, any normal matrix A can be diagonalised by j [email protected]
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a similarity transformation using a unitary transformation matrix S . Example Diagonalise t he matrix
1 0
A=
3
0 -2 0 3 0 1
Solution: The matrix A is symmetric and so may be diagonalised by a t ransformation of the form A' = stAS where S has the normalised eigenvectors of A as its columns. We have already found t hese eigenvectors in a previous example.
X
1
1 = -
J2
1
0 1
X
2
= -
0
1
J2
x3 = -
1
1
J2
0
- 1 0
1
So t he S matrix which is obtained by writ ing t he eigenvectors in the columns
s=
1
v'2
1
0
- 1
0
v'2
0
1
0
1
Alt hough the eigenvalues of A are degenerate, its t hree eigenvectors are linea1·ly independent and so A can still be j [email protected]
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diagonalised. Thus, calculating 1
0
1
0
v'2
0
- 1
0
1
4
0
stAS we obtain
1 0 3 0 -2 0 3 0 1
1
0
- 1
0 1
v'2
0 1
0
0
0 -2 0 0 0 -2 You can see that the matrix has the eigenvalues of A in its diagonal elements.
6
Function of matrices
Function of matrices can be obtained by Two methods. 1) By using similarity transformation properties and 2) Expanding t he matrix in series and using Cayley Hamilton t heorem. We will discuss both.
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6.1
Linear Algebra
Function of matrices by similarity transformation
If a m atrix A is diagonalized by t h e similarity t ransforma1AS , so that A'= diag (A1, A2, .. . , AN), then t ion A'=
s-
we know
N
Tr A' = Tr A =
N
L Ai
A' I = IA I =
i =l
IT Ai i =l
If a m atrix can be diagonalized by similarity transfor1AS, t h en we h ave m ation A' =
s-
Now we illustrate t h e use of equat ion (6.1) to find funct ion of matrices. but b efore we begin we need to know one important result
Important: If a matrix is diagon al, say A
=
X
0
0 y
then j ahir@physicsguide. in
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f (A)
=
0
0 f (y)
0
X
Proof: Let A =
J(x)
0 y xn 0 0 yn
We can prove An=
by induction on n
In case n = 1 it is true by t he definition of A Suppose that Ak =
X
. Then we have
0
X
0 y
k+l
0
Thus our hypothesis is true by induction method. Then for any function series
f (A)
which can be expanded in
•
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The expansion in series is valid for matrix also
Hence X
2
x 0 0 y2
0
0 y
x3 0 0 y3
+ ...
or
=
f( A) 2 3 ao + a1x + a2x + a3x + ...
0 2 ao + a1y + a2y + a3y 3 + ...
0
Which can b e easily seen as
J(A) =
J(x) 0
0 J(y)
Example: Consider the matrix
A=
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ou can easily verify t hat t he eigenvalues are 1 and 6 and the corresponding eigenvect ors are 1
-2
v'5
v'5
2
1
v'5
v'5
So upon diagonalizing we get
s- As = A' 1
S = l
where
)
and
v'5
A' =
1 - 2 2 1
1 0 0 6
Then for any function consisting of sums of powers of A , we have
J(A) = J (sn s- ) = Sf(A')s= S diag (Ai, A2, ... , An) s-1 1
1
where t he Ai are t he eigenvalues of A , and diag is a diagonal matrix, whose diagonal entries are given as t he arguments. T he second step above follows from t he observat ion t hat
(SA' s- )n = $A' s- s A' s- s A' s1
1
1
1
...
SA' s- : = SA'ns-
1
V
n terms j [email protected]
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A pp lying t his to any function of matrix A we get f(A) = S
1 5
1 -2 2 1 1
1 -2
5
2
-
f (1)
0
0
f(6)
s - 1
f (1)
0
0
f(6)
1
2
-2 1
f (1)
2f (1) -2f(6) f(6)
1
f(l) + 4f (6) 2f( l ) - 2f (6) 2f (1) - 2f (6) 4f (1) + f(6)
1 5
We apply t his result in two cases
A4
=~ 5
1
4
+4 · 6
2 . 14
4
4
2·1
2 . 64
-
l e =5 A
4.
14
2·6
4
1037 -518 - 518 260
+ 64
e + 4e 2e - 2e 1 1 6 6 2e - 2e 4e + e 1
6
1
6
e -
5
Now we will prove trace formula j [email protected]
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Linear Algebra
Trace formula
From equation (6. 1) we get exp A' =
s-
1
(expA)S.
By choosing t he similarity transformation which diagonalizes A , we get A' = diag ( A1, A2, . .. , AN) , and so
exp A = exp A' = exp [diag (A1 , A2 , .. . , AN)]I N
=
ldiag (exp A1, exp A2 , .. . , exp AN) =
II exp Ai i=l
Rewriting the final product of exponentials of the eigenvalues as the exponential of the sum of t he eigenvalues, we get
N
exp Al =
II exp Ai = exp
= exp(Tr A)
i =l
Hence we have proved the trace formula exp A = exp(Tr A ) I j [email protected]
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6.2
Linear Algebra
Function of matrix by using power series expansion and Cayley Hamilton the ore m
Let us st art with an example. You will get everything.
0 1 find, sin kA, cos kA, For the matrix A = 1 0 ei kA , where k is a scalar number .
Solution: Given that, A = 2
equation of the matrix is A
0 1 1 0
•
Characteristic
=1
0 1 1 0
0 1 1 0
1 0 0 1
1 0 0 1
0 1
0 1 =A 1 0
and so on ... So, A 2 ... = A
=
A4
1 0 =
ekA,
A6
= ... =
=I
J ; A3 = A5 = A7 =
Now expand the functions in power series.
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1. 3 5 kA) (kA ) ( sin kA =kA - - - - + - - + · · · 3! 5! k 3A k 5A =kA - + - + ... 3! 5!
=k
3
k 0 l 3! 1 0
O l 1 0
ks + 5!
0 1 1 0
0 k3
k5
k -JT+ w+ ·· · 0 sink sink 0 2. 2 4 (kA) kA) ( cos kA = I - --+--+ · .. 2! 4! k2I k4J = I - - + - + ··· 2! 4! k4 1 0 k2 1 0 2! 0 1 + 4! 0 1
1 0 0 1 0
k2
k4
l -21 + 41+ · ··
cos k 0 0 cos k jahir@physicsguide. in
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3. ekA
- I
-
kA
(kA)2 (kA)3 (kA)4 (kA) 5 ... + + 2! + 3! + 4! + 5! + 2 (kA) (kA) 4 ! +--+--+ ··· 2! 4! (k A)3 (k A)5 + kA + - - + - - + ... 3! 5! k3 A k5A + kA + - 3! + - 5! + · · ·
1 0 0 1
+
k
k
2
+ 2! O l 1 0
+
1 0 0 1
k3 3!
4
+
1 0 0 1
k 4!
ks 5!
0 1
1 0
+
0 1 1 0
+ coshk 0 cosh k 0
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4.
i kA
4
3
5
(i kA) (ik A ) (ikA ) (i kA) - + + 2! + 3! + 4! + 5! + (ikA )2 ( i kA) 4 ! +--+--+ ·· · 2! 4! 3 5 .kA (ikA) (ik A ) + 'l + - - + - - + •· · 3! 5! ik 3 A ik 5 A + i kA - - - + - - + · · · 3! 5!
eikA _
I
2
k2 1 0
1 0 0 1
+
ik
2!
3!
1 0
+
0 1 3 ik
O 1
k4
...
1 0 0 1
4!
5
ik + 5!
0 1 1 0
0 1 1 0
0 k2
0
+
k4
l - 21+41+ ·· · . k k3 k5 i -3T+51+ · ··
0
cos k 0 0 cosk j ahir@physicsgui de. in
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cos k i sin k i sin k cos k You could have solved t he problem by similarity transformat ion. That could have been easier also. As we see t he
1 0 diagonalized matrix A' in this case would b e A' 0 - 1 and t he S matrix S
=
F2
1
1
1 - 1
Hence
f( A)
/ (1) 0 0 / (-1)
=~
1
1 2
1
1 2
/(1) + /( -1) / (1) - / (-1) /(1) - /(- 1) /(1) + /(- 1)
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1
1 -1
1
1 - 1
1
1
1 -1
/(1) /(1) /(- 1) - /(- 1)
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Linear Algebra
Now when J(A )
.
1
=-
s1nkA
2
= sin kA we get sin k - sin k sin k + sin k sin k + sin k sin k - sin k 0 sink sink 0
We see that we have got the same result quite easily. Now do the rest three sums cos kA , ekA, eikA . Verify your result from previous example. Personally I prefer the similarity transformation method while evaluating any function of a matrix
Example: You are given two Matrices A and B. Consider t he matrices eA, eB . expand some of the t erms in the infinite series of eA, eB and eA+B. Multiply the two series of eA , and eB . Now show that
Solution: We know, eA
eB j [email protected]
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A3
2!
3!
B2
B3
2!
3!
= I + A + - + - + ··· = I + B + - + - + ··· 123
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t hen
Solution:
\A/e can prove by induction
Now suppose that J(A) is a function of A defined by a Taylor series in non negative powers of A , where the coefficients of the Taylor series are assumeed to commute with both A and B. It is easy to see that:
[J(A), B] = J'(A) [A, B] where J' denotes the formal derivative of an operator argument A . ex A
f
applied to
= L n -¾(xA)n is such a function. Therefore, n.
Now define a operator
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t hen
Solution:
\A/e can prove by induction
Now suppose that J(A) is a function of A defined by a Taylor series in non negative powers of A , where the coefficients of the Taylor series are assumeed to commute with both A and B. It is easy to see that:
[J(A), B] = J'(A) [A, B] where J' denotes the formal derivative of an operator argument A . ex A
f
applied to
= L n -¾(xA)n is such a function. Therefore, n.
Now define a operator
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Linear Algebra
By construction, G (x) is invertible, with
c-1 (X) = e- xBe- xA We differentiate w.r.t. x
dG (x) dx
dexA --exB
dexB + exA__ dx dx = AexAexB + exAexBB
= (A + B )exAexB + [exA, B ] exB = (A+ B + x[A , B] )G(x) Integrating by separation of variable
logG(x)
+ k = xA + xB +
1 2 -x [A , B ] 2
Now define
So the equation can be written
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We get from the definit ion of F (x)
[F' (x), F (x)] = A + B + x[A, B ], xA + xB + ~x [A , BJ 2
1
= x [A + B , [A, B]] + x [[A, B], A + B]
2
2
2
=0 Since [A, [A, B]] = [B , [A, B]] = 0 For any operator satisfying [F' (x), F (x)] t o show t hat:
= 0, it 's easy
d e- F(x) = - F' (x )e-F(x) dx Our differential equation becomes
ddx [ e-F(x) G (x)]
G (x)
=
=0
eF(x) c o
where G 0 is an op erat or which does not depend on x . Checking t he special case x = 0, we find G = 1 and F = 0. j [email protected]
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Therefore Go = 1. Eliminating F and G in favor of A and B , we find:
Setting X
=1
since [A , B ] commutes with A and B , this is equivalent to:
More generally it can be proved that when a, j3
,.. 1 a ,.. 1 a ,.. ) · m. 1ett1ng '±' = a 1 2 3 in v '±' = h1 8u1 e1 + h2 8u 2 e2 + h3 8u3 e3 and substit uting into the above equation, we find
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Vect ors
Repeating the analysis for V · (a2e2) and V · (a 3e3 ) adding t he results we get the final result.
2.8.3
Laplacian
In the expression of divergence use 1 8 ,-.
a = V = h
1 OU1
e1
1 8
+
1 a h2h3 a h1h2h3 8u1 h1 8u1 a h1h2 a I'.:\
7
-I'.:\-
1 8 ,-.
+ h 2 OU2 e2 + h 3 OU3 e3
You easily get the expression
v 2 =
A
,
and
11 oo : 04: 12 You easily get the expression
v 2 =
1 a h2h3 a h1h2h3 OU1 h1 OU1 fJ h1h2 fJ +OU3 h3 OU3 (2.73)
2.8.4
Curl
The curl of a vector field a = a 1e1+a2 e2 +a3e3 in orthogonal curvilinear coordinat es is given by
(2.74)
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This can be proved as follows Let us consider t he subexpression V x (a 1e1) . since e1 = h 1V u 1 we have y7 X
(a1e1)
= y7 X (a1h1 V u1) = V (a1h1) x V u1 + a1h1 V x V u1
But V x V u 1 = 0, so we obtain "
'\7 X (a1e1)
= '\7 (a1h1) X ::
· ( nm. 1 8 e" 1 + -h 1 8 e2 " + -h 1 8 e3 "' ) a 1h1 1n v '±' = -h 1 8 U1 2 8 U2 3 8 U3 and substit uting into the above eauation, we find · Lett1ng
m.
'±'
=
11 oo : 04: 1 s A
y'
X (a1e1) =
y'
(a1h1) X ::
. '±' m. · ( nm. l 8 "' 1 8 "' 1 8 "' ) Lett1ng = a 1 h1 1n v '±' = -h e 1 + -h e 2 + -h e3 8 8 8 1 U1 2 U2 3 U3 and substituting into the above equation, we find
Similarly find V x (a2e 2) . you get t erms in e 3 and e1 , . Find y7 X ( a3e3) and you get terms in e1 and e 2. Just add and you get the result.
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2. 9
Vectors
Exersizes
2.1. For the twist ed space curve y 3 given parametrically by
+ 27axz -
8l a2y = 0,
show that t he following hold: (a) ds/du = 3,J2a (I + u 2 ), where s is t he distance along t he curve measured from the origin
11 oo : 04: 17 x = au
'
show that t he following hold: (a) ds / du= 3yl2a ( 1 + u 2 ) , where s is t he distance along t he curve measured from the origin (b) the length of the curve from t he origin to the Cartesian point (2a, 3a, 4a) is 4vl2a (c) the radius of curvature at the point with parameter u is 2 2 3a(l +u ) (d) the torsion T and curvature K, at a general point are equal (e) any of t he Frenet-Serret formulae that you have not already used directly are satisfied. 2.2. The shape of the curving slip road joining two motorways, that cross at right angles and are at vertical heights z = 0 and z = h, can be approximated by the space curve
vl2h r = --ln cos 7r
Z 1r
h 2
•
1
+
vl2h l 1r
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Show that the radius of curvature p of the slip road is (2h/rr) csc(zrr / h) at height z and t hat t he torsion T = - 1/ p. To shorten the algebra, set z = 2h0/ 1r and use 0 as the parameter. 2.3. Parameterising the hyperboloid x2
y2
z2
-+---= 1 a2 b2 c2
11 oo : 04: 20 rameter. 2.3. P aramet erising t he hyperboloid
x2
y2
z2
+ = l a2 b2 c2 by x = a cos 0 sec N Un
converges
However , if I: Vn diverges and Un > Vn for all n great er than some fixed number then I: Un diverges.
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11 oo : oo : 44 some fixed number t hen
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1.3.3
Infinite Series
D' Alembert's ratio test
The ratio t est det ermines whether a series converges by comparing the relative magnitude of successive t erms. If we consider a series
I: U n p
and set
= lim
(1.17)
n➔ oo
t hen if p < 1 t he series is convergent; if p > 1 the series is divergent; if p = 1 then the behaviour of t he series is undetermined by this t est.
1.3.4
Ratio comparison test
Consider t he two series L U n and I: Vn and assume t hat we know the latter to be convergent. It can b e proved that if
for all n greater than some fixed value N then convergent . Similarly, if
I: U n
is also
11 oo:oo:47 for all n great er than some fixed value N then convergent. Similarly, if
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for all sufficiently lar·ge n , and diverges.
1.3.5
I: Un is also
I: Vn diverges t hen I: U n
also
Quotient test
Consider the two series limit p
I: Un and I: Vn, = lim
n ➔ oo
and define p as the
Vn
Then , it can be proved that: (i) if p =I= 0 but is finite then I: Un and L Vn eit her both converge or both diverge; (ii) if p = 0 and I: vn converges t hen I: U n converges (iii) if p = oo and I: v11, diverges then I: Un diverges.
1.3.6
Integral Test
We will replace t he n in the general n th term of t he series by x and write it as a function f (x) . Then it can be shown t hat , if the limit of the integral
11 oo : oo : 49 We will replace t he n in the general n th term of t he series by x and write it as a function f (x) . Then it can be shown t hat , if the limit of the integral N
f (x) dx
lim
N ➔ oo
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Infinite Series
exists, t he series I: Un is convergent. Otherwise the series diverges. Not e that the integral defined here has no lower limit.
Example: Determine whet her t he following series converges: 00
1
L (n - 3/ 2) n =l
2
4
4
4
4
= + + 9 + 25 + · · ·
Solution: Replacing n by x we get 1
f (x) = (x - 3/ 2)2 Applying the integral t est , we consider N
lim
N➔oo
1
(x - 3/2)2dx =
J~oo
- 1 N-3 / 2
=0
since t he limit exist s the series converges. Note, however , t hat if we had included a lower limit , equal to unity, in t he integral t hen we would have run into problems, since t he integrand diverges at x = 3/ 2
11 oo : oo : s2 since the limit exists the series converges. Note, however , that if we had included a lower limit, equal to unity, in the integral t hen we would have run into problems, since t he integrand diverges at x = 3/ 2
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1.3. 7
Cauchy's root test
Cauchy's root test may be useful in testing for convergence, especially if the n th terms of t he series contains an n th power . If we define t he limit p
= lim (Un)l / n n➔oo
t hen it may be proved that t he series I: Un converges if p < 1. If p > 1 t hen t he series diverges. Its behaviour is undetermined if p = 1.
1.4
Alternating Series
An alt ernating series can be written as 00
L (-1 n=l
) n+lun
= U1
-
U2
+ U3 -
U4
+ U5 -
···
11 oo : oo : s4 An alt ernating series can be written as 00
L (-l) n+lun n=l
=
U1 -
U2
+ U3 -
U4
+ U5 -
···
with all U n > 0. Such a series can be shown to converge provided
(i) U n
➔ 0
as n ➔ oo and
(ii) Un < Un- l for all n > N for some finit e N . If t hese conditions are not met t hen the series oscillates. [email protected]
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2
Infinite Series
Examples on Series Convergence Find whether of the following series converges or diverges?
CX)
Solution: Let us take,
L
Un
where, Un
=
n! l OOn.
n=l
Now, we will check t he ratio test. p
= lim
n➔ oo
(n + 1)! 100n n+ l - - x - - = lim = lim -1oon+l = n! 100 n ➔ oo
n➔ oo
CX)
1
L 2n-n2
n=5
00
11 oo :oo : s1 n➔ oo
Un
n,➔ oo
oon+1
1
00
L 2n
n ➔ oo
1 -n2
n=5
Solution: We will use a Special Comparison Test here, it says, 00
If
L bn is a convergent series of positive terms and an~ 0 n=l 00
and
abnntends to a (finite) limit, then L anconverges. n=l
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Infinite Series
1 Now, here we take, bn = n which is a convergent series. 2 . 1 1 l1m -;n➔ oo 2n 2n - n 2
Now, oo ~
So, L..t
. l1m
n2
= n➔ oc 1 -2'-n
= 1
I
n.
loon converges.
n =l
00
n(n + 1)
Solution: Use t hese propert ies 00
(a) If
Lb
1i
n=l
is a convergent series of positive terms and
11 oo : 01 : oo Solution: Use t hese prop ert ies 00
(a) If
L bn is a convergent series of positive terms and n =l 00
an > 0 and an/bn tends to a (finit e) limit , then L
an con-
n =l
verges. 00
L dn is a divergent series of positive terms and
(b) If
n =l
an > 0 and an,/ dn tends to a limit greater t han O (or tends 00
to + oo), t hen L
an diverges.
n =l
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Infinite Series
Here t he n t h term of t he series series go to
n 00
Now L 1 as n
➔
oo·
'
1 n+7
1
1
n
n
- is a divergent series wit h dn = - and an/ dn ➔
n= l ➔ oo .
Hence t he series diverges. oo (-2) n
11 oo : 01 00
Now
: 02
1
1
L -n is a divergent series with dn = -n and an/ dn -+ n=l
1 as n-+ oo . Hence t he series diverges.
Solution: This is an alternating series and an alternating series converges if the absolute value of t he terms decrease steadily to zero, i.e. , if
lan+l ~ an and lim an = 0 n --t
Now, here limn-too an = limn-too
(-2)n
=
n2
oo [Check using
L'Hopital Rule]. So, t he series will diverge. [email protected]
23
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@Sk J ahiruddin, 2020
Infinite Series
Find t he interval of convergence of the series
.
Solution: We can writ e 4
3
l -x/2+x /4-x / 8+ ... +(-x)n/2n+ ....
I n,
.I
00
I rr
.
=
I
CX)
00
n =O
n=O
L anXn = L Un
11 oo : 01 : os 4
3
l-x/2+x / 4-x /8+ ... +(-x)n /2n+ ... . =
L anXn = L Un n =O
n= O
CX)
=
p
. 1Im
Un+l Un
X
=L
2
n=O
Now, if the series converges in any region, then p < l. So, here,
3
X
2
0 is unity. Hence, 32
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y(t) =
t et' /a
e - t/a
- -dt' o a
= e - t/a et/a - 1 = 1 - e - t/a
(b) With f (t) = b(t), t he integration will be trivial:
y(t ) =
(c) For
e - t/a
t 6 (t') et' /a
- - - -dt' a
f (t) = 13- 1e - tff3 H (t), y(t) =
1
= e - t/a X -
a
= e-t/a
= --
a
with f3 < a , we have
t et' I a e - t' I f3
e - t/a
e - t/a
----dt' o a/3 e (a-1 - ,e-1)t'
t
a/3 (a -1 - 13-1) 0
e - t/(3
e - t/a
f3 -a
f3 - a
11 oo : 01
: 31
e- t/ fJ
e- t/ a
(3 -a
(3 - a
e-t/a _ e- t/fJ
a-
/3
As f3 ➔ 0, f (t) becomes very strongly peaked near t = 0, but with the area under the peak remaining constant at unity. In the limit, t he input f (t) becomes a 6 -fuction, the [email protected]
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33
Diff eq and Sp funs
@Sk J ahiruddin ) 2020
same as that in case (b) . It can also be seen that in the same limit t he solution y(t) for case (c) tends to that for case (b), as is to be expected. Solution: 2.4.. We see that t he equation is homoge-
neous in x and y and so we set y and obtain
=
vx, wit h
g; =
v + xg~,
av 2 + 3v v+x-=---
8x v- l 8v - 2 - 3v - v 2 + V x-=------v- l 8x dx (l - v)dv X v 2 + 2v + 2 2 (v+ 1)2 + 1
2
v + 2v + 2 v- l
v
+l
(v+ 1) 2 + 1 1 1 2 => lnAx = 2tan- (v + 1) - ln [1 + (v + 1) ] 2 2 2 1 ln { B x [1 + (v + 1) ]} = 4tan- (v + 1)
11 oo : 01
: 33
2 (v+ 1) 2 + 1
⇒ lnAx 2
ln { B x [1 + (v + 1)
2
]}
= =
V
+1
(v+ 1) 2 + 1 1 1 2tan- (v + 1) ln [1 + (v + 1) 2 ] 2 4tan - 1 (v + 1)
On setting v = y / x t his b ecom es
B [x 2 + (y + x) 2] = exp 4tan- 1
y+ x X
Solution: 2.5 .. As x and y only appear in the [email protected]
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34
Diff eq and Sp funs
@Sk J ahiruddin , 2020
n ation x + y we set v = x + y wit h dv / dx = l + dy / dx . The equation and its solution t hen b ecome
dv = 1 _ v dx 3v - 4 3v - 4 dx = 2v - 4 dv = ⇒
ln ( x
x+ k
+y -
2)
= =
3 2 2 + 2v - 4
3 v + ln(v - 2) 2 1 k - ( x + 3y) 2
=
dv
3 (x + y) + ln(x + y - 2) 2
Solution: 2.6 .. (a) Writing p = dy/dx, the equation b ecomes y
=
xp2
dy 2 p = - = p dx
+p
+ 2x pp' + p'
11 oo : 01
: 36
Solution: 2.6 .. (a) Writing p = dy/dx, the equation becomes y = xp2 + p dy 2 p = - = p + 2xpp' + p' dx 2 dp = p +(2xp+l)dx (b) In different ial form , this equation becomes
df
=
p( l - p)dx - (2xp + l )dp = 0
(c) We now apply t he standard test for the existence of an IF for f (x, p)dx + g(x, p)dp: 35
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l
8g
f
ax
8f 8p
1
1
p(l -p)[-2p- (1-2p)] = -p(l -p)
As t his is a function of p alone, an IF exists and is given by
exp
- 1 p-(l---p-) dp
= exp =
1
1
p
1 -p
-----
dp
exp [- ln p + In (1 - p)] =
With this IF, the equation becon1es
1
- p p
11 oo : 01
: 38 p
=
- p
exp [- lnp + ln(l - p)] = _l _-_ P p
Wit h t his IF, t he equation becomes
(1 - p)2dx - (2xp + 1)(1 - p) dp = 0 p
dp d[( l-p) x ]--+dp = O 2
p
( 1 - p) x - ln p + p 2
=c
This gives x = (c + lnp - p)( l - p)- 2 and, together with y = p + p 2x, gives a full paramet erisation, x = x(p) , y = y(p), of the solut ion. [email protected]
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Diff eq and Sp funs
Solution: 2. 7 .. Writ ing dy / dx have
dv -dx = 2YP,
du_ dx - 2x,
=
p and dv / du
dv yp q--- du - x'
x
X 2 xy 2q -(u+v- 1)-q +xy= O y y
We now multiply by 2
uq
-
Y X
and substitute again:
(u + v - l )q + v = 0
v (l - q) - uq + q + u q2 = 0 q
X
p= -q
Making the substitution gives 2
=
y
q, we
11
00:01 :41
We now multiply by
uq
2
-
Y
X
and substitute again:
(u + v - l )q + v = 0 2
v (l - q) - u q + q + u q
v
=0
= uq +
q
q- 1
, Clairaut 's form
As t he equation now has Clairaut 's form it has two solutions. (i) The first is
C
v =cu+
or
c- 1 In all cases a > 13 > 0
y
2
cx
-
2
C
= --
1
c-
2
2 2
2
• For c > 1, t his is a hyper bola of t he form y - a x = 13 . • For 1 > c > 0, it is a hyp erbola of the form x 2 - a 2 y 2 = 132 • For c < 0, t he conic is an ellipse of the form y 2 + a 2x 2 = [email protected]
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(ii) The second (singular) solution is given by
d dq
q
q- l - 1 (q - l )2
or or
+u= O
+u =
0
1
q
= l ± VU
Subst ituting t his into thje equation converted in Clariot form, expressed in t erms of x and y t hen gives
/32
.
11 oo : 01 OT
q
: 44 1
= l ±-
vu
Substituting t his into thje equation converted in Clariot form , expressed in t erms of x and y then gives y2
=
x2
1±
1
-
X
2
=x ±x±x+l =(x±l) y
2
= ±(x ± l )
These lines are the four sides of the square t hat has corners at (0, ± 1) and (±1, 0) Solution: 2.8 ..
(a) With dy / dx appearing in the first term and y in the second (and nowhere else), t his is a linear first-order ODE and therefore has an IF given by j [email protected]
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µ(x) = exp
Diff eq and Sp funs
X
a2
+ x2
= (a2 + x2) 1; 2 When multiplied t hrough by t his, t he equation becomes d dx
(a2
+ x2) 1/2 y =
x ( a2
+ x2) 1/2
11 oo : 01
: 46
When multiplied t hrough by this, t he equation becomes d (a2 + x 2) 1/2 y dx ⇒
= x (a2 + x2) 1/2
~ ~ (a2 + x2) 3/2 + A
(a2 + x2 ) 1/ 2 y =
a +x A y - - -+---3 (a2+x2)1; 2 2
⇒
2
(b) Again, an IF is needed ; this t ime given by
µ(x) = exp
- tan x dx
=
exp(ln cos x) = cos x
The equation now becomes
:x (y cos x) = cos x => y cos x = sin x + A The given boundary condition is that 3 / v'2 = 1/ v'2 + A , establishing A as v'2 . The final answer is therefore y = tan x + y'2secx [email protected]
39
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Diff eq and Sp funs
Solution: 2.9 .. At first sight this non-linear equation may appear to be homogeneous, but t he t erm A/ x rules this
out. since it is non-linear, we set dy / dx = p and rearrange t he equation to make y, which then appears only once, the subject: 2 y A p - - p+ -=0 X
X
11 oo : 01
: 49
out. since it is non-linear, we set dy / dx = p and rearrange t he equation to make y, which then appears only once, t he subject: 2 y A p - - p+ -=0 X
X
A
xp-y + -
p
y
=0 A
= xp+ -
p
This is now recognised as Clairaut's equation with F (p) = A / p. Its general solut ion is therefore given by y = ex + for arbitrary c
4
It also has a singular solut ion (containing no arbitrary constants) given by
d dp
A P
+X
=
0, :::;> p =
A X
Hence
y=x
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A x
A + -------;:::= = 2ffx A /x
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physicsguide CSIR NET I GATE
Diff eq and Sp funs
The final result was obtained by substit ut ing for p in y = xp + A /p Solution: 2.10 .. The equation is not homogeneous and the two variables x and y appear in different linear corn-
11
oo : 01 : s1
The final result was obtained by substituting for p in y = xp + A /p
Solution: 2.10 .. The equation is not homogeneous and the two variables x and y appear in different linear com-
binations on the two sides of the equation. We t herefore seek shifts in their origins that will make the expression for t he derivative homogeneous, i.e. remove the constant terms from both its numerator and denominator. To do this we set x = X + a and y = Y + /3 We then require
3a + 3/3 + 3 = 0 and
5a + /3 - 7 = 0
These have the straightforward solut ion a= 2 and /3 = -3; with these values the original equation reduces to This is now homogeneous and to solve it we set Y = vX and obtain
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Diff eq and Sp funs
@Sk J ahiruddin ) 2020
dY _ dX 7
-v+
X dv dX
11 oo : 01 : s4 @Sk J ahiruddin, 2020
Diff eq and Sp funs
dY _ X dv dX - V + dX X dv _ dY _ _ 3X + 3Y _ dX - dX v - 5X + Y v 3 + 3v - 5v - v 2 5+v
We now separate the variables and use method (iii) for a partial fraction expansion, obtaining dX
5+v A B = = + 2 x 3 - 2v - v 3 + v 1 - v 1 3
=+ 2(3+v) 2(1- v)
ln X =
1
2
ln (3 + v) -
3
2
ln (1 - v)
+k
Re-substituting for v, X and Y, gives
+3
1/ 2
x-2 = A 3+-Yx- 2
l -y+3 x-2
-3/2
1 2
= A-(3_x_+_y_-_3)_1_ (_x_-_2_) (x - y - 5)3/2 Finally, t his result can be re-written as [email protected]
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Diff eq and Sp funs
11 oo :01 : s1
@Sk J ahiruddin, 2020
Diff eq and Sp funs
3
(x - y - 5) = B (3x
+ y - 3)
Solution: 2.11 .. After being divided through by x, this equation is in t he form of a Bernoulli equation with n = 2, i.e. it is of the form
dy dx
+ P (x)y =
Here, P (x) = x- 1 and Q(x) y 1 - 2 = y - 1 and obtain
dy dx
d dx
I v
Q(x)yn
=
x - 512 . So we set v
I dv v 2 dx
The equation becomes I dv I ----+-= v2 dx vx dv v --dx x d V dx x
j [email protected]
(c)Sk J a,hir11ddin. 2020
1
-2 2 v x5/
I for which t he IF is 1/ x x 512 ' I x7/ 2
43
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Diff ea and Sn fi1ns
11 oo : 01 : sg 43
j [email protected]
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Diff eq and Sp funs
@Sk J ahiruddin , 2020
2 1 3 - = + , using y ( 1) X 5 x 512 5 1 2 1 3x - = ---+y 5 x 312 5 5x3/2 y = 2 + 3x 5/ 2 V
= 1
Solution: 2.12 .. Since x appears only in the combinat ion xdy / dx it will probably make t he solution simpler to take y as t he independent variable and x as the dependent one. With t his in mind, we re-arrange t he equation as
dx tan y dy
+x =
. 2 sin y
In stanadard form
dx dy
+ x cot y = 2 cosy
The IF is clearly exp(ln sin y) can be written
= sin y, and
the equation
d (x sin y) = sin 2y dy . 1 x sin y = - cos 2y + k 2 [email protected]
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11 oo : 02 : 02 -(x sin y) = sin 2y dy
1 cos 2y + k 2
x sin y = 44
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If y(O)
= 1r / 2
X=
Diff eq and Sp funs
t hen k
= -
~ and the solution is
- 2 cos 2 y . 2 Sln y
- 1 - cos2y ----2siny
= - cos y cot y
Solution: 2.13 .. As the equation contains only derivat ives, we write dy/dx = p and d2 y/dx 2 = dp/dx; t his will
reduce the equation to one of first order: dp dx
+ P2 + P = 0
Separating the variables we get dp p(p + 1)
= -dx
We now integrate and express the integrand in partial fract ions: 1
1
- - - - dp= p p+l ln(p) - ln(p + 1)
=A-
dx x
= B e-x
p
p+ l
e- x p = C -e Now p
= dy/dx and so
x
11 oo : 02 : 04 --=
p+ l
B e-x
e- x p = C -e - x
= dy/dx and so
Now p
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Diff eq and Sp funs
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-x
dy dx
e
C - e-x y = ln ( C - e-x )
+D
= ln ( C - e-x ) - ln(C - 1), since we require y(O) = 0, =
C -e- x ln - - C- 1
This is as far as y can be det ermined since only one boundary condition is given for a second-order equation. As C is varied the solution generates a family of curves satisfying the original equation. A variety of other forms of solut ion are possible and equally valid, t he actual form obtained depending on where in t he calculation the boundary condit ion is incorporated. They include
11 oo : 02 : 01
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3
Diff eq and Sp funs
0 DE of second and higher order
If f (x) = 0 t hen the equation is called homogeneous; otherwise it is inhomogeneous. In order to solve any equation of t his form we must first find the general solution of the complementary equation, i.e. t he equation formed by setting f (x) = 0.
dny an (x) dxn
dn-ly + an- 1 ( x) d xn- 1
dy + · · · + a1 ( x) dX
+ ao (x)y = 0
If t he n solutions of the equation are Y1 (x), Y2 (x) , .. . , Yn (x), t hen t he general solut ion is given by the linear superposition
11 oo : 02 : 1o If t hen solutions of the equation are Y1 ( x), Y2 (x), ... , Yn (x), t hen t he general solution is given by the linear superposition
If the equation has f (x) =/- 0 (i.e. it is inhomogeneous) then Yc(x) is only one part of the solution. The general solution is t hen given by y(x) = Yc(x) + Yp(x) [email protected]
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Diff eq and Sp funs
where yp(x) is t he particular integral, which can be any function that satisfies the equation (3. 1), provided it is linearly independent of Yc(x) .
3.1
Linear equations with constant coefficients
If the am in equation ( 3 .1 ) are constants rather than funct ions of x t hen we have
This is called linear equation with constant coefficient and are very common in Physics.
Finding CF: We do find the CF by putting y = Ae>-x. The auxiliary equation then
11 oo : 02 : 12 Inear equa I n are ve1~y common in Physics.
Wl v......
Finding CF: We do find the CF by putting y The auxiliary equation t hen
= A e.\x.
This equation have n roots. We t hen have t hree main possibilities.
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(i) All roots real and distinct. The CF is then .
(ii) Some roots complex. If one of the roots of t he auxiliary equation is complex , say a + i/3, then its complex conjugate a - i/3 is also a root. The solut ion for that part icular two roots
This can also be written as in two other very useful form y
= eax (c1 sin f3x + C2 cos f3x) ;
y
= ceax sin(,Bx + 1')
(iii) Some roots repeated If, for example, ..\ 1 occurs k t imes (k > l ) as a root of t he auxiliary equation, t hen we
11 oo : 02 : 1 s y
= ceax sin(,Bx + --y)
(iii) Some roots repeated If, for example, A1 occurs k t imes (k > l ) as a root of the auxiliary equation, then we need to find k - l solutions that are linearly indep endent of those already found and also of each other. Complementary function is then
Example:
Solve
I y'' - 6y' + 9y = 0 I
49
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Diff eq and Sp funs
Solution: We write the equation as
(D
2
-
6D
+ 9) y =
0
or
(D - 3)(D - 3)y
=
0
Since t he roots of the auxiliary equation are equal, we write t he result y
= (Ax+ B) e3x
Finding PI: There are many methods to find t he PI. (i) Substutution of trial functions. (ii) D operator method. (iii) Variation of parameter .
11 oo : 02 : 11 (i) Substutut ion of trial functions. (ii) D operator method. (iii) Variation of parameter. (iv) Inverse laplace and fourier transformation. (v) Green 's function method. Among these t he first two methods are useful for some particular functions in the RHS, not for all. The last three methods are more powerful and can be applied for many different types of functions in t he RHS of the different ial equation. [email protected]
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Diff eq and Sp funs
Firstly, we discuss the trial solution method. We are attempting to solve equations like t his
(1) If f (x) = aerx then try
IYp(x) = berx I (2) If f (x) = a 1 sin rx ) t hen t ry
+ a 2 cos rx (a 1 or a2
may be zero
11 oo : 02 : 20 IYp(x) = berx I (2) If f (x)
= a 1 sin rx + a 2 cos rx (a 1 or a 2 may be zero
) t hen t ry
IYp(x) = b1 sin rx + b2 cosrx I (3) If f (x) = ao + a1x + · · · + aNxN (some a,,.,i may be zero) t hen try
(4) If f (x) is t he sum or product of any of t he above t hen try Yp ( x) as the sum or product of the corresponding individual trial functions. Now we go into more details with examples j [email protected]
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Diff eq and Sp funs
ICase - 11We are trying to solve eq like (D - a)(D - b)y
=
F (x)
=
kecx
Take PI
cecx C xecx C x2ecx
if c is not equal t o eit her a or b if c equals a or b, a i- b
(3.3)
if c =a= b
ICase - 21To find a particular solution of (D -
a) (D -
11 oo : 02 : 22 C x ecx C x 2 ecx
if c equals a or b, a =I= b if c = a = b
(3.3)
ICase - 21To find a particular solution of ( D b)y =
a) (D -
k sin ax k cos a x
First solve ( D - a) (D - b)y
= k eio:x, then t ake real or
imaginary part .
- 3 I A particular solution Yp of (D - a)(D - b)y = ecx Pn(x) where Pn(x) is a polynomial of degree n is I Case
ecxQn(x) xecxQn(x) x 2ecxQn(x)
if c is not equal to eit her a or b if c equals a or b, a =I= b if c = a = b
(3.4) [email protected]
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Diff eq and Sp funs
@Sk J ahiruddin ) 2020
where Qn(x) is a polynomial of the same degree as Pn(x) with undet ermined coefficients to be found to sat isfy t he given different ial equation.
Example: Solve
ly'' + y' - 2y = l 8x ex I
11
oo : 02 : 2s
Example: Solve
ly''
+ y' -
2y
= l 8xex I
Solution: We write the eq as (D - l )(D + 2)y = l 8xex We have a= l , b = - 2, c = 1; also Pn(x ) = l 8x = P1 (x) is a polynomial of degree 1. We also see that c = a -=I= b. So follow equation (3.4) to see that t he form of t he solution is
We substitute this into the cliff eq to find A and B
+ B) 2 y; = ex (Ax +Bx+ 4Ax + 2B + 2A) 2yp = ex(6Ax + 3B + 2A) = l 8xex 2
y~ = ex (Ax +Bx+ 2Ax
y; + y~ -
So we must have
6A
=
18, 3B + 2A
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=
0,
or 53
A = 3 ' B = -2 ' physicsguide CSIR NET, GATE
Diff eq and Sp funs
@Sk J ahiruddin, 2020
Hence the solut ion
Example Solve I y''
+ y' -
2y = x
2
-
xI
Solution: We take v,., = Ax + B x + C. substituite the 2
11 oo : 02 : 2a
Example Solve I y''
+ y' -
2y
=x
2
-
xI
Solution: We take Yp = Ax + B x + C, substituite the t rial function into the cliff equation and get the particular solution Yp = _ x2 + 1) 2 2
!(
Example Solve d2
dx~
+ 4y = x
2
sin 2x
Solution: CF is easy m 2 + 4 = 0
Ye ( X )
= C1e2ix
+ C2e -2ix
= d1 cos 2X
· 2X + d2 sin
we first guess at a suitable trial function for t his case should be
( ax
2
2
+ bx + c) sin 2x + (dx + ex + f) cos 2x
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Diff eq and Sp funs
However, we see that t his trial function contains t erms in sin 2x and cos 2x , both of which already appear in t he complementary function .
11 oo : 02 : 30 However, we see t h at t his trial function contains t erms in sin 2x and cos 2x , b oth of which already appear in the complementary function. We must t hen multiply the trial function by the smallest integer p ower of x which ensures t h at n one of the resu lting terms appears in CF. So t he new t rial function is
3
2
2
3
(ax + bx + ex) sin 2x + (dx + ex + f x) cos 2x Put t he t rial function into t h e diff equation and get
Yp(x) =
x3 - - cos 2x 12
x2 + - sin2x 16
x + - cos 2x 32
The general solut ion is t hen
2
3
x x x sin 2x + cos 2x = d 1 cos 2x + d2 sin 2x cos 2x + 16 32 12
Example:
Solve
y'' + y' - 2y = (D - l )(D + 2)y =(ex)+ (4 sin2x) [email protected]
@Sk J ahiruddin, 2020
55
+ (x
2
-
x)
physicsguide CSIR NET, GATE
Diff eq and Sp funs
We see the solut ions separately. Then add.
11 oo : 02 : 33 @Sk J ahiruddin , 2020
Diff eq and Sp funs
We see the solutions separately. Then add.
(D - l )(D
+ 2)y = ex
(D - l )(D has PI Yp2
=-
1
5
has PI
Yp1
l X = 3x e
+ 2)y = 4sin 2x
cos 2x -
3 .
5
sin 2x
1 2 (D - l )(D + 2)y = x - x has PI yp3 = - (x + 1) 2 Adding these three solutions, we see that 2
Yp
= Ypl + Yp2 + Yp3
=
l x eX - l cos 2x - 3. 1( 2 ) sin 2x x +l 2 5 3 5
Example: Solve
(D
2
+ 2D + l 7)y = 60e- 4 x sin 5x
Solution: The Complementary function is easy m 2 + 2m + 17 = 0. Hence C.F = e- x(A sin 4x + B cos 4x) To get the part icular integral solve ( D 2 + 2D + l 7)y = 60e(- 4x+ i 5x) first and t hen t ake only the imaginary part of [email protected]
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1' T
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1
11 oo : 02 : 3s [email protected]
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t he solution. Now as know
J( D)y =
eax
Hence the P.I is the imaginary part of e-4x+i5x
(x(-4 + 5i))2 + 2(x(-4 + 5i)) + 17 Do some algebra to arrive at t he answer .
3.2 3.2.1
Variable coefficients and other forms Euler and Legendre form
T his type of equation called Legendre form. ( a, (3 and t he an are constants ) T his type of equation can be solved by making t he substit ut ion a x + (3 = et . We then have dy dx j [email protected]
a
dt dy dx dt
dy ax + (3 dt 57
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1
f"1
r
11 oo : 02 : 38 dx
ax + f3 dt
dx dt
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and
d dy dxdx
a2
(a x
+ /3)2
d2y dt 2
dy dt
A special case of Legendre's linear equation, for which a = 1 and /3 = 0, is Euler 's equation ,
ndny anx dxn
dy + · · · + a 1 x dX
+ aoy = f (x)
(3. 6)
it may be solved in a similar manner to the above by substituting x = et . And if f (x) = 0 you can substit ute y = x>- which will give you same solution . We will give you an example now
Example Solve
2d2y x dx2
dy + x dx - 4y = 0
Solution: First we make t he subst itution x = et , which, after cancelling et, gives an equation with constant coefficients,
d dt
d -- 1 y dt
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+
dy dt - 4y = 0 58
d2y - 4 = 0 dt 2 y physicsguide CSIR NET, GATE
11 d dt
d -- 1 y dt
+
00:02:41
dy dt - 4y = 0
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58
⇒
d2y - 4 = 0 dt 2 y
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The general solution is t hen ,
Since gives
f (x) =
0 here, you can substitute y
= x>- which
•
This has the solutions ,\ eral solution
3.2.2
= ± 2, so we get t he same gen-
Variable coefficient of order 2
Consider the equation
d2 y dx 2
dy + a1(x) dx
+ ao(x)y = f (x)
We focus special case when the coefficients a0 ( x) and
a1 (x) are upto order 2, means they are algebric functions and can contain terms uoto x 2 not of hiQ:her orders.
11 oo : 02 : 43 dx-9
+ a1 x dX + ao x y = f x
We focus special case when the coefficients a0 (x) and a 1 ( x) are upto order 2, means they are algebric functions and can contain terms upto x 2 not of higher orders. [email protected]
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TO solve the equation,
substit uite
y
= u(x)v(x);
1 -2
u(x) = exp
Solve for unkonwn v, t hen put the value of v in the value of y. Example: Solve
Solution: Divide t he equation by 4x 2 . We get 2
d y dx-?
l dy
l
2
+ -X dX + 4X 2 (x -
We get a 1 (x) = l /x and a0 (x) f (x) = 0. Now substit uite y
= vu = vexp
1) y = 0
= (x
2
-
Av
vx
1) / 4x
2
and
11 oo : 02 : 46 We get a1 (x) = 1/x and a0 (x) f (x) = 0. Now substituite
(x
2
-
1) / 4x
2
and
Av
y = vu = vex p
yx
We get
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T his equation is easily solvable.
=
V
1
C1
sin 2X
1
+ C2 cos 2X
So t he solut ion is y = -
V
yx
3.2.3
=
+
· 1 1 C1 Sln X C2 COS x 2 - - - - 2---
yx
Dependent variable y missing.
P ut
y1
= P,
II
I
y =p
Then the equation reduces to first order. Example:
Solve
d2 y dx2
dy + 2 dx = 4x
Solution: This is transformed by t he substitution p =
00: 02: 48 d2 y dx2
Solve
dy + 2 dx = 4x
Solution: This is transformed by the substitution p = dy / dx to t he first-order equation
dp dx
+ 2p = 4x 61
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This is now linear equation. Solution is
p
=
dy dx
=
ae- 2x
+ 2x -
l
where a is a constant . Thus by direct integration t he solution to the original equation
3.2.4
Independent variable x missing.
Put
y
I
= P,
dp dp dy dp y = dx = dy dx = P dy II
Example:
Solve
dy dx
2
=0
11 oo : 02 : s1 Example:
dy Solve
dx
2
=0
Solution: Put dy/dx = p and d2 y/dx 2 = p(dp/dy) and hence get t he first order equation
dp 1 + YP dy
+ P2 = 0
62
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This is separable. The technique to solve has been discussed before. Solution is
Using p = dy /dx we get
dy p =-=± dx
cy -y2
Simplifying we get
3.2.5
Variation of parameters
y2
11 oo : 02 : s3
3.2.5
Variation of parameters
Variation of paramet er is a very powerful technique in finding particular integrals for linear OD Es with variable (and constant) coefficients. We need to know all of the complementary functions to use variation of parameters to find the PI. General technique. We need to find the solution of, dny an(x) d
+ xn
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dy · · · + a1 ( x) d X
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Diff eq and Sp funs
The CF is
We transfer the constants to unknown functions and thus assume PI of this particular form
We choose k~ (x)y1(x) + k; (x) y2 (x) k~ (x) y~ (x) + k; (x) y; (x)
And so on ....
+ · · · + k~(x)yn(x) = 0 + · · · + k~(x) y~ (x) = 0
11 oo : 02 : s6 k~ (x)y1(x) k~ (x) y~ (x)
+ k;(x)y2(x) + · · · + k~(x)yn(x) = 0 + k; (x) y; (x) + · · · + k~(x) y~(x) = 0
And so on ....
2 k~(x)yin- )(x)
2 + k;(x)y~n- )(x)
2 + · · · + k~(x) yti- )(x) = 0 an X
So, remember that t he unknown coefficients ~rill be always as derivatives. The order of derivatives of t he CF's will be increased by every equation. Notice t he last equat ion carefully. RHS has been changed ...! We will give example of 2nd order different ial equation. Say you have
d2y dx2
dy + P (x) dx
+ Q(x)y = J(x) 64
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The CF of t he equation is
We then take t he PI as
Then we get two equations
+ k;(x )y2 (x) = 0 k~ (x )y~ (x) + k;(x )y; (x) = J (x)
k~ (x )Y1 (x)
('
1
11
7
\/
1
7/ /
\
('
I
1
I
11 oo : 02 : sg + k;(x )y2(x) = 0 k~ (x )y~ (x) + k;(x )y;(x) = f (x) k~ (x )Y1 (x)
Now you can find t he k(x )' and k~(x) from t hese two equations and t hereaft er by integrating find k1 and k2. Example: solve
Use the variation of parameters method to
d2y dx 2
+ y = cosec x
subj ect to the boundary conditions y(O) = y(w/2) = 0
Solution: The CF is Yc(x) =
c1
sin x
+ c2 cos x
We assume t he PI
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Now look at t he general procedure to solve
ki (x) sin x + k~ (x) cos x = 0 ki (x) cos x - k~(x) sin x = cosecx Solving these equations for k~ (x) and k~ (x) gives
k~ (x) = cos X CSCX = cot X k~ (x) = - sin x csc x = - 1 TT
•
.
'
,
1
I
\
,
11 oo : 03 : 02 k~(x) k~ (x)
cotx
= COSXCSCX = = -
sin x csc x
=
- 1
Hence, ignoring the constants of integration, k1 ( x) and k2 ( x) are given by k 1 ( x) = ln (sin x)
k2(x ) = -x The general solution is t hen y (X)
= [C1 + ln (sin X) ]sin X+ (C2 - X) cos X
Apply the boundary conditions y(O) find c1 = c2 = 0 and so
y (x)
= ln (sin x) sin x
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=
y(n/2)
=
0 we
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Diff eq and Sp funs
Example: Given that one solution of t he following
differential equation is x
Find t he other solution.
Solution: Solution:
We need to substitute y
11 oo : 03 : 04 Find the other solution.
Solution: Solution: We need to substitute y = u(x )v(x) as another solution. So, we get y' = xv'+ v, y'' = xv'' + 2v' . So the differential equation becomes
x 3 ( xv'' + 2v') + x (x v' + v) - xv = 0 or 2 4 3 x v'' + (2x + x ) v' = 0 Separating the variables and integrating, we get
dv'
2
1
X
2 X
-+
v'
dx,
1
In v' = - 2lnx + -
X
+ lnK
Solving for v' , integrating again, and writing y gives
uv
•
67
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v'
K l/x X
2e
physicsguide CSIR NET, GATE
'
v = - Kel fx,
y = - K xel/x
Thus the general solution of t he given equation is y A x+ B xe 1 /x
=
11 oo : 03 : 06 v =
X
e 2
,
v
= - Ke ,
= - K xe
y
Thus the general solution of t he given equation is y = Ax+ B xe 1/x We can find the second solut ion in another way using Wronskian which will learn now.
3.2.6
Wronskian: Linear independence of functions and Second solution
If f 1 (x) , f 2 (x), · · · , fn(x) have derivatives of order n- 1, and if t he determinant
W=
!1(x) f{ (x) !{' (x)
! 2(x) f~(x) !~' (x)
• •
• • •
•
• • •
• • •
• • • • •
fi n- l)(x) JJn- l)(x)
fn(x) f~(x) Ji(x)
(3.7)
• •
•
•
• • •
t o
JJn- l) (x)
Then the functions are linearly independent . The deterj [email protected]
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minant W is called the Wronskian of the functions. It should be noted, that W = 0 does not guarantee t hat t he functions are linearly dependent. Only if the Wronskian •
'
,
•
,
1
I
1
,,
r
11 oo : 03 : 09 minant W is called t he Wronskian of the functions. It should be noted, that W = 0 does not guarantee t hat t he functions are linearly dependent. Only if the Wronskian is zero over the ent ire range of the variable, t hen the funct ions are linearly dependent over t his range Example: Show that t he functions 1, x, sin x are linearly independent. Solution: We write and evaluate t he Wronskian, 1
W =
X
•
Sln X
0 1
cosx
0 0
•
=-
•
Sin X
- Sln X
since - sin x is not ident ically equal to zero, t he functions are linearly independent . Example: Check linear dependency of t he t hree funct ions cp 1 = ex , 0 vanish, and we obtain the indicial equation 1
11 oo : 04 : 40
L
(n + a) (n + a - 1)
n=O
+ 2(n + a ) + 4x
11 a nX '
=0
If we set x = 0 t hen all terms in the sum with n > 0 vanish, and we obtain the indicial equation 1 a (a-1) + a = O 2 j [email protected]
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This has roots a = 1/ 2 and a = 0. since these roots do not differ by an integer, we expect to find two independent solutions to the equation. T he coefficients of xn vanish separately
If we choose the larger root, a = 1/ 2, of the indicial equation then previous equation becomes - an -1
an = - - - -
2n(2n + 1)
Setting ao = 1, we find an = (- 1)n/ (2n + 1) !, and so the solution becomes
11 oo : 04: 42
oo
(- l )n
vx- (VX) 3!
=
3
n
+ (VX) 5!
5 -
·· · =
To obtain the second solut ion we set a [email protected]
102
sin VX
= 0 (the smaller
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Diff eq and Sp funs
root of t he indicial equation) which gives an-l an= - - - - 2n(2n - 1)
Setting a0 = 1 now gives an = (- l )1i/(2n) !, and so the second (independent ) solution becomes
00 (
y 2 (x) = ~ -
l )n
L..J (2n) !
(Jx)2
(y14)4
2!
4!
xn = l - - - + - -
n=O
- · · · = cosVX
We check that y1 (x) and y2(x) are linearly indep endent by computing the Wronskian:
vV = Y1Y; - Y2Y~
= sin VX 1
= -2VX
1
- 2VXX sin VX ('
Slll
2
cos VX
1
VX cos VX 2 X
c. 2 C.) 1 y X + cos y X = -2VX-:/ 0
11 1
= sin VX 1
= - 2,/x
oo : 04: 4s
- 2VXX sin VX ( .
2
r::.
Sln y X
+ COS
2
1
cos VX
VX cos VX 2 X
C.) 1 y X = - ,/x-/ 0 2
since W # 0, the solutions y1(x) and y2(x) are linearly independent. Hence, t he general solution is
y(x) =
c1
sin VX + c2 cos VX
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Example: Find series solut ion about x = 0 11
x(x - l )y
+ 3xy + y = 1
0
Solution: Dividing through by x (x- 1) to put the equat ion into standard form, we obtain II
3
1
I
y + (x- l )y + x(x- l )y=O We see p(x) = 3/(x - I ) and q(x) = 1/[x(x - 1)] . As x = 0 is a singular point but xp(x) = 3x/(x - 1) and x 2 q(x) x / (x - 1) are finite there, it is a regular singular point We put CX)
y
= x(J
L an x n n= O
Take derivative and get
3
00
\. ' r~ , _\~ ~n+u-1
11 oo : 04 : 48
Take derivative and get 3 00 2 L (n + a )(n + a - l )anxn+a- + x _ l L (n + a)anxn+a-l n=O n=O 00
l
00
n+a
Divide by xa- 2 , and get oo
L n=O
3x X (n + a )(n + a - l ) + --(n +a) + - x- l x- l
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multiply by x - l to get 00
L [(x - l )(n + a )(n + a - l ) + 3x(n +a) + x]anxn = 0 n=O
If we set x = 0 t hen all t erms in t he sum with t he exponent of x greater t han zero vanish, and we obtain t he indicial equation
a(a - 1) = 0 which has the roots a = l and a = 0. since the roots differ by an integer, it may not be possible to find two linearly independent solut ions. We are guaranteed, however, to find
11 oo : 04 : so which has the roots CJ = l and CJ = 0. since t he roots differ by an integer, it may not be possible to find two linearly independent solut ions. We are guaranteed , however, to find one such solution corresponding to the larger root, CJ = l. The coefficient s of xn vanish separately we get t he recurrence relation
+ CJ) ( n - 2 + CJ ) an-1 - (n + CJ) ( n + CJ + 3(n - l + CJ )an- 1 + an-1 = 0
(n - l
l ) an
simplify to get
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On substit ut ing
CJ
= l into this expression, we obtain n +l
n and on setting ao = l we find an = n + l ; so one solut ion is given by
00
Y1 (x) = x
L (n + l ) xn =
x ( 1 + 2x
+ 3x + •••) 2
n=O X
(1 -
X)2
,
11 oo : 04: s3 Y1 (x) = x
L (n + l ) xn =
x (1 + 2x
+ 3x + •••) 2
n=O X
(1 -
X) 2
If we attempt to find a second solut ion (corresponding to the smaller root of t he indicial equation) by setting a = 0 we find
n
n -1 But we require ao -=/- 0, so a1 is formally infinite and the method fails. We can find a second solution by Wronskian.
Example: Find a second solution of t he above equation by Wornskian method
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Solution: As we have seen t he method of obtaining a second solution in subsection 3.2.6 One solution we have already derived X
Y1 ( X)
So now using
= (1 - X) 2
11 oo : 04: 55 So now using
X
X
Y2 ( X)
(1
- 2 X2
= (1 - X) 2 X
X
X)2
(1 -
X
(1 -
X)2
exp -
X2) 4 - -- exp [-3 ln(x2 2 X2
3 ---dx1 X1 - 1 - l )]dx 2
- 1
X2
ln x
1
+-
X
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4.2
Special equations and functions
4.2.1
Legendre equation and Polynomials
This is t he Legendre equation (1 - x
dx2
--2-dx2
X)2
(1 -
X2
(1 -
x X2
X
)4
X2
2
)
y'' - 2xy'
+ /!, (/!, + 1)y = 0
x = 0 is an ordinary point. So substitute
11 oo : 04 : sa This is t he Legendre equation
(1 - x
2
)
y'' - 2xy' +f(f + l )y
=0
x = 0 is an ordinary point. So substitute
You get CX)
L
2 [n(n - l )anxn- - n(n - l )anxn - 2nanxn + f(f + l )anxn]
n =O
=0 From here 00
L {(n + 2) (n + l )an+2 -
[n(n + 1) - f (f + l )]an } xn = 0
n=O
We now get the recurrence relation
[n(n + 1) - f(f + 1)) an+2 = (n + l )(n + 2) an [email protected]
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for n = 0, 1, 2, ... . If we choose a0 we obtain the solution
=
1 and a 1
0 then
=
-
...
11 oo : os : oo
-
.. .
whereas on choosing a0 = 0 and a 1 = 1 we find a second solution
y2(x) = x-(R- 1)(£+ 2)
x3
x5
+(R- 3)(R- l )(R+ 2) (R+4) ! -· · · 31 5
In many physical applications the paramet er Rin Legendre's equation is an integer, i.e. R = 0, 1, 2, .... In t his case, the recurrence relation gives
_ [R(R + 1) - R(R + 1)] a i+2 -
(R+ l )(l + 2)
_ ai -
0
t he series terminates and we obtain a polynomial solut ion of order R . These solutions, after normalization, are called the Legendre polynomials of order R they are written Pi(x) and are valid for all finite x . It is conventional to normalise Pi(x ) in such a way t hat Pi(l ) = 1 and as a consequence j [email protected]
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Pi( - 1) = (- 1)i . The first few Legendre polynomials are easily constructed and are given by
Po(x) = 1, P1 (x) = x 2 P2(x) = ~ (3x - 1) , P3(x) = ~ (5x 3 - 3x ) 4 2 P4(x) = ~ (35x - 30x + 3) , Ps(x) = ~ (63x 5 - 70x 3 + 15x)
11 oo : os : 03 •
easily constructed and are given by
Po(x) = 1, P1(x) = x 2 3 P3(x) = ½(5x - 3x) P2(x) = ~ (3x - 1), 4 2 3 5 P4(x) = ~ (35x - 30x + 3) , Ps(x) = ~ (63x - 70x + l 5x) First few Legendre polynomials are shown here graphically. 2
y
Po
- 1
1
-0.5
X
- 1
-2
Second solut ion of the Legendre equation is Qe(x ) is, an [email protected]
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infinite series that converges only for
lxl < 1
l +x l- x
11 oo : os : 06 IxI
2
11 oo : os : 29 find
[(a
[(a+ 1)
2
2
an
+ n)
-
v
2
]
v
-
2
] a1
=0
+ an- 2 = 0
for n > 2
Substituting a = ±v you get
(1 ±2v)a 1 = 0 n(n±2v)an+an-2= 0
forn > 2
We consider now the form of the general solut ion to Bessel's equation for two cases: (i) v is not an integer [email protected]
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(ii) v is an integer (including ZERO)
( i) v is not an integer We need to set 1
an = - n(n ± 2v) an-2
=0 It is a general convention (have physical insight which you will learn while doinf advanced electrostatics problems) t o set 1 a0 -- ') ±vl' / 1 -!- /JI\
11 oo : os : 32 It is a general convention (have physical insight which you will learn while doinf advanced electrostatics problems) to set 1 ao = 2±vr (1 ±v) Then we write the solut ion as 1
l v(x) = r (v + 1) X
2
V
l X 1--v +l 2
2
1
1
X
+ (v +l)(v+ 2) 2!
2
oo (- l )n x =L n!f(v+n+ l ) 2
4 -
...
v+2n
n=O
(4.1) We replace v by -v which give us J_v(x)
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The functions l v (x) and J_v (x) are called Bessel funct ions of the first kind, of order v . You must know that Bessel functions of half-integer order are expressible in closed form in terms of trigonometric functions. Example: Calculate
J±1;2(x) .
Solution: Use t he general formula of Bessel function for non integer v as derived above. You get
11 oo : os : 34 Example: Calculate J±1; 2 (x) .
Solution: Use t he general formula of Bessel function for non integer v as derived above. You get
Using the fact that r(x + 1) find that , for v = l / 2
=
121
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xr (x) and
r (½)
=
fa,
we
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Diff eq and Sp funs
(½x)5/2
(½x)9/2
1,r (~) + 2,r (;) - · · · (½x)5/2 (½x)9/2 1! ( ~) (t) fa + 2! ( ~) ( ~) (1) fa - ... x2 x4
1 - - + - - ··· 3! 5!
fT
11 oo : os : 36 -x 2
-x 2
-x 2
(~) fi - 1! (~) (~) fi + -2,-(~-)-(~-)-(t_)_fi_1r - ... (t x)1/ 2 x2 x4 1 - - + - - ··· (~) fi 3! 5! 2
1x ) 1/ 2 Sln . X
(2
(~) \j'ff
.
-s1nx
X
7rX
For v = -1/ 2 we get
2 - cosx 7rX
(ii) v is an integer Firstly, let us consider the case v = 0, so that t he two solutions to the indicial equation are equal, and we clearly obtain only one solution in the form of a Frobenius series. j [email protected]
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Use equation 4. 1 to get (- l )nx2n
oo
x2
x4
x6
= 1 - 22 + 2242 - 224252 + ... In e-eneral. however. if v is a oositive intee-er then the
11 oo : os : 39 n =O
22nn!I'(l x2
= 1-
22
+ n)
x4
+
x6
2 24 2 -
2 24 25 2
+ ...
In general, however , if v is a positive integer then the solutions of the indicial equat ion differ by an integer. For t he larger root , o-1 = v, we may find a solution l v (x), for v = l , 2, 3, ... , in the form of the Frobenius series. First three Bessel functions are plotted below J(x) 1
0.5
- 0.5 123
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For the smaller root, o-2 = -v, the recurrence relation (n(n ± 2v )an+ an-2 = 0 for n > 2) b ecomes
n(n - m)an + an-2 = 0 IYYl
-
')
I) !
.
, c,
Ti l'"\TT T
t"'ITI
CIT TCI TI
for n > 2
.+.
'Y"\ l'"\C, 1
1T TCI
.
, n + orro r
.
1
Cl
IYYl
-
11 oo : os : 42 For t e sma er root,
(n(n
± 2v)an + an- 2 = 0
= -v, t e recurrence re ation for n > 2) becomes
2
m = 2v is now an even posit ive integer , i.e. m = 2, 4, 6, .... Starting with ao -=/=- 0 we can calculate a2, a4, a5, ... , but when n = m t he coefficient an is infinite, and the method fails to produce a second solution . Replace v by -v in the definition of Jv(x) given in equat ion 4. 1, you will be able to show t hat, for integer v,
So, Jv(x) and J_v (x) are linearly dependent. In this case, we cannot write the general solution to Bessel's equation in t he form y(x)
= c1 Jv(x) + c2 J-v(x)
So we define the function
Yv(x) = Jv(x) cos_v1r - J_v(x) Sln V1r
(4.2)
This is called a Bessel function of the second kind of order v. It is also called Weber or Neumann function. As [email protected]
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Bessel's equation is linear , Yv(x) is clearly a solut ion , since it is just t he weighted sum of Bessel functions of the first kind.
11 oo : os : 44 Bessel's equation is linear , Yv(x) is clearly a solut ion, since it is just t he weight ed sum of Bessel functions of the first kind. It can be shown that t he Wronski an of Jv(x) and Yv(x) is non-zero for all values of v . Hence Jv(x) and Yv(x) always constitute a pair of independent solutions.
Example: If n is an integer , show that
Solution: From equation 4.2 Yn+ l / 2(X )
=
Jn+1;2(x) cos (n + ½) 7f
If n is an integer , cos (n
-
. ( + 1) Sln n ?. 1f
J-n-1;2(x )
-
+ ½) 1r =
0 and sin ( n
+ ½) 1r =
(-l)n
The Neumann function (equation 4.2 ) becomes an indeterminate form 0/ 0 when v is an integer , as for integer v we have cosv1r = (-l )v and J_v(x) = (-l) vJv(x) . This j [email protected]
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indeterminate form can be evaluated using l'Hopital's rule
11 oo:os:47
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Diff eq and Sp funs
indeterminate form can be evaluated using l'Hopital's rule . Therefore, for integer v, we set
Jµ(x) cos µ1r - J_µ(x)
Yv(x) = lim - - -.- - - µ ---tv
Sl n
µ 7r
which gives a linearly indep endent second solution for t his case. Thus, we may write the general solut ion of Bessel's equation, valid for all v as
The functions Y0 ( x), Y1 ( x) and Y2 ( x) are plotted in figure below
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Y(x)
Yo
0.5
ox
1
-0.5
-1
Finally, in some applications, it is convenient to work with complex linear combinations of Bessel functions of t he first and second kinds given by
t hese are called, respectively, Hankel functions of t he first and second kind of order v .
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11 oo : 05 : 53 •
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Orthogonality: b
x l v(ax)Jv(f3x)dx = 0
for a-:/
/3
a
Recurrence relations: Try to prove these relations. You may not need to remember all. But prove yourself at
least once.
(i) d~ [xv Jv(x )] = Xv l v- l (x) d (ii) dx [x- vl v(x)] = -x- vl v+1(x)
(iii) x J~(x)
+ v l v(x) = x l v-1 (x)
(iv) xJ~(x) - v l v(x) = - xlv+1(x)
(v) l v-1(x) - l v+1(x) = 2J~(x) (vi) l v-1 (x)
+ l v+1 (x) =
2v
- Jv(x) X
Relation (i) and (ii) can also be written as integral form
Jxv l v-1(x )dx = xvl v(x) Jx-v l v+l (x )dx = -x-v l v(x) How to prove these? Let me give you some hints. Relation (i) and (ii) comes directly from the definition
11 oo : 05 : 55 How to prove these? Let me give you some hints. Relation (i) and (ii) comes directly from the definition
128
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Diff eq and Sp funs
of Bessel function. The prove of relation (i) is t he following.
00
=
L
n= O
(- l )nx2v+2n-1 2v+2n-ln!f(v+n)
00
=
l )nx(v- l )+2n
( -
x v l v- 1(x)
Expand out the derivative on t he LHS of (i) and divide by xv-I , you obtain relation (iii) Similarly,expand the derivative on t he LHS of (ii) and divide by x v+I , you get relation (iv) Add relation (iii) and (iv) to get (v) Subtract (iv) from (iii) to get (vi) 1 2
Example: You are told t hat = (2/ 1rx) 1 sin x 1 2 and J_ 1; 2 (x) = (2/ 1rx ) / cos x,. Now express J3; 2 (x) and J1; 2 (x)
J_ 3; 2 ( x) in trigonometric functions.
11 oo : 05 : 5a Subtract (iv) from (iii) to get (vi) Example: You are told t hat J1; 2 (x) = (2/ 1rx ) 112 sin x and J_ 1; 2 (x) = (2/ 1rx ) 1/ 2 cos x,. Now express J3; 2 (x) and
J_ 3; 2 ( x) in t rigonometric functions. Solution: From relation (iv) 129
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1;12(x)
J3;2 (x) = ~ J1;2(x) 2
2
1
2x 2
2
1/ 2 •
Sln X -
1fX 1/2
1 .
- Sln X
1fX
1
1/ 2 COSX
1fX
+-
2x
2 1fX
1/2 •
Sln X
COS X
-
X
From relation (iii)
J_3;2(x) = - ~ l-1; 2(x) + J~1;2(x) 2
1 2x 2 1fX
2 1fX 1/2
2
1/ 2
cosx -
1 - - COS X
-
1fX
1/ 2
1 • Sln X - 2x
2 1fX
1/2
cosx
. Sln X
X
We see that, by repeat ed use of t hese recurrence relat ions, all Bessel functions Jv(x) of half integer order may be expressed in terms of trigonometric functions. So as the
11 oo : 06 : oo We see that , by repeat ed use of these recurrence relat ions, all Bessel functions l v (x) of half integer order may be expressed in terms of trigonometric functions. So as the Yv(x) . Generating function: The Bessel functions l v (x), where v = n is an integer, can be described by a generating [email protected]
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funct ion in a way similar to that discussed for Legendre polynomials
G(x, h) = exp
00
h- .!.
X
2
(4.3)
h
n=-oo
Integral representations: Bessen function can be written as integral for n an integer. 7f
1
Jn(x) = -
7r
cos( n0 - x sin 0)d0 0
Special case n = 0 J0 (x) =
4 .2.5
1
21r
exp(ixsin0)d0
n
2
O
21r
1 =
exp (ix cos 0)d0
n
2
Spherical B essel functions
0
11 oo : 06 : 03 21r 4.2.5
21r
O
0
Spherical Bessel functions
When solving Helmholtz' equation (V + k ) u = 0 in spherical polar coordinates, t he radial part R(r) of t he solut ion satisfy the equation 2
131
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where f is an integer. You see that This equation looks very much like Bessel's equation and can be reduced to it by writ ing R(r) = r - 1!2 S(r), (Do t his !) in which case S(r) then satisfies 1 f+2
Doing change of variable x we obtain
=
2
S= O
kr and letting y ( x)
1 f+2
=
S (kr),
2
y= O
where t he primes now denote d/ dx . This is Bessel's equat ion of order f + and has as its solutions y(x) = J p_+ 1; 2 (x) and Yt+ 1; 2 ( x) . The general solution of spherical Bessel equa-
1
11 oo : 06 : os
where t he primes now denote d/ dx . This is Bessel's equa-
t
t ion of order f + and has as its solut ions y(x) = Jt+ 1; 2 (x) and Yt+ 1; 2 (x). The general solution of spherical Bessel equat ion (4.4) can then be written
1 2
1 2 /
The functions x- ! Jt+i ; 2 (x) and x- ½+ 1; 2 (x), after normalization, are called spherical Bessel functions of t he
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first and second kind, respectively, and are defined as follows:
For f
=0 •
. ( )
Jo
X
=
Sln X X
Recurrence relations:
t ion is given by
COS X
n0( x) = - - x t he f th spherical Bessel func-
11 oo :06 : oa t he I!, th spherical Bessel func-
Recurrence relations:
t ion is given by I
l d
xdx
fo(x)
where fg(x) denotes either jg(x) or ng(x) First two spherical Bessel functions are
= sin x x2 n (x) = _ cosx
. (x) J1 1
X
_
.'
s1n x
x2
j a [email protected]
X
= (l.. _ l ) sin X _ 3 cos x x3 x x2 3 n2(x) = - (; 3 - ; ) cos x - ~~ x
(x) . J2
cosx '
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4.2.6
Diff eq and Sp funs
Hermite equation and function
Hermite equation is y'' - 2xy'
+ 2v y = 0
This has an essential singularity at x = oo. The paramet er v is a given real number , although it nearly always takes an integer value in physical applications. The Hermite equation appears in the descript ion of the wavefunction of t he harmonic oscillator. Any solution of Hermite equat ion is called a Hermite function.
11 oo : 06 : 1 o The Hermite equation appears in the description of the wavefunction of t he harmonic oscillator. Any solution of Hermite equat ion is called a Hermite function . Do t he series solut ion about x == 0 yourself. First few Hermite polynomials are
Ho(x) == 1, H 1 (x) == 2x, 2 H2(x) == 4x
2
-
H3(x) == 8x - 12x H4(x) == 16x 4 - 48x 2 + 12 2, H5(x) == 32x 5 - 160x 3 + 120x
The Rodrigues' formula for t he Hermite polynomials is given by Rodrigues' formula:
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Orthogonality: 00
- 00
Generating function:
Recurrence relations:
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Recurrence relations:
Hn+1(x) = 2x Hn(x) - 2n Hn-1 (x) H~(x) = 2n Hn-1 (x)
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4.3
Exercises
4.1. Find a series solution of Hermite equation about x
y'' - 2xy' + 2vy = 0 choose a0 = (- l )nl 2n !/(n/2)! , for odd n take 1
2
(n- l ) !
=0
11 oo : 06 : 16 choose a0 = (- 1)nl 2n !/ (n/2) !, for odd n take
4.2. Solve Laguerre's equation about x
=0
xy'' + (1 - x )y' + vy = 0 choose a0
=1
4 .3. Find power solutions in z of t he differential equation
about z = 0
+ 9z 5y =
zy'' - 2y'
0
Ident ify closed forms for the two series, calculate their Wronskian , and verify that they are linearly independent . Compare the Wronskian wit h t hat calculated from t he different ial equation. 4.4. Investigate solut ions of Legendre's equation at one of
its singular points as follows.
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(a) Verify t hat z = I is a regular singular point of Legendre's equation and that t he indicial equation for a series solution in powers of ( z - I) has a double root a = 0 . (b) Obtain the corresponding recurrence relation and show t hat a polynomial solut ion is obtained if R is a positive integer. (c) Determine t he radius of convergence R of t he a = 0
11 oo : 06 : 1 a solution in powers of ( z - l ) h as a double root a
=0
.
(b ) Obtain t he corresp onding recurrence relation and show t h at a polyn omial solut ion is obtained if R, is a p osit ive integer. (c) Determine the radius of convergence R of the a
=
0
series and relat e it to the posit ions of the singularities of Legendre's equation. 4.5. ( a ) Find series solutions of the equat ion y'' - 2zy' - 2y
0. Identify one of the series as y 1 ( z)
=
=
2
exp z and verify this
by direct substitution. (b ) By setting y2 (z) = u( z )y1 (z ) and solving the resulting equ ation for u(z), find an explicit form for y2 (z) and deduce t h at X
I
(X)
~
n . ?. +I - - - - (2x)~n
n= O
2(2n
e- v dv = e- x ~ 2
2
o
+ 1) !
4.6. Find the radius of convergence of a series solution ab out
t he origin for t he equation ( z
+ az + b) y'' + 2y =
2
0 in the
following cases:
a = 5 b = 6·
'
'
(b) a = 5, b = 7 2
Show t h at if a and b are real and 4b > a then the radius of convergence is always given by 6112 . [email protected]
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+ z-
3
y = 0, show t he origin b ecomes a regular singular point if the indep endent variable is ch anged from z t o x = l/ z . Hen ce find a series solution of the form y 1 (z) = anz-n .B y setting Y2( z ) = u(z )y1 (z ) an d expanding t he result ing expression for du/ dz in p owers 4. 7. For t he equation y''
~r
11 oo : 06 : 21 4. 7. For t he equation y'' + z- 3 y = 0, show t he origin be-
comes a regular singular point if the indep endent variable is changed from z to x of the form Y1( z ) =
=
l / z . Hence find a series solution
I:r anz-n.By setting Y2( z) = u(z )y1(z)
and expanding t he result ing expression for du/ dz in powers of z-
1
,
show that y 2 (z) is a second solut ion with asymptotic
form y2 ( z)
=
1 c z + In z 2
+0
ln z z
where c is an arbitrary constant 4.8. Use t he explicit expressions of Spherical Harmonics as
given inside the notes to verify for
.e = 0, 1, 2 t hat
and so is independent of the values of 0 and for any
cp. This is true
.e , but a generatly proof is more involved. This result
helps to reconcile intuit ion with t he apparently arbitrary choice of polar axis in a general quantum mechanical system. 4.9. Express the function 2
2
2
f(0 , cp) = sin 0 [sin (0 / 2) cos cp + i cos (0 / 2) sin ] +sin (0 / 2) as a sum of spherical harmonics. jahir@physicsguide. in
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4.10. Use the generating function for t he Legendre polynomials Pn (x) to show t hat
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Diff eq and Sp funs
4.10. Use the generating function for t he Legendre polynomials Pn (x) to show t hat
(2n) ! o P2n+1(x)dx = (- 1) 22n+ln!(n + 1)! n
l
and except n = 0 1
P2n(x)dx = 0 0
4.11. Do step by step to prove the result
2
1
Pn(z) Pn(z)dz = - -1 2n + 1 (a) Square both sides of the generating-function definition of the Legendre polynomials, (X)
(1 - 2zh + h
2
)-
112
=
L Pn(z) hn n=O
(b) Express the RHS as a sum of powers of h, obtaining expressions for the coefficients (c) Integrate the RHS from -1 to 1 and use the orthogonality property of t he Legendre polynomials. (d) Similarly integrate t he LHS and expand t he result in powers of h. (e) Compare coefficients.
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4.12. By substituting y(x) as x 112 _f (x) reduce Stokes' equa-
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2 4.12. By substitut ing y(x) as x 1 f (x) reduce Stokes' equation, d2y dx2 + Axy = 0 1
to Bessel's equation. Hence show t hat a solut ion t hat is 1 2 finite at x = 0 is a mult iple of x / J1; 3 j ~ 4.13. The hypergeometric equation is
x( l - x)y'' + [c - (a+ b + l )x]y' - aby = 0 Solution of this equation is
ab x
+ F (a, b, c; x) = 1 + - c 11.
=
f (c) r (a)f (b)
2 a(a + l )b(b + 1) x
f n=O
( l) + · · · 21. cc+ f (a + n)f (b + n) xn r (c + n)
n!
F (a, b, c; x) is known as t he hypergeometric function or hypergeometric series. Now, Identify the series for the following hypergeometric functions, writing them in terms of better-known functions. (a) F (a , b, b; z) (b) F (l , 1, 2;-x) (c) F (½, 1, ~; -x 2 ) 2 (d) F ( ~, ~, ~; x ) [email protected]
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4.14. If y(x, a) is a solut ion of the equation
1 2 -x + a y 4
=0
Determine which of the following are also solutions: (i) y( a, -x), (ii) y( -a, x), (iii) y( a, ix) and (iv) y( -a, ix)
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4 .3 .1
Ans keys
2 2m ([n/ 2] denotes t he inte4 .1. '\:""[ri/ ](l )m n! (2x)nD m=O m!(n-2m)! ger part of n/2)
4.2. I:: =o(- l )m (m!)2(~- m)!Xm
4 .3.
Y1 =
a 0 sin z 3 and y 2 = b0 cos z 3
4.4. t he series converges in a circle of radius 2 cent red
on z
=
l.
4 .5. ( b ) ex2 A
Je- v dv 2
4.6. (a) R = 2
4 7
• • YI
( )Z
-
ao
(b)R = v'7 '\:""oo (-l)n D n=O (n+ l )(n!) 2 zn
4.8. This is a proof. 4.9. ~
Yao -
.!. Y,0 3 1
4 .10. This is a proof. 4 .11 . This is a proof. 4 .12. This is a proof.
.1. y - 1 15 2
11 oo : 06 : 34 4.10. This is a proof. 4.11. This is a proof. 4.12. This is a proof. j [email protected]
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4.13. (a) (1 - z)-a, (b) ; ln(l x- 1 sin- 1 x
Diff eq and Sp funs
+ x),
4.14. (i) Yes (i) No (iii) No (iv) Yes
(c) x- 1 tan- 1 x, (d)
11 oo : 06 : 36
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4.3.2
Diff eq and Sp funs
Solutions
Solution: 4.1. Since x
0 is an ordinary point of t he
equation, put 00
substituting you get 00
L [(m + l )(m + 2)am+2 + 2(v -
m)am] xm = 0
m=O
Demanding that t he coefficient of each power of x vanishes, we obtain the recurrence relation
a 2 m+
2(v - m) - - - - - - - -a
-
(m + l )(m + 2)
m
In nearly all physical applications, the parameter v takes integer values. Therefore, if v = n, where n is a non-negative integer, we see that a n+ 2 = a n+4 = · · · = 0, and so one solution of Hermite's equation is a polynomial of order n. Fo1~ even n, it is convent ional to choose a0 = (- l )nf 2n!/(n/2) !,
11 oo : 06 : 39 teger values. Therefore, if v = n, where n is integer, vve see that an+2 = an+4 = · · · = 0, lution of Hermite's equation is a polynomial even n, it is conventional to choose a0 = (whereas for odd n one takes
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a non-negative and so one soof order n. For 2 1)n/ n !/ (n / 2) !,
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Diff eq and Sp funs
These choices allow a general solution to be written as
Hn(x) = (2x)n - n(n - 1)(2x)n- l
+ n(n -
l )(n - 2)(n - 3) ( x) 71__ 4 _ 2 2! [n/2] I = ~ (-l)m n. (2x)n-2m ~ m !(n- 2m)!
...
m= O
where Hn(x) is called the nth Hermite polynomial and the notation [n/2] denotes the integer part of n/2. We should note t hat Hn(-x) = (- l )nHn(x) .
Solution: 4.2. Since t he point x = 0 is a regular singularity, put 00
y(x) =
L
amXrri+a
m=O
Substit ute t his into the equation and divide t hrough by xa- l, we get 00
11 oo : 06 : 42 y(x) = .__ amx m=O
Substitute this into the equation and divide through by xa- l , we get 00
L ((m + a )(m + a - l ) + (1 - x)(m + a )+ vx]amxm = 0 m=O
Setting x = 0, so that only the m = 0 term remains, we obt ain the indicial equation a 2 = 0, which trivially has a = 0 as its repeated root. Thus, Laguerre's equation has only [email protected]
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one solution and it, in fact , reduces to a simple power series. Substituting a = 0 into above equation and demanding that t he coefficient of xrri+l vanishes, we obtain the recurrence relation m-v am+l = (m + l )2 am In nearly all physical applications, the paramet er v takes integer values. Therefore, if v = n , where n is a non-negative integer, we see t hat an+l = an+2 = · · · = 0 , and so our solution to Laguerre's equation is a polynomial of order n. It is conventional to choose a0 = l , so that t he solution is polynomial
•
1
1 \n __
,l
11 oo : 06 : 44 po ynom1a
Ln(x)
n
Check that first few Laguerre's polynomials are
Lo(x) = 1
L 1 ( x)
= - x +1
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L 2 ( x)
146
= (x
2
-
4x
+ 2) / 2!
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Solution: 4.3. Putt ing the equation in its standard form shows t hat z = 0 is a singular point of the equation
but , as -2z/ z and 9z 7 / z are finite as z ➔ 0, it is a regular singular p oint . We t herefore substit ute a Frobenius type solut ion , 00
y(z ) = zo-
L anzn wit h ao -/= 0 n=O
We get 00
00
n=O
n=O 00
+9L n=O
anzn+o-+5 = 0
11
00:06:47
n=O
n=O 00
+9L
5 a nzn+a-+
=0
n=O
Equating the coefficient of za-- l to zero gives the indicial equation as a(a - l )a0
-
2aao
=0
a = 03
'
These differ by an integer and may or may not yield two independent solut ions. The larger root , a = 3, will give a solution; the smaller one, a = 0, may not.
(a) a = 3 j [email protected]
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2
Equating the general coefficient of zm+ to zero (with a = 3) gives
(m + 3)(m + 2)am - 2(m + 3)am + 9am - 6 = 0 So the recurrence relation is am=⇒
9am - 6
m(m + 3) 9
a5 = - - - - -a 5 P
6p(6p + 3)
6 = -
p-
The solution is t hen 00
111
(.r,)
=
Q,n z
3)
a6p- 6
2p(2p + 1)
11 oo : 06 : 49 9 a5 = - - - - -a 5 P 6p(6p + 3) p -
⇒
-1 Pao
a6p- 6
6 = -----
2p(2p + 1)
(2p + 1)!
The solution is t hen
=
(b)
o-
oo
(- l )n 3 2 a0 ~ - - - - z ( n+l )
~ (2n + 1)!
=
a0 sin z 3
=0
Equating t he general coefficient of a-= 0) gives
[email protected]
zm-l
to zero (with
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Diff eq and Sp funs
Hence t he recurrence relation is 9am - 6
am= - - - - -
m(m- 3) 9
a 6P = -
a 6p-6
6p(6p - 3) a 6p- 6
= - 2p(2p - 1)
So the solution
oo (- l )n 6n
3
11 oo : 06 : s2 e so ution oo (- l )n 6n
3
We see that a = 0 does, in fact, produce a (different) series solution. This is because the recurrence relation relates an to an+6 and does not involve an+3 the relevance here of considering the subscripted index m + 3 is t hat 3 is the difference between t he two indicial values . We now calculate the Wronskian of the two solutions, Y1
= ao sin z 3 and Y2 = bo cos z 3
W (y1, Y2)
:
= Y1Y~ - Y2Y~ 2 3 3 = ao sin z ( -3boz sin z ) 2 = -3aoboz i= 0
jahir@physicsguide. in
149
-
bo cos z
3
( 3aoz
2
cos z
3
)
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Diff eq and Sp funs
The fact t hat the Wronskian is non-zero shows t hat the two solutions are linearly independent. We can also calculate the Wronskian from the original equation in its standard form, 2 ' y '' - -y
z
as
+ 9z 4 y = 0
11 oo : 06 : s4 equation in its standard form,
y
11 -
2 1 4 - y + 9z y = 0 z
as
W = C exp
z
-
- 2
-
u
du
2
C exp (2ln z) = Cz
=
This is in agreement with the Wronskian calculated from t he solutions, as it must be.
Solution: 4 .4 . (a) In standard form, Legendre's equat ion is ii 2z i R( I!+ l ) - 0 y l - z2y + l - z2 y This has a singularity at z
-2z(z - 1) l - z2
➔
=
l , but, since
d R(f! + l )(z - 1) 1~ l -z2
2
➔
0
asz ➔ l
bot h limits are finite, the point is a regular singular point. We next change the origin to the point z j [email protected]
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Diff eq and Sp funs
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u
=
z - l and y(z) 111 _
= l by writing
=
j(u) . The t ransformed equation is
2(u + 1)
- u(u + 2)
1 1
+
/!(/! + l ) y = 0 - u(u + 2)
or
-u(u + 2)f
11 -
2(u + l )f
1
+ R(R + l )f =
0
11 oo : 06 : s1
or
-u(u + 2)f'' - 2(u + l )f'
+ f(f + l )f =
0
The point u = 0 is a regular singular p oint of this equation and so we set f(u) = u(T I:~=oanun and obtain 00
-u(u + 2) L (a
+ n)(a + n
- l )anu s0 )
I 1(s)
I so
C
c/s
0
ctn
en!/ sn+l
0
2
2 a ) 2 a )
eat tneat
a/ (s + 2 s/ (s + 1/(s - a) n! /(s - a)n+l
- : ........ 1-.. _ ..,_
~
sin at cos at
I ( ~2
~.2\
0 0
a a
I I
.
11 oo : oo : 26 a/ (s + a ) 2 2 cos at s/ (s + a ) eat 1/(s - a) tneat n! /(s - a)n+l 2 2 sinh at a/ (s - a ) coshat s/ (s 2 - a 2 ) sin at
1.3 1.3.1
2
2
0 0 a a la la
Properties of Laplace 'Iransformation Linear property
Laplace Trans£ormation is linear
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1.3.2
physicsguide CSIR NET I GATE
Integral Transformations
Shifting property
We must know t he shifting property of Laplace transformat ion CX)
0 CX)
J(t)e- (s- a)tdt . n
(1.1)
11 oo : oo : 29 00
£ [eat f (t )]
=
f (t )eate- stdt 0 00
f (t )e- (s- a)t dt
(1.1)
0
= f (s - a) 1.3.3
Laplace transformation of derivatives
Laplace t ransform of first derivative
£
oo
df dt
d+
_'.I
0
dt
e- st dt
(1.2)
= [f (t)e -st]; + s 0
= - f (O) + sf(s),
for s > 0
Laplace transform of second derivative
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Integral Transformations
2
= s f (s) - sf (O) General nth Derivative
dt (0),
for s > 0
(1.3)
11 oo : oo : 31
General nth Derivative
nf n-If() n-2df( ) dn-Ij () =s -s O -s - 0 - ··· - - - 0 dt dtn-I for s > 0
(1.4)
1.3.4
Laplace transform of integral t 0
1.3.5
f( u)du = ~£ [! ] s
(1.5)
Laplace transformation of Heavyside function
if t < 1 if t >
0
H (t - c) =
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12
C
C
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@Sk J ahiruddin , 2020
Integral Transformations
Then £ { H (t - c) f (t)} =
e - cs £{ f
(t + c) };
_c-I {e-cs f (s) } = H (t - c) f (t - c)
(1.6)
11 oo : oo : 34 + c) };
£ { H (t - c)f (t)} = e- cs £{ f (t
(1.6)
£ - 1 { e- cs f( s) } = H (t - c)f (t - c)
Example: Write t he following function in terms of Heavyside function and find its Laplace t ransformation
f(t) =
2
if O < t < 1
lt2
if 1 < t < ~ 7r if t > ½1r
2
cost
Solution: In terms of heavyside function 1
f (t) = 2(1 - H (t - 1)) + t
2
2
+ (cos t) H
H (t - 1) - H
1 t - -7r 2
1 t - 1r 2
Indeed, 2(1 - H (t - 1)) gives
f (t)
for O < t < 1, and so
on We must write each term in f (t) in t he form j(t-a) H (ta) . T hus, 2(1 - H (t - 1)) remains as it is and gives t he t ransform 2 ( 1 - e- s) / s. Then
j [email protected]
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~
1 2 - t H (t - 1) 2
13
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Integral Transformations
00: 00: 37 Integral Transformations
1 2 - t H (t - 1)
2
2
1
=2
1
2
-(t - 1) + (t - 1) + 2 2 1
H (t - 1) = 1 2H -t 2
1
1
e + + s 3 s 2 2s
- s
1 t - - Jr 2
1 1 t - -Jr 2 2 1
7r
7r2
s3
2s
8s
_ +_2 +_
2
1 t - -Jr 2
7r
+ -2
1 t - -1r 2
H
e - 1rs/2
and
=
1 (cos t) H t - - Jr 2 1 2 - sin t - - Jr 2
1
H t-
1
- 1r
2
s
2
e - 1rs/2
+1
so t oget her
Y (J) = 2 - 2 e- s + s
s
1
s3 [email protected]
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we know
7r
1
s3 7r2
+ -2s-2 + -8s 14
1
1
+ -s2 + -2s
e
e- 1rs/2 _
l s2
-s
+1
e - 1rs/2
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Integral Transformations
11 oo : oo : 39 @Sk J ahiruddin , 2020
Integral Transformations
we know
2 {J (t - a)H(t - a)} = e- as F (s) write f (t - a) = g(t ), hence again write f for g So you get
f (t)
=
g(t + a) and then
2{J(t) H (t - a)}= e- as 2{J(t + a )} Thus
1 2 -t H (t - 1) 2
as before. Similarly for 2 { ~ t 2 H (t - ~ 1r) } . Finally, by 2 {J (t) H (t - a) } = e-as2{J(t + a)} we get
2
costH
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t-
1
1r
2
?
,...2
=
e-1TS /
=
e - 1rs/ 22
=
-e- 1rs/2
15
COS
1 t + -Jr 2
{ - sin t} 1
s2 + 1
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Integral Transformations
11 oo : oo : 42 15
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1.3.6
Int egral Transformations
Some more properties
£ [J (at)] =
1- s
-f a
a
for n
= 1, 2, 3, .. .
(1.8)
00
j(t) t 1.3. 7
(1. 7)
j (u)du
(1.9)
s
Examples:
Example: Find the Laplace transform of J(t) = t sin bt Solution: Using equation (1.8)
-
f (s)
= (-
d
d
b
1) ds 2 [sin bt] = - ds
32
+ b2
2bs
for s > 0 Example: (1r/S) l/2 j [email protected]
1 2
Prove L [t 1 ]
= ~(
16
112 3 1r / S )
and L [t - 112 ]
=
physicsguide CSIR NET, GATE
11 oo : oo : 44 Example: (1r I S )l/2
1 2
Prove L [t 1 ]
j [email protected]
~
=
112 3 ( 1r / S )
16
and L [t- 112 ]
=
physicsguide CSIR NET, GATE
@Sk J ahiruddin , 2020
Integral Transformations
Solution: We know t he standard integral 00
exp (-x
2
1
dx
)
0
= \in 2
2
Put x = ts in that integral. Hence 2xdx dx = sdt/ (2\l'st), we get OO
sdt and
Is In e- st _V _ t -l f2dt = _v_ 0
11
2
0
2 00
⇒
2
t - l/2 e- stdt
t - 1/2
7r
=
s
0
Integrating the LHS of t his result by parts gives
e-st2t1; 2
00
00
0
7r
st (-s) e- 2t 112 dt =
s
0
The first term vanishes at both limits, and the second is a multiple of the required Laplace transform of t 112 . Hence,
2
00
tl/2 0
1
7r
2s
s
e- sttl/ 2dt = -
Example: Use the propert ies of Laplace transforms to prove the following without evaluating any Laplace integrals
11 oo:oo:47 e 0
dt=2s
s
Example: Use t he properties of Laplace transforms to prove the following without evaluating any Laplace integrals [email protected]
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Integral Transformations
explicitly:
.z [t5/2]
= 1/fos- 7/2 (b) .Z [(sinh at) / t] = ~ In [(s + a)/ (s - a)) , s > a 1 2 2 2 2 2 (c) .Z[ sinh at cos bt] = a (s - a + b ) [ ( s - a) + b J-
(a)
[(s + a) 2 + b2] - 1 Solution: (a) We use the general result for Laplace transforms that for n
=
l, 2, 3, . ..
If we take n = 2, then f (t) becomes t 112 , for which we found t he Laplace t ransform in t he previous example that 1 2
L [t 1 ]
=
3 112 ½(1f / S ) .
_z t5/2 = _z y0r 2
3 2
So Now t2tlf2
= (- 1)2 d2
ds 2
__5 s -1/ 2 = _15_y0r_1f s-7/2 2
8
(b) Here we apply a second general result for Laplace transforms which states that
11 oo : oo : 49 (b) Here we apply a second general result for Laplace t ransforms which states that 00
J(t) t
f(u )du s
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provided limt➔o[f (t )/t] exist s, which it does in t his case. 00
sinh (at)
t
s
00
1 2
=
a 2 2du, u - a 1
u-a s+ a s- a '
s
1 - ln 2
u> al 1
du
u+a s > la
(c) T he t ranslation prop erty of Laplace t ransfo rms can be used here t o deal with the sinh(at) factor , as it can be expressed in terms of exp onent ial funct ions:
.
2[s1nh (at) cos(bt)]
=2
1
eat
2
cos(bt) - 2
s-a l s+a - 2 ( s - a) 2 + b2 2 ( s + a)2 + b2 l
1
(s2 -
a 2 ) 2a + 2ab2
2 [(s - a)2 + b2 ] [(s + a)2 2 2 2 a (s - a + b )
+ b2 ]
[(s - a) 2 + b2] [(s + a) 2 + b2]
11 oo : oo : s2 1 2 [(s - a) 2 + b2 ] [(s + a) 2 + b2 ] 2 2 2 a (s - a + b )
[(s - a) 2 + b2 ] [(s + a) 2 + b2 ] The result is valid for s > a
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1.4 1.4.1
Integral Transformations
Convolution Convolution of two functions
Convolut ion of t he functions f * g is defined by
f
and g which is written as
X
f(x) * g(x) =
f(t)g(x - t)dt
(1.10)
0
Example Find convolution of 1/ vt and t 2 Solution: Here 1
f (t) = -Ji so
1
f(x) = .,fi H PnrP
and
and
g(t) = t
2
g(t - x) = (t - x)
2
11 oo : oo : s4 f (t ) = ,Ji so
J (x)
=
1
Jx
g(t)
and
9(t - x)
and
=
=
t
2
(t - x)
2
Hence
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1
,./i * t
2
=
Integral Transformations
f (t) * g(t)
= t
x-
1 2 /
[t
2
+x
2
-
2tx
-
2tx 112 + x 312 dx
]
dx
x=O t
t 2 x - 1l 2 x=O
=
2 2 t t22x 1; 2 - 2t-x3/2 + - x 5/ 2 3 5 x=O
= 2t2 . tl/2 -
it . t3/2 + 2 t5/2 = 3
5
16 t5/2
15
You can prove the following properties for convolution
f*9 = 9 * f
f * (91 + 92) = f (j * 9) * V = j
* 91 + f * 92
* (9 * V)
(1.11)
11 oo : oo : s1 f*9 ==9*f
f * (91 + 92) == f * 91 + f * 9 2 (j * 9) * V == f * (9 * V)
(1.11)
f*O==O*f==O 1.4.2
Convolution theorem for Laplace transformation
If .sf{f(x) } == F (s) and .sf{9 (x) } == G(s), then
.sf{f (x) * 9(x) } == .sf{f(x)}.sf{g (x) } == F (s)G(s) (1.12) j [email protected]
21
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Integral Transformations
If t he functions and 9( s) t hen
f
and 9 have Laplace transforms
f (s)
t
f (u)9(t - u)du == f (s)g(s)
(1.13)
0
where the integral in t he brackets on t he LHS is t he convolution of f and 9 denoted by f * 9. The convolution defined above is commutative, i.e. f * g == g * f, and is associative and distributive. We also see t hat t 1
.sf- [J(s)g(s)] ==
f (u)g(t - u)du == f * g 0
From the definition of Laplace Transformation
11 oo : 01 : oo t 1
2 - [J(s)g(s)]
f (u)g(t - u)du = f * g
= 0
From the definition of Laplace Transformation 00
J(s)g(s) =
00
e-suJ(u)du 0
e- stg(v) dv
0 00
00
dve-s(u+'u)J (u )g(V)
du 0
0
Now letting u + v = t changes t he limits on the integrals, with t he result that 00
J(s)g(s) =
00
dtg( t - u )e- st
duf (u) 0
u
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Integral Transformations
Now, look at the picture below. The shaded area of integration may be considered as t he sum of vertical strips. However, we may inst ead integrate over t his area by summing over horizontal strips as shown in figure . Then the integral can be written as I
I
I = II
, ll
I
I
/ / / / /
/ / /
/ /
/ / /
~ II
ll
11 oo : 01
: 02
ll
(h)
t
J(s)g(s) =
ll
00
dtg(t - u)e- st
duf (u) 0
0
t
00
dte- st 0
f (u)g(t - u)du 0
t
J(u)g(t - u)du 0
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1.5
23
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Integral Transformations
Inverse Laplace transformation
Example : Find inverse Laplace transformation of s+3
f (s) =
s(s + 1) -
3
Breaking into partial fraction f(s) = - 2 s - -. Hence J(t) = 3 - 2e- t s+ l
Solution:
Example: Find
11 oo : 01 : os .
2 - -. Hen ce s+ l
rea 1ng into part1a
ract1on
s
f (t) = 3 - 2e- t
Example: Find
3s - 137
L-l s
2
+ 2s + 401
Solution:
f
3(s + 1) - 140 = £-l (s + 1) 2 + 400 s+ l = 3_2-l
20
_7_2-l
(s + 1)2 + 20 2
Hen ce using first shifting t heorem we get
f (t) = e- t(3 cos 20t -
j [email protected]
24
7 sin 20t)
physicsguide CSIR NET, GATE
Integral Transformations
@Sk J ahiruddin , 2020
Example: Find the inverse t ransformation of
se-4s
(3s
+ 2)(s -
2)
Solution : We rewrite t he function as e - 4s (3s
s
+ 2)(s - 2) = e-4s F (s)
11 oo :01 : oa Solution : We rewrite t he function as s
- 4s _ _ _ _ _
e
(3s
+ 2) (s - 2) = e
- 4s p () 8
Now write F (s) into partial fraction
A
B
F (s) = 3s+ 2 + s- 2 s(A + 3B) + (2B - 2A) (3s + 2)(s - 2)
A(s - 2) + B (3s + 2) (3s + 2)(s - 2)
= 1/ 4
Hence we find A= B Now
F (s) =
-1 4
3(s+~ )
+
-1 4
s- 2
Inverse Laplace t ransformation of F (s) is
J(t) =
1
2t
- e - ""f
12
25
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1
+ -e
2t
4
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@Sk J ahiruddin , 2020
Integral Transformations
The main problem was
Y (s) = e- 4s F (s) So now using t he property of LT of hevisid e function we get
](s) = H (t - 4)f(t - 4) F,y::. rnnlP~ Finrl t h P in, rPr ~P
T,.: :i nl .:::irP tr.:::in~fn r m.:::i t.inn nf
11 oo : 01
: 11
So now using the property of LT of heviside function we get
f (s) = H(t -
4)f(t - 4)
Example: Find t he inverse Laplace transformation of 1 d ----an
s (s 2 + w 2 )
1
s 2 (s 2 + w 2 )
Solution: WE have from t he basic definitions 1 sin wt 2-1 s2 +w2
w
So using t he property, equation ( 1. 9)
2 -1
1
t
s (s 2 + w 2 )
o
sin wt --dT
=
W
1 W
(1 cos wt) 2
Using this result an applying the property again we get 1 t
1 w2
(1 - cos wT)dT = 0
Sln WT
w2
w3
@Sk J ahiruddin , 2020
T
Sln WT
w2
w3
t 0
•
t [email protected]
•
26
physicsguide CSIR NET, GATE
Integral Transformations
We could have got t he result by breaking t he funct ion in partial fractions also.
1.6
Solving Differential equation by Laplace
11 oo : 01
: 13
We could have got t he result by breaking t he function in part ial fractions also.
1.6
Solving Differential equation by Laplace transformation
Example: Solve using t he concepts of Laplace t ransformation y1 - 5y
=
y(O) = 0
e5x;
Solution: Taking t he Laplace transform of both sides of this differential equation we find t hat
T hen we obtain
[sY(s) - O] - 5Y(s) =
1
1 _ 8
Y( s) = (s - 5)2
5
Finally, taking t he inverse transform of Y ( s), we obtain
y(x)
1
2 - {Y(s)} = 2 -
=
j [email protected]
1
1
(s - 5) 2
27
=
xe5x
physicsguide CSIR NET 1 GATE
@Sk J ahiruddin, 2020
Integral Transformations
Example: Solve using the concepts of Laplace transformation y
II _
y - t, _
y(O)
=
1,
1
11 (0)
=
1
11 oo : 01
: 1s
@Sk J ahiruddin, 2020
Integral Transformations
Example : Solve using the concepts of Laplace transformation
y'' - y = t,
y(O) = l ,
y' (0) = 1
Solution: Taking Laplace transformation both sides we get, [ with Y = 2(y)] 2
s Y - sy (0) - y' (0) - Y ~~>
= 1/ s
2
2
(s - l)Y =s+ l + l /s
,
2
Y =s+ l + 1 s2 - l s 2 ( s 2 - 1) Simplification of the first fraction and an expansion of t he last fraction gives Y =
1
s- l
+
s2
1
1
- 1
s2
we obtain
y(t) = 2 - 1(Y )
=2-1
1
+2-1
s-l = et + sinh t - t_
1
s2
l
-
-2-1
1 s2
Example : Solve the set of equations
y' - 2y + z = 0 z' -y - 2z = 0 [email protected]
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Integral Transformations
By using the technique of Laplace transformation. Given
11 oo :01
:1
a
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Integral Transformations
By using the t echnique of Laplace transformation. Given t he init ial conditions y 0 = 1, z 0 = 0
Solution:
Taking Laplace transformation of both the equations and calling L(z) = Z and L (y) = Y pY - Yo - 2Y + Z
=0 pZ - zo - Y - 2Z = 0 Substituting the initial condit ions
(p - 2)Y + Z = 1 Y - (p - 2)Z = 0 We get by simply solving t hese two equations by algebric method
p-2
Y = ---· (p - 2) 2 + 1 ' Hence y
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2
2
z = e t sin t
= e t cost;
29
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Integral Transformations
11 oo : 01 [email protected]
: 20
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1. 7
Integral Transformations
Exercises
1.1. Find
(a)£ { e- 2x sin 5x}
(b )£ {x cos V7x}
1.2.
sin 3x
(b)£ x1;2
(a)£ { e- xx cos 2x}
X X
(d)£
sinh 2tdt 0
1.3. Find X
0
l 4 -e- t sin 3tdt t
1.4. Find
s
(a) £-1 s
(c) £-1
2
+6
s+l s2
-
9
5s
(b) £-1 (d)
.z-1
s2
-
2s + 9
1.5. Find
(a)
.z-1
(c)
.z-1
s
s 2 - 3s
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(b)
.z-1
s+4 s 2 + 4s + 8
+4 30
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11 oo : 01
: 23
s+ 2 (c) _z- 1 s2 - 3s + 4 j [email protected]
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30
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Integral Transformations
1.6. Use partial function to decompose
1
(a) (s + l ) (s 2 + 1) s+3 (c) (s - 2) (s + 1) 1. 7. Find
(a).z-1 (c) _z- 1
s+3 (s-2) (s + l ) 1
(b) _z- 1
1
1.8. Find J(x) * g(x ) when (a) J(x) = e3x and g(x) = e2x
(b) J(x)
=
x and g(x)
=
x2
1 6 1 1 1.9. Find (a) .z- , (b) .zs 2 - 5s + 6 s2 - 1 convolut ions
by
1.10.
Find .Z{g(x) } if
0 x< 4 (x - 4)2 X > 4 '
(a) g(x) =
(b) g(x) =
0
x
•
X
2
X
4
11 oo : 01 : 2s (x - 4) 2 (b) g(x)
0 2 x
=
X X
31
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X
>4
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Integral Transformations
1.11. Solve t h e initial value problem by using t he con cept of Laplace transformation 9 =, 0 y+y+y II
1
y(O) = 0 .16,
I
y (0)
=0
1.12. Solve the following differential equation by using t he con cept of Laplace transformation
y
I
+ y = s1n x; •
y(O ) = 1
1.13. Solve by using t he concept of Laplace transformation 8 sin t
+ 9y =
yll
if O < t < if t >
0
1r
1r
1
The initial conditions are y(O) = 0, y (0) = 4
1.14. Solve by using the con cept of Laplace transformation W
y
+ y = Sln X
I
1
.
-
z = ex
1
z +w+y= l w(O) = 0, y(O) with t he boundary condition
=
1,
z (O)
=
1
11 oo :01 : 2a z'+w+y= l w(O) = 0, y(O)
=
1,
z (O)
=
1
=
1,
z(O)
=
1
with the boundary condition
w(O ) = 0,
y(O)
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1.15.
w'' - y + 2z = 3e-x -2w' + 2y' + z = 0 2w' - 2y + z' + 2z'' = 0
with t he boundary condition
w (0)
=
1,
w' (0)
=
1,
y ( 0) = 2,
z (0)
=
2,
z' (0)
= -
2
11 oo : 01
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1. 7.1
Ans keys
1.2. (a)
(d)
2
(s+1) -4 [(s+1) 2 +4]2 '
(b)
105 r,;; -9/2 16 V 7r S '
(c) =
Z!: 2 -
arctan §..3,
s(s; - 4)
1.3.
2(s - 4) 2
1
v18
s 3 (s - 4) 2 arctan 3 + (s - 4) (s 2 + 9)
1
1r
-
1.4. (a) cos v'6x, (b) ~xsin x, (c) cosh 3x+~sinh 3x, (d) ex sin y'8x
1.5. (a) e2xcos 3x+je 2xsin3x, (b) e- 2xcos 2x+e- 2xsin2x, (c) e(3/ 2)x cos {; x + v'7e(3/ 2)x sin {; x
11 oo : 01
: 34
2 2 2 2 1.5. ( a) e x cos 3x+je x sin 3x, (b) e- x cos 2x+e- x sin 2x, (c) e( 3/ 2)x cos {; x + ,/7e(3/ 2)x sin x
f
1.7. (a) ~e 2x - ~e-x, (b) ½e-x - ½cosx + ~ sin x, (c) 2x cos 2x + 1 e- 2x sin 2x - .i. cos x + 1sin x + -1.e65 65 65 130 1.8. (a) e3x - e2x' (b) 112X4 1.9. (a) e3x - e2x [email protected]
,
(b) 3ex - 3e- x 34
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1.11. e- 0 -5t(0.16 cos 2.96t + 0.027 sin 2.96t) 1.12. ~e-x - ½cosx + ½sin x 1.13. sin t -1 sin 3t + [sin (t - 1r) - ; sin(3(t - 1r))] u(t-
1r) +; sin3t 1.14. 1 - ex, ex + sin x, cos x 1.15. w(x) = ex
y(x) ==ex + e- x z(x)
=
2e- x
11 oo : 01
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: 36
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@Sk J ahiruddin , 2020
1. 7.2
Solutions
Solution: 1.1. (a) Setf (x) =sin5x
F (s)
=
.2'{f(x) }
=
.2'{sin5x} =
5 s 2 + 25
t hen apply shifting property (1.1)
(b) Set f (x) = cos v!7x s s2 + 7
t hen apply equation (1.8) you get
11 oo : 01 (b) Set f (x)
: 39
= cos v!7x
F(s)=2{cosv'7x}=
s
3
../7 s2 + ( 7)2
s
2
+7
t hen apply equation (1.8) you get
s
d -2' { x cos v'7x} = - ds
s
2
+7
Solution: 1.2. (a) Let f(x) = xcos2x. You can easily see t hat s2 - 4 F(s) - - - - (s2+ 4)2 Then use shift ing property
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Integral Transformations
(b) Define f (x) =
VX. Then
7 2
x1 =x
3
VX = x
3
f (x)
You can easily see that
1 2 3 F( s) = .2'{f(x) } = -2'{ VX} = ~s- / 2 t hen apply equation (1.8) you get
-2' {x3VX} = (c) Taking
f (x) = sin 3x,
F (s) •
d3 (- 1)3-3 ds
, .
3
s
2
+9 /-a
r.\
l
-~s- 3/2
= 105 ~s- 9/2
2
16
then or
F (t ) -
3
t
2
+9
11 oo : 01 (c) Taking J(x)
F (s) -
: 42
= sin 3x, then 3 s2
F (t ) -
or
+9
3
t2 + 9
Then using equation (1.9) we get 00
sin3x X
s
3
t2
+9
dt = lim
R-too
lim arctan
R-too
= lim
R-too
t
R
3
8
arctan
s = - - arctan2 3
R
3
-
3
R
s
t2 + 9 dt
s arctan 3
7r
(d) Taking J (t ) = sinh 2t, t he Laplace transformation is
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Integral Transformations
Then using Laplace t ransformation property of integral, equat ion ( 1.5), we get X
sinh 2tdt 0
1 s
2 s2
-
2 4
s (s 2
-
4)
Solution: 1.3. We have already proved
.2 sin3x X
Use shifting theorem for a
7r
=- 2
s arctan -
3
= - 4 in equation (1.1), on t his
11 oo : 01
: 44
Solution: 1.3. We have already proved
sin3x
2
1r
=- 2
X
Use shifting theorem for a result to get 1
-
e- 4 x
s arctan 3
= - 4 in equation (1.1) , on t his 1r
=- -
sin3x
2
X
s +4 arct an - 3
Now we also know X
j(t)dt
=
0
1 - F (s) s
So X
1
1 s+4 - - - arct an - 2s s 3 1r
4
-e- t sin 3tdt
t
0
Now we know
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Integral Transformations
Use the above formula in our Laplace transformation for n = 1, we get X
,2 X 0
1
1r
2
2s
s
2
l 4 -e - t sin 3tdt t s+4
arctan
3
3
+ s [9 + (s + 4)2]
Finally using the shifting property again for a= 4 -,
11 oo : 01 n
1
2s2
s2
: 46
s+4
arctan
3
3
+ s [9 + (s + 4)2·]
Finally using the shifting property again for a
n 1 s - - - - - - - arctan 2 (s - 4) 2 ( s - 4) 2 3
=4
3 + - - - -2 - (s - 4) (s + 9)
Solution: 1.4. (a) s s2
+6
Hence s
g -1 s2
+6
= cos .J6x
(b) 2 {sinax} =
~>
. d 2 { x sin ax } = - ds
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a
2
2
s +a
2sa
a
82
+ a2
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Integral Transformations
@Sk J ahiruddin, 2020
Hence
g-1
5s
=2-1
~(2s) (s2
2s
+ 1)2
5 . = xs1n x 2
11 oo : 01
: 49
2s (s2 + 1)2
(c)
2 - 1 s+ l s2
-
9
1
s
=2- 1 +21 2 s
9
-
= cosh 3x + 2
-
9
3 ) _ ( 3 82 3 2 3 ) _ ( 82 32 1
- 1
1 = cosh 3x + 3 2
s2
-1
1
= cosh 3x + sinh 3x
3
(d) We see
2 2 2 2 s - 2s + 9 = (s - 2s + 1) + (9 - 1) = (s - 1) + ( v'8) Hence 1
1
1
Jg
s 2 - 2s + 9
(s - 1)2 + ( vf8)2
Jg
(s- 1)2+(vf8)2
40
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Integral Transformations
So using first shift ing t heorem
2-1
s2
-
1 2s + 9
= 1 2-1
J8
Jg
11 oo : 01 : s2 So using first shifting theorem
2-1 s2 -
1 2s + 9
v'8
= 1 2-1
v'8
1 . = ,./8 ex sin v'Sx
Solution: 1.5. (a)
2 -l
= 2 _l
S
(s- 2)2 + 9
= 2 -1
(s-2)+2 (s-2) 2 +9
+2 -1
s- 2
(s- 2) 2 + 9 cos 3x + 2-
1
= e x cos 3x + 2 -
1
=
2 ex 2
(s- 2) 2 + 9
3
=
2 e x cos 3x + - 2 - 1 3
=
2 ex
cos 3x + ~ 3
(s- 2)2 +9 2 (s - 2) 2 + 9 2 3
2
2 ex
2
3 (s - 2) 2 + 32
sin 3x
(b) Completing the square in t he denominator, we have
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Hence
s+ 4 s2 + 4s + 8
Integral Transformations
s+4
(s + 2)2 + (2) 2
11 oo : 01 : s4 @Sk J ahiruddin, 2020
Hence
Integral Transformations
s+4
s+4
s2 + 4s + 8 Now decompose the function as
s+4 s+2 2 - - - - = - - -2- - -2 + - - -2- - -2 s2 + 4s + 8 (s + 2) + (2) (s + 2) + (2) hence we get
s+4 £ -1 s 2 + 4s + 8
s+2 +£- 1
(s + 2) 2 + (2) 2
= e- 2x cos 2x + e- 2x sin 2x (c) Completing the square in t he denominator, we get 2
s -3s+4 =
2
s - 3s
9
9
4
4
3 s- 2
+ - + 4- -
vl7
2
+
2
So we have
s+2 s 2 - 3s
s+ 2
+4
We now rewrite the numerator as
3 7 s+2=s--+-= 2 2 [email protected]
@Sk J ahiruddin, 2020
3 s-2 42
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Integral Transformations
2
11 oo :01 : s1
@Sk J a hiruddin , 2020
Integral Transformations
so t hat 3 s-_ _ _ _2_ _
s+ 2 s 2 - 3s + 4
v17
2
(s-~ ) +
2
v'7
+ V7
2
v'7
2
(s-~) +
27
2
27
Then finally
.z-1
s+ 2 s 2 - 3s + 4
s--32 v'7
v'7
+ v7.z-1
2
2
v'7
2
=
2
ft + V7e(3/ 2)x Sln . -ftX
e( 3 / 2 )x COS - X
2
2
Solution: 1.6. (a) To t he linear s + l , we associate the fraction A/ (s + l ); whereas to the quadratic factor s 2 + 1, we associate t h e fraction ( B s + C) / (s 2 + 1) . We then set l A Bs + C - - - -2- - = - - + -2 - ( s + l ) ( s + 1) - s + l s +1 Clearing fractions, we obtain
or j a [email protected]
(c)Sk J a,h ir11ddin. 2020
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Intee:ral Transforma,t ions
2
11 oo : 02 : oo or 43
j ahir@physicsguide. in
physicsguide CSIR NET I GATE
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s
2
Integral Transformations
(0) + s(O) + 1 2
= s (A + B ) + s( B + C) +(A+ C) Equating coefficients of like powers of s, we conclude t hat A + B = 0, B + C = 0, and A + C = l . The solution of this set of equations is A = ~, B = and C = ½· 1 ·t t· t h 1 · t A B s+C Su b s t 1 u 1ng ese va ues 1n o (s+l)(s2+i ) -= s+l + 8 2+ 1 we obtain the part ial-fractions decomposition -
-½,
1
(s
+ l) (s 2 + 1)
1 2
- s
+l
+
_ls + l 2 s2
+1
2
(b) To t he quadratic factors s 2 + 1 and s 2 + 4s + 8, we associate the fractions (A s + B )/ ( s 2 + 1) and (Cs + 2 D )/ (s + 4s + 8) . We set A s+ B l ) ( s 2 + 4s + 8) - s 2 + l l
(s2 +
+
Cs+ D s 2 + 4s + 8
and clear fr actions to obtain
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11 oo : 02 : 02 or 2
3
s (0)+s (0)+s(O)+ 1
2
3
s (A+C)+s ( 4A+ B + D )+s(8A+4B+C )
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Integral Transformations
Equating coefficients of like powers of s, we obtain A + C =
0, 4A + B + D = 0, 8A + 4B + C = 0, and 8B + D = 1 The solution of this set of equation is
A=-4
B= 7
C= 4
D= 9
65
65
65
65
Therefore 4
7
4
9
-65S + 65 + 65 8 + 65 (s2 + 1) ( s2 + 4s + 8) s2 + 1 s 2 + 4s + 8 1
(c) To the linear factors s - 2 and s + 1, we associate respectively the fractions A/(s - 2) and B /(s + 1) . We set
s +3 A B = + (s-2)(s+l)- s-2 s+l and, upon clearing fractions, obtain
s+3
A (s + 1) + B(s - 2) = s(A + B ) + A - 2B
Hence
A - 2B = 3·
'
A+B= l
We immediately obtain A= 5/3 and B
s+3 _ 5/3 (s- 2)(s+ l )-s- 2
= -2/3. Thus 2/3 s+ l
11 oo : 02 : os A - 2B = 3·,
A+B= l
We immediately obtain A= 5/3 and B
s+3 _ 5/ 3 (s- 2)(s+ l ) - s-2 [email protected]
= -2/3. Thus 2/3 s+ l
physicsguide CSIR NET, GATE
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Integral Transformations
Solution: 1.7. (a) Using results of problem 1.6. (c) we
get
s+3 (s - 2)(s + 1)
1
1
s-2
s+ l
2 -x 5 2x = -e - -e 3 3
(b) and noting t hat
--s 2 + 2 2 s +1 1
1
l
s
- -
2
s
2
1
+2
+1
1 s2
+1
And t hen using results of problem 1.6. (a) we get 1
2 -1 (s
+ 1) (s 2 + 1)
= ~2-1 2 1
= -e 2
1
s - x
1
+1
- - COS X 2
- ~2-1 2
s s
2
+1
1 +-2-1 2
1 s
1 .
+ - SID X 2
(c) From t he results of problem 1.6. (b) we get
2 -1
1 (s 2 + 1) ( s 2 + 4s + 8)
2
+1
11 oo : 02 : 01 (c) From t he results of problem 1.6. (b) we get 1
,2- 1
(s 2
+ l) (s 2 + 4s + 8) 4
4
7
= ,2- 1 -65s + 65 + ,2-1 s
2
65 8
+l
s 46
j [email protected]
2
9
+ 65
+ 4s + 8
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@Sk J ahiruddin , 2020
The first term can b e evaluated easily if we note t hat 4 65
7
s
2 -+ s +1
1
s
65
2
+1
To evaluate the second inverse transforms , we must first complet e the square in the denominator, s 2 2 2) + (2) , and t hen note that 4
s2
9
+ 65
4
s+2
+ 4s + 8
65
(s + 2)2 + (2) 2
65
8
2
+ 4s + 8 = (s + 2
Therefore 1
,2-1
(s2
+ 1) (s 2 + 4s + 8)
4 s 7 = - - . 2 -1 - + -.2-1 2 65 s +1 65
1
s
2
+l
s+2 1 2 - - - - - + -.2-1 65 (s + 2) 2 + (2) 2 130 (s + 2) 2 + (2) 2 4 7 4 2 1 2 =cos x + sin x + e- x cos 2x + e- x sin 2x 4 + _ _z-1
65
65
..
,....
65
I
\
TT
130
.
\
11 oo : 02 : 1o + - 2-1 - - + 2-1 2 2 65 (s + 2) + (2) 130 4 7 4 2 e- x cos 2x sin x + cos x + 65 65 65
=-
Solution: 1.8. (a) Here f (t )
47
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(s + 2) 2 + (2) 2
1 + e- 2x sin 2x 130
e3t, g(x - t)
=
=
e 2 (x- t),
physicsguide CSIR NET, GATE
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Integral Transformations
and X
f (x) * g(x)
X
e3te2(x- t) dt
= 0
0 X
=
e3te2xe- 2tdt
=
e2x
et dt
=
e2x
[et]!=; = e2x (ex -
1)
0
= e3x _ e2x (b) Here f (t ) = t and g(x - t ) = (x - t) 2 = x 2 T hus,
-
2xt
X
f (x ) * g (x) =
t (x
2
-
2xt
+t
2
)
dt
0 X
=x
2
X
tdt - 2x 0
=X
2
x2
2
2 t dt +
X
0
x3 x4 -2x-+3
4
Solution: 1.9. (a) Note t hat
1
1
1
1
t 3dt
+ t2 .
11 oo : 02 : 13 -
2
3
4
12
Solution: 1.9. (a) Note t h at
1
1 (s -3)(s- 2)
s 2 -5s+6
1
1
s -3 s- 2
Defining F(s) = 1/(s - 3) and G(s) = 1/(s - 2), we h ave from inverse Laplace transformation f(x ) = e3x and g(x ) = e2 x and using the result of problem 1.8. (a)
2 -1
s2 -
1
=
5s + 6
f (x) * g(x) 48
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e3x * e2x = e3x - e2x
=
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Integral Transformations
@Sk J ahiruddin , 2020
(b) Note that
2 -1
6 (s - l )(s + 1)
6 s2 -
1
1
1
(s - 1) (s + 1) Defining F(s) = 1/ (s - 1) and G(s ) = 1/ (s + 1), we h ave from inverse Laplace transformation f (x) = ex and g( x) = e- x Now
2 -l
~ 82
1
= 62'- 1{F(s)G(s)} = 6ex * e- x X
=
6
X
et e- (x- t)dt
e 2t dt
6e-x 0
0
= 6e-x
=
e2x - 1 2
= 3ex - 3e- x
Solution: 1.10. We know t hat, If F(s ) = 2{f(x)}, +'h e n
11 oo : 02 : 1 s = 6e-x
e2x - 1
= 3ex - 3e-x
2
Solution: 1.10. We know t hat, If F(s ) = 2{f(x) },
t hen
2{u(x - c)f(x - c)} =
e - cs F(s )
Conversely,
0 x c (a) If we define f( x) = x 2 , t hen g(x) can be given compactly as g(x ) = u(x -4)f(x -4) = u(x - 4)(x - 4) 2 . Then, 49
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Integral Transformations
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noting that 2 { f (x) } = F( s) t heorem we conclude that
=
2/ s3 and using the above
2{g(x)} = 2 { u(x - 4) (x - 4)
2
2 = e- s s3 4
}
(b) We first determine a function f (x) such t hat f (x 4) = x 2 . Once t his has been done, g(x ) can be written as g(x) = u(x -4)f (x - 4) and then t he t heorem stated at t he beginning of t he solut ion can be applied. Now, f (x -4) = x 2 only if
f( x ) = f(x
+4-
4)
=
(x
+ 4)
2
=
x
2
+ 8x + 16
•
since, 2
1fi
11 oo :02 : 1a f(x)
f(x
=
+4 -
4)
=
(x
+ 4)
2
x
=
2
+ 8x + 16
•
since,
2 {f (x)}
=
2 {x
2
}
+ 82{x} + 162{1} =
2
16
8
+ + s3 s~ s ?
it follows that
2{g(x)}
=
2{u(x - 4)f(x - 4)}
= e-
4
2
s
8
16
-+-+s3 s2 s
Solution: 1.11. Taking Laplace transformation both side 2
s Y - 0.16s + sY - 0.16 + 9Y = 0 ~~>
(
s
2
+ s + 9) Y = 0.16 (s + l )
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The solution is t hen
y _ O.l6 (s + 1) _ 0.16 (s + ½) + 0.08 - s2 + s + 9 (s + ~) 2 + 345 Using first shift ing theorem we get 1
y(t ) = 2 - (Y)
= e- t/
2
0.16 cos
35 t 4
0.08
.
+ ~ v'35 Sill
35 t 4
= e- o. 5t(O.l6 cos 2.96t + 0.027 sin 2.96t) Solution: 1.12. Taking the Laplace t ransform of both sides of the differential equation, we obtain
11 oo : 02 : 20 = e- o. 5t(O.l6 cos 2.96t + 0.027 sin 2.96t) Solution: 1.12. Taking the Laplace transform of both sides of the differential equation, we obtain
2 {y'}+2{y} = 2 {sin x}
or
[sY (s)- l ]+Y( s) =
1
s2 + 1
Solving for Y ( s), we find 1
1
Y (s) = - - + (s + 1) (s 2 + 1) s + l Taking the inverse Laplace transform , and using the result
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of Problem 1. 7. (b) we obtain
y(x) = 2- 1{Y(s)} 1
1
= 2 -1 - - - - + 2 -1 ( s + 1) (s 2 + 1) s+ 1 - x - -1 COS X -e
2
3 = -e 2
2
x
1 - - COS X 2
l
+ -1 Sln . X + e-x
+
2 1 . - Sin X 2
Solution: 1.13. The RHS (P.I part) of t he differential
11 oo : 02 : 23 -
2
2
3 -x 1 = -e - - COS X 2 2
In X
1 .
+ - Sln X 2
Solution: 1.13. The RHS (P .I part) of t he different ial equation can be written as 8[sint - u(t - 1r) sint] where
u( t - 1r) = H (t - 1r) The Heaviside function The subsidiary equation is
s Y - 0 · s - 4 + 9Y = 8£ [sin t - u(t - 1r) sin t] 2
= 8£ [sin t + u(t - 1r) sin(t - 1r)] 1
= 8 (1 + e- 7rS) - S2
52
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Integral Transformations
simplified (s
2
+ 9) Y = 8 ( 1 + e-1rs)
1
s
2
+1
+4
The solution of t his subsidiary equation is 8 (1
+ e- 7rS)
4
Y =---+2 2 2 ( s + 9) (s + 1) s + 9 Apply partial fraction reduction
11
oo : 02 : 2s
The solution of t his subsidiary equation is 8 (1 + e-7rS)
4
Y = 2- - - + ( s + 9) ( s 2 + 1) s 2 + 9 Apply part ial fraction reduction
8 (s
2
1
+ 9) (s + 1) 2
s
2
1
+1
1 s2
+9
1
+1 -
since the inverse transform of 4/ tain y
s
2
4
+ 9 + s2 + 9 ( s 2 + 9) is i sin 3t, s2
we ob-
= £-l (Y)
=
+
. 1 . 3 Sln t Sln t
3
sin (t - 1r) -
1
3
sin (3 (t - 1r))
U (t
- 1I") +
4 .
3
Sln 3t
Hence, if O < t < 1r, then
y(t ) = sin t + sin 3t [email protected]
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and if t > 1r, then
y(t) =
4
3
sin 3t
Solution: 1.14. DenoteY{w(x) }, Y {y(x) }, and Y {z(x) } by W ( s), Y (s), and Z ( s), resp ect ively. T hen , taking Laplace t ransforms of all t hree different ial equations, we have
11 oo : 02 : 2a =-
3
Sln
Solution: 1.14. DenoteY{w(x) }, Y {y(x) }, andY{ z(x) } by W (s), Y (s), and Z (s), respectively. Then, taking Laplace t ransforms of all t hree different ial equations, we have
[sW(s) - O] [sY (s) - 1] - Z(s)
+ Y (s) = -
1
+ sW(s) + Y(s) 82 1
=
l
[sZ(s) - 1] + W (s)
=
+ 82 1
s sY (s) - Z(s) = s- l
or
s-l
1
1
+ Y(s ) = -W(s) + Y( s) + sZ(s) s
s+ l s The solut ion to this last system of simultaneous linear equations is 2
- 1
W( s) - - - s(s - 1)
8)
y(
s +s = -(s---1-)-(s2_ +_ 1)
Z(s) - -
s2
8 -
+1
Using the method of partial fractions and t hen taking in-
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verse transforms, we obtain
w(x)
=
y -1{W (s) } = y -1
y(x) = y -1 {Y(s)} = y -1 [
_1 - _1_ s s- l l ..
+
l)
1_
11 oo : 02 : 31 verse t ransforms, we obtain
w(x)
z(x)
=
=
1
2 - {W(s) } =
1
2 - {Z(s)}
=
1 1 2 - - - - - = 1 - ex s s -1 1 1 --+ = ex +sin x s - 1 s2 + 1 1
2-
s
1
s
2
= cosx
+1
Solution: 1.15. Taking Laplace t ransforms of all t hree
differential equations, we find that 2
[s W (s) - s - 1] - Y (s)
3
+ 2Z(s) = - +-
1
8
-2[sW (s) - 1) + 2[sY(s) - 2) + Z(s) = 0 2[sW(s) - 1) - 2Y (s) + [sZ(s) - 2)
+ 2 [s Z(s) - 2s + 2] = 0 2
s+
+ 2sY(s) + Z(s ) = 2 2 2sW (s) - 2Y(s) + (2s + s) Z(s) = 4s - 2sW(s)
The solution to t his system is
28 1 W(s) Y(s) s- 1 (s - l )(s + 1)
2 Z(s) = -s-+-1
Hence,
w(x)
=
ex y(x)
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=
2 -
1
1
1
s- 1
s+ l
--+-55
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z(x) = 2e-x
00: 02: 33 Integral Transformations
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z(x) = 2e-x
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Dirac Delta function
2 2.1
Definition
The 8 function is different from most functions encountered in the physical sciences but we will see that a rigorous mathematical definition exist s. The 8 function can b e visualised as a very sharp narrow pulse (in space, time, density, etc.) which produces an integrated effect having a definite magnitude. Basic definition of Delta function
f (t) 8(t - a) dt = f (a)
(2.1)
provided t he range of integration includes the point t = a; otherwise t he integral equals zero. This leads immediately t o two further useful results: b
8(t)dt = 1
for all a,b > 0
(2.2)
- a
and
8(t - a)dt = 1 j [email protected]
57
(2.3)
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T
,
,
,..,.,
r
,•
11 oo : 02 : 38 [email protected]
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provided t he range of integration includes t = a We need to know b
cp (t)fJ (t - to) dt =
(to) , a < to < b otherwise
0,
a
(2.4)
And
u(x - a) =
1,
X
0,
X
>a 0
(2.7)
0
2.2.3
Fourier Transform of a 5 Function
g(a)
1
00
5(x - a) e- iaxdx
= --
vt2i
.
- oo
=
1
.
--e- iaa
vt2i
(2.8)
Then the inverse definit ion
5(x - a) =
2.2.4
1
00
vf2i 21r
(2.9) -00
Delta function in three dimensional Carte•
s1an 00
00
00
if>(x, y, z)5 (x - xo) 5 (y - Yo) 5 (z - zo) dxdydz -ex::
-oo
-oo
= if> (xo, Yo, zo ) (2.10)
11 oo : 02 : 44 = (xo, Yo , zo ) (2.10)
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Delta function in three dimensional spherical polar
b(r - ro) = b(r -ro)b(0-_ 0o) b(cf> - o) r 2 Sln 0
(2.lla)
f (r, 0, cp) 6 (r - ro) dT = f (ro, 0o , o)
(2.llb)
6 (r - ro) = 6 (r - ro) 6 (cos 0 - cos 0o) 6 (cp - cf>o ) / r
2
(2.llc)
2.2.6
Delta function as Divergence and Laplacian
(2.12) In some Books 6(r) is written as
2.2. 7
.t H (t) - e- >.(t+z)H (t _
00
+ z)dt
>.z e zo
where z0 = 0 for z > 0 and z0 = z for z < 0; so
>.z 2>.t e e
a(z ) = ,\2
00
-2,\
The Fourier transform off (t) is given by ~
1
f(w ) = ~ [email protected]
00
o
1 - >.t - iwtdt -e e ,\ 99
1 =------
~ , \(,\
+ iw)
physicsguide CSIR NET, GATE
00: 04: 35 1 f(w) = y'2n
00
~
1 - >.t - iwt dt -e e
1 =------
y'2n,\(,\ + iw)
A
o
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T he special case of the Wiener-Kinchin t heorem in which both functions are t he same shows that t he inverse Fourier 2 t ransform of t he energy spectrum, v'2n j (w) , is equal to the auto-correlation function, i.e.
from which t he stated result follows immediat ely.
4
More Exercises
4.1. Find t he Fourier t ransformation of t he function
f(x ) =
0
forx
e-x/r sin( k 0 x)
forx
0
4.2. Find Inverse Laplace transformation of
6-s 2 s + 4s + 20 Solve these differential equation by Laplace Transformation 4.3.
d2 y
dy
dx
dx
-
+ 4 2
+ 4y = x 2e- 2t
wit h t he boundary condit ion y(O)
=
y' (0) = 0 1
•
•
1
r"'1 rt TT"'\.
,,_, T ,-,, rr,
,......
I,
~ T""\
11 oo : 04: 38 4 .3.
d2 y - 2 dx
dy +4 dx
+ 4y = x 2e- 2t
with the b oundary condition y(O) = y'(O) = 0 [email protected]
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4 .4.
dY dy dx 2 + 4 dx + 5y = 8(x - x 0 )
with t he boundary condition y(O) = y'(O) = 0 4 .5. Evaluate
1T
coshx8''(x - l )dx 0
4 .6. Expand the function in Fourier Series.
0,
J(x) =
sin x,
2
J(x) =
j [email protected]
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Find t he cosine t ransform of f(x) and use it to write f(x) as an integral. Use your result to evaluate 2
2
00
cos a sin a/2 da 2 o a 4.9. What would be the apparent frequency of a sound wave
represented by 00
_
cos60nnt
7
4.10. Find inverse Laplace Transformation of the functions
(a)
l +p (p + 2)2
2p - 1 p 2 - 2p + 10
(b)
(c)
p e- p1r
(p2
+ 1)
4.11. Find t he Laplace Transformation of
J(t) =
sin(x - vt) , t > x/v 0. t < xiv
11 oo : 04: 43 (p2
+ 1)
4.11. Find t he Laplace Transformation of
J(t) =
sin(x - vt), t > x/v 0, t < x/v
4.12. Solve t he following sets of equations by the Laplace
t ransform method [email protected]
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z' + 2y = 0 y' - 2z = 2
(a)
y' + z = 2 cost z' - y = l
(b)
4.13. Evaluate following integrals, 1
(a)
1r/2
1 3 e x 0 ( X) dx
- 1
4.1 4
- 1r/2
Ans Keys 1
·1·
2
1 k + ko - i / T
1 k - ko - i / T
4.2. e- 4t(2 sin 4t - cos 4t) x4e- 2t
4.3. y 4.4 .
cos xo(sin X )dx
(b)
= -12
(
11
oo : 04: 4s
o4.2. e- 4 t(2 sin 4t - cos 4t) x4e- 2t
= --
4 .3. y
12
4.4.
y=
e- 2(x-xo) sin (x - xo)
for x > Xo
0
for x < x 0
4.5. : cosh 1 4.6.
f (x ) =
1 1 2 -+sin x-rr 2 1r
j [email protected]
cos 2x 22 - 1
cos 4x
+ 42 -
103
1
cos6x
+ 62 -
1
+ ...
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Integral Transformat ions
4. 7. ; e-lxl 7r
4.8. -
8
4.9. 90
4.10. (a)
e- 2t -
te- 2t ;
(b)
cos(t - rr)
for
t>
1r
0
for
t
1r
for
t
0) we have f (z) = x - iy so that
av = - 1
A1,
11 oo : oo : 34 secon auc y- 1emann re at1on 1n .5 1s sat1s e everywhere. Turning to the first Cauchy-Riemann relation, in the first quadrant (x > 0, y > 0) we have J(z)
8u = l ax '
= x - iy so t hat
8v = - l ay
which clearly violates the first relation in (1.5) Thus J(z) is not analytic in the first quadrant. Following a similiar argument for t he other quadrants, we find
au -- -l or + 1 for x < 0 and x > 0, resp ectively ox and = - 1 or + 1 for y > 0 and y < 0, respectively.
t
Therefore 8u/8x and 8v/8y are equal, and hence J( z ) is analytic only in t he second and fourth quadrants. since x and y are related to z and its complex conjugate z* by
x=
~ (z + z*)
and
y =
;i (z -
z*)
we may formally regard any function f = u+iv as a function of z and z*, rather t han x and y. If we do t his and examine
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8f /8z* we obtain aJ aJ ax aJ ay - = -- + -8z* ax 8z* 8y oz*
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00: 00: 37
aJ /az* we obtain aJ aJ ax aJ ay - = -- + -az* ax az* ay az* au .av 1 au .av ax + i ax 2 + ay +iay l au av av au + +2 ax ay 2 ax ay
1 -2i
(1.6)
•
1,
Now, if f is analytic then the Cauchy-Riemann relations must be satisfied, and these immediately give that af / az* is ident ically zero. Thus we conclude that if f is analytic then f cannot be a function of z* and any expression representing an analytic function of z can contain x and y only in the combination x + iy, not in the combination x - iy We conclude this section by discussing some propert ies of analytic functions that are of great practical importance in t heoretical physics. These can be obtained simply from t he requirement that the Cauchy-Riemann relations must be satisfied by t he real and imaginary parts of an analytic function. The most important of these results can be obtained by different iating the first Cauchy-Riemann relation with respect to one independent variable, and the second with [email protected]
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respect to t he other independent variable, to obtain the two
11 oo : oo : 39 @Sk Jahiruddin, 2020
Complex Analysis
respect to the other independent variable, to obtain the two chains of equalities
a ax a ax
au ax av ax
a av a av a au ay ay ax ay ay ax a au a au a av ay ay ax ay ay ax
Thus both u and v are separately solutions of Laplace's equation in two dimensions, i.e.
a2u a2u ax2 + ay2 = 0
(1. 7)
and
Those functions satisfies the Laplace written above are called Harmonic function. Example: Consider the function
u(x, y)
= x2
-
y2.
Show t hat u(x, y) is a harmonic function. Find t he function v(x, y) such t hat u + iv is an analytic function of z Solution:
au au V u = ax2 + ay2 2
2
2
=
2- 2= 0
So u satisfies Laplace's equation hence u is a harmonic function . [email protected]
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11 oo : oo : 42 15
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Complex Analysis
Now by Cauchy - Riemann equations av = au = 2x 8y ax
Integrating part ially with respect to y, we get v (x, y)
= 2x y + g(x)
where g( x) is a function of x to be found. Differentiating partially with respect to x and again using the CauchyRiemann equations, we have av = 2y ax
I
+ g ( X)
OU
= - -
8y
=
2y
Thus we find g' (x) = 0,
or
g = const .
Then
f (z ) = u +iv= x
2
-
y
2
+ 2ixy +
const.
=
z
2
+
const
The pair of functions u, v are called conjugate harmonic functions.
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1.2
Complex Analysis
Power series in a complex variable
We can expand a function of complex variable in a power series like 00
(1.8)
where z is a complex variable and the an are, in general, complex. We now consider complex power series in more detail. Expression (1.8) is a power series about t he origin and may be used for general discussion, since a p ower series about any other point z0 can be obtained by a change of variable from z to z - zo . If z were written in its modulus and argument form , z = r exp i 0, expression (1.8) would become 00
f (z ) =
L anrn exp(in0)
(1.9)
n=O
This series is absolutely convergent if (1.10) which is a series of positive real terms, is convergent. Thus tests for the absolute convergence of real series can b e used
11 oo:oo:47 (1.10)
which is a series of positive real terms, is convergent. Thus tests for the absolute convergence of real series can b e used [email protected]
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in the present context, and of these t he most appropriate form is based on the Cauchy root t est. With the radius of convergence R defined by
1 R
lim an
(1.11)
l/n
n -tOO
t he series (1.8) is absolutely convergent if z < R and divergent if lz > R If lz = R then no particular conclusion may be drawn, and this case must be considered separately. A circle of radius R centred on the origin is called the circle of convergence of the series I: anzn. The cases R = 0 and R = oo correspond, resp ectively, to convergence at the origin only and convergence everywhere. For R finite the convergence occurs in a restricted part of the z - plane (the Argand diagram) . For a power series about a general point z 0 , the circle of convergence is, of course, cent red on t hat point. Find the parts of the z-plane for which t he following series are convergent: Example:
oo
00
(ii) ) . n! zn
(iii)
n
>. -z
11 oo : oo : so Find the parts of the z-plane for which t he following series are convergent: Example:
n
oo
00
(i) ~ _z L..t n !
(ii)
~
zn (iii) L..t n
L n!zn
n=O
n=O
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n=l
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Complex Analysis
Solution: (i) since (n !) 1fn behaves like n as n ➔ oo
we find lim(l/n!) 1/n convergent for all z .
0. Hence R
=
(ii) Correspondingly, lim(n!) 1/n t he series converges only at z = 0.
=
oo and t he series is
oo . Thus R
=
=
0 and
(iii) As n ➔ oo, (n) fn has a lower limit of 1 and hence lim(l/n) 1/n = 1/ 1 = 1. Thus the series is absolutely convergent if t he condit ion zl < 1 is satisfied. 1
Case (iii) in the above example provides a good illust ration of the fact t hat on its circle of convergence a power series may or may not converge. For this particular series, the circle of convergence is Iz = l , so let us consider the convergence of the series at two different points on t his circle. Taking z = l , t he series becomes 00
1
1
1
1
n
2
3
4
L -=l+-+-+-+ ··· n=l
which is easily shown to diverge . Taking z = - l , however, .
,
,
11 oo : oo : s3 1
1
1
1
n=l n
2
3
4
oc
L - = l+-+- + -+· ·· which is easily shown to diverge . Taking z = - l , however , t he series is given by
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which is an alternating series whose terms decrease in magnitude and which t herefore converges. The ratio test may also be employed to investigate the absolut e convergence of a complex power series. A series is absolut ely convergent if z ln+l lim _a_n_+_l _ __ n----+oo an z ln
(1.12)
and hence t he radius of convergence R of the series is given by _!_ = lim lan+l R n----+oo an For instance, in case (i) of the previous example, we have 1 R
=
n! 1 lim - - - = lim - n----+oo ( n + l ) ! n ----+oc n + l
=
0
Thus the series is absolut ely convergent for all (finite) z, confirming t he previous result.
11 oo : oo : 55 1 R
=
n! 1 lim - - - = lim - n➔ oo ( n + 1) ! n➔oo n + 1
=
0
Thus the series is absolutely convergent for all (finite) z, confirming t he previous result.
1.3
Some elementary functions
In the example at the end of t he previous section it was shown t hat the function exp z defined by [email protected]
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Complex Analysis
oo
exp z
=
n
L zn.
I
(1.13)
n=O
is convergent for all z of finite modulus and is t hus, by t he discussion of t he previous section, an analytic funct ion over the whole z-plane. You may note that functions that are analytic in the whole z-plane are usually called integral or ent ire functions. Like its real-variable counterpart it is called the exponential function; also like its real counterpart it is equal to its own derivative. The multiplication of two exponential functions results in a further exponential function, in accordance with t he corresponding result for real variables.
11 oo : oo : sa in a further exponential function, in accordance with the corresponding result for real variables.
We will Show that exp z 1 exp z 2
= exp (z1 + z 2)
From the series expansion of exp z1 and a similar expansion for exp z2 , it is clear t hat the coefficient of z1z~ in t he corresponding series expansion of exp z1 exp z2 is simply 1/(r·!s!)
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Complex Analysis
From the definit ion of exp z we also have
In order to find the coefficient of z 1z~ in this expansion, we clearly have to consider the term in which n = r + s, namely (z 1
+ z2)r+s
(r+s) !
= (r +1 ) ! (r + s) ! (r 0 zr+s + ... +r+s C zr zs + ... +r+s C zr+s) 1 s l 2 r+s 2 8 The coefficient of z1z~ in this is given by 1
1
1
11 oo : 01 : oo 1
= ---(r + s) (r zr+s + . .. +r+s C zr zs + ... +r+s C zr+s) (r + ) ! ! 0 1 s I 2 r+s 2 8
The coefficient of z1z~ in this is given by
= (r + s) !
l
r+s c s
(r+s)!
1
s!r ! (r+s) !
1
r !s!
Thus, since t he corresponding coefficients on the two sides are equal, and all t he series involved are absolutely convergent for all z , we can conclude t hat exp z1 exp z2
= exp (z1+
z2) You can easily see exp z
= (exp x) (cos y + i sin y)
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If we express z as
z
= rexp i0
where r is the (real) modulus of z and 0 is its argument (-1r
< 0 < 1r), t hen multiplying z b)r exp(2ik1r), where k is
an integer, will result in t he same complex number z. Thus we m ay write
z
= r exp [i(0 + 2k1r) )
where k is an integer. If we denote w in expw
=z
11 oo : 01 z
: 03
= r exp [i(0 + 2k1r)]
where k is an integer. If we denote w in exp w
=
z
by w
ln z
=
=
ln r
+ i (0 + 2k1r)
(1.15)
where ln r is t he natural logarit hm (t o base e ) of the real positive quantity r, then Ln z is an infinitely multivalued function of z . Its principal value, denoted by ln z is obtained by taking k = 0 so that its argument lies in the range -1r to
1r.
Thus
ln z
= ln r + i0,
with -
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1r
< 0 < 1r
(1.16)
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ft(# 0) and z are both complex, then the z th power oft is defined by tz = exp (z Ln t) (1.17) We will show that t here are exactly n distinct nth roots oft. t l fn
= exp
1 - Lnt n
On the RHS let us write t as follows: t
=
r
PYn
fi ( A + ? k1r) 1
11 oo : 01 : os t 1f n = exp - Ln t n
On the RHS let us writ e t as follows:
t = r exp[i(0 + 2k1r)] where k is an integer. We then obtain 1 (0 + 2k1r) t ln = exp - ln r + i - - - n n I/n .(0+2k1r) =r exp i n 1
where k = 0, 1, ... , n - 1; for other values of k we simply recover the roots already found. Thus t has n distinct n th roots.
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1.4
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Multivalued functions and branch cuts
In the definition of an analytic function, one of the condit ions imposed was t hat the function is single-valued. However, as shown in the previous section, the logarithmic function, a complex power and a complex root are all multivalued. Nevertheless, it happens that the properties of analytic functions can still be applied to these and other multivalued
11 oo : 01 : oa ever, as shown in t he previous section , the logarithmic funct ion, a complex power and a complex root are all mult ivalued. Nevertheless, it happens that the properties of analytic functions can still be applied to these and other multivalued functions of a complex variable provided that suitable care is taken. This care amounts to identifying the branch points of the multivalued function f (z) in question. If z is varied in such a way that its path in the Argand diagram forms a closed curve that encloses a branch point, then, in general, f (z) will not return to its original value. For definiteness let us consider t he multivalued function 1 2 f (z) = z 1 and express z as z = r exp i0. From figure 1.1 (a), it is clear t hat, as t he point z traverses any closed contour C that does not enclose the origin, 0 will return to its original value after one complete circuit. However , for any closed contour C' that does enclose the origin, after one circuit 0 ➔ 0 + 21r (see figure 1.1 (b)) . Thus, for the function 25
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f (z) =
1 2
z l
,
after one circuit
r 1 exp (i0 / 2) ➔ r 1 exp [i( 0 + 21r) / 2] = -r 1 exp (i0 / 2) 1 2
1 2
1 2
In other words, the value of f (z) changes any closed loop enclosing the origin; in t his case f (z) ➔ - f (z) . Thus z = 0 1 2 is a branch point of the function f (z) = z 1 We note in ,,
'
·r-
1
1
1
•
•
•
•
11 oo : 01
: 11
In other words, the value of f (z) changes any closed loop enclosing the origin; in t his case f (z) ➔ - f (z) . Thus z = 0 is a branch point of the function f (z) = z 1l 2 We note in t his case t hat if any closed contour enclosing the origin is 1 2 t raversed twice then f (z) = z 1 returns to its original value. The number of loops around a branch point required for any given function f (z) to return to its original value depends on the function in question, and for some functions (e.g. Ln z, which also has a branch point at the origin) t he original value is never recovered.
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y
y
y
,. 0
0 X
X
X
11 oo : 01
: 13
y
y
y
,. 0 X
(a)
-
X
(b)
X
(c)
Figure 1.1: (a) A closed contour not enclosing the origin; (b) a closed contour enclosing the origin; (c) a possible 1 2 branch cut for f (z) = z / In order that f (z) may be treated as single-valued, we may define a branch cut in the Argand diagram. A branch cut is a line (or curve) in the complex plane and may be regarded as an artificial barrier t hat we must not cross. Branch cuts are positioned in such a way t hat we are prevented from making a complete circuit around any one branch point, and so t he function in question remains single-valued. For the function f (z) = z 112 , we may take as a branch cut any curve starting at the origin z = 0 and extending out to lz = oo in any direction, since all such curves would j [email protected]
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equally well prevent us from making a closed loop around t he branch point at t he origin. It is usual, however, to take t he cut along either the real or the imaginary axis. For
11 oo : 01
: 16
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equally well prevent us from making a closed loop around t he branch point at the origin. It is usual, however, to t ake t he cut along either the real or the imaginary axis. For example, in figure 1.1 (c) , we t ake the cut as t he positive real axis. By agreeing not to cross t his cut , we restrict 0 to lie in the range O < 0 < 21r, and so keep f (z) single-valued. These ideas are easily extended to functions with more t han one branch point. Example: Find the branch points off (z) = J z 2 + 1, and hence sket ch suitable arrangements of branch cuts.
Solution: We begin by writ ing
f (z)
=
✓z2 + l
=
f (z) as
✓(z - i)( z + i)
As shown above, the function g(z) = z1l 2 has a branch point at z = 0. Thus we might expect f (z) to have branch points at values of z that make t he expression under the square root equal to zero, i.e. at z = i and z = -i As shown in figure 1. 2 (a), we use the notation z - i = r 1 exp i01
We can therefore write
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f (z)
and
z
+ i = r 2 expi02
as
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11 oo : 01
: 1a
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y.
i
l
Complex Analysis
z
rI
·· ·· 0I
y
y
l
l
r2
.,x
-,
.
02
.
- /
(b)
(a)
Figure 1.2:
X
-,. (c)
(a) Coordinates used in the analysis of the
112 2 branch points of f (z) = (z + 1) ; (b) one possible ar-
rangement of branch cuts; (c) anot her possible branch cut , which is finite. Let us now consider how f (z) changes as we make one complete circuit around various closed loops C in t he Argand diagram. If C encloses (i) neither branch point, t hen
01 -+ 01 , 02 -+ 02 and so f (z) -+ f (z) (ii) z = i but not z = - i, then 01 -+ 01 + 21r, 02 -+ 02 and so f (z) -+ - f( z) (iii) z = -i but not z = i, then 01 -+ 01, 02 -+ 02 + 21r and so f (z) -+ - f (z) (iv) both branch points, t hen 01 -+ 01+ 21r, 02 -+ 02+ 21r and so f (z) -+ f (z) Thus, as expected , f (z) changes value around loops containing either z = i or z = -i (but not both) . We must t herefore choose branch cuts that prevent us from making a complete Joop around j [email protected]
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either branch point; one suitable choice is shown in figure 1.2 (b) For this f (z), however, we have noted t hat after traversing a loop containing both branch points t he function ret urns to its original value. Thus we may choose an alternat ive, finite, branch cut that allows t his possibility but still prevents us from making a complete loop around just one of the points. A suitable cut is shown in figure 1.2 (c)
1.5
Singularities and zeros of complex functions
1.5.1
Singular point
A singular point of a complex function f (z) is any point in t he Argand diagram at which f (z) fails to be analytic. We have already met one sort of singularity, the branch point , and in t his section we will consider other types of singularity as well as discuss the zeros of complex functions. If f (z ) has a singular point at z = z 0 but is analytic at all points in some neighbourhood containing z0 but no other [email protected]
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If f (z) has a singular point at z = z 0 but is analytic at all points in some neighbourhood containing z0 but no other j [email protected]
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singularities, t hen z = z 0 is called an isolated singularity. (Clearly, branch points are not isolated singularities.) The most impo1~tant type of isolated singularity is t he pole. If f (z) has the form
f (z) = (Z
g(z) -
)' n Zo
(1.18)
where n is a positive integer, g( z) is analytic at all points in some neighbourhood containing z = z 0 and g (z 0 ) =I- 0, then f (z) has a pole of order n at z = z 0 . An alternative (though equivalent) definition is that lim [(z - zo)n f (z)] = a
z --+ zo
(1.19)
where a is a finite, non-zero complex number. We note t hat if t he above limit is equal to zero, then z = z 0 is a pole of order less than n, or f (z) is analytic there; if the limit is infinite then the pole is of an order greater than n. It may also be shown that if f (z) has a pole at z = zo, then f (z ) ➔ oo as z ➔ z 0 from any direction in the Argand diagram.' If no finite value of n can be found such that equation (1.19) is satisfied, then z = z 0 is called an essential singularity.
11 oo : 01
: 26
0
diagram. ' If no finite value of n can be found such that equation (1.19) is satisfied, then z = z 0 is called an essential singularity.
31
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Example: Find the singularities of the functions 1
1
(ii)f (z) = tanh z
(i)f (z) = 1- z - l +z'
f (z)
Solution: (i) If we write 1
f (z ) =
as 2z
1
1- z
1+ z
(1 -
z)(l + z)
we see immediately from (1. 19) t hat f( z) has poles of order 1 (or simple poles) at z = 1 and z = - 1 (ii) In this case we write
f (z) = tanh z =
sinh z = exp z - exp(-z) cosh z exp z + exp(- z)
Thus f (z) has a singularity when exp z = - exp (-z) or, equivalently, when exp z = exp[i(2n + l )1r] exp( -z) where n is any integer. Equating t he arguments of the exponent ials vve find z = (n + ½) 1ri for integer n Furthermore, using l'Hopital's rule we have ( r
I
1 \
.7
•
,
11 oo : 01
: 29
where n is any integer. Equating the arguments of the exponentials we find z = (n + ½ ) 1ri for integer n Furthermore, using l'Hopital's rule we have
[z - (n
lim
+ ½) 1ri] sinh z cosh z
z--+(n+½)1ri 32
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lim
Complex Analysis
[z - (n + ½) 1ri] cosh z + sinh z sinh z
z--+(n+~)1ri
= 1
Therefore, from (1.19) , each singularity is a simple pole. Another type of singularity exists at points for which the value of j(z) takes an indet erminate form such as 0/ 0 but limz--+zo f (z) exists and is independent of the direction from which z 0 is approached. Such points are called removable singularities.
Example: Show that j(z) singularity at z = 0
= (sin z)/z has a removable
Solution: It is clear that f (z) takes the indeterminate form O/ 0 at z = 0. However , by expanding sin z as a power series in z, we find 1
f( z) = z
z3 z - 3!
+
z5 5!
-
z2
...
z4
= 1--+--·•· 3! 5!
Thus limz--+O f (z) = 1 independently of t he way in which ,
, ,
.
,
..
11 oo : 01 1
f( z) = -
z
z3
: 31
z5
z4
3!
5!
= 1--+-- ·•·
z - - + - - ·•· 3!
z2
5!
Thus limz-+O f (z) = 1 independent ly of t he way in which z ➔ 0, and so f (z) has a removable singularity at z = 0.
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1.5.2
Behaviour of function at infinity
An expression common in mathematics, is 'z tends to infinity' . For complex z t he behaviour off (z) at infinity is given by t hat off (1/ ~) at ~ = 0, where ~ = 1/ z Example: A Find the behaviour at infinity of (i) f (z) 2 2 a + bz - , (ii) f (z) = z (l + z ) and (iii) f (z) = exp z
=
Solution: (i) f (z) = a+ bz - 2 : on putting z = l / ~, f (l / ~) a+ b~ 2 , which is analytic at ~ = 0 thus f is analytic at z = oo
(ii) f( z) = z (l + z of order 3 at z = oo
2
) :
f( l / ~)
=
1/~ + 1/~
3
;
thus a pole
(iii) f (z) = exp z : f (1/ ~) = I:r(n!)- 1 ~ -ri; thus f has an essent ial singularity at z = oo .
1.5.3
Zeros of a function
=
11 oo : 01
: 34
(iii) f (z) = exp z : f (1/~) = I::r(n!)- 1 ~-n ; thus an essential singularity at z = oo.
1.5.3
f
has
Zeros of a function
We conclude t his section by briefly mentioning the zeros of a complex function. As the name suggests, if f (z0 ) = 0 t hen z = z0 is called a zero of t he function f (z) . Zeros are
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classified in a similar way t o poles, in t hat if
where n is a positive integer and g (zo) -=/= 0, then z = zo is called a zero of order n off (z ). If n = l t hen z = z0 is called a simple zero. It may further be shown that if z = z 0 is a zero of order n of f (z) then it is also a pole of order n of the function 1/ f (z) We will return in to the classification of zeros and poles in terms of t heir series expansions after we discuss the series expansion of complex functions.
11 oo : 01
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1. 6
Exercises
1.1. Find an analytic function of z
= x +iy whose imaginary
part is
(y cos y + x sin y) exp x 1.2. (a) Find a function
f (z),
analytic in a suitable part of t he Argand diagram, for which
Ref=
sin2x cosh 2y - cos 2x
(b) Where are the singularit ies of J(z)? 1.3. Det ermine t he types of singularities (if any) possessed
by the following functions at z = 0 and z = oo : (b) ( 1 + z 3 ) / z 2 ( c) sinh ( 1/ z ) (a) ( z - 2)- 1 -
•
I)
,
'
1
/()
•
/
.
"\
1/2
(d)
11 oo : 01
: 39
(b) Where are the singularit ies of f( z )? 1.3. Determine t he types of singularities (if any) possessed
by the following functions at z = 0 and z = oo : (b) (1 + z 3 ) /z 2 (c) sinh(l/z) (a) (z- 2) - 1 ez/z3 (e) z1/2/ (1 + z2)1;2
(d)
1.4. Identify t he zeros, poles and essential singularities of
t he following functions: (a) tan z (b) [(z - 2)/z 2] sin[l/ (1 - z)] (c) exp(l /z) (e) z 2I 3 (d) t an (1/ z)
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Ans keys
1.1. zez
1.2. (a) f( z ) = sin2x-isinh2y (b) y = 0 and x = n1r cosh2y-cos 2x ' 1.3. (a) Analytic, analytic; (b) double pole, single pole;
(c) essential singularity, analytic; (d) t riple pole, essent ial singularity; (e) branch point, branch point. 1.4. See solution
11 oo : 01
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1.6.2
Solutions
Solution: 1.1. If the required function is
f (z) = u + iv,
with v = (y cosy + x sin y) exp x, t hen from the CauchyRiemann equations,
av . . au ax = ex(y COSY+ Sln y + Sln y) = - By X
Integrating with respect to y gives u = -ex
(ycosy
+ xsin y + siny)dy + f(x)
11 oo : 01
: 44
x( . . ) U -8-x = e y cosy + x Sln y + sin y = - 8y V
Integrating with resp ect to y gives
u = -ex
(ycosy
+ xsiny + siny)dy + f(x)
•
ys1n y -
sinydy - xcosy - cosy
+ j(x)
= -ex(y sin y + cosy - x cosy - cosy) + f (x) = ex(x cosy - y sin y) + f (x) We det ermine f (x) by applying t he second Cauchy-Riemann equation , which equates 8u/8x with 8v/8y : ~~ = ex(x cosy - y sin y +cosy)+ f'(x )
i~ = ex (cos y By comparison, [email protected]
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J'(x) = 0 38
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where k is a real constant that can be t aken as zero. Hence, the analytic function is given by
j (z) = u +iv= ex(xcosy -ysiny + iycosy + ixsin y)
= ex [(cos y + i sin y) (x + i y)] . = exeiY(x + i y)
The fin al line confirms explicitly that this is a function of z
11 oo : 01 =
: 47
ex [ ( cos y + i sin y) (x
+ i y)]
The final line confirms explicit ly that this is a function of z alone (as opposed to a function of both z and z* ) .
Solution: 1.2. Let the required function b e f (z ) = u + iv, wit h u = sin 2x / (cosh 2y - cos 2x) since y appears less often t han x in t he given expression, it will probably be easier to consider 8u/8y rather than 8u/8x . This indicates that the relevant Cauchy-Riemann equation is
8u 8y
- 2 sin 2x sinh 2y (cosh 2y - cos 2x ) 2
av ax
Having differentiat ed w. r.t y , we now int egrate w.r.t x :
v=
2 sin 2x sinh 2y d X= (cosh 2y - cos 2x ) 2
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By insp ect ion, or by subst it ution in the other C - R equation , f (y) can b e seen to b e an ignorable constant. The required function is t herefore
f (z) =
sin 2x - i sinh 2y cosh 2y - cos 2x
To det ermine what funct ion of z t his is, consider it s form
11 oo : 01
: 49
required function is there£ore
f (z) =
sin 2x - i sinh 2y cosh 2y - cos 2x
To determine what function of z t his is, consider its form on the real axis where y = 0
f(x) =
2 sin x cosx -----=cotx 2 2 sin x
sin 2x 1 - cos 2x
f (z) =
cot z
This can be verified as follows. c~s(x + ~y) s1n(x + i y ) cos x cosh y - i sin x sinh y sin x cosh y + i sin x sinh y (cos x cosh y - i sin x sinh y) (sin x cosh y - i cos x sinh y)
f (z) =
sin x cosh y + x sinh y 2 2 2 _ sin x cos x (cosh y - sinh y) - i cosh y sinh y (cos x 2
2
2
cos2
2
cosh y - 12 - cos 2 x
+ sin x )
+ 12
sin 2x - i sinh 2y cosh 2y - cos 2x since f (z) = cot z the poles can only occur at t he zeros n1r where n is an integer ; cos n1r =I= 0 of sin z, i.e at z j [email protected]
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and so there will be a (simple) pole at each such point. The same conclusion is reached by studying cosh 2y - cos 2x . since cosh 2y 2:: 1 and cos 2x ~ 1, this denomiator can only vanish if both t erms equal 1; this requires y = 0 and x = n1r
2
11 oo : 01 : s2 and so there will be a (simple) pole at each such point. The same conclusion is reached by studying cosh 2y - cos 2x . since cosh 2y ~ 1 and cos 2x :S 1, this denomiator can only vanish if both terms equal 1; t his requires y = 0 and x = n1r Solution: 1.3. 1
(a) Although (z - 2)- has a simple pole at z = 2, at both z = 0 and z = oo it is well behaved and analytic. (b) Near z = 0, J (z) = (1 + z3 ) / z 2 behaves like 1/ z 2 and so has a double pole there. It is clear t hat as z ➔ oof (z) behaves as z and so has a simple pole there; this can be made more formal by setting z = 1/~ to obtain g(~) = ~2 + ~- 1 and considering ~ -+ 0. This leads to the same conclusion . (c) As z ➔ oo, J(z) = sinh(l / z) behaves like sinh ~ as ~ -+ 0, i.e. analytically. However , the definition of the sinh function involves an infinite series - in t his case an infinite series of inverse powers of z . Thus, no finite n for which lim [zn f (z)] is finite
z ~O
can be found, and J (z) has an essent ial singularity at z = 0 (d) Near z = 0, J(z) [email protected]
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ez/ z3 behaves as 1/ z3 and has a 41
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pole of order 3 at t he origin. At z = oo it has an obvious essential singularity; formally, t he series expansion of e1/~
11 oo : 01
: 55
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pole of order 3 at t he origin. At z = oo it has an obvious essential singularity; formally, t he series expansion of e1/( about ( = 0 contains arbitrarily high inverse powers of ( 2 112 z )
behaves as z 112 (e) Near z = 0, f( z) = z 1 / (1 + and t herefore has a branch point t here. To investigate its behaviour as z ➔ oo, vve set z = 1/ ( and obtain 1 2
f (z)
(-1
=
Hence
g(()
=
1/ 2
(
--
--
1+ (-2
f (z ) also has
1/ 2 f"'..J
(2 + 1
a branch point at z
(
1 2 /
as ( ➔ 0
= oo
Solution: 1.4. •
(a) This function tan z = sin z has zeroes when cosz 1 ( e ix-y . . ) . . sin z = 2i - e-ix+y = 0 ⇒ e 2 x e-y = e-1,XeY The two t erms can only be equal if t hey have equal mag. nitudes, i.e. e- YI = eY ⇒ y = 0. We also need eix = . e-ix ⇒ x = n 1r , where n is an integer. Thus t he zeroes of tan x occur at z = n1r The poles of tan z will occur at t he zeroes of cos z . By a similar argument to t hat above, t his needs y
= 0 and
e ix
=
jahir@physicsguide. in
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-e- ix
=
e i( - x+1r+ 2n1r) ⇒ 42
2x
= (2n + 1)1r
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Thus, the (simple) poles of tan z occur at z = (n
+ ~) 1r
We note that both sin u and cos u have Maclaurin series t hat contain arbitrarily large powers of u and that they are not mult iples of each other; we can conclude that their rat io will also have a Maclaurin series containing arbitrarily large powers of u. The same conclusion is reached by differentiating tan u and so constructing its Maclaurin series directly. Thus, when z ➔ oo is replaced by z = 1/ ~ with ~ ➔ 0, there will be arbitrarily large inverse powers of ~ in t he series expansion; this establishes that ~ = 0 (i.e. z = oo ) is an essential singularity of tan z (b) This function,
z- 2
z2
1 sin-- , has obvious zeroes at
1 -z
z = 2 and z = oo . Equally clearly, at z = 0 it has a 2 nd-order pole. Further zeroes will occur when the sine term factor is zero; from the analysis in part (a), this will be when (1- z)- 1 = n1r i.e. at z = 1- (n1r)- 1 . The remaining singularity to classify is that at z = l. By a similar argument to t hat given in part (a), the Laurent expansion of t he function about the point will have no largest negative power of 1 - z; t he point is therefore an essential singularity. (c) since exp(0) = 1, the function is well behaved and anj a [email protected]
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alytic at oo. The only nonanalytic point is the origin, z = 0, where t he defining series for t he exponential function generates a Laurent expansion with no largest negative power of z; the point is therefore an essent ial singularity. 1
(d) The singularities of tan (z- ) follow from those of tan z in part (a) . They are therefore zeroes at z = oo and 1 1 (n1r) - , simple poles at z = (n1r + t1r) - and an essential singularity at z = 0 (e) The origin, z = 0 is both a zero and a branch point of t he function z 213 To determine its behavior at oo we have 2 3 to consider 1/ ~ / near ~ = 0. There is clearly a singularity t here, and, since the function cannot even be expressed as a Laurent series, the singularity is an essential singularity.
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Integration and Series Expansion
2 2.1
Integration of a Complex function
If a complex funct ion
J(z)
is single-valued and continuous
in some region R in the complex plane, t hen we can define t he complex integral of f (z) between two points A and B along some curve in R ; its value will depend, in general, upon t he path taken b etween A and B Let a particular path C be described by a continuous (real) paramet er (a :S t :S /3) t hat gives successive posit ions on C by means of t he equations X
= x(t) ,
y
= y(t)
wit h t = a and t = f3 corresponding to the points A and B , respectively. Then the integral along path C of a continuous function f (z) is written
J(z)dz C
And can be given explicit ly as a sum of real integrals as follows:
11 oo : 02 : os J(z)dz C
And can be given explicit ly as a sum of real integrals as follows: [email protected]
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J(z)dz = C
(u
+ iv) (dx + idy)
C
udxC f3 a
v dy C
u
dx dt
+i
udy
+i
C f3
dta
dy V d dt + i t
vdx C
(3
u a
!yt dt + i
(3
a
dx V dt dt
(2.1) We will see how to evaluate integral of complex function manually from the definit ion of integration.
Example: Evaluate t he complex integral of f (z) = z- 1 along the three paths mentioned below and the figure (i) the circle z = R, starting and finishing at z = R : figure 2 .1 (a) (ii) the contour C 2 consisting of the semicircle z = R in the half-plane y > 0 : figure 2.1 (b) (iii) the contour C 3 made up of the two straight lines C3a and C3b :
figure 2 .1 ( c)
11 oo :02 : oa figure 2 .1 (c)
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y
y
y
C1
C2 C3b
R .·
R .· .
t
·,
-R
X
s= 1
{
R
(a)
Figure 2.1:
iR
X
(b)
t=O
R
-R
X
(c)
Different paths for an integral off (z)
= z- 1
Solution: (i) The path C 1 is circular. In t hat path t he distance from center is fixed . Hence t he path is parameterised as follows; see figure 2.1 (a)
z(t) = x(t)
+ iy(t)
and X(
t) = R cos t'
z(t) ,
. ..
= rl
x (t) \
•
y(t)
=
Rsin (t);
+ iy(t) = R cost+ iR sin t, •
,
0 < t < 21r
11 oo : 02 : 11 X(
y(t ) = R sin(t );
t ) = R cos t'
z(t) while
+ iy(t)
x (t)
=
f (z)
=
0 < t < 21r
R cost+ iR sin t,
is given by
f
z = ( )
l
X
+ iy
=
x 2 + y2
47
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iy
X -
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Thus the real and imaginary parts of
Rcos t R2
X
U=---
x2
+ y2
and
f (z)
are R sin t
-y
v=---= - --x2 y2 R2
+
Hence, using expression (2.1) 2 71"
1
- dz = C1
z
271"
cos t
R (-R sin t )dt -
-sint
R
R cos tdt
0
0 27!"
+i
o
cos t
- -R cos tdt + i R
2 71"
-sint R
o
(- R sin t) dt
= 0 + 0 + i1r + i1r = 21ri You could have done it directly 2
dz
1r
0
- R sin t + i R cos t d t= R eos t + iR sin t
271"
idt = 21ri (2.2) 0
(ii) This is just as in t he previous example, except that now O ::; t :::; 1r . Wit h this change, we have
11 oo : 02 : 13 0
idt = 21ri
- - - - - -dt = R cos t + i R sin t
2.2
0
(ii) This is just as in t he previous example, except that now O :S t :S 1r. Wit h this change, we have
dz
- = 7r'l •
(iii) The straight lines that make up the countour C3 may be parameterised as follows :
z = (l - t)R + itR
C3a,
for
48
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z = -sR + i(l - s) R
C3b ,
for
O< s < 1
With t hese parameterisations the required integrals may be written
1
dz 0
R + iR d ----. - t+ R +t(-R +iR )
1 0
- R - iR ( ) ds_ iR+ s - R -iR
Evaluation of these integral are time consuming. You can evaluate the first integral like this 1 0
- R + iR d t= R (l - t) + itR 1 0
1
2t - 1 1 - 2t+ 2t 2dt + i
= - rln (1 -
?.t + ?.t
2\ 1 1 1
(-1 + i) (1 - t - it) - - - - - -dt (1 - t)2 + t2 1 1 - - - - d2 t
1
0
0
1 - 2t + 2t
i.
+ -
r?.
t.::l.n
-1
(t -
2l
'11
11 oo : 02 : 16 i
0
-----dt = R (l - t) + itR 1 2t - 1 2 dt + i 1 2t + 2t 0
1 [ = ln ( 1 2
0 1
O
1 l - 2t + 2t 2 dt
2]1 i 2t + 2t ) + 0
2
.
i
(1 - t)2 + t 2
2 tan-
1
1 t--2
2
1
0
•
7r
7r
7r'l
2 2
2
2
= 0 +- - -
The second integral can also be shown to have the value wi/2. Thus dz
- = 7r'l •
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2.1.1
Cauchy's theorem
Cauchy's theorem states that if f (z) is an analytic function, and f' (z) is cont inuous at each point within and on a closed contour C, then
f (z) dz = 0
(2.3)
C
To prove t his theorem we will need the Green's t heorem in a plane . This says that if p and q are two functions with continuous first derivatives within and on a closed contour C (bounding a domain R ) in t he xy -plane, then
rr
8p "I
+
8q -"I-
dx dy =
£(pdy -
qdx)
11 oo : 02 : 1 a cont inuous first derivatives within and on a closed contour C (bounding a domain R ) in t he xy -plane, then 8p ax
R
With f (z) to
=
I=
u
+
8q ay
+ iv and dz
f( z )dz
=
dxdy =
dx
= c (pdy - qdx) + idy , this can be applied
(udx - vdy)
+i
C
C
(vdx
+ udy)
C
to give
I=
R
8(-u)
8(-v)
8y
8x
--+--
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dxdy+i
50
R
8(-v)
au
8y
8x
--+-
dx dy
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Now, recalling t hat f (z) is analytic and t herefore that t he Cauchy-Riemann relations apply, we see that each integrand is identically zero and t hus I is also zero; this proves Cauchy's theorem. Whenever you are asked to find to evaluate t he integration of a complex variable over a closed path (called a contour) you check that if the function is analytic over the bounded region or not (find whether the function diverges inside any point of the bounded region). If the function remains analytic then the integration over the closed contour is ZERO. That 's the Cauchy's theorem.
11 oo : 02 : 21 bounded region or not (find whether the function diverges inside any point of the bounded region). If the function remains analytic t hen t he int egration over the closed contour is ZERO. That 's the Cauchy's t heorem.
2.2
Cauchy's integral formula
Another very important theorem in t he theory of complex variables is Cauchy's integral formula, which st at es t hat if f (z) is analytic wit hin and on a closed contour C and z0 is a point within C t hen
f (zo) =
1
f (z) dz
.
27r1, C
Z -
(2.4)
Zo
This formula is saying t hat the value of an analyt ic funcj ahir@physicsguide. in
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t ion anywhere inside a closed contour is uniquely determined by its values on the contour. We will skip t he proof. Consult any good text book. An extension to Cauchy's integral formul a can be made, yielding an integral expression for f' (z 0 ) :
f( z) c (z - zo)
2
d
z
(2.5)
11 oo : 02 : 23 f( z ) d 2 z c (z - zo )
(2. 5)
Furt her , it may be proved by induction that the n th derivative of f (z) is also given by a Cauchy int egral,
f( z) dz C ( Z - Zo )
(2.6)
n+ l
Example: Evaluate:
fi
·
2
+ cos 1rz c (z - l)(z - 2) is t he circle Iz = 3 (a)
sin 1r z
2
dz
52
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e 2z
(b)
fc (z + l )4 dz where C
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1
1 - -, we have
z -2 2
+ cos 1r z dz c (z - l)(z - 2) 2 2 sin 1rz + cos 1rz d zsin 1rT
c
z -l
2
z- 2
sin 1rT + cos 1rz dz c z - l 2
2
By Cauchy's int egral formula wit h a = 2 and a = l , respect ively, we have
11 oo : 02 : 26 sin 1rz
c
2
+ cos 1rz dz -
sin 1rT + cos 1r z dz z - 1
2
z- 2
2
c
2
By Cauchy's integral formula wit h a = 2 and a = l , respect ively, we have sin 1r z 2 + cos 1r z 2
c
------dz
z- 2
= 21ri {sin 1r(2)
sin 1r z 2 + cos 1r z 2
2
+ cos1r(2)
2
}
= 21ri
- - - - - -dz = 21ri { sin 1r(1) + cos 1r(1) } = - 21ri c z -1 2 2 since z = 1 and z = 2 are inside C and sin 1r1r z + cos 1r z is analytic inside C. Then , the required integral has the value 21ri - (- 21ri) = 41ri (b) Let
2
f (z) = e
2
z
and a
= -
2
l in t he Cauchy integral
formula
f (z)
dz
(z - a)n+l 111 and f ( - 1) = 8e- 2 . Hence the C
If n = 3, then
f
111
(
z) = 8e2z
above equation becomes 2
8e- = [email protected]
e2z
3! -
- - -dz 21ri C (z+ 1)4 53
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from which we see that t he required integral has t he value
81rie- 2 / 3
2.2.1
Cauchy's inequality
11 oo : 02 : 2a
2.2.1
Cauchy's inequality
Suppose that f (z) is analytic inside and on a circle C of radius R centred on the point z = z0 . If If (z) < M on the circle, where M is some constant, we can show that
f
(n)
(z ) < Mn! o - Rn
(2.7)
We start from n! 2w
f (z)dz C (Z -
Zo )
n +l
We will state a result on complex integration without proof. Let is consider the integral of a function f (z) along some path C . If M is an upper bound on the value of f (z) on t he path, i.e. f (z) < M on C, and L is the length of the path C then
f( z) dz < M
f (z) dz < C
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Using the above result we get
f
(n)
(2.8)
C
C
j [email protected]
dl =ML
(z ) < n ! M 2w R 0 - 2w Rn+l
= Mn! Rn
This result is known as Cauchv's ineoualitv
11 oo : 02 : 32 Using the above result we get
(n)(z ) < n ! M 21rR = M n ! f 0
-
21r Rn+l
Rn
This result is known as Cauchy's inequality
2.3
Taylor Series
Taylor's theorem for functions of a complex variable is the following.
If f (z) is analytic inside and on a circle C of radius R centred on the point z = zo, and z is a point inside C, t hen 00
f (z) =
Lan(z -
zo)n
(2.9)
n= O
where
The Taylor expansion is valid inside the region of analyticity and, for any particular z 0 , can be shown to be unique.
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To prove Taylor 's theorem we note that, since f (z) is analytic inside and on C, we may use Cauchy's formula to
.
.
,..
/
\
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To prove Taylor's theorem we note t hat, since f (z) is analytic inside and on C, we may use Cauchy's formula to write f (z) as 1 (2.10) f (z) = . 21ri
where~ lies on C. Now we may expand the factor (~ - z)as a geometric series in (z - z0 ) / (~ - z0 ) Z -
1
n
Zo
~ - Zo
so
f (z)
becomes
f (z) =
!(~)
1
oc
21ri
1
= 2 . 1ri
z - zo
n
~ - Zo
f(~)
00
L (z -
zo )n
n=O
d
t )n+l ( C '::. - Zo
(n) ( ) . 1 """" ( 21ri 1 _ _ )n Zo -~ z-zo n.1 21ri oo
n=O
~
oo _ """" - ~
)n an z-z0 (
n=O
(2.11)
we have used equation (2.6) Example: Let j(z ) = ln(l + z), where we consider the branch that has the zero value when z = 0.
(a) Expand j(z) in a Taylor series about z = 0. [email protected]
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1,
\
~
I
.
'
,
f'
'
,
11 oo : 02 : 37 [email protected]
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(b) Determine t he region of convergence for the series in (a).
(c) Expand ln(l + z/1 - z) in a Taylor series about z = 0 Solution: (a)
(a) f (z) = ln (1 + z) , 1 1 f (z) = = (1 + z) - , f I I ( z) = - (1 + z)- 2' 11 J (z) = (- 1)(- 2)(1 + z) - 3 ,
f (O) = 0 f '(O) = 1 1 f ' (0) = - 1 1 f ' ' (0) = 2!
• •
•
i!z
•
•
•
T hen
z2 z 3 z4 = z - - + - - -+ •·· 2
3
4
We could have proceed in another wayIf z < l 1
2
3
- - = l -z+z -z + ··· l +z j [email protected]
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1
A.
1
•
11 oo : 02 : 39 - - = l -z+z -z + ··· l +z [email protected]
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Then integrating from O to z yields z2 z3 z4 ln(l + z) = z - - + - - - + · · · 2 3 4 (b) Then t h term is Un = (- l )n-Izn/n . Using t he ratio test, nz Un+l 1 · . 1Im - - = Im = z n➔ oo Un n----+oo n + l and the series converges for Iz I < 1. The series can be shown to converge for z = l except for z = - 1 This result also follows from t he fact t hat the series converges in a circle t hat extends to the nearest singularity (i.e., z = - 1 ) of
f( z) (c) From the result in (a) we have, on replacing z by - z z2
z3
z4
1nl ·· ( +z ) = z--+2 3 --+· 4 z2 z3 z4 1n ( 1 - z ) = -z - 2 - 3 - 4 - · · ·
both series convergent for z < 1. By subtraction, we have oo 2 z2n+l z3 z5 l +z ln =2 z+-+-+ ···
1 -z
3
5
which converges for z < 1. We can also show t hat t his series converges for z = 1 except for z = ± 1 [email protected]
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11 oo : 02 : 42 n=O
n+
which converges for z < l. We can also show t hat t his series converges for z = l except for z = ± l [email protected]
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2.4
Laurent Series
So far we have assumed that f (z) is analytic inside and on t he (circular) contour C. If, however, f (z) has a singularity inside C at t he point z = z0 , then it cannot be expanded in a Taylor series. Nevert heless, suppose that f (z) has a pole of order p at z = z 0 but is analyt ic at every other point inside and on C. Then the function g (z) = (z - z0)P f (z) is analytic at z = z 0 , and so may be expanded as a Taylor series about z = zo : 00
g(z) =
L bn (z -
zo)n
(2.12)
n=O
Thus, for all z inside C, f (z) will have a power series representation of the form
wit h a _ P i= 0. Such a series, which is an extension of the Taylor expansion, is called a Laurent series. By comparing
11 oo : 02 : 44
wit h a _ P #- 0. Such a series, which is an extension of t he Taylor expansion, is called a Laurent series. By comparing [email protected]
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t he coefficients in (2 .12) and (2.13), we see that an = bn+p· Now, t he coefficients bn in the Taylor expansion of g(z) are seen from (2.11) to be given by
g(z) (Z -
Zo )
d n+l
Z
and so for the coefficients an in (2. 13) we have g(z)
dz
(z - zo)n+l+p
= I 21ri
f (z )n+ldz
(z - zo )
an expression t hat is valid for both positive and negative n. The terms in the Laurent series with n -> 0 are collectively called the analytic part , whilst the remainder of t he series, consisting of terms in inverse powers of z - z 0 , is called the principal part. Depending on t he natu1·e of t he point z = z 0 the principal part may contain an infinite number of terms, so that +oo
f (z) =
L
n = - oc
an (z - zo) n
(2.14)
11
00:02:47
+oo
f (z) =
L
an (z - zo)n
(2.14)
n = - oo
In this case we would expect the principal part to con1 verge only for (z - z 0 ) - less than some constant, i.e. outj [email protected]
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side some circle centred on z 0 . However , t he analytic part will converge inside some (different ) circle also cent red on z 0 . If the latter circle has t he greater radius t hen the Laurent series will converge in t he region R between the two circles (see figure 2.2 below). Ot herwise it does not converge at all. y
R
• zo
X
Figure 2.2: The region of convergence R for a Laurent series off (z) about a point z = zo where f( z ) has a singularity.
11 oo : 02 : so X
Figure 2.2: The region of convergence R for a Laurent series off (z) about a point z == zo where f( z) has a singularity. In fact, it may be shown that any function f (z) that is analytic in a region R between two such circles C 1 and C2 centred on z == z 0 can be expressed as a Laurent series about z 0 that converges in R. We note that , depending on [email protected]
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the nature of t he point z == z 0 , the inner circle may be a point (when the principal part contains only a finite number of terms) and t he outer circle may have an infinite radius. We may use the Laurent series of a function f (z) about any point z == z0 to classify the nature of t hat point. If f (z) is actually analytic at z == z 0 , t hen in (2.14) all an for n < 0 must be zero. It may happen that not only are all an zero for n < 0 but a0, a 1 , .. . , am- l are all zero as well. In t his case, t he first non-vanishing term in (2.14) is am (z - zo)m, with m > 0, and f (z) is then said to have a zero of order m at z == z0 If f (z) is not analytic at z == z0 , t hen two cases arise, as discussed above (p is here taken as positive): (i) it is possible to find an integer p such t hat but a-p-k == 0 for all integers k > 0
a _p -=I=
(ii) it is not possible to find such a lowest value of - p.
0
11
oo : 02 : s2
(i) it is possible to find an integer p such t hat a _p -=f. 0 but a-p-k = 0 for all integers k > 0 (ii) it is not possible to find such a lowest value of -p. In case (i), f (z) is of the form (2 .13) and is described as having a pole of order pat z = z0 ; the value of a _1 ( not a -p ) is called t he residue of f (z) at t he pole z = z0 , and will play an important part in later applications. For case (ii) , in which t he negatively decreasing powers 62
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of z - z0 do not terminate, singularity.
f (z)
is said to have an essent ial
Example:
Find t he Laurent series of 1
f (z) = z (z -
2)3
about the singularities z = 0 and z = 2( separately ) . Hence verify that z = 0 is a pole of -order 1 and z = 2 is a pole of order 3, and find the residue of f (z) at each pole.
Solution: To obtain t he Laurent series about z = 0, we make the factor in parentheses in t he denominator take the form (1 - a z ), where a is some constant, and t hus obtain
f (z ) = -
1 Q,.,/ 1
,., / ')\3
11 oo : 02 : 55 Solution: To obtain the Laurent series about z = 0, we make the factor in parentheses in t he denominator take the form (1 - a z ), where a is some constant, and t hus obtain 1
f (z ) = - 8z( l - z/2)3 z
= _ 1 l + (-3) - ~ + (-3)(-4) 8z
2
+ (-3)(-4)(-5) _ z
2! 3
2
2
+ ...
3!
2 2 1 3z 5z 3 - . .. 8z 16 16 32 since the lowest power of z is - 1, t he point z = 0 is a pole of order 1. The residue of f (z) at z = 0 is simply the 63
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coefficient of z- 1 in the Laur ent expansion about that point and is equal to - 1/8 The Laurent series about z = 2 is most easily found by letting z = 2 + ( (or z - 2 = ( ) and substituting into the expression for f (z) to obtain
f (z) =
1 (2 + ()(3 1 12(3
=
1 1 2(3 - 4( 2 1
1
2(3(1 + (/2)
( 2
+
+
( 2
1 1 8( - 16 1
2
+
( 2 ( 32
3
+
( 2
...
1
1
4
-
...
11 oo :02 : s1 1-
~ 2
+
~ 2
2
~
-2
3
~
+ -2
1 1 1 1 ~ = 2~3 - 4~ 2 + 8~ - 16 + 32 - ... 1 1 1 1 = 2(z - 2)3 - 4(z - 2) 2 + 8(z - 2) - 16
4
+
- ···
z- 2 32 - · · ·
From this series we see t hat z = 2 is a pole of order 3 and t hat the residue of J(z) at z = 2 is 1/ 8
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2.5
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Calculation of Residue from Laurent Series
Finding the residue of a funct ion at a singularity is of crucial importance in t he evaluation of complex int egrals. Specifically, formulae exist for calculating t he residue of a function at a part icular (singular) point z = z 0 without having to expand the function explicitly as a Laurent series 1 about z 0 and identify the coefficient of (z - z 0 ) - . The type of formula generally depends on the nature of t he singularity at which the residue is required.
11 oo : 03 : oo a function at a particular (singular) point z = z 0 without having to expand the function explicitly as a Laurent se1~ies about z0 and identify the coefficient of (z - z 0 ) -
1
.
The type
of formula generally dep ends on the nature of the singularity at which the residue is required. Suppose that f (z) has a pole of order m at t he point z = zo. By considering t he Laurent series off (z) about zo . We will derive a general expression for the residue R (z0 ) of
f( z) at z = zo If f (z ) has a pole of order m at z series about t his point has t he form
= z0 , then its Laurent
a_m a-1 ! (z ) =---+ ··• +--(z - z0 )m (z - zo) 2
+ ao + a1 ( z - zo) + a 2 ( z - zo) + · •· which , on multiplying both sides of the equation by (z - z0 )m, j [email protected]
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•
gives
(z - zo)m J(z) = a_m + a-m+l (z - zo) + · · ·
+ a_ 1 ( z - z0 )
m- 1
+ ···
Different iating both sides m - 1 t imes, we obtain ~
-1
00
dzm- l [(z - zo)m f (z) ] = (m - l )!a_1 +
L bn (z -
zo)n
n=l
for some coefficients bn. In the limit z
➔
z0 , however , the
11 oo : 03 : 02 d m- 1
dzm- l [(z - zo)m f (z)] = (m - l )!a_1 +
oo
L bn (z -
zo )n
n=l
for some coefficient s bn. In the limit z ➔ zo, however , the t erms in t he sum disappear , and aft er rearranging we obtain t he formula
R (zo)
=
a-1
1 =
dm-1
(m - 1)! dzm-l [(z - zo)m f (z)]
lim
z➔ zo
(2. 15) which gives t he value of t he residue of f (z) at the point z = z0 . This is t he general residue formula for mth order pole. An import ant special case of (2.15) occurs when has a simple pole (a pole of order 1 ) at z residue at z 0 is given by [email protected]
66
f (z)
= z 0 . Then the
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= lim [(z - z0 ) f (z) ]
R ( z0 )
(2 .16)
z ➔ zo
If f (z) has a simple pole at z = zo and, as is often t he case, has t he form g(z)/h(z), where g(z) is analytic and non-zero at z 0 and h (z0 ) = 0, then equation (2 .16) becomes . ( z - zo) g(z ) _ ( ) . ( z - z0 ) R (Zo) - 11m g Zo 1I m _
7
/
\
-
7
/
\
11 oo : 03 : os If f (z) has a simple pole at z = zo and, as is often the case, has t he form g( z) /h( z), where g(z) is analytic and non-zero at z 0 and h (zo) = 0, then equation (2. 16) becomes
. (z - z0 ) g(z) ( ) . (z - zo) R (Zo) = 11m = g Zo 11m z➔ zo
=
z➔zo
h(z)
. 1 g ( zo) z➔zo l1m hI ( z )
h(z )
g (zo)
(2.17)
( ) h' Zo
where we have used l'Hopital's rule. This result often provides t he simplest way " of det e1~mining the residue at a simple pole.
Example: Evaluate the residue of the function , by t he help of above prescription , at t he point z = i
f (z) =
•
exp iz (z2 + 1)2
Solution: •
•
exp iz
f( z) = exp iz (z2 [email protected]
@Sk J ahiruddin ,
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we see immediately t hat it h as poles of order 2 (double poles) at z = i and z = -i. To calculate t he residue at (for example) z = i, we may apply the residue formula (equation (2.15)) with m = 2 P erforming the required different iation, we obtain
11 oo :03 : oa po es at z = i an z = -i. To ea cu ate t e resi ue at or example) z = i, we may apply the residue formula (equation (2.15)) with m = 2 P erforming t he required differentiation , we obtain
d
d
2
exp iz
dz[(z - i) f (z)]= dz (z+i) 2 1
= (
2
") 4 [(z + i) iexp iz - 2(exp iz)(z + i)]
z+i
Setting z
= i, we find the residue is given by 1 1 i 1 1 R (i) = 1! ( - 4ie- - 4ie- ) = - e 2 16
Example: Find R (-~) (Residue at-~) and R (5) (Residue at 5) for z f (z) = (2z + 1)(5 - z)
Solution: Multiply f (z) by (z + ½), [ Caut ion: not by (2z + 1)] , and evaluate t he result at z =
68
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-~ . We find
z+
1 2
f( z) = /
1\
1 z +2 1
z (2z + 1)(5 - z) 1
z 2(5 -
z)
11 oo : 03 : 1 o -~ - We find 1 z +2
R
1 z +2
J( z) =
(2z
+ 1)(5 - z)
- -1
1 -2
2(5 - z )
1 22
2
2 (5 +
z
z
½)
Similarly (z - 5)f (z)
=
z (z - 5) ( 2z + l )( 5 _ z)
Hen ce
z 2z+ 1
5
R(5) = - 11 Example: Find the residue of cot z at z
=0
Solution:
.
R (0) = l1m
z cos z .
z ---tO Sill Z
.
= cos O · l1m .
z
z ---tO Sill Z
Example: Find t he residue of (sin z )/
=
1·1= 1
(1 - z4 )
at z
=i
Solution: By (2. 17) we have, sin z R(i) = -4z3 .
z=i
jahir@physicsguide. in
@Sk J ahiruddin , 2020
sin i e- 1 - e 1 = - 4i3 = (2i)(4i) = 8 69
_1
(e - e ) =
1 .
4 s1nh 1
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Example: Find the residue of the following functions
11 oo : 03 : 13 @Sk J ahiruddin , 2020
Complex Analysis
Example: Find t he residue of t he following functions
(i)
f (z)
•
__
Z Sln Z
(z - 1r) 3
at z
= 1r
(ii) 2
ez / sin z at z = m 1r Solution : (i) We know that the residue at p ole of order m , situated at a
a_1 = lim ( z➔a
1
dm- 1
_1 )1 m - 1 . dzm
{ (z
- a)m f (z) }
As z sin z is finit e at 1r, t he order of pole is 3 or m ay be less t han 3. Now applying t he formula of residue a1
1 d2 1 = !dz 2 (z sin z ) lz=7r = [- zsin z +2 cos z]z=1r = - 1 2 2 2
(ii) ez/ sin z has double poles at z = 0, ± 1r, ±21r, . .. Residue at z
=
m1r is
1 d lim - z➔m1r 1! dz j [email protected]
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ez (z -m1r)2. 2 Sln Z
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00 : 03 : 15 [email protected]
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=
Complex Analysis
ez [(z - m 1r) 2 sin z + 2 (z - m 1r) sin z - 2 (z - m 1r) 2 cos z]
.
l1m - - - - - - - - - - - - - - - - - - - - - - -
sin3 z
z -+ni
Put z - m1r = u
a1
=
2
lim eu+ m1r
u sin u
+ 2u sin u -
2
2u cos u
3
sin u
u -+0
. u 2 sin u + 2u sin u - 2u 2 cos u 1Im - - - - - - -3 - - - - - 11,---+0 sin u T he limit can be evaluated by using L 'Hospital's r ule. First, note t hat
u3 lim - 3- = lim u-+0 sin u u-+0
u •
3
Sln U
= 1
So
+ 2u sin u - 2u cos u u ------------- . --2
2
u sin u
3
u3 sin 3 u 2 2 u sin u + 2u sin u - 2u cos u = em1r 1n1r . = e 1Im - - - - - - -3 - - - - - u -+0 u
Alternat ive method - By expanding in Laurent Series [email protected]
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00 : 03 : 18 Alternative method - By expanding in Laurent Series 71
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f (z) = ez sin z in a Laurent 2
In this method, we expand
1
= m1r and obtain t he coefficient of ( ) z-m1r as required residue. Let z = u + m1r . Then, the funct ion to be expanded in a Laurent series about u = 0 is
series about
z
sin ( m1r + u) = em1r eu / sin pansions for eu and sin u, we find em1r+u /
2
(1 u 2
m1r
e
u2
= so t he residue is
[email protected]
6
+
u.
Using t he series ex-
u4 120 -
...
)2
u2
l +u+?T+ ··· ~-
(1 -
em1r
u2
2
1 u2
u2
3
+ 2u4 + ... ) 45
1 +u
5 u +- +6 3
+ ...
em1r
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11 oo : 03 : 20
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2.6
Complex Analysis
Residue theorem
Suppose t he function f (z) has a pole of order m at the point z = zo, and so can be written as a Laurent series about zo of the form 00 (2.18)
n=-m Now consider t he integral I of f (z) around a closed contour C that encloses z = z 0 , but no other singular points. Using Cauchy 's t heorem, t his integral has t he same value as the integral around a circle ry of radius p centred on z = z 0 since f (z) is analytic in the region between C and ry. On the circle we have z = zo + p exp i0 (and dz = ip exp i0d0) , and so
f (z)dz
I = 'Y
00
n=-m 21r
00
i pn+l exp[i(n + 1)0]d0
n=-m For every t erm in the series with n =/= - 1, we have 27r . /
i
p
n+l
f "(
exp i n
+
1)0ld0
=
ri pn+l exp [i (n .1
.
.. \
+ 1)0]l 27r =
0
11 oo : 03 : 23 exp i n
+
0
n=-m
For every term in the series with n =/= - 1, we have 21r
o
i
. n+ l
p
[ .(
exp i n
+
l )0]d0
j [email protected]
. n+l
=
ip
73
[ .(
exp i .(
+ 1)0] n
in+ l
21r
)
-_ 0
0
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but for the n = - l term we obtain 21r
id0 = 21ri 0
1
Therefore only t he term in (z - zo)- contributes to the value of the integral around 'Y ( and t herefore C ) , and I takes the value
I=
J(z)dz = 21ria_1
(2.19)
C
Thus the integral around any closed contour containing a single pole of general order m (or, by extension, an essential singularity) is equal to 21ri times the residue of f (z) at z =
zo If we extend t he above argument to the case where f (z) is cont inuous wit hin and on a closed contour C and analytic, except for a finite number of poles, within C, then we arrive at t he residue theorem (2.20)
11 oo : 03 : 26 except for a finite number of poles, within C, then we arrive at the residue theorem (2.20)
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, • ,-,, , -
•
•
,,
' • ''
, ,,
I I
• C
(a)
,
,,
,
/
.
'
'
'
'• ,I' , , , , ,,
,,
C'
• ''
'
--- -·-------
---
,,
(b)
Figure 2. 3: The contours used to prove the residue theorem: (a) the original contour; ( b) t he contracted contour encircling each of t he poles.
In figure 2. 3, (a) shows the original contour C referred to in t he residue theorem (2.20) and (b) shows a contour C' giving the same value to t he integral, b ecause f is analytic between C and C'. Now t he contribution to the C' integral from t he polygon (a triangle for t he case illustrated) joining the small circles is zero. since f is also analvtic inside C' . Hence the whole value
11 oo :03 : 2a between C and C' . Now t he contribution to t he C' integral from t he polygon (a triangle for t he case illustrated) joining the small circles is zero, since f is also analytic inside C'. Hence the whole value of t he integral comes from t he circles and, by result (2.19) each of t hese contributes 21ri times the residue at the pole it encloses. All t he circles are traversed in their positive sense if C is thus traversed and so t he residue theorem follows. [email protected]
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Formally, Cauchy's theorem is a particular case of (2.20) in which C encloses no poles.
2.6.1
General expression of Integral around a pole in terms of residue
Finally we prove another important result , for later use. Suppose that f (z) has a simple pole at z = z0 and so may be expanded as the Laurent series
f (z) = cp(z) + a _1 (z -
zo)-
1
where cp(z) is analytic within some neighbourhood surrounding z0 . We wish to find an expression for t he integral I of f (z) along an open contour C, which is the arc of a circle of radius p centred on z = zo given by
IZ
-
Zn I
=
0.
(} ,
3 (d) z < l
(c) 0 < z
+l
1 3.2. P rove t hat if J (z) has a simple zero at z0 then 1/ J(z)
has residue 1/ J' (zo) there. Hence evaluat e
11
oo : 04: s4
for a real and > 1 3.2. Prove t hat if
f (z)
has residue 1/ f' (z0 )
has a simple zero at z0 t hen 1/ f (z) t here. Hence evaluate 1r
-7r
sin 0 dB a - sin 0
where a is real and > 1 3.3. P rove t h at, for a > 0, the integral 00
0
t sin at dt 1 + t2
has t he value (1r / 2) exp ( -a) 3.4. Prove t hat 00
0
cos mx dx=1f - m/2 -m 4 e -e 4x 4 + 5x 2 + 1 6
j [email protected]
for m> 0
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3.5. Evaluate 00
o
3.6. Evaluate
dx x6
+1
00
2
cos (x )dx 0
3. 7. Show t hat the principal value of t he integral 00
cos (x /a) dx
- oo x2 -
is - (1r /a) sin 1
a2
11 oo : 04: s6 00
cos(x/a) dx
- oo x2 -
a2
is - (1r /a) sin 1 3.8. Using a suitable cut plane, prove t hat if a is real and
0 0, all t he condit ions for Jordan 's lemma to hold are satisfied and the integral 4
11 oo : os : 11 z+i z-i around a semicircular contour like figure 3.2 . 4 As f (z) z - as z --+ oo and m > 0, all the conditions for Jordan's lemma to hold are satisfied and the integral f"'--1
j [email protected]
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around the large semicircle cont ributes not hing. For t his integrand there are two poles inside the contour, at z = i and at z = ~i The respective residues are - 2ie-m/ 2
e-m/2
e-m
6
2i3ii
and
3i
2
(-i) 2i 2
The residue theorem therefore reads . 00 eimx ie- m dx + 0 = 21ri 4 2 _ 00 4x + 5x + 1 6
3
2ie- m/ 2
3
and the stated result follows from equating real parts and changing the lower integration limit, recognising that the integrand is symmetric about x = 0 and so the integral from O to oo is equal to half of t hat from -oo to oo. Solution: 3.5. We consider c
closed contour. Here, z 6
x
dx , where C is the 6
+1
+ 1 = 0 gives us the poles of t he function.
That are,
These are simple poles, amongst t hem only
00: 05: 20
That are,
These are simple poles, amongst t hem only • 1r
·
31r
· 57r
z = e 6 e e; e e; ' ' 2
2
2
lie inside t he contour.
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Residue at
Similarly, Residue at .1m [- 1 ] 1 - 31r 6z 5 z ➔e 6
1
= -e
- i
151r
6
i
6
Residue at
So,
dz C
= =
'Tri
3 'Tri
3
[cos( [cos(
51r
6 51r
6
z6
+1
) +i sin( .
)-is1n(
51r
6 57r
1 ·51r = 21ri-[e-ie;
6
) + cos(
151r
6 1r
· 151r
+ e-1.
) +i sin( .
1r
151r
6
6
· 251r
+ e-i
) +cos( 57r
6 ]
251r
6
) +i sin( .
57r
251r
6
)]
27r
)+cos( )-isin( )-cos( )+is1n( )] = 2 2 6 6 6 3
11 oo : os : 23 = =
wi
3 wi
3
[cos( [cos(
5w 6
5w 6
) +i sin (
)-isin(
- oo
6
5w 6
dx
00
So,
5w
x6 + 1
) + cos(
15w
15w
6
6
) +i sin(
) +cos(
25w 6
) +i sin(
w w 5w 5w )+cos( )-i sin( )-cos( )+i sin( )] 2
2
6
dx
00
21r 3
6
o x6 + 1
25w 6 =
)]
2w 3
7r
3
Solution: 3.6. [email protected]
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Consider t he contour drawn in t he figure. First con. 2
sider t he complex integral
e iz
dz . There is no pole in the
C
contour. So the value of t he integral would be zero. Now
y B
C
0
A
R
X
breaking t he integral along the lines of t he contour . . 2
e
2
z
dz
. 2
+
OA
eiz
AB
dz
+
e
. 2
2
z
dz
=0
BO
Now on OA , z = x( from x = 0 to x = R); on AB , z = R ei 0 ( from 0 = 0 to 0 = 1r / 4); on BO , z = re1ri/4 (from r = R to r = 0). Hence
00: 05: 25
Now on OA , z = x( from x = 0 to x = R); on AB , z = R ei8 ( from 0 = 0 to 0 = 1r / 4); on BO , z = re1ri/4 (from r = R t o r = 0). Hen ce
R
. 2
e ix
dx
1r/4
+
. i R2 e2iB
e
R i0 d0 i e .
0
+
e
1ri/4d 0 e r =
R
0
0
ir2e1ri/2
From h ere j [email protected]
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R
( cos x
2
+ i sin x
2
)
dx
0 R
1r/4
2
e-r dr -
eiR2 cos 20- R2 sin 20 i R ei0 dB
0
0
First integral b ecomes oo
e1ri/4
e
- r2 d
r =
0
•
,Jir 1ri/ 4 1 - -e = 2
11"
'l
-2 + -2
2
11"
2
Second int egral b ecomes
1r/4
2
e'iR
2
cos 20- R sin 20 i R ei0 d0
7r
< 0
0
R f 1r12
0 2 ..,; .... ,J.. • •
/4
2
e- R
sin20 Rd0
11 oo : os : 2a 7i/4
eiR2 cos 20-R2 sin 20 i R ei0 d0
7i/ 4
.. , µ are constant real numbers. Show that C is an isotropic tensor. 3.12. Consider a coordinate syst em (u, v, w)
x=vw
'
'y
= uw,
Z
= UV
0 btain the metric in the new coordinate system (u , v, w) . 3.2.1
Ans Keys 2
3.3 (a) Principal moment of intertia's are 20ma and (20
2v'5)ma
2
±
11 oo : 02 : 33 3.2.1
Ans Keys
3.3 (a) Principal moment of intertia's are 20ma2 and (20 ±
2v'5)ma
2
(b)
2
1
- 1
- 1
2
-2 v'5
'
0
'
v'5
3.5 (a) 2
ds = du dA
2
+ h~dv
, hu = 1, hv = 112 2 v ) - dudv, ds = 2 112 j ( 2v - v ) , 2 112 v) j ( 1 - v)
u (2v -
=
2
eu = i (1 - v) + ev = - i ( 2v + au= eu = au, av = hvev, [email protected]
2
u (2v - v
)-
112
+ hvevdv
eudu
av = ev/ hv
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(b) m
u-
aV = Fu au
uv2
v(2 - v)
uv 2(v - 1)
uii + 2uv m
[v(2 -v)]1/ 2
= -u-
1
[v(2 -
+ [v(2 -v)] 3/ 2
1 2 aV v)] l - =
Fv
av
(c) v u = euaU/ au
+ evu-
1
1
J v(2 - v) aU/ av
1
V. V = u - a (uVu) / au+ u - J v(2 - v) aVv / av n 2u v
=
! Aa 11,
11.
au uA 11.
1 ✓,----- a v(2 -v) A 2
+ 11.
V
✓,---- au
v(2 -v) ,,V ~
11 oo : 02 : 36 1
VU = euaU/ au + evu- J v(2 - v)au/ av 1 1 V. V = u- a (uVu) ; au+ u- J v(2 - v) a-vv; av n 2u V
=! a
au 1 ✓~-- a a Ua + v(2 -v) a 2 u u u u V
(d) u - 1 ,
l k u '
✓~-- au
v(2 -v) ;::i
uV
0
3.10
T'=
[email protected]
1 0 0 0 2 0 0 0 4
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physicsguide CSIR NET ) GATE
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Green's function Basic reading and solved examples
This is again an advanced topic. I'm guiding you step by step to understand the topic. First, you need to underst and Laplace Transformation, Dirac Delta function and How to to solve differential equation by Laplace t ransform ation when t here is a delta function in t he RHS of t he equation. For
11 oo : 02 : 38 1s 1s again an a vance op1c. step to understand the topic. First, you need to understand Laplace Transformation, Dirac Delta function and How to to solve differential equation by Laplace t ransform ation when t here is a delta function in the RHS of the equation. For example you need to be able to solve differential equation like this
Now proceed to Boas Chapter 8, Section 12, A Brief Introduction to Green 's Functions. The section is brief, t here are only four pages. You need to understand the examples.
I'm writing t he general procedure to solve an equation by Green 's function. You will be given a general equat ion
d dx
dy p (x) dx
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+ q(x )y = f (x) 57
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This equation is to be satisfied on t he range within a boundary x 1 < x < x 2 . Boundary condit ion will be given.
Our Green 's function G(x, x') will be the solution of the equation
d dx
dG(x, x' ) ( ) p x dx
+ q(x)G(x, x') = b(x -
x')
11
00:02:41
Our Green's function G(x, x') will be the solution of the equation d dG(x, x' ) ( ) dx p x dx
+ q(x )G(x, x') = 6(x -
x')
G(x, x') needs to satisfy that boundary conditions. General solution of the equation will be X2
y(x) =
G(x.x')J (x') dx' X1
Getting the Green's funct ion may not be so easy. I'm giving you step by step procedure. Step - 1: Identify p(x) from t he given equation. Step - 2: Get the boundary condition (it will be given) Step - 3: Find t he solution of the equation (homogeneous part) d dy dx p(x) dx + q(x) y = O Step - 4: You get the solution as y(x) . Now write the same function as two different functions: y1( x) and Y2 (x) . j [email protected]
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Step - 5: Apply given two boundary conditions on t hese two functions y 1 ( x) and y 2 ( x) and get two solutions. Ignore any common constants. Step - 6: Find G(x,x') by using (
.
I
'.
I
I '.
I
11 oo : 02 : 43 two functions y 1 x and y 2 x and get two solutions. Ignore any common constants. Step - 6: Find G (x,x') by using
G(x, x') =
Ay1 (x )y2 (x' ), Ay2 (x)y1(x'),
X X
< >
X X
1
' 1
(4.1)
'
Where
Step - 7: After you get the Green 's function you can easily get the general solution of t he equation by using X2
y(x) =
G(x .x') f (x')dx' X
= A Y2(x)
X2
Y1(x')j(x')dx'
+ Y1(x)
Y2(x')f(x')dx' X
Now follow t he next example carefully to find t he Green 's function.
Example 4.1. Consider t he ODE -~:;= f (x). W hat will be a suitable Green 's function for the equation with t he [email protected]
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boundary condition y(O) = y( l ) = O? Solution: We are trying to find t he Green 's function for t he equation t-!
I
t-!,, ,\
11
00:02:47
boundary condition y(O) = y( l ) = O? Solution: We are trying to find t he Green 's function for t he equation
d~
as
d~
(- )
j~
=
p(x) :;
f (x) .
+ q(x)y = f( x)
So p( x) = ( - 1).
First we need to make solution of the homogeneous part
d:y
which is = 0. The general solut ion of this equation dx can be written as y = a+ bx, where a and bare constants. Now we need to apply t he boundary condit ions on the general solution t o get two separate solutions. let me explain in more details. Lets take Y1 (x) = c1 + c2x and Y2(x) = d1 + d2x . Bot h of Y1 and y 2 are solut ions of the homogeneous part. We will not apply both of t he boundary conditions on a single equation. R ather we will use one Boundary condit ion on one solution. Let 's say we apply y(O) = 0 on Y1 = c1 + c2x . As y(O) = 0 c1 must be zero. y 1 = c2 x. While constructing Green 's funct ion we ignore the common constant , so we get one solution [email protected]
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Yl = x .
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11 oo : 02 : 49 @Sk J ahiruddin, 2020
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Yl = x .
Now we apply y(l) = 0 on y2 = d1 + d2x. As y( l ) = 0 d1 + d2 = 0. So y 2 = d1 - d 1 x . As mentioned before while constructing Green 's function we ignore the common constant, so we get t he other solut ion y 2 = 1 - x.
G(x, x') =
Ay1 (x )y2(x'), Ay2(x)y1(x'),
X
X
< >
X
X
1
' 1
(4.2)
'
Where
A=
l p( x') [y; (x')y1 (x') - y~ (x')y2 (x')]
Here, p(x') = - 1, So 1
A=--------= l - 1[- 1 X X - l X (1 - x)] Hence
G(x , x' ) =
x( l - x'),
0 < x < x',
(1 - x)x',
x' < x < l ,
(4.3)
Example 4.2. Find the solution of the equation
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11 oo : 02 : s1 61
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(k is a constant) Solution: We have already got the Green 's function for t he general equation - ~;; = f (x) in t he previous problem
G(x , x') =
x( l - x'),
0 < x < x' ,
(4.4)
(1 - x)x' , x' < x < l ,
So we can directly formulate the solution X2
y(x) =
G(x .x') J( x')dx'
1
G(x.x') sin(kx')dx' 0
1
X
Y1(x') sin(kx')dx' + Y1(x)
= A y2(x) 0
X
1
X
x' sin(kx')dx' + x
= (1 - x)
Y2(x' ) sin(kx')dx'
0
(1 - x') sin(kx')dx' X
Example 4.3. Find the solution of the equation
d2y
- 2 = coskx d x
(k is a constant) Example 4.4. Construct Green's function for the differential equation
d2y 2 dx2 +w y = J(x) jahir@physicsguide. in
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11 oo : 02 : s4 d y dx2
+w
2
y
(
=f x
62
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given t he boundary condit ion y(O) = y( 1r /2) = 0 Solution: We are trying to find the Green 's function for the equation d dx
dy p (x) dx
d2y Our equation is d2 x
equation as d~
:~
+w
+w
2
+ q (x )y = f (x) 2
y
y =
= f (x) . We can write the j(x) . So p(x)
2
= 1, q(x) = w .
First we need to make solut ion of t he homogeneous part which is ~;; +w 2y = 0. T he general solution of this equation can be written as y = a sin x + b cos x, where a and b are constants. Now we need to apply the boundary conditions on the general solution to get two separate solut ions. Lets take YI (x) d2 cos x.
= CI
sin x
+ c2 cos x
and y2(x)
=
dI sin x
+
Let's say we apply y(O) = 0 on YI (x) = c I sin x + c2 cos x . As y(O) = 0 c2 must be zero. YI = cI sin x. While constructing Green 's function we ignore t he common constant , so we get one solution YI
= sin x .
Now we apply y(1r / 2) = 0 on y 2 (x) = dI sin x + d 2 cos x. j [email protected]
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11 oo : 02 : s6 get one solut ion y 1
= sin x .
Now we apply y(1r/2) = 0 on y 2 (x) = d 1 sin x [email protected]
63
+ d2 cosx.
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Advanced Topics
As y(1r / 2) = 0 d1 = 0. So y2 = d2 cos x . As mentioned before while constructing Green 's function we ignore the common const ant, so we get t he ot her solution y 2
G(x, x' ) =
A sin (x) cos (x'), A cos(x) sin(x' ),
X X
< >
X X
= cos x .
1
'
(4. 5)
1
'
Where
A= Here, p(x')
A=
1 p(x') [y~ (x')y1(x') -
Yi (x' )y2(x')]
= 1, So l
1 [- sin x' x sin x' - cos x' x cos x']
=
-1
Hence
G(x, x') =
- sin(x) cos(x'),
0 < x < x',
- cos(x) sin(x'),
x' < x ::; 1r / 2,
(4.6)
Example 4 .5. Solve
d2y 2 dx 2 +w y = cosec(x) by using Green 's function. Sol11tion: W P- 8.lrP-Fl,.. ,T ) =
81rhc )..5
1 ehc/>.kT _ 1
(2.7)
total energy density which is expressed in terms of Stefan- Boltzmann's total power per unit surface area is 00
81rh u(v, T )dv = 3 o c
00
O
) W:
hv = W + K
(2.9)
where K represents the kinetic energy of the electron leaving t he material
11 oo : oo : 29 hv = W + K
(2.9)
where K represents the kinetic energy of the electron leaving t he material The above equation gives the proper explanation to the experimental observation that the kinetic energy of t he ejected electron increases linearly with the incident frequency v,
K = hv - W = h (v - vo)
(2.10)
where v 0 = W / h is called the threshold or cutoff frequency of the metal. Moreover , this relation shows clearly why no electron can be ejected from t he metal unless v > v 0 : since the kinetic energy cannot be negative, the photoelectric effect cannot occur when v < v 0 regardless of the intensity of t he radiation. The ejected electrons acquire t heir kinetic energy from t he excess energ)' h (v - v 0 ) supplied by t he incident radiation. [email protected]
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Compton effect
Compton found that when X-rays are make incident on free electrons, the wavelength of t he scattered radiation is larger t han the wavelength of the incident radiation. How do we explain it? According to classical physics, the incident and scattered radiation should have the same wavelength. since
11 oo : oo : 32 Compton found that when X-rays are make incident on free electrons, the wavelength of t he scattered radiation is larger t han the wavelength of t he incident radiation. How do we explain it? According to classical physics, t he incident and scatt ered radiation should have the same wavelength. since t he energy of the X-ray radiation is too high to be absorbed by a free electron, the incident X-ray would then provide an oscillatory electric field which sets t he electron into oscillatory motion, hence making it radiate light with t he same wavelength as you have studied in radiation chapter in ED course. The experimental findings of Compton reveal that t he wavelength of the scattered X-radiation increases by an amount ~A , called the wavelength shift, and t hat ~A depends not on t he intensity of the incident radiation, but only on t he scattering angle. If we t reat t he incident radiation as a stream of particles which are named as photons - colliding elastically with individual electrons. In t hat scattering process, we use the laws of elastic collisio11s - t he conservation of energy and 12
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moment um. You can derive t he Compton scatt ering formula (hence . I ' m not der1v1ng . . .......''') ,\
\
\ I
\
h
i1
In
1 1 \
11 oo : oo : 34 momentum. You can derive the Compton scattering formula (hence .. I ' m notd er1v1ng . . .... ..''')
.6A = A
1
-
A= h (l - cos 0) = 2Acsin
2
ffieC
2.4
e2
(2.11)
Pair production
When high-frequency electromagnetic radiation passes through a foil, individual photons of this radiat ion disappear by producing a pair of particles consisting of an electron, e- , and a positron, e+ : photon This process is called pair production
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e
E=hw
------Incoming
►
ffi
11 oo : oo : 36
e
E=hw
------Incoming photon
►
EB
Nucleus
Figure 2.2: Pair production: a highly energetic photon, interacting with a nucleus, disappears and p1~oduces an elect ron and a posit ron
Due to charge, momentum, and energy conservation, pair production cannot occur in empty space. For the process photon ➔ e- + e+ to occur, the photon must interact with an external field such as the Coulomb field of an atomic nucleus to absorb some of its momentum. In t he figure you see that an electron-positron pair is produced when the photon comes near (interacts with) a nucleus at rest; energy conservation dictates that
nw =
+ E e+ + E N 2 2 = (mec + ke-) + (mec + ke+) + K N Ee-
:::'. 2mec
2
(2.12)
+ ke- + ke+ 2
where hw is the energy of the incident photon, 2mec is the [email protected]
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Old Qt1antum Theory
sum of the rest masses of t he electron and positron, and
11 oo : oo : 39
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Old Quantum Theory
sum of the rest masses of t he electron and positron, and ke- and ke+ are the kinetic energies of the electron and posit ron, respectively. As for EN = K N, it represents the recoil energy of the nucleus which is purely kinetic. since t he nucleus is very massive compared to t he electron and t he positron, K N can be neglected to a good approximation. Note the photon cannot produce an electron or a a positron alone, for electric charge would not be conserved. Also, a massive object, such as the nucleus, must participate in t he process to t ake away some of t he photon's momentum. The minimum energy E min of a photon required to produce an electron-posit ron pair must be equal to the sum of rest mass energies of the electron and positron; t his corresponds to t he case where the l.. t he momentum of t his particle will be rather high. Formally, this means that if a particle is accurately localized (~ x -+ 0) , there will be total uncertainty about its moment um (i.e., ~Px -+ oo ) . To summarize, since all quantum phenomena are described by waves, we have no choice but to accept limits on our ability to measure simultaneously
11 oo : oo : s1 mally, this means that if a particle is accurately localized (~ x ➔ O) , t here will be total uncertainty about its moment um (i.e., ~ Px ➔ oo ) . To summarize, since all quantum phenomena are described by waves, we have no choice but to accept limits on our ability to measure simultaneously any two complementary variables. Heisenberg's uncertainty principle can be generalized to any pair of complementary, or canonically conjugate, dynamical variables: it is impossible to devise an experiment t hat can measure simultaneously two complementary variables to arbitrary accuracy (if this were ever achieved, the j [email protected]
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t heory of quantum mechanics would collapse) . Energy and t ime, for inst ance, form a pair of complementary variables. Their simultaneous measurement must obey t he time-energy uncertainty relation:
n ~E~t -> -2
(4.2)
This relation states that if we make two measurements of the energy of a system and if t hese measurements are separated by a time interval ~ t, t he measured energies will differ by an amount ~E which can in no way be smaller than n/ ~t . If t he time interval between the two measurements is large, the energy difference will be small. We also have the ane::ular momentum and ane::le uncer-
11 oo : 01 : oo differ by an amount ~E which can in no way be smaller than Ii/ ~t . If t he time interval between the two measurements is large, the energy difference will be small. We also have the angular moment um and angle uncertainty
(4.3)
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Old Quantum Theory
Bohr's Model
According to Rutherford the positive charges of an atom is centered in very small spaces, called nucleus, and t he negative charges are orbiting around the nucleus just as do planets do orbit around the sun. The main problem in t his model is according t he classical electromagnetics the negat ive charges do emit radiation as the are charged particle accelerating with acceleration v2 /r, so after some time the electrons will exhaust its energy and collapse on the nucleus. Bohr proposes solution to this problem by taking some
11 oo : 01
: 02
accelerating with acceleration v2 / r, so after some time the electrons will exhaust its energy and collapse on the nucleus. Bohr proposes solution to this problem by taking some extra assumptions. (1) Instead of a contimunm of orbits, which are possible in classical mechanics, only a dis- crete set of circular stable orbits, called stationary states, are allowed. Atoms can exist only in certain stable states with definite energies:
E1, E2, E3, etc. (2) The allowed (stationary) orbits correspond to those for which t he orbital angular momentum of t he electron is an integer multiple of n(n = h/21r) :
23
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L
=nn
(5.1)
This relation is known as t he Bohr quantization rule of t he angular momentum (3) As long as an electron remains in a stationary orbit , it does not radiate electromagnetic energy. Emission or absorption of radiation can take place only when an electron jumps from one allowed orbit to another. The radiation corresponding to the electron's transition from an orbit of
11 oo : 01 : os 3 As long as an electron remains in a stationary orbit , it does not radiate electromagnetic energy. Emission or absorption of radiation can take place only when an electron jumps from one allowed orbit to another. The radiation corresponding to t he electron 's transition from an orbit of energy E n to another E m is carried out by a photon of energy
(5.2) So an atom may emit (or absorb) radiation by having t he electron jump to a lower (or higher) orbit. Now we use
e2
v2
- - - = m e47rcor2
(5.3)
r
and
nn
L = m evr = j [email protected]
(5.4) Physicsguide
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@Sk J ahiruddin, 2020
Old Quantum Theory
to get
(5.5) where
47rcon
2
ao = - -
(5.6)
m ee2
is the Bohr radius, a0 = 0.053nm. The speed of the orbiting electron can also be derived
nn
(
2
e
~ 1
11 oo : 01
: 01 (5.6)
ao = is the Bohr radius, a0
=
m ee 2 0.053nm. The sp eed of t he orbiting
electron can also b e derived e2
nfi
Vn=--
1
(5.7)
41rco nfi
m ern
The ratio b etween the sp eed of the electron in the first Bohr orbit , v 1 , and the sp eed of light is equal to a dimensionless constant a, known as the fine structure constant: e2 41rc 0 fie 1
C
1
(5.8)
137
Total energy of electrons
E=
1
1 e2
2
(5.9)
-meV - - - -
41rc0 r
2
The kinetic energy, ~ m ev 2 , is equal to ~ e 2 / ( 41rc 0 r) Hence e2
1 E =-2 [email protected]
(5.10)
41rcor 25
Physicsguide
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Old Q t1antum Theory
The total energy t h en , in quantized form m e
e2
2/i2
41rc 0
2
1
n2
R n2
(5.11)
known as t he Bohr energy, where R is the Rydberg constant:
11 oo : 01
: 1o
2
41rEo
2/i
n
2
n2
(5.11)
known as t he Bohr energy, where R is the Rydberg constant:
R
=
2
me 2n2
= 13.6eV
(5.12)
In deriving t he energy expression we have neglected the mass of the proton. If we include it, the expressions become 2
41rEori rn = µe2
E - - µ n 2/i2
1 + m e a 0n 2 , mp 2 1 1
n2 e2
n2
47rEo
R
1 + m e/m p n
(5.13)
2
where µ = mpme/ (mp + me) = m e/ (1 + m e/mp) is the reduced mass of the proton-electron system. We should note t hat rn and En, which were derived for t he hydrogen atom, can be gene1~alized to hydrogen-like ions where all electrons save one are removed. To obtain the radius and energy of a single electron orbiting a fixed nucleus 26
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Old Quantum Theory
of Z protons, we need simply to replace e2 in (1.75) by Z e2
'
I_
m f'. \ ao
'>
z2
11 oo : 01
: 13
radius and energy of a single electron orbiting a fixed nucleus j [email protected]
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Old Quantum Theory
of Z protons, we need simply to replace e2 in (1.75) by Ze 2
'
m e ao
1 +M
2
z n,
Z2 R En = - - - - - -2 1 + m e/ M n
(5.14)
where M is the mass of the nucleus; when m e/ M we can just drop t he term m e/ M
j [email protected]
27
(x) is given by (1jJ, c/> ) =
'lj; * ( x) c/> (x) dx
(1.3)
In quantum mechanics we deal with square integrable vvave functions so t he integral converges. A function 1j; (x) is said tn h P ~n,,~rP intPo-r~hl P if th P ~r~l~ r nrnrl11rt. nf 1/, "\7iTith it~Plf
11 oo : oo : 29 ('lj; , cp) =
'lj; * ( x) cp (x) dx
(1.3)
In quantum mechanics we deal with square integrable wave functions so t he integral converges. A function 'lj; (x) is said to b e square integrable if t he scalar product of 'lj; wit h itself, 1.e 2 'lj;(x) dx = finite (1.4) ('lj; ' 'lj; ) = •
The linear vector space which contains square integrable funct ions and a posit ive inner product is a Hilbert space. We must note t hat the dimension of the Hilbert space of squareintegrable functions is infinite, since each wave function can be expanded in t erms of an infinite number of linea1~1y independent functions. The dimension of a space is given by t he maximum number of linearly independent basis vectors required to span t hat space.
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Dirac Notation: Bras and Kets
All the information about a syst em is stored in a stat e vector. That is called a ket. Kets belong to Hilbert space. There exits some elements which give real or complex numbers, while taking inner products with t he ket vetors.
11 oo : oo : 31 All the information about a system is stored in a stat e vector. That is called a ket. Kets belong to Hilbert space. There exit s some elements which give real or complex numbers, while taking inner products with t he ket vetors. These elements are said to belong to bra space. For every ket l'l/J) there exists a unique bra ('lf; . The scalar product (cp, 'lj;) is denoted by the bra-ket (cp 'lj;)
- >
(2.1)
(c/> 'lp)
The wave functions are basically t he projections of ket vectors on posit ion or momentum axis.
(r, t 'lfJ) = 'l/J (r, t)
(2.2)
In the coordinate representation , the scalar product (cp 'lj;) is given by ( c/>
'lj;) =
c/> * (
r, t )'lj; (r, t )d r 3
(2.3)
The momentum space wave function is t hen [email protected]
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Nl athem atical Backg1·ound
(p, t 'lj;) = 'lj; (p, t )
2.1
Properties of Bra and kets
(2. 4)
11 oo : oo : 34 (p, t 7P) = 7P (p, t )
2.1
(2.4)
Properties of Bra and kets
(1) To every ket 1/;) , there corresponds a unique bra (1/;
(2. 5) (2) There is a one-to-one correspondence between bras and kets: a 7/J) + b q;) < > a* (7/J +b*(
(3.4)
or
V1
a11
a12
• • •
a1n
V1
V2
a ??
--
a22
•
• •
a2n
V2
• • •
• • •
•
• • •
• • •
>
• • •
Vn
V
•
•
• • •
Vn
(3.5) V
11 oo : oo : 55 >
•
•
• •
•
•
•
• •
•
•
• •
•
(3.5)
• • •
•
V
V
On the tensor product space, the same matrix can still act on the vectors, so t hat iJ 1--t Ail, but 1--t unt ouched . This matrix is written as A @ I , where I is t he identity mat rix. In the previous example of n = 2 and m = 3 ,
w w
t he matrix A is two-by-two, while A @ I is six-by-six, [email protected]
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Mathematical Background
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A @I =
a11
0
0 0
a 11
a21
0 0
0 0
0 0 a11
a 21
0 0
0
a 21
a1 2
0
0 0
a12
a 22
0 0
0 0
0 0 a12
a 22
0 0
0
a22
(3.6)
The reason why this expression is appropriat e is clear once you act it on iJ @ w,
(A @I )(iJ@w) =
a 11
0
0 0
a 11
a21
0 0
0
a 21
0 0 a 11
0 0
a12
0
0 0
a12
a22
0 0
0
a22
0 0
V1W1
a12
V1W3
0 0
V2W1
V1W2
V2W2
11 oo : oo : ss
(A ® I )(v ® w) =
0 0
a1 1
0
0 0
a12
0
V1W2
0
a1 1
0
a 12
V1W3
a21
0
a22
0
0
a22
0 0
V2W1
a21
0 0
0 0
0
a 21
0
0
a22
V2W3
+ a12V2) W1 ( a11 v 1 + a12v 2) w2 ( a11 V1 + a 12V2) W3 ( a21 V 1 + a22V2) W1 ( a21 V 1 + a22V2) W2 ( a21 V 1 + a22V2) W3
V2W2
( a11 V1
j [email protected]
= (Av) ® w
(3.7)
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:Nl athematical Background
Clearly the matrix A acts only on w E W untouched.
v
E V and leaves
Similarly, the matrix B : W ➔ W maps w can also act on V ® W as I ® B , where
l ®B =
It acts on
b11
b 12
b13
0
0
0
b21
b22
b23
0
b31
b32
b33
0
0 0
0 0
0 0
0 0
0 0
b11
b12
b13
b21
b22
b23
0
0
0
b31
b32
b33
v ® w as
H
Bw. It
(3.8)
11 oo : 01 : oo 0 0
0 0
0 0
b11
b12
b13
b21
b22
b23
b31
b32
b33
0 0 0
0 0 0
0 0 0
It acts on v © was
b21
(I 0 B )(v 0 w) =
[email protected]
b 22
b 23
b31
b32
b33
0 0 0
0 0 0
0 0 0
b11
b1 2
V1W1 V1W2 V1W3
V2W1
b13
V2W2
b21
b22
b23 V 2W3
b31
22
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+ b12 W2 + b13W3) V1 ( b21 W 1 + b 22W2 + b23W3) V1 ( b31 W1 + b32W2 + b33W3) V2 ( b11 W1 + b12W2 + b 13W3)
b32
b33
P hysicsguide
Jvi athematical Background
V1 ( b11 W 1
= v © (Bw)
(3.9)
+ b22W2 + b23W3) V2 ( b31 W 1 + b32W2 + b33W3) V2 ( b21 W1
In general, (A © I )(v ©w) = (Av) © w, and (I © B )(v© w) = v0 (B w) If you have two matrices, their multiplications are done on each vector space separately,
11 oo : 01
: 02
w) = v 0 (Bw) If you have two matrices, t heir multiplications are done on each vector space separately,
(A1
0 I ) (A2 0 I ) =
(A1A2)
0 I
(J 0 B 1) (I 0 B 2) = I 0 (B1B2) (A 0 J)(J 0 B ) = (I 0 B )(A 0 I ) = (A 0 B )
(3.10)
T he last expression allows us to write out A 0 B explicitly,
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Nl athematical Backg1·ound
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A 0 B=
P hysicsguide
a11b11
a11b12
a 11 b13
a 12 b 11
a12 b1 2
a12b13
a11b21
a1 1b22
a 11 b23
a12b21
a 12b22
a12b23
a11 b31
a11b32
a 11 b33
a12b31
a12b32
a12b33
a21b11
a2 1b12
a21 b13
a22b11
a22b12
a22b13
a21 b21
a21b22
a21 b23
a22 b21
a22b22
a22b23
a21 b31
a2 1b32
a21 b33
a22b31
a 22 b32
a22b33
(3.11)
It is easy t o verify t hat (A 0 B )(i10 w) = (Ail) 0 (Bw)
11 oo : 01 : os 22 23
a21b31
a21b32
a21b33
a 22b31
a22b32 a22b33
(3.1 1)
It is easy to verify t hat (A 0 B )(v0 w)
= (Av) 0 (B w)
Not e t hat not every matrix on V 0 W can be written as a t ensor product of a matrix on V and anot her on W . Ot her useful formul ae are
det (A 0 B ) = (
p2= f>
'
(4.14)
,,.,
The unit operator I is a simple ex ample of a projection ,,., ,,., operator, since J t = I , j2 = j A
The product of two commuting projection operators, P 1 ,,.,
and P 2 , is also a projection operator, since
(4.15) and
The surn of two projection operators is generally not a proj ection operator.
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projection operator.
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T wo proj ection operators are said to be orthogonal if t heir product is zero "
"
"
For a sum of projection operators P 1 + P2 + P3 + · · · to be a projection operator, it is necessary and sufficient that t hese projection operators be mut ually orthogonal (i.e., the cross-product terms must vanish).
4. 7
Commutator of operators "'
"'
The commutator of two operators A and B , denoted by [A, B] , is defined by A
A
AA
AA
[A , B ] = AB - BA
(4.16)
and t he antiCommutator A
A
A
A
A
A
{ A , B} = AB + BA
(4.17)
Two operators are said to commute if t heir commutator is equal to zero and hence AB = BA . A
A
A
A
Any operator commut es with itself: "'
"'
[A, A] = 0
(4.18)
11 oo : 01 A
: 26
A
A
A
equal to zero and hence A B = B A . Any operator commutes with itself: (4.18) 31
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Mathematical Background
If two operators are Hermit ian and t heir product is also Hermit ian, these operators commute: (4.19)
and since (A.B)t
=
AB we have AB = BA
The basic commutation relation between position and moment um operators A
A
A
in,
X , Px =
A
in,
Y , Py =
(4.20)
Some propert ies:
A
A
A
A
[A, B] = - [B , A]
(4.21)
[A., B + 6 + b + ... ] = [A., B] + [A., 6] + [A., b ] + . . . (4.22) [A., B]t = E t, At A
A
A
A
A
A
A
(4.23) A
A
+ B[A , C] [AB ,6] = A[B,6] + [A, C]B [A , BC]
=
[A , B ]C
J acobi ident ity r
A
..
A
r .....,,
A
~, ,
A
r .....,,
A
r ,,,-..,
A
..
11
A
r ,,,,.........
r
A
,.
A
T""""I, , ,
(4.24)
11 oo : 01
: 29
[A, BC] = [A, B]c + B[A, c ] [AB, 6] = A[B, 6] + [A, C]B
(4.24)
J acobi ident ity A
A.
A
A
A
A
[A , [B ,C]] + [B, [C,A]]
A
A
A
+ [C, [A ,B]] = 0
32
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4.8
(4.25)
Uncertainty Relation between Two Operators
Let (A) and (B) denote t he expectation values of two Hermit ian operators A and B with respect to a normalized state vector 'l/J) = ('l/J A 'l/J) and (B) = ('l/JIB 'l/J) Introducing the operators ~ A and ~B , A
A
A
A
~A = A -
(A),
we have (~ A.)2 = A.2 - 2A(A) 2-8 (B) + (-8) 2, and hence
~B = B + (A.) 2
(B)
(4.26)
and (~ B) 2
=
-8 2
-
(4.27) and
(~ -8)2 =
_a2 -
(-8)2
where A. = 'ljJ A 'ljJ and -8 = uncertainties ~ A and ~ B are defined by 2
2
A
,1
_
flrA
-1 \? \ _
2
f I :i?\
'ljJ
A
/
(4.28)
,1 \
?
B2
'ljJ . The
11 oo : 01
: 31
where A. 1/; A 1/; and B uncertainties ~A and ~B are defined by 2
2
2
1/;
B
2
1/J . The
(4.29)
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We can prove after doing some algebra t hat (4.30)
Check using the formula that (4.31)
4.9
Functions of Operators
Let F(A ) be a function of an operator A. If A is a linear "' operator, we can Taylor expand F(A) in a power series of ,,.. A: 00 (4.32) n= O
where an is just an expansion coefficient. As an illustration of an operato1~ function, consider eaA, where a is a scalar ~
11 oo : 01
: 34 00
( 4.32) n=O
where an is just an expansion coefficient. As an illust ration of an operator function, consider eaA , where a is a scalar which can be complex or real. We can expand it as follows: ~
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Nl athematical Backg1·ound
Commutators of functions of operators: A
A
A
If A commutes with another operator B, t hen B corn" mutes with any operator function that depends on A :
[A, B] = o ~~> A
[F (A), B] = [A, F (B )] = 0
( 4.34)
A
A
F (A) commutes with A and with any other function, G(A), of A : A
[A, F (A)] =
o,
An, F (A)
=
o, [F (A) , G(A)] = o ( 4.35)
Hermitian adjoint of function of operators:
11 oo : 01 [A , F (A)] =
o,
: 36
o, [F (A) , G(A)]
An, F (A) =
=
o
(4.35) Hermitian adjoint of function of operators:
The adjoint of F (A) is given by
lF(A.)J t = F *(A.t) "
(4.36)
"
If A is Hermitian, F (A) is not necessarily Hermitian; F (A) will be Hermitian only if F is a real function and A
35
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is Hermitian. Examples -
ei·11 t = e- iA.t
(4.37)
'
Commutators of function of operators: Non zero commutators: You must know "
"
[A, B] # 0
~;>
"
"
[B , F (A)] # 0
(4.38)
We can easily show by expanding the function in series t hat
(4.39)
11 oo : 01
: 39
t hat
(4.39)
We can prove (4.40)
The corollary of which is If A and B commute wit h t heir commutators ([A, [A, B]] = [B , [A, B]] = 0) then
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:Ni athematical Backg1·ound
We can also prove Baker-Campbell-Hausdorff Lemma A " -A
e Be
"
" "
l
"
" "
= B+ [A, B] +2! [A, [A, B]] +
1 "
"
" "
[A, [A, [A, B ]] I] + · · · 31 (4.42)
4.10
Inverse operator "
1
"
The inverse A- of a linear operator A is defined by the relation (4.43) "
where I is the unit operator
11
00:01 :41
1
The inverse A_- of a linear operator
A is
defined by the
relation (4.43) A
where I is the unit operator You can easily prove the following relations 1) - 1
(i) (A=A (ii) (AT) - 1 = (A-1) T
= (A- 1) t (iv) (AB) - 1 = B- 1 A- 1 (V) (AB · · · G)- 1 = G- l (iii) (At) - i
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· · ·
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@Sk J ahiruddin, 2020
4.11
(4.44)
Unitary operators
A linear operator () is said to b e unitary if its inverse ()- 1 is equal to its adjoint ()t :
or
(4.45)
There are few important results about unitary op erators you need to know
11 oo : 01
: 44
or
(4.45)
There are few important results about unitary operat ors you need to know
(1) The product of two unitary operators is also unitary as you can easily see for two op erators
(uv)(uv)t = (uv) vtut = u vvt ut = uut =
1
(4.46) This result can be generalized for any number of unitary operators A
A
(2) If
E
is real and G is Hermitian , t he operator
eicG
would b e unitary as you see A
-- e- icG --
(4.47)
(3) The det erminant of a unitary matrix has unit modj ahir@physicsguide. in
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lVIathem atical Background
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ulus. (4) A unit ary matrix represent s, in a part icular basis, a linear operator that leaves t he norms (lengt hs) of complex vectors unchanged.
5
Eigenvalues, Eigenvectors and ....
11 oo : 01
: 46
vectors unchanged.
5
Eigenvalues, Eigenvectors and Basis Transformation
5.1
Eigenvalues and Eigenvectors of operators
A state vector 'ljJ) is said to b e an eigenvector (also called an eigenket or eigenstate) of an operator A if the application "
of A to 'ljJ) gives
(5.1) "
where a is a complex number , called an eigenvalue of A . T his equation is known as t he eigenvalue equation, or eigenvalue "
problem , of the operator A. Its solutions yield t he eigenvalues and eigenvectors of A .
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Mathematical Background
We must know A n 'l/J)
= an 'l/J)
(5.2a) ~
"
F(A) 'l/J) = F (a) 'l/J)
=~►
;1- 1 'l/J) = ~l'l/J) ,
,
'ljJ) =
eia
'ljJ)
(5.2b) (5.2c)
a
,
eiA
. ,
.,
,
11 oo : 01
: 49
~
A
F(A) 1/;) = F(a) 1/;)
===>~>
eiA
1/J) =
eia
A-11/J) = ~ 1/J)
1/J)
(5.2b) (5.2c)
a
You already know t hese theorems about t he eigenvalues and eigenvectors of various operators (1) For a Hermitian operator, all of its eigenvalues are real and the eigenvectors corresponding to different eigenvalues are orthogonal.
If A_t
=A
'
(5.3)
(2) The eigenstates of a Hermitian operator define a complet e set of mutually orthonormal basis states. The operator is diagonal in this eigenbasis with its diagonal elements equal to the eigenvalues. This basis set is unique if t he operator has no degenerate eigenvalues and not unique (in fact it is infinite) if t here is any degeneracy. A
A
(3) If two Hermitian operators, A and B , commute and if A has no degenerate eigenvalue, then each eigenvector of A
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A
A
A is also an eigenvector of B . In addition, we can const ruct a common orthonormal basis that is made of t he joint eigenvectors of A and B. "'
A
(4) The eigenvalues of an ant i-Hermitian operator are
11 oo : 01 : s2 A
A
A is also an eigenvector of B . In addition , we can construct a common orthonormal basis that is made of t he joint eigen,..,
A
vectors of A and B. (4) The eigenvalues of an ant i-Hermit ian operator are eit her purely imaginary or equal t o zero. (5) The eigenvalues of a unit ary operator are complex numbers of moduli equal to one; the eigenvectors of a unitary operator t hat has no degenerat e eigenvalues are mutually ort hogonal.
5.2
Parity operator
The space reflection about the origin of the coordinate system is called an inversion or a parity operation. This trans,.., formation is discrete. The parity operator P is defined by its action on t he kets f) of t he posit ion space: A
(5.4)
P r) = -r), so that
JJ'lj; (r) = 'lf; (-r)
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The parity operator is Hermit ian ,
P hysicsguide
Jvi athematical Background
pt = P, since
11 oo : 01 : s4 @Sk J ahiruddin, 2020
lVIathem atical Background
A
A
The parity operator is Hermit ian , p t = P , since
d3rcp*(r) ['P'lj; (r)]
d3rcp*(r)'lj;( - r)
=
d3rcp*(- r) 'lj;(r)
=
d r[Pcp (r)] *'ljJ (r) 3
(5.6) (5.7) hence A
2
A
P =I
p = p-1
or
(5.8)
The parity operator is therefore unitary, since its Hermitian adjoint is equal t o its inverse:
(5.9) Now, since P2 = f , the eigenvalues of P are + 1 or - 1 with t he corresponding eigenstates A
A
P'lj;+ (r)
=
"P+ (- r)
=
"P+(r),
P'lj;-(r)
=
"P-(- r)
=
- 'lj;-(r) (5.10)
The eigenstate "P+) is said to be even and "P- ) is odd . Therefore, the eigenfunctions of t he parity operator have definite parity: t hey are eit her even or odd. Since "P+) and "P- ) are joint eigenstates of the same Hermit ian operator P but with different eigenvalues, t hese eigenstates must b e A
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rir t
h f"IO'f"ln ~ l ·
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@Sk Jahiruddin, 2020
Mathematical Background
ort hogonal: 3
(1/J+ 1/J_) = =
d r'lj;~( - r)'l/J_(-r)
-
3
d r'ljJ~(rJ'l/J- (r)
- (1/J+ 1/J-) = 0 (5.11)
The states 1/J+) and 11/J- ) form a complet e set since any function can be written as 1/J(r) = 'l/J+(rJ + 1/J_(r), which leads to
'l/J- (r) =
1 2 ['ljJ (r) - 1P ( -r) ]
(5.12) Parity operator anti-commutes with position operator "'
"' -+
"'
PR r) = rP r) = f1 - f) "'
"'
RP r;
=
RI -
f)
=
(5.13)
- r1 - f)
hence we say "'
{P , R} = o
(5.14)
Similarly we can check t hat "'
{P , f} = o
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(5.15)
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5.3
Basis transformation
In quantum mechanics we deal with unitary transformat ions. When the elements are real the unitary transformat ion becomes orthogonal transformation. Nature does not allow us non unitary transformation in our present discussion of non relativistic quantum mechanics. Kets 'l/J) and bras ('l/J transform under unitary transformation as follows:
l'l/J') = Ul'l/J),
(5.16)
The action of operator under old and new basis A
A
'l/J) =
A' 'l/J') =
)
')
In new basis we write
A' l'l/J') =
') ->
A'U 'l/J) = u ) = uA 'l/J) -> A'U = uA
Multiplying both sides of [![rt =[rt[!= i we have
A'U =
uA by ut and since
'
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In summary we write (5.18)
and
11/J) = (;t 1/J'),
(1/J
=
(1/J' u, A= utA'u
(5.19)
Properties of unitary transformation: A
A
(1) If an operator A is Hermitian , its transformed A is also Hermitian, since
(5.20)
(2) The eigenvalues of
A and A' are same: (5.21)
(3) Commutators that are equal to (complex) numbers remain unchanged under unitary transformations, since the t ransform ation of [A, B] = a , where a is a complex number, is given by (5.22)
11 oo : 02 : os remain unchanged under unitary transformations, since t he t ransformation of [A, .BJ = a, where a is a complex number, is given by "' 1 "' 1 (5.22) A ,B = A,B A
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A
(4) (1/J A x), remain unchanged under unitary transformations A
1/J' A1 X1
f)t f)
A.
f)tu
(5.23)
x = (1/J A. x)
1 1 (1/J x )
= (1/J x)
(5.24)
(1/J' 1/JI)
= (1P 7P)
(5.25)
(5) f) _Af)t n = f) _An(Jt
(5.26)
(5.27)
A unitary transformation does not change the physics of a system; it merely transforms one description of the system to another physically equivalent description.
11 oo : 02 : 01 scription of the system to another physically equivalent description.
46
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5.4
Infinitesimal unitary transformation A
Consider an operator U which depends on an infinitesimally small real parameter E and which varies only slightly from t he unity operator I : A
Uc(G) = f + iEG
(5.28)
A
where G is called the generator of t he infinitesimal transformation. Clearly, Uf; is a unitary transformation only when the paramet er E is real and G is Hermitian, since A
A
where we have neglected t he quadratic t erms in E The transformat ion of a state vector l'lf') is
7P') = (i + icG) 'lf') = 1P) + bl'lf')
(5.30)
where A
(b*1 b*2 b*3
1/J)
A1 1 A1 2
A13
• • •
a1
A21
A22
A23
• • •
a2
A31
A32
A33
• • •
a3
• • •
• • •
• • •
• • •
•
•
•
• • •
(6.10)
6.4
Trace of matrix representation
The trace Tr(.A) of an operator A is given , within an ort honormal basis { 0 is little bit complicated. it is enough to remember is the reflection and t ransmission amplitude. Remember the reflection only. The t ransmission can be calculated by T = l - R R=
l
1 + (2n2 E /ma 2 )
( 5 15 )
·
The reflection and transmission coefficients for Delta potential barrier (V (x) = ab(x)) are same as delta function well scattering state. i.e for barrier
1 R = - -2 - - l + (2n E / ma 2 )
(5.16)
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Angular Momentum Sk J ahiruddin Assistant Professor Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay M.Sc Physics 2009-2011 batch He ranked 007 in IIT JAM 2009 and 008 (JRF) in CSIR NET J t1ne 2011 He has been teaching CSIR NET aspirants since 2012
1
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Angular Momentum
11 oo : oo : 04 1
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Angular JVIomentum
Contents 1 Angular momentum
4
1.1
Basic definition and commutation relation .
4
1.2
General formul ation of t he problem
•
•
•
•
•
5
•
•
•
•
•
8
Analytical expression of angular momentum operator in t hree dimension . . . . . . . . .
11
Matrix representation of angular momentum
14
Eigenvalues of J 2 and Jz ..
1.2.1 1. 3
1.4
1.5 Geometrical representation .
•
•
•
•
•
•
•
•
•
15
1.6 Spin
•
•
•
•
•
•
•
•
•
18
•
•
•
•
•
•
19
•
•
•
•
•
•
23
•
•
•
•
•
•
•
•
•
•
•
•
•
1.6.1 Stern Gerlach experiment 1.6.2
General theory of S = 1/ 2
1.6.3
P auli Matrices . . . . . . . . .
•
•
•
•
28
1.6.4
Spin part icle in magnetic field
•
•
•
•
31
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Spin part icle in magnet ic field . . . .
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2
31
Angular :NI01n entum
Addition of angular momentum 2. 1
Addition of t wo spin 1 / 2
2.2
Addit ion of t wo l
=1 . .
38
•
•
•
•
•
•
•
•
•
•
•
38
•
•
•
•
•
•
•
•
•
•
•
42
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1 1.1
P hysicsguide
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Angular JVIomentum
Angular momentum Basic d efinition and commutation relation
Angular moment um is defined as
L= R x P
(1.1)
You can evaluate t he commut ators like
[Lx, Ly] = [Y Pz - Z Py, Z Px - X Pz]
= [YPz,ZPx] + [Z Py, XPz] = Y [Pz, Z] Px + X [Z, Pz] Py
(1.2)
= -inY Px + inX Py = inLz Similarly evaluate the ot her commutators
[L x, Ly] = inLz [Ly, L z] = inLx [Lz, Lx] = inLy we write t his is
(1.3)
11 oo : oo : 11 [Lx, Ly] = inLz [Ly, L z] = inLx [L z, Lx] = inLy
(1.3)
we write t his is
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Angular Moment um
Prove
(1.5) and
(1.6) where ~ are the position vectors, i,e x , y , z and Pi are the momentum operators, i.e Px, Py, Pz
1.2
General formulation of the problem
We call angular momentum operator a set of variables which satisfy [Jx, Jy] =iriJz
[Jy, Jz] = inJx
(1. 7)
[Jz, l x] = inly We define an operator
J2 = J2 + J2 + J2 X y Z
(1.8)
11 oo : oo : 14 We define an operator
=
J 2
J 2 X
+
J 2 y
+ 1 Z2
(1.8)
Prove that 2
[J , l x] = [J;+ J;+ J;, l x] [J;, lx] + [J;, l x] = 0 j [email protected]
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Angular Niomentum
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2
So t he 1 operator commutes with all components of angular momentum operators. So we can take any of the component to construct a simultaneously commuting observable or CSCO with 1 2 . We take
= j(j + l )fi k , j, m) l z k,j , m) = mri k , j, m) 2
2
J k,j, m)
(1.10)
Inst ead of taking j (j + 1), we could have taken any constant like a, f3 etc. But as know already know what will happen so to make our calculation little bit easier we write our eigenvalue like t his. Now define
1+ =Ix+ i l y
(1.11)
J_ = l x - i l y
Like t he operators a and at of the harmonic oscillator, J+ - ..........1
7
- .... -
--L
l l - .... -.-- :.4....:
- .... . .
J. 1..... -- .. ___.., _
_
...J..: - . : .._...,4... _
_
,e - - -1..... -.l.- 1.... - ....
11 oo : oo : 16 Now define
l + = l x + i Jy
(1.11)
J_ = l x - il y
Like t he operators a and at of the harmonic oscillator, l + and J _ are not Hermitian : t hey are adjoints of each other.
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Angular Momentum
Prove these commutation relations first
[Jz, J+] = nJ+ [Jz, J_] = -nJ_
(1.12a) (1 .12b)
[J+, J_] = 2nJz [J 2, l +] = [J 2, J_] = [J2, Jz]
(1.12c) =
0
(1.12d)
Calculate the products l +l- and J_J+ to get
l +J- = (l x + ily) (l x - il y)
= J;+ J;-i[Jx, l y] = J; + J; + n,Jz
(1.13)
J_J+ = (Jx - i Jy) (Jx + i J y)
= J;+ J;+i[Jx, l y]
= 1:. + J~~ - nJ
7
(1.14)
11 oo : oo : 19 J_J+ = (Jx - i Jy) ( Jx + i J y)
= J; + 1i + i [J1: , Jy]
=
1 X2 + 1 y2
_
(1.14)
nJ Z
So we can write J+ l J _J+
=J
2
=J
2
-
J; + nJz J; - n,Jz
(1.15)
and also
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1 .2.1
Angular Moment um
Eigenvalues of J 2 and l z
As we have already calculat ed [Jz, l +]
=
[Jz, J _]
= - nJ _
hJ+
(1.17)
We easily see l zl + j, m)
and
= l +l z j , m) + nJ+ j, m) = m riJ+ j , m) + riJ+ j, m) = (m + l )riJ+ IJ, m)
(1.18)
11 oo : oo : 21 and
l zJ-lj, m) = J_JzlJ, m) - nJ_lj, m)
= mnJ_ lj, m) - nJ_ j, m) = (m - l )nJ_ j, m)
(1.19)
So l + j, m) is therefore an eigenvector of l z with the eigen-
+ 1) n and
J _ j, m) is there£ore an eigenvector of l z wit h the eigenvalue (m - l )n
value (m
So l + j, m) is proportional to j, m proportional to
lj, m
+ 1)
and J_ j, m) is
- 1) . So we write
j [email protected]
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8
Angular Niomentum
@Sk J ahiruddin , 2020
l + j , m)
= C+IJ, m + 1);
J_ j, m) =
c_ j, m -
1) (1.20)
Now, consider the vectors l + IJ , m) and J_ j, m) , and note t hat the square of their norms is posit ive or zero:
J+ j, m) I J_ j, m)
11
2
1 1
2
(j, m IJ_ J+ j, m) ~ 0 2 (j, m 1+ J_ j, m) ~ o
=
C+ 2
=
c_
=
We use the expression of l +J- and J_J+
(1.21)
11 oo : oo : 24 II J_
j, m) II =
c_
(j, m J+l- j , m)
We use t he expression of J+ J_ and J_ J+
(j,m J_J+ j,m)
= (j,m (J =
j(j
2
+ l)h
2
J; - fiJz) j , m)
-
2 2
m fi
-
-
2
mfi
=
2
C+ (1.22)
and (j,m lJ+J- j,m)
=
(k,j,m
2
(J
= j(j + l)fi
2
-
J; + fiJz) k,j,m) 2 2
m fi
-
+ mfi = 2
2
C_ (1.23)
Substituting these relations into t he expression of norms,
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Angular JVIoment um
equation (1. 21) , we get
+ 1) - m(m + 1) = (j - m)(j + m + 1) j(j + 1) - m(m - 1) = (j - m + l) (j + m) j(j
Hence
- (j + 1) -j
~ m ~
j
~
j
m
~
~
0
(1.24)
(1.25)
+l
We have calculated the values of C+ and write J+ j, m)
~ 0
IC- .
= nJj(j + 1) - m(m + l )lj, m + 1)
So we
(1.26)
11 oo : oo : 26 We have calculated the values of C+ and write
J+ j, m) = nJj(j
IC- .
So we
+ 1) -
m(m + l )lj, m
+ 1)
(1.26)
+ 1) -
m(m - 1) j, m - 1)
(1.27)
And
J_l j, m) = nJj (j
As you see the value of m increases as st ep of ONE, and t he value of m varies from -l to l, then we must have
- l +n= l;
(1.28)
Where n is an integer
So
2l = n
(1.29)
Hence l is integer or half integer And so m is also an integer or half integer j [email protected]
10
Physicsgt1ide
Angular :NI01nentum
@Sk J ahiruddin , 2020
1.3
Analytical expression of angular momentum operator in three dimension L = (n/i)(r x V )
(1.30)
and as
r
= rf·' and·'
a V = f 8r
A1 a 0 + ~a0
1 a + cp rsin08cp A
3 (l. l )
11 oo : oo : 29 and as r
" 8 0" 1 8 ;, 1 8 = r - + -- + 'f' ar r 80 r sin 0 8cp
n
= rf·' and·,
V
(1.31)
So L
n (
i
=
a r 8r
A
rr
")
X
( + r A
X
a 0") 80
( + r A
1 a '+' Sin 0 8cp ;,)
X
(1.32)
We know A
(f x f)
A
A
0, (f x 0) = cp, and (f x cp) = - 0
=
Hence
L = ~ i A
A
J a_ 0 80
1
a
(1.33)
(1.34)
sin 0 8cp
A
Write 0 and cp in cartesian components:
A
A
0 = (cos 0 cos cp)i + (cos 0 sin cp) j - (sin 0)k A
cp = -(sin cp)i + (cos cp)j jahir@physicsguide. in
(1.35)
Physicsguide
11
@Sk J ahiruddin , 2020
Angular JVIomentum
Hence
Ii
L =•
'l
(1.36)
So we get
n I
A\
00: 00: 32 •
i
(1.36)
So we get (1.37)
n
(1.38)
L y= -. i
(1.39) You can also write
L± = Lx ± iLy
(1.40) As cos cp
± i sin cp = e±i 8
8
( 1.41)
80 ± i cot 0 8cp You can easily calculate
8
2
8
2
+ cot 0 802 80
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12
L +L-
= - ri
2
8 8 + cot 0 O
4
-
½
2 (2V0 -
2 E
Va ) 2
2
=
~ [3 ± J1 + 4c ] ::::; 2
~o [3 ± (1 + 2c
2
)]
So the 2nd and 3rd eigenvalues are
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Perturbation and Variational
(b) The unperturbed Hamiltonian is
Ho = Vo
(1) 0 0 O l 0
0
0 2
And t he p ert urbation
H'
= EVo
(- 1) 0 0 0 0 1 0 1 0
There are t hree states
0
1
x1=
0 0
•
'
x2 =
1
0
•
'
X3
=
0 0 1
11 oo : oo : s6 There are t hree states
0
1
x1 =
0 0
•
'
x2 =
1
•
X3
'
0
0 0 1
=
First order correction on state 3 is ZERO as you know the matrix element of the perturbation matrix (X3 H' X3) = 0. So the result is zero. However if you want to do the matrix
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Perturbation and Variational
multiplication explicitly then ... - 1 0 0
Ej = (X3 H' X3) = EVo(O O 1)
=
EVo (O O 1)
0 1 0
0 0
=
0 0 1
0 1 1 0
0
The second order correction is
and hence TT
T
T
11 oo : oo : s1 m =l ,2
and hence
So
2
Ej = (x1 IH' X3) 1 + (x2 H' X3) Vo
2
Vo
You know the matrix elements, right???
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21
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Perturbation and Variational
@Sk J ahiruddin, 2020
So the result 2 _
E3
-
(x1
IH'
X3)
2
+ l(x2 H'
Vo
X3)
2 _
-
Vo
( 17)2 ; 17 _ E vo
vo -
217
E
vo
So t he energy aft er correction
E 3 = Eg
+ E~ + E ~ =
2V0 + 0 +
This is correct upto the order values of t he matrix .
2
E
2
E
Vo
=
V0 (2 +
2
E )
as found in t he exact eigen-
(c) The state 1 and state 2 are degenerate. So we need to find the matrix element of t he perturbation Hamiltonian first . And as t he matrix elements are already given. You have - 1 0 0 (\
(\
1
11 oo : 01
: 01
find the matrix element of the perturbation Hamiltonian first. And as t he matrix elements are already given. You have
-1 0 0 0 0 1 0 1 0
H' = cV0 Hence
(x1
H'I x1) =
- E½ ;
(x2 H' x2) = O; (x1 H ' x2)
=
(x2 H' x1)
=
o
So the W matrix will be - 1 0
W = cV0
0 0 22
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This is already diagonalized. So the eigenvalues are -E½, So to the first order the energy correction is
0.
A schematic diagram of t he problem is l3>
2~
~o
r
ll,)
,J
12>
l
l>
t2,>
'')
-2-Vo ~Vo
e1-
\lo Vo-Vo t:=-
Figure 1.2: Energies of t he Harmonic oscillator after t he perturbation
11 oo : 01
: 03
Figure 1.2: Energies of t he Harmonic oscillator after the perturbation
Exercise:
Do t he same problem for
(1 -E) E 0 H = Vo
1.2
E
1
E
0
E
2
Feynman Hellman theorem
Suppose t he Hamiltonian H , for a particular quantum syst em , is a function of some paramet er .\; let En (A) and 1/Jn (A) be [email protected]
@Sk J ahiruddin, 2020
23
P hysicsguide
Perturbation and Variational
the eigenvalues and eigenfunctions of H (A) . The FeynmanHellmann theorem states t hat
(1.12) We will apply it t o t he one-dimensional harmonic oscillator, (i) using A = w (this yields a formula for the expect ation value of V )
11 oo : 01
: 06
tor ,
=w
(i) using ,X
( t his yields a formula for the expectation
value of V ) (ii) using ,X
= n (this
(iii) using ,X
= m - this yields a relation b etween (T )
yields (T )), and (V) .
For Harmonic oscillator
En=
1
+ 2 nw;
n
n
2
H
d
2
1
= - - -2 + -mw x 2m dx
2 2
2
(1.13)
=w
(i) Take .X
8En
n
aw
8H
1
+ - n· 2
-
aw
'
=mwx
2
' ·
so Feynmann Helman theorem gives
1 n +2
ri
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2
= \ n mwx n) Physicsguide
24
@Sk Jahiruddin, 2020
Perturbation and Variational
2 2
As V = ! mw x 2 ' (V) =
1
2 2
n -mw x 2
n
=
n+
1 2
so
1 -w n+2 2 1
Hence
(V) = (ii) Take .X
=n
!2
(1.14)
11 oo : 01 : os (V) = (ii) Take ,,\
1
!2
n+ -
(1.14)
2
=n
aEn
1
an
2
- - = n +-
w·
'
aH an
n
d2
mdx 2
2
n2 d2
n
2mdx 2
-
So Feynman Hellman t h eorem gives 1
n+2
(T ) =
!2
n+
1 2
nw
(1.15)
(iii) Take ,,\ = m
aEn = O
am
and
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aH am
n2 d2 1 ----+-w2x2 2m2 dx 2 2 25
P hysicsguide
Perturbation and Variational
@Sk Ja.h iruddin, 2020
n2 d2 1 1 + 2 m 2mdx m So Feynman Hellman Theorem gives
0 = - l (T) m
+
l (V) m
(1.16)
or ( 1.17)
11 oo : 01
: 11
So Feynman Hellman Theorem gives
+
0 = - l (T) m
l (V) m
(1. 16)
or
(T ) = (V)
(1.17)
Feynman Hellman theorem can also be used to determine the expectation values of 1/ r and 1/r 2 for hydrogen. The effective Hamiltonian for the radial wave functions is
n d n l(l + 1) e 1 --- +--- ---2
H =
2
2m dr 2
2
2
2m
r2
41rE0 r
(1.18)
The energy is 4
me E - - - - - - - - -2 n 321r 2 E5n2 (Jmax + l + l )
(1.19)
(i) We take A = e
and
8H
2e 1 --41rEo r
ae 26
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Perturbation and Variational
So the t heorem says
4E _ _
e Hence
n-
e
1
21rEo
r
11 oo : 01
: 13
So t he t heorem says 1
4 E __ e ne 21rEo
r
Hence 1
81rEo E _ _ 81rEoE1 n e2 e2n2
r
87rEo e2
m
e2
2/i2
47rEO
2
e2m 1 41rE0 1i2 n 2
1
n2
and as
- - 2- = a me then 1
(1.20)
r
(ii) We take A = l then 8En
az -
2me
4
3 2 2 E ~li 321r (Jmax + l + 1)
and
8H
az
/i2 -2m_r_2 (2l
-
2En n
+ 1)
So from the t heorem
2En n [email protected]
2
/i (2Z + l ) 2m 27
@Sk J ahiruddin, 2020
or
1 r2
1 r2 P hysicsguide
Perturbation and Variational
n (2Z + l )/i2
n 3 (2l
+ 1)/i2
11 oo : 01
: 16
@Sk Jahiruddin, 2020
or
1 r2
Perturbation and Variational
4m E n n (2l + l )n2
as hence
4mE1
2
n,2
a2
1 r2
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4m E 1 n 3 (2l + l )n2
n3
1 (l +
t) a2
28
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2
Variational Principle
(1.21)
P hysicsguide
Perturbation and Variational
11 oo : 01
: 1s
@Sk Ja.h iruddin, 2020
2
Perturbation and Variational
Variational Principle
We want to calculate the ground state energy,Egs, for a system described by the Hamiltonian H, but we are unable to solve t he (time-independent) Schrodinger equation. The variational principle will get us an upper bound for E 98 which is sometimes all we need, and if we guess the wavefunction properly very close to the exact value. Here's how it works: Pick any normalized function 'l/; and we will get
(2.1 ) In t he exam we will be given a potential and t he trial wavefunction 'ljJ
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29
P hysicsguide
Perturbation and Variational
11 oo : 01 [email protected]
: 21
Physicsguide
29
@Sk Jahiruddin, 2020
Perturbation and Variational
You will need to remember few integrals +oo
e -oo CX)
2
- ax +bxd
b2
1r
x=
- exp a
4a
an ? 2 2 X'"'ne - a x dx = (-l)n __ 8a2n
-oo
00
-oo
fi(2n - 1) !! 2na 2n+l CX)
_
00
(- l) n an
dx (x2 + a2)n+l -
a>O
' CX)
n ! 8a2n _ 00 x2
1r(2n - 1)!! 2nn!a2n+l '
+ a2
a>O
bl -----7r - J (x - a)(b - x)dx =-(a+ b - 2vioJy), 2
a X
dx
O 0) in the new potent ial well; it s energy levels and wave function are now given by [email protected]
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P hysicsguide
Perturbation and Variational
@Sk Ja.h iruddin, 2020
n21r2n 2
E' = - - -2 n ~ 2m(8a)
2 . - Sln 8a
n1rx
8a
(0
a.
(a) Calculate t he total cross section in the low-energy limit. (b) Calculate the total cross section in t he high-energy limit. Solution:
(a) As the scattering is dominated at low energies by s-waves, l = 0, t he radial Schrodinger equation is
11 oo : 01 : os Solution: (a) As t he scattering is dominated at low energies by s-waves, l = 0, t he radial Schrodinger equation is
d2u(r) = Eu(r) 2m dr 2
-
where u(r)
n,2
( ) r>a
= r R (r) . T he solut ions of t his equation are
u(r) =
r a
2m E /n2 . The cont inuity of u(r) at r = a
where k 2 leads to sin ( ka
u1(r) = 0,
+ 60) = 0
~~>
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tan 50
= - tan (ka)
23
@Sk J ahiruddin , 2020
P hysicsguide
WKB and Scattering
since sin a = 1/ ( 1 + cot a) . The lowest value of the phase shift is 50 = -ka; it is negative, as it should be for a 2 2 repulsive potential. Insert sin 50 = sin (ka) into t he total scattering cross section formula 2
2
41r . 2 41r . 2 o-o = k 2 sin 50 = k 2 sin (ka)
For low energies, ka
1, the number of part ial waves contribut ing to the scattering is large. Assuming t hat lrriax ~ ka, we write t he total scattering cross section formula
since so many values of l cont ribute in t his relation, we 2 may replace sin bz by its average value, ~; hence
j a [email protected]
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Physicsguide
24
WKB and Scattering
where we have used I:~0 (2l+ 1) = (n+ 1) . since 1 we have 2
lmax
>>
(2.9)
11 oo : 01 where we have used 1 we have
: 1o
I:~0 (2l + 1) =
2
(n+ 1) . since lmax
21r 2 21r 2 2 a~ k2 l max = k2 (ka) = 21ra
>>
(2.9)
In conclusion, the cross section from a hard sphere potent ial is ( a) four times t he classical value, 1ra 2 , for lowenergy scattering and (b) twice the classical value for highenergy scattering. Example: Consider t he elastic scatt ering of 50 MeV neutrons from a nucleus. The phase shifts measured in
t his experiment are 60 = 95°, 61 = 72°, 62 = 60°, 63 = 35°, 64 = 18°, 65 = 5°; all other phase shifts are negligible ( i.e., 6t ~ 0 for l > 6) (a) Find the total cross section. (b) Estimate the radius of the nucleus.
Solution: (a) As 6z ~ 0 for l > 6 25
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@Sk J ahiruddin, 2020
4
6
k: L(2l + 1) sin 81 = 1: (sin 80 + 3 sin 81 + 5sin 8 + 7 sin 8 +9 sin 8 + 11 sin 85) = 1: x 10.702 2
a=
l=O
2
2
2
4
2
2
2
2
3
11 oo : 01 a = k2
: 12
2l + 1 sin Ol
..__
l=O
=
47f 2 2 2 2 ? (sin 80 + 3 sin 81 + 5sin 82 + 7 sin 83 k+9sin2 04
+ 11 sin
2
05)
=
1: x 10 .702
2
To calculate k , we need simply to use the relation E = n 2 k 2 / (2mn) = 50MeV, since the neutrons move as free part icles before scattering. Using mnc2 197.33MeV fim , we have
= 939.57MeV and
nc =
2
k 2 = 2mnE = 2 (mnc ) E = 2(939.57MeV) (50MeV) n2 (nc) 2 (197.33MeVfm) 2
= 2.41fm - 2 hence 47f a = _ lfm_ 2 x 10.702 24
2
= 55.78fm = 0.558barn
(b) At large values of l , when t he neutron is at its closest approach to t he nucleus, it feels mainly the effect of t he centrifugal potential l(l + l )n2 / (2m nr 2 ); t he effect of the nuclear potential is negligible. We may thus use the approximations E ~ l (l + l )n jahir@physicsguide. in
@Sk J ahiruddin , 2020
2
/
(2m nr~) ~ 42n
2
26
/
(2mnr~) where Physicsguide
"\i\'KB and Scattering
we have taken l ~ 6, since Oz ~ 0 for l > 6. A crude value of the radius of t he nucleus is t hen given by 21(197.33MeVfm) 2 (939.57MeV) (50MeV )
11 oo : 01 j [email protected]
26
: 16
Physicsguide
WKB and Scat tering
@Sk J ahiruddin , 2020
we have taken l ~ 6, since bz ~ 0 for l > 6. A crude value of the radius of t he nucleus is t hen given by
21/i2 mn E
21 (lie ) 2 (mn,c2 ) E
21 (197.33MeVfm)2 (939.57Me V) (50MeV)
= 4. 17fm
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27
Physicsguide
11 oo : 01 j [email protected]
26
: 11
Physicsguide
WKB and Scat tering
@Sk J ahiruddin , 2020
we have taken l ~ 6, since bz ~ 0 for l > 6. A crude value of the radius of t he nucleus is t hen given by
21/i2 mn E
21 (lie ) 2 (mn,c2 ) E
21 (197.33MeVfm)2 (939.57Me V) (50MeV)
= 4. 17fm
j [email protected]
27
Physicsguide
11 oo : 01 j [email protected]
26
: 20
Physicsguide
WKB and Scat tering
@Sk J ahiruddin , 2020
we have taken l ~ 6, since bz ~ 0 for l > 6. A crude value of the radius of t he nucleus is t hen given by
21/i2 mn E
21 (lie ) 2 (mn,c2 ) E
21 (197.33MeVfm)2 (939.57Me V) (50MeV)
= 4. 17fm
j [email protected]
27
Physicsguide
11 oo : oo : oo
Basic Therillodynaillics Sk J ahiruddin Assistant Professor Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay M.Sc Physics 2009-2011 batch He ranked 007 in IIT J AM 2009 and 008 (J RF) in CSIR NET J t1ne 2011 He has been teaching CSIR NET aspirants since 2012
1
@Sk J ahiruddin, 2020
Basic Thermodynamics
11 oo : oo : 03 1
@Sk J ahiruddin, 2020
Basic Thermodynamics
Contents 1
Introduction
4
2
Basic concepts
7
3
4
2. 1
System . . .
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9
2.2
P arameters
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11
2.3
Thermodynamic State
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13
2.4
equilibrium
. . .
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14
2.5
Equation of State
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17
Thermodynamic transformation
18
3.1
irreversible process
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19
3.2
reversible process .
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20
Work Heat Internal energy
j [email protected]
2
22
Physicsguide
11 oo : oo : 06
j [email protected]
2
P hysicsguide
@Sk J ahiruddin , 2020
5
Basic Thermodynamics
4. 1
Work.
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22
4.2
Heat .
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25
4.3
Internal energy
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27
Laws of thermodynamics
29
5. 1
First Law . .
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5.2
Carnot engine
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5.3
Second Law
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5.4
Third Law .
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Introduction
Statistical mechanics is the art of turning the microscopic laws of physics into a description of Nature on a macroscopic scale. Suppose you've got theoretical physics cracked. Suppose you know all t he fundamental laws of Nature, t he properties of the elementary particles and the forces at play between t hem. How can you t urn this knowledge into an understanding of t he world around us? More concretely, if I give you a box containing 1023 particles and tell you t heir mass, t heir charge, t heir interactions, and so on, what can you tell me about t he stuff in the box? There's one strategy t hat definitely won't work: writ2 ing down the Sehr·· odinger equation for 10 3 particles and solving it. That's typically not possible for 23 particles, let alone 102 3. What's more, even if you could find the wavefunction of the system , what would you do with it? The posit ions of individual particles are of little interest to anyone. We want answers to much more basic, almost childish, questions about t he cont ents of the box. Is it wet ? Is it hot? What colour is it? Is the box in danger of exploding? Wh:::i.t h:::i.n nPn ~ if wP ~n11 PP'ZP it. n1111 it . h P.:::i.t it 11n? Hnw r :::i.n
11 oo : oo : 11 positions of individual particles are of little interest to anyone. We want answers to much more basic, almost childish, questions about the contents of the box. Is it wet? Is it hot? What colour is it? Is the box in danger of exploding? What happens if we squeeze it, pull it, heat it up? How can
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we begin to answer these kind of questions starting from the fundamental laws of physics? For centuries from the 1600s to the 1900s scientists were discovering laws of physics that govern different substances. There are many hundreds of these laws, mostly named after their discovers. Boyle's law and Charles's law relate pressure, volume and temperature of gases (t hey are usually combined into the ideal gas law); the Stefan-Boltzmann law tells you how much energy a hot object emits; Wien's displacement law tells you t he colour of that hot object; the Dulong-Petit law tells you how much energy it takes to heat up a lump of stuff; Curie's law tells you how a magnet loses its magic if you put it over a flame ; and so on and so on. Yet we now know t hat these laws aren't fund amental. In some cases they follow simply from Newtonian mechanics and a dose of statistical thinking. In other cases, we need to throw quant um mechanics into the mix as well. But in all cases, we're going to see how derive them from first principles A large part of this course will be devoted to figuring
11 oo : oo : 13 1n 1ng. n o er cases, we nee quantum mechanics into the mix as well. But in all cases, we're going to see how derive them from first principles A large part of this course will be devoted to figuring out the interesting things that happen when you throw 102 3 particles together. One of t he recurring themes will be that 2 10 3 =f. 1. More is different: there are key concepts t hat j [email protected]
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are not visible in the underlying laws of physics but emerge only when we consider a large collection of particles. One very simple example is temperature. This is not a fundamental concept: it doesn't make sense to talk about the temperature of a single electron. But it would be impossible to talk about physics of the everyday world around us without mention of temperat ure. This illustrates the fact that the language needed to describe physics on one scale is very different from that needed on other scales. We'll see several similar emergent quantities in this course, including the phenomenon of phase transitions where the smooth continuous laws of physics conspire to give abrupt, discont inuous changes in the structure of matter . Historically, the techniques of statistical mechanics proved to be a crucial tool for understanding the deeper laws of physics. Not only is the development of the subject int imately tied with the first evidence for the existence of atoms, but quantum mechanics itself was discovered by ap-
11 oo : oo : 11 to be a crucial tool for understanding t he deeper laws of physics. Not only is the development of t he subj ect int imat ely t ied with t he first evidence for the existence of atoms, but quantum mechanics itself was discovered by applying statistical methods to decipher the spectrum of light emitted from hot objects. (We will study t his derivation in Section 3). However, physics is not a finished subject. There are many important systems in Nature from high temperature superconductors to black holes which are not [email protected]
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yet understood at a fundamental level. The information t hat we have about t hese systems concerns their macroscopic properties and our goal is to use these scant clues to deconstruct t he underlying mechanisms at work. The tools t hat we will develop in t his cour·se will be crucial in t his task.
2
Basic concepts
Few definit ions to start wit h 1. A system that is completely isolated from all outside influences is said t o be contained in adiabatic walls. We will also refer to such syst ems as insulated 2. Walls that are not adiabat ic are said to be diathermal
11 oo : oo : 19 1. A system that is completely isolated from all outside influences is said to be contained in adiabatic walls. We will also refer to such systems as insulated 2. Walls that are not adiabatic are said to be diathermal and two systems separated by a diathermal wall are said to be in t hermal contact . A diathermal wall is still a wall which means that it neither moves, nor allows part icles to transfer from one system to the other. However, it is not in any other way special and it will allow heat to be transmitted between systems. If in doubt , think of a thin sheet of metal. [email protected]
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3. An isolated system , when left alone for a suitably long period of t ime, will relax to a state where no further change is noticeable. This state is called equilibrium For a gas, the only two variables t hat we need to specify are pressure P and volume V : if you know the pressure and volume, then all other quantities - colour, smell, viscosity, t hermal conductivity all are fixed. For other syst ems, furt her (or different) variables may be needed to describe t heir macrostate. Common examples are surface tension and area for a film; magnetic field and magnetization for a magnet ; electric field and polarization for a dielectric. In what follows we'll assume that we're dealing with a gas and use P and V to specify the state. Everything t hat we say can be
11 oo : oo : 22 for a film ; magnetic field and magnetization for a magnet ; electric field and polarization for a dielect ric. In what follows we '11 assume that we 're dealing with a gas and use P and V to specify the state. Everything that we say can be readily extended to more general settings. In t he case of a magnetic solid the appropriate variables are t he magnetic field H, the magnetization M, and the temperature T . In more complicated situations, such as when a liquid is in contact wit h its vapor, addit ional variables may be needed: such as the volume of both liquid and gas VL , Va, the interfacial area A, and surface t ension (j. If t he t hermodynamic variables are independent of t ime, the system is said to be in a steady state. j [email protected]
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2.1
8
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System
Any macroscopic material body could be considered as a t hermodynamic system. Macroscopic system means a system composed of atoms or molecules of t he order of one Avogadro number (NA ~ 6.022 x 10 23 ) per mole. The examples of t hermo- dynamic system could be a wire under tension, a liquid film, a gas in a cylinder, radiation, a solid material, magnetic material, dielectrics, and many others. The thermodynamic systems should have a boundary which separates the systems from t he surroundings. Consider a drop of liquid as a thermodynamic system. The surface of t he liauid is the boundarv between the liauid and air. In t he
11 oo : oo : 24 material, magnetic material, dielectrics, and many others. The thermodynamic systems should have a boundary which separates the systems from t he surroundings. Consider a drop of liquid as a thermodynamic system. The surface of t he liquid is the boundary between the liquid and air. In t he language of t hermodynamics, the boundary is considered as a wall. This has been demonstrated in t he following Figure.
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SYSTEM
Figure 2.1: Schematic representation of a thermodynamic system. The shaded area is the surroundings or the universe. The thick line represents the boundary. The cent ral white space is the system.
11 oo : oo : 21 Figure 2.1: Schematic representation of a t hermodynamic system. The shaded area is the surroundings or the universe. The thick line represents the boundary. The central whit e space is the system.
The nature of the wall classifies t he thermodynamic system in different categories. (i) If the wall is such t hat energy or matter (atoms or molecules) cannot be exchanged between t he system and its surroundings, t hen the system is called isolated . Total energy E , and total number of part icles N are conserved for t his system. (ii) The wall is such t hat only energy could be exchanged between the system and t he surroundings. If the system is in thermal contact with a heat bath in t he surroundings, j [email protected]
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heat energy will be exchanged however total number of part icles will remain constant. This syst em is known as closed system. (iii) If t he wall is porous, t hen, beside energy, matter (atoms or molecules) can also be exchanged between t he system and the surroundings. If the system is in contact wit h a heat bath as well as with a part icle reservoir, heat energy and number of part icles both will be exchanged. Neither
11 oo : oo : 29 (atoms or molecules) can also be exchanged between t he system and the surroundings. If the system is in contact with a heat bath as well as with a particle reservoir, heat energy and number of particles both will be exchanged. Neither energy nor number of particles is conserved in this system. The system is called an open system.
2.2
Parameters
Thermodynamic paramet ers are measurable macroscopic physical quantities of a system. Consider a gas in a cylinder. Measurable physical quantities are t he pressure (P ) , t emperature (T ) and the volume (V ) of t he gas. These physical quant ities are called thermodynamic parameters or thermodynamic variables. The t hermodynamic variables are macroscopic in nature. They are divided in two categories, intensive and extensive parameters. [email protected]
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Extensive quantit ies are proportional to the amount of matter present - V , U, N, total magnetization etc Intensive quantities are independent of t he amount of matter present - P , T , µ magnetic field etc. Every intensive parameter has a corresponding indepen-
11 oo : oo : 32 Intensive quant ities are independent of t he amount of matter present - P , T , µ magnetic field etc. Every intensive parameter has a corresponding independent extensive parameter. They form a conjugate pair of t hermodynamic variables. Since they are independent of each other , one could be changed wit hout effecting the other. Keeping the pressure constant the volume of the gas can be changed and vice versa. A partial list of conjugate thermodynamic parameters are given in Table Syst em Wire Liquid film Fluid Charged particles Magnetic mat erial Dielectrics
Intensive paramet er Tension (T) Surface tension (,) P ressure (P) Electric pot ential ( cp) Magnetic field (B) External electric field (E)
Extensive paramet er Length (L) Surface area (A) St1rface area (A) Electric charge ( q) Total magnetization (M: Elect ric polarization (P)
For each system, there always exists one more pair of conjugate intensive and extensive parameters. They are temperature (T ) and ent ropy (S). Temperature is the ot her [email protected]
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intensive parameter and ent ropy is the corresponding conjugat e extensive parameter.
2.3
Thermodynamic State
11 oo : oo : 34 intensive parameter an entropy 1s t e correspon 1ng conjugate extensive parameter.
2.3
Thermodynamic State
Position and momentum coordinates are used to specify the state of a particle in mechanics. Similarly, the state of a t hermodynamic system can be specified by the given values of a set of thermodynamic parameters. For example, the state of a fluid system can be specified by t he pressure P, volume V, and t emperature T and specified as (P, V, T ). For an dielectric of polarization P at temperature T under an external electric field E , t he state is defined by (E , P , T). For a magnetic system t he state can be given by ( M , B , T ). For every thermodynamic systems there always exists t hree suitable thermodynamic parameters to specify t he st ate of t he system. It is important t o notice that thermodynamic parameters are all macroscopic measurable quantities. On t he other hand, microscopic quantities like position or moment um of the constit uent particles are not used for specifying t he state of a thermodynamic system.
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2.4
13
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equilibrium
The equilibrium condition in mechanics is defined as: in
00: 00: 37 Basic Thermodynamics
2.4
equilibrium
The equilibrium condition in mechanics is defined as: in absence of external forces, if a particle is slightly displaced from its st able equilibrium posit ion it will come back to its original position aft er some time. Consider a thermodynamic system like gas in a cylinder. Suppose the gas is in a state defined by the given thermodynamic parameter values (P, V, T ) . The downward force (W ) due to the weight of the piston is just balanced by the upward force exerted by the pressure (P ) of the gas, and the system is in equilibrium. If the piston is slightly depressed and released , it will oscillate around the equilibrium posit ion for some t ime and slowly come to rest at t he original equilibrium position. It means that if a small external force is applied to the system and released , the system would come back to the thermodynamic st at e it was in originally, i.e., the values of all the extensive and intensive parameters would recover. This definition is very similar to the definition of equilibrium given in mechanics and known as mechanical equilibrium of a thermodynamic system. Apart from mechanical [email protected]
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equilibrium, the syst em should have thermal and chemical
11 oo : oo : 39 @Sk J ahiruddin, 2020
Basic Thermodynamics
equilibrium, the system should have thermal and chemical equilibrium as vvell to achieve thermodynamic equilibrium of a system. Consider an isolated system with two partial systems. Initially each of them are in equilibrium at different t emperatures. Temperature at all points of each system are the same. They are now taken into thermal contact, only exchange of heat and no exchange of matter, with each other. Heat would flow from the system of higher temperature to t he system of lower temperature until uniform temperature is attained t hroughout t he combined system. The system is then in thermal equilibrium. Experience shows, all systems which are in thermal equilibrium with a given system are also in thermal equilibrium with each other. This principle defines t he temperature of a t hermodynamic system and known as zeroth law of thermodynamics. Hence systems which are in t hermal equilibrium with each other have a common intensive property, i.e., t emperature. Suppose the system is a mixture of several different chemical components. When t he composition of t he syst em remain fixed and definite, t he system is said to be in chemical equilibrium. Generally, chemical equilibrium takes a long [email protected]
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t ime to achieve. Sometimes the system appears to be in chemical equilibrium, having fixed amount of components but the chemical reaction may continue with an extremely slow reaction rate. The mechanical equilibrium t herefore refers to uniformity of pressure, t he thermal equilibrium refers to uniformity of t emperature and the chemical equilibrium refers to t he constancy of chemical composit ion. If t here exist in t he system gradients of macroscopic parameters such as pressure, temperature, density, etc such a state of t he system is referred as a non-equilibrium state. A system which satisfies all possible equilibrium conditions is said to be in thermodynamic equilibrium. Thermodynamic equilibrium is thus correspond to the sit uation when t he t hermodynamic state does not change wit h time.
T = const. P = const. µ = const.
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Thermal equilibrium Mechanical equilibrium Chemical equilibrium
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2.5
Physicsguide
Basic Thermodynamics
Equation of State
In equilibrium t he stat e variables are not all independent and are connected by equations of state. The role of statist ical mechanics is the derivation, from microscopic interact ions, of such equations of state. Simple examples are the ideal gas law,
where N is the number of molecules in the system and kB is Boltzmann's constant; the van der Waals equation,
where a , b are constants; the virial equation of state
1+
N B2(T) N 2 B3(T ) V + v2 + ... = 0
where the functions B 2 (T ), B 3 (T ) are called virial coef-
11 oo:oo:47 1+
NB2(T ) V
2
+
N B3(T ) v2
+ .. .
= 0
where t he functions B 2 (T ), B 3 (T ) are called virial [email protected]
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ficients; and in the case of a paramagnet, the Curie law, M- C H = 0
T
where C is a constant called t he Curie const ant. These equations of st ates are approximations, and we shall use t hem primarily to illust rate various principles. The virial equation of state is, in principle, exact, but calculation of more t han a few of the virial coefficients is very difficult.
3
Thermodynamic transformation
A thermodynamic transformation is a change of state. If one or more of t he paramet ers of a syst em are changed, the state of the system changes. It is said t hat the system is undergoing a transformation or process. The transformat ion is generally from an initial equilibrium state to a final equilibrium st ate. T hermodynamic processes are classified into two groups (i) irreversible and (ii) reversible.
11 oo : oo : so t ion is generally from an initial equilibrium state to a final equilibrium state. Thermodynamic processes are classified into two groups (i) irreversible and (ii) reversible.
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3.1
Physicsguide
Basic Thermodynamics
irreversible process
The water from the slopes of the Himalayas flows down the Ganges into the Indian Ocean. The water in the Indian Ocean will never go back to the hill on itself even if t he total energy loss during the down flow producing heat and sound energy is supplied back to the water at the Ocean. This means t hat the work done in the forward process is not equal to the work done in the backward process. Such natural flow of liquid downward is spontaneous and is irreversible. Almost all natural spontaneous processes are irreversible, just reversing the direction of the process it is not possible to get back the initial state. Consider free expansion of a gas. It does no work during t he free expansion however to compress it back to the original volume a large amount of work has to be performed on t he gas. A pendulum without a driving force will by itself cease to swing after some t ime, since its mechanical energy is transformed into heat by friction. The reverse process, that a pendulum starts swin~ bv itself while the surroundin~s cool. has never been
11 oo : oo : s3 work has to be performed on t he gas. A pendulum without a driving force will by itself cease to swing after some t ime, since its mechanical energy is transformed into heat by friction. The reverse process, that a pendulum starts swing by itself while the surroundings cool, has never been occurred. It is characteristic of irreversible processes that they proceed over non-equilibrium states dissipating energy in various forms during the transformation from one state j [email protected]
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to t he other. Ferromagnets are magnetized by applying external magnetic field. If the external filed is reduced the magnetization curve does not follow the original path and forms a hysteresis loop because during magnetization the system dissipates energy in the form of heat and sound.
3.2
reversible process
In a reversible process, the change of states occurs only over equilibrium intermediate states. That is to say, all steps between the final and init ial states are in equilibrium during a reversible process. A reversible process is t hen an idealizat ion. Because, if a system is in thermodynamic equilibrium, t he parameters should not change with time. On t he contrary, in order to change the state one needs to change the parameter values. However, a reversible process could be realized in a quasi static manner.
11 oo : oo : 55 t he parameters should not change with time. On t he cont rary, in order to change the state one needs to change the paramet er values. However , a reversible process could be realized in a quasi static manner. In a quasi static process, infinitesimal change in t he paramet er values are made sufficient ly slowly compared to the relaxation time of t he system . Relaxation t ime is t he t ime required for a system to pass from a non-equilibrium state to an equilibrium state. Thus, if t he process rate is [email protected]
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erably less t han t he rate of relaxat ion, there will be enough t ime for the parameters to equalize over the entire system and t he system could be considered at equilibrium. The process will represent a cont inuous succession of equilibrium states infinitely close to each other and could be considered as a reversible process, reversing t he direction of t he process one could reach to the init ial state from the final state following the same path. The reversible change could be performed under different conditions. Consider a t hermally insulated system where no heat exchange is possible, any process under this condition is an adiabatic process. Reversible adiabatic process are also known as iso-entropic process. If a system undergoes a change keeping temperature constant , it is called an isothermal process, it is an isochoric process if volume kept constant and it is an isobaric process if pressure remains constant.
11 oo : oo : sa Reversible adiabatic process are also known as iso-entropic process. If a system undergoes a change keeping temperature constant, it is called an isothermal process, it is an isochoric process if volume kept constant and it is an isobaric process if pressure remains constant.
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4 4.1
Basic Thermodynamics
Work Heat Internal energy Work
Work app ears during a change in state. The definition of work in t hermodynamics is borrowed from mechanics and it is given by -+ ➔ bW=-F-dR. (4.1) -+
where F is the force acting on the syst em during a small ➔ displacement dR.. The negative sign is a convention in t hermodynamics and it is decided by the fact that: work done by t he system is negative and work done on the system is positive. In mechanics, doing work t he potential or kinetic en_ __ ......,__ _ £J..L ,....
- --- J.. - . . . . -- .: _
- 1--- .. -- ............. .-l
Q ,: _____ .:l - . ..... 1 __
___ _ ___ , _ ,: _
---· -- =--- 1 - .... . . J..
11 oo : 01 : oo ➔
displacement dl!, . The negative sign is a convention in thermodynamics and it is decided by the fact that: work done by t he system is negative and work done on the system is posit ive. In mechanics, doing work t he potential or kinetic energy of the system is changed. Similarly, work is equivalent to energy exchange in t hermodynamics. Energy exchange is positive if it is added to a system and it is negative if it is subtracted from a system. Not e t hat, only macroscopic work is considered here, and not on an atomic level. Consider a gas enclosed in a cylinder at an equilibrium t hermodynamic state (P, V, T ). Assuming that there is no friction between the piston and the cylinder, the force acting on the gas, i.e., the weight on the piston, is F = PA [email protected]
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Basic Thermodynamics
where A is the cross sectional area of the piston. In order to compress the volume by an infinitesimal amount dV, the piston is pushed down by an infinitesimal amount di!,. The corresponding work done is bW
-I
=
➔
- F · df
=
PAdi!,
=
- PdV
(4.2)
since pressure is acting in a direction opposite to the displacement and Adi!, = -dV during compression. The same definition is also valid for expansion. In case of expansion, t he pressure will act in the same di1·ection of the displacement and Adi!, = dV
11 oo : 01
: 03
definition is also valid for expansion. In case of expansion, the pressure will act in the same direction of the displacement and AdR = dV Thus, work is the product of an intensive state quantity (pressure) and the change of an extensive state quantity (volume) . One could easily verify that the same definition can be applied to t he other thermodynamic syst ems. For example, in case of dielectrics and magnetic mat erials, in order to change the electric polarization P or magnetization M by a small amount dP or dM in presence of electric field E or magnetic field H , the amount of work has to be performed on the syst ems are
JW
= E · dP
or
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JW
= B · dM
(4.3) Physicsguide
Basic Thermodynamics
where E and B are the intensive parameters and P and M are ext ensive parameters. In order to change t he particle number by dN, one should add particles those have energy comparable to the mean energy of other particles otherwise equilibrium will be lost. Let us define
JW
= µdN
(4.4)
as the work necessary to change the particle number by dN. The intensive field quantity µ is called the chemical
11 oo : 01 : os (4.4)
V1 , t hen L).Q 1 > 0, i.e., the amount of heat L).Q 1 is absorbed by the gas from t he surroundings. Step 2: Adiabatic expansion of t he gas from ½ to V3 . The temperature decreases from T1 to T2 (T1 > T2 ). The 312 equation of state is ½/V2 = (T1 / T2) . since L).Q = 0
(5.10)
Step 3: Isothermal compression from ½ to ¼ at temperature T 2 . The equation of state is: P3 V3 = P 4V4 = RT2 . Again one has,
!1E3 = \
A /"'I
L). W 3 A
r ;r ,
+ L).Q2 = nrri
1 • ·~
0 (
V3\
11 oo : 01 : 3s Again one has,
(5.11)
since ½ > V4 , ~Q 2 < 0, the amount of heat is released by t he gas. Step 4: Adiabatic compression from V4 to V1 . Here temperature increases from T2 to T1 and the equation of state [email protected]
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(5.12)
The net change in internal energy ~E = ~E1 + ~E2 + ~E3 + ~E4 = 0 as it is expected. Consider the amount of heat exchanged during the isothermal processes:
and
½ ~Q2 = - RT2ln ¼
(5.13)
During t he adiabatic processes, one has combining t he equation of st ates
11 oo : 01
: 37 (5.13)
During t he adiabatic processes, one has combining t he equation of states (5.14)
This implies (5.15)
If the Carnot 's cycle is made of a large number of infinitesimal steps, the above equation modifies to 6Q
T j [email protected]
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=0
(5.16) Physicsguide
Basic Thermodynamics
This is not only true for Carnot 's cycle but also t rue for any reversible cyclic process. Suppose that t he state of a t hermodynamic system is changed from state 1 to stat e 2 along a path C1 and the system is taken back to t he initial state along another reversible path C2 , as shown in Figure 5.2. Thus, 1 - C 1 - 2C2 - 1 forms a closed reversible cycle and one has
X
1
11 oo : 01
: 40
1
y Figure 5.2: A reversible cyclic process on a XY diagram where X and Y form a conjugate pair of thermodynamic variables. 2
or 1
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8Q T
+
l
2
8Q T
=0
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Basic Thermodynamics
Since, the paths are reversible, one also has
8Q T
l 2
so hence 2 1
8Q T
2 1
2
1
8Q T 8Q T
(5.18)
(5.19)
Thus t he integral J 8Q/ T is path indep endent, i .e., independent of t he process of heating or cooling the system. The integral depends only on the init ial and final states of t he syst em and t hus represents a state function whose total differential is 8Q / T. since heat is an extensive quantity,
11 oo : 01
: 42
pendent of the process of heating or cooling the system. The integral depends only on the initial and final states of t he system and t hus represents a state function whose total differential is bQ /T. since heat is an extensive quantity, t his st ate function, say S, is also extensive whose conjugate intensive parameter is temperature T. This extensive state function is t he entropy S and defined as dS
= bQ T
(5.20)
and
Note that, only ent ropy difference could be measure, not the absolute entropy. The statistical mechanical definit ion of entropy will be given in t he section of st atistical mechanics.
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5.3
Basic Thermodynamics
Second Law
The second law of thermodynamics introduces the entropy S as an extensive state variable and states that for an infinitesimal reversible process at temperature T , t he heat given to t he system is
aQ rev
=
TdS
11 oo : 01 : 4s
aQ rev
=
TdS
while for an irreversible process
a Q irrev
~ TdS
If we are only interested in thermodynamic equilibrium states we can use a Q rev = T dS and t reat the ent ropy S as t he generalized displacement which is coupled to the 'force' T. The above formulation of t he second law is due to Gibbs We present next two equivalent statements of the second law of t hermodynamics. The Kelvin version is: There exists no thermodynamic process whose sole effect is to extract a quantity of heat from a j [email protected]
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Basic Thermodynamics
system and to convert it entirely to work.
The equivalent stat ement of Clausius is:
No process exists in which the sole effect is that heat flows from a reservoir at a given temperature to a reservoir at a higher temperature.
11 oo : 01
: 47
No process exists in which the sole effect is that heat flows from a reservoir at a given temperature to a reservoir at a higher temperature.
A corollary of t hese statements is that the most efficient engine operating bet ween two reservoirs at temperatures T1 and T2 is the Carnot engine. The first law of t hermodynamics tells us about t he conservation of energy in a t hermodynamic process during its change of st at e. The second law tells us about t he direction of a natural process in an isolated system. The ent ropy S = 8Qrev/T is t he amount of heat reversibly exchanged with the surroundings at temperature T . since t he amount of heat 8Qirr exchanged in an irreversible process is always less than that of 8Qrev exchanged in a reversible process, it
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Basic Thermodynamics
is t hen always true that 8Qirr < 8Qrev = T dS
(5.21)
For an isolated system, 8Qrev = 0. Therefore, in an isolated system t he ent ropy is const ant in t hermodynamic equilibrium and it has an extremum since dS = 0.
11 oo : 01 : so 6Qirr < 6Qrev = TdS
(5.21)
For an isolated system, 6Qrev = 0. Therefore, in an isolated system t he ent ropy is constant in t hermodynamic equilibrium and it has an extremum since dS = 0. It is found t hat in every sit uation this extremum is a maximum. All irreversible processes in isolated system which lead to equilibrium are t hen governed by an increase in ent ropy and t he equilibrium will be reestablished only when t he entropy will assume its maximum value. This is t he second law of thermodynamics. Any change of state from one equilibrium st at e to another equilibrium state in an isolated syst em will occur naturally if it corresponds to an increase in entropy. For reversible process the ent ropy change is zero.
dS= 0
(5.22)
and for irreversible processes the entropy change is al-
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ways posit ive
dS > 0
(5.23)
Note t hat, entropy could be negative if there is heat exchange wit h the surroundings i.e., t he system is not an iso-
11 oo : 01 : s2 ways posit ive dS > 0
(5.23)
Note t hat, entropy could be negative if there is heat exchange with the surroundings i.e., the system is not an isolated system. It is posit ive only for an isolated system. The first law for reversible changes now can be rewritten in terms of entropy: Fluid system: Surface film:
dE = T dS - P dV dE = T dS + ,ydA
Strained wire: Magnetic materials:
If there is exchange of energy of several different forms, the first law should take a form
dE = TdS - P dV
5.4
+ H dM + µdN + · · ·
(5.24)
Third Law
The third law of thermodynamics deals with t he entropy of a system as the absolute t emperature tends to zero. It is [email protected]
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already seen t hat
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c
c
11 oo : 01
: 55
@Sk J ahiruddin, 2020
Basic Thermodynamics
already seen t hat
(5.25) and one can only measure the ent ropy difference between two states. The absolute value of ent ropy for a given thermodynamic states remains undetermined because of the arbitrary additive constant depending on the choice of the initial state. The t hird law enables us to determine t he additive constant appearing in the definition of entropy. It states t hat: t he ent ropy of every system at absolute zero can always be taken equal to zero,
lim S=O
T➔ O
(5.26)
The zero temperat ure ent ropy is t hen independent of any other properties like volume or pressure of the system. It is generally believed t hat the ground state at T = 0 is a single non-degenerate state. It is therefore convenient to choose this nondegenerate state at T = 0 as the standard initial state in the definition of entropy and one could set t he entropy of t he standard state equal to zero. The entropy of any state A of t he system is now defined, including the [email protected]
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Basic Thermodynamics
additive constant , by the integral
S(A) =
A
T =O
,SQ T
(5.27)
where the integral is taken along a reversible transformation from T = 0 state (lower limit) to the state A . since dQ = G(T) dT , the entropy of a system at temperature T can also be given as T
Gv(T ) dT
T
0
T
or
Gp(T ) dT
T
0
(5.28)
when the system is heated at constant volume or constant pressure. As a consequence of the third law S (0) = 0, the heat capacities Gv or Gp at T = 0 must be equal to zero otherwise the above integrals will diverge at the lower limit. Thus, one concludes
Gv
or
Gp
➔
0
as
T
➔
0
(5.29)
The results are in agreement with the experiments on t he sp ecific heats of solid
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Basic Thermodynamics
additive constant , by the integral
S(A) =
A
T =O
,SQ T
(5.27)
where the integral is taken along a reversible transformation from T = 0 state (lower limit) to the state A . since dQ = G(T) dT , the entropy of a system at temperature T can also be given as T
Gv(T ) dT
T
0
T
or
Gp(T ) dT
T
0
(5.28)
when the system is heated at constant volume or constant pressure. As a consequence of the third law S (0) = 0, the heat capacities Gv or Gp at T = 0 must be equal to zero otherwise the above integrals will diverge at the lower limit. Thus, one concludes
Gv
or
Gp
➔
0
as
T
➔
0
(5.29)
The results are in agreement with the experiments on t he sp ecific heats of solid
j [email protected]
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11 oo : oo : 01
Thermodynamic potentials and applications Sk J ahiruddin
Assistant P rofessor· Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay lVI. Sc Physics 2009-2011 batch He ranked 007 in IIT JAlVI 2009 and 008 (JRF) in CSIR NET June 2011 He has been teaching CSIR NET aspirants since 2012
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Thermodynamic potentials and applications
11 oo : oo : 04 1
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Thermodynamic potentials and applications
Contents 1 Thermodynamic Potentials
4
1.1
Entropy as a t hermodynamic potential: . . .
5
1. 2
Enthalpy as a t hermodynamic potential: . .
6
1.3
Helmholtz free energy as a t hermodynamic potential: •
•
8
Gibbs free energy as a thermodynamic pot ential: . . . . . . . . •
9
•
1.4
•
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•
•
•
•
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•
•
1.5
Grand potential as a thermodynamic potential: 11
1.6
Maxwell's relations
•
•
•
•
•
•
•
•
•
•
•
•
•
•
13
1.7
Response functions
•
•
•
•
•
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•
•
•
•
•
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•
•
14
•
•
•
•
•
•
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•
•
•
•
•
14
1.8
1.7.1
Definit ions .
1.7.2
Relations between t he response function 15
Thermodynamics of a magnetic system 1.8.1
j [email protected]
Definition of potentials 2
•
•
•
•
•
•
•
•
17
•
•
•
18
Physicsguide
11 oo : oo : 06 1.8.1
Definition of potentials
j [email protected]
•
•
•
•
•
•
•
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2
•
Thermodynamic potent ials and applicat ions
1.8.2
Maxwell relations for magnetic systems 20
1.8.3
Response functions for magnetic systems . . . . . . . . . . . . . . . . . .
Some applications of thermodynamics
20
22
2.1
TdS equations . . . . . . .
•
•
•
•
•
•
•
•
•
•
22
2.2
Internal energy equations .
•
•
•
•
•
•
•
•
•
•
23
2. 3
Heat transfer of isothermal expansion of van der Waals gas . . . . . . . . . . . . . . . . .
24
2.4
Reversible isothermal change of pressure . .
25
2.5
Internal energy of van der Waal gas
•
•
•
•
•
27
2.6
Heat capacities of ideal gas .
•
•
•
•
•
•
•
•
•
28
2.7
Gibbs paradox:
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
30
2.8
Radiation: . . .
•
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•
31
11 oo :oo :oa •
•
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•
•
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•
•
•
•
•
•
•
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1
•
Thermodynamic potent ials and applications
Thermodynamic Potentials
By the second law of thermodynamics, an isolated system during a spontaneous change reaches an equilibrium st ate characterized by maximum entropy:
dS = 0,
S = Smax
(1.1)
On the other hand, it is known from mechanics, electrodynamics and quantum mechanics t hat a syst em which is not isolated minimizes its energy. An interacting t hermodynamic syst em always exchanges heat or perform work on t he surroundings during a spontaneous change to minimize its internal energy. However, t he entropy of the system plus the surroundings, which could be thought as a whole an isolated syst em, always increases. Thus, a non-isolated system at constant entropy always leads to a st at e of minimum energy.
11 oo : oo : 11 energy.
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1.1
P hysicsguide
4
Thermodynamic potentials and applications
Entropy as a thermodynamic potential:
Both entropy and t he internal energy are state funct ions. If t hey are known as function of state variables of an isolat ed syst em then all ot her thermodynamic quantities are complet ely known. Consider t he internal energy E = E (S, V, N) t hen , t he different ial form of energy is
dE = TdS - P dV
+ µdN
(1.2)
So, t he temperat ure and pressure are known as fun ctions of other state variables
T=
oE as
V,N
'
- P=
oE av
S ,N
'
µ=
oE 8N s,v
(1.3)
Similarly, consider t he entropy S
TdS = dE
= S(E, N , V ), then
+ P dV
- µdN
and t he t emperature and pressure can be found as
(1.4)
11 oo : oo : 14 Similarly, consider the entropy S = S(E, N , V ), then
TdS = dE + P dV - µdN
(1.4)
and t he temperature and pressure can be found as 1
T
as aE
V,N
'
P=T
as av
E ,N
'
µ=- T
as aN
EV '
(1.5) j [email protected]
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Thermodynamic potentials and applications
The entropy and the internal energy then can be calculated as functions of t he state variables form t he equation of state. Since t he equilibrium state of the system is given by a maximum of the entropy as a fun ction of (E , V), it gives information about the most stable equilibrium state of the system as potential energy does in mechanics. As the difference in potential energy defines the direction of a natural process in mechanics, t he entropy difference determines the direction of a spontaneous change in an isolated system. Thus, the ent ropy can be called as a t hermodynamic potential.
1.2
Enthalpy as a thermodynamic potential:
The enthalpy of a syst em is defined as
11 oo : oo : 16 tial: The enthalpy of a system is defined as
(1.6)
H =E+PV
t he differential form is dH = dE+ P dV + VdP
=:>~>
dH = TdS
+ V dP + µdN (1.7)
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Knowing t he enthalpy H = H (S, P , N) , the state variables can be calculated as T =
8H
as
P ,N
'
V =
8H BP
, S,N
µ=
8H 8N
S,P
(1.8) Consider an isolated system at constant pressure. Process at constant pressure are of special interest in chemistry since most of the chemical reactions occur under constant atmospheric pressure. In an isolated-isobaric system, bQ = 0 and P is constant , thus
dE+PdV = 0
=:>~>
dH
=0 (1.9)
In a spontaneous process of an adiabatic-isobaric system, t.h P Pn11ilihri11m rnrrPsnnnrls t.n t.h P minim11m nf t.hP Pnt.h::i.lnv
11 oo : oo : 19 ===>~►
dE+PdV = 0
+ PV ) = 0
d( E
===>~►
dH = 0 (1.9)
In a spontaneous process of an adiabatic-isobaric system, t he equilibrium corresponds to t he minimum of the enthalpy
dH = 0,
H (S, P ) =
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1.3
(1.10)
H min
P hysicsguide
Thermodynamic potent ials and applications
Helmholtz free energy as a thermodynamic potential:
The Helmholtz potential (free energy) is defined as
F=E-TS
(1.11)
t he differential form
dF = dE-SdT-TdS
dF = -SdT - P dV +µdN (1.12) since dE = T dS- P dV +µdN . Thus, knowing F = F (T , V, N), S, P and µ could be determined as
8F 8T
V,N
'
~~►
- P=
8F
av
T ,N
'
µ
=
8F 8N
T ,V
11 oo : oo : 21 since dE = T dS - PdV +µ dN . Thus, knowing F S, P and µ could be determined as
- S=
8F 8T
V,N
'
8F
- P =
av
T ,N
'
µ=
=
F (T , V, N),
8F 8N
TV
'
(1.13) The Helmholtz potential is useful in defining t he equilibrium
of a non-isolated system in contact with heat bath at constant temperature T . The system is interacting wit h t he heat bath through heat exchange only. Consider an arbit rary isothermal t ransformation of this system from a state A to state B. By t he second law , one have (1.14) j [email protected]
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Since T is constant (1.15)
where ~Q is the amount of heat absorbed during the transformat ion and ~ S = S(B) - S(A). Using t he first law, the inequality could be written as (1.16)
where ~ W is the work done by t he system. Thus, the equilibrium of an isothermal system which does not perform work (mechanically isolated) always looks for a minimum of Helmholtz potential. Irreversible process happen spontaneously, until t he minimum
11 oo : oo : 24 librium of an isothermal system which does not perform work (mechanically isolated ) always looks for a minimum of Helmholtz potential. Irreversible process happen spont aneously, until t he minimum dF
=
0,
F
(1.17)
= Fmin
is reached.
1.4
Gibbs free energy as a thermodynamic potential:
The Gibb's potential (free energy) is defined as G = E - TS + PV = F + PV [email protected]
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(1.18)
Thermodynamic potentials and applications
t he differential form dG = - SdT + V dP
+ µd N
(1.19)
since E = T S - PV + µ N for a system attached wit h heat bath as well as wit h a bariost at . System exchanges heat and does some work due to volume expansion at constant pressure. The thermodynamic variables can be obtained in terms of G (P, T , N) as
- S=
ac 8T
P,N
'
V=
ac 8P
T ,N
'
µ=
ac 8N
T ,P
11 oo : oo : 21 pressure. The thermodynamic variables can be obtained in terms of G(P, T , N) as
ac
-S=
8T
'
P,N
ac
V=
8P
T ,N
'
µ=
ac 8N
T ,P
(1.20) Notice that the chemical potentialµ can be defined as Gibb's free energy per particle. Consider a system at constant pressure and temperature. For isothermal process, (1.21)
as it is already seen. If the pressure remain constant P~V, t hen
~W
=
(1.22)
Thus, a syst em kept at constant temperature and pressure, t he Gibb's free energy never increases and the equilibrium j [email protected]
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Thermodynamic potent ials and applicat ions
state corresponds to minimum Gibb's potent ial. Irreversible spontaneous process in an isothermal - isobaric system dG
=
0 G
= G roin
(1.23)
are always achieved.
1.5
Grand potential as a thermodynamic J
J
•
1
11 oo : oo : 30 are always achieved.
1.5
Grand potential as a thermodynamic potential:
The grand potential is defined as
qJ
=
E - T S - µN
=
F - µN
= -
PV
(1.24)
since E = TS - PV + µN . The syst em attached with heat bath as well as with a part icle reservoir. System exchanges heat with heat bath and exchanges particle with the particle reservoir. Differentially the grand potential can be expressed as
dqJ = dE - TdS - SdT- µdN - Ndµ
(1.25)
= -SdT - P dV - Ndµ
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Thermodynamic potentials and applications
since dE = TdS - PdV + µdN. The thermodynamic variables are then obtained in terms of qJ (V, T , µ) as
-S =
8T
, -P = V ,µ
av
T ,µ
'
-N=
aqJ
8µ
TV '
(1.26)
Consider an isothermal system at constant chemical potent ial. For an isothermal svstem
11 oo : oo : 32 -S=
ar
, V,µ
-P=
av
, -N= T ,µ
a 8µ
TV ,
(1.26) Consider an isothermal system at constant chemical potent ial. For an isothermal system (1.27)
and ~ W = - µ~N since µ is constant and the inequality leads to (1.28) Thus, a system kept at constant temperature and chemical potential, the grand potential never increase and the equilibrium stat e corresponds to minimum grand potential. Irreversible spontaneous process in an isothermal system with constant chemical potential correspond to (1.29)
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1.6
12
Physicsguide
Thermodynamic potentials and applications
Maxwell's relations
A number of relations between the thermodynamic state variables can be obtained since the thermodynamic potent ials E, H , F and G ( also ) are state functions and have
11 oo : oo : 3s axwe A number of relations between the thermodynamic state variables can be obtained since the thermodynamic potent ials E , H , F and G ( also ) are state functions and h ave exact different ials. We know for exact differentials dcp
aM ay
= Mdx + Ndy
aN ax
X
(1.30) y
• From dE
aT av
= T dS- P dV s
•
aT aP
dH = TdS + VdP
• dF
= -SdT - PdV
av as
8
as aP
= -SdT + VdP
(1.31)
v
as av
• dG
aP as
p
aP aT
r
r
(1.32)
v (1.33)
av aT
p
(1.34) j [email protected]
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1. 7 1. 7 .1
13
Thermodynamic potent ials and applications
Response functions D efinit ions
Physicsguide
00: 00: 37 Thermodynamic potent ials and applications
1. 7 1. 7 .1
Response functions Definitions
A great deal can be learned about a macroscopic system t hrough its response t o various changes in externally cont rolled parameters. Important response functions for a PVT syst em are t he specific heats at constant volume and pressure,
Cv= Gp=
aQ 8T
as
=T
8T
V
aQ BT p
as
=T
V
(1.35)
8T p
The isothermal and adiabatic compressibilities, 1 K T= - -
V
av 8P
av
1 K s=- -
T
(1.36)
8P s
V
and the coefficient of thermal expansion 1
av
v
fJT
O'.= -
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t his can be writ t en as
14
(1.37) P,N P hysicsguide
Thermodynamic potent ials and applications
11 oo : oo : 39 @Sk J ahiruddin, 2020
Thermodynamic potentials and applications
t his can be written as a,p
1. 7.2
=
1 V
av aT
P
1 V
(1.38)
Relations between the response function
Intuit ively, we expect the specific heats and compressibilit ies to be posit ive and Gp > Gv, K r > K 8 . In this section we derive relations between t hese response functions. The intuition that the response functions are positive will be justified in the following section in which we discuss t hermodynamic stability. We begin with t he assumption t hat the entropy has been expressed in t erms of T and V and t hat the number· of particles is kept fixed . Then
as as dS = ar V dT+ av dV T as as as av T aT p =T aT +T av r fJT V as av Gp - Gv = T fJT p av
p
(1.39)
T
We now use t he Maxwell relations and the chain rule [email protected]
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11 oo : oo : 43 15
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Thermodynamic potentials and applications
az ax
y
ay az
X
ax ay
=-1
(1.40)
z
We get
as av r Gp -
aP ar v aP Cv = -T av
T
aP av av r ar p 2 av = -rv 0'.2 aT P Kr
(1.41)
In a similar way we obtain a relation between the compressibilit ies and Assume that t he volume has been obtained as function of S and P.
Kr
Ks .
V
Then we get
-
av av dV= aP dP+ as p dS s 1 av 1 av 1 av as - - V aP T V aP s V as p aP 1 av as Kr- Ks=- -V as p aP
(1.42) T
T
The Maxwell relation and
av as j [email protected]
av p
8T 16
p
as
- l
aT
p Physicsguide
11 oo : oo : 4s av
av as
8T
p
j [email protected]
p
as aT
-i
p
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Thermodynamic potent ials and applications
•
gives
TV 2 Kr -Ks = Gp a
(1.43)
Now we have useful relations
Gp (K r - K s) And
1.8
=
Kr (Gp - Gv)
Gp Gv
Kr Ks
=
TV a 2
(1.44)
(1.45)
Thermodynamics of a magnetic system
In order to study magnetic properties of matter one requires t he expression for the work of magnetizing a m at erial. One needs to be careful in defining precisely the system and the processes in order to calculate m agn etic work done. Let us assume that the effect s of pressure and volume on a magnetic system is negligible.
11 oo:oo:47 processes In or er o ea cu a e magne IC wor e us assume that t he effects of pressure and volume on a magnetic system is negligible.
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1.8 .1
17
P hysicsguide
Thermodynamic potentials and applications
Definition of potentials
T he thermodynamic parameters of a magnetic system are going to be t he external magnetic induction B, total magnetization M and t emperature T instead of P, V and T of a fluid system. The first law of t hermodynamics: t he differential change in internal energy E for a reversible change of state can be writ t en in two different but equivalent forms as
dE = T dS-MdB
(1.46)
dE = TdS+BdM
(1 .47)
or
we will use t he SECOND form We define other state functions and t hermodynamic p otentials such as ent halpy H (N, S, B) , the Helmholtz free energy F (N, M , T) and t he Gibbs free energy G(N, B , T ). T he definit ions of t hese t hermodynamics state functions and differential change in a reversible change of state are given by
11 oo : oo : so energy F (N , M , T) and the Gibbs free energy G(N , B , T ). The definit ions of t hese t hermodynamics state functions and differential change in a reversible change of state are given by
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Thermodynamic potent ials and applicat ions
H (N , S , B ) = E - M B F (N , M , T ) = E - TS G(N , B , T ) = E - TS - MB
and and and
dH = TdS-MdB dF = -SdT + BdM dG = - SdT - MdB (1.48)
where explicit N dep endence is also avoided. If one wants to take into account of number of part icles t here must be another t erm µdN in all different ial forms of t he state functions. It can be noticed t hat the thermodynamic relat ions of a magnetic syst em can be obtained from t hose in fluid system if V is replaced by - M and P is replaced by
B. T he thermodynamic parameters can be obtained as
T =
S = -
aE as
or
T =
NI
aF
ar
( aE\
or M
S= -
aH as
ac ar ( aF\
(1.49) B
(1.50) B
11 oo : oo : s3 as
or
as
NI
(1.49) B
aF ac or S =S =aT M aT B aE aF or B = B= aM s aM T aH ac or M =M =aB s aB T j [email protected]
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1.8.2
(1.50) (1.51) (1.52) Physicsguide
19
Thermodynamic potentials and applications
Maxwell relations for magnetic systems
• dE = TdS + BdM
aT aM s
aB
as
M
(1.53)
• dH = TdS-MdB
• dF = -SdT + BdM
aT aB
aM 8
as
B
(1.54)
as
aB
aM r
ar
M
(1.55)
• dG = -SdT - MdB
as
aM
aB r
ar
B
(1.56)
11 oo : oo : 55 •
as
dG = -SdT - MdB
f)B
8M T
f)T
B
(1.56)
1.8.3
Response functions for magnetic systems
The specific heats CM and CB are t he measures of the heat absorption from a temperature stimulus at constant magne20
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tization and constant external magnetic field respectively. The definition of heat capacities are :
as 8T
= -T JVI
as 8T
(1.57)
=-T P
f)T2 B
In case of magnetic systems, instead of isothermal and adiabatic compressibilities one have the isothermal and the adiabatic magnetic susceptibilities
xr= and Xs
=
8M f)B
(1.58)
T
8M 8B
s
f)B2
s
(1.59)
where M is the total magnetization and B is the external magnetic field. Note that t he normalizing factor of 1/V is
11 oo : oo : sa 8M (1.59) Xs == 8B 8 8B 2 8 where M is the total magnetization and B is the external magnetic field. Note t hat t he normalizing factor of 1/V is absent here. T he change of magnetization M with respect to temperature T under constant external magnetic field a.B, is defined as
8M
ar j [email protected]
(1.60) B 21
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P hysicsguide
Thermodynamic potentials and applications
The response functions are not all independent of one another. It can be shown t hat (1.61)
T wo other relations you can prove are XT (CB - CM)== T a.~
2
and
CB (XT - xs) == T a.1 (1.62)
Some applications of thermody•
nam1cs .
.
11 oo : 01 2
: 01
Some applications of thermody•
nam1cs 2.1
TdS equations
The entropy of a pure subst ance can be considered as a function of any two variables, such as T and V; thus,
__ as d as d ds T + V ar v av r as dT T as dV Tds = T ar + av V T [email protected]
@Sk J ahiruddin, 2020
since TdS follows that
22
(2.1)
Physicsguide
Thermodynamic potentials and applications
aQ for a reversible isochoric process, it
T as ar
= Cv
v and from t he Maxewells relation (
TdS = CvdT + T
(2.2)
gt )T = (if)v aP
ar v
dV
(2.3)
This is t he firs TdS equation. Similarly t ake entropy as a function of T and P and derive the second TdS equation /f"\yr'\
11 oo : 01
: 03
This is t he firs TdS equation. Similarly t ake ent ropy as a function of T and P and derive the second TdS equation TdS
2.2
== CpdT - T
av aT
dP
(2.4)
p
Internal energy equations
For the internal energy equation we start from dU
== TdS - PdV
(2.5)
and take partial derivative with respect to V keeping T constant
au
av
T
jahir@physicsguide. in
== T
as av
- P T
23
@Sk J ahiruddin , 2020
(2.6) Physicsguide
Thermodynamic potentials and applications
Using Maxwell's t hird relation, (aS/ 8V )T == (aP/ aT)v, we get
au
av
== T aP T
ar
_P v
(2.7)
This equation is very useful. Similarly t aking partial derivative wit h respect to P both side of the equation dU == T dS - P dV prove that I
~ TT\
11 oo : 01
: 01
This equation is very useful. Similarly t aking partial derivative wit h resp ect to P both side of the equation dU = T dS - P dV prove that
au
av
=-T
8P r
2.3
8T
_ P av 8P r
P
(2. 8)
Heat transfer of isothermal expansion of van der Waals gas
Start with
TdS = Cv dT + T
8P BT
dV V
From van der Waal's equation
P = RT V-b j [email protected]
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24
(2.9) P hysicsguide
Thermodynamic potentials and applications
8P 8T v Hence
a V2
R V - b
dV T dS = CvdT + RT V _ b
(2.10)
since T is constant, Cv dT = O; and , since the process is reversible, q = J Tds Therefore, V1
q = RT
½ ?.
dV V-b
(2.11)
11 oo : 01
: 01
This equation is very useful. Similarly t aking partial derivative wit h resp ect to P both side of the equation dU = T dS - P dV prove that
au
av
=-T
8P r
2.3
8T
_ P av 8P r
P
(2. 8)
Heat transfer of isothermal expansion of van der Waals gas
Start with
TdS = Cv dT + T
8P BT
dV V
From van der Waal's equation
P = RT V-b j [email protected]
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24
(2.9) P hysicsguide
Thermodynamic potentials and applications
8P 8T v Hence
a V2
R V - b
dV T dS = CvdT + RT V _ b
(2.10)
since T is constant, Cv dT = O; and , since the process is reversible, q = J Tds Therefore, V1
q = RT
½ ?.
dV V-b
(2.11)
11 oo : 01
: 1o
since T is constant, CvdT = O; and, since the process is reversible, q = J T ds Therefore, q = RT
(2.1 1)
V-b
½
Finally V1- b q= RTlnV:-b
(2.12)
'l
2.4
Reversible isothermal change of pressure
When T is constant TdS
=
-T
oV 8T
dP
and so
Q
=
p
-T
av 8T
dP p
(2.1 3)
The volume expansivity is
av 8T [email protected]
p
25
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Thermodynamic potentials and applications
t hen Q
=-T
V f3dP
(2.14)
In t he case of a solid or liquid, V and /3 do not change much with pressure. So we can take t hem out of the integration to get ,........
,
✓
-
___ ,
11 oo : 01
: 13
In t he case of a solid or liquid , V and f3 do not change much with pressure. So we can take t hem out of the integration to get
It is seen from t his result that, as t he pressure is increased isothermally, heat will flow out if /3 is insositive but t hat , prossunce with a negative expansivity (such as water between O and 4°C, or a rubber band) , an isothermal increase of pressure causes an absorption of heat. If the pressure on 15cm3 of mercury at 20°C is increased reversibly and isothermally from O to lOOOatm, the heat t ransferred will be approximately
where T = 293K, V = 2 x 10- 5m 3 , f3 and P1 = 1.01 x 108 Pa Hence, [email protected]
1.81 x 10- 4 K- 1 ,
26
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5
Q = - (293K) (1. 5 x 10- m
= -80.3N · m = -80.3J
=
3
)
4
(1.81 x 10- K-
1
)
(1.01 x 108 Pa)
11 oo : 01 Q
= - (293K) (1.5 x = -80.3N · m = -80.3J
2.5
: 1s
5 3 10- m )
4
(1.81 x 10- K-
1
)
(1.01 x
8 10 Pa)
Internal energy of van der Waal gas P = RT _ a V-b V 2
and
8P 8T
R v
V - b
so
au
av
R
T
=
RT
TV - b - V - b +
so dU
=
a V2
=
a V2
a CvdT + V 2 dV
(2.16)
(2.17)
Finally
U= j ahir@physicsguide. in
@Sk J ahiruddin , 2020
2.6
a CvdT - V
+ const .
27
(2.18) Physicsguide
Thermodynamic potentials and applications
Heat capacities of ideal gas
Th P. rliffP.rP.nr.P. hP.t,wP.P.n t.h P. hP.::i.t. r.::i.n::i.r.it.i P.s is Q'ivP.n hv
11 oo :01
:1
@Sk J ahiruddin , 2020
2.6
a
Thermodynamic potentials and applications
Heat capacities of ideal gas
The difference between t he heat capacit ies is given by
Gp -
2 Cv = TVn /""T
(2.19)
t
where a = (i~) P is the volume expansion coefficient and ""T = ~~) T is the isothermal compressibility. Using t his relation, t he difference between heat capacities of an ideal gas could be easily obt ained. For one mole of an ideal gas, t he equat ion of state is P V = RT where R is the universal gas constant . T hus,
-t (
1
Q=-
V
av aT
l R -VP
p
1
(2.20)
-
T
and 1 ""T = - -
V
av 8P
1
--
V
T
V -p
1 -
p
(2.21)
PV =R T
(2.22)
so the differences
Cp- Cv
=
TV
-
1
T
2
1
1/ P
--
As T -+ 0, by third law of t hermodynamics, the ent ropy S -+ S0 = 0 becomes independent of all parameters like [email protected]
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Physicsguide
Thermodynamic potentials and applications
pressure, volume and temperature. T herefore, in t he limit
11 oo : 01
@Sk J ahiruddin, 2020
: 21
Thermodynamic potent ials and applications
pressure, volume and temperature. Therefore, in the limit T --+ 0 the heat capacities also tend to zero,
Cv=T
as ar
v
--+ 0
and
Cp=T
as aT
p
--+ 0 (2.23)
This is because of the fact that as T --+ 0, the system tends to settle down in its nondegenerate ground state. The mean energy of t he system then become essentially equal to its ground state energy, and no further reduction of temperature can result in a further reduction of mean energy. Not only t he individual heat capacity goes to zero but also t heir difference goes to zero as T --+ 0. Because, the volume expansion coefficient a also goes to zero 1 a=V
av 8T
as aP
P
r
--+ 0
(2.24)
However, the compressibility ""T, a purely mechanical property, remains well-defined and finite as T--+ 0. Thus,
j [email protected]
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Thermodynamic potent ials and applicat ions
11 oo : 01
: 23
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Thermodynamic potent ials and applications
as
T
➔
Gp - Gv
0
➔
0
(2.25)
This is not in contradiction to the ideal gas result Gp Gv = R because as T ➔ 0, the system approaches its ground state and quantum mechanical effects become very important. Hence the classical ideal gas equation PV = RT is no longer valid as T ➔ 0.
2. 7
Gibbs paradox:
An isolated syst em with two parts of equal volume V each contains N number of molecules of the same monatomic perfect gas at the same t emperature T , pressure P . Init ially, t he two parts were separated by a membrane and t hen the membrane was removed. The syst em is allowed to equilibrate. The change in entropy is given by
oQ T
~
T
oQ
(2.26)
From first law, for a perfect gas dE(T) = oQ + oW = 0 and t hus oQ = -oW = P dV. The volume changes from V to
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: 26
t hus bQ = -bW = P dV. The volume changes from V to
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Thermodynamic potentials and applications
2V for the each part . Therefore ,
b,.S = 2_ T
PdV 2V
= NkB
V
dV V
2V
+ NkB
dV
V
(2.27)
V
= 2NkB ln 2 > 0 The entropy of the system then may increase indefinitely by putting more and more membranes and removing them. However , the process is reversible. By putting back the membranes one would recover the initial state. According to Clausius t heorem , the change in entropy must be b,.S = 0 in a reversible process. This discrepancy is known as Gibb's paradox. The paradox would be resolved only by applying quant um statistical mechanics. The same problem will b e discussed again in the next section.
2.8
Radiation:
According to electromagnetic t heory, the pressure P of an isotropic radiation field is equal to 1/ 3 of the energy density:
P
=
u(T )/3 = U(T )/3V
(2.28)
11 oo : 01 : 2a According to electromagnetic theory, the pressure P of an isotropic radiation field is equal to 1/ 3 of the energy density:
(2.28)
P = u(T )/3 = U(T )/3V 31
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where V is the volume of t he cavity, U is the total energy. Using the thermodynamic principles, one could obtain Stefan 's law: u = aT4 , where a is a const ant. To do t hat we will start from t he radiation pressure and also t ake t hat the total energy of photon gas is proport ional t o volume. By the second law dU = TdS - PdV, thus
au
av
=T r
as
av
au
av
- P
r
av
r
(if)v
· (av 8S )T since u(T )V
au
= u (T)
=T
r
and
ar
v
ar
-P
v (2.29)
by Maxwell equation.
aP
aP
1 du 3dT
Since U
=
(2.30)
Hence
u(T ) =
T du -3 dT
-
l -u 3
where a is a const ant.
T du = 4 dT u
4
u = aT (2.31)
11 oo : 01
: 31 (2.28)
P = u(T)/3 = U(T )/3V 31
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where V is the volume of t he cavity, U is the total energy. Using the thermodynamic principles, one could obtain Stefan's law: u = aT4 , where a is a const ant. To do t hat we will start from the radiation pressure and also take that t he total energy of photon gas is proport ional to volume. By the second law dU = TdS - P dV, thus
au = T as - P av r av r
au = T aP - P av r ar v (2.29)
(if) v
· (av 8S )T since u(T )V
au = u(T ) av r
by Maxwell equation.
aP ar
and
v
l du 3dT
Since U
=
(2.30)
Hence u(T)
=
Tdu -3 dT
-
l -u
3
Tdu =4 dT u
(2.31)
where a is a constant.
j [email protected]
32
Physicsguide
11 oo : 01
: 33 (2.28)
P = u(T)/3 = U(T )/3V 31
j [email protected]
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Physicsguide
Thermodynamic potentials and applications
where V is the volume of t he cavity, U is the total energy. Using the thermodynamic principles, one could obtain Stefan's law: u = aT4 , where a is a const ant. To do t hat we will start from the radiation pressure and also take that t he total energy of photon gas is proport ional to volume. By the second law dU = TdS - P dV, thus
au = T as - P av r av r
au = T aP - P av r ar v (2.29)
(if) v
· (av 8S )T since u(T )V
au = u(T ) av r
by Maxwell equation.
aP ar
and
v
l du 3dT
Since U
=
(2.30)
Hence u(T)
=
Tdu -3 dT
-
l -u
3
Tdu =4 dT u
(2.31)
where a is a constant.
j [email protected]
32
Physicsguide
11 oo : oo : 01
Basic concepts of Statistical Mechanics Sk J ahiruddin
Assistant P rofessor· Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay lVI. Sc Physics 2009-2011 batch He ranked 007 in IIT JAlVI 2009 and 008 (JRF) in CSIR NET June 2011 He has been teaching CSIR NET aspirants since 2012
1
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Basic concepts of Stat Mech
11 oo : oo : 03 1
Basic concepts of Stat Mech
@Sk J ahiruddin, 2020
Contents 1
Basic concepts of Statistical Mechanics 1.1
Specification of states: .
1.2
3
•
•
•
•
•
•
•
•
•
•
4
Counting of Microstates: .
•
•
•
•
•
•
•
•
•
•
5
1.3
Equal a priori probability:
•
•
•
•
•
•
•
•
•
•
8
1.4
Statistical ensembles:
•
•
•
•
•
•
•
•
•
•
•
•
•
9
1.5
Phase point density:
•
•
•
•
•
•
•
•
•
•
•
•
•
11
1.6
Statistical average and mean values: .
•
•
•
•
12
1. 7
Condition of Equilibrium:
•
•
•
•
13
j [email protected]
2
•
•
•
•
•
•
•
Physicsguide
11 oo : oo : 06
j [email protected]
2
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1
P hysicsguide
Basic concepts of Stat Mech
Basic concepts of Statistical Mechanics
Consider a physical system composed of N identical pa1·t icles confined in a volume V. For a macroscopic syst em, N is of the order of Avogadro number NA~ 6.022 x 1023 per mole. In this view, all the analysis in statistical mechanics are carried out in the so-called thermodynamic limit. It is defined as: both number of particles N and volume V of t he syst em t ends to infinity whereas the density of particles p = N / V remains finite.
N
➔
oo,
V
➔
oo ,
p
= N / V = finite.
(1.1)
In t his limit , the extensive properties of t he system become directly proportional to t he size of the system (N or V ), while the intensive properties become independent of the size of the system. The particle density becomes an important parameter for all physical properties of the system. In order to develop a microscopic t heory of a macroscopic syst em, it is necessary to specify t he state of micro-particles, atoms or molecules as a first step. Next is to construct the macro-st ate from t he micro-states of N number of parti-
11 oo :oo :oa In order to develop a microscopic theory of a macroscopic system, it is necessary to specify t he st ate of micro-part icles, atoms or molecules as a first step. Next is to construct the macro-state from t he micro-states of N number of [email protected]
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Basic concepts of Stat Mech
@Sk J ahiruddin, 2020
cles when N is very large. Finally, one needs to extract t he macroscopic properties in terms of the micro-states of a macroscopic system. In this section, all t hese essent ial definitions will be given.
1.1
Specification of states:
The specification of t he state of a part icle depends on t he nature of the particle, i .e. whether the particle's dynamics is described by classical mechanics or by quantum mechanics. The dynamics of a classical system is determined by its Hamiltonian 1-l(p, q) where q and P are the generalized posit ion and momentum conj ugate variables. The motion of a particle is described by the canonical Hamilton 's equation of motion
and
.
Pi = -
81-l(p, q)
aqi ;
i = 1 2 ··· N
' '
' (1.2)
The state of a sine:le oarticle at anv time is t hen e:iven bv
11 oo : oo : 11 and
.
Pi = -
81i(p, q)
aqi ;
i = 1 2 ··· N
' '
' (1.2)
The state of a single particle at any time is t hen given by t he pair of conjugate variables (qi, Pi ) . The state of a system [email protected]
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Physicsguide
4
Basic concepts of Stat Mech
composed of N particles is then completely and uniquely defined by 3N canonical coordjnates q1 , q2 , · · · , q3N and 3N canonical momentum P1 , P2, · · · , P3N. These 6N variables constitute a 6N -dimensional r space or phase space of the system and each point of t he phase space represents a state of t he system. Each single particle constitutes a 6 -dimensional µ -space. r -space is evident ly built up of N such 6 -dimensional µ -space of each particle. The locus of all t he points in r -space satisfying the condition 1i(p, q) = E , total energy of t he system , defines a surface called energy surface. Specification of states of quantum particles will be discussed later.
1.2
Counting of Microstates:
It is important to enumerate t he number of microstates for a given macrostate (N, V, E ) of a system. For a classical
11 oo : oo : 14 1.2
Counting of Microstates:
It is important to enumerate t he number of microstates for a given macrostate (N, V, E) of a system. For a classical particle, t he microstate is specified by a phase point . Consider an infinitesimal change in t he position and moment um coordinates in the phase space. T he phase point of this parj [email protected]
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@Sk J ahiruddin , 2020
Basic concepts of Stat Mech
t icle then under go a small displacement in the phase space and the microscopic state of t he system will be modified. However, if t he change in t he microstate is so small that it is not measurable by the most accurate experiment , then it can be assumed that t here is no change in t he macroscopic state of the system. T hus, t he state of a syst em is given by a small volume element Jn of t he r -phase space. If bqbp = h , t hen
By Heisenberg's uncertainty principle in quantum mechanics it can be shown that h is the P lanck's constant. Thus, t he number of states of a system of N particles of energy < E is given by
r=
1 h 3N
dn =
1 h3N
d3N qd3Np
(1.4)
11 oo : oo : 16 energy < Eis given by
r=
1 h3}l
dD.
=
1 h3N
d3N qd3Np
(1.4)
The number of microst at es is then proportional to the volume of the phase space. The counting of number of microstates however depends on the distinguishable and indistinguishable nature of t he [email protected]
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6
Basic concepts of Stat Mech
@Sk J ahiruddin, 2020
particles. If the constituent particles are distinguishable, t he two microst ates corresponding to the interchange of two particles of different energy a1~e distinct microstates. On the other hand, if t he constituent particles are indistinguishable, these two microstates are not distinct microstates. Thus, for a syst em of N indistinguishable, identical classical particles having different energy states, the number of microstates is then given by
r=
1 h3NN!
d3N qd3Np
(1.5)
Consider a system of N part icles of t otal energy E and corresponds to a macrostate (N , V, E ). If the particles are
11 oo : oo : 19 r ==
d3N qd3Np
h3N N!
(1.5)
Consider a system of N particles of total energy E and corresponds to a macrostate (N, V, E ). If the particles are distributed among t he different energy levels as, ni particles in t he energy level Ei, the following conditions has to be satisfied
(1.6)
and .
.
'l
'l
The total number of possible distributions of N such [email protected]
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Physicsguide
Basic concepts of Sta t lVIech
@Sk J a hiruddin , 2020
cles is then given by
(1. 7) If the particles are distinguishable, then all t hese permutations would lead to distinct microstate whereas if the particles are indistinguishable, these permutations must be regarded as one and t he same microstate.
1.3
Equal a priori probability:
As it is already seen that for a given macrostate (N, V, E) , t here is a large number of possible microstates of t he system. In case of classical non-interacting system, t he total T7
1
1•
,
.,
i
1
I
1
1\ T
, •
1
•
11 oo : oo : 21 As it is already seen that for a given macrostate (N, V, E ), t here is a large number of possible microstates of t he system. In case of classical non-interacting system, t he total energy E can be distributed among the N particles in a large number of different ways and each of these different ways corresponds to a microstate. In case of a quantum system, the various different microstates are identified as the independent solutions \JI (r 1 , r 2 , · · · , r N) of the Schrodinger equation of the system, corresponding to an eigenvalue E . In any case, to a given macrostate of the syst em there exists a large number of microstates and it is assumed in statistical mechanics that at any time t the system is equally likely j [email protected]
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Physicsguide
Basic concepts of Stat Mech
to be in any one of these microstates . This is generally referred as the postulate of equal a priori probability for all microstates of a given macrostate of the syst em. Equal a priori probability postulate is one of t he two fundamental postulates in Statistical Mechanics.
1.4
Statistical ensembles:
The microstate of a macroscopic system is specified by a point in 6N -dimensional phase space where N is of the 23 order of Avogadro number ( 10 ) . At any time t the system
11 oo : oo : 24 The microstate of a macroscopic system is specified by a point in 6N -dimensional phase space where N is of the order of Avogadro number ( 1023 ) . At any t ime t the system is equally likely to be in any one of the large number of microstates corresponding to a given macrostate ( N, V , E) . As t he dynamical system evolves with time, the system moves from one microst ate to another. After a sufficiently long time, the syst em passes through a large number of microstates. The behaviour of the system t hen can be obtained as averaged over those microstates through which the system passes. Thus, in a single instant of time, one could consider a
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Physicsguide
Basic concepts of Stat Mech
collection of large number of replicas (mental copies) of the original system characterized by the same macrostate but could be in any of t he possible microstates to obtain the average behaviour of the system. This collection of large number of copies of the same system is called an ensemble. It is expected t hat t he ensemble averaged behaviour of a system would be identical with the time averaged behaviour of the given system. Since the values of (q, p) at any instant are different for
11 oo : oo : 21 averaged behaviour of a system would be identical with the time averaged behaviour of the given system. Since the values of (q, p) at any instant are different for a syst em of an ensemble, they are represented by different points in the phase space. Thus, the ensemble will be represented by a cloud of phase points in the phase space. As t ime passes, every member of the ensemble undergoes a continuous change in microstates and accordingly the system moves from one place to another on the phase space describing a phase trajectory. The density of the cloud of phase points at a particular place t hen may vary with time or at a given time the density may vary place to place. It must be emphasized here t hat the systems of an ensemble are independent systems, t hat is, there is no interaction between them and hence the
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Basic concepts of Stat Mech
trajectories do not intersect.
1.5
Phase point density:
Since a phase point corresponds to a microstate of a system and its dynamics is described by the phase trajectory, the density of phase points p(p, q) is t hen determining the number of microstates per unit volume, that is, the probability
11 oo : oo : 29 Since a phase point corresponds to a microstate of a system and its dynamics is described by t he phase t rajectory, the density of phase points p(p, q) is t hen determining the number of microstates per unit volume, that is, the probability to find a state around a phase point (p, q). The phase point density p(p, q) is given by Number of states pp ( q) = - - - - - ' Volume element
(1.8)
At any t ime t , t he number of representative points in the volume element d3N qd3N p around the point (p, q) of t he phase space is then given by p(p, q)d3JVqd3Np
(1.9)
The density function p(p, q) then represents the manner in which t he members of t he ensemble are distributed over all possible microstate at different instant of t ime.
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1.6
11
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Basic concepts of Stat Mech
Statistical average and mean values:
Consider any physical p1~operty of t he system X(p , q), which may be different in different microstates for a system. The macroscopic value of X must be t he average of it over all possible microstates corresponding to a given macrost ate. The ensemble average or t he statistical average (X ) of the
11 oo : oo : 32 Consider any physical property of t he system X (p , q), which may be different in different microstates for a system. The macroscopic value of X must be the average of it over all possible microstates corresponding to a given macrostate. The ensemble average or the statistical average (X) of the physical quantity X at a given instant of time, is defined as (1.10)
where p(p, q) is the density of phase points. Note that the integration is over the whole phase space. However , it is only t he populated region of phase space that really contribute. -
The mean value X of the physical quantity X depends how it evolves with time t In a sufficiently long time, the phase trajectory passes through all possible phase points. If t he duration of time is T , t he mean value is defined as
-
X
1
= T--+oo lim T
T
o
X (t)dt
(1.11)
The statistical average (X ) and the mean value X are [email protected]
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equivalent. This is known as ergodic hypothesis. Ergodic hypothesis is the second fund amental postulate of statistical mechanics.
11 oo : oo : 34 equiva ent . Ergodic hypothesis is the second fund amental postulate of statistical mechanics.
1. 7
Condition of Equilibrium:
By Liouville's t heorem, t he total t ime derivative of t he phase point density p(p, q), in absence of any source and sink in the phase space, is given by
dp dt where
=
8p 8t
+ {p , 1-l} =
(1.12)
O
3N
{p , 1-l} =
L
(1.13)
i =l
is known as Poisson bracket of t he density function p and t he Hamiltonian 1-l of the system. Thus, the cloud of phase points moves in the phase space like an incompressible fluid . T he ensemble is considered to be in statistical equilibrium if p(p, q) has no explicit dependence on t ime at all points in the phase space, i .e.,
(1.14) j [email protected]
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Under the condition of equilibrium is then 3N
{p, 1-l} = ),
8p 81-l
8p 81-l
=0
(1.15)
11 oo : oo : 37 -at =
(1.14)
0
13
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Physicsguide
Basic concepts of Stat Mech
Under the condition of equilibrium is then 3N
{p,1-l} =
L
=0
(1.15)
i =l
and it will be sat isfied only if p is independent of P and q. That is p(p, q) = constant (1.16) which means that the representative points are distributed uniformly over the phase space. The condition of statistical equilibrium then requires no explicit time dependence of t he phase point density p(p, q) as well as independent of the coordinates (p, q) .
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11 oo : oo : oo
Random -walk and Probability distributions Sk J ahiruddin
Assistant P rofessor· Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay lVI.Sc Physics 2009-2011 batch He ranked 007 in IIT J AlVI 2009 and 008 (JRF) in CSIR NET J une 2011 He has been teaching CSIR NET aspirants since 2012
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Random walk and P robabilities
11 oo : oo : 03 1
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Random walk and P robabilities
Contents 1 Introduction
3
2 Random Walk and Probabilities
4
2.1
Random walk in one dimension
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•
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•
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5
2.1.1
Mean value
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•
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•
7
2.1.2
Dispersion .
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•
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•
•
•
•
•
•
8
3 Probability distributions
9
3. 1
Gaussian Distribut ion
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•
•
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•
•
•
•
•
•
9
3.2
Poisson's distribution .
•
•
•
•
•
•
•
•
•
•
•
•
10
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1
2
P hysicsguide
Random walk and P robabilities
Introduction
Thermodynamics and statistical mechanics deals with many particle system. In thermodynamics, for a system in equilibrium, we try to make some general statements about t he relationship between different macroscopic parameters of these systems. Thermodynamics requires very few number of postulates without any detailed assumption about t he microscopic properties of the system. Statistical Mechanics, for a system in equilibrium, starts with the microscopic properties of t he particles in t he system and the laws of classical and quantum mechanics Statistical mechanics requires input from the laws of statistics.Statistical Mechanics reproduces all t he results of t hermodynamics, plus a large number of general relations for calculating the macroscopic properties of system from the knowledge of its microscopic elements. For more details on probabilities please go to the mathematical physics part
11 oo :oo :oa mathematical physics part
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3
Random vvalk and Probabilities
Random Walk and Probabilities
The problem of random walk is one of the classic problems in physics having a wide range of application in various fields. A drunk start out from a lamppost located on a street. Each steps he takes is of equal size of length l. The man is however so drunk t hat t he direction of each step he takes , whet her it is left or right, is completely independent of t he preceding step. All we can say, t hat the probability of taking a step to right is p and t he probability of taking a step to left is q = l - p. Let's choose x-axis is t he road and t he lamppost is located at x = 0. The question we ask, after t he man takes N steps, what is the probability of t he man being located at x = ml? The problem can be reformulated in terms of N man rather than N steps. One can also study t he problem in more t han one dimension. The random walk problem illust rate some very elementary concepts of probability theory. It relates to many problem in physics. Think of two examples from Magnetism and Diffusion. .
.
1
11 oo : oo : 12 more t han one dimension. The random walk problem illust rate some very elementary concepts of probability t heory. It relates to many problem in physics. Think of two examples from Magnet ism and Diffusion. Magnetism: An atom has a spin [email protected]
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i and magnetic moPhysicsguide
4
Random walk and P robabilities
ment mu. In accordance with quant um mechanics t he spin can be either '' up'' or '' down'' with respect to a given direct ion. If bot h t his probabilities are equal likely,what is the net total magnetic moment of N such atoms.? Diffusion of a molecule in a gas: A given molecule t ravels in three dimension a mean distance l between colli-
sions with other molecules. How far it is likely to have gone after N collisions?
2.1
Random walk in one dimension
We are going to calculate the probability of finding the walker at distance x = m l after N steps wit h
-N 6 . So U ( ) becomes (under t he infinitesimal change of )
U ( ) + 6U () = U ()
+
dU de/> 6
d = [1 + 6 de/> ] U ()
= (1 + X(b) )U() -+ As given in t he problerr
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P hysicsguide
Advanced topics in Mathematical Physics
By comparing last two lines we get
Now you need to apply your knowledge of Quantum mechanics to ident ify a relation between t he X operator and . d £ 2 operator. You know £ 2 = -in de/> rl
•
'I
I
rl \
•
'I
11 oo :01 : oa ow you nee to app y your now e ge o uantum mechanics to identify a relation between t he X operator and . d L 2 operator. You know L 2 = -in dcp . d -in dcp •
x t(81>) = - !_81>i1 n
So the X operator is Anti-Hermitian. Solution 2.8. According to the group property each element must have an inverse. a and b has self Inverse. And t he group also follows the closure rules - multiplication of any two elements belongs to the group. Hence a 0 b also belongs to t he group. Now the group has been given as abelian. So a 0 b = b 0 a. Now a 0 b 0 b 0 a= a(0 b 0 b) 0 a= I. Hence a 0 b has self inverse too. Then the smallest group contains a, b, ab, I . So t he order is FOUR.
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Advanced t opics in :Nlathem atical Physics
Tensors
Prob 3.1. Under a certain rotation of coordinate axes, a rank- 1 tensor va(a = 1, 2, 3) transforms according to the orthogonal transformations defined by the relations v{ =
fi(v1 + v2); v~ = fi(-v1 + v2); v~ = v3 . Under the same rotation, a rank-2 tensor T ab would transform such t hat: l
11 oo : 01
: 1o
. . n er a certain rotation o coor inate axes, a rank- 1 tensor va(a = l, 2, 3) transforms according to t he orthogonal transformations defined by the relations vf = fi(v1 + v2); v~ = J2(-v1 rotation, a rank-2 tensor
+ v2); T ab l
v~ = v3. Under the same would transform such that:
(GATE 2008] (a) T{,1
= T1,1T1,2
(c) T{,1
= T1,1 + 2T2,2 - T2,1
T1,2 + T2,1)
Prob 3 .2 . Let
T ij =
L
Eijk a k
and
f3k =
L
Eijk1ij,
where
IJ
k
is the Levi-Civita d ensity, defined to be zero if two of t he indices oincide and +1 and - 1 depending on whether Eijk
i,j, k is even or odd permutation of 1, 2, 3. The
/33
is equal
[GATE 2009]
to:
(a) 2a 3
(d)
-a3
Prob 3 .3 . Consider an ant i-symmetric tensor P ij wit h indices i and j running from 1 to 5. The number of independent component of t he tensor is: (a) 3 (b) 10 (c) 9 (d) 6 [GATE 2010]
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Prob 3 .4 . The no. of independent components of the symmetric tensor A ij with indices i, j = l , 2, 3 is [GATE
2012]
(a) 1
(b) 3
(c) 6
(d ) 9
11 oo : 01
: 13
Prob 3.4. The no. of independent components of the sym[GATE metric t ensor A ij with indices i, j = l , 2, 3 is 2012] (a) 1 (b) 3 (c) 6 (d) 9 Prob 3.5. In t he most general case, which one of the following quantities is NOT a second order t ensor? [GATE 2013] (a) Stress (b) Strain (c) Moment of Inertia (d) Pressure Prob 3. 6. Let ¼ be the i th component of a vector field V , which has zero divergence. If
ai = aa , the expression for X·
.7
(c) 8J¼
[GATE 2016] (d) -8}¼
Prob 3. 7. The rank 2 tensor X i Xj where X i are the Cartesian coordinates of t he position vector in three dimensions, has 6 indep endent elements. Under rotation, these 6 elements decompose into irreducible sets (that is, the elements of each set transform only into linear combinations of ele[NET June 2015] ments in that set) containing (a) 4 and 2 elements (b) 5 and 1 elements (c) 3,2 and 1 elements (d) 4,1 and 1 elements
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Prob 3. 8. The lengt h element ds of an arc is given by
11 oo : 01 @Sk J ahiruddin , 2020
: 16
Advanced topics in Mathematical Physics
Prob 3. 8. The lengt h element ds of an arc is given by
The metric t ensor
(a)
2
v'3
v'3
1
[GATE 2014]
is
(b)
2 v'3 2 v'3 1 2
(c)
2
1
3
1
(d)
9 ij
2
2 Prob 3. 9. The scale factors corresponding to t he covariant
metric tensor 2018] 2
2
9 ij
2
(a) 1, r ,r sin 0
[GATE
in spherical coordinat es are 2
2
(d) l ,r,rsin0
(b) 1, r ,sin 0 (c) 1, 1, 1
Prob 3.10. A fourt h rank Cartesian tensor t he following identities
~jkl
satisfies
= T jikl (ii) ~ jkl = ~ jlk (iii) ~ jkl = T klij
(i)
~ jkl
Assuming a space of t hree dimensions (i.e i, j, k = l , 2, 3) what is the number of independent components of T ijkl [TIFR 2018] Prob 3.11. An array Thas elements [email protected]
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1, 2, 3, 4. It is given that
28
~ jkl
where i , j , k, l
=
Physicsguide
Advanced topics in Mathematical P hysics
00 : 01 : 18
@Sk J ahiruddin , 2020
Advanced topics in :Nlathematical P hysics
1,2 , 3, 4. It is given that Ti jkl = T jikl = Ti jlk = - T klij
for all values of i, j, k, l. The number of independent components in this array is [TIFR 2019] (a) 256 (b) 1 (c) 45 (d) 55
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3.1
: 21
P hysicsguide
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Ans keys
3.1. d
3.5. b
3.2. a
3.6. d
3.9. d
3.10. 21 3.3. b
3.7. b
3.4.
3.8. b
3.2
C
3.11.
C
Solutions
Solution 3.1. First we write the transformation equation
for vector in matrix form v'1 v'2 v'3
1
1
J2
J2
1
../2 0
0
J2
0
0
1
Transformation equation for t ensor is Tlt
= akiazj'Ti,j
1
Ji
For this problem our a matrix is a =
~
1
Ji
~ 0
0 0 So t he Tensor t ransformation equation is
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30
0 •
1
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j [email protected]
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T{ 1
: 24
Advanced topics in :Nlathematical Physics
=
a11a11T11
+ a11a12 T 12 + a11a13T13
+
a12a11T21
+
a12a12T22
+
a12a13T23
+
a13a11T31
+
a13a12T32
+
a13a13T33
Now put the elements of the a matrix in the equation
,_ llT + llT 1 T11 11 12 + /n 2 J2J2 J2J2 VL1
+
1
J2 J2T21
+ 0
X
1
+
O T31 + 0
1
J2 J2T22
1 X v'2T32
X
OT
13
1
+
v'2 x
+ 0
X
O T23
O T 33
So we get , T 11
=
1 1 1 1 T11 + T12 + T21 + T22 2 2 2 2
Solution 3.2. /33 = L i,j Eij3~j Now from the definition of Levi-Civita Tensor Eij3 will give you ZERO when any of i,j will b e THREE. Now when i,j will both be ONE or TWO then also the Tensor give you ZERO. So the only non ZERO values are
11 oo : 01
: 26
i , j will be THREE. Now when i, j will both be ONE or
TWO then also the Tensor give you ZERO. So the only non ZERO values are
31
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Now wit h t he same logic ~ j
=
L
Eijk a k k
Solution 3.3. Second Order Tensor with dimension n is a
n x n matrix. You can easily verify t hat the no of indepenn n
.
.
.
+1
. n n - 1
matrix is anti-symmet ric and n = 5. Hence t he ans is 10. Solution 3.4. As t he general formula is given in the previous problem, here the matrix is symmetric and n = 3. Hence t he ans is 6. Solution 3.5. The direct product of two vectors and division of two vectors is a Tensor. (Refer to Quotient rule) . We all know Moment of Inertia is a tensor (L = I w). St ress and Force . .. pressure are A . So as d1v1s1on of two vectors they are r ea tensors. Strain is division of two scalars (elongation/length).
So Strain can't be a t ensor.
11 oo : 01
: 29
all know Moment of Inertia is a tensor ( L = I w). Stress and Force . .. pressure are A . So as d1v1s1on of two vectors they are rea tensors. Strain is division of two scalars (elongation/length). So Strain can't be a tensor.
Solution 3.6. Levi Civita symbol index can be changed in cyclic permutation . Eijk = Ejki = Ekij . Similarly Ezmk = [email protected]
32
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Emkl
=
Physicsguide
Advanced topics in Mathematical P hysics
Eklm
Now we write t he given identity as
We know t he relation between Levi-Civita and Kronekar Delta function
Hence Ekij Eklm Oj Ol V m
= (Oil Ojm - OimOjz) Oj Ol V m = 8 rri 8 zV m - 8z8z V m = 8 z8 m V m - of V m = - 8[Vm ➔
as
1.v =
0
a
- - Vm =O OXm
Solution 3.7. Any three dimensional vector in cartesian coordinate has t hree independent elements. As we have already seen that the direct product of two vectors is a tensor, so the most generalized rank 2 tensor XiXj where Xi are the .
.
.
.
11 oo : 01
: 32
Solution 3. 7. Any t hree dimensional vector in cartesian coordinate has t hree independent elements. As we have already seen that t he direct product of two vectors is a tensor, so t he most generalized rank 2 tensor X i Xj where Xi are the Cartesian coordinates of the position vector in t hree dimensions should have nine independent elements. The given case in the problem is a special case as t he t ensor has 6 independent element. The main catch is to identify from t he [email protected]
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Advanced topics in 1\ll athematical Physics
ment ion of 6 independent elements is t he tensor as SYMMETRIC tensor. Now from our previous knowledge we get that tensor product of two spaces of dimension 3 can be seen as direct sum of two spaces of dimension 1, 3, 5 wit h j values 0, 1, 2 respectively. Here again we need to understand t hat the j value of 1 which leads to dimension 3 is ANTI-SYMMETRIC. The given tensor is symmetric. So we omit the ant i-symmetric combination and conclude t he direct sum to contain dimension 1 and 5 only ( j values O and 2 ) . Solution 3.8. Metric Tensor is a symmetric tensor defined as a property of length element of a space. 2
ds = du dv
E F FE
du dv
To explain more lets your length element is c1 dx I
•
I
2
+ c2dx~ + c3dx1dx2
., , ,
11 oo : 01
: 34
FE
To explain more lets your length element is 2 c1 dx + c2dx~
+ c3dx1 dx2
Your metric tensor will be
In this problem 1 2 2 2 2 ds = 2(dx ) + (dx )
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Advanced topics in Mathematical Physics
So = 2, C2 =
C1
1,
C3
=
,/3
Hence the ans will be 2 v'3 2 v'3 1 2
You can easily generalize t he formulation in three and four dimensions. Solution 3.9. Obvious from the definit ion of scale factors you have studied. Solution 3.10. (i) and 2nd indices
1ijkl
(ii) T ijkl = T tjlk (iii) T ijkl = T klij of indices
~►
1'T
I
•
I
•
=
T jikl ===;:~>
Symmetric under l st
Symmetric under 3rd and 4th indices ~> Symmetric under l st and 2nd pair 1
• 1
•
1
11 oo : 01
: 37
and 2nd indices (ii) ~Jkl = ~ j lk ~► Symmetric under 3rd and 4th indices ~► Symmetric under 1st and 2nd pair (iii) T i.ikl = T ktij of indices Now we can partition ~ jkl as T (i j) (kl) where we now t hink of T (ij )(kl ) as a m x m symmetric matrix with ~m(m + 1) independent components. We can also t hink the pairs (ij) and ( kl) as individual matrices which are symmetric wit h ½n(n + l ) independent components.
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So we can write. 1
m (m + 1) =
2
1
1
n (n + 1)
2 2
(As n = 3) 1 1 - . - ♦ 2 2 1
1
-
=
2 1 2
-
2
X
6
X
X
3 ♦ 4
1 - ·3·4 + 1 2
1 12 - X 12 + 1 2 42 7 = - = 21 2
Solution 3.11. All we need to do some count ing. For a general rank-4 t ensor in four dimensions, since each index can take any of four possible values, the number of indep endent components is simply
11 oo : 01
: 39
Solution 3.11. All we need to do some counting. For a general rank-4 tensor in four dimensions, since each index can take any of four possible values, the number of independent components is simply 4
independent components = 4 = 256 Taking into account the first symmetry relation, the first part ~.ikl = T jikl indicates t hat the tensor is antisymmetric when the last pair of indices is switched. Thinking of t he last pair of indices as specifying a 4 x 4 symmetric matrix, t his means instead of having 42 = 16independent elements, we actually only have 4( 4 + 1) / 2 = 10 independent choices for the last index pair (t his is the number of elements in an symmetric 4 x 4 matrix). Similarly, the second part of the [email protected]
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first relation 1iJkl = T i.i lk indicates that the tensor is symmetric when the last pair of indices is switched. Hence by similar argument we get 10 independent choices. So with the three relations we have 10 x 10 = 100 independent choices and e can take it as 10 x 10 matrix. We are now able to handle the last condit ion 1iJkl = - T klij
By now, it should be obvious that t his statement indicates that the tensor is ant isymmetric when the first index pair is interchanged with the second index pair. The counting of independent components is t hen the same as t hat for 10 x 10 antisymmetric matrix. So we get t he final ans
11
00:01 :41
cates that the tensor is antisymmetric when the first index pair is interchanged with the second index pair. The counting of independent components is t hen the same as t hat for 10 x 10 antisymmetric matrix. So we get t he final ans 10 x (10 - 1)/2 = 45 independent choices.
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4
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Green's functions
8(x), where k is a constant. Which of t he following state[JEST 2013] ments is true?
(a) Both G(x) and G'(x) are continuous at x = 0 (b) G(x) is continuous at x = 0, but G' (x) is not (c) G' (x) is discontinuous at x = 0 (d) The continuity properties of G (x) and G' (x) at x = 0 depend on the value of k
11 oo : 01
: 44
(b) G(x) is continuous at x = 0, but G' (x) is not (c) G' (x) is discont inuous at x = 0 (d) The cont inuity propert ies of G (x) and G' (x) at x depend on the value of k
=
0
o(x - x 0 ) with the boundary conditions g(- L , x 0 ) = 0 = g(L, x 0 ) is [NET June 2017]
(a)
iL(xo - L) (x ½,(xo
(b)
(c)
+ L ),
+ L )(x -
L ),
- L < x < x0 x0 < x < L
2i(xo+L)(x+L) ,
-L ~x:Sx 0
ft (xo - L ) (x - L) , 2i (L - x0 )(x + L), 2i (xo + L ) (L - x ),
x0 < x < L - L
< x < x0
x0 < x < L
(d) 2i(x- L )(x+ L ), - L < x < L - - - - - - - j [email protected]
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Prob 4.3. The Green 's function G (x, x') for t he equation
a2 y
X
with t he boundary values y(O) = y (; ) = 0, is June 2018]
[NET
11 oo : 01
: 47
dx with t he boundary values y(O) = y (; ) June 2018]
(a) G (x, x') =
(b)G(x ,x') =
(x' - ;) , (x - ; ) x',
X
0
E 1 , if the particle crosses t hose points then from above eqution we will get r 2 = -ve. From the picture t here is only such point , the particle has only one turning point. From point r = a the potential energy start to decrease,
11 oo : oo : 55 From the picture there is only such point , the particle has only one turning point. From point r == a the potential energy start to decrease, so K.E start to increase, hence the velocity also start to increase till t he t he particle reaches point c, where the potential in minimum. At that point velocity and K.E become maximum. •
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physicsguide CSIR NET, GATE
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Basic Newtonian Niechanics: Part-2
At points r = a and b the potential energies are equal, so are velocities and K.E. Sol 1.12.
x(t) = y(t) x(t) = iJ = -x(t) and,
y(t) = -x(t) y(t) = -x == -y(t) The equation of motion,
F
=
m(xx + yf))
=
- (xx + yf))
If we take t he potential as V(x, y) = ~ (x 2 + y 2 ) then, 1 . ,.,
,., .
11 oo :oo : s1 F
=
m(xx + yy)
=
- (xx + yy) 2
If we take the potential as V(x,y) = ~(x +y
2
)
then ,
Sol 1.13. The equation of the surface on which the bike is moving 1s, •
•
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21
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Basic Newtonian 1\/[echanics: Part-2
@Sk J ahiruddin , 2020
A unit normal vector to the surface at (x, y, z)is,
" V cp(x, y,z) n = ---cp(x, y,z) 2 2 v' [z - a(x + y ) ] v' [z - a(x 2 + y 2 ) 2a(xx + yy) + z 2 2 2 J4a (x + y ) + 1
TcosB
" (\
sin8
p
'-------+ . . . . . . . . . . mg
Zo
11 oo : 01 : oo mg
Switching to a cynlindrical coordinate syst em , x 2 + y 2
=
p2 .
At z = zo, p
2
[email protected]
=
Zo a
22
physicsguide CSIR NET ) GATE
Basic Newtonian Nlechanics: Part-2
@Sk J ahiruddin ) 2020
if n makes an angle 0 with z axis , then
n can be written
as, A
n
2app
2
✓4a p + 1
✓4a p + 1
= ----::========== + ----::========== 2 2 2 2 = sin 0p + cos 0z
So, tan0
= 2ap
The Normal Force at z 0 acts in the direction of the normal vector to the surface at that point. The z direction of t he Normal Force N z
=
.....
N cos 0, bal-
11 oo : 01
: 02
The Normal Force at z 0 acts in the direction of the normal vector to the surface at that point. .....
The z direction of t he Normal Force N z = N cos 0, balances the downward gravitational force -mgz , so t hat the bike doesn 't slip down. And the radial component of the ..... Normal Force N r = N sin 0, produces the required centripetal force for the bike to circle around. So, Ncos0 Nsin0
= mg mv
2
= -p
Now we have got, 2
mv 2ap = - pmg v
=
2zog
23
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physicsguide CSIR NET , GATE
@Sk J a hiruddin, 2020
B asic Newtonia n Nlechanics: P ar t-2
Sol 1.14.
1
E = -mv 2
2
1
+ -mw
2
2
Then 1
- m4 2
2
1
+ - mw
2?
1
?
1
22
2- = -m2~ + - mw 6 2 2 2 2 2 16 + 4w = 4 + 36w 2 3 w = 8
11 oo : 01 : os 16 + 4w = 4 + 36w 2 3 w = 8 2
2
Equation for S.H.O
x
= a cos (wt - cp)
v = -aw sin(wt - cp)
At any point
a= j [email protected]
@Sk J ahiruddin, 2020
2
24
140 3 physicsguide CSIR NET, GATE
Basic Newtonian lVIechanics: Part-2
Momentum
Prob 2.1. A jet of gas consists of molecules of mass m, speed v and number density n all moving co-linearly. This jet hits a wall at an angle 0 to t he normal. The pressure exerted on the wall by t he jet assuming elastic collision will be [JEST 2012] (a) p = 2mnv 2 cos20 (b) p = 2mnv 2 cos 0 ( c) v
= " fim nvcos20
(d )v=mnv 2
11 oo : 01
: 01
jet hits a wall at an angle 0 to t he normal. The pressure exerted on the wall by t he jet assuming elastic collision will be [JEST 2012] (a) p = 2mnv 2 cos 20 (b) p = 2mnv 2 cos 0
(c) P =
3 mnvcos 20 2
(d) p = mnv 2
Prob 2.2. A cylindrical rod of length L has a mass density
distribut ion given by p(x) = Po( l +~),where x is measured from one end of t he rod and p0 , is a constant of appropriate dimensions. The center of mass of the rod is [JAM 2016]
5 (a) L 9
4 (b) L 9
(c)
!9 L
(d)
!2L
Prob 2.3. A block of mass M is free to slide on a frictionless horizontal floor. The block has a cylindrical cavity of
radius R in the middle of it. The center of mass (CM) of the block lies on t he dashed line passing t hrough t he center of t he cavity (see figure) . Initially t he CM of t he block is at a horizontal distance X 1 from t he origin, Now a point part icle of mass m is released from point A into t he cavity. There is negligible friction bet~reen t he particle and t he cavity [email protected]
@Sk J ahiruddin, 2020
25
physicsguide CSIR NET, GATE
Basic Newtonian :tvlechanics: Part-2
face. When t he particle reaches point B, t he CM of the block is at a distance X 2 from t he figure. Find (X 2 - X 1 ) [JAM 2009]
11 oo : 01
: 11 [JAM 2009]
A
B
I I
I I
I I
X1
(Cent er of mass of the t otal system should be conserved as there is no external force) Prob 2.4. A homogeneous semi-circular plate of radius R = 3m is shown in t he figure. The distance of t he COM of t he
plate (in meter) from the point O is?
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26
[JAM 2010]
physicsguide CSIR NET, GATE
Basic Newtonian l\/Iechanics: Part-2
@Sk J ahiruddin , 2020
O
3m
11 oo : 01
: 13
O
3m
(If you are getting 2R/3 then you are totally wrong on t he definition of cent er of mass) Prob 2.5. A chain of mass M and length L is suspended vert ically with its lower and touching a weighting scale. The chain is released and falls freely onto the scale. Neglecting t he size of the individual links, what is the reading of t he scale when a length x of the chain has fallen? [JEST 2015]
(a) Mix
(b) 2~gx
(c) 3~gx
(d) 4~gx
(Falling chain prob is very interesting dude!) Prob 2.6. Sand falls on a conveyor belt at the rate of 1.5 kg/s. If the belt is moving with a constant speed of 7m/ s, t he power needed to keep t he conveyor belt running is: [JAM 2017] (Specify your answer in Watts to two digits after t he decimal j [email protected]
@Sk J ahiruddin, 2020
27
physicsguide CSIR NET 1 GATE
Basic Newtonian l\llechanics: Part-2
point .) Prob 2. 7. A water cannon starts shooting a jet of water
11 oo : 01
: 16
@Sk J ahiruddin , 2020
Basic Newtonian l\llechanics: Part-2
point .) Pro b 2. 7. A wat er cannon start s shooting a jet of water horizont ally, at t = 0, into a heavy trolley of mass M placed on a horizontal ground . The nozzle diameter of the water cannon is d, t he density of water is p , and the speed of water coming out of the nozzle is u . Find t he sp eed of t he trolley as a function of t ime. Assume t hat all t he water from the jet is collected in t he t rolley. Neglect all frictional losses. [JAM 2012]
I
\VATERCANNON
I>---..,
M
-------- ----- - -- ---
-- - - - - -0
•
0
Truly speaking t he problem is incomplet e. It is not given whether t he cannon is also moving wit h t he t rolley. If you assume t hat the cannon is also moving with t he t rolley and maintain a constant difference wit h t he t rolley then you get t his ans. Prob 2.8 . A rain drop falling vertically under gravity [email protected]
@Sk J ahiruddin , 2020
28
physicsguide CSIR NET, GATE
Basic Newtonian l\llechanics: Part-2
ers moisture from t he atmosphere at a rate given by
d;: =
11 oo :01
:1
a Basic Newtonian :tvlechanics: Part-2
@Sk J ahiruddin, 2020
ers moisture from the atmosphere at a rate given by dd"; = 2 kt , where m is the instantaneous mass, t is time, and k is a constant. The equation of motion of the rain drop is m~ + vd:; == mg. If the drop starts falling at t == 0, with zero init ial velocity and initial mass m 0 (given m 0 == 2gm, k == l2gm/s 3 and g == lOOOcm/s 2), the velocity (v) of the drop after one second is [JAM 2011] (a) 250cm/ s (b) 500cm/ s (c) 750cm/ s (d) 1000cm/ s Prob 2.9. A particle with time-varying mass m(t) == m 0 (1;), where m 0 and T are positive constants, moves along the x-axis under the action of a constant positive force F for 0 ::; t < T. If t he particle is at rest at time t == 0, then at time t == t, its velocity v will be [TIFR 2013] TF t Ft t (a)- -log 1 - (b)-- log-
mo
Ft
t
(c)- 1 - m0
mo
T
T
-i
(e) r F mo
T
t
l -T
Prob 2.10. An aircraft, which weighs 12000 kg when unloaded, is on a relief mission, carrying 4000 food packets weighing 1 kg each. The plane each is gliding horizontally with its engines off at a uniform speed of 540 km/h when t he first food packet is dropped. Assume t hat the horizon-
j [email protected]
@Sk J ahiruddin , 2020
29
physicsguide CSIR NET, GATE
Basic Newtonian 1\/[echanics: Part-2
11 oo : 01 [email protected]
: 21
29
@Sk J ahiruddin, 2020
physicsguide CSIR NET, GATE
Basic Newtonian l\/Iechanics: Part-2
tal air drag can be neglected and t he aircraft keeps moving horizontally. If one food packet is dropped every second, t hen the distance between the last two packet drops will be [TIFR 2016] (a) l .5 km (b )200 m (c) 150 m (d)lOO m
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30
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11 oo : 01
: 23
j [email protected]
30
@Sk J ahiruddin, 2020
2.1
Basic Newtonian l\llechanics: Part-2
Ans Keys
2.1. a
2.5.
2.2. a
2.6. 73.5
2.3. 2.4.
physicsguide CSIR NET 1 GATE
-2mR
m+M 4 1r
2. 7. u ln
C
V
M+kt M
where k 2.8. b 2.9.
C
2.10.
C
=
1rd2
4 up
11 oo : 01
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2.2
: 26
31
physicsguide CSIR NET, GATE
Basic Newtonian Niechanics: Part-2
Hints
2.1. what is the change in momentum Px? .....
2.2. definit ion of c. o. m for a continuous rigid body is R
=
ITrdm ITdm . 2.3. 2.4.
2.5. 2.6. what is the required energy ~ W to keep the belt mov-
ing at t he same speed for t ime ~ t? 2. 7. use conservation of momentum. 2.8. you are already given all information, so just solve the
differential equation. 2.9. after doing t he last two problem, this one should be
easy.
11 oo :01 : 2a differential equation.
2.9. after doing the last two problem , this one should be easy.
2.10. Ans key is right .... !!
32
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@Sk J ahiruddin , 2020
2.3
physicsguide CSIR NET I GATE
Basic Newtonian 1\/[echanics: Part-2
Solutions
Sol 2.1. For a single particle t he change in x component of moment um, ~P.'[; = 2mv cos 0 •
~
Foi~ce Fx apllied on the surface
F _ ~Px _ 2mvcos 0 X -
~t -
~t
11 oo : 01
: 31
Fo1·ce Fx apllied on t he surface
F _ ~ P.x _ 2mvcos 0 ~t -
X -
~t
The number of particle striking in t ime ~ t on t he area
A is N = nv~ tA cos 0. [email protected]
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33
@Sk J ahiruddin ) 2020
Basic Newtonian Nlechanics: Part-2
Hence t he pressure
P = N ·
i
2
2
= 2nmv cos 0
Sol 2 .2 .
f xdm X =-T_ _ f dm f0L p(x )xdx 7
foL p(x) dx Po foL (1+ f ) xdx -
L
Po fo (1 + I,) dx 5L 9
Sol 2 .3 . Center of m ass of t he total system should be conserved as there is no external force. The COM of t he combined syst em init ially is
MX1
+ m (X1 - R) '/\If+ rn
11 oo : 01
: 33
Sol 2.3. Center of mass of the total system should be conserved as there is no external force.
The COM of t he combined system initially is MX1
+ m(X 1 - R) M+m
After the ball reaches t he other side the COM is MX2 + m(X2 + R)
m+M Equating t he initial and final coordinate of COl\!l
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34
@Sk J ahiruddin, 2020
Basic Newtonian Nlechanics: Par t-2
Hence we get
Sol 2.4 .
COM
=
R = L i miri mi
Center of mass is a vector. keep t his t hing in mind while calculating t he COM. You need to find x coordinate and y coordinate of the COM separately in some types of problems particularly where t he are no spherical / cylindrical symmetry. x coordinate of t he COM is at ZERO (taking t he center of the semicircular plate as origin ) . Now y coordinate is
" \/
f ydm
I. CM=
Now, substituting the values y
f dm =
r sin 0 and dm
=
e7rdrd0 ,
11 oo : 01
: 36
of the semicircular plate as origin ) . Now y coordinate is y;
c1v1
J ydm J dm
=
Now, substituting the values y = r sin 0 and dm = ardrd0, we get YcM
=
f R f 1r ar 2 sin 0d0dr 0 0 R 0 01r ar d0dr
f J
R
_
fr0
2
(2)r dr
f oR ( 1r )rdr
_
R
(J0 sin 0d0) r 2 dr R f0 (J0 d0) rdr
f R 0
=
1r
fr0 r dr 1r JR rdr 0
2
1r
2
2
_
r
3
3
R 0 _
2
1r [ r 2
] :
3 4R -31r R
2
4R 31r
4
Put R = 3 and get ans== -
1f
35
j [email protected]
@Sk J ahiruddin , 2020
physicsguide CSIR NET, GATE
Basic Newtonian lVIechanics: Part-2
Sol 2.5. The reading of t he weight machine will come from two parts. The first is obviously the weight of the chain accumulated on the machine. As the chain has fallen t he distance x on the weight machine, t he weight due to t his part is
W1 = Mgx/ L The second component contributing to t he weight is due to t he momentum delivered by the chain to the weight machine. As the chain is falling freely the velocity of t he chain parts at t he instant of hitting t he scale is simply found from 2
v = 2gx
So the momentum dp delivered in t ime dt
11 oo : 01
: 39
at t he instant of hitting t he scale is simply found from v
2
=
2gx
So t he momentum dp delivered in t ime dt dp
=
d(mv)
=
vdm
where t he v has taken to be constant as the momentum is delivered in very small time dt. The chain is moving in velocity v, the length of the chain falls down to t he machine in t ime dt is dx = v dt . The mass per unit length is M/ L. So t he mass dm which falls on the weight machine in time dt is nothing but
M
M
L
L
dm = - dx = - vdt [email protected]
36
@Sk J ahiruddin , 2020
physicsguide CSIR NET, GATE
Basic Newtonian :tvlechanics: Part-2
So t he momentum dp delivered in t ime dt M M 2 dp = d(mv) = vdm = v L vdt = L v dt
So the force which will be exerted by t he chain to the weight machine is the rate of change of this momentum delivered dp = lim fl.p = M v 2 dt ~t➔O i:l.t L
As v
2
=
2gx, so t he force due to momentum delivered be-
comes W _ dp 2 - dt
M
X
L v2=2MgL
The total force on t he scale during the fall of t he chain
11 oo : 01 As v 2
= 2gx,
: 42
so t he force due to momentum d elivered b e-
comes W - dp 2 -
dt
The total force on t he scale during the fall of the chain is t her e£ore X
X
X
L
L
L
Mg -+ 2Mg - = 3Mg -
Sol 2.6. Say, l:!,,.m m ass has been added on t h e already existing sand of mass m, in t ime 1:!,,.t. According to t he question we have to keep the b elt moving at the same speed. So t he reqired extra energy in t ime 1:!,,.t I:!,. W
1
= - (m + l:!,,.m) v 2
=
2
1
- - mv
2
2
1 - l:!,,.mv 2 2
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Basic Newtonian l\/Iechanics: Part-2
@Sk J ahiruddin , 2020
Hence the required p ower ,
P = lim
I:!,.
w
b.t➔O I:!,.
t
. 1 l:!,,.m 2 = 11m ---v b.t➔O 2 i:!,,.t l dm 2 = --v 2 dt
=
!2 · 1.5 · 7
=
73.5
2
kg • m
2
•
s-
3
Watt
Sol 2. 7. Let us fo cus on t he short time interval t and t + 1:!,,.t.
11 oo : 01
: 44
-
2 dt
= !2 · 1.5 · 72 kg • m 2 • s- 3 = 73.5 W att Sol 2. 7. Let us focus on t he short t ime interval t and t + ~ t . Let water of mass ~ m added to t he t rolley in t ime ~ t. At any t ime t he system consist of the mass of the t rolley M (t) and ~ m. The initial momentum is
P (t) = M(t)v + ~mu. T he final momentum is
P (t + ~ t) = (M(t) + ~ m) · (v + ~v) . The change in momentum is (ingnoring the negligible products like ~ m~v) ~p
= P (t + ~ t) - P (t) = M(t) ~v + (v - u )~ m.
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physicsguide CSIR NET 1 GATE
38
@Sk J ahiruddin, 2020
Basic Newtonian l\llechanics: Part-2
The rate of change of momentum is approximately ~p ~t
In t he limit
~m
~v
= M(t) ~ t + (v - u) ~ t
~t ➔
0, we have
dP dv dt = M (t) dt
+ (V
-
dm u) dt
11 oo : 01 .6.t ➔
In t he limit
dP dt
: 47
0, we have dv M (t ) dt
=
+ (V
-
dm u) dt
In this case since there are no external force cZJ: = 0. dm dM 1rd2 = = up = k (say ). dt dt 4
So t
M(t)
=
dM' M
kdt' 0
M (t)
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= kt + M
39
physicsguide CSIR NET, GATE
Basic Newtonian Niechanics: Part-2
@Sk J ahiruddin , 2020
Then, (u - V)k
k
dt
=
dv (kt + M) dt dv
kt + M (u - v) M + kt (u - v) ln - - - = - ln M u
11 oo : 01
: 49
(u - V)k
= (kt + M) dt dv (u - v)
dt
k
kt + M
In M
+ kt = -
In (u - v)
M
u
M+kt M
u
(u -v) M
V
= u 1- M + kt
A frequent error!! If u start with F = (d/dt)(mv ) you would get m(dv/dt) + v(dm/ dt). Which is simply wrong! Because in that case you are not considering the moment um change of the water. Sol 2.8. The mass of t he water droplet at any time t is,
m
dm = kt2 dt t
kt'2 dt'
dm' = mo
0
1 3 m = kt + mo 3 From the eq. of motion of t he rain drop, j [email protected]
@Sk J ahiruddin , 2020
40
physicsguide CSIR NET I GATE
Basic Newtonian 1\/[echanics: Part-2
dv dm md +v = mg t dt dv 2 m dt + v kt = mg
11 oo : 01 : s2 dv dm m dt + v dt = mg dv 2 m-+vkt = mg dt kt 2 dv - + v - =g dt m kt 2 dv - + v - -3 - - = g dt 1kt + m 0 We can solve t his linear first order ODE by multiplying, exp
kt' 2 - - - -dt' }kt'3 + m 0
d (1kt'3 + m )
= exp
o dt'
3
}kt'3 + m 0
= exp In
1 3
kt
3
+ mo
on both side, thus turning it into an exact differential. Hence,
[email protected]
@Sk J a hiruddin ) 2020
41
physicsguide CSIR NET ) GATE
Basic Newtonian Nlechanics: Part-2
11 oo : 01 : s4 @Sk J ahiruddin, 2020
Basic Newtonian l\llechanics: Par t-2
dv kt 2 . -+v---- · dt ½kt3 + mo
or,
1 3 kt + mo g 3 d 1 3
or,
d V
dt v
Integrating
on
kt + m0 3
both
sides,
v,t
t
d
V
0
0 ,0
1 3
kt + m 0 gdt 3
4
t2kt + mat
·g
or,
v=
or,
+ mat v = -3 - - - ·g ½kt + mo
tkt3 + mo 112kt4
Putting the values k t = l s, we get,
=
12gm/s
v
3
,
m0
=
2gm, g
=
lOOOcm/s
2
= 500cm/s
•
[email protected]
@Sk J ahiruddin, 2020
42
physicsguide CSIR NET, GATE
B asic Newtonian l\llechanics: Part-2
,
11 oo :01 : s1 j [email protected]
physicsguide CSIR NET, GATE
42
@Sk J ahiruddin, 2020
Basic Newtonian Nlechanics: Part-2
Sol 2.9. Using t he same principle of problem Prob 2. 7. , we get
dm dv F = dt v + m dt mo dv or, F = --v + m(t) d
t
T
or,
t mo F +-v=mo 1 - T
T
dv dt
or,
Int egrating bot h side, we get Ft
v= mo
t
- 1
1- -
T
Sol 2.10. 540 km/ h is 150 m/s
Well you have thought quite right . The starting t hought would be the variable mass problem . But, t he velocity in horizontal direction do not change. The horizontal velocity of food packet is, when released, t he same before and after release. So t he momentum change is in t he vert ical direction only. Horizontal motion do not get effected. Please go t hrough the previous solutions of variable mass problems once again. j [email protected]
43
physicsguide CSIR NET, GATE
11 oo : 01 : sg j [email protected]
physicsguide CSIR NET, GATE
42
@Sk J ahiruddin, 2020
Basic Newtonian Nlechanics: Part-2
Sol 2.9. Using t he same principle of problem Prob 2. 7. , we get
dm dv F = dt v + m dt mo dv or, F = --v + m(t) d
t
T
or,
t mo F +-v=mo 1 - T
T
dv dt
or,
Int egrating bot h side, we get Ft
v= mo
t
- 1
1- -
T
Sol 2.10. 540 km/ h is 150 m/s
Well you have thought quite right . The starting t hought would be the variable mass problem . But, t he velocity in horizontal direction do not change. The horizontal velocity of food packet is, when released, t he same before and after release. So t he momentum change is in t he vert ical direction only. Horizontal motion do not get effected. Please go t hrough the previous solutions of variable mass problems once again. j [email protected]
43
physicsguide CSIR NET, GATE
11 oo : 02 : 02 j [email protected]
physicsguide CSIR NET, GATE
42
@Sk J ahiruddin, 2020
Basic Newtonian Nlechanics: Part-2
Sol 2.9. Using t he same principle of problem Prob 2. 7. , we get
dm dv F = dt v + m dt mo dv or, F = --v + m(t) d
t
T
or,
t mo F +-v=mo 1 - T
T
dv dt
or,
Int egrating bot h side, we get Ft
v= mo
t
- 1
1- -
T
Sol 2.10. 540 km/ h is 150 m/s
Well you have thought quite right . The starting t hought would be the variable mass problem . But, t he velocity in horizontal direction do not change. The horizontal velocity of food packet is, when released, t he same before and after release. So t he momentum change is in t he vert ical direction only. Horizontal motion do not get effected. Please go t hrough the previous solutions of variable mass problems once again. j [email protected]
43
physicsguide CSIR NET, GATE
11 oo : 02 : 04 j [email protected]
physicsguide CSIR NET, GATE
42
@Sk J ahiruddin, 2020
Basic Newtonian Nlechanics: Part-2
Sol 2.9. Using t he same principle of problem Prob 2. 7. , we get
dm dv F = dt v + m dt mo dv or, F = --v + m(t) d
t
T
or,
t mo F +-v=mo 1 - T
T
dv dt
or,
Int egrating bot h side, we get Ft
v= mo
t
- 1
1- -
T
Sol 2.10. 540 km/ h is 150 m/s
Well you have thought quite right . The starting t hought would be the variable mass problem . But, t he velocity in horizontal direction do not change. The horizontal velocity of food packet is, when released, t he same before and after release. So t he momentum change is in t he vert ical direction only. Horizontal motion do not get effected. Please go t hrough the previous solutions of variable mass problems once again. j [email protected]
43
physicsguide CSIR NET, GATE
11 oo : 02 : 01 j [email protected]
physicsguide CSIR NET, GATE
42
@Sk J ahiruddin, 2020
Basic Newtonian Nlechanics: Part-2
Sol 2.9. Using t he same principle of problem Prob 2. 7. , we get
dm dv F = dt v + m dt mo dv or, F = --v + m(t) d
t
T
or,
t mo F +-v=mo 1 - T
T
dv dt
or,
Int egrating bot h side, we get Ft
v= mo
t
- 1
1- -
T
Sol 2.10. 540 km/ h is 150 m/s
Well you have thought quite right . The starting t hought would be the variable mass problem . But, t he velocity in horizontal direction do not change. The horizontal velocity of food packet is, when released, t he same before and after release. So t he momentum change is in t he vert ical direction only. Horizontal motion do not get effected. Please go t hrough the previous solutions of variable mass problems once again. j [email protected]
43
physicsguide CSIR NET, GATE
11 oo : 02 : 1o j [email protected]
physicsguide CSIR NET, GATE
42
@Sk J ahiruddin, 2020
Basic Newtonian Nlechanics: Part-2
Sol 2.9. Using t he same principle of problem Prob 2. 7. , we get
dm dv F = dt v + m dt mo dv or, F = --v + m(t) d
t
T
or,
t mo F +-v=mo 1 - T
T
dv dt
or,
Int egrating bot h side, we get Ft
v= mo
t
- 1
1- -
T
Sol 2.10. 540 km/ h is 150 m/s
Well you have thought quite right . The starting t hought would be the variable mass problem . But, t he velocity in horizontal direction do not change. The horizontal velocity of food packet is, when released, t he same before and after release. So t he momentum change is in t he vert ical direction only. Horizontal motion do not get effected. Please go t hrough the previous solutions of variable mass problems once again. j [email protected]
43
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11 oo : 02 : 12 j [email protected]
physicsguide CSIR NET, GATE
42
@Sk J ahiruddin, 2020
Basic Newtonian Nlechanics: Part-2
Sol 2.9. Using t he same principle of problem Prob 2. 7. , we get
dm dv F = dt v + m dt mo dv or, F = --v + m(t) d
t
T
or,
t mo F +-v=mo 1 - T
T
dv dt
or,
Int egrating bot h side, we get Ft
v= mo
t
- 1
1- -
T
Sol 2.10. 540 km/ h is 150 m/s
Well you have thought quite right . The starting t hought would be the variable mass problem . But, t he velocity in horizontal direction do not change. The horizontal velocity of food packet is, when released, t he same before and after release. So t he momentum change is in t he vert ical direction only. Horizontal motion do not get effected. Please go t hrough the previous solutions of variable mass problems once again. j [email protected]
43
physicsguide CSIR NET, GATE
11 oo : 02 : 1 s j [email protected]
physicsguide CSIR NET, GATE
42
@Sk J ahiruddin, 2020
Basic Newtonian Nlechanics: Part-2
Sol 2.9. Using t he same principle of problem Prob 2. 7. , we get
dm dv F = dt v + m dt mo dv or, F = --v + m(t) d
t
T
or,
t mo F +-v=mo 1 - T
T
dv dt
or,
Int egrating bot h side, we get Ft
v= mo
t
- 1
1- -
T
Sol 2.10. 540 km/ h is 150 m/s
Well you have thought quite right . The starting t hought would be the variable mass problem . But, t he velocity in horizontal direction do not change. The horizontal velocity of food packet is, when released, t he same before and after release. So t he momentum change is in t he vert ical direction only. Horizontal motion do not get effected. Please go t hrough the previous solutions of variable mass problems once again. j [email protected]
43
physicsguide CSIR NET, GATE
11 oo : oo : oo
Problellls and Solutions in Angular Molllentulll and Rigid body Mechanics : Part-1 Sk J ahiruddin * Sandip Biswas
*Assistant Professor Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay iVI.Sc Physics 2009-2011 batch He ranked 007 in IIT JAM 2009 and 008 (JRF) in CSIR NET June 2011 He has been teaching CSIR NET aspirants since 2012
1
@Sk J ahiruddin , 2020
Angular momentum & Rigid body: Part-1
11 oo : oo : 03 1
Angular momentum & Rigid body: Part-1
@Sk J ahiruddin , 2020
Contents 1 Simple motion of rigid bodies 1.1
3
Angular Momentum and torque
1.2 Moment Of Inert ia . . . . . . .
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1.3
Rotational and transnational motion
1.4
Ans Keys
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1.5
Hint s . . .
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31
1.6
Solutions .
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35
j [email protected]
2
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11 oo : oo : os
j [email protected]
2
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Angular momentum & Rigid body: Part-1
physicsguide CSIR NET I GATE
Problems from NET, GATE, JEST, TIFR & JAM papers
1 1.1
Simple motion of rigid bodies Angular Momentum and torque
Prob 1.1. A solid sphere of mass m and radius a is rolling
with a linear speed v on a flat surface without slipping. The magnitude of angular momentum of t he sphere in t he surface is [JAM 2005] 2 (a) mav 5
7 (b) mav 5
3 (c) mav 2
(d) mav
Prob 1.2. A particle of mass 2kg is moving such t hat at a time t, its position in meter, is given by r(t) = 51 - 2t 23. The angular moment um of the particle at t = 2sec about 2 2 t he origin in kgm s- , is [GATE 2006]
(a) - 40k
(b) - 80k
(c) 80k
(d) 40k
Prob 1.3. A mass m is constrained to move on a horizontal frictionless surface. It is set in circular motion wit h radius r 0 and angular speed w0 by an applied force F communi-
cated through an inextensible thread that passes t hrough a
11 oo : oo : 09 mass m 1s cons ra1ne frictionless surface. It is set in circular motion with radius r 0 and angular speed w0 by an applied force F communicated through an inextensible thread that passes t hrough a [email protected]
3
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Angular momentum & Rigid body: Part-1
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hole on t he surface as shown in the fig. This force is t hen suddenly doubled. The magnitude of t he radial velocity of [GATE 2008] t he mass
F
(a) increases t ill the mass fall into the hole. (b) decreases t ill t he mass falls into the hole. (c) remains constant. (d) becomes zero at radius r 1 where O < r 1 < r 0 . Prob 1.4. Two bodies of equal mass m are connected by t he massless rigid rod of length l lying in the xy-plane with
t he center of the rod at the origin. If the system is rotating about the z-axis wit h a frequency w, its angular momentum is [NET Dec 2012] 2w 2w (a) ml -;; (b) ml ~ (c) ml 2w (d) 2ml 2w
11 oo : oo : 11 the massless rigid rod of length l lying in the xy-plane with the cent er of t he rod at the origin. If the system is rotating about the z-axis wit h a frequency w , its angular moment um [NET Dec 2012] is
(a) ml 2 ~
(b) m l2 ~
(c) ml 2w
(d) 2ml 2w
[email protected]
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Angular momentum & Rigid body: Part-1
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Prob 1.5. A t hin massless rod of length 2l has equal point masses m attached at its ends (see fig) . T he rod is rotat-
ing about an axis passing through its center and making an angle 0 with it. The magnitude of the rate of change of its angular momentum
2
2
-
~f (only absolute value) is
(a) 2ml w sin 0 cos 0 2
2
(c) 2ml w sin 0
2
2
[JAM 2009]
2
(b) 2ml w sin 0 2 2
2
(d) 2ml w cos 0
Prob 1.6. At an instant shown, t hree point masses m , 2m
11 oo : oo : 14 2
2
(a) 2ml w sin 0 cos 0 2
2
2
(b) 2ml w sin 0 2
(c) 2ml w sin0
2
2
(d) 2ml w cos0
2
2
Prob 1.6. At an instant shown, three point masses m , 2m
and 3m rest on a horizontal surface, and are at the vertices of an equilateral triangle of unit side length. Assuming that G is t he gravitational constant, t he magnit ude and direction j [email protected]
5
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Angular momentum & Rigid body: Part- I
physicsguide CSIR NET, GATE
of t he torque on t he mass 3m, about the point O, at t hat instant is [JAM 2014]
m
0
(a) zero 2 (b) ~ Gv'3m , going into t he paper 2 (c) 3Gv'3m , coming out of the paper 2 (d) !G-/3m , going into the paper
Pro b 1. 7. A spherical ball of ice has radius Ro and is rotating with an angular speed w about an axis passing t hrough its center. At time t = 0, it starts acq uiring mas because
11 oo : oo : 16
Prob 1. 7. A spherical ball of ice has radius Ro and is rotating with an angular speed w about an axis passing t hrough its center. At time t = 0, it starts acquiring mas because t he moisture (at rest ) around it starts to freeze on it uniformly. As a result its radius increases as R (t ) = Ro + at, where a is a constant. The curve which best describes its angular speed with t ime is [JAM 2014]
[email protected]
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Angular moment um & Rigid body: Part-1
(a)
physicsguide CSIR NET, GATE
(b)
t (c)
0>1....o--
t
I
(d)
(I).,___ t
Prob 1.8. MSQ: A particle moves in a circular path in t he xy-plane centered at the origin. If t he speed of the particle [JAM 2016] is constant, then its angular momentum (a) about t he origin is constant both in magnitude and direction (b) about (0,0 ,1) is constant in magnitude but not in direct ion (c) about (0,0,1) varies both in magnitude and direction
11 oo : oo : 19 rection (b) about (0,0,1) is constant in magnit ude but not in direction (c) about (0,0,1) varies both in magnit ude and direction (d) about (0,0,1) is a const ant in direction but not in magnit ude
Prob 1.9. A uniform circular disc of radius R and mass
M is rotating wit h angular speed w about an axis, passing t hrough its center and inclined at an angle 60 · with respect [email protected]
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Angular momentum & Rigid body: Part-1
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to its symmetry axis. The magnit ude of t he angular mo[GATE 2013] ment um of the disc is
(a) J}wM R
(c)
f wMR
2
2
(b) 1wMR (d)
f wMR
2
2
Prob 1.10. What is the change in the kinetic energy of
rotation of t he earth if its radius shrinks by 1%? Assume t hat the mass remains the same and the density is uniform. [JEST 2019] (a) increases by 1% (b) increases by 2% (c) decreases by 1%
(d) decreases by 2%
Pro b 1.11. If t he diameter of t he Eart h is increased by 4 % without changing t he mass, t hen t he length of t he day is
11 oo : oo : 21 (a) increases by 1%
(b) increases by 2%
(c) decreases by 1%
(d) decreases by 2%
Pro b 1.11. If t he diameter of t he Earth is increased by 4 % without changing t he mass, then t he length of the day is _____ hours. [JAM 2019]
(Take t he length of the day before the increment as 24 hours. Assume the Earth to be a sphere with uniform density.) (Round off to 2 decimal places) Prob 1.12. A wheel is rotating at a frequency Jo Hz about
a fixed vertical axis. The wheel stops in t 0 seconds, with constant angular deceleration. The number of t urns covered by the wheel before it comes to rest is given by: [JAM j [email protected]
8
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Angular momentum & Rigid body: Part-1
physicsguide CSIR NET, GATE
2020]
(A) Joto Prob 1.13. The angular momentum of a particle relative
to origin varies wit h t ime ( t ) as L = (4x + at 2 y) kgm 2 /s, where a = l kgm 2 /s 3 . The angle between L and the t orque acting on t he particle becomes 45° after a time of - - - - - - - - - - - - - - s [JAM 2020] --+
1.2
Moment Of Inertia
Prob 1.14. The moment of inertia of uniform radius r about ')
11 oo : oo : 24
1.2
Moment Of Inertia
Prob 1.14. The moment of inertia of uniform radius r about 2 5 an axis passing through its center is given by -~r p. A rigid sphere of uniform mass density p and racfius R has
two smaller sphere of radius ~ hollowed out of it , as shown in fig. The moment of inertia of t he resulting body about t he Y axis is [GATE 2007]
[email protected]
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Angular momentum & Rigid body: Part-1
physicsguide CSIR NET, GATE
y
X
11 oo : oo : 26 X
(b) 51rpR5 12
( C)
5 71rpR
12
(d) 31rpR5 4
Prob 1.15. Consider a uniform thin circular disk of radius R and mass M. A concentric square of side ~ is cut out from the disk (see fig.). What is moment of inertia of t he resultant disk about an axis passing t hrough the center of [JAM 2017] t he disk and perpendicular to it?
j [email protected]
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Angular momentum & Rigid body: Part-1
physicsguide CSIR NET I GATE
R
11 oo : oo : 29 R
(b) I
=
M R2 2
(d)I =MR2 1 2
1 1 - 481r 1
241r
Prob 1.16. A cylindrical shell of mass m has an outer radius b and an inner radius a. The moment of inertia of the shell about the axis of the cylinder is [JEST 2016]
(a) lm(b 2 (c) m(b
2
2
-
a
+a
2
2
)
)
(b) lm(b 2 ( d)
m(b
2
2
-
+a a
2
2
)
)
Prob 1.17. The three-dimensional object sketched on the [email protected]
11
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Angular momentum & Rigid body: Part-1
physicsguide CSIR NET, GATE
right is made by taking a solid sphere of uniform density (shaded) wit h radius R , and scooping out a spherical cavity (unshaded) as shown, which has diameter R. If this object has mass M, its moment of inertia about the tangential axis passing through the point where the spheres touch (as shown in the figure) is [TIFR 2019]
11 oo : oo : 31 unsh aded as shown, which has diameter R. If this object has mass M, its moment of inertia about t he tangential axis passing through the point where the spheres touch (as shown in the figure) is [TIFR 2019]
3 2 (a) MR 16
(c) 31 MR2 70
(b) 62 MR2 35
(d) 31 MR2
20
Pro b 1.18. Seven uni£orm disks, each of mass m and radius r, are inscribed inside a regular hexagon as shown. The moment of inertia of t his syst em of seven disks, about an axis passing through t he central disk and perpendicular to t he plane of the disks, is [JAM 2015]
[email protected]
12
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Angular moment um & Rigid body : Par t-1
physicsguide CSIR NET, GATE
11 oo : oo : 3s
(a) ;mr
2
(b) 7mr
1
2
(c) }mr2
(d)
55 2 mr 2
Prob 1.19. Consider the uniform solid right cone depicted in t he figure. This cone has mass M and a circular base of radius r . If the moment of inert ia of the cone about an axis parallel to t he X axis passing through t he center of mass Oc.m. (see fig) is given by, s15M( 4r 2 + h2 ) then the moment of inert ia about another axis parallel to t he X axis, but passing [TIFR 2013] t hrough t he point Or(see fig) , is
physicsguide CSIR NET, GATE
j a [email protected]
13
@Sk J a hiruddin , 2020
Angular momentum & Rigid body: Part-1
.
"
11 oo : oo : 37
--
••
T Soffd COfl•_. Man •M
(a) /o M (4r (c)
1 20 M
2
+h
2
(23r 2 + 2h2 )
)
+h ) 2 2 ~M(15r + 4h ) 3
(b) foM(2r
(d)
2
2
Prob 1.20. The mass density of a disc of mass m and radius R varies as
p(r)
r
p( l - R) for (0 < r < R)
0 (for r > R)
Find t he moment of inertia of t he disc about t he axis perpendicular to the plane of the disc and passing t hrough the [email protected]
14
@Sk J ahiruddin , 2020
Angular momentum & Rigid body: Part-1
physicsguide CSIR NET, GATE
center of t he disc in t erms of m and R.
[JAM 2013]
11 oo : oo : 40 Angular momentum & Rigid body: Part-1
@Sk J ahiruddin, 2020
center of t he disc in t erms of m and R.
[JAM 2013]
Prob 1 .21. The moment of inert ia of a disc about one of its diameters is IM. The mass per unit area of the disc is proportional to the distance from its cent er . If t he radius of t he disc is R and it s mass is M , the value of IM is [JAM 2014]
(a) 1MR
2
(b) ~M R
2
(c)
2 3 10 M R
(d) ~M R
2
Prob 1.22. T wo uniform t hin rods of equal length, L and masses M 1 and M 2 are joined toget her along t he length. The moment of inert ia of the combined rod of length 2£ about an axis passing through the mid-point and perpendicular to t he length of t he rod is [GATE 2012]
(a) (M1 + M2) f;
(c) (M1
+ M2 )!f
2
(b) (M1 + M2)~ (d) (M1
+ M2)¥
Prob 1 .23. The mass per unit length of a rod (length 2 m) varies as p = 3x kg/m. The moment of inert ia (in kg m 2 ) of t he rod about a perpendicular-axis passing the t ip of the [JAM 2019] rod (at x = 0) (A) 10 (B) 12 (C) 14 (D) 16
[email protected]
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Angular momentum & Rigid body : Part-1
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11 oo : oo : 42 [email protected]
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Angular momentum & Rigid body : Part-1
1.3
physicsguide CSIR NET, GATE
Rotational and transnational motion
Prob 1.24. A uniform cylinder of radius r and length l ,
and a uniform sphere of radius R are released on an inclined plane when their centers of mass are at t he same height . If t hey roll down without slipping, and if t he sphere reaches t he bot tom of the plane with a speed V, then the speed of [NET June t he cylinder when it reaches the bottom is 2013]
(a) V
l4rl 15r 2
(c) ~
(b) 4V ~
(d) V
14 15
Prob 1.25. Moment of inertia of a solid cylinder of mass
m , height h and radius r about an axis (shown in fig by dashed line) passing through its center of mass and perpen[JAM 2009] dicular to its symmetry axis is
j [email protected]
16
physicsguide CSIR NET, GATE
11 oo : oo : 44
j [email protected]
16
@Sk J ahiruddin , 2020
Angular momentum & Rigid body: Part-1
physicsguide CSIR NET, GATE
'I I I
•i • •t I
r J
•
I
h:
~
• ''I I I I I
I
2 1 (a) 4 mr 2 c) lmr ( 2
+ l12 mh
2 + lmh 12
2
(b) ~mr
2
+ ~mh
2
2 2 (d) lmr + lmh 2 4
Pro b 1. 26. A uniform rod of mass m and length L is hinged at one of its ends O and is hanging vertically. It is hit at
its midpoint with a very short duration impulse J so that it starts rotating about O. Find the magnitude and the direction of t he horizontal impulse t hat O applies on the rod [JAM 2014] when it is hit.
11 oo:oo:47
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17
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Angular momentum & Rigid body: Part-1
physicsguide CSIR NET, GATE
0
J
Prob 1.27. A mass m t ravels in a straight line with velocity v 0 perpendicular to a uniform stick of mass m and length L , which is initially at rest. The distance from t he center of the stick to the path of the t raveling mass is h (see fig). Now the traveling mass m collides elastically wit h the stick, and t he center of t he stick and t he mass m are observed to move with equal speed after the collision Assuming that t he t raveling mass m can be treated as a point mass, and t he 2 moment of inertia of the stick about its cent er is I = , it follows that t he distance h must be [TIFR 2010]
~t
m
L
11 oo : oo : 49 m
L
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Angular momentum & Rigid body: Part-1
(a)
t
(b)
i
(c)
}e
(d)
physicsguide CSIR NET, GATE
!h,
(e)
f
(f) zero
Prob 1.28. A scient ist is given two heavy spheres made of same metal, which have t he same diameter and weight , and is asked to distinguish and spheres, without damaging t hem in any way. Though the spheres look ident ical, one of t hem is actually a hollow spherical shell, while t he other is a set of concentric shells mounted on four thin rods of the same metal (see fig). [TIFR 2011]
To make t he distinction, the scientist must perform an experiment where each sphere is (a) set rotating under the action of a constant torque (b) made into t he bob of a long simple pendulum and set oscillating
11 oo : oo : s2 experiment where each sphere is (a) set rotating under t he action of a constant torque (b) made into t he bob of a long simple pendulum and set oscillating (c) immersed fully in a non-corrosive liquid and then weighed (d) given t he same electric charge Q and t he potential is measured
j [email protected]
19
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Angular momentum & Rigid body: Part-1
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Prob 1.29. A ring of mass m and radius R rolls (wit hout slipping) down an inclined plane starting from rest. If the
center of the ring is initially at a height h, t he angular velocity when the ring reaches the base is [NET December 2013]
h
(a)
(c)
g
(b)
h _ R tan0
g(h - R) R2
(d)
g
h- R
2g h-R
11 oo : oo : 55 (a)
(c)
g
(b)
h _ R tan0
g(h - R)
R2
g
h-R
2g h-R
(d)
Prob 1.30. MSQ: A rod is hanging vertically from a pivot. A particle traveling in horizontal direction, collides with the rod as shown in fig. For the rod-particle syst em, consider t he linear momentum and t he angular momentum about the [email protected]
20
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Angular momentum & Rigid body: Part-1
physicsguide CSIR NET, GATE
pivot. Which of the following statements are NOT t rue? [JAM 2015]
•
►
(a) Both linear momentum and angular momentum are conserved (b) linear momentum is conserved but angular momentum is not (c) linear momentum is not conserved but angular moment um is conserved (d) Neither linear momentum nor angular momentum are
11 oo :oo : s1 is not (c) linear momentum is not conserved but angular moment um is conserved (d) Neit her linear moment um nor angular momentum are conserved
Prob 1.31. A hoop of radius a rotates with constant angular velocity w about the vert ical axis as shown in the fig. A bead of mass m can slide on the hoop wit hout fiction. If g < w 2 a at what angle 0 apart from O and 1r is t he bead stationary (i.e.,!~=~:i =0)?
[JEST 2016]
j [email protected]
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Angular momentum & Rigid body: Part-I
(a) tan 0 = (c) cos 0 = ~
9
; 2a
physicsguide CSIR NET I GATE
(b) sin 0 = ( d)
tan 0 =
w·~a
1r!2a
11 oo : 01 : oo
(a) tan0 = C:fa
(b) sin0 = w.ga
(c) cos 0 = w~a
(d) tan 0 = 1r32a
Prob 1.32. Two cylinders A and B of t he same length L and outer radius R were placed at t he same height h on an inclined plane at an angle ef> with the horizontal (see fig) . Starting from rest, each cylinder was allowed to roll down t he plane without slipping. It was found t hat A reached t he end of t he inclined plane earlier t han B . which of the [TIFR 2015] following possibilit ies could be true.
[email protected]
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Angular momentum & Rigid body: Part-1
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h I
I
(a) A is solid and made of copper; B is hollow and made of copper. (b) A is solid and made of copper; B is solid and made of
11 oo : 01
: 03
(a) A is solid and made of copper; B is hollow and made of copper. (b) A is solid and made of copper; B is solid and made of aluminium. (c) A is hollow and made of aluminium; B is solid and made of aluminium. (d) A is hollow and made of copper; B is hollow and made of copper ; B is heavier than A. Prob 1.33. A t hin rod of length 2l and mass M is pivoted at one end P on a horizontal plane (see fig). A ball of mass m < < M and speed v 0 strikes t he free end of t he rod perpendicularly and bounces back with velocity VJ along the original line of motion as shown in the fig. If t he collision is [email protected]
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[TIFR 2015]
perfectly elastic the magnitude of v f is
1't _...,. Vo _____________ _
• I I I I I
M
21 I
t I I
I
t n
11 oo : 01 : os I I I I
I
♦
p
M (a ) M
+ 3m
- 3m M-4m (c) M + 4m Vo
b) M 3mv ( M + 3m o
Vo
(d) M + 4m vo M-4m
Prob 1 .34. A uniform solid wheel of mass M and radius r is halted at a step of height h as shown in t he fig . The
minimum force F , applied horizontally at the center of t he wheel, necessary to raise t he wheel over t his step is [TIFR 2016]
j [email protected]
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Angular momentum & Rigid body: Part- I
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F
r _ _.....:::::......a...:::::.........i.. - - - - - - -
(a ) (C)
M ✓h(2r-h) g
r+h
M ✓h(r+h) g
r- h
_t !i
b) M ✓h(2r+h) ( g r- h
d) M ✓h(2r-h) ( g r- h
11 oo :01 : oa
(a)
J h(2r- h) M g r+h
c) M J ( g
h(r+h) r- h
b) M J h(2r+h) ( g r- h d) M J h(2r- h) ( g
r- h
Prob 1.35. A uniform rigid met er-scale is held horizontally
with one of its end at the edge of a table and t he other supported by hand. Some coins of negligible mass are kept on [JAM 2017] t he meter scale as shown in the fig.
As t he hand support ing t he scale is removed, t he scale starts rotating about its edge on the table and the coins start moving. If a photograph of the rotating scale is taken soon after, it will look closest to:
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Angular momentum & Rigid body: Part-1
physicsguide CSIR NET, GATE
tA)
(B)
(C)
(D)
Prob 1.36. A pendulum is made of a massless string of
11 oo : 01
: 1o (0)
(C)
Prob 1.36 . A pendulum is made of a massless string of
lengt h L and a small bob of negligible size and mass m . It is released making an angle 00 ( r t hen v > vo .
Sol 1.4.
I =
L m ir; Z
2
+m
=m 2
l
2
2
ml2 2 hence,
z2
L = I w =m-w 2
Sol 1.5. In this case w is constant. Then from t he definition
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37
@Sk J ahiruddin , 2020
Angular momentum & Rigid body: Part-1
of angular momentum .
.
1,
.
'l
.
'l
dL
-
dt
=m
L. r -w -r· 2-
i
1,
- (r· i
i
-) ·w
physicsguide CSIR NET, GATE
11 oo : 01
: 44 .
1,
_,
L
dL 2 _, _, ( _, _,) -dt =m . r-i w-r·,,, r·,,, · w 'l,
.
'l,
•
1,
Now iJ =
w X r'
t hen
_, _, V ·W Therefore,
=
(_, W
X
;::;' \ . . . T; · W
=
0
_,
dL ~ _, ( _, _,) dt = -m ~ V i r i · w 'l,
j [email protected]
38
@Sk J ahiruddin, 2020
Angular moment um & Rigid body: Par t-1
physicsguide CSIR NET, GATE
Using (r, 0. Which of t he following statements is t rue for the coordinates of t he centre of mass of t he disc in t he reference frame of t he car? (a) Only t he x and the z coordinat es change (b) Only t he y and t he z coordinates change (c) Only t he x and the y coordinates change (d) all the t hree coordinates change. j [email protected]
21
@Sk J ahiruddin , 2020
Angular moment um & Rigid body: Part-2
physicsguide CSIR NET I GATE
Prob 2.12. T wo objects of unit mass are thrown up vert i1 cally wit h a velocity of 1m s- at latitudes 45°N and 45°S,
respectively. T he angular velocit y of t he rotat ion of Earth is given to be 7.29 x 10- 5 s- 1 . In which direction will t he objects deflect when t hey reach t heir highest point (due to Coriolis force)? Assume zero air resistance. [JEST 2019] (a) to t he east in Nort hern hemisphere and west in Sout hern Hemisphere (b) to the west in Northern hemisphere and east in Sout hern Hemisphere (c) to t he east in both hemisphere and east in Sout hern Hemisphere (d) to t he west in bot h hemispheres Prob 2.13. Consider an obj ect moving wit h a velocity v in
11 oo : 01 : oo (c) to the east in both hemisphere and east in Southern Hemisphere (d) to the west in both hemispheres Prob 2.13. Consider an object moving with a velocity iJ in
a frame which rotates with a constant angular velocity w. The Coriolis force experienced by the object is [JAM 2019] (a) along iJ (b) along w (c) perpendicular to both iJ and w (d) always directed towards t he axis of rotation Prob 2.14. A turn-table is rotating with a constant angu-
lar velocity w 0 - In the rotating frame fixed to the t urntable, [email protected]
22
@Sk J ahiruddin, 2020
Angular momentum & Rigid body: Part-2
physicsguide CSIR NET, GATE
a particle moves radially outwards at a constant speed v 0 . The acceleration of the particle in t he r0 -coordinates, as seen from an inert ial frame, t he origin of which is at the centre of the turnable is [NET June 2019]
(a) -rw5f (b) 2rw5f 2 ,. _ (d) -rw0 f + 2v 0w0 0
+ vowo0
(c) rw5f
+ 2vowo0
11 oo : 01
: 02
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23
@Sk J ahiruddin, 2020
Angular momentum & Rigid body: Par t-2
2.1
physicsguide CSIR NET, GATE
Ans Keys
2.1. b
2.8. a
2.2. b
2.9. b
2.3. 0.34
2.10. b , c
2.4.
C
2.11. d
2.5.
C
2.12. d
11 oo : 01 : os 2.4. c
2.11. d
2.5. c
2.12. d
2.6. b
2.13. c
2.7. c
2.14. d
j [email protected]
24
@Sk J ahiruddin , 2020
Angular momentum & Rigid body: Part-2
2.2
physicsguide CSIR NET, GATE
Hints
2.1. There can be no hints for this one. 2. 2. Use the expression of Coriolis force and centrifugal
force .
2.3. The weight are different in pole and equator due to different centrifugal force.
11 oo : 01
: 01
force . 2.3. The weight are different in pole and equator due to
different centrifugal force. 2.4. Use the expression of Coriolis force and centrifugal
force to find out the magnitude of t he forces. 2.5. Use t he expression of Coriolis force to find out the di-
rection and the magnitude of the force. 2.6. Use t he expression of Coriolis force to find out which direction t he particle goes. 2. 7. Use the expression of Coriolis force and centrifugal
force , then cornpare t he two. 2.8. Use t he expression of Coriolis force to find out the direction and the magnit ude of the force. 2.9. What is the direction of Coriolis force in southern [email protected]
25
@Sk J ahiruddin, 2020
Angular moment um & Rigid body: Part-2
physicsguide CSIR NET, GATE
sphere? 2 .10. Use t he expression of Coriolis force and centrifugal
force to find out the magnitude and t he direction of the forces . 2.11. The sit uation is analogous to spinning top.
11 oo : 01
: 1o
2.10. Use t e expression o
orce an centri uga force to find out the magnitude and t he direction of the forces . 2.11. The sit uation is analogous t o spinning top. 2.12. Same problem of a previous problem 2.13. Use the expression of Coriolis force and centrifugal force to find out the direction of the forces. 2.14. This one is easy, you don't need hints.
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Angular momentum & Rigid body : Part-2
2.3
physicsguide CSIR NET, GATE
Solution
Sol 2.1. Foucault 's pendulum is designed such a way so t hat it can swing back and forth in any direction. As earth rotate t he plane of the pendulum rotates. From which we
11 oo : 01 2.3
: 13
Solution
Sol 2.1. Foucault 's pendulum is designed such a way so t hat it can swing back and forth in any direction. As earth rotate t he plane of t he pendulum rotates. From which we can say that earth is not an inertial frame.
Sol 2.2. Direction of coriolis force
x
is along axis at Q. Then w x Similarly centrifugal force -w x ward.
2w x v-;, where w and v-;
v-; is along y axis. (w x r)
points radially out-
Sol 2.3. The centripetal force on the person at t he pole mwz x (w z x R z)
= 0,
at equator mwz
X
(wz
X
Rf)
= -mw
2
Rf.
So, t he force felt by t he person at equator We
GMm = - R2
=
9
+ mw= R 2
-mg + mw R
at the pole Wp
=
-mg
j [email protected]
27
@Sk J ahiruddin , 2020
Angular momentum & Rigid body: Part-2
physicsguide CSIR NET, GATE
Therefore
lOO Wp
I)
I!
We
= lOO -mg+ mg - mw _
/YYl
r,
2
R
11 oo : 01
: 16 Angular momentum & Rigid body: Part-2
@Sk J ahiruddin , 2020
Therefore
lOO Wp
-
We
= lOO -mg + mg -
2
mw R
- mg
Wp
= lOO w2R g
= 100
2
21r 86400
6400 X 10 3 10
= .338 ~
.34 •• ......
Sol 2.4. The acceleration of the ovserver R = 0. The coriolis force F cor
= m2w
x iJ
= .01 · 2 · 22 X .06( -f) " = .0024. So, t oward his right. Sol 2.6. For P the coriolis force F corp
= - m 2w x iJ = - 2mwvcf> A
i.e t oward west. For Q t he coriolis force F corQ
= - m 2w x iJ = - 2m wv cf> A
i. e also toward west. Sol 2. 7. Coriolis force F co
= - 2mwz x (-vJ) 2 = - 2mw rf
j [email protected]
29
@Sk J ahiruddin , 2020
Angular momentum & Rigid body: Part-2
physicsguide CSIR NET, GATE
11 oo : 01
: 21
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@Sk J ahiruddin , 2020
Angular momentum & Rigid body: Part-2
physicsguide CSIR NET, GATE
Centrifugal force
Fe = -mwz
= mw
2
X
(w z
X
rf)
rf
Sol 2.8. Coriolis force on t he bullet ,...
Fcor = - 2mDz x vcp 2
= - 2mD rf So, to his right. Sol 2.9. Cyclone rotates clockwise in southern hemisphere due to coriolis force. Sol 2.10. Coriolis force
Fcor = -m2w x iJ ,...
= -2mwz x (-wRcp) 2 = -2mw Rf Centrifugal force
Fcjg =
-mw X
(w
~
X
R)
= -mwz X (wz X Rf ) 2 = -mw Rz X J) 2 = mw Rf [email protected]
30
physicsguide CSIR NET, GATE
11 oo : 01
: 23 2
= -mw R z 2 = mw Rf
X
J)
j [email protected]
30
@Sk J ahiruddin, 2020
Angular momentum & Rigid body: Part-2
physicsguide CSIR NET, GATE
Hence -+
2
-+
F ear + F cfg
A
= -mw R r
Sol 2 .11. Due to the pseudo force the disc will move toward -x direction at first . Since the body coor·dinate is rotating about t he fixed car coordinate, t here will be a torque acting
on the body which will help it rotate in xy plane. Sol 2.12. Same problem of 3.6, so west in both hemisphere. Sol 2.13. The expression of corriolis force 2mw
always perpendicular to bot h
xv. So it's
w and v.
Sol 2.14. The acceleration in t he fixed coordinate •
ap = aM + w x
r + 2w xv+ w x
(w x r) .
•
Since v and w are constant aM and w are zero. Hence
ap = 2woz x vof + Wo Z x (wo z x rf)
=
A
2
2wovo - w0 rf
11 oo : 01
: 26 2
= mw Rf j [email protected]
30
@Sk J a hiruddin , 2020
Angular momentum & Rigid body: Part-2
physicsguide CSIR NET, GATE
Hence -+
-+
F ear
+ F cj9 =
2
A
-mw R r
Sol 2 .11. Due to the pseudo force the disc will move toward -x direct ion at first . Since t he body coordinate is rotating about t he fixed car coordinate, there will be a torque acting
on the body which will help it rotat e in xy plane. Sol 2.12. Same problem of 3.6, so west in both hemisphere. Sol 2.13. The expression of corriolis force 2mw x iJ. So it 's
always perpendicular to both w and iJ. Sol 2.14. The acceleration in t he fixed coordinate •
ap = aM+ w x r + 2w x iJ + w x (w x r) . Since
v and w
•
are constant
ap = 2woz
j a [email protected]
X
aM and w are zero.
Vof
+ Wo Z X
31
(woz
X
Hence
rf)
physicsguide CSIR NET , GATE
11 oo :01 : 2a 2
= mw Rf j [email protected]
30
@Sk J a hiruddin , 2020
Angular momentum & Rigid body: Part-2
physicsguide CSIR NET, GATE
Hence -+
-+
F ear
+ F cj9 =
2
A
-mw R r
Sol 2 .11. Due to the pseudo force the disc will move toward -x direct ion at first . Since t he body coordinate is rotating about t he fixed car coordinate, there will be a torque acting
on the body which will help it rotat e in xy plane. Sol 2.12. Same problem of 3.6, so west in both hemisphere. Sol 2.13. The expression of corriolis force 2mw x iJ. So it 's
always perpendicular to both w and iJ. Sol 2.14. The acceleration in t he fixed coordinate •
ap = aM+ w x r + 2w x iJ + w x (w x r) . •
Since v and w are constant aM and
ap = 2woz
j a [email protected]
X
Vof
w are zero.
+ Wo Z X
31
(woz
X
Hence
rf)
physicsguide CSIR NET , GATE
11 oo : 01
: 31 2
= mw Rf j [email protected]
30
@Sk J a hiruddin , 2020
Angular momentum & Rigid body: Part-2
physicsguide CSIR NET, GATE
Hence -+
-+
F ear
+ F cj9 =
2
A
-mw R r
Sol 2 .11. Due to the pseudo force the disc will move toward -x direct ion at first . Since t he body coordinate is rotating about t he fixed car coordinate, there will be a torque acting
on the body which will help it rotat e in xy plane. Sol 2.12. Same problem of 3.6, so west in both hemisphere. Sol 2.13. The expression of corriolis force 2mw x iJ. So it 's
always perpendicular to both w and iJ. Sol 2.14. The acceleration in t he fixed coordinate •
ap = aM+ w x r + 2w x iJ + w x (w x r) . •
Since v and w are constant aM and
ap = 2woz
j a [email protected]
X
Vof
w are zero.
+ Wo Z X
31
(woz
X
Hence
rf)
physicsguide CSIR NET , GATE
11 oo : 01
: 33 2
= mw Rf j [email protected]
30
@Sk J a hiruddin , 2020
Angular momentum & Rigid body: Part-2
physicsguide CSIR NET, GATE
Hence -+
-+
F ear
+ F cj9 =
2
A
-mw R r
Sol 2 .11. Due to the pseudo force the disc will move toward -x direct ion at first . Since t he body coordinate is rotating about t he fixed car coordinate, there will be a torque acting
on the body which will help it rotat e in xy plane. Sol 2.12. Same problem of 3.6, so west in both hemisphere. Sol 2.13. The expression of corriolis force 2mw x iJ. So it 's
always perpendicular to both w and iJ. Sol 2.14. The acceleration in t he fixed coordinate •
ap = aM+ w x r + 2w x iJ + w x (w x r) . •
Since v and w are constant aM and
ap = 2woz
j a [email protected]
X
Vof
w are zero.
+ Wo Z X
31
(woz
X
Hence
rf)
physicsguide CSIR NET , GATE
11 oo : oo : 01
Advanced Problems and Solutions in Ne-wtonian Mechanics Sk J ahiruddin* Sandip Biswas
*Assistant Professor Sister Nibedita Govt. College, Kolkata Aut hor was the topper of IIT Bombay NI.Sc Physics 2009-2011 bat ch He ranked 007 in IIT JAM 2009 and 008 (JRF) in CSIR NET June 2011 He has been teaching CSIR NET aspira11ts since 2012
1
@Sk J ahiruddin, 2020
Advanced Problems in Newtonian Mechanics
11 oo : oo : 03 1
@Sk J ahiruddin, 2020
Advanced P roblems in Newtonian Mechanics
Contents 1 Advanced Problems on Newtonian Mechanics •
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
18
1.2 Hints . . .
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
19
Solutions .
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
21
1.1
1.3
Ans Keys
3
j [email protected]
2
physicsguide CSIR NET, GATE
11 oo : oo : 06
j ahir@physicsguide. in
@Sk J ahiruddin , 2020
2
physicsguide CSIR NET I GATE
Advanced Problems in Newtonian Mechanics
Problems from NET, GATE, JEST, TIFR & JAM papers
1
Advanced Problems on Newtonian Mechanics
Do these problems only if you are a serious aspirant of JEST and TIFR. CSIR NET and GATE aspirants may skip this section
Prob 1.1. Two masses M 1 and M 2 (M1 < M 2 ) are suspended from perfect ly rigid horizontal support by a system of t hree taut mass-less wires W1 , W 2 and W 3 , as shown in t he figure. All t he three wires have identical cross-sections and elastic properties and are known to be very strong. If the mass M2 is increased gradually, but wit hout limit, we should expect t he wires to break in the following order: [TIFR 2012]
11 oo : oo : 09 •
[email protected]
3
Advanced P roblems in Newtonian Mechanics
@Sk J ahiruddin , 2020
(a) first W 2 , t hen W 1 (c) first W 2 , then W 3
physicsguide CSIR NET, GATE
(b) first W 1 , t hen W 2 ( d)
first W 3
Prob 1.2. A particle slides on t he inside surface of a frict ionless cone. T he cone is fixed wit h its t ip on the ground and its axis vert ical, as shown in t he figure. The semi-vertex angle of t he cone is a. If the particle moves in a circle of radius r 0 , without slipping downwards, the angular frequency w of this motion will be [TIFR 2015]
11 oo : oo : 11 1us r 0 , wit out s 1pp1ng w of this motion will be
e angu ar
[TIFR 2015]
[email protected]
Advanced P roblems in Newtonian Mechanics
_.
(a)
(c)
g
(b)
ro tan a g •
r 0 sin a
physicsguide CSIR NET, GATE
4
@Sk J ahiruddin , 2020
equency
(d)
----
g
ro cot a g
r 0 cos a
Prob 1.3. A stone is dropped is vertically from t he top of a tower of height 40 rn. At the same t ime a gun is aimed directly at the stone from the ground at a horizontal distance 30 m from t he base of the tower and fired. If the bullet from the gun is to hit t he stone before it reaches t he ground, t he minimum velocity of the bullet must be, approximately, [TIFR 2013] /,...\ 17 '7,yy-,,., -1
11 oo : oo : 13 tance 30 m from t he base of the tower and fired. If t he bullet from the gun is to hit t he stone before it reaches t he ground, t he minimum velocity of t he bullet must be, approximately, [TIFR 2013] (a)57.4ms- 1 (b)27.7ms-1 (c) 17.7ms -1 (d)7.4ms-1 Prob 1.4. A car st arts from rest and accelerates under a force F increasing linearly in time as F = at at where j [email protected]
5
physicsguide CSIR NET, GATE
Advanced Problems in Newtonian Mechanics
@Sk J ahiruddin , 2020
a is const ant . At t ime t 1 > 0, t he force F is suddenly switched off. At a later time t 2 > t 1 , brakes are applied resulting in a force F' whose magnit ude increases with time, F' = -a(t - t 2 ) where a is t he same constant as before . which of t he following graphs would best represent the [TIFR change in the posit ion of the car x(t) wit h time? 2015]
(a)
(b)
x(t)
x(t)
(c)
(d)
x(t)
0
x(t)
I I I I I I I I I I
t,
l2
12+ 11
t
0
,,
'2
f 2 +t ,
t
11 oo : oo : 16 x(t)
x6)
Prob 1.5. A ball is dropped vertically from the height H on to a plane surface and permitted to bounce repeatedly along a vertical plane. After every bounce, its kinetic energy
becomes a quarter of its kinetic energy before the bounce. The ball will come to rest after time [TIFR 2016] [email protected]
6
@Sk J ahiruddin, 2020
(a) infinity
physicsguide CSIR NET, GATE
Advanced Problems in Newtonian Mechanics
1
1
(b) 2H 2
(c) 2 2H 2
g
g
(d)3
1 2H 2 g
Prob 1.6. In a moving car , the wheels will skid if the breaks are applied too suddenly. This is because [TIFR 2016]
(a)the inertia of the car will carry it forward . (b )the momentum of the car must be conserved. (c)the impulsive retarding force exceeds the limiting force of static friction. (d )the kinetic friction will suddenly get converted to static friction.
3
Prob 1.7. A ball of mass 0.1 kg and density 2000 kg/m is suspended by a massless string of length O.5 m under water having density 1000 kg /m3 . The ball experiences a drag
force, Fd = -0. 2(vb -
ilw), where Vb and Vw are t he veloc-
11 oo : oo : 19 3 kg/m
Prob 1.7. A ball of mass 0.1 kg and density 2000 is suspended by a massless string of length O. 5 m under water having density 1000 kg /m3 . The ball experiences a drag _, force, Fd = - 0.2 (vb - vw), where Vb and Vw are t he velocit ies of the ball and water respectively. What will be t he frequency of small oscillations for the mot ion of pendulum, [JEST 2017] if t he water is at rest ?
Prob 1.8. A small elastic ball of mass m is placed at the apex of a 45 ° inclined plane as shown in the figure below. [email protected]
@Sk J ahiruddin , 2020
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physicsguide CSIR NET, GATE
Advanced Problems in Newtonian Mechanics
[TIFR 2017]
The ball is allowed to slip wit hout friction down t he plane (along t he dotted line), hit t he ground (as shown) and bounce along it. If the height of t he inclined plane in h and the coefficient of restit ution between the ball and t he ground is 0. 5, then t he distance AB , as marked on t he figure, will be
(a) 3h
(b) 2h
(c) (1 + v2)h
(d) 3v2h
11 oo : oo : 21 h and the coefficient of restitution between the ball and the ground is 0.5, t hen the dist ance AB , as marked on t he fig-
ure, will be
(a) 3h
(b) 2h
(c) (1
+ v'2)h
(d) 3\1'2h
Pro b 1. 9. A toy car is made from a rect angular block of mass M and four disk wheels of mass m and radius r . The car is attached to a vert ical wall by a massless horizontal spring wit h spring constant k and constrained to move
perpendicular to the wall. The coefficient of static friction between the wheels of the car and the floor is µ. The maximum amplitude of oscillations of the car above which the wheel start slipping is [JEST 2017] j [email protected]
@Sk J ahiruddin, 2020
(a)µ g(M
Advanced Problems in Newtonian Mechanics
+ 2m)(M + 4m)
mk 2 µg (M + m) ( ) c 2m k
physicsguide CSIR NET, GATE
8
(b)µg (M
2
-
m
2
)
Mk (d)µg (M + 4m)(M + 6m) 2mk
Prob 1.10. Consider a point part icle A of mass mA colliding elastically with another point particle B of mass mB at
rest , where m B/mA = ry . After collision, the ratio of the kinetic energy of particle B to t he initial kinetic energy of particle A is given by [JEST 2017]
,
,
(c)
2 i ry+ 2 - ,
(d).!_ "'(
Prob 1.11. A cyclist , weighing a t otal of 80 kg with the hir,rrlP n Prl~.1~ ~-t ~- ~n PPli nf 1 n m /~
ShP ~t n n ~ n P 0). What is the distance of the closest approach d? [JEST 2019] k (a) d = b2 + mv~
(c) d = b
1.4
1 2
1
(b) d = b2 -
(d) d =
k mv02
Inverse square Law
k mv~
z
11 oo : oo : 41 (c) d = b
1.4
(d) d =
k
m v02
Inverse square Law
Prob 1.29. A space station is moving in a circular orbit around the Earth goes into a new bound orbit by firing its engine radially outwards. This orbit is [GATE 2007] (a) a larger circle (c) an ellipse
(b) a smaller circle
(d) a parabola
Prob 1.30. A satellite moves around a planet in a circular orbit at a dist ance R from its center. The time period of revolution of the sat ellite is T. If the same satellite is taken to an orbit of radius 4R around t he same planet, t he t ime [JAM 2007] period would be
(a) 8T
(b) 4T
Prob 1.31. A satellite is moving in a circula1~orbit around t he Earth. If T , V and E are its average kinetic, average [email protected]
@Sk J ahiruddin , 2020
15
physicsguide CSIR NET, GATE
Central force
potential and total energies, respectively, then which one of [GATE 2015] t he following options is correct?
(a) V
- 2T; E = - T (c) V = - T / 2; E = T / 2 =
(b) V =-T; E = O (d) V = - 3T/ 2; E = - T / 2
11 oo : oo : 44 [GATE 2015]
t he following options is correct ?
(a) V = - 2T; E = -T (c)V=-T/2; E=T/2
(b)V =-T; E=O (d) V = -3T/ 2; E = - T/ 2
Prob 1.32. Consider a classical p article subj ected to an attractive inverse-squ are force field. The tota energy of the particle is E and t he eccentricity is
follow a p arabolic orbit if (A) E > 0 and E = 1 (B) E < 0 and (C) E = 0 and
E
=
1
The p article will
E .
(D) E < 0 and
[JAM 2019] E E
0 is a constant. If a system of such particles has reached virial equilibrium, the ratio of the kinetic to the total en er gy of the system is
(a) 1/ 2
(b) 1/3
(c) 3/ 4
[NET Dec 201 7] (d) 2/ 3
Prob 1.54. A p article of mass m is p laced in a potential well U(x)
= cxn ,
where c is a positive constant and n is
an even positive integer. If t he particle is in eq uilibrium at
11 oo : 01 (a) 1/ 2
(b) 1/ 3
: 1o (c) 3/ 4
(d) 2/ 3
Prob 1.54. A part icle of mass m is placed in a potential
well U(x ) = cxn, where c is a posit ive constant and n is an even positive integer. If t he part icle is in equilibrium at constant temperature, which one of the following relations between average kinetic energy (K ) and average potential energy (U) is correct? [JEST 2020]
(A) (K)
=
2 (U) n
(C) (K ) = ; (U)
(B) (K )
=
(U)
(D) (K ) = 2(U)
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25
physicsguide CSIR NET ) GATE
Central force
@Sk J ahiruddin ) 2020
1.6
Ans Keys
1.1. A
1.12. a
1.23. d
1.2. a
1.13.
C
1.24. 4
1.14. d
1.25. d
1.15. d
1.26. b
1.4.
C
-8l 2 a 2
1.5. (a) - - r5 , m
(b) 0
1.16. (a) L 3
rt
1
= lO , 1.27.
C
E = soo· (b) 1.28. a L 2 = unchanged , 1
1
v15
E2 = , ,. .,. .
1.29.
C
11 oo : 01 1.4.
: 12
C
2 2
-8l a 5 1.5. (a) - - r , m
(b) 0
1.16.
1.7. a 1.8. b 1.9. a,c,d 1.10. a 1.11. d
1.27.
C
3
E = soo· (b) 1.28. a L 2 = unchanged , 1 400
1.29.
C
1.17. (a) 1 m
1.30.
C
1.18. d
1.31. a
E2 =
1.6. d
J5 (a) L = lO ,
1.19.
1.32.
C
C
1.20. B
1.33. b
1.21.
1.34.
C
1.35. 2
1.22. a 26
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Central force
1.43. b
1.50.
1.37. d
1.44. 121
1.51. b
1.38. b
1.45. 85
1.52. C
1.39. d
1.46.
1.53. b
1.40.
C
1.47. d
1.41.
C
1.48.
1.36.
C
1 .42. d
C
C
1.49. b
C
1.54. C
11 oo : 01
: 1s
1.40.
C
1.47. d
1.41.
C
1.48.
1.42. d
j [email protected]
1.54. C
C
1.49. b
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1.7
Central force
Hints
1.1. This is a standard basic conceptual question. Try to
prove t hat the motion is indeed restrict ed in a plane for central force. 1.2. You should be able to do it, since the trajectory is very
familiar. 1.3. Write the energy equation replacing
r by
1~-
11 oo :01
:1
a
1.2. You should be able to do it, since the trajectory is very
familiar. 1.3. Write the energy equation replacing
r by j; .
1.4. (You are given r, put it into the equation of motion to
get J(r )) 1.5. You don't need hint for this problem. 1.6. Use the concept of problem ??. Although be careful in
using the given equation as it is not correct!! 1. 7. Almost a same as the previous one.
1.8. The locus would be lissajous figure. 1.9. Same problem as the previous one. 1.10. You should be able to do it.
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1.11. Almost a same as the previous one. 1.12. You don't need hint for t his problem. 1.13. Conservation of angular momentum.
1.14. We have done this kind of problem before.
Central force
11 oo : 01
: 21
1.13. Conservation of angular momentum. 1.14. We have done this kind of problem before. 1.15. Very easy problem. 1.16. Use conservation of energy and angular moment um. 1.17. Write t he equation of centripetal force and find out r
from t here. 1.18. Write t he equation of centripetal force and find out l
from t here. 1.19. Find out w from cent ripetal force. 1.20. Find out radius from cent ripetal force. 1.21. 1.22. 1.23. Find if r 0 is stable equilibrium. [email protected]
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1.24. 1.25. You don 't need hint for t his one. 1.26.
Central force
11 oo : 01
: 23
1.25. You don 't need hint for t his one. 1.26. 1.27. Repulsive potential do not attract. 1.28. Use conservation of energy 1.29. 1.30. Use Kepler's law. 1.31. Use virial t heorem. 1.32. What is t he eccentricity and energy for a parabolic
orbit? 1.33. Use Kepler's law. 1.34. You don't need hint for t his one. 1.35. Use conservation of energy and angular moment um. 1.36. You don't need hint for t his one.
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1.37. Use conservation of energy and angular momentum . 1.38. Use conservation of energy and angular momentum .
11 oo : 01 : 2s 1.37. Use conservation of energy and angular momentum. 1.38. Use conservation of energy and angular momentum. 1.39. Use conservation of energy and angular moment um. 1.40. You don't need hint for t his one. 1.41. Use Kepler's law. 1.42. You don't need hint for t his one. 1.43. Find out radius from centripetal force. 1.44. You don't need hint for this one. 1.45. gravitational force on t he man will be due to t he imaginary sphere of radius r . 1.46. What is gravitational self energy? The energy the attractive system t ake to assemble itself .. .. ! 1.47. Use the result of Sol 1.45 .. 1.48. Use conservation of energy 1.49. [email protected]
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1.50. See when condition of stable equilibrium are met.
00 : 01 : 29
Central force
@Sk J ahiruddin, 2020
1.50. See when condition of stable equilibrium are met. 1.51. 1.52. Use Kepler's law. 1.53. Use virial t heorem. 1.54. Use virial t heorem.
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Central force
11 oo : 01
: 31
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1.8
Solutions .....
Sol 1.1. The form of a central force is F Hence the equation of the motion
diJ
= f (r )f.
...
m-=F dt
!( ) rxmdt=rx rr ...
diJ
...
A
(multiplying bot h side by rx)
d (..... .....) 0. . . dt r x mv = ..... .r. . x mv. . . = L .....
where L is angular moment um vector and is a constant. Multiplying both side by f .,
r
So we see that is always perpendicular to contant angular ..... moment um vector L , so the motion takes place in a plane because angular momentum is conserved/ constant. Sol 1.2.
Hence
mi= m [email protected]
(xx + yy) = m (-wfxx + -w~yy) 33
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11 oo : 01
: 33
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Central force
So we can see t hat only if w1 was equal to w2 we could write it as F = = -wf i.e in the form of a central force. -+
••
mr
r,
Sol 1.3. The total energy eqation,
1 . 2 2 2 -m r +r 0 + V(r) =E 2 For a central force the angular moment um is conserved,
Using chain rule
r
=
dr . d0 = d0 dt
0dr
d0
•
replacing 0 , we get, . dr L r = - • - -2 d0 mr
replacing
r in energy equation we get
L dr 2 2 mr 4 (d ) + r + V(r) = E 2 0
(you have to remember this eqt
We are given that E=O and r = r 0 exp( k0) . Hence
dr de = kro exp(k 0) = kr.
11 oo : 01
: 36
We are given that E=O and r = r 0 exp ( k0) . Hence
dr dB = kro exp( k0) = kr.
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Central force
Replacing this to the energy equation
L 2 2 - -4 [k r 2mr
or, Sol 1 .4 . Use
V (r) = d2u d02
+r ] = 2
L 2mr 2
(k
2
-V(r)
+ 1).
m + u = - z2u2 f
There are lot's of examples in the cent ral force notes. Solve t hem first.
Sol 1 .5 . (a) From fig.
u
=
r = 2acos 0 1 1 - = -sec 0 r 2a
Hence,
du l - sec 0 tan 0 d0 2a d 2u l 2 3 or, dB 2 = a [sec 0 tan 0 + sec 0] 2
11 oo : 01 or,
: 39
d2u l 2 3 d02 = a [sec 0 tan 0 + sec 0] 2
35
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Replacing ~~~ and u in t he given orbit equation 1
2
2
2
a sec 0 [t an 0 + sec 0]
1
+ 2a sec 0 = -
1 2 2 a sec 0 [tan 0 + sec 0 + 1] 2
1
2
m
l2 u 2 f
= - l~ 2 f m
a sec 0 · 2 sec 0 = l 2u 2 f 2 (putting expression of u) gz2a2
f = - mr5 (b) r
f (r' )dr'
V(r) = 00
2 2
8l a m
1
r 00
r
,5 dr'
2l 2a 2 mr 4 Putt ing t he expression of r, and ~ in the energy equa-
11
00:01 :41 mr 4
Putting the expression of r, and ~ in the en ergy equa-
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t ion,Sol 1.3.
L dr 2mr4 (d0)2 + r2 + V(r) = E L
sin 0) [ (2a 2mr 4
2
+ (2a cos 0)
2 2
2
2l a ] -
= E mr 4 2 2
L 2l a 2 - - 4(2a) - - 4 = E 2mr mr 2l 2a 2
2l 2a 2
--=E 4
mr 4
mr
:. E = 0 Sol 1.6. R educing the problem into equivalent one body problem , we can say t hat a massµ= 1; moving in a circular 2
orbit of radius 2R under the influence of forcce (~ ~ 2 • Hence to find the angular velocity we equate, 2
µw R = 1
w=-
2R
Gm2 (2R )2
Gm R
11 oo : 01
: 44 Gm2 µw R = (2R)2 2
1 w=2R
Gm R
The t ime period T
R Gm
21r
= = 41rr w
When the two masses stop orbiting and start to fall onto each other. the equation of motion gives us, [email protected]
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Central force
dv Gm 2 µv dr = - r 2 dv 2Gm v - = - -2dr r V
r
v'dv' = -2Gm
2r
0
v2
l dr' r'2
1
1
-=2Gm - - 2 r 2R 2R -r 1 2 v = (2Gm) / Rr dr = (2Gm )1; 2 2R-r dt R r
t= Now replace r by x 2 ,
R
2Gm
0
r'l/2 ----:::========== dr'
2R
J2R - r'
11 oo : 01
: 46
t =
Now replace r by x
2
0
R 2Gm
r'l / 2
--;:========= dr' 2R
J2R - r'
,
dr = 2x dx
•
• •
But we will use the formula (which we will prove in t he next problem) ,2
d X ' = -a Sln . Ja - x'2 2 x
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1
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Central force
o x2 R -----;:::===== dx = t 2Gm -/2R, J2R- x 2
R a . - Slll 2Gm 2
1
X
'1m,
,----- 0 -J2R -x 2 X
-
2
=t
v'2R R 2Gm =t T
t= 4✓2
Sol 1. 7. The equation of motion gives us dv GMm m v dr = - r2 dv GM v dr r2
11 oo : 01
: 49
dv GMm mvd = 2 r r dv GM v - = - -2dr r V
v'dv' = -Gm
/2 r 2R
0
v2
l dr'
r
1
1
- = Gm - - 2 r 2R v = (Gm) 1; 2
2R - r
R
r 2R-r r 0
t=
Gm
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r'l/ 2 -----;::==== dr'
2R ✓2R-
39
r'
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Central force
Now replace r by x
2
,
• • •
2
dr
= 2xdx
o x2 R ----;::::========== dx = t Gm -/2R, ✓2R-x 2
Let
= vl2R,sin0 :. dx = ✓2R, cos 0d0 x
Substit uting t=2
R Gm
2
2R sin 0\/2Rcos 0 dB
\/2Rcos0
11 oo : 01 : s1 Substituting
t = 2 = 2
2
R Gm
2R sin 0-/2R cos 0 dB -/2Rcos0
RR Gm
(1 - cos 20) d0
v'R
=2
R R Gm
=2
R3 Gm
R3
sin- 1
1r
4+
X
-/2R 1 2
1r
Gm 2+l
Sol 1.8. The coordinate of the particle at any time t can 40
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be written as
= a cos(wt) y = bsin(wt + c5)(where c5 is the phase different between
x
Therefore,
t = cos(wt) cos 8 y
X
b
a
- = -cosc5 From there we get
sin(wt) sin 8 x2
1 -a2
x and 1,
11 oo : 01 : s4 b
-
y
X
b
a
x2
- = -cosb -
l -a2
From there we get
y - +a b X
-
2xy ab
COS ()
.
= Sln
2 ()
Which is equation of an ellipse centered at t he origin. Sol 1 .9 . (a)we prooved it in the previous problem. Only in .
t his case 8 =
7r
2
(b)
v=
-wA sin(wt)i + wB cos(wt)J
So speed of the particle is not constant. (c) Force
F = ma= -w
2
A cos(wt)i + B sin(wt)J
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Central force
So the force is toward the origin. (d)Angular moment um _, _, L =r x mv _,
= mw A cos(wt)i + B sin(wt)J x - w A sin(wt)i + wB cos(wt)J
= mw AB cos (wt)k + AB sin (wt)k 2
A
=mwABk
2
11 oo :01 : s1 - w A sin(wt)i + wB cos(wt)3
= m w AB cos2 (wt )k + AB sin 2 (wt )k A
=mwABk
Sol 1.10. Cent ripetal force acting on the first particle
Similarly for the second particle 2
4
L1 = mr2
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Therefore L1
2
r1
L1
4
r2
22
=
r1 r2
V2 =
r1 r2
4
11 oo : 01 : sg r2 4
r1
22 =
r2
~=
r1 r2
Sol 1.11. If we take corn as t he origin, then mr 1f
+ 2mr2 f = 0
3m r1 + 2mr2 And we are given that r 1
-
r2
= r Then
r2
=
r1
=0
2r 3
r
= --
3 Since same force is acting on the part icle we can write
mvr
2mv~
43
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Sol 1.12. Equating the cent ripetal force 2
mw R w2
= 4kR 3 V"\
R2
11 oo : 02 : 02 Sol 1.12. Equating the centripetal force 2
mw R
w2
= 4kR V"\
3
R2
Sol 1.13. Conservation of angular momentum gives us
0 = constant k0 80 = constant r
2
08 d0 = constant 1 9 -0 t 9 0 -1i
dt
V"\
V"\
Sol 1.14. Writing the total energy equation
1 2
.2
mr
!mr 2
2
+ U(r ) = E = E - U(r)
Since t he K.E > 0, the particle is bounded by the condition E > U(r) . So, region I and III are forbidden for the particle.
44
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Sol 1.15.
V (r) = -
r
.
()()
k - -dr r3
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Central force
Sol 1.15. r
V (r) = -
00
k - - dr r3
k
2r 2 Sol 1.16. T he force acting on t he body
-F =-V-V 3 (- ; )
-exp = V---
5r2
-
1
1
5r
5
3r
= v e- 2
-
2 + -e2
1
3r -
V
3 e - ?f
2 e- ?f
10 r
5 r
r2
" - ---r " = -----r 2 3
(a) T he equation of motion 3r
or,
e- 2
?
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@Sk J ahiruddin , 2020
=
32r
- 3r -
L- = m (3r + 4)
or,
putting r
2e _
3 -mw R = 10 r 2 5 r3 L2 3 e- ?f 2 e- ?f mr 2 10 r 2 + 5 r 3 2
2 and m
e
2
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Central force
=
1 we get,
11 oo :02 : oa Central force
@Sk J ahiruddin , 2020
putting r
=
2 and m
=
1 we get, -3
£ 2 = (6 + 4) elO
L=
or,
l
Jw
J5
or,
10
The total energy equation
E=
1
£ 2
2 mr 2 1
-
.
+ V (r) e- 3
1
2 20 · 4 1 1
160
5·4
400
3 800
(b )Angular momentum of the body which falls toward the center , -+
-+
L 1 = r 1 x mv1
= r1f1 =0
x
mv(-r1)
Using conservation of angular momentum before and afj [email protected]
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ter parting away
~ lO = L1
+ L2
~ - L2 10 -
Let T1 and E 2 be the K.E of the 1st and t otal energy of the 2nd body respectively. Then conservation of energy gives us •
E = T 1 + E2 +
1 V (r) 2
3 1 2 800 = 2 . .5 x . l + E 2 1 400 = E 2
e- 2 -4-0
-
Sol 1.17. (a) The cent ripetal force on t he part icle mv 2 r
=
3 r2
£2
+
1 r
- = 3r + r m 2 r + 3r - 4 = 0
or, or,
2
r = l
or, •
(b) For a circular iJ = r0. So, t he t angential equation of mot ion r0 + 2r0 = Ar 0 ••
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11 oo : 02 : 12 motion ••
r0
•
•
+ 2r0 = >..r0
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multiplying both sides by mr
m(r
2
or,
0 + 2rr0) = >..mr
0 d . . 2 2 dt (mr 0) = >..mr 0 dL = >..L dt dL = >..dt L L ln = >..t Lo
or, or, or, or, or, Sol 1.18. Force
F
= _
av
ar
= -~
r
Then from equation of motion
L2
k
mr3 r~ ~l = mkro Sol 1.19. Force F = _ 8V = _ nk
8r
2
rn+l
11 oo : 02 : 1 s Sol 1.19. Force
F
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= _
8V 8r
nk rn+I
= _
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Central force
Then from equation of motion
1
2
nk m
1 R ~+I 1 2
R~+l
Similarly -1 2
t herefore
Sol 1.20. Equation of motion
(2R) ;+1
11 oo :02 : 1a
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Central force
L2
av
mrg L2
ar
3
mr0
L2
-
mr03
e - oro
= - k a--+ r 0
=k
e-oro 2
r0
L2
-- = mr0
e - or0
ke - oro
2
r0
(1 + ar0 ) (1 + aro)
Sol 1.21. This is a inverse square force where the effective potential looks like t he the figure Sol 1.21. below.
00: 02: 20
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Central force
I
' ,2
'
.., \ 2mr" \ \
V'
',
\
V'
,- -
--
\.. /~
I
/
I _
k
I V- - , I
I
Figure 1.1: Effective potential for inverse square force
If we ignore the effect of atmosphere and the energy loss due to friction, as t he initial energy is positive for the approaching meteor there is no capture like the previous problem ( ~ potential ) . But still the meteor can hit the earth,
11 oo : 02 : 23 If we ignore t he effect of atmosphere and t he energy loss due to friction, as t he initial energy is positive for the approaching met eor t here is no capt ure like t he previous problem ( -fs potential ) . But still the meteor can hit the earth, rather touch t he earth in a hyperbolic orbit . When it is at a large distance, it has velocity v00 and impact paramet er b. So if we take the mass of t he met eor 1 as m , t hen initial energy to be mv~ . 2 The angular momentum L is mbv00 which remains [email protected]
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Central force
stant in due course of motion. The distance b between the initial t rajectory and line aa' is called t he impact paramet er of the t rajectory. The largest value of b for which t he trajectory hits the planet is indicated by b' in the figure.
b' --
a
----
a
Figure 1.2: capture cross section for inverse square force
So if t he particle touches the earth (in case of b' impact oaramet er ) t hen t he kinetic enere:v when touched is ZERO.
11 oo : 02 : 2s Figure 1.2: capture cross section for inverse square force
So if t he particle touches t he earth (in case of b' impact parameter ) t hen t he kinetic energy when touched is ZERO . So the energy when t he particle touches the eart h is
GmM Re
+
L2 2mR~
This equals to the total energy when the particle was at long distance.
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Central force
Substitute L = mv00 b and get 1
b'
2GM
= R e (1 + R 2 voo
)
2
Now use escape velocity to be ~w you get
For the particle to be captured this b' is the maximum value of b. Hence
11 oo : 02 : 2a of b. Hence
'Veff
r)
-----------£ - - -£ -+---+-----+--►
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@Sk J ahiruddin , 2020
Sol 1.22. See figure above. The effective potential is
z2
Ve11(r) = 2 2 mr
k 3r 3
Take derivative make it to zero. Check double derivative is negative. v max e ff -
z6
6m3k2
(b) If the energy E of t he particle is less than ½,jJx, then t he particle will reach a minimum value of r , and t hen head back out to infinity (see fig). If E is greater than ½,jJx, t hen the particle will head all t he way in to r = 0, and will never return. The condition for capture is t herefore Veffx < r
I
•
1
l
1 •
I
r
11 oo : 02 : 30 t he part icle will reach a minimum value of r , and t hen head back out to infinity (see fig). If E is greater than ½7Jx, t hen the particle will head all t he way in to r = 0, and will never return. The condition for capt ure is t herefore Veffx < E. Now for a particle at very large distance from cent er of attraction l = mub and E = becomes
½mu . 2
So the condition
Extra: The cross section for capt ure is
(J"
= 1rb~ax = 1r
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Sol 1.23. For stable orbit In this case
82V:
areft > 0.
82½11 - 3L2 12k r5 8r 2 mr6 0 3L l0 12kL 10 (3km) 4m (3km) 5 L lO which is less t han zero, i.e r 0 is unstable equilibrium point. So the particle will never move toward r 0 .
11 oo : 02 : 34 L IO
(8 1k 4 m 5
which is less t han zero, i.e r 0 is unstable equilibrium point. So the particle will never move toward r 0 . Sol 1.24. Sol 1.25. If N is the flux of the incoming particle then the total number of scattered part icle.
Ntotal =
Nda
N(a
2
+ b cos 2
2
0)dw
b2
=41rN a2+-
3
. da Sol 1. 26. The formula you know 1s for dD,, here you are dN . . da asked dB which is proportional to dB . Read the Rut herford [email protected]
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Central force
scattering carefully once again. Sol 1.27. For a repulsive central potent ial closed orbit can
not form . So (b) and (d) are wrong. In case of (a) the part icle moved toward the origin, which will not happen for a repulsive cent ral potent ial. In case of (c) t he particle moved away from t he origin , which will happen for a repulsive cent ral potential.
11 oo : 02 : 3s t icle moved toward t he origin, which will not happen for a repulsive central potent ial. In case of (c) t he particle moved away from the origin, which will happen for a repulsive cent ral potential. Sol 1.28. The angular momentum of the proj ectile L mv0 b. For distance of closest approach r = 0. The central potent ial V (r) = ~
The conservation of energy gives us 1
2
2mvo =
L2 2md2
k
+d
0 = !mv2d2 - kdmv5b2 2 2 k b2 + k2 d =-+ 2 mv 0 mv5
°
Sol 1.29. Angular momentum remains same and energy is decreased. Obviously it will b e a ellipse.
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Sol 1.30. From Kepler 's law 2
T ex R 2
3 3
T = kR T2 k = R3 If R becomes 4R, then
11 oo : 02 : 39 2
3
T = kR T2 k= -R 3 If R becomes 4R, t hen 2
T' = k(4R)
3
T ' = ST Sol 1.31. From virial theorem
n+l T = - -V. 2 In case of gravit at ional potent ial n = - 2. Then, -
- 1T = V 2 V = -2T -
-
And E = T
-
+V
=
-
-T
Sol 1.32. For attractive inverse square force field -~ ec-
centricity is given by (you have to remember it). E j [email protected]
1+ mK2 57
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For a parabola
=
2EL2
E
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Central force
= 1. Then t he energy E becomes O.
Sol 1.33. The major axis of t he second sat ellite orbit is SR
t hen semi major axis is 4R. Then t he problem becomes same as Sol 1.30. , and t he t ime period of the second orbit
11
00:02:41
For a parabola
E
= 1. Then t he energy E b ecomes O.
Sol 1.33. The m ajor axis of the second satellite orbit is 8R t hen semi major axis is 4R. Then the problem b ecomes sam e as Sol 1.30. , and t he t ime period of the second orbit is 8days. Sol 1.34. The gravitational force at height h
99h
2
GM
99 GM
(R + h) 2
100 R 2
+ 2 · 99hR -
R
2
h
=0 2
= - 2 · 99 + J (2 · 99) + 4 · 99 R 2 · 99 - 2 · 99 + 2 · 99
h=
h=
2 · 99 100 _
99
R 1
1 -1 h= (1 - 100 ) T
-
1 R
1 h ~ 200R
h = 32km [email protected]
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Sol 1.35. Conservation of angular momentum gives us
11 oo : 02 : 44 Central force
@Sk J ahiruddin, 2020
Sol 1.35. Conservation of angular momentum gives us mvar a = mvprP rp ra Vp
ra rp
=8
ra
The cent ripetal force on t he eart h
T he conservation of energy gives us 1
2
Gm M s
-mv - - - 2 a ra
=
l
Gm Ms
2
-mv - - - 2 P rp
Sol 1.36. T he escape velocity on a planet of radius R and [email protected]
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@Sk J ahiruddin , 2020
acceleration due to gravity _q is
Vescave
= ,J2iiR. Hence
11 oo : 02 : 46
@Sk J ahiruddin, 2020
Central force
acceleration due to gravity g is
V escape
= v12iR,. Hence
V escapeearth
9 earth R earth
V escapeniars
9 mars R mars
Sol 1.37. The aphelion distance of t he probe is RE and
t he perihelion distance is RM. The conservation of energy at t hese two distances gives us
£2
GmM8
2mR1 L2 1 2m R2E
RE
E =--
1 - 2-
RM L2
£2 2m Ri
= GmMs
1 RE
RM
2
= 2Gm M sRERM RM+ RE
j [email protected]
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11 oo : 02 : 49 [email protected]
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Central force
The conservation of energy at ant distance r gives us 2
2Gm MsRERM GmMs ------ - --2m(R M + RE) RE
=
1 2 GmMs - mv - --2 r 1 2 -v 2
2GMsRERJ\IJ GMs GMs ----- - -- + -- = 2(R1V1 + RE) RE r 2GMsRM + RE-r =v2 (RM+ RE)r
Sol 1.38. For attractive inverse square force field - ~ eccentricity is given by (you have to remember it). E=
2EL 2 1+ -mK2
Conservation of angular moment um reads
If v 0 is the initial velocity t hen
L 0 = mv0 R 0
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11 oo : 02 : s1
61
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Central force
Total initial energy
Gm M - --- = 0 Ro 2GM 2 or, v 0 = Ro
1 Ea = -mv02 2
So,
2 GM Ro Lo= m Ro = mJ2GMRo Now, L1 = mv1R 1 = m J2GMRo VJ=
1
2KRo
R1
m
Total final energy
1 GmM 2 E1 = -mv1 - - - - = 0 2 R1 For circular orbit
E
= 0. So, from the first equation mK 2 E1 = - 2L2
11 oo : 02 : s4 For circular orbit
= 0. So, from the first equation
E
E1 =
-
mK2 2£2
62
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Central force
So, 1 2 Gm M - mv1 2 R1 2 2K K 2 V --f - m R _r £ 2
mK
2
2£2
Putting t he expression of v I and L as we have got previously 1 2KRo
R}
m
1 2KR0
R}
m 2R0 R}
2K K2 m R 1 2GMRo K 2 1 m R1 2Ro 2 1 R 1 2Ro
4R6 = 4R oR 1 - R 1 R 1 = 2Ro Sol 1.39. Let the equation of mot ion of the planet around t he st ar is 1
- = B + A cos0 r
where B =- GMm £ 2
11 oo : 02 : s6 1 r
= B + Acos0
where
B =-GMm
£2 B2 2mE + £2
A=
and, j [email protected]
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Cent ral force
We can write it as l
r=----
l l
wher e,
l= B A E=B
and, For a parabola
E
= 1. So, l r= - - - l - cos 0.
at pericenter distance 0
For a parabola
E
= 1r. Hence
= 0. So, re=
Therefore
l.
ECOS
0
L2 GMm
11 oo : 02 : sg re = l. Therefore
Sol 1.40. Weight GMm
mg= [email protected]
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Central force
@Sk J ahiruddin, 2020
if R
➔
2R , t hen
l GMm ---➔---R2 4 R2 GMm
w
w
➔ 4
decreases by factor of 4.
Sol 1.41. From Kepler 's law Tmax - Tmin
E=-----
+
Tmax Tmin Tmax - Tmin
2a Tm ax - Tmin
2T ~ And,
+1
Tmax
1
Tmin
- -Tmin
= Tmax
E
E-
E+ l
E- 1 For B this equation reads
11 oo : 03 : 01 €
+1
Tmax
1
T min
€ -
c+ l r rrii n E- 1 For B t his equation reads E+ l E - 1R
= r m ax
= r B ma,x
And for A E- 1 1 - --
2
3
2TA E=TAmax
c+ l
2
Tjc(E + 1) = j [email protected]
TAmax
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Central force
@Sk J ahiruddin, 2020
The m aximum possible distance between t he planets A and B
-2
Tj ( 2 TBmax 2TB
-
R)
+
R
l
E-
2
Tj (E + 1 R _ R ) + R 2 2T3 E - 1 E - 1 B
2
c+ l
E-1
T3
1 +1 T3 B
Sol 1.42. We are given QPn1111fnr
=
1 ~ O r 1nlP
00: 03: 04 E-1
1 + 1 T -3 B
Sol 1.42. We are given 9 eq·uator
or,
=
1 9pole 2
GMm
2 Vequator
GMm
R2
R2
2R2
GMm
2
or,
R 2
= Vequator
GMm - R
or,
2
-
2 v equator
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Then 9 pole
=
GMm
2v;quator
R2
R
So, t he escape velocity at pole
4v;quator Vesc
=
2 Vequator
Sol 1.43. The frequency of the planet relative to earth year •
1S
0 w=-
t
4.2
X
lQ - 3
11 oo : 03 : 01 •
1S
0 w=-
t
4.2 X 10- 3 l year The centripet al force on 2 m ewe R e
GM m e
=
R2 e
R3 = GM w2
e
e
67
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Central force
The centripet al force on the planet
rnw2 R = G Mm R2 R3 = W e R3 w2
e
R = 121Re
Sol 1.44. Weight at equator ,
GMm
W equator = 2
R2
-
mw
2
R =
0
GM
w =--
R3
putt ing R ~ 6370 ~ 6400km and g
lOms- 2 , the time
11 oo : 03 : 09 2
W'equator= - R2- m w R = O GM 2 w =-R3 putting R ~ 6370 ~ 6400km and g period of the planet
21r
GM
T 21r T
R3
lOms- 2 , the time
g
-
R
T = 21r
R g
= 5026s ~
84min
Sol 1.45. If r is t he distance of the p erson at any time from t he center of earth. Then the gravitational force on t he man j [email protected]
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will be due to the imaginary sphere of radius r. if p is the mass density of earth t hen t he mass of the sphere of radius r
or, or, Equation of motion
11 oo : 03 : 12 - - -1rr 11rR 33 3 r3 = MR3
or,
Equation of motion r3 m
mx = - GM R3 r 2 sin 0 .. X
=-
Hence t he frequency w =
GM R3 X
~If .and t he t ime period
T = 21r w
= 21r
R3 GM
Sol 1.46. Imagine a mass dm is brought in from infinity,
where its potent ial energy is zero, to t he surface of a part ially const ructed star, where its potent ial energy is - Gm (dm )/r. [email protected]
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The star 's gravitational potential self energy is t he sum (integral) of all such mass t ransfers in building up t he star. T he total change in Uz is t he integral of dU8 over all shells dm : rnd m = - G m pdV Ua = - G r r R m (r) 2 =-G - - p(r) 41rr dr o r Here, p(r) is the density of t he sun at a dist ance r from its
11 oo : 03 : 1 s rnd m = - G mpdV Ua = - G r r R m(r) =-G --p(r)41rr 2 dr o r Here, p(r) is the density of t he sun at a distance r from its center , m(r) is the mass inside a sphere of radius r, 41rr 2 dr is the volume of the spherical shell of mass dm , and R is the radius of t he star. Doing the integral requires knowing the functions m (r) and p(r) . We can calculate m (r) if we know p(r) because r
m(r) =
2
dm =
41rr' p (r') dr' 0
If we imagine t hat the sun is layered like an onion, each 2
layer has a volume dV = 41rr' dr' and a mass dm = p (r') dV = 41rr' 2 p (r') dr' . The mass m( r) is t he sum of dm for all layers up to t he radius r . Now, do the rest of the integrals yourself.
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Central force
Sol 1.47. From problem Sol 1.45. we get the t ime p eriod
T = 21r Now
R3 GM
11 oo : 03 : 11 T = 21r
GM
Now 4
M = 31rR3p
R
3
or,
3 41rp
M
So,
T = 21r
1 3 G41rp
31r pG
Sol 1.48. As t he energy is conserved
E =
~mv2-
Gm M
2 1 2GM =-m-2 R
R
GmM
R
=0 Therefore
E
= 1, hence t he orbit is a parabola.
Sol 1.49. Solution: We know K.E + P.E = E nergy, So . k we get from t he potential t hat K.E - - = E. Hence K.E =
r
k E + -. r j [email protected]
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Central force
k Now when K.E will be m aximum? When - will be maxr imum, i.e when r will be minimum. Now we a11~eady know (you can just derive from the given solution of r( 0) ) t hat t he m aximum and minimum r for el-
11 oo : 03 : 20 Now when K.E will be m aximum? When k will be m axr imum, i.e when r will be minimum. Now we already know (you can just derive from the given solut ion of r( 0)) that the m aximum and minimum r for ellipt ic orbit is
Tmax =
a( l
Just put t he value of k
+ e)
and
r min ·
Tmin =
a(l - e) .
So we get
K.Emax
=
E
+
a( l - e)
k
Now as E =
we get - = - 2E. 2a a
So K.Emax
=
k
E -
2E -1- e
K .E max
=
e+ l
E --
e- l
Sol 1.50. If
V (r) = -k - /3
r3
r
t hen t he effective potent ial
k /3 Vef f = 2mr2 - r - r3
L2
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Central force
hence, 8Veff
L
2 -i..
k
-i..
3/3 = n
11 oo : 03 : 22 Central force
@Sk J ahiruddin, 2020
hence, 2
8 Veff
_L_
ar
mr 3
kr
+ _k + _3/1 = O
£2
2
r2
r4
- r + 3/1 = 0
-
m
£2
-m +
4
L m2
- 4· k3ffJ-
·2
(d) m(~
·2
+ rJ)(~ + ~)
Prob 1.26. The parabolic coordinates (~, rJ) are related to t he Cartesian coordinates (x, y) by x = ~rJ and y = ~ ( ~ 2 rJ 2 ). The Lagrangian of a two-dimensional simple harmonic [NET oscillator of mass m and angular frequency w is
Dec 2016]
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Lagrangian iVIechanics
11 oo : oo : 39 @Sk J a hiruddin , 2020
(a) (b) (c) (d)
Lagra ngia n Mechanics
tm[~2 + iJ2 - w2(~2 + 772)] ½m(~2 + 77 2) [~2 + 7]2 _ ¼w2(~2
+ 77 2)]
~m(~2 + 772)[~2 + 7]2 - ~w2~77] ½m(~2 + 772) [~2 + 7]2 - t w2]
Prob 1.27. The dynamics of a particle governed by t he La•
grang1an 1
·2
1
2
.
L = - mx - - kx - kxxt 2 2 [NET Dec 2016]
describes
(a) an undamped simple simple harmonic oscillator (b) a dumped harmonic oscillator with a time varying damping factor (c) an undamped harmonic oscillator with a t ime dependent frequency (d) a free part icle
Prob 1.28. A double pendulum consists of two equal masses m susp ended by two strings of length l. What is the Lagrangian of t his system for oscillations in plane? Assume t he angles 01 , 02 made by t he two strings are small (you can use cos 0 = 1 (given , wo
0 ;)
[JEST 2014]
= Jg/l)
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Lag1·angia n l\/Iechanics
11 oo : oo : 42 [email protected]
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Lagrangian Mecl1anics
@Sk J a hiruddin , 2020
(a) L
r--.J r--.J
m l2(02 + 1.02 _ 1w202) 1 22 _ w202 01 2 02
(b) L ~ ml
(0f + ~0~ + 0102 - w50i - ~w50~) 2 (c) L ~ m l (0i + ~ 0~ - 0102 - w50r - ~w50~) 2 (d) L ~ ml (0t + -;0~+ 0102 - w50r - w50~) 2
Prob 1.29. A classical particle with total energy E moves 2 3 under the influence of a potential V (x, y) = 3x + 2x y + 2 2xy + y 3 . The average potential energy, calculated over a long time is equal to [JEST 2015]
Prob 1.30. The Conservation Principles for energy, linear momentum and angular momentum arise from t he necessity t hat [TIFR 2014] (a) the laws of physics should not involve infinite quantit ies. (b) internal forces on a body should cancel out, by Newton's (third) law of action and reaction. (c) physical measurements should be independent of the origin and orientation of the coordinate system. (c) t he laws of physics should be independent of the state of rest or motion of the observer. Prob 1.31. If t he Lagrangian Lo j [email protected]
16
= ~m(!~)
2
-
2 2
~mw q is
physicsguide CSIR NET , GATE
11 oo : oo : 44 2 2 2 Prob 1.31. If t he Lagrangian Lo = tm(!j) - ~mw q is j [email protected]
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Lagrangian JVIechanics
modified to L = Lo + aq( !i), which one of t he following is TRUE? [GATE 2017] (a) Both t he canonical momentum and equation of motion do not change (b) Canonical momentum changes, equation of motion does not change (c) Canonical momentum does not change, equation of motion changes (d) Both the canonical momentum and equation of motion change Prob 1.32. A rob of mass m and length l is suspended from two massless vertical springs with a spring constants k1 and K 2 . What is the Lagrangian for the system, if x 1 and x 2 be t he displacements from equilibrium posit ion of the two ends of the rod? [JEST 2017]
(a) ?f (xi + 2x1x2 + X§) - ½k1Xi - ½k2X§ (b) ~(xi + x1x2 + x~) - ! (k1 + k2)(xr + x~) (c) W (xr + X1X2 + X§) - ½k1Xi - ½k2X§ (d) 7(xi - 2x1x2 + x~) - !(k1 - k2)(xi + x~) Prob 1.33. A possible Lagrangian for a free particle is [JEST 2017] 2 2 2 (a) L = q - q (b) L = q - qq
11 oo:oo:47 Prob 1.33. A possible Lagrangian for a free part icle is [JEST 2017] (a) L = q2 - q2 (b) L = q2 - qq [email protected]
17
@Sk J ahirudd in, 2020
2
(c) L = q
-
q
physicsguide CSIR NET, GATE
Lagrangian iVIechanics
(d) L = q2 - .!q
Prob 1.34. A particle of mass m moving in one dimension x is subject ed to t he Lagrangian L = 12 m(x - Ax) 2 where A is a real constant. If it starts at the origin at t = 0, its motion corresponding to t he equation (a is constant) [TIFR 2018] (a) x = a exp At (b) x a sin At (c) x (d) x
a( l - exp(- At)) a sin hAt
Prob 1.35. A block of mass M is moving on a friction-less inclined surface of a wedge of mass m under the influence of gravity. The wedge is lying on a rigid frictionless horizontal surface. The configuration can be described using t he radius vectors ri and r2 shown in the figure. How may [JEST constrains are present and what are t he typ es? 2018]
11 oo : oo : so t he radius vectors ri and r2 shown in t he figure. How may [JEST constrains are present and what are the types? 2018]
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Lag1·angian l\/Iechanics
@Sk Jahiruddin, 2020
(a) One constrain: holonomic and scleronomous (b) Two constrains: Bot h are holonomic; one is scleronomous and rheonomous (c) Two constrains: Bot h are scleronomous: one is holonomic and t he other is non-holonomic (d) T wo constrains: Both are holomic and scleronomous Prob 1 .36 . Consider t he Lagrangian
L=
~. q2 l - ✓1 - q2 - 2
of a particle executing oscillations whose amplitude is A. If p denotes t he momentum of the particle, t hen 4p2 is [JEST ?01
~l
11 oo : oo : s2 L = l - ✓1 - q2
2 -
q
2
of a particle executing oscillations whose amplitude is A. If p denotes the momentum of the particle, then 4p 2 is [JEST
2018]
(a) (A2 - q2)(4 + A2 - q2) (c) (A2 - q2)(4 + A2 + q2 ) j ahir@physicsguide. in
(b) (A 2 + q2 ) (4 + A 2 - q2 ) (d) (A2 + q2)(4 + A2 + q2) 19
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Lag1·angian Mechanics
Prob 1.37. The motion of a p article in one dimension is 2 2 described by the Lagrangian L = ½ ( in suitable - x
¥t)
units. The value of the action along the classical path from
x = 0 at t = 0 to x = x 0 at t = t 0 , is
[NET Dec 2018] 2
Xo
2 cos 2 to
Prob 1.38. A p endulum is created by hanging a heavy bob of mass m from a rigid support (taken as zero level of potent ial! symmetrically using two massless inextensible strings, each of length 1!, making an equilateral triangle as shown in the figure below.
[TIFR 2019]
11 oo : oo : s6
A correct Lagrangian for the angular oscillations of the bob in t he plane perpendicular to t he paper is (a) L = ~mf
0
2 2
2 2
(b) L = v0,mR 0
-
v0,mgRsin ~ 2
+ 4mgf sin
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2
~
20
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@Sk J ahiruddin , 2020
2
(c) L = !m0 (d) L = ~mf
-
Lagrangian Nlechanics
)3~- sin ~ 9
2
0 + mgf 0 sec 0 + v0,mgRsin
2 2
2
2
~
Prob 1.39. Which of the following terms, when added to t he Lagrangian L(x, y, x, y) of a system with two degrees of
freedom , will not change the equations of motion? Dec 2019] (A) XX - yy (B) xy - yx (C) xy - yx (D) yx 2 + xy 2
[NET
Technically t he qs is wrong. If you use gauge invariance of Lagrangian, L' = L + A C,
a\
11 oo : 02 : 1o
Sol 1.38. The radius of oscillation
.J,f'z. Lagrangian
1 ·2 v'3 L = I 0 - mg l ( 1 - cos 0) 2 2 1 32 2 . 2 2 0 = -m-l 0 - v'3mgl sin 2 4 2
= [email protected]
@Sk J ahirudd in, 2020
3
2 ·2
- ml 0 8
-
V
47
~
3mgl sin
2
0
-
2
physicsguide CSIR NET, GATE
Lagrangian iVIechanics
Sol 1.39. Technically the qs is wrong. If you use gauge invariance of Lagrangian , L' = L + df:, you get option B , but L can't b e function of acceleration.
00 : 02 : 13 3mgl sin j [email protected]
47
@Sk J ahiruddin , 2020
2
physicsguide CSIR NET, GATE
Lagrangian Mecl1anics
Sol 1.39. Technically the qs is wrong. If you use gauge invariance of Lagrangian, L' = L + O
= oo
for x < 0
Which of the following schematic diagrams best represents t he phase space plot of the ball? [GATE 2019]
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Hamiltonian Mechanics & Phase Space
@Sk J ahiruddin , 2020
+.J21nE
physicsguide CSIR NET, GATE
(a)
+.J2mE r---_ E rng
r--------it--"-- X
- ../2mE
- ../2mE ~
00 : 01 : 13 (a) E
E mg
mg
r-----~- x
X
-✓2mE
- ✓2111£ ~
(c)
+,/2m£
(d) E mg
!-----3==---_:::.._- x
E
E mg
mg
-+----1-----+-...;a..._- x
- ,/2m£
- ,/2m£
j [email protected]
27
@ Sk J ahiruddin, 2020
1.3 1.1. b
physicsguide CSIR NET, GATE
Ham iltonian :Nlechanics & Phase Space
Ans Keys 1.12. a
1.23. a
1.34.
C
11 oo : 01
: 1s Hamiltonian :Nlechanics & Phase Space
@Sk J ahiruddin, 2020
1.3
Ans Keys
1.1. b
1.12. a
1.23. a
1.34.
1.2. b
1.13. d
1.24. a
1.35. d
1.3. b
1.14. b
1.25.
C
1.36. a
1.4. b
1.15. a
1.26. b
1.37. d
1.5. a
1.16.
1.27.
C
1.38. b
1.6. b
1.17. a
1.28. a
1.39. a
1. 7.
C
1.18. a
1.29. a
1.40.
1.8. a
1.19. a
1.30. a
1.41. b
1.9. d
1.20. b
1.31.
C
1.42. a
1.10. a
1.21. a
1.32. a
1.43. b
1.11. d
1.22. b
1.33. b
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1.4
Solutions
C
28
C
C
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Hamiltonian :Nlechanics & Phase Space
11 oo :01
:1
a Hamiltonian Mechanics & P hase Space
@Sk J ahiruddin, 2020
1.4
Solutions
Sol 1.1. The electric field .....
E =-V cp- 8A 8t and
.....
.....
.....
B =V x A The Lorentz force
can be derived from a velocity dependent potential .....
U = qcp - qA · v. So the Lagrangian
L=
1 2 .......... -mv - qcp + qA · v . 2
conjugate momentum
. . . 8L ... p = oil = miJ + qA .....
=p+qA
j [email protected]
@Sk J ahiruddin , 2020
29
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Hamiltonian Mechanics & Phase Space
11 oo : 01
: 21
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Hamiltonian Mechanics & Phase Space
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Therefore the Hamiltonian _,
H = P· v- L _,
_,
P -.... l _, -- 2 A .... -= - · (P - qA) - - (P - qA) + q..qq .
q=
p m-
>..q
The Hamiltonian
H =pq-L p2 1 p2 ).. p2 =-----m----+ -q--2 m - >..q 2 (m - >..q) 2 m - >..q p2 2(m - >..q) Sol 1.15. Volume of the phase space I'
11 oo : 01
: 39
q
mp2
2(m - >..q)
Sol 1.15. Volume of the phase space
V
If E 0 is the m aximum available energy t hen p
= ✓2m(Eo - a q4 ).
Hence
✓2m(E0
V =
aq4 )dq
-
V
aq4 l - 2mE )dq
2mEo
0
V
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Hamiltonian Mechanics & P hase Space
@Sk J ahiruddin , 2020
1
If we take
a 2mEo
4
q
= t , then --1 4
dq =
dt
2mEo
Hence t he volume of the phase space
V
1
1
= (2mE0 ) 2 (2mE 0 ) 4
V
Sol 1.16. T he conjugate moment um Px
= mx+miJ
11 oo : 01
: 42
Sol 1.16. The conjugate momentum Px
= mx+miJ •
Py =mx
The Hamiltonian
H = PxX + PyY -
=
Py Pxm
Py ( ) + x- Px - Py m
p2 -m Y2 2 m 1
p
+ mY (Px -
P) Y
2
PxPy l P y ----
2m
m
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37
Hamiltonian Mechanics & Phase Space
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Sol 1.17. From Hamilton 's equation
.q =oH op
.p= - -oH oq
CJ. = 2pq2
= -2p2q
Using separation of variables •
~=-P=k or,
q
q
•
•
p
q
- = -- =
k
(say)
11 oo : 01
: 44
Using separation of variables •
p -
•
q p p
--
p
- -
q q
•
or,
-k (say)
•
--
- - -k -
q
Solving p
= Poekt
and q = qoe-kt_ But
p = -2p5e2ktqoe-kt Pokekt = - 2p5e2kt qoe-kt Pokekt = - 2p5qoekt equating the coefficient
k Po = 2qo j [email protected]
Hamiltonian :Nlechanics & Phase Space
@Sk J ahiruddin , 2020
If we take -
k 2-qo
physicsguide CSIR NET, GATE
38
=Band
_k 2 A
= A, then
p = B e-2At and
A q = -e2At _ B
Sol 1.18. The conjugate momentum 'Pl
= 2d,
11 oo : 01
: 47 =
q
A
-e2At _
B
Sol 1.18. The conjugate momentum
= 2q1 P2 = 2q2
Pl
The Hamiltonian H
= P1 ri1 - P2ri2 - L
PI
p~ 2
2
PI
p~ 4
4
PI +P2 4
Sol 1.19. The Lagrangian
L = >..q1 ri2 The conjugate momentum Pl P2 [email protected]
= Aq2 = >..q1 physicsguide CSIR NET, GATE
39
@Sk J ahiruddin, 2020
Hamiltonian :Nlechanics & P hase Space
The Hamiltonian H
= P1ri1 = Pl ri1
- P2ri2 - L
- P2ri2 - Aq1 ri2 2P1P2 P1P2 )..
P1P2 )..
)..
11 oo : 01
: 49
= Pl ci1 - P2 ci2 2P1P2
P1P2
;\
;\
Aq1 ci2
P1P2
A Sol 1.20. The K.E 1 2 T = -mv 2 1 • 2 = m rf + rsin0 + r00 2 1 2 2 2 2 2 2 = -m r + r sin 0~ + r 0 2 The potential energy A
•
A
k V (r) = -r
The Lagrangian
L = T-V =
~m
r +r 2
2 The conjugate moment um Pr = mr
r
= Pr
m j [email protected]
@Sk J ahiruddin , 2020
2
2
sin 0~
2 .
P0 = mr 0
B=
2
P0
mr
2
40
+r 0
2 2
P
k +-r 2
2
= mr sin 0~ .
P
= - 2 mr2 sin
0
physicsguide CSIR NET I GATE
Hamiltonian Mechanics & P hase Space
The Hamiltonian
H = T+V 1
= -m 2
k r
11 oo : 01 : s2 The Hamiltonian
H = T+V k r
1 =-m 2
P~
P~
P~
k 2m + -2_ mr_2 + -2_ mr_ 2 _0 - r 2 -si_n_
Sol 1.21. We see that sin2 q
8L
p
= 8q
2
. pq
•
'
2
. sin q q 2
=
hence we see
H =pq -L = O Now the question is can t he Hamiltonian of a physical system be ZERO? Ans is NO! To explore more let's t ry to calculate the Lagrangian equation of motion of t he given Lagrangian. 2
8L
p
d dt
= 8q =
8L 8q
=
•
2·q .
8L
sin q 2 ;
2
8q
•
q Sln q COS q;
Sln
q COS q
d 8L _ 8L = O dt
8q
8q
So we did not get a valid Lagrangian equation of motion of the particle. [email protected]
@Sk J ahiruddin, 2020
41
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Hamiltonian Mechanics & Phase Space
Hence we conclude that the Hamiltonian is NOT DEFINED . And we see that t here may not always b e a valid
11 oo : 01
: 55 Hamiltonian :Nlechanics & Phase Space
@Sk J ahiruddin, 2020
Hence we conclude that the Hamiltonian is NOT DEFINED . And we see that t here may not always be a valid Hamiltonian for a given Lagrangian. Sol 1.22. If t he Hamiltonian is independent of t ime
Thus H is a constant. By Euler 's theorem on homogeneous functions the K.E
If the potential V does not contain velocity dependent term, d we h ave .. , t hen Pi = 8f}LQi.. = 8f}T an Qi H
=
L
PiQi - L
=
2T - (T - V ) = T
+V
=
E
Sol 1.23. From Sol 1.9. we se that under Gauge transfor-
mation the Lagrangian changes by a total derivative.
[email protected]
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Hamiltonian Mechanics & Phase Space
11 oo :01 : s1
Hamiltonian :Nlechanics & Phase Space
@Sk J a hiruddin , 2020
So the action which is defined as S =
S
➔ S' = S' = S'
=
S' =
J L dt changes as
(L + !~)dt Ldt + S+
!;dt
dx
s + X (t) I~~
The last t erm is a constant and a surface element term i.e depends only on t he end points. Sol 1.24. Let
x+y
X+=--
\1'2
x-y
and,
X-
= y'2
Therefore
+x_ x=--X+
v'2 X+ - X_ y = v'2
and,
j a [email protected]
(c)Sk J a,h ir11ddin. 2020
43
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Ha.miltonian Nlechanics & Phase Sna.ce
11 oo : 02 : oo 43
j ahir@physicsguide. in
physicsguide CSIR NET I GATE
Hamilt onian Mechanics & P hase Space
@Sk J ahiruddin , 2020
t hen x2
+
(x+
= =
+ y2
xy
+ x_)
2
+
2
3
2
+
x_)
(x+ -
2
1
+ -x2 + 2 -
-x2
2
xt - x~
2
Hence 1 2
\x
+ xy + y
J e- f3H (x 2 + x y + y 2 )dpxdpydx dy J e- f3HdPxdPyd dY 3 x 2 + lx 2 )dx dx ef3H ( J 2 + 2 + -
2) _
-
X
Sol 1.25. We write the equations in matrix form
d dt
q
0
p
-3 - 4
1
q p
Eigenvalues of the matrix are -3, - 1 Eigenvectors (1, -3) and (1, -1) for the eigenvalues -3 and - 1 respectively. General solution will be
q(t ) p(t )
1
=a
e- 3t
-3
+ /3
1 - 1
e
-t
or q [email protected]
=
ae- 3t 44
+ f3e - t physicsguide CSIR NET, GATE
11 oo : 02 : 03 or
q = ae-3t
+ f3e - t
44
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Hamiltonian Mechanics & P hase Space
@Sk J ahiruddin , 2020
and
p = -3ae- 3t - f3e - t a and
/3
can be found :&·om the initial conditions which has not been given. Area is
dA = dqdp dq = -3ae- 3tdt - f3e - tdt dp = 9ae- 3t dt
+ f3e - tdt
dA = dqdp
= (-3ae- 3t - f3 e- t) (9ae- 3t + f3e - t) dtdt Till now is OK. Without t he initial conditions we can 't proceed further in t his way. Fortunat ely there is another way in which we do not solve p q explicitly, rather we go for the increment.
oq = O·
oq
,
Elementary are in phase space
dA = dqdp They are all funct ions of time. dq = dq(t). Now suppose at t ime t = 0 dq = dq(O ). What will be dq at time dt ? The
11 oo : 02 : os dA = dqdp They are all funct ions of t ime. dq = dq(t ). Now suppose at time t = 0 dq = dq(O ). What will be dq at t ime dt ? The [email protected]
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45
Hamiltonian Mechanics & P hase Space
@Sk J ahiruddin , 2020
increment in differential elements will be dq'(dt ) dp' (dt)
= dq(O) + = dp(O) +
a·
a! dqdt a·
a:dpdt
dA = dqdp dA' (dt )
= (dq (O) +
a·
a! dqdt )
X
(dp(O)
8q
+ 8p
a·
a: dpdt)
= dq (O)dp (O) + Bq dqdtdp + Bp dpdtdq Where we ignored the second order term. We get
dA' = dA
+
a· a· _q + p 8q
8p
dA · dt
Using the results
dA' = dA - 4dtdA
= dA (l - 4dt) The increment in time is dt . Total t ime is t. Say we have N number of increments. So we can write t = N dt or
11 oo :02 : oa dA = dA - 4dtdA
= dA (l - 4dt) The increment in time is dt . Total time is t. Say we have N number of increments. So we can write t = N dt or dt = t / N or
46
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Hamiltonian :Nlechanics & P hase Space
@Sk J ahiruddin , 2020
This is for one increment from O t o dt . For t he next increment it will be
t l- d N
dA'' = dA'
2
= dA 1 - d t
N
and so on .... For N increment
t
N
l -dN At limit N ➔ oo t he limit in t he RHS becomes e- 4t. Hence
A(t) = A(O)e-4t Sol 1.26. We are givenF(x, y) = x 2 + y 2
fJF
U=-
OV
= 2x+y
and and
+ xy .
Then
fJF v=fJy
=
2y+x
11 oo : 02 : 1o o 1.26. We are givenF x, y
oF u=ov = 2x+y
and
oF v=-
and
= 2y + x
oy
From there we get x = ½(2u - v)and y = -½(u - 2v) .
47
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Hamiltonian Mechanics & Phase Space
@Sk J ahiruddin, 2020
From Legendre's t ransformations
G (u, v) = F (x , y) - ux - vy
=
1
3 2 (2u
l
-
32
7
- v )2
( 2u
1
+ 32 ( u -
2v )2
- v) (u - 2v) -
2
=- (u -uv+v 9
2
u
( 2u
3
- v) +
)
Sol 1.27. From Hamilt on's equation
.
8H X=Op m . p = -x X
The Lagrangian
L = px- H m ,:,. ;.,.
( x ( m ;.,. \
2
,
1 ,A""\
V 3
(u
- 2v)
11 oo : 02 : 13 The Lagrangian L = px- H m
•
-xxx 2 mx -- -
m.
X
•
2
-X
2m
X
lk +-2 X
1 -kx
2x
2
48
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@Sk J ahiruddin , 2020
Hence
8L
ax
d dt
mx
8L
•
X
ax
..
mx mx 2
fJL
ax
=
1 - -k -
2
mx2 2x 2
x2
X
Euler-Lagrange equation in this case
..
mx X
8L
d dt
mx
ax
2
1
8L
-
ax mx
= 0
2
+-k + = 0 2 x2 2 2x 2 2 2mxi - mx + kx = 0
Sol 1.28. The total time derivative of a dynamical system Q is defined as dQ _ rlf -
{ Q,
H
}+
8Q At
11 oo : 02 : 1 s Sol 1.28. The total time derivative of a dynamical system Q is defined as
dQ _
8Q
H
dt - {Q, } + 8t
So, in t his case dp + v'2x = {
0 H} p + V L-X,
dt
= {p , H} + v'2{ x, H} = _ 8p8H + v'28x8H 8p ax
ax 8p
= -x + v'2p 49
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Hamiltonian :Nlechanics & Phase Space
@Sk J ahiruddin , 2020
Sol 1.29. From Hamilton's equation .
8H 4 q= = aq p 8p •
q p= aq4
The Lagrangian
L =pq- H
q . aq4
q-
•
aq4
q
2
a q4
q2
/3
2aq4
q2
Sol 1.30. Since t he force is conservative, t he Hamiltonian
11 oo :02 : 1a •
-
q-
aq4
+ q2
2
q2
/3
2aq4
q2
Sol 1.30. Since t he force is conservative, the Hamiltonian
From Sol 1.20. for spherical coordinate 2
T = Pr
2m in t his case r = a, and The potential energy
V
+ r
2
p2
P0 + d
-
Q 2f 41rcor
--
= 0 : a d ~ 41rc r 2
r < a·'
: b < r < c;
- Q f 41rc 0 r 2 •
0
9
physicsguide CSIR NET, GATE
Electrostatics: Part-1
(d)E = 0 : r < a· '
E = 4ncor Q r 2
=0 : a d - 41rcor2 •
Prob 1.16. In order to have equal surface charge densit ies on t he outer surface of the bot h shells, the following conditions should be satisfied
(a) d = 4b and c = 2a (c) d = v"2,b and c > a
(b) d = 2b and c =-J2a (d) d > b and c = v"ia
00: 00: 26 t ions should b e satisfied (b) d = 2b and c =0a (d) d > b and c = v'2,a
(a) d = 4b and c = 2a (c) d = v'2,b and c > a
Prob 1.17. A thin hollow cylinder of radius and length
bot h equal L is closed at the bottom. A disc of radius L / 2 is removed from t he bottom as shown in t he figure. [JAM 2008]
p L
-------L
I I
- : L/2 'I
j [email protected]
I
L
i+I
10
physicsguide CSIR NET I GATE
@Sk J ahiruddin , 2020
Electrostatics: Part-1
This object carries a uniform surface chaarge density a. Calculat e the electrostatic potential at t he point P on the axis of the cylinder as shown in the figure. (use
dx
J ✓x2 + a2
= ln (x + ✓x2 + a2))
Pro b 1.18. Consider a charge Q at the origin of 3-dimensional coordinate syst em. The flux of the electric field through the curved surface of a cone that has a height h and a circular
11 oo : oo : 29
Prob 1.18. Consider a charge Q at the origin of 3-dimensional coordinate syst em. The flux of the electric field through the curved surface of a cone t hat has a height h and a circular
base of radius R (as shown in figure) is [NET Dec 2015]
h ---'Q
R
(b)
(c)
Q
2c 0
h,Q
Rea
E=
Prob 1.19. For an electric field k,Jxx where k is a non-zero constant, total charge enclosed by the cube as
[JEST 2017]
shown below is
[email protected]
11
physicsguide CSIR NET ) GATE
Electrostatics: Part-1
@Sk J ahiruddin ) 2020
z l
l
y l "' •
-----------
11 oo : oo : 31 l
o: - --
l
--- 4 - - - - 4
y l "
-----------
•
l
X
(a) 0 (b) kc0 l~(v'3 - 1) kcal ~( v'2 - 1)
(c) kc0 l~(vb - 1)
(d)
Prob 1.20. Five sides of a hollow metallic cube are grounded and the sixt h side is insulated from the rest and is held at a potential (see figure) [TIFR 2012]
--
The potential at the centre O of t he cube is
(a) 0
(b)
! (c) ! (d) f
[email protected]
2
12
physicsguide CSIR NET, GATE
@Sk J ahiruddin , 2020
Electrostatics: Part-1
Prob 1.21. A point charge q sits at a corner of a cube of side a, as shown in t he figure below. The flux of the electric
[TIFR 2013]
field through the shaded side is
71
11 oo : oo : 34 1.21. point c arge q sits at a corner o side a, as shown in t he figure below. The flux of t he electric [TIFR 2013]
field through the shaded side is
q
q
(a) 8 co
q
(b) 16co
q
(c) 24c
(d) 0
q
6co
Prob 1.22. A charge q is at the cent re of two concent ric sp heres. The outward electric flux through t he inner sphere is 4? while t hat t hrough t he outer sphere is 4?. The am ount of charge contained in t he region between t he two spheres
[JAM 2015]
is
(a) 2q
(b) q
(c) -q
(d) -2q
Prob 1.23. A closed Gaussian surface consisting of a hemisphere and a circular disc of radius R , is placed in a uniform electric field ,as shown in the figure.The circular disc
£
makes an angle 0 = 30° with the vertical. The flux of the electric field vector coming out of the curved surface of the j [email protected]
13
physicsguide CSIR NET, GATE
@Sk J ahiruddin , 2020
Electrostatics: Part-1
[JAM 2011]
hemisphere is , , ,
00: 00: 36 Electrostatics: Part-1
hemisphere is
[JAM 2011] , , ,
,
, , ,
, , ,
,
, , ,
E
I , I , I ,
Prob 1.24 . A cube has a const ant electric potent ial V on its surface. If t here are no charges inside t he cube, t he potential at t he centre of t he cube is [JEST 2012]
(a) V
(b)
r (c) 0
(d)
r
Prob 1.25. T he charge per unit length of a circular wire of radius a in t he xy plane, with its cent re at t he origin, is A= ,X. 0 cos0, where ,X. 0 is a constatnt and t he angle 0is measured from t he posit ive x-axis. T he electric field at t he cent re of t he circle is [NET Dec 2016] (a) = - 4 2 (b) 15 = 4 2 (C) £ = - 4 d)
15 ;~a 15=41rcoa Ao k
;~a
[email protected]
14
@Sk J ahiruddin, 2020
;~a]
(
physicsguide CSIR NET, GATE
Electrostatics: Part-1
Prob 1.26. T wo uniformly charged insulat ing solid spheres
11 oo : oo : 39
@Sk J ahiruddin , 2020
Electrostatics: Part-1
Prob 1.26. Two uniformly charged insulating solid spheres A and B, both of radius a, carry total charges + Q and -Q , respectively. The spheres are placed touching each other as [NET Dec 2016] shown in the figure.
+
+ +
A
-
+
-
B
-
+ +
If t he potential at t he centre of the sphere A is VA and t hat at t he centre of B is VB, then t he difference VA-VB is
(a)
4~a
(b)
2--;~a
( C) 2~a
( d) 4--;~a
Prob 1.27. Consider a sphere S 1 of radius R which carries
a uniform charge of density p. A smaller sphere S2 of radius a < ~ is cut out and removed from it. The centres of the two spheres are seperated by t he vector b = nf ,as shown in [NET June 2016] t he figure. [email protected]
@Sk J ahiruddin, 2020
15
physicsguide CSIR NET, GATE
Electrostatics: Part-1
11 oo : oo : 42 15
[email protected]
physicsguide CSIR NET, GATE
@Sk J ahiruddin , 2020
Electrostatics: Part-1
The electric field at a point P inside S 2 is (b) pR (7 -na) (c) pRn (a) pRn 3 £o
3£oa
6 £0
(d)
7 3£o R pa
Prob 1.28. Four equal charges of + Q each are kept at the
vertices of a square of side R. A part icle of mass m and charge + Q is placed in t he plane of the square at a short distance (a < < R) from t he centre. If the mot ion of the particle is confined to the plane, it will undergo small oscillations with an angular frequency [NET June 2016]
(a)
21r£~~ 3m
(b)
1r£o~3m
(
C)
1r~~~
(
d)
Q2
j [email protected]
16
physicsguide CSIR NET, GATE
11 oo : oo : 44 Q2
j [email protected]
16
@Sk J ahiruddin , 2020
physicsguide CSIR NET, GATE
Electrostatics: Part-1
Prob 1.29. A sphere of radius R has a unifor charge density p. A sphere of smaller radius ~is cut out from the original sphere, as sho,vn in t he figure below. The centre of
t he cut out sphere lies at z = ~. After the small sphere has been cut out, the magnitude of t he electric field at z = - ~ is .e!i. What is the value of integer n? [JAM 2017] nt0
R/2
R
Prob 1.30. A hollow, conducting spherical shell of inner
radius R 1and outer radius R 2 encloses a charge q inside, which is located at a distance (d < R 1 ) from t he cent re of t he spheres. The potential at the cent re of the cell is [JAM 2015]
11 oo:oo:47
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17
physicsguide CSIR NET, GATE
@Sk J ahiruddin, 2020
(a) Zero
(b)
Electrost atics: Part-1
i 41rt 0
q
d
Prob 1.31. A solid, insulating sphere of radius 1 cm has 7 charge 10- C distributed uniformly over its volume . It is surrounded concent rically by a conducting thick spherical shel of inner radius 2 cm , outer radius 2.5 cm and is charged with -2x10- 7 C. What is the electrostatic potential in Volts on t he surface of t he sphere? [JEST 2017]
Prob 1.32. Consider an infinitely long cylinder of radius R , nlA.c:P-o 2r 02
[NET
(d) _ 2ecoc/>o
(c) _ Eoc/>o er02
r2 0
Prob 1.38. A solid sphere of radius R has a charge density, given by p(r) = p0 (1 - ~) where r is the radial coordinate
and p 0 , a and R are positive constants. If the magnitude of [email protected]
20
physicsguide CSIR NET, GATE
@Sk J ahiruddin, 2020
t he electric field at r t he value of a is (a) 2
Electrost atics: Part-1
=
(b) 1
R/ 2 is 1.25 t imes that ar r = R , the [NET Dec 2014] (c) 1/2
(d) 1/4
Prob 1.39. Consider a hollow charged shell of inner radius a and outer radius b . The volume charge density is
p(r) = r\(k is constant) in the region a< r < b. The magnitude of the electric field produced at a distance r > a is [NET Dec 2012] (a) (b) (c) (d)
k (b- 2a)
for all r > a
k (b~i) k (r - 2a )
for a < r < b and Eor kb2 for r > b fora < r < b and k(b- 2a ) for r > b
k (r- a)
for a < r < b and
Eor
Eo
Eor
Eo a 2
Eor
k(b- a)
Eor 2
for r > b
Prob 1.40. The shape of a dielectric lamina is defined by t he two curves y = 0 and y = 1- x 2 . If the charge density
11 oo :oo : s1 (d)
k(r - 2a) Eo a
for a < r < b and
k(b- 2a) b
Prob 1.40. The shape of a dielectric lamina is defined by t he two curves y = 0 and y = 1- x 2 . If the charge density
of t he lamina a = l 5y C / m 2 , then the total charge on t he [JAM 2016] lamina is ___________ C.
Prob 1.41. A sphere of inner radius 1 cm and outer radius 2 cm , centered at origin has a volume charge density
Po =
~r,
4
where K is a non zero constant and r is the radial
j [email protected]
21
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Electrostatics: Part-I
distance . A point charge of magnitude 10- c is placed at 2 t he orgin. For what value of Kin units of C/ m , the electric field inside the shell is constant? [JEST 2017] 3
Prob 1.42. A point particle of mass m carrying an electric
charge q is attached to a spring of stiffness const ant k. A constant electric field E along the direction of the spring is switched on for a time interval T ( where T i. l Eo
11 oo : 01 : s1 Sol 1.13. We can solve this by applying Gauss' Law. We consider a cylindrical Gaussian surface or length land radius r :
EdS = Al
Eo
⇒
A E · 21rr l = -l
Eo
A l " ⇒ E = --r 21rEo r We can now note f =~and r= xx+yy with r = ✓x 2 + y 2 . To find the field the value of t he field at (a, b), we plug in t he values of (x, y) in the expression for f and obtain
E= A ax+by 21rEo a2
+ b2
Sol 1.14. The electric field on t he axis is
E=
z
A(21rR)z
1
41rEo ( R 2 + z2)3/2
This will be maximum when the derivative of the electric field with respect to z will be zero.
8E
3z 2
1
az
(R2 + z 2 )3/ 2 This will equals to ZERO , hence 47rEo
3z
2
=
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R
2
+z
2
z= -
R
v'2
41
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@Sk J ahiruddin ) 2020
Electrostatics: Part-1
Sol 1.15. Fi1·stly, for r < a , No charge is enclosed , and, as we know, inside the conductor all charges reside at t he C' 11rft'.l d, Total charge Enclosed
2Q, so,
Applying Gauss Law,
E
2 = Q 2r 41rc 0r
. So, option (d) is correct.
Sol 1.16. In order to have,
ab
=
ad ,
where
charge density of the shell of radius b and, charge density of t he shell of radius d Q 41rb2
ab
=surface
ad
=surface
2Q 41rd2
Hence,
Hence,option (c) is correct. [email protected]
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Electrost atics: Part-1
11 oo :01 : s1
@Sk J a hiruddin , 2020
Electrostatics: Part-1
Sol 1.17. For the disc, t he electrostatic potent ial at point P, dV = dq 41rc:0Jr2 + 4£2
d-V =
a(21rrdr) 4nc:o J r 2 + 4£ 2 1
or, L
rdr
J r + 4£ Hence, 2
L/ 2
Now consider, r 2
+ 4£ 2 = z 2 .
v = 2c:o a
~ L
m2
V = o-L (v'5 -
2so Now for t he curved path, dV =
2
dz
v'17) 2
dq 4nsoJ (2L - r) 2 + L2
[where we considered a ring at a distance r from t he bot tom of the cylinder.] Here, dq = a(21rrdr )
V= j a [email protected]
(c)Sk J a,h ir11ddin. 2020
a (21rLdr) 4nc:o o J (2L - r) 2 + £ 2 l
L
43
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Electrosta,t ics: Part-1
11 oo : 01 : sg 43
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Electrostatics: Part-1
L
dr
Now, if we consider (2£- r) t he question, we get,
V=
= z From the formula given in
- o-L [ln((2 L -r)+ J (2L-r )2 + L 2 )] 2co
in t he limit Oto L. Hence,
1/ = ;c~ln [(2 + v'5) /( 1 + v/2)]
Sol 1.18. We know from Gauss law, Electric flux t hrough
a closed surface
ff E. dS =
Q enc co Now make it a close surface imagining another cone below the original one. Through that closed surface, electric flux coming out will be :- co Q Hence, flux coming out through
one cone is nothing other t han, Q . Hence, option (b) is 2co correct. ~
Sol 1.19. Since the electric field is given, we can use v7 · E .P... to determine the volume charge density p €Q
=
Now, we can calculate the total charge enclosed by t he cube [email protected]
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11 oo : 02 : 02 = kEo-Jx =
ax
o
2,Ji
Now, we can calculate the total charge enclosed by t he cube 44
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@Sk J a hiruddin ) 2020
Electrost atics: Part-1
be performing the following volume integral with appropriate as indicated by the figure. Look carefully at the limit of x axis. •
Q=
pdxdydz V
2z dx
kEo 2
= = =
k Eo l
l 2
2
Jx
vx
l
l
dy 0
dz 0
2l
2 l 2 kEol +~(V2- 1) kEol~ ( V2 - 1)
Sol 1.20. There are two ways to solve the problem. One
is the longer way where you need to apply some advance techniques called boundary value problem. The other is t he short way which I will discuss now. We will apply superposition principle of electric field and potential plus the idea of conductor. First consider what if all the sides have potential q> ? Consider two arbitrary points inside the cube. Let's say t he points are a and b. The potential difference between a
11 oo : 02 : os Consider two arbitrary points inside the cube. Let's say the points are a and b. The pot ential difference between a
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Electrostatics: Part-1
and b will be equal to b
(b) -(a)= -
-
-
E-dl a
As t here is no charge inside the cube, application of Gauss law gives that t he field inside t he cube will be ZERO. Hence
cp(b) - (a) = 0
⇒ (a) =
(b)
There is no harm if one of t he points a or b touches the surface. As the potential at the surface is , potential at all the points inside the cube will be same which will be . From symmetry of the problem we see that at t he center of t he cube all sides will have equal influences. If we say t hat the contribut ion to t he potent ial at t he center from each side is c, then the potential at the center will b e
We have already proved t hat the pot ent ial everywhere inside the cube is same. Hence,
= 6c·
'
C
1
=-
6
11 oo : 02 : 01 We have already proved t h at the potent ial everywhere inside the cube is sam e. Hence,
=
1
6c ·
C
'
=-
6
Now in our problem a single side is at potential , all other sides are grounded.
So the contribut ion to t he cent er is 46
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Electrostatics: Part-1
coming from one single side. Hence the potential at t he center is ½ enter
= c = 6
Sol 1.21. To m ake the charge closed , imagine 8 more cubes and electric flux coming out of these 8 cubes Q enc
Eo
Hen ce, flux coming out of 1 cube
Q 8Eo
This flux can be out only through 3 sides as other 3 sides are p erpendicular to electric field lines. Hence electric flux coming out of shaded surface
Q 24c 0
Hen ce, option (c) is correct
11 oo : 02 : 1o Q 24c: 0
Hence, opt ion (c) is correct Sol 1.22. Electric flux t hrough a closed surface is defined as
E . dS =
Qenc
(1.1)
Eo 47
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Let q2 be t he amount of charge in the region between t he two spheres. Then t he total flux t hrough t he out er sphere 2
= q + q2 EO
2 = + q2 EQ
q
q2
Eo
Eo
q
= q2
Sol 1.23. Since there are no enclosed in t he volume , so from 1.1 t he total flux coming out of t he volume is zero. Hence
-+
-+
E · dS = 0 -I-
-I
~
E · dScurved surface
+
~
E · dSflat surface
=0
f E · d S curved surface + f E · ndSflat surface = 0
11 oo : 02 : 12 _,
_,
E · dS = 0 _,
_,
E · dScurved surface _,
_,
+
_,
E · dSflat surface
=0
_,
+
E · dScurved surface
where
_,
E · ndSflat surface
=0
n is the normal unit vector to the fl at surface.
48
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So
n
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Electrostatics: Part-1
@Sk J ahiruddin , 2020 _,
makes an angle 150° with E. Hence _,
_,
E · dScurved surface +
_,
E
COS
150°dSflat surface
_,
E · dScurved surface
v'3 -
_,
2
7r R
2
=0
E = 0
_,
E · dScurved surface
=
v'3 2
7r R
2
E
Sol 1.24. Since there are no charges inside t he cube, Laplace's equation gives us
(1.2) Since cp = V at the surface. So, one solution of t he Laplace's equation is cp = V inside. Uniqueness theorem warrant us t hat this is t he only solut ion. -
......
...
..
.....
.
.
..
-
..
......
..
11 oo : 02 : 1 s Since = V at the surface. So, one solution of t he Laplace's equation is = V inside. Uniqueness theorem warrant us t hat this is the only solut ion. Sol 1.25. The electric field for a line charge A is defined as -
R =r-r
where
49
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-'I
(1.3)
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Electrostatics: Part-1
Hence, in this case
-
21r \
1
/\ QCOS :;, 2
E = - --
41rEo
r
0
A
41rEo r'
0 21r
Ao Ao
'd0' , . , r r
21r
1 Ao
4nE0 a
0'
A
cos 0'i
+ sin 0' j cos0' d0'
cos2 0'i
+ cos 0' sin 0'3 d0'
0
~
--i
4Eoa Sol 1.26. In t his problem , we can first find the potential at
t he center of A due to t he charge + Q distributed over the first sphere A and the charge - Q distributed ove1~ the second sphere. We can do a similar calculation for the potential at t he center of B and subtract the two to obtain the required
11 oo : 02 : 11 Sol 1.26. In this problem, we can first find t he potential at
t he center of A due to the charge + Q distributed over the first sphere A and t he charge - Q distributed over the second sphere. We can do a similar calculation for the potential at t he center of B and subtract t he two to obtain the required potential difference. (In fact, we shall see that we don't have explicitly calculate the potential at B ). First , we consider t he center of the sphere A as our origin and calculate t he electric field at a point r > a and at a point r < a using Gauss' law. These come out to be
for r > a, and
E = Q r 41rEo a 3 [email protected]
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Electrost atics: Part-1
for r < a We can now calculate t he potential at the center of A (i.e., at t he origin ) using t he above fields. 0
Vi = -
Edr 00
a
o Q
Qrdr
00
+ 41rEoa3 r2
a
a
1
Qdr 47rEor2 00
47rEO 2a 3 0 r a 3 Q --2 41rEoa Now the sphere B will also have a contribution to the potential at the center of A. Here, if we calculate the field or
11 oo : 02 : 20 47rEO 2a 3 0 r a 3 Q 2 41rEoa Now the sphere B will also have a contribution to the potential at the center of A. Here, if we calculate the field or potential for a point outside the sphere, it would appear as if t he ent ire charge is located at the center of t he sphere (and not just distributed t hrough the entire volume of the sphere) . Now the center of t he sphere B is at a distance of 2a from the center of of the sphere A. So we imagine that t he charge -Q is placed at a distance 2a from the origin, and calculate the potential at t he origin. We can directly
right is off as
4
1r~%a), and so adding t his to Vi 3 2
-
1 2
we get
Q 41rE0 a
Now we have to repeat the same calculation of the potent ial j [email protected]
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Electrostatics: Part-I
at t he center of the sphere B but wit h charge - Q instead of + Q , so we merely flip t he sign of Q in the above expression to obtain VB = 4-7rcoQa So the potential difference becomes
VA - VB= 2 Q 41rE0 a
Q 21rE0 a
Sol 1.27. In this problem, a portion of the sphere has been
removed. Since t he portion that has been removed is a sphere as well, we can think of solving t he problem in the following manner: We can first find the field due to the hi O'O'Pr l cm physicsguide CSIR NET, GATE
Electrostatics: Part-1
@Sk J ahiruddin, 2020
So, t he electrostatic p otential
1 T
T
{ 2.5x 10- 2
dr
1
{1 x 10- 2
dr
11 oo : 02 : 33 @Sk J ahiruddin, 2020
Electrostatics: Part-1
So, t he electrostatic potential V
=
l
2.5x 10- 2
41rEo
=9
=9
X
X
oo
10
9
10
4
X
10-
7
dr
1
r2
41rEo 10
X
1
1
2
2.5
--
1x 10- 2 2 x 10-
1
2
2.5
dr
r
2
+9 X
2
10
9
X
10-
5
1 1 -2
= 9000 Sol 1.32. Using Gauss's law the electric field inside the cylinder at distance r, E · 21rrh
=
r
1 -21rh
p(r')r' dr'
EQ r
0
E= k r' 2dr' Ear o E
=
k r2
3Eo
So, the electrost atic potential V = - k
R
r' 2dr
3Eo r k ex - (r 3 - R3 ) 3Eo k r3 ex - - - 1 3Eo R 3 [email protected]
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Electrostatics: Part-1
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I
,
,
.,
11 oo : 02 : 36 [email protected]
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Electrostatics: Part-1
Sol 1.33. Let t he sheets are parallel to xy plane , and the posit ive charged plate is at z = 0 and the negative charged
on is at z = l m. Then , t hen t he electric field in the region between the sheets for t he positive charged plate is ;~ z and for the negative charged one is ; EQ- z. Using super position principle a+
a+
E=-
+ 2Eo 2E0
6.8
X
10- 6
4.3
X
10- 6
----+---2Eo
~ 6.27
X
2Eo
10 5
Sol 1.34. Using Gauss's law the electric field for electron
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T""l
,
I
,
•
T""\
•
-1
00: 02: 38
57
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Electrostatics: Part-1
distribut ion , at a dist ance a a
Ee41ra2 = - qo
41r
Eo
1ra3
2r
2
e- 0 r dr o
a
Ee = -
E e-- E e-- Ee-
Ee =
qo
Eo1ra 5
r2
r
dr
0
qo Eo7ra5
qo Eo7ra5
qo Eo7ra5
5qo 41rE0a 2 e 2
5a 3 - 4e 2
+
a3 4
1 qo 41rEo a2
The electric field for the proton at a distance a is E _ P -
1 qo 41rEo a 2
Hence the total electric field at a distance a is
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11
00:02:41
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58
@Sk J ahiruddin, 2020
Electrostatics: Part-1
Sol 1.35. _,
_,
p
p
_,
V-E = Eo -= V· Eo A
=O'.
A
_, ( r) r ( r) _, r V 1 - e- R • r 2 + 1 - e- R V · r 2 r
1 e- R ( r) 3 + 1 eR 41rb (r) =a - R r2 r aEo e - R P = R r2
Hence the charge contained wit hin a sphere of radius R , centred at t he origin is R
q
2
=
p41rr dr 0
41r0'.Eo q= R 41raEo q= R q
0
= 41raE0 (1 -e- 1 )
11 oo : 02 : 43
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Electrostatics: Part-1
Sol 1.36. Using Poisson's equation, charge density
n n = -Eo V
·
V
n = - A Eo v •
-Ar
Ae r ,e
-Ar
- /\
rA
r
e
- Ar
r2
r A
A r e --r r
Ar n e - v · - -2 r r 1 8 e- Ar (re - Ar)+ A - - e- Arv. = -Aco -A-2 2 A
r
ar
A
r
A
r r2
1 e -Ar 3 (r) A-(eAr Are-Ar)+ A 41ro = - Aco r2 r2
The total charge enclosed within a sphere of radius of
½,
11 oo : 02 : 46
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Electrostatics: Part-1
with its origin at r = 0 is 1
>.
q=
2
p41rr dr 0
= - A E0 41r 1
>.
= - A Eo 41r
re- >.r dr - 1
0
e
Sol 1.37. Using Poisson's equation, charge density
_£__ = V2 R2)
0
12
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Electrostatics: Part-2
The electrost atic energy
allspace
R2
EO
-
')
n
.\ 1
2
21rsds
11 oo : oo : 34
allspace
R2
,\ 1 21rc0 s
EQ 2 Ri 1 ,\2
2
21rsds
R2 1
---
-ds
2 27rEo
Ri
S
,\2
=-
47rEQ
ln(R2/R1 )
Prob 1.9. For a conductor all charges will be on the sur-
face. So, the electrostatic energy in that case
q2 1
E =-81rco R
9
X
109 100
2
9 = 7
X
8.5
X X
10- 6 10- 2
1000
= 529 Prob 1.10. For a uniformly charged spherical shell the elec-
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Electrostatics: Part-2
t rostatic energy is given by q2 1 E =-R7rFn
R
00: 00: 36 Electrostatics: Part-2
t rostatic energy is given by q2 1
£ =--
=
81rco R 2 15 24 25 8 3.125
Prob 1.11. This problem is similar to Prob 1.3. . The electrostatic potential energy of the configuration is then in
this case £
= Ll1 + Ll2 + L'.l3 q2 1
2q 2 1 = 0+ + 41rco L 41rc0 L 3q 2 1 41rco L
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2
Conductors
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Electrostatics: Part-2
11 oo : oo : 39 @Sk J ahiruddin, 2020
2
Electrostatics: Part-2
Conductors
Prob 2.1. A solid spherical conductor encloses 3 cavities, a cross-section of which are shown in the figure. A net charge +q resides on the outer surface of t he conductor. Cavities
A and C contain point charges +q and -q respectively. +q
The net charges on the surfaces of t hese cavities are [TIFR 2015] (b) A = - q, B = q, C = 0 (a)A = - q, B = 0, C = -q (c) A= +q, B = 0, C = -q (d) A= -q, B = O,C = + q Prob 2.2. An arbitrarily shaped conductor encloses a charge q and is surrounded by a conducting hollow sphere as shown in t he figure. Four different regions of space, 1,2,3,4 are indicated in t he figure.W hich one of t he following statements is correct? [JAM 2016]
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Electrostatics: Part-2
11 oo : oo : 41 15
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Electrostatics: Part-2
-
1
3 4
')
(a) The electric field lines in region 2 are not affected by t he posit ion of the charge q (b) The surface charge density on the inner wall of the hollow sphere is uni£orm (c) The surface charge density on the outer surface of the sphere is always uniform irrespective of t he position of charge q in region 1 (d) T he electric field in region 2 has a radial symmetry Prob 2.3. A solid spherical conductor has a conical hole
made at one end, ending in a point B , and a small conical projection of the same shape and size at the opposite side, ending in a point A. A cross-section t hrough the center of t he conductor is shown in the figure below.
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11 oo : oo : 44
j [email protected]
16
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Electrostatics: Part-2
A
If, now a posit ive charge Q is t ransferred to t he sphere, t hen [TIFR 2014] (a) the charge density at bot h A and B will be undefined (b) t he charge density at A will be the same as t he charge density at B. (c) the charge density at A will be more than the charge density at B. (d) the charge density at B will be the more than t he charge density at A.
Prob 2.4. A spherical conductor, carrying a total charge Q, spins uniformly and very rapidly about an axis coinciding with one of its diameters. In t he diagrams given below, t he equilibrium charge density on its surface is represented by the t hickness of t he shaded region. Which of these dia[TIFR 2014] grams is correct?
11 oo:oo:47 ing with one of its diameters. In t he diagrams given below, t he equilibrium charge density on its surface is represented by the t hickness of t he shaded region. Which of these dia[TIFR 2014] grams is correct? [email protected]
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Electrostatics: Part-2
(b)
(c)
Prob 2.5. A point charge q < 0 is brought in front of a
grounded conducting sphere. If t he induced charge density on t he sphere is plotted such t hat the thickness of the black shading is proportional to the charge density, t he correct [TIFR 2019] plot will most closely resemble
11 oo : oo : 49
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Electrostatics: Part-2
(a) 0
q
(c)
0
q
0
q
--
(b)
0
q
(d)
Prob 2.6. MSQ: Out of the following statements, choose [JAM t he correct option(s) about a perfect conductor. 2019] (a) The conductor has an equipotential surface (b) Net charge, if any, resides only on the surface of conductor (c) Electric field cannot exist inside the conductor (d) Just outside the conductor, t he electric field is always perpendicular to its surface Prob 2. 7. A conducting sphere of radius lm is placed in air. The maximum number of electrons that can be put on
11 oo : oo : s2 perpendicular to its surface Prob 2. 7. A conducting sphere of radius l m is placed in air. The maximum number of electrons that can be put on t he sphere to avoid electrical breakdown is about 7 x 1on,
where n is an integer. The value of n is
[GATE 2020]
Assume: j a [email protected]
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Electrostatics: P art-2
-+
Breakdown electric field strength in air is E 6
10 V / m Permitti,rity of free space co = 8.85 x 10- 12 F / m Electron charge e = 1.60 x 10- 19 C
3
X
11 oo : oo : s4
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2.1
Ans keys
2.1. d 2.2.
C
2.3.
C
2.4. b 2.5. b 2.6. a,b,c,d
2.7. 14to 15
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Electrostatics: Part-2
11 oo :oo : s1 2. 7. 14 to 15
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2.2
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Electrostatics: Part-2
solution
Sol 2.1. The +q charge will induce -q charge on the surfac e of the conductor and vice versa. So (d) is the right answer. Sol 2.2. This problem is discussed in details in Griffit hs Electrodynamics book. Sol 2.3. A conductor mus always be equipotential. For a charge configuration V rv ; . But for Ar > R (the radius of t he spherical conductor) so the charge accumulation at A must be more than the average charge density to balance it out. Similarly for B r < R so the charge accumulation at A must be less than the average charge density.
11 oo : 01 : oo t he spherical conductor) so t he charge accumulation at A must be more t han the average charge density to balance it out. Similarly for B r < R so t he charge accumulation at A must be less t han the average charge density. Sol 2 .4. Since the charges are rotating around t he axis in a circular path very rapidly. The magnetic field for t his
rotation of charge is along z direction . So the force on the charge is
= v B cp x z A
=pw B p
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Electrostatics: Part-2
vector from t he axis and is proportional to radial distance. So (b) is t he right answer. Sol 2.5. This is obvious that more charge will be induced on the near side of the conductor from the charge than the far side. But it 's not t hat far side will have no induced
charges. Sol 2.6. All four statements are basic properties of a con-
ductor. The proofs are rigorous and you can find it in any book. Sol 2.7.
11 oo : 01
: 02 p r
1.. ,.,
ductor. The proofs are rigorous and you can find it in any book.
Sol 2.7.
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3
23
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Electrostatics: Part-2
Dipole
Pro b 3 .1. Four point charges are placed in a plane at the following positions:
+ Q at (1,0), -Q at (-1,0), + Q at (0,1), -Q at (0,- 1). At large distances the electrostatic potential due to t his charge distribution will be dominated by the [GATE 2007] (a) monopole moment (b) dipole moment (c) quadrupole moment (d) octopole moment Prob 3.2. Three point charges q, q and -2q are located
11 oo : 01 : os charge distribution will be dominated by the [GATE 2007] (a) monopole moment (b) dipole moment (c) quadrupole (d) octopole moment moment Prob 3.2. Three point charges q , q and -2q are located at (0,-a,a), (O ,a, a) and (0,0, -a) respectively. The net dipole moment of t his charge distribution is [GATE 2006
(b) 2qak
(a) 4qak
(d) -2qa]
(c) -4qai
Prob 3.3. A circular disc of radius a on the xy plane has a surface charge density a = aor ~osB. The electric dipole
[GATE 2008]
moment of t his charge distribution is 4
3
3
4
(a) o-o;a X (b) o-o;a X (c) - o-o;a X (d) o-o;a X Prob 3.4. An insulating sphere of radius a carries a charge
density p(r) = p0 ( a 2 - r 2 ) cos 0 : r < a. The leading order term for the electric field at a distance d, far away from the charge distribution, is proportional to [GATE 2010] j [email protected]
24
physicsguide CSIR NET, GATE
@Sk J ahiruddin, 2020
(b) d- 2
Electrostatics: Part-2
(c) d- 3
(d) d-4
Prob 3.5. A charge -q is distributed uniformly over a sphere,
with a positive charge q at its centre in (i). Also in (ii), a charge -q is distributed uniformly over an ellipsoid with a positive charge q at its center. With respect to the origin of the coordinate syst em , which one of the following state[GATE 2015] ments is correct?
11 oo : 01
: 01
charge -q is distributed uniformly over an ellipsoid with a posit ive charge q at its center. With respect to t he origin of t he coordinate system, which one of the following statements is correct? [GATE 2015] X
X
~t--
y
z
----- z y
(i)
(ii)
(a) The dipole moment is zero in both (i) and (ii). (b) The dipole moment is non-zero in (i) but zero in (ii). (c) The dipole moment is zero in (i) but non-zero in (ii). (d) The dipole moment is non-zero in both (i) and (ii). Prob 3.6. Charges Q, Q and - 2Q are placed on the vert ices of an equilateral t riangle ABC of sides of length a, as shown in the figure. The dipole moment of this configurat ion of charges, irrespective of t he choice of origin, is [NET June 2012]
25
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physicsguide CSIR NET, GATE
@Sk J ahiruddin , 2020
Electrostatics: Part-2
C
-2Q
t:,
J
A.____.________. B Q a Q . __ _ _ _ _ 11,
I
(a) + 2aQz
(b)
+v13aQ]
(c) -v'3aQ]
(d) 0
11 oo : 01
: 1o
A------------- B Q a Q
(a) +2aQi
+v'3aQ]
(b)
(c) -v'3aQ]
(d) 0
Prob 3. 7. Consider an axially symmetric st atic charge dis-
[NET June 2013]
t ribution of t he form, p
=
p0
ro
2
-
r
2 e - ro cos cp
r T he radial component of t he dipole moment due to t his charge distribution is (a) 21r port (b) 1r port Pro b 3. 8. Three charges (2C , - 1C, - 1C) are placed at
t he vert ices of an equilateral t riangle of side lm as shown in t he figure.T he component of t he electric dipole moment about t he marked origin along they direction is [GATE 2017]
________________c m. 26
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physicsguide CSIR NET, GATE
Electrostatics: Part-2
@Sk J ahiruddin , 2020
y
';:::::===~ ===-==--___, _ _. x 1.5 -1C
m
11 oo : 01
: 12
y
-1C L - - - - -......:....:...1---_____.____;__;:__.:..... x 1.5 m
•
Prob 3.9. A particle of charge e and mass m is located at
the midpoint of t he line joining two fixed collinear dipoles with unit charges as shown in t he figure. ( The particle is constrained to move only along the line joining the dipoles). Assuming that t he length of the dipoles is much shorter than t heir separation, the natural frequency of oscillation of the [NET June 2013] particle is
R
I( I I I I I I I
I I
>i
> d) is [JEST 2017] (a) zero (b) proportional to d 1 1 (c) proportional to r 3 ( d) proportional to r 4 Prob 3.13. The electrostatic potential (r) of a distribution of point charges has the form (r )ex r- 3 at a distance r from the origin (0, 0, 0) , where r > a. Which of the following distributions can give rise to this potential? [TIFR 2015]
Prob 3.14. Four charges (two +q and two -q) are kept fixed at the four vertices of a square of side a as shown [NET Dec 2012] j [email protected]
@Sk J ahiruddin, 2020
29
physicsguide CSIR NET, GATE
Electrost atics: P art-2
11 oo : 01 29
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: 20
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Electrostatics: Part-2
@Sk J ahiruddin, 2020
q
q
-q
q (0,0,0)
a
q
a from the centre of the square. Then V(2r)/ V(r) is
(a) 1
(b) ~
(c) ~
(d)
1
Prob 3.16. The electrostatic lines of force due to a system
of four point charges is sketched b elow. [NET Dec 2014]
At a large distance r , t he leading asymptotic behaviour of the electrostatic potential is proportional to
11 oo : 01 : 2s
At a large distance r , t he leading asymptotic behaviour of the electrostatic potential is proportional to 31
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(a) r
Electrostatics: Part-2
(c) r- 2
(b) r- 1
(d) r- 3
Prob 3.17. T wo electric dipole P 1 and P 2 are placed at (0, 0, 0) and (1, 0, 0) respectively, with bot h of them pointing in the +z direction. Wit hout changing t he orientationsof t he dipoles P 2 is moved to (0, 2, 0). The ratio of t he electrostatic potential energy of t he dipoles after moving to that before moving is [JAM 2006]
(a)
1 16
(b) ~
(c)
!
(d) ~
Prob 3.18. MSQ: Multiple Select Question For an elect ric dipole with moment = p 0 ez placed at t he origin , ( Po is a const ant of appropriat e dimensions and ex, ey and ez are unit vectors in Cartesian coordiante system) [JAM 2016] (a) potent ial falls as r12 , where r is the distance from origin (b) a spherical surface cent ered at origin is an equipotential surface (c) electric flux through a spherical surface enclosing the • or1g1n 1s zero (d) radial component of £ is zero on the xy-plane.
P
•
•
11 oo :01 : 2a surface (c) electric flux t hrough a spherical surface enclosing t he or1g1n 1s zero •
•
•
(d) radial component of
15 is zero on t he xy-plane.
32
j [email protected]
physicsguide CSIR NET I GATE
@Sk J ahiruddin , 2020
Electrost atics: Part-2
Prob 3.19. MSQ: For a point dipole of dipole moment P = pz located at t he origin, which of the following is (are) correct? [JAM 2017] (a) The electric field at (0, 0, 0) is zero. (b) T he work done in moving a charge q from (O,b,O) to (0, 0, b) is qp b2 47rEo (c) The electrostatic potential at (b,0,0) is zero. (d) If a charge q is kept at (0,0,b) it will exert a force of magnit ude qp on t he dipole. 41rEob3 .....
,...
Prob 3.20. An electric dipole of dipole moment P = qbi is placed at t he origin in the vicinity of two charges +q and - q at (L , b) and (L , - b), respectively, as shown in the figure [NET Dec 2018] below. The electrostat ic potential at t he point (L/ 2, 0) is
(a) (c)
;! (p +
qb 1rcoL2
(d)
L2_;4b2) 3qb 1rc0 L 2
(b)
1rco[L;~~b2]3/2
11 oo : 01
: 31
The electrostatic potential at t he point ( L / 2, 0) is ( a)
(c)
qb ( 1 7rco V
qb 1re:0L2
+
2 ) L2+ 4b2
(d)
3qb 1re:0L2
(b)
4qbL 1re:o[L 2+ 4b2]3/ 2
Pro b 3. 21. Consider a system of t hree charges as shown in t he figure below:
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33
@Sk J a hiruddin ) 2020
Electrostatics: Part-2
y
+q(L.b)
p·
• I
I
A
---0
•
I
•.
(Ll2.0)
--- -• X
I
' I I
•
-q (L,-b)
z
--t-- -2
!
I I I
q 0
d
q
.,
(r, 0)
q
--
2
,., I I
d
I
d
y
I I
)I I
For r = 10m; 0 = 60degrees; q = 10- 6 Coulomb, and d = 10- 3m, the electric dipole potential in volts (rounded off to t hree decimal places) at a point (r, 0) is _____ (Use 2 1 9 : = 9 X 10 Nm2 ) [GATE 2019] 47rEQ C
11 oo : 01
: 33
or r = m; egrees; q = ou om , an d = 10- 3m , the electric dipole potential in volts (rounded off to three decimal places) at a point (r, 0) is _____ (Use 2 1 9 : = 9 X 10 Nm [GATE 2019] ) 4~EQ C2
Prob 3.22. Charges are placed as follows: q at (a, a, 0) and ( -a, -a, 0), and - q at (a, - a, 0) and ( -a, a, 0) At large distances, how does the electrostatic potential behave as a function of the distance r from the centre (0,0,0)? [JEST
2020] [email protected]
34
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@Sk J ahiruddin , 2020
(A) 1/ r 3
(B) 1/ r 2
Electrostatics: Part-2
(C) 1/r
(D) 1/r 4
11 oo : 01
j [email protected]
35
@Sk J ahiruddin , 2020
3.1
: 36
physicsguide CSIR NET, GATE
Electrostatics: Part-2
Ans keys
3.1. b
3.9. d
3.17. d
3.2. a
3.10. b
3.18. a,c,d
3.3. b
3.11.
3.19. b ,c
3.4.
C
3.12. d
3.20.
3.5. a
3.13. a
3.21. 0.0045
3.6.
3.14.
3.22. a
C
3.7. a
C
C
3.15. d
C
11 oo : 01 3.6.
3.14.
C
3.22. a
C
3.7. a
3.15. d
3.8. 1.735
3.16. d
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: 38
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@Sk J ahiruddin, 2020
3.2
Electrostatics: Part-2
Solutions
Sol 3.1. The monopole moment
.
'l
=0 The dipole moment
. 'l
= (Q + Q)x + (Q + Q) = 2Qx + 2QiJ
11
00:01 :41
= (Q + Q)x + (Q + Q) = 2Qx + 2QiJ Hence t he dominant term is dipole potential.
Sol 3.2. The dipole moment
= q(-a] + ak) + q(a] + ak) - 2q(-ak) A
= 4qak Sol 3.3. The dipole moment for a cont inuous charge distri-
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@Sk J ahiruddin , 2020
Electrostatics: Part-2
bution
p=
o-(r')dr' R
a
a r=O 4 a-or a l a 4 2
271"
cos 0' (cos 0' x + sin 0' i)) r' dr' d0'
0'=0 21r
(1 - cos 20' )d0' x
0'=0
o-01ra3,...
--x 4
Sol 3.4. The monopole moment
11 oo : 01
(1 - cos 20' )d0' x
-
a 4 2 ao1ra3
: 44
0'=0
A
--x 4
Sol 3.4. The monopole moment Q
= 21r
p0 ( a
2
-
r
2
2
)
cos 0 cos 0 sin 0r d0dr
r
2
The dipole moment a
21r
p=
po (a r=O
2
-
)
2
cos 0 sin 0' r d0drdcp
0'=0 a
p = Po
21r
r'
3
(a
2
-
r
2
cos 0 sin 0'
)
r=O
0'=0
[sin 0' cos cp' x + sin 0' sin ' fJ
r
16 a
6
cos 0'
3
0
+ cos 0' z] d
-l
1
= non zero So the dipole moment exist, and the electric field at a disj [email protected]
38
@Sk J a hiruddin , 2020
physicsguide CSIR NET 1 GATE
Electrostatics: Part-2
•
tance d , far away from t he charge distribution, 1s proport ional to rv d- 3
Sol 3.5. In both of t he configuration for every charge q at rthere is a char·ge q at So t he dipole moment is zero in both (i) and (ii) .
-r.
Sol 3.6. Since the monopole moment Li Qi = 0. The dipole moment becomes indep endent of the origin. So,
11 oo : 01
: 46
rthere is a charge q at -r' . So t he dipole moment is zero in both (i) and (ii) .
Sol 3.6. Since the monopole m oment L i Qi = 0. T he dipole moment becom es independent of t he origin. So,
_v;;:;aQ 0_ . Q. J+ i 2 A
=2X
Q1,.
A
A
-
= v3aQ] Sol 3.7. 21r
00
ro Pr= 2 Po r r=O 0 . The region x < 0 is filled uniformly wit h a met al. The electric field at t he point (~, 0, 0) is [NET June 2014] (a) -
(c)
lOe 97rEo e
7rEO
d2 ( 1, 0, 0)
d2 (1, 0, 0)
(b) g
lOe
7rEo
e
(d) -
7rEO
d2 ( 1, 0, 0)
d2 (1,0,0)
+q is
placed at (0, 0, d) above a grounded infinite conducting plane defined by z = 0. There are no charges present anywhere else. What is t he magni[JEST 2012] t ude of electric field at (0, 0, -d)? Pro b 4. 7. A point charge
(a) ,, q ,,,
(b ) -oo
( c) 0
( d ) ~,, q ,,,
11 oo : 02 : 1o +q is placed at (0, 0, d) above a grounded infinite conducting plane defined by z = 0. There Prob 4. 7. A point charge
are no charges present anywhere else. What is t he magnit ude of electric field at (0, 0, - d)? [JEST 2012]
(a)
(b) -oo
B1r:od2
(c) 0
(d)
161r~od2
Prob 4.8. A point charge q of mass m is released from rest
47
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@Sk J ahiruddin, 2020
physicsguide CSIR NET, GATE
Electrostatics: Part-2
at a distance d from an infinite grounded conducting plane (ignore gravity) . How long does it take for the charge to hit t he plane? 3 3 (a) J21r Eomd q
~-3 (b) J21r Eomd q
[JEST 2016] 3 3 (c) J1r Eomd (d) q
3 E-om-d ✓~1r~
q
Prob 4.9. A hollow metallic sphere of radius a, which is kept at a potential V0 has a charge Q at its centre. The potential at a point outside the sphere, at a distance r from
[NET Dec 2015]
t he cent re, is
(a) Vo (c)
Q 41rEor
(b)
+ Voa r
2
Q 41r > d t he electrostatic potent ial of t his charge configuration would approximately be [NET Dec 201 7]
j [email protected]
@Sk J ahiruddin , 2020
51
physicsguide CSIR NET I GATE
Electrostatics: Part-2
Prob 4.16. Two point charges + 2Q and -Q are kept at
points with Cartesian coordinates (1,0,0) and (2,0,0) respect ively, in front of an infinit e grounded conducting plate at x= O. The potent ial at (x,0,0) for x > > l depends on x as [NET June 2018] (a) x - 3 (b) x - 5 (c) x - 2 (d) x - 4
11 oo : 02 : 22 points with Cartesian coordinates 1,0,0 and 2,0,0 respect ively, in front of an infinite grounded conducting plate at x=O. The potent ial at (x,0,0) for x > > l depends on x as [NET June 2018] (a) x - 3 (b) x - 5 (c) x- 2 (d) x - 4 Pro b 4.1 7. Consider two concentric spherical metal shells of radii r 1 and r 2 ( r 2 > r 1 ) . The outer shell has a charge q and t he inner shell is grounded. What is the charge on t he
[JEST 2019]
inner shell?
(a) _ r1 q
(c) 0
r2
Prob 4.18. An infinitely long wire parallel to the x -axis
is kept at z = d and carries a current I in the positive x direction above a superconductor filling the region z < 0 (see figure). The magnetic field B inside t he superconductor is zero so t hat t he field just outside the superconductor is parallel to its surface. The magnetic field due to t his con[GATE 2019] figurat ion at a point (x, y, z > 0) is --+
52
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@Sk J ahiruddin ) 2020
Electrostatics: Part-2
I d
superconductor
/
/
/
/
/
11 oo : 02 : 2s I d A
----r----,..---r----r---r---r...___,,.....----r--r---r-~
X
superconductor
//// A
-(z- d)j+yk [y2 + (z - d)2]
(a) (b)
- (z - d)j + yk + (z + d)] - yk y2 + (z - d)2 y2 + (z + d)2
(c)
-(z- d)j+yk y2 + (z - d)2 y J + (z - d) k
(z + d)] - yk y2 + (z + d)2
( z + d) k ----- + -----
(d)
y2
+ (z - d) 2
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y3y2
+ (z + d) 2
physicsguide CSIR NET , GAT E
53
@Sk J a hiruddin , 2020
4.1
Electrostatics: Part-2
Ans keys
4.1. a
4.10. b
4.2. 11
4.11.
4.3. b
4.12. d
C
11 oo : 02 : 2a 4.1. a
4.10. b
4.2. 11
4.11.
4.3. b
4.12. d
4.4. a
4.13. 0.9142
4.5. b
4.14. 3.9
4.6. d
4.15. b
4.7. c
4.16. d
4.8. a
4.17. a
4.9. d
4.18. b
j [email protected]
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4.2
54
C
physicsguide CSIR NET, GATE
Electrostatics: Part-2
Solutions
Sol 4.1. Place an image charge - q at x = -.5m and -2q at x = -l.5m. Hence the force on q for this charge config1,ri::1 +1()n
11 oo : 02 : 30 4.2
Solutions
Sol 4.1 . Place an image charge -q at x = -.5m and -2q at x = -l .5m. Hence the force on q for this charge configuration 2 1 F = - -q2 - 2q2 - _q
4
41rEo
7q2 47rEO
Sol 4.2. For an angle 0 which is integer divisor of 180°, the required number of image charges is given by 360°
n =-- -l
0
Here 0 is 30°, hence 360° n=---l 30°
= 11 Sol 4.3. Let x be the horizontal axis and y be t he vertical axis. Place an image charge -q at (d,-d), q at (-d,-d) -q at
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55
physicsguide CSIR NET, GATE
Electrostatics: Part-2
(-d,d) . Hence the force on q for this charge configuration
11 oo : 02 : 33 @Sk J ahiruddin, 2020
Electrostatics: Part-2
(-d,d) . Hence the force on q for this charge configuration 1 q2 F = - - -2 41rEo d 1 q2 41rEo d 2 1 q2 41rEo d 2
t he minus sign indicates that the force is towards the corner
1
2
(2v'2 - 1) 2 (8y'2)
47rEo -
IF
2
41rE08
Sol 4.4. For equilibrium the the equation of motion is 1
q2
41rEo (2d) q2 1
2
=mg
- - - - = d2 47rEO 4mg
d=
q 4 ✓~m-g-1rE-o
Sol 4.5. Sol 4.6. P lace an image charge e at (-d, 0,0) . Hence t he [email protected]
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56
@Sk J ahiruddin , 2020
,
,
•
,_a
11
I
1
Electrostatics: Part-2
1
•
I
,rl
I"'\
I"'\\
•
11 oo : 02 : 3s [email protected]
physicsguide CSIR NET, GATE
56
@Sk J ahiruddin, 2020
Elect rostatics: P art-2
electric field at the point ( g, 0, 0) is _, 1 E =-
4?TEo
1 41TEo d2 9
9
lOe
d2 (1, 0, 0)
1TEo
Sol 4. 7. Since the V = 0 at z = 0 plane and there are no charges in z < 0 , so t he magnitude of electric field at z < 0
is 0 Sol 4.8. Let z is t he distance of the point charge from t he infinite grounded conducting plane at any t ime . By New-
t on's law
dv q2 1 m v - = - - - -2 dz 41TEo z v2
dz v= -
dt
=
q2 1 ---41TEo 2m q
1
1
z
d
-- 1
1
2
d- z
~~
z
Let
j [email protected]
57
physicsguide CSIR NET , GATE
T""l
,
I
,
•
T""\
•
II"'\
11 oo : 02 : 38 [email protected]
57
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@Sk J ahiruddin , 2020
Electrostatics: Part-2
⇒
z
d 0
0
II
0
2
dz = 2dsin 0cos 0d0
Hence 1
z d- z 2
dsin 0 d - d sin 2 0
2
dz =
q 1 ~~
dt
1 2
41rEo 1r
2
2d
sin2 0d0
0
d
1T
2
=
q
2md 1
~~ qt J81rEomd
dt t
J21r 3 Eomd3
t= - - - q
Sol 4.9. The potential at the surface of the sphere is given by
Vo=
!
q
41rEo a q
V0 a = - 41rEo
Hence he potential at a point outside the sphere, at a [email protected]
58
physicsguide CSIR NET, GATE
11
00:02:41 q
V0 a= - 41rEo
Hence he potential at a point outside the sphere, at a [email protected]
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58
@Sk J ahiruddin, 2020
Electrostatics: Part-2
tance r from the centre, is
l Vout= 41rEo r q
Voa r
Sol 4 .10. Since the sphere is grounded , we must find an image dipole for which potential at any point on the surface A
of the sphere is zero. Let us take that point as ak, and p' be the image dipole. Then the potential for these dipoles at A
ak
Sol 4 .11 . Let an image line charge per unit lenght - ,,\ runs parallel to x-axis at y = -d. Then the potential for any
pointy> 0 is V (x, y, z)
=
1 '),rr C
,,\ In ~
3
+
11 oo : 02 : 43 Sol 4.11. Let an image line charge per unit lenght - A runs parallel to x-axis at y = -d. Then the potential for any pointy> 0 is
V (x,y,z) = [email protected]
1 - A ln 21rEo 59
S+
s_
physicsguide CSIR NET, GATE
@Sk J ahiruddin , 2020
Electrostatics: Part-2
where s+ = (x 2+z 2 +(y-d) 2)½ and s_ = (x 2+z2 +(y+d) 2)½. t hen surface charge density on t he conducting plane is
oV(x ,y,z) oy a - A y - d Eo 21rEo (x2 + z2 + (y _ d)2) ~ - Ad a =-----(x2 + d2 + z2) a
y+d y=O
Sol 4.12. Since the inner cylinder is insulated , no electric field penetrate the conductor ,and Since t he outer cylinder is grounded no electric field penetrat e the out er conductor either . Hence d is the right answer. Sol 4.13. We have found in Sol 4.3. t hat
1
q2 FA= -8d2 (2v'2 - 1) 41TEQ
And we know
q2 1
FB = - - - -2 41TEo 4d
11 oo : 02 : 46 And we know
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Electrostatics: Part-2
Hence
= .9142 Sol 4.14. Then the potential for any point at a distance sfrom the axis is
1 s V (x, y , z ) = --\ln R 21TEQ
On t he surface of inner conductor 10 10 ,\ ln ( ;)
=
1 27rEo
,\ ln
2
5
1 ,\ 21rEo
Then
V (3.5mm )
1
= - -,\ ln 2 7rEO
10 \lr1/2\
ln
3.5 5 3.5 c;
11 oo : 02 : 48 Then 1
3.5 V (3.5mm ) = - -,\ In 2 7rEO 5 10 ln 3.5 ,\ ln ( 5
i)
= 3.90 Sol 4.15. Place image charges - 3Q and + Q at (0, 0, -d) and (0, 0, - 2d), respectively. The monopole monement for 61
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Electrostatics: Part-2
@Sk J ahiruddin, 2020
t his charge configuration is zero. The dipole moment A
p = (3Qd - 2Qd + 3Qd - 2Qd)k "' = 2Qdk Hence t he dominating term in t he potential will be t he dipole term. So, for z > > d the electrostatic potent ial of t his charge configuration would approximately be V( z)
= 1 2Qd 41rEo z 2
Sol 4.16. Place image charges - 2Q and +Q at (- 1, 0, 0) and (-2, 0, 0), respectively. The monopole monement for
t his charge configuration is zero. The dipole moment
p=
(2Q - 2Q
+ 2Q -
2Qd)i
=0 So the dipole moment is also zero. The quadrupole domin::i.t.in P- t.Prm . Sn nnt.Pnt,i::i.l ::i.t. ( r..n.n1 fnr r. '> '> 1 rl PnPnrl~ nn
11 oo : 02 : s1 p=
(2Q - 2Q
+ 2Q -
2Qd)i
=0
So the dipole moment is also zero. The quadrupole dominating term. So potential at (x,0,0) for x > > l depends on x as Sol 4.17. Let q' be the charge on the inner shell. So t he
potential of outer shell is
62
j [email protected]
physicsguide CSIR NET I GATE
@Sk J ahiruddin , 2020
Electrostatics: Part-2
Since the the inner shell is grounded potential on the inner •
IS
0=
~-~ +q+ q'~
q' 41rEo
0=
q'
r1
q'
q
q
l
2
r2
q'
+ + r1 r 41rEo r r 2
I
41rEo
r2
2
r1
= --q r2
Sol 4.18. The magnetic field inside and on t he plane of t he superconductor is zero. Using the method of image charges
we replace the superconductor with an infinitely long wire parallel to the x -axis is kept at z = -d and carries a current I in the negative x direction . The magnetic field for t his current configuration at a point (x, y, z > 0) is ".
"
i X S-
11 oo : 02 : s4 parallel to the x -axis is kept at z = -d and carries a current I in the negative x direction . The magnetic field for t his current configuration at a point (x, y, z > 0) is ".
"
S+
i X
".
"
i X S-
S_
S+
3+ (z -
l
µof
3 + y k (z + d) 3- y k -----+----
21r
x (y
d) k)
3+ (z + d) k)
µ 0I 21r
i
y 2 + (z - d) 2
x (y
y 2 + (z
+ d) 2
-(z - d) y2
+ (z -
d) 2
y2
+ (z + d) 2
where s+ = y] + (z - d)k and s_ = y] + (z + d)k) is position vector of any point from the upper and lower wire [email protected]
@Sk J ahiruddin ) 2020
respectively.
63
physicsguide CSIR NET ) GATE
Electrostatics: Part-2
11 oo : 02 : s6
64
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@Sk J ahiruddin, 2020
5
Electrost atics: Part-2
Boundary Value Problems
Prob 5.1. Equipotential surface corresponding to a particular charge distribution are given by 2
2
2
4x + (y - 2) + z = ¼
E
, where the values of ½ are constants. The electric field at t he origin is [JAM 2011] (c) = 4i) (d) (a) = o (b) = 2x
E
E = -4iJ
E
E
11 oo : 02 : sg (a)
E=
E =-411
o
(b)
E
2x
(c)
E
4y
(d)
Prob 5.2. A spherical conductor of radius a is placed in a
E
uniform electric field = Eok. The potential at a point P(r, 0) for r > a, is given by ( r, 0) = constant - E 0 rcos0
+
E 0a 3
cos0
r2 where r is the distance of P from the centre O of the sphere and 0 is the angle OP makes with the z-axis. [GATE 2011] p
9
j [email protected]
-k
physicsguide CSIR NET, GATE
65
@Sk J ahiruddin, 2020
Electrostatics: Part-2
The charge density on the sphere at 0 = 30° is
(d)
EoEo 2
Prob 5.3. Consider a spherical shell with radius R such
t hat the potential on the surface of the shell in spherical coordinates is given by,
V(r
=
R , 0, )
=
2
Vocos 0
11 oo : 03 : 01 Prob 5.3. Consider a spherical shell with radius R such t hat the potential on the surface of the shell in spherical coordinates is given by, 2
V(r = R , 0, cp) = Vocos 0
where t he angle Bis shown in the figure.There are no charges except for those on the shell. The potential outside the shell at the point Pa distance 2R away from its centre C (see figure) is [TIFR 2017] I
I I
p
C
(a) V = ~0 (1 + cos 2 0) (b) V = 1t-(l + 2cos 2 0) (c) V = ~(l - cos 2 0) (d) V = ~(-2 cos 0 + cos 30) [email protected]
66
physicsguide CSIR NET, GATE
@Sk J ahiruddin, 2020
Electrostatics: Part-2
2
Prob 5.4. Solving Poisson's equation \/ cp = -Po/t=.0 for the electrostatic potent ial cp(x) in a region with a constant charge density p0 , two students find different answers, viz.
and The reason why these different solutions are both correct is because [TIFR 2014]
11 oo : 03 : 04 and The reason why these different solutions are both correct is because [TIFR 2014] (a) space is isotropic and hence x and y are physically equivalent . (b) we can add solutions of the Laplace's equation to both
c/J1(x) and c/J2(x) (c) the electrostatic energy is infinite for a constant charge density (d) t he boundary conditions are different in t he two cases.
Pro b 5. 5. For a scalar function cpsatisfying the Laplace equat ion, "\lcp has [GATE 2013] (a) zero curl and non-zero divergence (b) non-zero curl and zero divergence (c) zero curl and zero divergence (d) non-zero curl and non-zero divergence Prob 5.6. A charge distribution has the charge density given 67
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Electrost atics: P art-2
by p = Q{o(x-x 0 )- o(x+x 0 ) }. For this charge distribution t he electric field at (2x 0 ,0, 0) [GATE 2013]
(a)
2Qx
91rcox5
(b)
Qx
41rcox5
(c)
Qx
41rcox5
(d)
Qx
l 61rcoX6
Prob 5. 7. A static, spherically symmetric charge distribut ion is given by p(r) = 1e-Kr where A and K are positive
11 oo : 03 : 06 c r1c
(a)
2Q A 9n d, where d is t he spacing between the plates, is charged to a potential V and then disconnected from t he
charging circuit. If now the plates are slowly pulled apart (keeping them parallel) so that t heir separation is increased [TIFR 2013] from d to d t he work done will be 1
,
1rcor ( )
a
2v 2
2d 2v2
1rEor ( C) 2d
(l _ .4.) d' , ( 4_ _ d
l
)
(b)
2v 2 , 1rcor 4_ 2d d
(d) 1rEor
2v2
2d
.4. d'
Prob 1.3. A parallel plate capacitor with square plates of
11 oo :oo :oa 2d ( C)
1rEor
2
2d
V
2
( d' _
d
)
1
d) (
1rEor
2v2
2d
.!i d'
Prob 1.3. A parallel plate capacitor wit h square plates of [email protected]
@Sk J ahiruddin ) 2020
3
physicsguide CSIR NET ) GATE
Capacitance and Dielectrics
side l m separated by l micro meter is filled wit h a medium of dielectric constant of 10. If t he charges on t he two plates are lC and - l C, the voltage across t he capacitor is __ _kV. (up to two deceimal places)(Eo = 8.854 x 10- 12F/m) [GATE 2017]
Prob 1.4. A rectangular piece of dielect ric material is in-
serted part ially into t he (air) gap between t he plates of a parallel plate capacitor. The dielectric piece will [NET Dec 2017] (a) remain stationary where it is placed (b) be pushed out from the gap between the plates (c) be drawn inside the gap between the plates and its velocity does not change sign (d) execute an oscillatory motion in the region between t he plates
Prob 1.5. A parallel plate capacitor is formed by two circular conducting plates of radius a separated by a distance d, where d < < a . It is being slowly charged by a current t hat is nearly constant . At an instant when the current is
11 oo : oo : 11 Prob 1.5. A parallel plate capacitor is formed by two circular conducting plates of radius a separated by a distance d, where d < < a. It is being slowly charged by a current t hat is nearly constant. At an instant when t he current is l , the magnetic induction between the plates at a distance [email protected]
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4
@Sk J ahiruddin, 2020
C apacitance and Dielectrics
a/2 from t he centre of the plate, is
(a) µal
(b) µal
1ra
21ra
(c) µal a
[NET Dec 2016]
(d) µ0 1 41ra
Prob 1.6. Two conductors are embedded in a material of conductivity 10- 4 ohm-m and dielectric constant E = 80E 0 . The resistance between the two conductors is 106 ohm. What is the capacitance (in pF) between the two conductors? Ig[JEST 2018] nore the decimal part of the ans. (Next Two questions are linked) In a hydrogen atom, consider that t he electronic charge is uniformly distributed in a spherical volume of radius a(= 0, 5 x 10- 10m ) around the proton. The atom is placed in a 5 uniform electric field E = 30 x 10 V/m . Assume that the spherical distribution of the negative charge remains undistorted under the electric field. [GATE 2012]
Prob 1. 7. In t he equlibrium condition, the sepatration between t he positive and the negative charge centers is? (a) 8.66 x 10-16 m (b) 2.60 x 10- 15m
11 oo : oo : 13 Prob 1. 7. In t he equlibrium condition, the sepatration between t he posit ive and the negative charge centers is? 16
15
(a) 8.66 x 10- m (b) 2.60 x 10- m (c) 2.60 x 10-16m (d) 8.66 x 10- 15 m
Prob 1.8. The polarizability of t he hydrogen atom in unit j [email protected]
5
physicsguide CSIR NET, GATE
@Sk J ahiruddin , 2020
of (C
2
m/N)
C apacitance and Dielectrics
is
(a) 2.0 x 10- 40 m (b) 1.4 x 10-41 m (c) 1.4 x 10-40 m (d) 2.0 x 10-39 m
Prob 1.9. A conducting sphere of radius RA has a charge Q. It is surrounded by a dielectric spherical shell of inner radius RA and outer radius RB ( as shown in the figure below) having electrical p ermitivity c:(r ) = c: 0r [JAM 2006]
(a) Find t he surface bound charge density at r
= RA
(b) Find the total electrostatic energy stored in t he dielectric (region B)
11 oo : oo : 16 (a) Find t he surface bound charge density at r = RA (b) Find the total electrostatic energy stored in t he dielectric (region B) Prob 1.10. A dielectric sphere is placed in a uniform elec-
t ric field directed along the positive y axis. Which of the following represents t he correct equipotential surfaces . [GATE 2008]
[email protected]
6
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Capacitance and Dielectrics
@Sk J ahiruddin, 2020
(b)
(a)
.---.... --..
.. ----.
--------- ....._____ _ -·-----·
(d)
(c)
.........
--------
-.... --. --.
Prob 1.11. The space between two plates of a capacitor carrying charges +Q and - Q is filled with two different
materials, a shown in t he figure. Across t he interface of t he two dielectric materials, which one of t he folllowing statements is correct? [GATE 2015]
11 oo :oo :1a materials, a shown in t he figure. Across the interface of the two dielectric materials, which one of t he folllowing statements is correct? [GATE 2015]
•
-
•
• •
•
-
(a) E and D are cont inuous jahir@physicsguide. in
7
@Sk J a hiruddin , 2020
physicsguide CSIR NET , GATE
Capacitance and Dielectrics
(b) E is continuous and D is discontinuous (c) E- is discontinuous and D- is continuous (d) E and D are discontinuous. Pro b 1.12. Two infinitely extended homogeneous isotropic dielectric media (medium-I and medium-2 wit h dielect ric constant E1 / E2 = 2 and E2 / Eo = 5, respectively) meet at t he z = 0 plane as shown in t he figure. A uniform electric field exists everywhere. For z > 0, t he electric field is given by E 1 = 2i - 3j = 5k. The interface separating the two media is charge free. The electric displacement vector in t he [GATE 2012] medium-2 is given by. A
A
A
11 oo : oo : 21 arge ree. medium-2 is given by.
[GATE 2012]
medium- I
medium - 2
(a)
D2 =
z- 0
Eo[l02 + 15] + 10k]
(b)
10k] (c) D2 = Eo[42 - 6] + 10k] j [email protected]
(d)
D2 =
D 2 =
Eo [l02 - 15] +
Eo[42 + 6] + 10k]
physicsguide CSIR NET 1 GATE
8
Capacitance and Dielectrics
@Sk J ahiruddin, 2020
Prob 1.13. Consider an electromagnetic wave at the int erface between two homogeneous dielectric media of dielectric constants E1 and E2 . Assuming E2 > E1 , and no charge on the surface, t he electric field vector E and the displacement vector D in the two media satisfy t he following inequalities -I
-I
[NET June 2014]
E2 < IE1I and (c) IE2I < IE1 and (d) E2 > IE1I and D2 < D1 (b)
-I
-,,
-I
-I
-J
-,,
Prob 1.14. Suppose t he yz - plane forms a chargeless bound, , ,. ,. •
I
♦
♦
I
♦
11 oo : oo : 24 ~
~
~
-+
(c) IE2I < IEi and D2 > Di -+
-+
-+
-+
(d) E 2 > IEi l and D2 < Di Prob 1.14. Suppose t he yz- plane forms a chargeless bound-
ary between two media of permitivities Eze ft and Eriglit where Eze ft : Eriglit = 1:2. If t he uniform electric field on t he left -+ is E zeft = c( i + j + k) ( where c is a constant) , t hen t he _, electric field on t he right E right is [NET June 2015] (a) c(22 + 3+ k) (b) c(l + 23 + 2k) (c) c(l / 22+] + k) (d) c(2i + 1/ 2] + 1/ 2k) A
A
A
Prob 1.15. Suppose yz plane forms t he boundary between two linear dielectric media I and H with dielectric constant
3 and E2 = 4 respectively. If t he electric field in region 1 at t he int erface is given by E i = 4x + 3i) + 5z , t hen the Ei
=
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9
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Capacitance and Dielectrics
@Sk J ahiruddin , 2020
electric field E 2 at the interface in region 2 is [JEST 2016]
(a) 4x+3i)+5z (d) 3x
(b) 4x+3/ 4i)5/ 4z
(c) -3x +3i) +5z
+ 3i) + 5z
Prob 1.16. A ray of light inside Region 1 in the xy plane is incident at the semicircular boundary that carries no free
charges. The electric field at the point P (r 0 , 1r / 4) in plane _, polar coordinates is E i = 7e"-r - 3e4>, where er and e4> are the unit vectors. The emerging ray in Region 2 has the electric _, field E 2 parallel to x- axis. If Ei and E2 are the dielectric
11 oo : oo : 26 ary a carries no ree charges. The electric field at the point P (r 0 , 1r / 4) in plane -+ polar coordinates is E 1 = 7e"-r - 3e, where er and e are the unit vectors. The emerging ray in Region 2 has the electric -+ field E 2 parallel to x- axis. If E1 and E2 are the dielectric constants of Region 1 and Region 2 respectively then E1 / E2 is??? [GATE 2014]
r
•
0 1 - - - - + - - - - - - -... C
Region
c •. Region_
Pro b 1.1 7. The half space regions x > 0 and x < 0 are filled with dielectric media of dielectric constants E1 and E2 j [email protected]
physicsguide CSIR NET I GATE
10
Capacitance and Dielectrics
@Sk J ahiruddin, 2020
respectively. There is a uniform electric field in each part. In the right half, t he electric field makes an angle 01 to the interface. The corresponding angle 02 in t he left half satis[NET June 2016] fies
xO
11 oo : oo : 29 xO
-----------
sin 0 2 = E2 sin 01 (c) E1 tan 01 = E2 tan 0 2 (a)
E1
(b) E1 tan 02 = E2 tan 01 (d) E1 sin 01 = E2 sin 02
Prob 1.18. Two semi-infinite slabs A and B of dielectric constant EA and EB meet in a plane interface, as shown in t he figure below. [TIFR 2016]
[email protected]
11
physicsguide CSIR NET ) GATE
Capacitance and Dielectrics
@Sk J ahiruddin ) 2020
.-l B
-- - - _( - - - .
(a)
e.4 e,.
- -
. .. .
cos0A = EA /EBCOS 0B (b) sin0A = EA/EB Sin0B f~\ -i- ~-n ~ _ ; _ -i- ~-n f ;i \ ~:-n ~ _ ;_ ~:,~n
11 oo : oo : 31 .
(a) cos0A
- -
.
(b) sin0A
= EA/EECOS0E
(c) tan0A = EA/EE tan0E
..
= EA/EESin0E
(d) sin0A = EE/EA sin0E
Prob 1.19. A charge q is placed at the center of an other-
wise neutral dielectric sphere a and relative permitivity Er. 2 We denote the expression q/ 41rE0 r by E (r). Which of the following st at ements is false? [JEST 2013] (a) The electric field inside the sphere, r < a, is given by
E(r)/Er (b) The field outside the sphere r > a, is given by E(r) (c) The total charge inside a sphere of radius r > a is given by q (d) The total charge inside a sphere of radius r < a is given by q. Prob 1.20. A cylindrical rod of length L and radius r made of an inhomogeneous dielectric, is placed with its axis along
t he z direction with one end at the origin as shown below. [GATE 2009] 12
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Capacitance and Dielectrics
@Sk J ahiruddin, 2020
X
I I
r
. > - - t - _ _ , _ _ - - t - - t - --+
=
11 oo : oo : 34 X
I I
r
. , > - - l f - - ' - - - - t - - - - t - -- .
-
=
L
If t he rod carries a polarization P = (5z volume bound charge inside the dielectric is -
(a) Zero
(b) 101rr2 L
2
+ 7) k, A
the
(c) -51rr 2 L
Prob 1.21. A dielectric sphere of radius R carries a polarization P = kr 2f, where k is a constant and r is t he distance from t he center. Take t he system spherical polar coordinate.
-
[GATE 2006]
(i) The bound volume charge density inside t he sphere at a distance r from the center is 2
(a)-4kR
3
(b)-4kr (c)-4kr (d)-4kr (ii) The electric field inside t he sphere at a distance d from t he center is
(a) -kd2 f / Eo
(c) - kd 20/ Eo j [email protected]
@Sk J ahiruddin, 2020
(b) -kR 2 f / Eo
(d) -kR 2 0/ Eo 13
physicsguide CSIR NET, GATE
Capacitance and Dielectrics
Prob 1.22. A sphere of radius R carries a polarizat ion P = kr, where k is a constant and r is measured from t he center of the sphere.
[GATE 2007]
00: 00: 37 apacitance and Dielectrics
-I
Prob 1.22. A sphere of radius R carries a polarization P = kr, where k is a constant and f is measured from t he center
[GATE 2007]
of the sphere.
(i) The bound surface and volume charge densit ies are given, respectively, by
(a) -k f1 and 3k
(b) -k f1 and - 3k (c) k f1 and - 41rkR (d) -k f1 and 41rkR (ii) The electric field E at a point r outside the sphere is given by (a) 0 (b) kR(R2 - r2) f -I
Eor3
(c) k R (R
2
2
-
(d) 3k(r - R ) f
r )f
Eor5
Eor4
Prob 1.23. A spherical shell of inner and outer radii a and b, respectively, is made of a dielectric material with frozen polarization P (r) = whe1~e k is a constant and r is the distance from the its center. The electric field in the region
;r,
a< r < bis. k (a) E = - f Eor
(c)
E= o
[JEST 2015]
(b)
E= -
(d) E =
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@Sk J ahiruddin, 2020
k f Eor
k f
Eor2
14
physicsguide CSIR NET, GATE
Capacitance and Dielectrics
Prob 1.24. A long solid dielectric cylinder of radius a is
11 oo : oo : 39 @Sk J ahiruddin, 2020
Capacitance and Dielectrics
Prob 1.24. A long solid dielectric cylinder of radius a is permanently polarized ao that the polarization is every-
where radially outward, with a magnitude proportional to .... t he distance from the axis of the cylinder, i.e P = ½Porf. t he bound charge density in t he cylinder is given by [TIFR 2016]
(a) - Po
(b) Po
(c) - Po/2
(d) Po/2
Prob 1.25. An infinite conducting slab kept in a horizontal plane carries a uniform charge density er. Another infinite
slab of thickness t, made of a inner dielectric material of dielectric material of dielectric const ant k, is kept above the conducting slab. The bound charge density on t he upper [GATE 2016] surface of t he dielectric slab is (a) cr/2k (b) cr/k (c) cr(k - 2) / 2k (d) cr(k - l )/k Prob 1.26. Assume that z = 0 plane is t he interface between two linear and homogenous dielectrics (see figure).
The relative permitivities are Er = 5 for z > 0 and Er = 4 .... for z < 0. The electric field in the region z > 0 is E 1 = (3i - 5] + 4k)kV/m . If t here are no free charges on t he interface, the electric field in the region z < 0 is given by [JAM 2010]
[email protected]
@Sk J ahiruddin , 2020
15
physicsguide CSIR NET, GATE
Capacitance and Dielectrics
11 oo : oo : 42 15
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Capacitance and Dielectrics
@Sk J ahiruddin, 2020
s, - s
..--o -
E2 = (3/4i - 5/ 4] + k)kV/m (b) E2 = (3i - 5] + k)kV/m (c) E2 = (3i - 5] - 5k)kV/ m (d) E2 = (3i - 5] + 5k)kV/m (a)
Prob 1.27. A conducting spherical shell of radius shell of radius R 1 carries a total charge Q. A spherical layer of a linear , homogenous and isotropic dielectric constant K and outer radius R 2 ( R 1 ) covers the shell as shown in t he figure. [JAM 2010]
(a) Find t he electric field and polarization vector P inside the dielctric. From this P calculate t he surface bound charge density, j [email protected]
a b.
on the outer surface of t he dielectric 16
physicsguide CSIR NET, GATE
11 oo : oo :44
side the dielctric. From this P calculate the surface bound charge density,
on the outer surface of t he dielectric
a b.
j [email protected]
16
physicsguide CSIR NET, GATE
C apacitance and Dielectrics
@Sk J ahiruddin , 2020
layer and the volume bound charge density Pb, inside the dielectric. (b) Calculate t he electrostatic energy stored in the region
R1 < r
~
R2
Prob 1.28. a Two concentric, conducting spherical shells of raddi R 1 and R 2 (R 1 < R 2 ) are maintained at potent ials
V1 and ½, respectively. Find t he potential and electric field [JAM 2012] in t he region R 1 < r < R 2 .
b A polarized dielctric cube of side 1 is kept on x - y plane as shown. If t he polarization in the cube is P = kxx, where k is posit ivwe constant, then find all t he bound surface charge densities and volume charge density.
-
G\1---------,. F
••
•
B X
Prob 1.29. A conducting solid sphere of radius a, carrying a charge q is kept in a dielectric of dielectric constant k,
such that half of the sphere is surrounded by t he dielectric
11 oo:oo:47 Prob 1.29. A conducting solid sphere of radius a , carrying a charge q is kept in a dielectric of dielectric constant k,
such t hat half of the sphere is surrounded by t he dielectric [email protected]
17
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Capacitance and Dielectrics
@Sk J ahiruddin, 2020
as shown in the figure. Find t he surface charge densit ies in t he upper and lower hemispherical surfaces. [JAM 2013] q
Prob 1.30. In a parallel plate capacitor t he distance between the plates is 10 cm. T wo dielectric slabs of t hickness
5 m each and dielectric constant K 1 = 2 and K 2 = 4 respect ively, are inserted between t he plates. A potential of lOOV is applied across t he capacitor as shown in t he figure. The value of the net bound surface charge density at t he [JAM 2014] interface of t he two dielectric is
IOcrn
K2 = 4 K1 = 2
(a) -200to / 3
(b) - lOOto/ 3
-- I00 V
(c) - 250Eo
(d) -2000to/3
11 oo : oo : 49
(a) - 200Eo/3
(b) - 100Eo/ 3
(c) - 250Eo
(d) -2000Eo/3
Prob 1.31. MSQ: A unit cube made of a dielectric material has a polarization P = 3i + 4j units. The edges of -
[email protected]
@Sk J ahiruddin , 2020
A
18
A
physicsguide CSIR NET, GATE
Capacitance and Dielectrics
t he cube are parallel to t he cartesian axes. Which of the following statements are t rue? [JAM 2015] (a) The cube carries a volume bound charge of magnit ude 5 units. (b) There is a charge of magnitude 3 units on both t he surfaces parallel t o t he y - z plane. (c) There is a charge of magnit ude 4 units on both the surfaces parallel t o t he x - z plane. (d) There is a net non-zero induced charge on the cube. Prob 1.32. MSQ: Consider a spherical dielctric material of radius a centered at origin. If the polarization vector P =
-
P0e"'x, where Po is a constant of appropriate dimensions, t hen (e"'x, ey, ez ) are unit vectors in Cartesian coordinate sytstem. [JAM 2016] (a) The bound volume charge density is zero. (b) The bound surface charge density is zero at (0, 0, a) (c) The electric field is zero inside t he dielectric. (d) The sign of the surface charge density changes over the surface.
11 oo : oo : s2 (b) The bound surface charge density is zero at (0 , 0,a) (c) The electric field is zero inside t he dielectric. (d) The sign of the surface charge density changes over t he surface. Prob 1.33. T wo parallel plate capacitors, separated by dist ances x and 1. l x respectively, have a dielectric material
of dielectric constant 3. 0 inserted between t he plates and j [email protected]
19
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Capacitance and Dielectrics
@Sk J ahiruddin, 2020
are connected to a battery of voltage V . T he differences in charge on t he second capacitor compared to t he first is [NET June 2016] (a) 66% (b) 20% (c) - 3.3% (d) - 10% Prob 1.34. An atom of atomic number Z can be modeled
as a point posit ive charge surrounded by a rigid uniformly negatively charged solid sphere of radius R. The electric polarisability a of t his system is defined as a = P.J where PE is t he dipole moment induced on application of electric field E which is small compared to the binding elect ric field inside the atom. It follows t hat a = [TIFR 2018]
Prob 1.35. MSQ: A dielectric sphere of radius R has con= P0z so t hat t he field inside the stant polarization
sphere is
>
E iri
P
= - ~~ z T hen which of t he following is( are)
11 oo : oo : 55 Prob 1.35. MSQ: A dielectric sphere of radius R has constant polarization = P0z so that the field inside the
sphere is correct?
> E in
P
- £:~z Then which of the following is(are) [JAM 2017]
(a) The bound surface charge density is P0 cos 0 (b) The electric field at a distance r on t he z - axis varies as r~ for r > > R [email protected]
20
physicsguide CSIR NET, GATE
@Sk J ahiruddin, 2020
Capacitance and Dielectrics
(c) The electric potent ial at a distance 2R on the z-axis is PoR 12Eo
(d) The electric field outside is equivalent to that of a dipole at t he origin.
Prob 1.36. Two dielectric spheres of radius R are separated by a distance a such that a >> R. One of the spheres
(sphere 1) has a charge q and t he other is neutral. If the linear dimensions of the systems are scaled up by a factor two, by what factor should the charge on the sphere 1 be changed so that the force between t he two spheres remain unchanged? [JEST 2018]
(a) 2
(b) 4\1'2
(c) 4
(d) 2\1'2
Prob 1.37. A dielectric interface is formed by two homo-
11 oo :oo : s1 [JEST 2018]
unchanged?
(a) 2
(b) 4v'2
(c) 4
(d) 2v'2
Prob 1.37. A dielectric interface is formed by two homo-
geneous and isot ropic dielectrics 1 and 2 with dielectric constants 4/ 3 and 1 respectively and it carries no residual free charge. A linearly polarized electromagnetic wave is incident on t he interface from dielectric 1 at a point where the 1 10 ,_ unit normal to t he surface is ii = ( v 3i + k ) pointing into 2 t he dielect ric 1. The incident wave, which is incident from 1 into 2, just before it reaches t he interface, has electric vector j [email protected]
physicsguide CSIR NET I GATE
21
Capacitance and Dielectrics
@Sk J ahiruddin , 2020
[TIFR 2019] E.. .1
. t + y'Ei z = "E i o exp iw C
where E 0 is a real constant . The electric vector just after it er ..... from E 1 by an angle (a)
1r
/6
(c) sin-
(b) tan1
1
v'l9
2 19
1
2
v'3
5 3
(d) csc- 1 3
4 -19
Prob 1.38. During the charging of a capacitor C in a series R C circuit, the typical variations in t he magnitude of t he
charge q( t) deposited on one of the capacitor plates, and t he current i (t) in the circuit, respectively are best represented ,
11 oo : 01 : oo Prob 1.38. During the cha1~ging of a capacitor C in a series
RC circuit, the typical variations in t he magnit ude of t he charge q(t) deposit ed on one of the capacit or plates, and t he current i (t) in the cir·cuit, respectively are best represented by [JAM 2019]
[email protected]
22
physicsguide CSIR NET ) GATE
@Sk J ahiruddin ) 2020
Capacitance and Dielectrics
Fig. I
q
•
1
t
0
Fig. III
q
0
t
(a) Fig. I and Fig. II (c) Fig. III and Fig. II
0
Fig. II
t
Fig. IV
•
l
0
t
(b) Fig. I and Fig. IV (d) Fig. III and Fig. IV
P rob 1.39. A oarallel olat e ca.o acitor. with 1cm seoara.t ion
11 oo : 01 (a) Fig. I and Fig. II (c) Fig. III and Fig. II
: 02
(b) Fig. I and Fig. IV (d) Fig. III and Fig. IV
Prob 1.39. A parallel plate capacitor, with 1cm separation between t he plates, has two layers of dielectric with dielec-
t ric constants K = 2 and K = 4, as shown in the figure below. If a potential difference of lOV is applied between the plates, t he magnitude of the bound surface charge density (in units 2 of C/m ) at the junction of the dielectrics is [NET D ec 2019]
[email protected]
@Sk J ahiruddin, 2020
23
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Capacitance and Dielectrics
=2
1cm
(a) 250E0
• 4
(b) 2000c0 / 3
(c) 2000E 0
( d)
200c0 /3
Prob 1.40. Which one of the following relations determines
t he manner in which the electric field lines are refracted
11 oo : 01 : os (a) 250E0
(c) 2000E 0
(b) 2000E0 / 3
( d)
200E0 /3
Prob 1.40. Which one of the following relations determines
t he manner in which the electric field lines are refracted across t he interface between two dielectric media having dielectric const ants c 1 and e: 2 (see figure)? [GATE 2020]
(A) E1 sin 0 1 =
E 2 sin 02
(C) c 1 tan 0 1 = c 2 tan 0 2 j [email protected]
(B ) E1 cos 01
= E2 cos 02
(D ) e: 1 cot 0 1 = e: 2 cot 0 2
physicsguide CSIR NET, GATE
24
C apacitance and Dielectrics
@Sk J ahiruddin , 2020
Prob 1.41. What is t he charge stored on each capacitor C 1 and C 2 in t he circuit shown in the given figure? [JEST
2020] eo 2
l O
llf'
1 i,f
- - ,....------11.,_____. Cl
Cl
+ 11 V
lO
11 oo :01 : oa eo
)0
2 µF
--,-- - - - - 1 - Ct
Cl
+ 12 V
(A) 6µC , 6µC
(C) 3µC, 6µC
lO
(B) 6µC , 3µC (D) 3µC , 3µC
Prob 1.42. MSQ: A spherical dielectric shell with inner where radius a and outer radius b, has polarization P = -4r, r k is a constant and f is t he unit vector along the radial direction. Which of the fallowing statements is/ are correct? [JAM 2020] (A) The surface density of bound charges on the inner and outer surfaces are - k and + k respectively. The volume density of bound charges inside the dielectric is zero. (B) The surface density of bound charges is zero on both t he inner and outer surfaces. The volume density of bound charges inside t he dielectric is +k [email protected]
@Sk J ahiruddin , 2020
25
physicsguide CSIR NET, GATE
Capacitance and Dielectrics
(C) The surface density of bound charges on the inner and outer surfaces are :~ and -ffe respectively. The volume density of bound charges inside the dielectric is zero. (D) The surface density of bound charges is zero on both t he inner and outer surfaces. The volume density of bound charges inside t he dielectric is 41r(b~k_a3 )
11 oo : 01
: 1o
sity of bound ch arges inside the dielectric is zero. (D) The surface density of bound ch arges is zero on both t he inner and outer surfaces. The volume density of bound charges inside t he dielectric is
41r(b~k_a3 )
Prob 1.43. Using a battery, a lOpF capacitor is charged to 50V and then the battery is removed. After that, a second uncharged capacit or is connected to the first capacitor in parallel. If the final voltage across the second capacitor is 20V, its capacitance is - - - - - - - - - - pF? [JAM
2020]
Prob 1.44. A sinusoidal voltage of the form V (t) = V0 cos(wt) is applied across a p arallel plate capacitor placed in vacuum. Ignoring the edge effects, the induced emf within the region b etween the capacitor plates can b e expressed as a power series in w . The lowest non-vanishing exponent in w is - - - - - - - - - [GATE 2020]
jahir@physicsguide. in
26
@Sk J ahiruddin , 2020
1.1
physicsguide CSIR NET, GATE
Capacitance and Dielectrics
Ans keys
1.1. 48 nF
1.4. d
1. 7.
1.2. b
1.5. d
1.8. b
C
11 oo : 01 1.1
: 13
Ans keys
1.1. 48 nF
1.4. d
1. 7.
1.2. b
1.5. d
1.8. b
1.3. 11 .30
1.6. 007
1.9. (a)
Q2
RA - l Q
,,, , (b) 4
R~
/I
2
47rE0
l
1
1
3
R3B
R~
C
1.10. d
1.16. 2.35
1.22. (i) b, (ii) a
1.11.
C
1.17.
C
1.23. b
1.12. b
1.18.
C
1.24. a
1.13.
C
1.19. d
1.25. d
1.14.
C
1.20. d
1.26. d
1. 21. (i) b , (ii) a
1.15. d 1.27. (a)
E =
~ f 2 41r Ear Q
41rE0 r
j [email protected]
for R1
< r < R2
f for r > R2 2
27
physicsguide CSIR NET 1 GATE
Capacitance and Dielectrics
@Sk J a hiruddin , 2020
p =
(k-l ) Q f· k 41rr 2 '
O"l,
=
(k - 1) Q
.
.
-r-.'l ;
11 oo : 01
: 1s
@Sk J ahiruddin , 2020
p =
(k-l ) Q
41rr 2
k
Capacitance and Dielectrics
f· '
at t he outer surface
At the inner surfacce Pb =
_ _ 29 1 • • ab -
1 0 (b) _Q __ l - k2 41rE5R§
EOXeVo R '
Vo / R) , on northern hemisphere ( Eo Vo / R) (1 + Xe), on southern hemisphere ( Eo
1.30. a
1.35. a,c,d
1.40. d
1.31. b ,c
1.36. b
1.41. a
1.32. a,b ,d
1.37.
1.42. C
1.33. d
1.38. a
1.43. 15
1.34.
1.39. b
1.44. 2
C
[email protected]
@Sk J ahiruddin , 2020
1.2
solution
C
28
physicsguide CSIR NET, GATE
Capacitance and Dielectrics
11 oo :01
:1
a
@Sk J a hiruddin , 2020
1.2
Cap acitance and Dielectrics
solution
Sol 1.1. This is Jackson Chapter 1 problem no 7.
Two long, cylindrical conductors of radii a 1 and a 2 are parallel and separated by a distance d, which is large compared with either radius. Show that t he capacitance per unit length is given approximately by
C
d In a
~ 1fEo
-l
where a is the geometrical mean of the two radii. Use t his formula. Here d = 20 cm. a = 2 cm. Total capacitance approximately
C ~ 3.14 x 8.85 x 10-
12
x 2.3 x 750
We get approximately 48 x 10- 9 F
which does not match any ans. Sol 1.2. When t he capacitor is charged to a potential V and seperation between the plates is d, then , capacitance
C= EoA d j [email protected]
@Sk J a hiruddin , 2020
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physicsguide CSIR NET , GATE
Capacitance a nd Dielectrics
11 oo : 01 [email protected]
: 21
29
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@Sk J ahiruddin , 2020
Capacitance and Dielectrics
and, on the other hand, when t he seperation is d ', capacitance is C' = EoA
d'
. Now, when t he batte1·y is fully charged and t hen removed, no. of charges remain t he same on the capacitor, so,
Q connected = Q removed
Or, CV= C'V' Hence, V
V'
d
d'
or,
V' = ~d' d Now, energy stored in a capacitor, as we know,
Hence,work done to increase the seperation from d to d' is,
Or, W = !(CV)V' - ~(CV )V 2
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2
30
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11 oo : 01
: 23
Or,
W = !(CV)V' - ! (CV )V 2
2
j [email protected]
physicsguide CSIR NET 1 GATE
30
Capacitance and Dielectrics
@Sk J ahiruddin, 2020
Or,
w = 2 cv v - v ) 1
(
I
Hence
w = !cv 2 (d' 2
W
=
Eo1rr
2v
- 1)
d 2 d'
(- -
1)
d
2d
Hence option (c) is correct. Sol 1.3. The potential difference between the plates in the
presence of dielectrics
10
=
X
8.854
11.3
3
X
10
X
10-12
Sol 1.4. You have t o picturize t he problem. The field out-
side the capacitor go from positively charged plat e to negat ively charged plat e, and in t his processes t hese fields polarize the dielectric and induces negative bound surface charge on the upper surface of the dielectric. So the dielectrics
11 oo : 01
: 26
tively charged plate, and in t his processes t hese fields polarize the dielectric and induces negative bound surface charge on the upper surface of the dielectric. So the dielectrics
31
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Capacitance and Dielectrics
@Sk J ahiruddin , 2020
always gets pulled inside the capacitor, and executes an oscillatory motion. Sol 1.5. From Maxwell's equation _,
_, \7
X
_, aE B = µoEo at
The electirc field
E = a(t) Eo
Q(t) 1ra2Eo
8E
dQ 1ra2Eo dt 1
at aE 8t
I 1ra2 Eo A
Magnetic field is along cp . We take a circular loop of radius a / 2 between the plates. Hence
_, _,
B · dl
= µoEo
I
1r 2 1ra Eo 2
a I B21r- = µ oEo2 4Eo B = µo f Jlnrrr
a
2
11 oo :01 : 2a ... ... B · dl
=
µ oEo
I
1r 1ra Eo 2 I 2
a B21r- = µoEo2 4Eo B = µo f 41ra
Sol 1.6. 32
j [email protected]
physicsguide CSIR NET I GATE
Capacitance and Dielectrics
@Sk J ahiruddin , 2020
Sol 1.7. In equilibrium
E =
1 qd 41rc0 a 3
E
41rco x - = d q
1
5 9 x 109 ( · x lO
30
-lO 3
10
X
5
) 1.6 x 10- 19 = d 2.60
X
10- 16
=d
Sol 1.8. p=a.E
qd = a.E qd a.= E 40 a.= 1.4 X 10Sol 1. 9. T he electric field in t he region B
E= _ l_
Q 41rc0 r 3
The polarization
11 oo : 01
: 31
Sol 1.9. The electric field in the region B
E=
_ l_ Q 41rEo r 3
The polarization
-p = EoXeEf = Eo(r - l )E
33
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Capacitance and Dielectrics
@Sk J ahiruddin ) 2020
At the surface r = RA ab
-
= p ·f = Eo (r - l ) x
l Q 41rEo r 3
Q (RA - l) R3A 47f The electrostatic energy in region B
- -
1 D. EdT 2 l Q 2 47f R B Q - -2- - - r dr 2 RA 41rr 41rEo r 3 RB l Q2 - dr 3 81rEo RA r Q2 1 1
R2B Sol 1.10. None of the picture is correct. Electric field inside the linear dielectric will be proportional to the external
11 oo : 01
: 33
1
Q2
l61rE0 R~
Sol 1.10. None of the picture is correct. Electric field inside t he linear dielectric will be proportional to t he external electric field. So it will be uniform as well . But (D) is close. _,
_,
Sol 1.11. D only dep ends on the free charge, while E de_,
pends on the free charge as well as the medium. So, E is _, discontinuous and D is continuous.
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34
Capacitance and Dielectrics
@Sk J ahiruddin, 2020
Sol 1.12. In medium-I _,
D 1
= E1E1 = 2E0 (22 - 3] + 5k) -+
The z component of t he D is continuous, since there are no surface charge. Hence _,
D 2z
_,
= =
D 1z
10Eo _,
The parallel component of the E is continuous
Hence
11 oo : 01
: 36
Hence -+
(2) A
(2)A
+ E y J.) + D2z = 5c0 (22 - 3]) + 10c0 k = Eo[l02 - 15] + lOk]
D2 = E2(EX i
Sol 1.13. From previous problem we can write ---+
A
D 1 = E1(E xi
A
A
+ Eyj) + D zk an d
D1I = I Ei(E;, + E;) + D; I j [email protected]
35
physicsguide CSIR NET, GATE
Capacitance and Dielectrics
@Sk J ahiruddin , 2020
And ---+
A
D 2 = E2(E xi
D2 =
A
A
+ E yj) + Dzk and
E~(E; + E;) + D;II -+
-+
-+
-+
Since E2 > E1, E 2 < E1 and D 2 > D 1 Sol 1.14. E r'ight z
=
Eright
=
= Hence
l
D left z
Eleft E left z Eright
1 -c 2
11 oo : 01
: 39
=
Eleft E left
z
Eright
=
1 -c 2
Hence ~
A
E riglit
A
A
= c(l / 2i + j + k)
Sol 1.15. At t he boundary
nl
1
Dl2) =
)
=
ni
2
).
So
E1 Ei l )
= 12
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36
Capacitance and Dielectrics
@Sk J a hirudd in , 2020
Now
12
4
=3 Hence
Sol 1.16. The radial component of electric field .....
E1 · e"-r = E1 cos 01
11
00:01 :41
Sol 1.16. The radial component of electric field .....
E1 · er 7
=
E1 cos 01
J588 = cos01 3
J58 58
cos01 = tan0 1 =
3
7
And tan0 2 = 1 tan 01 tan 02
7
3
tan01 =
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7 3 physicsguide CSIR NET, GATE
37
@Sk J ahiruddin , 2020
Capacitance and Dielectrics
Sol 1.1 7. Since t here are no ch arge on the surface
D1 cos 01 = D2 cos 02 1 E 1 cos 01 1 E 2 cos 02
tan 02 E1 tan 01
E1
=
E2 t an 02
Sol 1.18. If the electric field in slab A makes an angle 0A with the normal to t he boundary and the electric field in slab B m akes an an~le 0 n with the same normal (see fi~ure )
11 oo : 01 : 4s
Sol 1.18. If the elect ric field in slab A makes an angle 0A with t he normal to t he boundary and t he electric field in
slab B makes an angle 0B with the same normal (see figure) t hen Sol 1.19. Just like the previous problem. tan0A = EA/EB t an0B Sol 1.20. (a), (b), (c) are basic. The t otal charge inside a sphere of radius r < a is also due to bound charges. Sol 1.21. Total bound charge inside the cylinder
-+
A
P · kdx dy 2
2
= -(5£ + 7)1rr + 7 2 2 = - 5L 1rr j [email protected]
physicsguide CSIR NET 1 GATE
38
Capacitance and Dielectrics
@Sk J ahiruddin, 2020
Sol 1.22. -+
-+
Pb= - V. p
= _ 1 8 ( 2k 2) r2 or r
r
= - 4kr Sol 1.23. Since t here are no free charge, D is zero inside
t he dielectric. Hence
11 oo : 01
: 47
= - 4kr Sol 1.23. Since there are no free charge, D is zero inside
t he dielectric. Hence _,
EoE
_,
+P = 0
_,
_,
p
E =-
Eo 2 = - kd f / Eo
Sol 1.24. (i) bound surface charge density _,
Clb
=f ·p = f · kRf =kR
bound volume charge density _,
_,
Pb= - \7 · p l a 8 (r 2 kr ) 2
=-r r = -3k [email protected]
39
physicsguide CSIR NET, GATE
Capacitance and Dielectrics
@Sk J ahiruddin, 2020 _,
_,
(ii) Outside the sphere , D is zero, and P is zero as well
Sol 1.25. Since there are no free charge, D is zero in the
region a < r < b. Hence _,
_,
r,
...
11 oo : 01 : so
Sol 1.25. Since there are no free charge, D is zero in the
region a < r < b. Hence -+
.....
EoE + P = 0 .....
...
p
E=-
Eo k -r A
Eor
Sol 1.26. bound volume charge density .....
.....
Pb = - V. p
1a 1 - - - r-P0 r r8r 2
= - Po Sol 1.27. We know for a dielectric .....
.....
D = EokE O"
=E
Eok
j [email protected]
40
physicsguide CSIR NET I GATE
Capacitance and Dielectrics
@Sk J ahiruddin, 2020
For linear dielectric .....
.....
P = Eo(k - l )E
(J"
= Eo(k - 1) k EQ
11 oo : 01 : s2 For linear dielectric
P = Eo(k -
l )E a = Eo(k - 1) Eok (k - 1) = - - -a k
(k -
1)
ab = - - -a k
⇒
Sol 1.28. Since t here are no free charges on t he interface, E zO E2
= 54 4
=5 Hence t he electric field in t he region z < 0 is
Sol 1 .29. If we consider a concentric spherical shell of radius shell of radius r, t he displacement vector at r 2
D 41rr = Q
D=
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41
Q 41rr 2 physicsguide CSIR NET ) GAT E
Capacitance and Dielectrics
@Sk J ahiruddin ) 2020
The electric field at r _,
-+
D
11 oo : 01
: 55
@Sk J ahiruddin , 2020
Capacitance and Dielectrics
The electric field at r -+
D E=-
coK
=
Q - - - -r 4c0 K 1r r 2 1
A
-+
and polarization vector P -+
-+
P = co(K - l )E (K - 1) Q = - - - -r 4K 1r r 2
A
bound surface charge density at R 1 -+
O"b
=f ·p A
(K - 1) Q
A
=r· - - - -r
4K1r Rt (K - 1) Q 4K1r Ri
bound surface charge density at R 2 -+
O"b
=f. p A
(K - 1) Q
A
=r · - - - - r
4K1r R~ (K - 1) Q 4K1r R~
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physicsguide CSIR NET, GATE
Capacitance and Dielectrics
11 oo : 01 : sa
@Sk J ahiruddin , 2020
Capacitance and Dielectrics
bound volume charge density .....
.....
Pb = - V. p
Q(K-1) --
r"
= - - - -V·
r2
4K1r
= 0 (for points away from origin) (b) electrostatic energy stored in t he region R 1 < r < R 2
1
1
Sol 1. 30. a The electric field in the region R 1 < r < R 2
E=-Q_l 41rEo r 2
Hence
Q
j [email protected]
(c)Sk J a,h ir11ddin. 2020
1
43
physicsguide CSIR NET, GATE
Ca,n aci ta.nee and Dielectrics
11 oo : 02 : oo 43
j ahir@physicsguide. in
physicsguide CSIR NET I GATE
Capacitance and Dielectrics
@Sk J ahiruddin , 2020
The potent ial at any point r
b bound surface charge density at ABCD -+
aABcn=x ·P
= x · kx =k bound surface charge density at OEFG aoEFG
=
-+
x·P
= x · kOx =0 bound volume charge density -+
-+
Pb= - v7 · p
= -k Sol 1.31. Since the sphere is made of conductor it's surface 1 is an equipotential surface, having potent ial V0 = 41rc 9.r . The 0
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11 oo : 02 : 03 Sol 1.31. Since the sphere is made of conductor it's surface 1 g_ . The is an equipotential surface, having potent ial V0 = 41rc r 0
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Cap acitance and Dielectrics
free surface charge density q O"J -
- 41ra 2 Vaca a
.....
The displacement vector D = 41rr q 2 • Hence t he electric field at a distance r from t he sphere .....
... D E =-
cok
Polarization vector
P = c0 (k - l )E bound surface charge density in t he dielectric .....
= - P .f
O"b
co(k - l)Vo a Hence t he total surface charge density
a=
O"j
+ O"b
Vaca
Eo(k - l )Vo
a
a
VoEok n
11 oo : 02 : 06 (J = (JJ + (Jb
VoEo
Eo(k - l )Vo
a
a
VoEok a [email protected]
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Capacitance and Dielectrics
@Sk Jahiruddin , 2020
Sol 1.32. Between the parallel plate capacitors D = (J. The potential difference between t he plates
V = (Jd
+
(Jd
2E1
2 E2
(Jd
1
1
2Eo
+ K1 K2
(Jd
1
1
2Eo 2 + 4 3(Jd
8Eo 8VE0
(J = - -
3d
The electric field is the region 1
E = :!_ E1
8VE0 3Eok1d
4000 3 Similarly electric field is the region 2
E = :!_ E2
8VEo 0
1
1
11 oo :02 : oa Similarly electric field is the region 2
E = !!_ E2
8Vco 3cok2d 2000 3 46
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Capacitance and Dielectrics
@Sk J ahiruddin, 2020
P1 = Eo(k1 - l )E1 4000Eo
3 hence
Similarly for region 2
Hence the total surface charge
2000Eo 3 Sol 1.33. bound volume charge density -+
-+
Pb= -V · P
=0 bound surface charge density at surfaces parallel to t he y - z
11 oo : 02 : 11 -+
-+
Pb = -V. p
=0 bound surface charge density at surfaces parallel to t he y- z plane
47
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Capacitance and Dielectrics
@Sk J ahiruddin, 2020
There is a charge of magnit ude 4 units on both t he surfaces parallel to the x - z plane -+
"'
a = P -j = 4
total charge Pb + a' s
= 0+ 3 - 3+4 - 4
Sol 1.34. bound volume charge density -+
-+
Pb= - V. p
=0 The bound surface charge density at (0, 0, a)
=0
11 oo : 02 : 13
=0 -I
The electric field is E
=
-
P .
Xe€o
t he surface charge density
-I
O'= P· f
= Pocos0
48
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@Sk J ahiruddin , 2020
Capacitance and Dielectrics
Sol 1.35.
X
Similarly Q
2
_
Qi
=
3Eo V X
=
l _ 1 1.1
-9.1
Sol 1.36. We have done this problem before. The answer is 41rE0 R 3 . Sol 1.37. The bound surface charge density is
11 oo : 02 : 16 Sol 1.36. We have done this problem before. The answer is 41rc0R 3 . Sol 1.37. The bound surface charge density is
= P0 cos0 Since t he electric field is due to dipole, it should vary as
49
j [email protected]
@Sk J ahiruddin,
1 r3
physicsguide CSIR NET 1 GATE
2020
Capacitance and Dielectrics
The potent ial due to dipole is at a distance 2R
p·r V = --41rco(2R)2 -
A
-
41rR3 P · f 3 41rE0(2R) 2
PoR 12Eo Sol 1.38. Sol 1.39. Sol 1.40. Using Kirchhoff's voltage law we can write for a
series RC circuit Vo= Rf +
Q 11
11 oo : 02 : 19 Sol 1.40. Using Kirchhoff's voltage law we can write for a
series RC circuit Q
Vo = RI + V Vo dQ 1 Q R = dt + RCV
d
Vo
t
t
= -R e Rc
dt QeRc
Vo
t
t
QeRC = R
t
eRC dt
O t
QeRc = Q(t ) =
t
VoC eRc VoC 1 Vo _ t
dQ I =- =- e dt R [email protected]
1
t e - RC
Re
physicsguide CSIR NET, GATE
50
@Sk J ahiruddin , 2020
Capacitance and Dielectrics
So option (a) Sol 1.41. Between the parallel plate capacitors D The potential difference between t he plates V =ad + ad 2E1
2E2
ad
1
-K1 + -K2
2Eo ad 1 1 2Eo 2 + 4 3ad 8Eo 8V Eo
1
= a.
11 oo : 02 : 21 ad 1 1 -+2Eo 2 4 3ad 8Eo 8VEo a = 3d The electric field is t he region 1
E = :!_ El
8VEo 3Eok1d 4000 3
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physicsguide CSIR NET I GATE
51
Capacitance and Dielectrics
@Sk J ahiruddin , 2020
Similarly electric field is the region 2
E = :!_ E2
8V E0 3Eok2d 2000 3
P1 = Eo(k1 - l )E1 4000Eo 3 tlPnr-P
11 oo : 02 : 23 3
P1 = Eo(k1 - l )E1 4Q00Eo
3 hence
Similarly for region 2 O'b2
= -
6Q00Eo
3
Hence t he total surface charge
2Q00Eo
3
Sol 1.42. We have done t his before. Option (D) Sol 1.43. jahir@physicsguide. in
52
physicsguide CSIR NET ) GATE
@Sk J ahiruddin ) 2020
Capacitance and Dielectrics
Sol 1.44. bound surface charge density at inner radius a O"a
=
-+
-f · P
-k a2
bound surface charge density at outer radius b -+
0-a
=f . p k 1 ')
11 oo : 02 : 26 a2
bound surface charge density at outer radius b
bound volume charge density
Pb= - V. p =0 Sol 1.45. The initial total charge on t he first capacitor
Q = C1V
= 50 X 10 = 500 t he final voltage across both capacitor is 20V. Then
+ Q2 500 = C1 V + C2 V 25 = 10 + C2 Q = Q1
C2 = 15 [email protected]
53
physicsguide CSIR NET, GATE
C apacitance and Dielectrics
@Sk J ahiruddin , 2020
Sol 1.46. the induced emf
£= a
at
-B · dS-
Now if the distance between t he parallel plate capacitor is x then R
= _ Vo cos(wt)
11 oo : 02 : 29 j [email protected]
physicsguide CSIR NET, GATE
53
C apacitance and Dielectrics
@Sk J ahiruddin , 2020
Sol 1.46. the induced emf _,
£= a
_,
B · dS
at
Now if t he distance between t he parallel plate capacitor is x then E =_ Vo cos(wt) X
From Maxwell's equation
_,
aE
curl B = µoEo at
= µoEow
Vo sin(wt ) X
So, B ex w sin(wt), so is £. Hence lowest non-vanishing ex3 ponent of w is in the expansion of w sin (wt ) = w (w - tiw + ...) is 2.
j [email protected]
54
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11 oo : 02 : 31 j [email protected]
physicsguide CSIR NET, GATE
53
C apacitance and Dielectrics
@Sk J ahiruddin , 2020
Sol 1.46. the induced emf _,
£= a
_,
B · dS
at
Now if t he distance between t he parallel plate capacitor is x then E =_ Vo cos(wt) X
From Maxwell's equation
_,
aE
curl B = µoEo at
= µoEow
Vo sin(wt ) X
So, B ex w sin(wt), so is £. Hence lowest non-vanishing ex3 ponent of w is in the expansion of w sin (wt ) = w (w - tiw + ...) is 2.
j [email protected]
54
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11 oo : oo : oo
Problellls and Solutions in Magnetostatics: Part-1 Sk J ahiruddin * Shreya Mitra Sandip Biswas
*Assistant Professor Sister Nibedita Govt. College, Kolkata Author was t he topper of IIT Bombay NI.Sc Physics 2009-2011 batch He ranked 007 in IIT J AM 2009 and 008 ( JRF) in CSIR NET J une 2011 He has been teaching CSIR NET aspirants since 2012
1
@Sk Jahiruddin, 2020
:Niagnetostatics: Part-1
11 oo : oo : 03 1
@Sk J ahiruddin , 2020
Magnet ost atics: P art-1
Contents 1
Lorentz Force, Bio Savart law, Ampere's Law etc 1.1
Ans Keys
1.2 Solutions .
j [email protected]
3
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•
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•
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•
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2
Physicsguide
11 oo : oo : 06
j [email protected]
Physicsguide
2
@Sk J ahiruddin , 2020
Magnetost atics: P art-1
Problems from NET, GATE, JEST, TIFR & JAM papers
1
Lorentz Force, Bio Savart law, Ampere's Law etc
Prob 1.1. A charged particle is released at time t
= 0,
from the origin in the presence of uniform static electric and magnetic fields given by E = E 0 y and B = B 0 z respectively. Which of t he following statements is true fort> O? [JEST 2015] (a) The particle moves along the x-axis. (b) The particle moves in a circular orbit. (c) The particle moves in t he (x,y) plane. (d) particle moves in the (y,z) plane. Prob 1.2. The value of the magnetic field required to maintain non-relativistic protons of energy 1 MeV in a circular orbit of radius 100mm is __________Tesla.
(Given : mp= 1.67 2014]
X
10- 27 kg, e = 1.6
X
10- 19 C ) [GATE
11 oo :oo :oa (Given : mp= 1.67
X
10- 27 kg, e = 1.6
X
10- 19 C ) [GATE
2014]
[email protected]
3
@Sk J ahiruddin, 2020
Physicsguide
Nl agnetostatics: P art-1
Prob 1.3. An alpha particle in accelerated in a cyclotron. It leaves the cyclotron with a kinetic energy of 16 MeV . The potential difference between the D electrodes is 50 kilovolts. The number of revolutions the alpha particle makes in its spiral path before it leaves the cyclot ron is ------------· [GATE 2016]
Prob 1.4. A charged particle in a uniform magnetic field = Boezstarts moving from the origin with velocity ef' = (3e"x + 2ez) m / s. The trajectory of the particle and the t ime t at which it reaches 2 met ers above the x-y plane are [JAM 2016] (e"x , e"y, and ezare unit vectors in Cartesian-coordinate system) (a) Helical path; t = l s (b) Helical path; t = 2/ 3s (c) Circular path; t = l s (d) Circular path ; t = 2/ 3s
B
B
Prob 1.5. A magnetic field = B 0 ('i + 2] - 4k) exists at point. If a t est charge moving with a velocity, ef' = v 0 ( 32 - 3+ 2k) experiences no force at a certain point , the electric field at that point in SI units is [JEST 2012] (a) = -voBo(3'i - 2] - 4k)
E \
11 oo : oo : 11 Prob 1.5. magnetic eld = Bo i + 2j - 4k exists at point. If a test charge moving wit h a velocity, -i}' = v0 ( 3i - + 2k) experiences no force at a certain point, t he electric field at that point in SI units is [JEST 2012]
3
(a) (b)
E = -voBo(3i - 23 - 4k) E = -voBo(i + 3+ 7k )
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4
Physicsguide
:Niagnet ost atics: Part-1
(c) E = voBo(l4J + 7k ) (d) E = -voBo(l43 + 7k ) Prob 1.6. A charged particle of mass m , charge q and constant velocity -;J enters a uniform magnetic field = B 0ex,(B 0 > 0) , at an angle 0to the direction of magnetic field . Find the angle 0,if in one revolution of t he helical motion, the particle advances along the direction of the magnetic field a distance equal to the radius of the helical path? [JAM 2013]
B
Prob 1.7. A charged particle moves in a helical path under t he influence of a constant magnet ic field. The initial velocity is such t hat t he component along the magnetic field is twice the component in the plane normal to t he magnetic field. The ratio 1 / R of the pitch 1 to t he radius R of t he [NET Dec 2014] helical path is
11 oo : oo : 14 [NET Dec 2014]
helical path is
j [email protected]
5
P hysicsguide
@Sk J ahiruddin , 2020
Magnetostatics: Part-1
J,..
To< (a) 1r/ 2
(b) 41r
(c) 21r
(d)
7r
Prob 1.8. A posit ively charged part icle, wit h a charge q,
E
enters a region in which t here is a uniform electric field and a unifrom magnetic field both directed parallel to t he positive y-axis. At t = 0, t he particle is at t he origin and has a speed v0 directed along the positive x axis. The orbit of the particle, projected on the x-z plane, is circle. Let T be t he time taken to complete one revolut ion of t his circle. The y-coordiante of the particle at t = T is given by [JAM 2015]
B,
11 oo : oo : 16 Let T be t he t ime taken t o complete one revolution of t his circle. The y-coordiante of the particle at t = T is given by [JAM 2015]
(d)
21rmvo
qB
Prob 1.9. A proton moves with a sp eed of 300m/s in a circular orbit in t he xy-plan in a magnetic field 1 tesla along [email protected]
P hysicsguide
6
@Sk J ahiruddin, 2020
Magnetostatics: Part-1
t he positive z-direction. When an electric field of 1 V / m is applied along the positive y-direction, t he centre of the [NET June 2015] circular orbit (a) remains stationary (b) moves at 1 m/s along the negative x-direction (c) moves at 1 m/ s along the positive z-direction (d) moves at 1 m/s along the positive x-direction
Prob 1.10. In t he laboratory frame two electrons are shot at each other with equal and opposite velocities ¼ and ½ respectively, but not along the same straight line, as shown [TIFR 2013] below z X
ii2 ◄ -- •
11 oo :oo :1a z X
ii2 • ◄ -- •
Each electron will be act ed on by the Coulomb repulsion due to the other, as well as t he Lorentz force due to its own motion in the magnetic field created by t he each ot her. Which of t he diagrams given below best describes the final velocities Vi and ½of these electrons? [You may assume t hat the electrons are distinguishable]
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Physicsguide
7
@Sk J ahiruddin , 2020
Magnetostatics: P art-1
z
...
.•
V1
_..,. X
z
X
• --◄... Vz
.....
Vz -•► •
• -◄-
X
X
-
Prob 1.11. If the dimensions of mass, length, t ime and charge are M,L,T , and C respectively, t he dimensions of is [JAM 2013] t he magnetic induction field (a) ML2r - 1 c-1 (b) Mr- 1c- 1
B
11 oo : oo : 21 Prob 1.11. If the dimensions of mass, length, t ime and charge are M,L,T , and C respectively, t he dimensions of is [JAM 2013] t he magnetic induction field
B
(a) ML2 r - 1 c- 1 (b) Mr- 1c- 1 (c) L 2r- 1c (d) L- 1r- 1 c Prob 1.12. In terms of the basic units of mass (NI), length (L), t ime(T ), and charge (Q) , the dimensions of t he mag[JAM 2007] netic permeability of vacuum ( µ 0 ) are
(a) M LQ- 2
(c) LTQ- 1
(b) ML2r- 1 Q- 2 (d) LT- 1Q- 1
Prob 1.13. The magnetic field at a distance R from a long j [email protected]
8
@Sk J ahiruddin , 2020
Physicsguide
Nlagnetostatics: Part-1
straight wire carrying a steady current I is proportional to [NET June 2012] 2 2 2 (a) IR (b) I/ R (c) 1 / R (d) I/ R Prob 1.14. The force between two long and parallel wires carrying currents 11and 12 and separated by a distance D is
proportional to (a) 1112/ D (b) (11+12)/ D 2 1112/ D
[NET Dec 2013] 2 ( d) (c) (1112/ D)
Prob 1.15. Two parallel infinitely long wires separated by a distance D carry steady currents 11 and 12 ( 11 > 12 ) flow-
ing in t he same direction . A positive point charge moves
11 oo : oo : 24 Prob 1.15. Two parallel infinitely long wires separated by a distance D carry steady currents 11 and 12 ( 11 > 12 ) flow-
ing in the same direction. A positive point charge moves between the wires parallel to the currents with a speed v at at distance D / 2 from eithe1~ wire. T he magnitude of an electric field t hat must be turned on to maintain the trajectory of t he part icle is proportional to
(a) (I1- I2)v/D (b) (I1 + I2)v/D 2 2 (d) (I1 + I2)v / D
[JEST Sample] 2 2 (c) (I 1-I2)v / D
Prob 1.16. In an ionization experiment conducted in t he laboratory, different singly-charged positive ions are pro-
duced and accelerated simultaneously using a uniform elect ric field along x-axis. If we need to determine the masses of
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9
Physicsguide
Magnetostatics: P art-1
various ions produced, which of t he following methods will NOT work [TIFR 2016] (a) Detect them at a fixed distance from the interaction point along x-axis and measure t heir time of arrival (b) Apply a uniform magnetic field along y-axis and measure the deviation (c) Apply a uniform electric field along y-axis and measure t he deviation (d) Apply a uniform electric field along y-axis and a (variale) uniform magnetic field along z-axis simulataniously and note t he zero deviation.
11 oo : oo : 26 (c) Apply a uniform electric field along y-axis and measure t he deviation (d) Apply a uniform electric field along y-axis and a (variale) uniform magnetic field along z-axis simulataniously and note t he zero deviation. Prob 1.1 7. Consider t hree ident ical infinite straight wires A,B and C arranged in par allel on a plane as shown in t he figure . [TIFR 2012]
X
I
I
I
-- d -
-
d-
'' • •
B
A
C
The wires carry equal currents I with directions as shown j [email protected]
10
@Sk J ahiruddin , 2020
Physicsguide
Magnetostatics: P art-1
in t he figure and have mass per unit length m. If the wires A and C are held fixed and t he wire B is displaced by a sm all distance x from its position, then it (B) will execute simple harmonic motion with a t ime period
(a) 21r
n;o (1)
(d) 21r
2:
0
(~)
(b) 21r
:;;i (1) (c) 21r
2
~
:; (
1)
(e) 21r : (~)
Prob 1.18. Consider a planar wire loop as an n -sided reg-
11 oo : oo : 29
Pro b 1.18. Consider a planar wire loop as an n -sided regular polygon , in which R is t he distance from the cent re to a vert ex. If a steady current I flows t hrough the wire, the magnit ude of t he magnetic field at t he centre of the loop is [NET June 2019] ( 21r) µof . ( a ) R sin -;:
2
(c)
i:~ tan( ~ )
Prob 1.19. A conducting wire is in t he shape of a regular hexagon , which is inscribed inside an imaginary circle of radius R , as shown. A current I flows through the wire. The magnit ude of t he magnetic field at the center of the circle [JAM 2015] is
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@Sk J ahiruddin, 2020
11
Physicsguide
Nl agnetostatics: P art-1
11 oo : oo : 31 I I c.
Prob 1.20. The strength of t he magnetic field at the cent re of a regular hexagon with sides of length a carrying a steady
current I is:
(a) µof
[JEST 2016] (b)
vl31ra
v16µol
(c) 3µol
1ra
1ra
(d)
-./3µo l 1ra
Prob 1.21. A current I is flowing through t he sides of an equilateral triangle of side a. The magnit ude of t he magnetic field at t he cent roid of the triangle is [JAM 2018] (a) 9µol (b) µo l (c) 3µol (d) 3µo l 21ra
1ra
21ra
1ra
Prob 1.22. Consider two infinitely long wires parallel to t he z-axis carrying t he same current I . One wire passes
t hrough t he point L with coordinates (-1 ,1) and the ot her t hrough M with coordinates (-1 ,1) in the XY planes shown [email protected]
12
@Sk J ahiruddin, 2020
P hysicsguide
:Niagnetostatics: Part-1
in the figure. The direction of t he current in both the wires [JAM 2009] is in the posit ive z-direction. "-(,
I
11 oo : oo : 34 in t e e 1rection o t e current 1n e [JAM 2009] is in the positive z-direction. '{ I
I
, I
B1
(a) Find t he value of p .d along the semi circular closed path of radius 2 units shown in the figure. (b) A third long wire carrying current I and also perpendicular to t he XYplane is placed at a point N with coordinates (x, 0) so that the magnetic field at the origin is doubled. Find x and t he direction of t he current in t he t hird wire? Prob 1.23. [MSQ]Three infinitely-long conductors carrying currents I 1 , I 2 and I 3 lie perpendicular to the plane of paper as shown in t he figure. If t he value of the integral p .d for the loops C1, C2 and C3 are 2µo , 4µoand µo in [JAM 2016] t he units of ~ respectively, then
B1
j [email protected]
@Sk J ahiruddin , 2020
13
Physicsguide
Magnetostatics: Part-1
00: 00: 37 :Nlagnetostatics: P art-1
C2
I
12.
•
c, .11
(a) 11 = 3A into the paper (b) I2 = 5A out of the paper (c) l 3 = 0 (d) 13 = lA out of the paper
Prob 1.24. A circular loop of fine wire of radius R carrying a current I is placed in a uniform magnetic field B p erpen-
dicular to the plane of the loop. If the breaking t ension of t he wire is Tb, the wire will break when the magnetic field [TIFR 2016] exceeds (d) µoTb Tb ( ) Tb ( ) µoTb ( ) a IR b 21rIR c 21rIR 41rIR Prob 1.25. A parallel plate capacitor is formed by two cir-
cular conducting plates of radius a separat ed by a distance [email protected]
@Sk J ahiruddin, 2020
14
Physicsguide
Magnetostatics: Part-1
d , where d < < a. It is being slowly charged by a current
11 oo : oo : 39 @Sk J ahiruddin, 2020
Nl agnetostatics: Part-1
d, where d < < a. It is being slowly charged by a current t hat is nearly constant . At an instant when t he current is I, t he magnet ic induction between the plates at a distance a/2 from the center of t he plate is [NET D ec 2016]
(b) ~~~
0 ;
(a) ~
(c)
µ~I
( d)
f~~
Prob 1.26. A set of N concent ric circular loops of wire each carrying a st eady current I in the same direction , is arranged in a plane. The radius of t he first loop is r 1 = a and the radius of t he nt h loop is given by r n = nrn - l · The magnit ude B of t he magnet ic field at t he centre of t he circles in t he limit N ➔ oo, is [NET June 2017] 2 (a) µoI (e - 1) (b) µol (e - 1) 4na na
(c) µo I (e
2
-
1)
8a
(d) µo I (e - 1) 2a
Prob 1.27. A const ant current I is flowing in a piece of wire that is bent into a loop as shown in t he figure [NET June 2017]
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@Sk J ahiruddin , 2020
15
P hysicsguide
Magnetostatics: P art-1
11 oo : oo : 42 15
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Physicsguide
@Sk J ahiruddin , 2020
:Niagnetostatics: Part-1
t
.
0-..
I,
I
?J..o..' ,
~b
I ..
lo
..
.
. )(
Prob 1.28. A system of two circular co-axial coils carrying equal currents I along same direction having equal radius R and separated by a distance R (as shown i the figure below) . The magnit ude of magnetic field at t he midpoint P is given by [JEST 2014]
~
'l
1
I
I
R
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16
Physicsguide
11 oo : oo :44
j [email protected]
Physicsguide
16
@Sk J ahiruddin , 2020
µal (a) 2v'2R
Magnetostatics: Part-1
(b) 4µo l
(d) 0
5v'5R
Prob 1.29. Consider a thin long insulator coated conducting wire carrying current I. It is now wound once around an insulating t hin disc of radius R to bring the wire back on t he same side,as shown in the figure.The magnetic field at t he cent er of the disc is equal to [JAM 201 7] I
Prob 1.30. A uniform surface current is flowing in t he positive y-direction over an infinite sheet lying in t he x-y plane. The direction of the m agnetic field is [GATE 2011] (a) along 1, for z > 0 and along -2for z < 0 (b) along k for z > 0 and along -k for z < 0 (c) along -2for z > 0 and along 1, for z < 0 (d) along -k for z > 0 and along k for z < 0 A
A
A
A
Prob 1.31. A conducting slab of copper PQRS is kept on t he xy plane in a uniform magnetic field along x-axis as in-
11 oo:oo:47 A
A
(d) along -k for z > 0 and along k for z < 0 Prob 1.31. A conducting slab of copper PQRS is kept on t he xy plane in a uniform magnetic field along x-axis as [email protected]
Physicsguide
17
@Sk J ahiruddin, 2020
Magnetostatics: Part-1
dicated in the figure. A steady current I flows t hrough the cross section of t he slab along the y axis. The direction of t he electric field inside t he slab, arising due to the applied [JAM 2014] magnetic field is along the
z
I
~ - + - - - ~- ----+ y B
X
(b) positive Y direction (a) negative Y direction (d) positive Z direction (c) negative Z direction Prob 1.32. A rectangular loop of dimension L and width
w moves wit h a constant velocity v away from an infinitely long straight wire carrying a current I in the plane of the loop as shown in the figure below. Let R be t he resistance of the loop. What is t he current in t he loop at the instant t he near side is at a dist ance r from t he wire? [JAM 2017]
11 oo : oo : so of the loop. What is t he current in t he loop at the instant t he near side is at a distance r from t he wire? [JAM 2017]
[email protected]
18
P hysicsguide
Magnetostatics: P art-1
@Sk J ahiruddin , 2020
fv
fw
R L
◄ ◄
µ 0 IL wv (a) 21rR r[r + 2w] µ 0 IL
wv
(c) 21rR r [r + w]
I
b µ 0 IL wv ( ) 21rR r[2r + w]
d µ 0 IL wv ( ) 21rR 2r [r + w]
Prob 1.33. The loop shown in t he figure below carries a
[NET June 2018]
steady current I .
0
11 oo : oo : s2 a
0
The magnitude of the magnetic field at the point O is (a) µof (b) µo f (a) µo f (d) µo f 2a 6a 4a 3a Prob 1.34. MSQ: In presence of a magnetic field B j and j [email protected]
19
Physicsguide
@Sk J ahiruddin , 2020
Nlagnetostatics: Part-1
"'
an electric field ( - E) k , a particle moves undeflected. Which of the following statements is (are) correct? [JAM 2018]
EA
(a) The particle has positive charge, velocity =
-Bj
(b) The particle has positive charge, velocity =
~3
E "' (c) The particle has negative charge, velocity= - Bj E "' (d) The particle has negative charge, velocity = B j
Prob 1.35. Two current-carrying circular loops, each of radius R, are placed perpendicular to each other, as shown in t he figure below.
The loop in the xy -plane carries a current I 0 while that in the x z -plane carries a current 210 . The resulting mag..... netic field B at t he origin is [NET Dec 2018]
z
y
11 oo : oo : 55 e or1g1n 1s
[2j + k] (d) µ2010 [-2j - k]
(a)
_µ2010
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20
Physicsguide
@Sk J a hiruddin , 2020
Magnetostatics: P art-1
Prob 1.36. Consider three straight, coplanar, parallel wires of infinite length where t he distance between adjacent wires
is d. Each wire carries a current J in the same direction. The p erpendicular distance from the middle wire (on either side) where the magnetic field vanishes is
(a) d )3
(b) 2d
3
(c) ~
[TIFR 2019]
(d) 2d
)3
3
Prob 1.37. Consider a hydrogenic atom in its ground state as conceived in Bohr's theory, where an electron of charge
-e is rotating about a central nucleus of charge + Z e in a circular orbit of radius a= 41rc 0 n2 / Ze 2 m. In this model, the magnetic field at a distance r from t he nucleus, p erpendicular to t he orbit, will be [TIFR 2019]
r
11 oo :oo : s1 [TIFR 2019]
ular to the orbit, will be
r a r2
1+
a
- 1/ 2
2
j [email protected]
41T"EQ
fi 5
2
1+
r 2 a
- 1
Physicsguide
Magnetostatics: P art-1
r2
1+
3 m2
21
@Sk J ahiruddin , 2020
(c)
z e2
a
2
-3/ 2
z e2
3 m2
41T"EQ
fi 5
(d)
r2
1+
a
2
Prob 1.38. A cyclotron can accelerate deuteron to 16 MeV . If the cyclotron is used to accelerate a -particles, what will be their energy? Take t he mass of deuteron to be twice the
mass of proton and mass of alpha particles to be fou1· times [JEST 2019] t he mass of proton. (a) 8 MeV (b) 16 MeV (a) 32 MeV (a) 64 MeV Prob 1.39. A wire with uniform line charge density ,\ per
unit length carries a current I as shown in the figure. Take t he permittivity and permeability of t he medium to be co = µ 0 = l. A particle of charge q . is at a distance r and is traveling along a trajectory parallel to the wire. What is
-3/
11 oo : 01 : oo unit length carries a current l as shown in the figure. Take t he permittivity and permeability of t he medium to be co = µ 0 = l. A particle of charge q . is at a distance r and is traveling along a trajectory parallel to the wire. What is [JEST 2019] t he speed of the charge?
I charge density = A current=
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r
22
Physicsguide
@Sk J ahiruddin, 2020
A (a) -
Nlagnetostatics: P art-1
A (c) 31
A (b) 21
l
(d) 4A l
Prob 1.40. Which one of the following is an impossible ...., magnetic field B ? [JAM 2019] ...., 2 2 (A) B = 3x z x - 2xz 3z
(B)
B=
(C)
B= ....,
- 2xyx + y z (xz
+ 4y) x
2
y+
- yx
3
2yz -
3
z3
z
y+
(D) B = -6xzx + 3yz y 2
Prob 1.41. Consider an annular region in free space containing a uniform magnetic field in the z -direction, schemat-
11 oo : 01
: 03
(D) B = -6xzx + 3yz y 2
Prob 1.41. Consider an annular region in free space containing a uniform magnetic field in the z -direction, schemat-
ically represented by the shaded region in the figure. A particle having charge Q and mass M starts off from point P (a, 0, 0) in the + x -direction with constant speed v . If t he radii of inner and outer circles are a and b, respectively, t he minimum magnetic field required so that the particle returns to the inner circle is [JAM 2020]
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23
P hysicsguide
@Sk J ahiruddin, 2020
:Niagnet ost atics: Part-1
y
p
, . . Mv
(
b2 - a 2 \
- l
,_ . Mv
X
(
b2 - a2\ -
l
11 oo : 01 : os
(A) Mv
b2 - a2 b
Q (C) M v Q
b2 - a2
3b
- 1
(B) Mv Q
-1
(D) Mv Q
b2 - a2
-1
2b
b2 - a2
-1
4b
Prob 1.42. A positively charged particle is placed at the origin (with zero initial velocity) in the presence of a con-
stant electric and a constant magnetic field along the posit ive z and x directions, respectively. At large times, the overall motion of the particle is adrift along the [NET Dec 2019] (a) positive y -direction (b) negative z -direction (c) positive z-direction (d) negative y -direction
j [email protected]
@Sk J ahiruddin, 2020
24
Physicsguide
Magnetostatics: Part-1
Prob 1.43. Consider three infinitely long, straight, and coplanar wires which are placed parallel to each other. T he dis-
tance between the adjacent wires is d. Each wire carries a current I in the same direction. Consider points on either side of the middle wire where t he magnetic field vanishes. What is the distance of these points from t he middle wire? [JEST 2020] (A) 2d (B) 2d (D) d 3 v'3 v'3
11 oo :01 : oa side of the middle wire where t he magnetic field vanishes. What is the distance of these points from t he middle wire? [JEST 2020] (A) 2d (B) 2d (D) d 3 v'3 v'3
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25
@ Sk J ahiruddin, 2020
1.1
Physicsguide
Magnetostatics: Part-1
Ans Keys 1.14. a
1.26. d
1.2. 1.44
1.15. a
1.27. b
1.3. 160
1.16.
1.28.
1.1.
C
C
C
11 oo : 01
: 11
1.2. 1.44
1.15. a
1.27. b
1.3. 160
1.16.
C
1.28.
1.4. a
1.17.
C
1.29. b
1.5. d
1.18. d
1.30. a
1.6. tan- 1 21r
1.19.
1.31. d
1.7. b
1.20. d
1.32.
1.8. b
1.21. a
1.33. b
1.9. d
1.22. (a) 2µ1r , (b) 1.34. b,d lm
1.10. a ,d
C
z '
1.35.
C
C
C
1.23. a,b 1.11. b
1.36. a 1.24. a
1.12. a
1.37. d 1.25. d
1.13. d
1.38.
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C
26
@Sk J ahiruddin , 2020
Physicsguide
Magnetostatics: P art-1
1.39. a
1.41. b
1.40. d
1.42. a
1.43. D
11 oo : 01 1.39. a
1.41. b
1.40. d
1.42. a
j [email protected]
@Sk J ahiruddin, 2020
1.2
: 13 1.43. D
27
Physicsguide
Nlagnetostatics: Part-1
Solutions
Sol 1.1. The particle will start moving in x direction first
11 oo : 01
: 16
@Sk J ahiruddin , 2020
1.2
Magnetostatics: Part-1
Solutions
Sol 1.1. The particle will start moving in x direction first because of Electric field. The magnetic field will give it a ..... force to q(v x B ) direction. So the magnetic force is towards
x direction. The particle will continue to move in xy plane, not in z as t he forces are in xy plane. Circular orbit is not possible due to t he presence of constant acceleration towards y direction because of electric field. Sol 1.2. For a charged particle in Magnetic field in circular
orbit
qvB
mv 2
= -r
Hence
1 2 1 mv = qv B r 2 2 Energy is 1 MeV. So the velocity is
2E V= m 2 X 1 X 106 X 1.6 X l0- 19 1.6 X 10- 27 7 = \l'2 X 10 Put the values 6
(1 X 10 )(1.6
X
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@Sk J ahiruddin, 2020
Hence
10-
19
)
19
7
= ~(1.6 X 10- )(,/2 X 10 )B (0.1 ) 28
Physicsguide
Magnetostatics: P art-1
11 oo :01
:1
a
@Sk J ahiruddin, 2020
Magnetost atics: Part-1
Hence
B = V2T Sol 1.3. In each revolut ions the a particle energy qV = 2eV 3
= 2 x 50 x 10 eV 5
= 10 eV , where e the charge of a proton. By the t ime it leaves the cyclot ron it has completed N
=
16MeV qV
16
X
106
105 = 160 number of revolutions.
B
Sol 1.4. The motion of the particle along i.e along e~ remains unchanged. And because of the velocity component perpendicular t o the magnetic field the particle will execute a circular mot ion with magnetic field providing the centripetal accceleration. Hence the trajectory of the part icle will be Helical. j [email protected]
@ Sk J ahiruddin, 2020
29
Physicsguide
Magnetostatics: P art-1
11 oo : 01 [email protected]
: 21
29
P hysicsguide
@Sk J ahiruddin , 2020
Magnetostatics: P art-1
Since t he z component of velocity remains unchanged, to t ravel 2 meters t he particle will take t ime
2
-2 = ls . Sol 1.5. The Lorentz force on the particle is zero. Hence ....
....
O=E +iJ x B
E = -vo(3i - 3+ 2k) x Bo(i + 23 = - vo B 0 (14] + 7k)
4k)
Sol 1.6. If T is the time for one revolut ion, t hen VzT Vz
=R
2
1rR = R
V > Ui
CUi
Magnetostatics: Part-2
(b)
Ug
=
CUi
(c)
Ug
O J-11 = 5 X 1Q-6H/m
_
K
= 100.x Alm
-B, =4.x - 6y+2z mT ~ ~ - - - - ~ ~ ---;-:::-::-;--::-:--.,.,..-..,.........,..""7
z =O
,.. X
-
,..
n
82 1-12 = IOx I0-6 Him
z< O 1. 20. Which of the following is not a correct boundary con-
dition at an interface between two homogeneous dielectric media? (In the following n is a unit vector normal to t he interface, a and Js are t he surface charge and current den[NET June 2019] sities, respectively.) (a) n x (D1 - D2) = 0 (b) n x (H1 - H2) = Js (c) n · (D 1 - D2) = a (d) n · (B1 - B2) = 0
j [email protected]
@Sk J ahiruddin, 2020
13
Physicsguide
Magnetostatics: Part-2
1. 21. The magnetic fields in tesla in the two regions separated by the z = 0 plane are given by Hi = 3x+ 5z and~ =
x+ 3y + 5z. The magnit ude of t he surface current density at
00: 00: 37 :Nlagnetostatics: Part-2
1.21. The magnetic fields in tesla in the two regions separated by the z = 0 plane are given by Hi = 3x+ 5z and~ =
x + 3y + 5z. The magnit ude of t he surface current density at t he interface between t he two regions is a x 106 A/m. Given 2 7 t he permeability of t he free space µ 0 = 41r x 10- N/A , t he value of a is - - - - - - - - - - (Round off to 2 decimal places) [JAM 2020 ]
1.22. Far from the Earth, the Earth's magnetic field can
be approximat ed as due to a bar magnet of magnetic pole strength 4 x 1014 Am. Assume this magnetic field is generated by a current carrying loop encircling t he magnetic equator. The current required to do so is about 4 x 10n A, where n is an integer. The value of n is - - - - - - - - [GATE 2020] (Earth's circumference: 4 x 107 m )
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@Sk J ahiruddin, 2020
1.1
Ans Keys
14
P hysicsguide
Magnetostatics: Part-2
11 oo : oo : 39 @Sk J ahiruddin, 2020
1.1
Nl agnet ost atics: Part-2
Ans Keys
1.1. do yourself
1.9. a
1.17.
C
1.2.
C
1.10. b
1.18.
C
1.3. d
1.11. b
1.19. 2
1.4. b
1.12. a
1.20. a
1.5. 4
1.13. d
1.21. 2.86 to 2.88
1.6. d
1.14. d
1.22. 7
1. 7.
C
1.15. b
1.8.
C
1.16.
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C
15
Physicsguide
Magnetost atics: Part-2
11 oo : oo : 42 15
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Physicsguide
@Sk J ahiruddin , 2020
1.2
:Niagnetostatics: Part-2
Solutions
1.1. For r
< r 1 using ,
pii.di
I enc
J1rr2 Hence,
H.21rr = J1rr 2 Hence, we get, H = Jr 2 Similarly for , r1r < r2 , J1rrr Hence,
pfi.Jz
I enc
Hence, H.21rr
H = Jrr 2r and sarnne in case of r > r 2 1.2. given ,
1 = Jork Or,
-H .dl- = Jo
d
rdrd > r) and is at a distance x(x > > r) above it . If the lower loop is held fixed and upper loop moves upwards with a uniform velocity v = ~; , then t he induced emf and t he direction of t he induced current in t his loop will be .
?
2 R 2v
.1 k .
x 4 ; antic oc wise 2 2 R v 2 2 (b) 2i µ 01r r x 4 ; clockwise 2
(a ) 3iµo'lr"-r
2 2 2
(c) 3iµ 0 7r r R
2x
2 2R
(d) 2iµ 01r r
~;
anticlockwise
2
v
x 3 ; clockwise
3
1.25. A rectangular metallic loop with sides L 1 and L 2 is
11 oo:oo:47 2x 3 ' 2 2R
(d) 2iµ 01r r
2
v
x 3 ; clockwise
3
1.25. A rectangular metallic loop with sides L 1 and L 2 is [email protected]
17
Physicsguide
@Sk J ahirudd in, 2020
Electromagnetic Induction
placed in the vertical plane, making an angle cpwith respect to the x axis, as shown in the figure, and a spatially uni-
B
form magnetic field = By is applied. The loop is free to rotate about the z axis (shown in the figure with a double
[TIFR 2017]
line)
2
B y
••
X
••
••
••
The magnetic field changes with t he time at a constant
~f
rate = k If t he resitance of the loop is R , the torque r required to prevent the loop from rotating will be 2 2 2 2 (a) -kB (b) kB ) sin 2cpz ) sin cp cos cpz
(L~i
(c) kB (L~i ) 2
2
sin cpz
(L~i
(d) - kB (L
i
1
2
2 )
sin cpz
1.26. A long solenoid is embedded in a conducting medium and is insulated from the medium. If the current t hrough t hP. solP.noi1
= - B ldx
And, induced emf e
=
de/> -dt
=
Blv
Again, in second case, At x = dx, c/>1 = B l (w - dx )
And x
= 4w, c/>2 = Blw
11 oo : 01 x
dx, c/>1
=
: 42
=
B l(w - dx )
And x = 4w , c/>2 = B lw
Thus chane in flux ,
and related induced ernf e
= - Blv 0 opt ion (c) is
Inside, Magnet ic field is uniform , so e correct .
1.32. As the magnet ic field is uniform, T here is no change in flux. [email protected]
P hysicsguide
37
@Sk J ahiruddin , 2020
Electromagnetic Induction
1.33.
V = IR + L dI dt P utting the values, we get, dI -dt = I A /s 1.34. For a long solenoid,
B inside
i
Hence, B
=
µ 0n K
tz
= Kt
= µonl z
11 oo : 01 1.34. For a long solenoid,
= µonI z
B inside
i Hence, B
: 44
=
Kt
= µ 0 n K tz
1.35. Bo = O.OlTr = 0.02m
t
-t
Bo -e o
to
R = 2, to = l
E.&=
e =
IR
0.01. (0.02)2 .w
=
I = w(0.01)(0.02) 2 Q = 6.28 j ahir@physicsguide. in
2
dQ dt
Physicsguide
38
@Sk J ahiruddin , 2020
Electromagnetic Induction
1.36. The flux t hrough the loop
=
-B · dxdyz L 2
= (30
3y tdx dy 0
= f3otL3 The emf
11 oo : 01
: 47 2
= f3o
3y tdxdy 0
= f3otL3 The emf £
= - a
at
=f3o L3 1. 3 7. Since t he loop is closed to t he solenoid , we can safely assume that the magnetic field through t he loop is also µo n l s(t) .
the induced emf in the wire loop £
= - a
at
dl s t 2
i(t) = _µonr dI.(t) R
dt
So, the in t he wire loop will at first take narrow and big spike t he a small and short spike. [email protected]
Physicsguide
39
@Sk J ahiruddin , 2020
Electromagnetic Induction
1.38. the induced emf
£
= _a
at
= Bvl 3 90 X 10 4 1 = .2 x 10- x x 2Tms3600 -
1
,.-yo
T/
11 oo : 01 E= -
: 49
ot
= Bvl = .2 x 10
-4
_1 90 X 103 x x 2Tms 3600
= lmV
j [email protected]
@Sk J ahiruddin , 2020
2
40
P hysicsguide
E lectromag11etic Induction
Magnetic Vector potential
2 .1. The vector potential in a region is given as
1 (X, y , Z) = - yi + 2x)
11 oo : 01 : s2 2
Magnetic Vector potential
2.1. The vector potential in a region is given as
1 (X, y, Z) = -yi + 2x J The associated magnetic induction is
(a) I+ k
(b)
3k
(c)
-i + 2]
B is
[GATE 2006] (d)
-i + 3+ k
2.2. A current I = lOA flows in an infinitely long wire along the axis of a hemisphere (see figure) . The value of =t ➔➔ J(V x B ) .ds over the hemispherical surface as shown in t he [JAM 2017] figure is:
I = 10 A
(a) 10 µo
(b) 5 µo
(c) 0
(d) 7. 5µo
2.3. Given that t he magnetic flux t hrough the closed loop
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@Sk J ahiruddin , 2020
PQRSP is cp. If
41
Physicsguide
Electromagnetic Induction
11 oo : 01 : s4 @Sk J ahiruddin, 2020
PQRSP is
Electromagnetic Induction
cp. If R
p
R
along PQR, the value of
J 1 .dl along PSR is
[GATE
p
2015]
Q
R
p
s
2.4. A rod of length L with uniform charge density .\per
unit length is in the xy plane and rotating about z-axis passing through one of its edge wit h an angular velocity cJ as shown in t he figure below. (f ,cp, z) refers to t he unit vect ors at Q, is the vector potent ial at a distance d from t he origin O along z-axis for d < < L and is current density due to the motion of the rod. Which one of t he following A
1
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1
42
Physicsguide
Electromagnetic Induction
rr-.A'T'R ?flfl~l
11 oo :01 : s1
@Sk J a hiruddin , 2020
Electromagnetic Induction
[GATE 2008]
statements is correct?
I
tl I
I
1 along f; 1 along z; 11 ex¼ (b) 1 along ~; 1 along ~; 11 ex J (c) J along f; 1 along z; 11 cxJ2 (d) 1 along J; 1 along J; 1 1ex! (a)
2
2.5. Which one of the following current densities , 1,can
generate the magnetic vector potential
2 (a) -(xi+yj) µo 2 -(xi - yj) µo A
A
A
2 (b) --(i+ j) µo A
A
A=
(y 2i + x 2]) ? [GATE 2009]
2 (c) -(i-j) µo A
A
(d)
A
j a [email protected]
(c)Sk J a,h ir11ddin. 2020
43
Physicsguide
Electromae:netic Induction
11 oo : 02 : oo 43
j [email protected]
P hysicsguide
@Sk J a hiruddin , 2020
Electromag11etic Induct ion
2.6. Which of t he following magnetic vector potent ials give rise to a uniform magnetic field B 0 k? [GATE 2016] A
(a) B 0 zk
(b) -B0 x]
(c) ,(- yi+x])
(d) ,(yi+x])
2. 7. The magnetic filed corresponding to the vector poten-
t ial
1 = !1 2
X
7 + lQ7 3 r
1 is a constant vector, is [NET June 2012] 3 3 (a) 1 (b) -1 (c) 1 + ~7 (d) 1- ~7 r r
where
A due to a magnetic moment rri 7 at a point ?is given by A = ~~ . If rriis directed along 2.8. The vector potential
positive z-axis, the x-component of t he magnetic field, at t he point 7, is [NET June 2012]
(a) 3m;z r
(b) _ 3myz r5
(c) 3m:z r
(d) 3m(z
2 r5
xy)
2.9. A thin infinitely long solenoid placed along the z-axis
contains a magnetic flux . Which of the following vector potentials corresponds to the magnetic field at an arbitrary point (x, y, z)? [NET June 2014]
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44
Physicsguide
11 oo : 02 : 02 _ _p_
y _p_ 21r x2 +y2 , 21r
X
x2+y2 ,
44
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@Sk J ahiruddin, 2020
(b) (Ax, Ay, Az) =
0
Electromagnetic Induction
_ _p_
_p_
y
X
21r x2+y2+z2, 21r x2+y2+z2, . 0 in free space ) t ravels t hrough an absorbing medium with di-
electric permit ivity given by E = ER + iEJ where Y... = v13. If ER t he skin depth is ~, t he ratio of the amplitude of the elect ric field E to t hat of t he magnetic field B, in the medium [NET June 2017] (in ohms) is (a) 1201r (b) 377 (c) 30v'21r (d) 301r Prob 1.10. A beam of light of frequency w is reflected from a dielectric metal interface at normal incidence. The refrac-
t ive index of the dielectric medium is n and t hat of t he metal is n 2 = n( l + ip). If the beam is polarised parallel to t he interface, t hen t he phase change experienced by the light ,...
..
.
r,
T...-,,......,
-Y
11 oo : oo : 16 a dielectric metal inter£ace at normal incidence. The refractive index of the dielectric medium is n and t hat of the metal is n 2 = n( l + ip). If the beam is polarised parallel to the interface, then the phase change experienced by the light [NET June 2014] upon reflection is (a) tan(2/ p) (b) tan- 1 (1/ p) (c) tan- 1 (2/ p)
(d) t an (2p)
Prob 1.11. An electromagnetically-shielded room is designed [email protected]
Physicsguide
6
@Sk J ahiruddin, 2020
ED Advanced: Part-1
7
so t hat at a frequency w = 10 rad/ s t he intensity of the external radiation that penetrates the room is 1% of t he in1 6 cident radiation. If a = 2~ x l0 (Dm)- is t he conductivity of t he shielding material, its minimum thickness should be (given t hat ln 10 = 2.3 ) [NET June 2014] (a) 4.60 mm (b) 2.30 mm (c) 0.23 mm (d) 0.46 mm
Prob 1.12. The magnetic field of t he T E 11 mode of a rectangular waveguide of dimensions a x b as shown in th figure is given by Hz = H 1cos(0.31fx)cos (0.41fy), where x and y are in cm. [NET June 2011]
11 oo : oo : 19
(i)The dimensions of the wave guide are (a) a = 3.33cm, b = 2.50cm (b) a = 0.40cm, b = 0.30cm [email protected]
7
@Sk J ahiruddin , 2020
Physicsguide
ED Advanced: Part-1
(c) a = 0.80cm, b = 0.60cm (d) a= 1.66cm, b = 1.25cm (ii) The entire range of frequencies f for which the T E 11 mode will propagate is (a) 6.0 GHz < f < 7.5GHz (b) 7. 5 GHz < f < 9.0GHz (c) 7.5 GHz < f < 12.0GHz (d) 7.5 GHz < f Prob 1.13. Consider a rectangular wave guide with transverse dimensions 2m x 1m driven with an angular frequency w
=
109rad/ s . Which transverse electric (TE) modes will
propagate in this wave guide? (a) TE10, TE01, and TE20 (c) T E o1 , T E10 , and T E11
[NET June 2015]
(b) TE10, TE11, and TE20 ( d) T Eo1, T E10, and T E 22
11 oo : oo : 21 propaga e 1n
1s wave gu1
(a) T E10, T Eo1, and T E 20 (c) TE01,TE10, and TE11
(b) T E10, T E11 , and T E 20 (d) T Eo1 , T E10, and T E 22
Prob 1.14. A waveguide has a square cross section of side
2a. For TM modes of wavevector k, the transverse electromagnetic modes are obtained in terms of a function 'l/J (x , y) which obeys the equation 32 32 w2 8 2 + 8 2 + ( 2 - k2) 'lf; (x,y) = 0 y
X
C
with t he boundary condition 'l/J (±a, y) = 'lf; (x, ± a) = 0. The frequency w of the lowest mode is given by [NET June j [email protected]
Physicsguide
8
@Sk J ahiruddin, 2020
ED Advanced: P art-1
2016]
(a) w2
= c2(k2 + ~2)
2 2 2 (c) w = c ( k +
;:2 )
2 2 2 (b) w = c (k + :: )
(d)
w
2
= c (k + ;:2) 2
2
Prob 1.15. The output intensity I of radiation from a sindJ wo gle mode of resonant cavity obeys dt = - Q I where Q is
t he quality factor of the cavity w 0 is the resonant frequency. The form of t he frequency spectrum of t he output is [JEST 2016] (a) Delta function (c) Lorent zian
(b) Gaussian (d) Exponential
11 oo : oo : 24 2016]
(a) Delta function (c) Lorent zian
(b) Gaussian (d) Exponential
Prob 1.16. The characteristic impedance of a co-axial ca[TIFR 2018] ble is independent of the
(a) dielectric medium between t he core and t he outer mesh (c) outer diameter (b) length of the cable (d) core diameter Prob 1.17. Consider a dipole antenna with length l, charge q and frequency w. The power emitted by the antenna at a large distance r is P. Now suppose the length l is increased
to v12l , the charge is increased to -/3q and the frequency is increased to ybw. By what factor is t he radiated power [email protected]
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increased ?
Physicsguide
9
ED Advanced: Part-1
[TIFR 2018]
Prob 1.18. A hollow waveguide supports transverse elec1 t ric (TE) modes wit h t he dispersion relation k = - Jw 2 - w~in, C where Wmn is t he mode frequency. The speed of t he flow of
electromagnetic energy at t he mode frequency is June 2018] (a) C (b) W mn / k (C) 0 (d) oo
[NET
Prob 1.19. The permittivity tensor of a uniaxial anisotropic
11 oo : oo : 21 electromagnetic energy at the mode frequency is June 2018] (a) c (b) Wmn/k (c) 0 (d) oo
[NET
Prob 1.19. The permittivity t ensor of a uniaxial anisotropic 4£0 0 0 medium, in the standard Cartesian basis, is O 4£0 0 0 0 9£o where £o is a constant.The wave number of an electromagnetic plane wave polarized along t he x -direction, and propagating along the y -direction in t his medium (in terms of t he wave number k0 of t he wave in vacuum is [NET June 2019]
(a) 4co
(b) 2co
(c) 9co
i
(d) 3co
Prob 1.20. A metallic wave guide of square cross-section of side L is excited by an electromagnetic wave of wavenumber k . The group velocity of the T E 11 mode is [NET j [email protected]
10
Physicsguide
@Sk J ahiruddin , 2020
ED Advanced: Part-1
Dec 2019] (a) ckL / ✓~k2_L_2 _+_n_2
(C) C ✓k2 £ 2 kL
7r2
(b)
C
kL
✓k2 £2- 21r2
(d) ckL/ ✓k 2 L2 + 2 1r2
Prob 1.21. A medium (er > 1, µr = 1, a> 0) is semi-transparer t o an electromagnetic wave when [GATE 2020] (a) Conduction current >>. Displacement current
11 oo : oo : 29 Prob 1.21. A medium (Er > 1, µr
= 1, a> 0) is semi-transparer
to an electromagnetic wave when [GATE 2020] (a) Conduction current >>. Displacement current (b) Conduction current
> which Elives.
EW
00: 00: 37
Now, For metals,
a>>
EW
which gives, K,
Hence, skin depth, d ex
µaw
=
2
a~
Hence, option (b) is correct.
Sol 1. 7. The electric field in the region a < r < b
E=
1 qr 41rEo r 2
Then the potential difference between the inner and outer sphere 1
1
1
Vo= --q - - 41rEo a b 1 b -a
=
q
41rEo ab 41rE0 Voab q= b- a
Total current I flowing from inner to outer sphere
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P hysicsguide
ED Advanced: Part-1
11 oo : oo : 39 @Sk J ahiruddin, 2020
ED Advanced: Part-1
I=
.....
J · da
a(r)E. da 1 A .!l_b2 f · drlf b2 41rEo b2
A 41rq b2 41rEo Aq 1 Eo b2 A 41rEo Voab b2 E0 b - a a
= 41rAb(b - a) Vo Vo=b(b - a) 1
41rAa
.
.
.
. b b- a
t Sol 1.8. We know, P1(t) = P1(0)e t ion, we get,
T
Here from t he ques-
t
which on calculation gives, t = 10s. Hence, option (d) is [email protected]
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15
Physicsguide
ED Advanced: Part-1
11 oo : oo : 43 15
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@Sk J ahiruddin , 2020
ED Advanced: Part-1
correct.
Sol 1.9. Given,
E = ER+ iE1
a
and, f.R ~ =
v'3.
Hence,
= -/3
EW
Again, skin depth,
5 = Ao 41r (Given) Putting the expression for skin depth, we can get
And, Wave-vector K is given by, K
✓(k2
+ A: 2 )
which
•
gives,
k = wfaµ[ Hence,
E0 w 1 =V= - = - Bo k flqi, Ea c Ho= 4µ 0
c
= -
4
which on calculation gives the answer 301r Hence option (d ) is correct.
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11 oo : oo : 4s IS
correc .
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16
ED Advanced: Part-I
@Sk J ahiruddin , 2020
Sol 1.10. Parallel Incidence, so,
Where,
Hence, path difference,
Hence,
or, 2 0 = arctan (-) p
Hence, option (c) is correct. Sol 1.11. As we know, intensity decays by the formula,
2z I= I 0 e 6 Hence, I
- = lo
2z
-e 6
11 oo:oo:47 e
Hence,
2z
l
- =e lo
[email protected]
..
-
at
= -ax
which on calculation gives,
>.. = - atx Hence,
4
A
=
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➔
A
A -
ati P hysicsguide
37
@Sk J ahiruddin , 2020
ED Advanced: Part-1
Hence, opt ion (d) is correct. Sol 2.17. Here, From, Gauge Transformation ,
(- \7 >..) Hence,
1' = 1 +
J = q> + 9af Hence, option (a) is correct. 1
B
1,
1
x3,
Sol 2.18. If we try, = v7 x We will get , = B0 which satisfies, Coulomb-Gauge. Hence, opt ion (c) is cor-
rect. Sol 2.19. We know,
B=
v7 x
1.
which on calculation
11 oo : 01
: 44 A
- ati j [email protected]
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37
ED Advanced: Part-1
@Sk J ahiruddin, 2020
Hence, option (d) is correct. Sol 2.17. Here, From, Gauge Transformation,
(- v7 A) Hence,
J = q; + i 1
A' = A+
Hence, option (a) is correct.
A,
1J
A
x3,
Sol 2.18. If we try, = v7 x We will get , = B0 which satisfies, Coulomb-Gauge. Hence, option (c) is cor-
rect. Sol 2.19. We know,
t ion,
1J
= v7 x
A, which on calculation
tkµ 0 A 0c Hence, option (b) is correct -
Sol 2.20. Given, A
Gauge,
=
1 -
B x
2
x Now,
we know,Coulomb
v7.A = o Calculating, we can prove t his, i.e., valid in coulomb Gauge. If option (b) is correct.
j [email protected]
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11 oo : 01
: 47 A
- ati j [email protected]
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37
ED Advanced: Part-1
@Sk J ahiruddin, 2020
Hence, option (d) is correct. Sol 2.17. Here, From, Gauge Transformation,
(- v7 A) Hence,
J = q; + i 1
A' = A+
Hence, option (a) is correct.
A,
1J
A
x3,
Sol 2.18. If we try, = v7 x We will get , = B0 which satisfies, Coulomb-Gauge. Hence, option (c) is cor-
rect. Sol 2.19. We know,
t ion,
1J
= v7 x
A, which on calculation
tkµ 0 A 0c Hence, option (b) is correct -
Sol 2.20. Given, A
Gauge,
=
1 -
B x
2
x Now,
we know,Coulomb
v7.A = o Calculating, we can prove t his, i.e., valid in coulomb Gauge. If option (b) is correct.
j [email protected]
38
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11 oo : 01
: 49 A
- ati j [email protected]
Physicsguide
37
ED Advanced: Part-1
@Sk J ahiruddin, 2020
Hence, option (d) is correct. Sol 2.17. Here, From, Gauge Transformation,
(- v7 A) Hence,
J = q; + i 1
A' = A+
Hence, option (a) is correct.
A,
1J
A
x3,
Sol 2.18. If we try, = v7 x We will get , = B0 which satisfies, Coulomb-Gauge. Hence, option (c) is cor-
rect. Sol 2.19. We know,
t ion,
1J
= v7 x
A, which on calculation
tkµ 0 A 0c Hence, option (b) is correct -
Sol 2.20. Given, A
Gauge,
=
1 -
B x
2
x Now,
we know,Coulomb
v7.A = o Calculating, we can prove t his, i.e., valid in coulomb Gauge. If option (b) is correct.
j [email protected]
38
Physicsguide
11 oo : oo : 01
Problellls and Solutions EM Wave
• Ill
Sk J ahiruddin * Shreya Mitra Sandip Biswas
*Assistant Professor Sister Nibedita Govt. College, Kolkata Author was t he topper of IIT Bombay NI.Sc P hysics 2009-2011 batch He ranked 007 in IIT J AM 2009 and 008 ( JRF) in CSIR NET J une 2011 He has been teaching CSIR NET aspirants since 2012
1
@Sk J ahiruddin, 2020
E:NI Wave
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@Sk J ahiruddin, 2020
EM Wave
Contents 1
Electromagnetic Wave 1.1
Ans Keys
1.2 solution .
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Physicsguide
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E:Nl Wave
Problems from NET, GATE, JEST, TIFR & JAM papers
1
Electromagnetic Wave
Pro b 1.1. The electric field of an electromagnetic wave prop-
agating t hrough vacuum is given by
E(7, t) = E 0 zcos(l00v'31rx -
[JAM 2009]
l001ry - wt )
➔
(a) What is t he wave vector k? Hence find t he value of w.
(b) At t he t ime t = 0 t here is a point charge q wit h velocity ef' = v 0 x at t he origin. What is t he instantanious Lorentz force acting on t he part icle?
Prob 1.2. The magnetic field associat ed wit h t he electric field vect or£ = E 0 sin(k z - wt)] is given by [JAM 2010]
(a) (c)
B = -~ sin(kz-wt)l B = ~ sin(kz - wt)] 0
B B
(b) = ~0 sin(kz-wt)i (d) = ~ sin(kz - wt)k
11 oo :oo :og ro . . e magne IC e. . ~ associa e WI u...... e e ectric field vector £= E 0 sin(kz - wt)] is given by [JAM 2010]
B= B=
(a) (c)
(b) B = ~ 0 sin(kz-wt)i (d) B = ~ sin(kz - wt)k
sin(kz-wt)i ~ sin(kz - wt)] -~0
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EM Wave
Prob 1.3. A plane polarized electromagnetic wave in free space at time t=O is_iven by E (x,y) = IO]exp[i (6 x+8z)] .
(x, z, t) is given by [GATE 2012] (a) B(x , z , t) = ¼(6k - 8i ) exp [i(6x + 8z - lOct)]
The magnetic field
B(x, z , t) = ~(6k + 8i) exp [i(6x + 8z - l Oct)] (c) B(x , z, t ) = ¼(6k - 8i) exp [i(6x + 8z - et)] (d) B(x, z, t ) = ~(6k + 8i) exp [i(6x + 8z + et)] (b)
Prob 1.4. T he electric filed of an electromagnetic wave is [NET Dec 2013] given by
E=
E 0cos[1r(0.3x
+ 0.4y -
The associated magnetic field
l OOOt)]k
B is
+ 0.4y - l OOOt)]k (b) 10- 4 E 0 cos[1r(0.3x + 0.4y - 1000t)](4i - 3]) (c) E 0 cos[1r(0.3x + 0.4y - 1000t)](0.3i + 0.4]) 2 (d) 10 E 0 cos[1r(0.3x + 0.4y - 1000t)](3i + 4]) (a) 10- 3 E 0 cos[1r(0.3x
Pro b 1. 5. An electromagnetic wave of frequency w t ravels in t h x-direction through vacuum. It is oolarized in the v-
11 oo : oo : 11 2
(d)10 E 0 cos[1r(0.3x + 0.4y - 1000t)](3l + 43)
Pro b 1. 5. An electromagnetic wave of frequency w travels
in t h x-direction through vacuum. It is polarized in the ydirection and t he amplitude of the electric field is E 0 . With [email protected]
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4
ENI Wave
@Sk J ahiruddin , 2020
k = w / c where c is the speed of light in vacuum, the electric and magnetic fields are then conventionally given by [JEST 2013] (a) = E 0 cos(ky - wt) x andB = ~ cos(ky - wt) z (b) £ = E 0 cos(kx - wt)y and = ~ 0 cos(kx - wt) z (c)E = E 0 cos(kx -wt) z and = ~cos(ky-wt)y (d) £ = E 0 cos(kx - wt)x and = ~ cos(ky - wt)y
E
B B B
Prob 1.6. Consider two electromagnetic plane waves propagating in vacuum wit h their electric field vectors 1 = E 0 cos(kz - wt)i and
E2 = E 0 cos(kz + wt)i
£
[JAM 2005] (a) Evaluate the magnetic field vector corresponding to the
superposition of these two waves.
(b) Calculate the t ime averaged energy density as well as t he t ime averaged pointing vector for the resultant wave(The t ime averaged is carried over one period of oscillations)
Prob 1. 7. The electric field of uniform plane wave prop-
11 oo : oo : 14 t he t ime averaged pointing vector for the resultant wave(The t ime averaged is carried over one period of oscillations)
Prob 1. 7. The electric field of uniform plane wave propa ating in a dielectric non-conducting medium is given by
= x lOcos(61r x 107 t - 0.41rz)V /m The phase velocity of t he wave is ____ l0 8 m/s
[GATE 2014]
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Prob 1.8. A monochromatic plane wave at oblique inciA
A
dence undergoes reflection at a dielectric interface. If ki, kr and n are the unit vectors in the directions of the incident wave, reflected wave and normal to t he surface respectively, which one of t he following expressions is correct? [GATE 2013] (a) (ki - kr) x n # 0 (b) ( ki - kr) .n = 0 A
A
A
A
A
A
(c) (ki x n) .kr = 0 Prob 1.9. A light source has a small filament at t he center
of a spherical glass bulb of radius 5cm and negligible thickness. If this source emits lOOWatts of power in the form of spherical electromagnetic waves, the r .m.s electric field E at t he surface of the bulb(in unit of Volt/ m) will be approximately [TIFR 2015] (a) 1094 (b) 109.4 (c) 15.47 (d) 1547 'II
-
...
_
........
,.
'
,
.
'
,
I"
, ,
,
.
•
11 oo : oo : 16 sp er1ca e ectrom agne 1c waves,
. .
at
t he surface of the bulb(in unit of Volt/m ) will be approxi[TIFR 2015] m ately
(a) 1094
(b) 109.4
(c) 15.47
(d ) 1547
Prob 1.10. Measurement of t he m agnitudes of the electric field ( E ) and t he m agnetic field (B ) in a plane-pola1·ized electromagnetic wave in vacuum leads t o t he following re[TIFR 2015] sults
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6
EM Wave
@Sk J ahiruddin, 2020
aB at
aE
ay
8B
ay
1 aE --c2
at
at all points where t he measurement is made,.In this case the electric vector the magnetic vector and t he wave ➔
E,
B
vector k ( t he m agnitud e k) can be written in terms of unit vectors (x, y, z) along the Cartesian axes as
E = Ex, B = Bz, k = -ky (b) E = Ex, B = - Bz, k = ky (c) E = Ex, B = Bz, k = ky (d) E = - Ey, B = - Bz,k = - k x (a)
Prob 1.11. A beam of plane microwaves of wavelength 12 cm strikes t he surface of a dielectric at 45°. If t he refractive index of the dielectric is ; , what will be t he wavelength , in units of mm, of the microwaves inside the dielectric?
11 oo : oo : 19 Prob 1.11. A beam of plane microwaves of wavelength 12 cm strikes the surface of a dielectric at 45°. If t he refractive index of t he dielectric is ~, what will be t he wavelength , in units of mm, of the microwaves inside t he dielectric?
[TIFR 2017]
Prob 1.12. A plane electromagnetic wave traveling in a vacuum is characterized by the electric and magnetic fields
[TIFR 2013]
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7
EM Wave
@Sk J ahiruddin , 2020
£ = 1(301rVm-
fl = ](HoAm-
1
1
)expi(wt + kz)
)expi (wt + kz )
If w, k > 0, the value of H 0 must be
(a) 21r (b) 0.67
(c) 0.25
(d) 0.94
Prob 1.13. An electromagnetic wave wit h
£ (z, t) = Eocos(wt -
kz)l
is traveling in free space and crosses a disc of radius 2cm placed perpendicular to z-axis. If E 0
= 60Vm- 1, t he
aver-
age power,in Watt, crossing the disc along the z-direction is
[GATE 2007]
(a) 30
(b) 60
(c) 120
(d) 270
11 oo : oo : 21 1s trave 1ng 1n
ee space an
crosses a
m
placed p erpendicular to z-axis. If E 0 = 60Vm- 1, t he average power ,in Watt , crossing the disc along t he z-direction is
[GATE 2007]
(a) 30
(c) 120
(b) 60
(d ) 270
Prob 1.14. A beam of unnpolarized light in a medium with dielectric constant E1 is reflected from a plane interface formed with anot her medium of dielectric constant E2 = 3 E1. The two media have ident ical m agnetic permeability. If the an[NET Dec gle of incidence is 60°, t hen reflected light 2015] (a) is plane polarized p erpendicular to the plane of incidence j [email protected]
P hysicsguide
8
@Sk J ahiruddin, 2020
Eivl Wave
(b) is plane polarized parallel to the plane of incidence (c) is circularly polarized (d) has same polarization as t he incident light
Prob 1.15. The Fresnel relations between t he amplit udes of incident and reflected electromagnetic waves at an interface b etween air and a dielectric of refractive indexµ are E reflected
=
COS r
- µ COS 1, E incident cos r + µ cos i 11
11
E reflected
= µ COS r µ cos r
..1 •
.
I I
,
•
.
~
.
COS 1, E incident
+ cos i ...
.
.
..1
.
11 oo : oo : 24 E reflected
=
11
COST - µ
cos T
E reflected ..1_
The subscript
•
=
COS 'l E incident
+ µ cos i
11
•
µCOST -
µ cos T
COS 'l E incident
+ cos i
..1_
and ..l refer to polarization, parallel and nor-
m al to t he plane of incidence resp ectively. Here, i and r are t he angles o incidence and refraction respectively. [GATE
2007] (a) The condit ion for reflected ray t o be completely polarized is
(i) µcosi = COST (iii) µcosi
=
-COST
(ii) cosi = µcosT (iv) cosi
=
-µcosT
(b) For n ormal inciden ce at an air-glass interface wit h µ
=
1.5 the fraction of energy reflected is given by
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9
EM Wave
@Sk J ahiruddin, 2020
(i) 0.40
(ii) 0.20
(iii) 0.1 6
(iv) 0.04
Pro b 1 .16. A plane electromagnetic wave incident normally on t he surface of a m at erial is partially reflected . Measurements on the standing wave in t he region in front of t he interface.Show t hat t he ratio of the electric field amplit ude at the m axima and t he minima is 5. T he ratio of t he reflected intensity to t he incident intensity is
2014]
(a) 4/ 9
(b) 2/ 3
(c) 2/ 5
(d ) 1/ 5
[NET Dec
11 oo : oo : 26 interface.Show that the ratio of the electric field amplitude at the maxima and the minima is 5. The ratio of t he reflected intensity to t he incident intensity is [NET Dec 2014] (a) 4/ 9 (b) 2/ 3 (c) 2/ 5 (d) 1/5
Pro b 1.1 7. A plane electromagnetic wave propagating in air wit h E = (8i + 63 + 5k)ei(wt+ 3x- 4y) is incident on a perfectly conducting slab positioned at x=O. E field of the reflected wave is [JEST 2017]
(a) (-8i-6J-5k) ei(wt+3x+4y)
(b) (-8i+6J-5k)ei(wt+3x+4y)
(c) (-8i+ 6J - 5k )ei(wt-3x-4y)
(d) (-8i- 6J -5k)ei(wt-3x- 4y)
Prob 1.18. The electric field in an electromagnetic (EM) 4 6 wave is = vl61ri sin[21r(l0 z - 3 x 10 t)] . What is the intensity of the EM wave and the number of photons per sec-
E
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10
P hysicsguide
E:Nl Wave
ond falling on t he unit area of a perfectly reflecting screen kept perpendicular to t he direction of propagatio11? When a photon in this beam is reflected from t he screen, what is the impulse it imparts to the screen? Use t his to find t he pressure exerted by t he EM wave on t he screen. [JAM 2014]
Prob 1.19. A plane electromagnetic wave of frequency 5 x
11 oo : oo : 29 pressure exerted by the EM wave on t he screen. 2014]
JAM
Pro b 1.19. A plane electromagnetic wave of frequency 5 x 4 10 H z and amplit ude 103 V/ m t raveling in a homogeneous dielectric medium of dielectric constant 1.69, is incident normally at t he interface with a second dielectric medium of dielectric constant 2.25. The ratio of the amplitude of the transmitted wave to that incident wave is ____ . [JAM 2015]
Prob 1.20. MSQ: For an electromagnetic wave travelling in free space, the elctric field is given by
E = 100 cos( l 08t + kx )3 mV . Which of the following statements are t rue? [JAM 2015] (a) The wavelength of t he wave in meter is 61r (b) The corresponding magnetic field is directed along t he [email protected]
@Sk J ahiruddin , 2020
11
Physicsguide
EM Wave
posit ive z direction (c) The Poynting vector is directed along the positive zdirection (d) The wave is linearly polarized
Prob 1.21. The electric field of a light wave is given by
11 oo : oo : 32 direction (d) The wave is linearly polarized Prob 1.21. The elect ric field of a light wave is given by
E = E0 [isin(wt -
kz) +] sin(wt - kz -
1r ) ]
4
The polarization state of t he wave is [JAM 2015] (a) Left handed circular (b) Right handed circular (c) Left handed ellipt ical (d) Right handed ellipt ical
Prob 1.22. The electric field of an electromagnetic wave is
given by
E = (2k- 3]) x 10- sin[l 0 (x+ 2y+3z-,Bt )] 7
3
[JAM 2017]
The value of (3 is (c is the speed of light) (a) v'Mc (b) v'l2c (c) v'Ioc (d) if7c
Prob 1.23. In a non-conduct ing medium characterized by E = c0 ,µ = µ 0 and conductivity a = 0, t he elect ric filed (in
vm- 1 ) field
E
is given by = 20sin [l0 8t - kz]] . The magnetic in (Am- 1 ) is given by [GATE 2009]
H
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12
ENI Wave
@Sk J ahiruddin, 2020
(a) 20k cos[108 t - kz]l 20 k sin[108t - kz]l (c) - 108µ0
6~~
8
(b) 1 0 sin[10 t - kz]] (d) - 20k cos[108t - kz]]
Prob 1.24 . Consider the propagation of electromagnetic waves in a linear , homogeneous and isotropic material medium , . , , .. . , ., .. . . . .
.
.
'
11 oo : oo : 34 a (c) -
cos 1 t - z i 20 8 t - kz]l k sin[10 108 µ0
sin 1 t - z j (d) - 20k cos[108t - kz]] 108µ a
/>
Prob 1.24. Consider the propagation of electromagnetic waves in a linear, homogeneous and isotropic material medium with electric permittivity E and magnetic permiability µ.
[GATE 2010] (a) For a plane wave of angular frequency w and propagation ➔ vector k propagating in the medium Maxwell's equations reduce to ➔➔ ➔➔ x = x = (i) k. E = O; k .H = O;
k E wEH;k H -wµ E
t.E = O; t.H = O; k x E = -wEH;k x H= wµE (iii) 7t. E = o; t. H = o; k x E = -wµH;k xH = wEE (iv)k.£ = O; k.H = O; kxE =wµH;kxfl = -wEE (ii)
(b) If E and µ assume negative values in a certain frequency range, t hen th directions of t he propagation vector and the Poynt ing vector in that frequency range are ➔ ➔ related as (a) k and S are parallel and are anti-parallel (b) ➔ ➔ (c) k and S are perpendicular to each other and makes an angle that depends on t he magni(d) t ude of E and lµI
f
S
k
S
k
S
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EM Wave
Prob 1.25. A plane electromagnetic wave t raveling in free space is incident normally on a glass plat e of refractive index
3/2. If there is no absorption by the glass, its reflectivity is
00: 00: 37 EM Wave
Prob 1.25. A plane electromagnetic wave t raveling in free space is incident normally on a glass plate of refractive index
3/ 2. If there is no absorption by the glass, its reflectivity is [GATE 2012] (a) 4% (b) 16% (c) 20% (d) 50% Prob 1.26. The intensity of a laser in a free space is 150m W/m The corresponding amplit ude of t he electric field of t his laser 12 2 2 [GATE 2014] is _____ ~ (Eo = 8.854 X 10- c /N.m )
Prob 1.27. How much force does light from a 1.8W laser
exe when it is totally absorbed by an object? [JEST 2016] (a) 6.0 x 10- 9 N (b) 0.6 x 10- 9 N (c) 0.6 x 10- 8 N (d) 4.8 x 10- 9 N Prob 1.28. A plane electromagnetic wave has t he magnetic field given by (x, y, z, t) = B 0 sin[(x + y) ✓'i + wt]k where k is t he wave
B
A
A
A
number and i, jandk are the Cartesian unit vectors in x,y and z directions respectively. [GATE 2011] (a) The electric field [email protected]
@Sk J ahiruddin, 2020
above wave is given by
E(x, y, z, t ) corresp onding to the 14
Physicsguide
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2
.
11 oo : oo : 40 @Sk J a hiruddin , 2020
EM Wave
above wave is given by A
A
(i) cBosin[(x + y) ~ + wt]i/2 A
A
+ y) ~+wt] i~ cBosin[(x + y) ~ + wt]i cBosin[(x + y) ~+wt]]
(ii) cB 0 sin[(x (iii) (iv)
(b) The average Poynting vector is given by
(i) cB
2 ~ -: 0 (i-J)
2µo v'2
(ii) _cB 2µo v'2
2 cB 0 (i+j)
2 cB 0 (i+j)
A
..111.) (
2 ~ -: 0 (i-J)
A
----
2µov'2
A
(.IV )
A
-----
2µov'2
Prob 1.29. The approximate force exerted on a perfectly reflecting mirror by an incident laser beam of power 1OmW at normal incidence is [JEST 2015] (a) 10- 13 N (b) 10- 11 N (c) 10- 9N (d) 10- 15N
Prob 1.30. A monochromatic plane wave in free space wit h electric field amplit ude of 1V / m is normally incident on a fully reflecting mirror . The pressure exerted on the mirror is ____ x 10- 12 P a (Upto two decimal places)(Eo = 8.854 x 12
2
10- C /N.m
2
[GATE 2017]
)
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Prob 1.31. A plane electromagnetic wave is propagating
in a lossless dielectric. The electric field is given by [NET June 2011] (x, y, z, t ) = E 0 (x + A z)exp[ik0 -ct + (x + J&z)] where c is the speed of light, E 0 , Aandk0 are constant and x and z are unit vectors along t he x and z axes. The relative dielctric constant of the medium Er and the constant A are
E
1
v'3 v'3
(a)
Er =
4 and A = -
(c)
Er =
4 and A =
(b) (d)
Er =
Er =
v'3 -v/3
4 and A =
4 and A =
1
Prob 1.32. An electromagnetic wave is incident on a waterair interface. The phase of the perpendicular component of t he electric field E _1_ of t he reflected wave into t he water is
found to remain the same for all angles of incidence. The phase of magnetic field H [NET June 2012] (a) does not change (b) changes by 31r / 2 (c) changes by
1r /
2
(d) changes by
1r
Prob 1.33. Consider the interference of two coherent elec-
t romagnetic waves whose electric field vector are given by £ 1 = iE0coswt and £ 2 = ] E0cos(wt + cp) where cp is the phase difference. The intensity of the resulting wave is given j [email protected]
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11 oo : oo : 4s E
1E 0 coswt and E2 = JE 0 cos (wt + ) where is the phase difference. The intensity of the resulting wave is given 1
=
j [email protected]
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EM Wave
by ~ (E ) , where(E intensity is 2
(a) 0 (b) E0 E5
2
)
2
is te time average of E . The total [NET Dec 2012]
(c) EoE5 sin
2
( d)
2
E0 E5cos cp
Prob 1.34. An electromagnetic wave is travelling in free space (of permittivity Eo) with electric field = kE 0 cosq(x-
E
ct). The average power (per unit area) crossing planes par[NET Dec 201 7] allel to 4x + 3y = 0 will be
(a) iEocE5 5
(b) EocE5
(c) ~EocE5 2
(d)
16 25
E0 cE5
Prob 1.35. A plane electromagnetic wave from within a dielectric medium (with E = 4E0 and µ = µ 0 ) is incident on its boundary with air, at z = 0. The magnetic field in the medium is 1f = ]Hocos(wt - kx - k./3z) , where w and k are posit ive constants.
The angles of reflection and refraction are, respectively, [NET Dec 2017] (a) 45° and 60° (b) 30° and 90° (c) 30° and 60°
(d) 60° and 90°
Prob 1.36. An electromagnetic plane wave is propagating with an intensity I = 1.0 x 105 W m - 2in a medium with
E
=
3Eo and µ
= µ0.
The amplitude of the electric field in-
11 oo : oo : 48 Prob 1.36. An electromagnetic plane wave is propagating with an intensity I = 1.0 x 10 5 W m - 2in a medium with
E
=
3Eo and µ
= µ0.
The amplitude of the electric field in-
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side t he medium is ___ x 103 Vm- 1 (up to one decimal place) . [GATE 2018] 2 7 (Eo = 8.854 x 10-12F/m, µ 0 = 41r x 10- N A- , c 3 x l0 8ms- 1 ) Prob 1.37. The electric field of an electromagnetic wave is
E
= Eocos (k z + wt)l + 2Eosin(kz + wt)], where w and k
are positive constants. This represents [NET Dec 2016] (a) a linearly polarised wave t raveling in the positive zdirection (b) a circularly polarised wave traveling in the negative zdirection (c) an elliptically polarised wave t1·aveling in t he negative z-direction (d) an unpolarised wave traveling in t he positive z-direction Prob 1.38. A quarter wave plate introduces a path difference of A/ 4 between the two components of polarization parallel and perpendicular to t he optic axis. An electromagnetic wave wit h = (x + y)E0 ei(kz - wt) is incident normally on a quarter wave plate which has its optic axis
£
making an angle 135° wit h the x-axis as shown. 2018]
[GATE
11 oo : oo : so para e an perpen icu ar to t e optic axis. n e ectromagnetic wave wit h = (x + y)E0 ei(kz - wt) is incident normally on a quarter wave plate which has its optic axis making an angle 135° wit h the x-axis as shown. [GATE
£
2018]
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''
'
The emergent electromagnetic wave would be (a) ellipt ically polarized (b) circularly polarized (c) linearly polarized with polarization as that of incident wave (d) linearly polarized but with polarization at 90 to that of t he incident wave
Prob 1.39. A light beam of intensity J0 is falling normally on a surface. The surface absorbs 20% of t he intensity and t he rest is reflected. The radiation pressure on t he surface is given by 0 , where X is _____ (up to one decimal place). Here c is the speed of light . [GATE 2018]
x:
11 oo : oo : s3 t he rest is reflected. The radiation pressure on t he surface is given by 0 , where X is _____ (up to one decimal place). Here c is the speed of light . [GATE 2018]
x:
Prob 1.40. The dispersion relation for electromagnetic wave where c t ravelling in a plasma is given by w 2 = c2 k 2 + and wp are constants. In this plasma, the group velocity is
w;,
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Eivl Wave
[JAM 2017] (a) proportional to but not equal to the phase velocity (b) inversely proportional to the phase velocity (c) equal to the phase velocity (d) a constant
• •
Prob 1.41. A plane electromagnetic wave, which has an electric field
E(";t, t)
= (P l + Q] ) exp iw(t - z) C
is passing through vacuum. Here P ,Q and w are all constants, while c is t he speed of light in vacuo. What is the average energy flux per unit time (in SI units) crossing a unit area placed normal to the direction of propagation of t his wave, in terms of t he above constants ? [TIFR 2018]
Prob 1.42. An electromagnetic wave of wavelength A is in-
11 oo : oo : 55 agation of t his wave, in terms of t he above const ants ? [TIFR 2018]
Prob 1.42. An electromagnetic wave of wavelength A is incident on a dielectric slab of t hickness t. If K is the dielectric constant of t he slab, t he change in phase of the emergent wave compared wit h the case of propagation in t he absence of the dielectric slab is [JEST 2018]
(a)
vK -
1
(b) 21r
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(d)
(c) 21rt/ A
2It (vK -
1)
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Prob 1.43. In t he region far from a source, t he t ime dependent electric field at a point (r, 0,
o
o
Which on calculation gives,
IE0
2
o 2
1
Hence, intensity is,
1 2 coE0 2 Sol 1.34. Given ,
E
"'
Rn
kE0 cos q(x - et ) From this, we get A
11 oo : 01
: 47
Which on calculation gives, E 0
Hence, intensity is,
1 2 -2 EoE 0 Sol 1.34. Given,
magnetic field, vector,
E
B= -
= kE0 cos q(x - et ) From this, we get Eo cos(qx - wt )] From t his, poynting C
-
E5 S = - cos(qx - wt)i µo c A
and t ime averaged value of this quantity gives,
E5 ~ - -'l, 2µoc This value on t he given component gives ~ EocE5 Sol 1.35. We know,
sin ei sin 0t
n2
n1
Again kx l tan0i = kz J3 Which gives, 0i = 30 and 0t = 90 option (b) is correct . [email protected]
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1
- vcE5 I 2 culation, we get , Eo = 6.6 x l0 8 V / m Sol 1.36. As we know, I
=
=
1 1
- - -EE6 On cal2ffe
Sol 1.37. Since, amplit udes are different , Hence elliptically
polarised light moving in z- direction.
11 oo : 01 : so culation, we get, Eo = 6.6 x l0 V /m 8
Sol 1.37. Since, amplit udes are different, Hence elliptically
polarised light moving in z- direction. 7[
Sol 1.38. Incident electric field along - with x-axis. Now,
>.. / 4 path difference will introduces
7[
4
phase difference and 2 will make it Oo with optic axis. Hence, wave will be plane polarised. L
- 0.8Io
Sol 1.39. Radiation pressure, -2. -
Hence, value
C
C
of x= 1.8 Sol 1.40. In the plasma media, given, w 2 which gives
w k = -[l C
Then,
w Vp = -
c2k2
+ w2p
1 w 2
w
P] 2 2 C
= ----::====
k
l - (Wp)2 w
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And, group velocity, Vg
=
C
Hence, option (b) is correct.
11 oo : 01 : s2 And, group velocity, Vg
=
C
Hence, option (b) is correct. Sol 1.41. Given, A
z
A
(Pi+ Qj) exp iw(t - - ) C
Using t he formula,
-+
A
k X E B =-c -+
, we get, -+
l
z
C
C
A
A
B = - expiw(t - -)(P j - Qi)
Hence, Poynting Vector,
l 2 2 2 S = -cos (wt - kz)(P + Q )k µoc -+
A
Hence time-average of t hat gives
and, on t he direction of propagation its component will be,
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Sol 1.42. Assuming t he wave propgating along z-axis,
d>,
=
wd
kz - wt = -
-
wt
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Sol 1.42. Assuming t he wave propgating along z-axis, wd
1 = k z - wt = -
- wt
C
Where d = thickness and wd wvfk 2 = k z - wt = - - wt = - -d - wt V
C
Changing in phase,
Hence, opt ion (d ) is correct.
Sol 1.43. Given , ➔
-6' (r, 0, )
A
= Eow
2
sin0 r
r cos w t - C
Now, calculationg the value of Poynting Vector, We get , 2
4
.
0
E 0w (sin ) 2 2 ( S = - - - - cos wt - k r ) µ 0c r Taking t he time average, we will get,
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Hen ce, total power =
E5w4 2µ 0 cr 2 Which on calculation gives the answer ,
Hence,
option (b ) is correct.
Sol 1.44. Given ,
£ (z, t) = So, we get,
""=
1
3a
l Eoe-zf3a
and k
tancp
=
z
---wt
cos
1
v'3a
And we know,
10 v3a K,
= (-) = -
1
J3
k
Hence, B lags behind E by 30° Sol 1.45. As we know, E -
= - v (k A
x B) Calculating we get -
option (a) as the answer.
Sol 1.46. Given , n(w)
~
w
1-
2
wo
'
C
w2
vp
w5
-= l - j a [email protected]
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E:Nl Wave
Which on calculation gives,
= ~ Again,
vp
Which gives, group velocity, v9 = 4c Hence, t he ratio gives t he answer 3. Hence, option (a) is correct. Sol 1.47. Given , .....
+ 4cos(kz -
E = 3 sin(kz - wt )x When, kz
wt )y
= 0, -,
A
E = -3 sin wti -,
A
+ 4 coswtj 1T
A
-I
A
,E When wt = 0, E = 4j , similarly, wt = -3i 2 Hence, elliptically polarized in counter-clockwise direction when seen travelling towards the observer -,
Sol 1.48. k
A
A
= 3j + 4k Hence, 3
-
4
Hence, 4
cos 0R = 5 [email protected]
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11 oo : 02 : 03 Hence,
cos 0R
4
=-
5
44
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Hen ce, power
21
P = - cos 0R C
Which gives on calculation the value 0.8 _,
Sol 1.49. Given , E = Ea exp [i (k 1 z - wt)]
x Hence,
Hen ce,
BR = - Ea exp [i (-k 1 z - wt)] y V1
Hence, option (c) is correct.
11 oo : 02 : os j [email protected]
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Hence, power
21 P = - cos 0R C
Which gives on calculation the value 0.8 -;
Sol 1.49. Given , E = Ea exp [i (k 1 z - wt)]
x Hence,
Hence,
B R= - Ea exp [i (-k1z - wt )] y V1
Hence, option (c) is correct.
j [email protected]
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11 oo : oo : oo
Problellls and Solutions in Vectors Sk J ahiruddin Arnab Choudhury Assistant Professor Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay 1\11.Sc Pl1ysics 2009-2011 batch He ranked 007 in IIT J Al\11 2009 and 008 (JRF) in CSIR NET J une 2011 He has been teaching CSIR NET aspirants since 2012
1
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Vect ors
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Vectors
Contents 1 Problems from NET , GATE, JEST, TIFR & JAM papers 1.1
Ans keys
1.2 Solutions .
j [email protected]
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•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
24
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•
•
•
•
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•
•
•
26
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1
Vectors
Problems from NET, GATE, JEST, TIFR & JAM papers A
Prob 1.1. (a) Consider a constant vector field iJ = v 0 k. _, Find any one of t he m any possible vectors il, for which V x _,
_,
U = V
(b) Using stokes theorem , evaluate t he flux associated with the field iJ through the curved hemispherical surface defined by x
2
+y +z = r 2
2
2
;
z > 0
[JAM 2005] Prob 1.2. For t he vector field
V = xz i-yz ]+z(x -y )k, 2
2
2
2
(a) Calculate t he volume integral of the divergence of out of t he region defined by -a
< x < a, -b
X
= 6.
4 6
6 9
•
Trace of the matrix is 13. So, sum of eigenvalues must be 13 and product must be 0. So, the eigenvalues must be, 0 and 13. 4 6 We can see t hat MT = = M. So, Mis a symmetric
6 9
matrix. 1 But as Det(M)=O, so, M- doesn 't exist. A matrix is invertible only when there exist the inverse matrix, So, M is not an invertible matrix. Finally, we know that eigenvectors of a symmetric matrix j [email protected]
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are always orthogonal. So, t hat must be true in this case also. Then the correct options are a, c, d. Sol 1.60. You need to know the following things (1) The
11 oo :03 : og are always orthogonal. So, t hat must be true in this case also. Then the correct options are a, c, d. Sol 1.60. You need to know the following things (1) The product of t he eigenvalues are the determinant of the matrix (2) Ort hogonal matrix means AAT = ~
So if Ax = AX (the eigenvalue equation)
So the eigenvalues will be of unit modulus, but t he eigenvalues could be real and imaginary. There are no problems on that. Now you can prove with the help of Mathematical Physics books t hat an orthogonal matrix with determinant 1 represents a rotation and has at least one eigenvalue = 1, whereas an orthogonal matrix with determinant = - 1 represents a reflection and has at least one eigenvalue = -1 . The other two eigenvalues in both cases are ei 0 and e-iB, . So wit h t hese informations in mind t he correct option is
(b) Sol 1.61. We know t he trace formula
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Now, So,
Trace[A] = a. Det[eA] = ea.
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11 oo : 03 : 12 Linear Algebra
@Sk J ahiruddin, 2020
Now, Trace[A] = a. So, Det[eA] = ea. So, product of the eigenvalues of eA is ea, So, if one of t he eigenvalues is eA, t hen the product of the other two eigenvalues will be 1, so t hat t he product of the t hree eigenvalues can be eA . Sol 1.62. As At A
= 4~, the matrix A/ 2 is unitary. Hence
t he no of independent parameters of A will be same as of 4 x 4 unitary matrix. Now N x N unitary matrix can be written in the following form
2
2
2
So, there are N of a' sand N of b's, and 2N of parameters altogether. Unitary condition is given by
(ut u) ij
=
u itk ukj
= (aki
+ ibkj) + i (akibkj
- ibki) ( akj
= a ki akj
+ bki bkj
-
a kj bki) = 8ij
Hence a ki a kj
+ bki bkj =
akibkj -
akjbki
8ij
=0
l st eq is symmetric under i and j excahnge, so this equat ion actually shows N j [email protected]
@Sk J ahiruddin, 2020
+ (N
2
70
-
N) / 2 independent condiP hysicsguide
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t ions. 2nd eq is antisymmetric under i and j exchange, so this shows (N 2 - N) /2 independent condit ions. Hence, t here are N2 - N N2 - N N2 N +---+---= 2 2 independent conditions in t his unitary condition. Therefore, number of independent parameters in a N x N unitary matrix is 2 2 2 2N - N = N Hence there are 42
= 16 independent parameters.
Sol 1 .63. From Similarity Transformation, we know that
P matrix is called a similar matrix to A having same eigenvalue, determinant and trace. Here P matrix is the matrix is formed by placing the eigenvalues of A matrix in diagonal positions. And S matrix is formed by placing the corresponding eigenvecto1·s of A as columns. s-1 AS = P Now, from
s [s- 1 As] s-1 = sps- 1
A = S P s- 1
Here, t he eigenvalues are
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and t he eigenvectors are
So, 1
1
1
1 - 1
1 -2
2
1
0
s-1
3-1= 1
2 1
6
2
0
0 - 1 0
0
Now, we have to calculate
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P=
&
0 0
1
2
1 -2 3 -3 0
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A =SPS-
1
1 1 1 -1 1 - 2 0 1
1
-
6
1
1
1
-
1
1 1 - 1
1
6
-
Linear Algebra
1 -2 1
-
6
0
2 0 0 0 -1 0
2
2
2
1
1
-2
0
3 - 3
0 1
0
4 4 4 - 1 - 1 2
3 -3 0
6 0 6 0 6 6 6 6 0
1 0 1
0 1 1 1 1 0 S ol 1.64 . The Solution is easy. As we know and the question also says that vector along the axis of rotation remains unchanged during the rotation. So lets t ake an arbitrary
vector (x, y, z) and see the action of the matrix on this vector.
0 1 0
X
y
0 0 - 1 - 1 0 0
y
-z -x
z
Now as the vector must remain same to be a vector along
11 oo : 03 : 23 0
1 0
0 0 - 1 - 1 0 0
X
y
y
-z -x
z
Now as t he vector must remain same to be a vector along t he rotation axis, so x = y, y = -z, z = -x. So the general j [email protected]
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pattern of this vector would be (1,, 1- 1), when normalised, t his becomes }a(l , , 1 - 1). So the ans is (b ). Sol 1.65. You must know t hat (A+ B )- 1 # A- 1 + B-l. If
(A + B) is invertible then
A-1
+ B-l = B-l + A-l = B- 1 AA- 1 + B- 1 BA- 1 = B- 1 (A + B )A- 1
So A- + B- is the product of invertible matrices and thus is invertible, with inverse equal to 1
1
(A-l
+ B-1) - 1 = (B-l(A + B)A- 1) -l 1
= A(A+ B) - B Now t hink of the ans. Sol 1.66. You need to check individually. And you must know that (A B )t = B tAt and (A- 1 )t = (At)- 1
Now check option (a)
Again check option (b)
11 oo : 03 : 2s Now check option (a)
Again check option (b)
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Now you can easily see that
(c- 1DC)(c- 1 n- 1 c) = ]_ Sol 1.67. T he matrix is not diagonalizable. So we can 't use t he similarity transformation method. Hence we use Caley Hamilton t heorem . The characteristric equation of t he matrix is
Using t his you can check that 8
A = 8A - 7 and so on. Hence
The m atrix also follow t he charact eristic equation.
A Hence
27
=
27A - 26li
11 oo :03 : 2a The matrix also follow t he charact eristic equation. A
27
=
27 A - 26 li
=
1 351
Hence A 21
0
1
Sol 1.68. If you can t hink just looking at the picture t hen
it is fine. Otherwise do as follows. [email protected]
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a b c Suppose the transformation matrix is d e f g h i This matrix transforms t he set of coordinates So a b c
1
0
f
1
g h i
0 0
a b c
0
f
1
0 0
g h i
0
1
a b c
0
2
d e f g h i
0
0 0
d e
d e
2
0
0 0 1 Hence you get the transformation matrix is S
=
1 0 0 .
0 1 0
11 oo : 03 : 30 it is fine. Otherwise do as follows. j [email protected]
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a b c Suppose the transformation matrix is d e f g h i This matrix transforms t he set of coordinates So a b c
1
0
f
1
g h i
0 0
a b c
0
f
1
0 0
g h i
0
1
a b c
0
2
f
0
g h i
2
0 0
d e
d e
d e
0
0 0 1 Hence you get the transformation matrix is S =
l O O .
0 1 0 Now we can further decompose the matrix but we need not to. We can see t hat t he determinant of S is - 1. And we also know t hat the determinant of rotaion mat rix is 1 and of reflection matrix is - 1. So S is composed of either two rotation or two reflection matrix but not of one rotation and one reflect ion. Only option (c) matches with the criteria. j [email protected]
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11 oo : 03 : 33 it is fine. Otherwise do as follows. j [email protected]
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a b c Suppose the transformation matrix is d e f g h i This matrix transforms t he set of coordinates So a b c
1
0
f
1
g h i
0 0
a b c
0
f
1
0 0
g h i
0
1
a b c
0
2
f
0
g h i
2
0 0
d e
d e
d e
0
0 0 1 Hence you get the transformation matrix is S =
l O O .
0 1 0 Now we can further decompose the matrix but we need not to. We can see t hat t he determinant of S is - 1. And we also know t hat the determinant of rotaion mat rix is 1 and of reflection matrix is - 1. So S is composed of either two rotation or two reflection matrix but not of one rotation and one reflect ion. Only option (c) matches with the criteria. j [email protected]
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Problellls and Solutions in Differential Equations and Special Functions Sk J ahiruddin Arnab Choudhury Assistant Professor Sister Nibedita Govt. College, Kolkata Aut hor was t he topper of IIT Bombay 1\II.Sc Physics 2009-2011 batch He ranked 007 in IIT JAM 2009 and 008 (JRF) in CSIR NET J t1ne 2011 He has been teachi11g CSIR NET aspirants since 201 2
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@Sk J ahiruddin, 2020
Diff. eq and Sp. ftmctions
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Diff. eq and Sp. functions
@Sk J ahiruddin, 2020
Contents 1 Problems from NET , GATE, JEST, TIFR & JAM papers
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1.1
j [email protected]
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j [email protected]
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Diff. eq and Sp. funct ions
Problems from NET, GATE, JEST, TIFR & JAM papers
Prob 1.1. A flat surface is covered with non-overlapping
disks of same size. W hat is the largest fract ion of t he area [JEST 2013] t hat can be covered?
(a)
3
(b)
5
(c)
1r
6
6 7 Prob 1.2. Given a function 1r
(d)
1T
2y3 f (x, t ) of both position x and
•
(where
j
(a) 02 f 8x 2
=
df (x, t ) x = dx) dt ' dt
(c) ~
(b) 8f
8x
X
[JAM 2012] (d) df dx
Prob 1.3. T he operator ( d - x)( d dx dx
d2 (a) d 2 X
d2 (c) dx~?
-
-
x2
d2 (b) d 2 X
-
is equivalent to
[JEST 2013]
x2 + 1 d2
d 2 x dX x + 1
+ x)
(d) d X 2
-
d 2 2x dX - x
Prob 1.4. Find t he solution of t he differential equation 2
d y dx 2
dy 5 + dx = O
11 oo :oo :og Prob 1.4. Find the solution of the differential equation
d2 y dx 2
dy 5 + dx = O
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Diff. eq and Sp. functions
@Sk J ahiruddin , 2020
with t he boundary condition
=2
y(O) = 2 and dy
dx
x=O
giving all steps clearly. Find the value of x where y = 0 [JAM 2014]
Prob 1.5. Consider the equation
!Yx =
2
y with the boundX
ary condition y( l ) = 1. Out of t he following t he range of x in which y is real and finite is [JAM 2015] (a) -oo < x < -3 (b) -3 < x < 0 (c) 0 < x < 3
(d) 3:::;
X
----------------------
X
t
t
(d)
(c) X
X
----------------------
t
t
Prob 1.10. Given the recurrence relation for the Legendre polynomials
Which of the following integrals has non-zero value? [GATE 2010] 1
(a) - 1 ,. 1
2 x Pn(x) Pn+1(x) dx
1
xPn(x )Pn+2(x )dx
(b) ,. 1
- 1
11 oo : oo : 16 Which of the following int egrals has non-zero value? [GATE 2010] 1
1 2
(b)
x Pn(x) Pn+1(x) dx
(a) - 1 1
- 1
1
xlPn(x )] 2 dx
(c)
xPn(x )Pn+2(x )dx 2
x Pn(x) Pn+2(x)dx
(d)
-1
-1
Prob 1.11. The solutions to the different ial equation dy
X
dx [email protected]
@Sk J ahiruddin,
y
+1
6
2020
P hysicsguide
Diff. eq and Sp. functions
are a family of [GATE 2011] (a) circles with different radii (b) circles with different centers (c) straight lines wit h different slopes (d) straight lines wit h different intercepts on t he y-axis
Prob 1.12. r (n + ½) is equal to [Given r (n + 1) = nf (n) and r (~) = y0r] [GATE 2013] 2n!~ c) (b) 2n!~ (d) n!~ ( 22n n !22n n !2n Prob 1.13. The solution of t he different ial equation d2 y dt2 - y = 0 subject to the boundary condit ions y(O) = 1 and y( oo) = 0 •
IS
[GATE 2014]
11 oo : oo : 19 dt2 - y = 0 subject to the boundary condit ions y(O) = 1 and y( oo) = 0 •
IS
[GATE 2014] (a) cos t + sin t (c) cost - sin t
(b) cos ht + sin ht (d) cos ht - sin t 2
2
Prob 1.14. If f( x) = e-x and g(x) = x e-x , then [GATE 2015] (a) f and g are differentiable everywhere (b) f is differentiable everywhere but g is not (c) g is differentiable everywhere but f is not [email protected]
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Diff. eq and Sp. functions
(d) g is discontinuous at x = 0
Prob 1.15. A function y(z) satisfies the ordinary different ial equation 2 1 ffi y + -y - -2y = 0 z z , where m = 0, 1, 2, 3, ... Consider the four stat ements P,Q,R,S as given below. II
I
• (P ) zm and z - m are linearly independent solut ions for all values of m
• (Q) zm and z - m are linearly independent solut ions for all values of m > 0 • (R) lnz and 1 are linearly independent solutions for m =O
11 oo : oo : 21 all values of m
• (Q) zm and z - m are linearly independent solut ions for all values of m > 0 • (R ) lnz and 1 are linearly independent solut ions for m = O
• (S) zm and lnz are linearly indep endent solutions for all values of m The correct opt ion for the combination of valid st at ement s •
lS
[GATE 2015] (a) P,R and S only (c) Q and R only
j [email protected]
@Sk J ahiruddin, 2020
(b) P and R only (d ) R and S only
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P hysicsguide
Diff. eq and Sp. ftmctions
!~
Prob 1.16. Consider the linear differential equation xy. If y = 2 at x = 0 , then the values of y at x=2 is given by (GATE 2016] (a) e- 2 (b) 2e- 2 (c) e2 (d) 2e2 Prob 1.17. If [x] denotes t he greatest integer not exceed00 [JEST 2012] ing x, t hen f 0 [x]e- xdx (a) 1 (b) 1 ( c) e - 1 (d) 2 e e- l e e - 1 Prob 1.18. What are solutions to J" (x) - 2f' (x) + J(x) = O? [JEST 2014] (a) c1 ex/ x (b) c1 x + C2 / x (c) c1xex +c2 (d) c1ex + c2xex
11 oo : oo : 24 Prob 1.18. What are solutions to j" (x) - 2j' (x)
[JEST 2014]
O?
(a) (c)
+ f (x ) =
c1ex
/x
C1Xex
+ C2
(b) c1x + c2/x (d) c1ex + C2X ex 00
9
Prob 1.19. The value of the integral
dxx exp(-x 0
(a) 20160
(c) 18
(b) 12
2
)
is
[TIFR 2013]
(d) 24
Prob 1.20. The differential equation
d2y - 2dy dx 2 dx
- 0
+y -
[TIFR 2013]
(a) Aexp x + B xexpx (c) A ex p x + B exp(-x)
[email protected]
(b) Aexpx + B xexp(-x) (d) x A exp x + B xexp(-x)
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Diff. eq and Sp. functions
@Sk J ahiruddin , 2020
Prob 1.21. Consider t he differential equation
d2y = -4 dx 2 y
+
dy dx
with t he boundary condition t hat y(x) = 0 at x = ~- When 5 plotted as a function of x, for x > 0, we can say with certainty that the value of y [TIFR 2015] (a) oscillates from positive to negative wit h amplitude decreasing to zero (b) has an extremum in t he range O < x < 1 (c) first increases, then decreases to zero (d) first decreases, then increases to zero
11 oo : oo : 26 (a) oscillates from posit ive to negative wit h amplitude decreasing to zero (b) has an ext remum in t he range O < x < 1 (c) first increases, then decreases to zero (d) first decreases, then increases to zero Prob 1.22. The generating function for a set of polynomials in x is given by [TIFR 2015]
f (x, t) = (1 - 2xt + t 2 ) - 1 The t hird polynomial (order x 2 ) in this set is (a) 2x 2 + 1 (b) 4x 2 + 1 (c) 2x 2 - x (d) 4x 2 - 1 Prob 1.23. The generating function 00
F (x, t ) =
L Pn(x)tn n=O
for t he Legendre polynomials Pn(x ) is F (x, t ) = (1 - 2xt + 2 t )½ . The value of P 3 (- 1) is [NET Dec 2011] (a) 5/ 2 (b) 3/ 2 (c) +1 (d) -1 j [email protected]
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Diff. eq and Sp. functions
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Prob 1.24. Let x 1 (t) and x 2 (t ) be two linearly independent solut ions of the differential equation 2
dx dx dt2 + 2 dt + f (t )x = 0 and let
dx 2 ( t ) W (t ) = X 1 (t ) dt -
If w(O) = 1 , then w(l ) is given by 2 (a) 1 (b) e ( c) 1/ e
( ) dx 1 ( t) X2
t
dt
[NET Dec 2011] 2 ( d) 1/ e
11 oo : oo : 29 and let
( ) dx 1 ( t) dx 2 ( t) W (t ) = X1 (t ) dt - X2 t dt If w(O)
= 1 , t hen w( l ) is given 2
(b) e
(a) 1
(
[NET Dec 2011] 2 ( d) 1/ e
by
c) 1/ e
Prob 1.25. Consider t h e different ial equation
2 d x dt2
dx + 2 dt
+X= 0
with the initial conditions x(O)
=
0 and !x
t
= 1.
The
t=O
solut ion (x( t )) attains it s maximum value when t is [NET
June 2014]
(a) l/2
(c) 2
(b) l
(d )oo
Prob 1.26. Given 00
L Pn(X )tn = (1 -
2xt + t
2
)-
1 2
l
,
for
t < l
n= O
t he value of P5 (-1) is
(a) 0.26
[NET June 2014]
(d) -1
(c) 0.5
(b ) 1
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Prob 1.27. The function X
-
2n+l
2
satisfies the differential equation 2
d f (a) x2 dx2 ,.()
n
df + x dx ,
+ (x2 + 1)f = 0 n
[NET Dec 2014]
11 oo : oo : 32
[NET Dec 2014]
satisfies the differential equation d2f df (a) x2 dx2 + x dx + (x2 + 1) f = 0
d2f 2 (b) x dx 2
df + 2x dx
+ (x
2
- 1) f = 0
d2f df (c) x2 dx2 + x dx + (x2 - 1) f = 0
Prob 1.28. Consider t he differential equation 2 dx dx dt2 - 3 dt
+ 2x =
0
. If x = 0 at t = 0 and x = 1 at t = 1, the values of x at
t = 2 is (a) e + 1 2
[NET June 2015]
(b) e + e 2
(c) e + 2
(d) 2e
Prob 1.29. Consider the differential equation!~ + ytan x = cos x . If y(O) 2017]
= 0, y(1r / 3) is - - -(two decimal) [GATE
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2020
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Diff. eq and Sp. ftmctions
Prob 1.30. The function y(x) satisfies the differential equa. dy cos 1TX t1on x d + 2y = - -. If y (l) 1, the value of y(2) X X 1S [NET June 2017] (a)1r (b) l (d) l/4 (c) 1/2 •
11 oo : oo : 34 e unction y x sat1s es t e i erent1a equa-
1.30.
. dy t1on x dx is
+ 2y
cos1rx_ If y(l) x
(b) l
(a)1r
1, t h e va1ue of y (2) [NET June 2017]
(c)l/2
(d) l/4
Prob 1.31. Consider two part icles moving along t he x-axis. In t erms of their coordinates x 1 and x 2 , t heir velocit ies . dxi dx2 are given as dt = x 2 - xi and dt = xi - x 2 , respect ively. When they st art moving from their initial locations of x 1(0) = 1 and x 2 ( 0) = - 1, the t ime dependence of both x 1 and x 2 contains a t erm of t he form eat, where a is a con-
[JAM 2017]
stant. The value of a (an integer) is:
Prob 1.32. Consider the differential equation y'' + 2y' +y = 0. If y(O) = 0 and y' (0) = 1, t hen the value of y(2) is ?? (Specify ans up to two digits after decimal point) [JAM 2017]
Prob 1.33. The number of lineraly independent power series solut ions, around x = 0, of the second order linear . d2 y dy . equation x dx 2 + dx + xy = 0, is [NET Dec 2017] (a) 0 (this eq doesn 't have a power series solution)
(b) 1
(c) 2
(d) 3
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Diff. eq and Sp. functions
Prob 1.34. Evaluate t he expression A
n!
Xn - 1
[TIFR 2017] Xn - 2
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Prob 1.34. Evaluate t he expression A
Xn - 1
[TIFR 2017] Xn - 2
n! 0
0
0
0
0
0
Prob 1.35. Write down x(t), where x(t) is the solution of the following differential equation [TIFR 201 7] d dt
+
d dt
2
+
1 x= l
with t he boundary condition
= O·
dx
dt
t=O
x (t) It=O
'
=-
1 2
Prob 1.36. Consider t he two equations [TIFR 2018] . x2 y2 (i) + = 1 (ii) x 3 - y = 1. 2 3 How many real simultaneous solutions does this pair of equations have? Prob 1.37. If y(x) satisfies t he differential equation y'' 4y' + 4y = 0, with boundary conditions y(O) = 1 and y'(O) = 0, t hen y(- 1/2) = [TIFR 2018] 2
(a) e
1 1 (b) 2 e + e
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1
(c) e
14
e
(d) - 2 Physicsguide
Diff. eq and Sp. func tions
Prob 1.38. The solution of t he differential equation for y(t) : , ')
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Diff. eq and Sp. functions
@Sk J ahiruddin , 2020
Prob 1.38. The solution of the differential equation for y(t) :
0 and
!~
= 0 is:
[GATE 2010]
t=O
(a) ½cosh( t )+tsinh (t) (d) t sinh (t)
(b) - sinh( t)+t cosh( t)
(c) t cosh(t)
Prob 1.39. \iVhat is t he maximum number of extrema of x 4 x2) t he function j(x ) = Pk(x )e- ( 4 + 2 where x E (-oo, + oo)
and Pk(x) is an arbitrary polynomial of degree k? 2015] (a) k + 2
(b) k + 6
(c) k + 3
[JEST
(d) k
Prob 1.40. The Bernoulli polynomials Bn(s) are defined by xs
xe = ~ B (s) £ . Which one of t he following relations ex - l LI n n! is true? [JEST 2015] xex(l-s) xn (a) ex - l = L Bn(s) (n + l ) ! x ex(l-s ) x'n (b) - - = I:Bn(s)(- l)n( )! ex - l n+l . x(l-s) n (c) xe = I:Bn(-s) (-l) nx ex - l n! x(l-s) n (d) xe = L Bn(s)(- l )nx I ex - l n. [email protected]
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Physicsguide
Diff. eq and Sp. functions
00: 00: 42
15
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Diff. eq and Sp. ftmctions
Prob 1.41. The solution of t he integral equation: X
f( x)=x -
dt
f (t)
0
has t he graphical form:
[TIFR 2014]
1
1
0.8
0.8
(a)
---._ 0.6
---._ 0.6
~
~
""-, 0 .4
~ 0.4
0.2
0.2
0 0
1
2
0
4
3
(b)
5
0
1
2
X
3
4
5
3
4
5
X
1.2
1
1 0.8
0.8
---._ 0 .6
---._ 0.6
~
(c)
~ 0 .4
~ 0 .4
0.2
0.2
0
0 0
1
2
(d)
~
3
4
0
5
1
2 X
X
Prob 1.42. Consider t he differential equation:
dy y+ dx
with t he boundary condition that y(x) = 0 at x = 1/5. When plotted as a function of x, for x > 0, we can say with j [email protected]
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Physicsguide
11 oo : oo : 44 with t he boundary condit ion that y(x) = 0 at x = l / 5. When plotted as a function of x, for x > 0, we can say with j [email protected]
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Physicsguide
Diff. eq and Sp. functions
@Sk J ahiruddin , 2020
certainty t hat the value of y
[TIFR 2015] (a) first increases, then decreases to zero (b) first decreases, then increases t o zero (c) has an ext remum in the range O < x < l (d) oscillat es from posit ive to negative with amplit ude decreasing to zero Prob 1.43. The function y(x) sat isfies the different ial equa-
t ion : x
!~= y(ln y -
ln x
+ 1)
wit h the init ial condit ion
y(l ) = 3. What will be the value of y(3) ? [TIFR 2016] Prob 1.44. Let Pn(x) (where n = 0, l , 2, ...) be a polynomial of degree n wit h real coefficient s, defined in the interval 4
2 < n < 4. If
Pn(X)Pm(x)dx = bmn, t hen 2
[NET June 2011]
(a) Po(x)
=
~ and P1(x) =
(b) Po(x)
=
~ and P1(x)
=
(c) Po(x) = ~ and P1( x) = (d) Po(x)
=
~ and P1(x) =
~( -3 - x) v'3(3 + x)
~(3 -
x)
11 oo:oo:47 (c) Po (x) = ½ and P1 ( x) = (d) Po (x) =
~ ( 3 - x)
}2 and Pl (x) =
[email protected]
~ ( 3 - x)
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Diff. eq and Sp. func tions
@Sk J ahiruddin, 2020
Prob 1.45. Let y(x) be a continuous real function in the range O and 21r satisfying the inhomogeneous differential equation:
. d2 y sin x dx 2
dy 1r + cos x dx = fJ x - 2
The value of :~ at the point x =
1r /
2 is
[NET June
2012] (a) is continuous (b) has a discontinuity of 3 (c) has a discontinuity of 1/ 3 (d) has a discontinuity of 1 Prob 1.46. Which of t he following is a self-adjoint operator in t he spherical polar coordinate system (r, 0, )? June 2012] in 8 in 8 (c) - sin 0 80 (a) - sin2 0 80
[NET
Prob 1.47. The function f (x) obeys the different ial equaand f (x) ➔ 0 as x ➔ oo. The value off (1r) is [NET Dec 20121
11 oo : oo : so Prob 1.47. The function f(x) obeys the differential equaand f (x) ➔ 0 as x ➔ oo. The value of f(1r) is [NET Dec 2012]
(a) e21r
(b) e- 21r
(
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c) -e- 21r
(
d) -e 21ri Physicsguide
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@Sk J ahiruddin , 2020
Diff. eq and Sp. functions
(We beleive there is a printing mistake. The given eq should d2f be dx 2 - ( 3 - 4i) f = 0) Prob 1.48. The graph of t he function f (x) shown below is best described by [NET Dec 2012]
1 r.------------, 0.75 0.5 0.25 ~
~
~
0.0 -0.25 -0.5 -0.75 -1 .0 .____ _ _ _ _ _ ___. o 1 2 3 4 5 6 7 8 9 10 X
(a) The Bessel function J0 ( x)
(b) cos x
(c) e- x cos x
(d) 1/x cosx Prob 1.49. The solut ion of the partial differential equation 32
.
-
32
-
11 oo : oo : s3 (a) The Bessel function J0 ( x)
(b) cos x
(c) e- x cos x
(d) 1/x cosx Prob 1.49. The solut ion of the partial differential equation
a2
a2
at 2 u(x, t) - ax 2 u(x, t) = 0 satisfying the boundary conditions u(O, t) = 0 = u(L , t) and
j [email protected]
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Diff. eq and Sp. ftmctions
@Sk J ahiruddin , 2020
initial conditions 21rx
t •
t=O
L
[NET June 2013]
1s:
L (a) sin(1rx/L ) cos(1rt/L ) + 1r sin(21rx/L) cos(21rt/L ) 2 (b ) 2 sin(1rx/L ) cos(1rt/L) - sin(1rx/L ) cos(21rt/ L ) L (c) sin(1rx/ L ) cos(21rt/ L ) + - sin(21rx/ L ) sin(1rt / L ) 7r
(d) sin(1rx / L ) cos( 1rt / L ) +
L 1r sin(21rx / L) sin(21rt / L ) 2
Prob 1.50. The solution of the differential equation : x 2 with t he initial condition x(O) to:
(a) 1
(b) 2
= 1 will blow up as t t ends [NET June 2013]
(c) 1/ 2 (d) oo
Prob 1.51. The function
0 is described by: 2 2 2 2 (a) e - (x - v t ) (b) e - (x- vt)
j [email protected]
[NET June 2015] (
C)
21
~
e-(x-vt)
2
Diff. eq and Sp. functions
+ e-(x+vt)
Prob 1.54. If y
(a) ln
(c) ln
y
+
1
y- l y- l y+ l
2
P hysicsguide
@Sk J ahiruddin , 2020
(d)
!e - (x - vt) + !e - (x+vt) 2
=
2
1
h
.
tan x
, t hen x 1s
[NET Dec 2015]
y- l (b) ln y+ l y+ l (d) ln y- l
Prob 1.55. The solution of t he different ial equation dx
dt
= 2J1 - x2
with initial condit ion x = 0 at t = 0 is (a) x
= sin 2t
O ::; t
~ and x = 1 t > ~
t ~ ~
11 oo : 01 : oo with initial condition x = 0 at t = 0 is [NET Dec 2015] (a) x = sin 2t O :::; t < ~ and x = sinh 2t t ?: ~ (b )x = sin 2t O :::; t < ~ and x = 1 t > ~ (c)x = sin2t O < t < j and x = 1 t > l (d)x =l -cos 2t, t >O
Prob 1.56. The Hermite polynomial Hn(x) satisfies t he differential equation: 2
d Hn dHn ( ) dx 2 - 2x dx + 2nHn x = 0 The corresponding generating function
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satisfies the equation: 2
8G BG BG (a) 8x2 -2x ox +2t at = 0
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Physicsguide
Diff. eq and Sp. functions
[NET Dec 2015]
(b)a2c - 2xac -2t2aG 8x 2 ox at
=0
2 2 8 G BG BG a G aG a G (C) f) x~? - 2x f) X + 2 f) t = 0 (d) 8x 2 - 2x OX + 2 axat = 0 Prob 1.57. The Gauss hypergeometric function F (a , b, c, z ) defined by the Taylor expansion around z = 0 as F (a, b, c, z) = 2
f n=O
a(a + 1) ... (a + n - l )b(b + l ) .. . (b + n - 1) zn c(c + 1) ... (c + n - l )n!
satisfies the recursion relation: [NET June 2016] d C (a) rl7. F (a, b, c, z) = n.h F( a - 1, b - 1, c - 1, z)
11 oo : 01
: 03
L a(a + I)... (a + n -
I )b(b + 1) ... (b + n - 1) zn
c(c + l ) ... (c + n - l )n!
n =O
satisfies the recursion relation: [NET June 2016] d C (a) dz F (a, b, c, z) = abF (a - 1, b - l , c - 1, z ) d C (b)dzF(a,b,c,z) = abF(a+ l , b+ l ,c+ l ,z) ab d (c) d F (a, b,c, z ) = -F(a - l , b - 1,c- l , z) Z C d ab (d)-F (a, b,c,z) = - F (a+ l , b+ l ,c+ l , z) c dz Prob 1.58. The integral equation:
cp(x, t) = A
dwdk e-ik(x-x')+iw(t-t') (21r )2 w 2 - k 2 - m 2+·iE qi(x' ,t')
dx' dt'
[NET June
is equivalent to the differential equat ion: 2016]
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Diff. eq and Sp. ftmctions
@Sk J ahiruddin , 2020
a2
(a) at 2 a2
(b) at 2
P hysicsguide
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a2
2
1
3 )
+ ax 2 - m + iE cp(x, t ) = - 6 Acp (x, t a2 2 . 2 ax 2 + m - iE cp(x, t) = +Acp (x, t )
Prob 1.59. A stable asymptotic solution of t he equation 3 X n +l = 1 + is x = 2. If we take X n = 2 + En and 1 +xn Xn+ l = 2 + En + l where En and En+l are both small, t he ratio E'Yl-1-1
.
•
r -
---
_
_
_
.
_,
11 oo : 01 : os 8t 2
-
8x 2
+m
-
•
1,E
X
'
t =-
Prob 1.59. A stable asymptotic solution of the equation 3 Xn+l = 1 + is x = 2. If we take Xn = 2 + En and 1 +xn Xn+l = 2 + En +l where En and En+l are both small, t he ratio En+ l is approximately: [NET Dec 2016] En
1 2 (c) -- (d)-3 3 Prob 1.60. Let f (x) be the solution of the heat equation 1 (a) -2
1 (b) -4
aatf
82
f . d. . Th . . . 1 d. . t 0 = ax2 1n one 1mens1on. e 1n1t1a con 1t1on at = is f (x, 0) = e- x for -oo < x < oo. T hen for all t > 0, D
2
00
1f]:
dxe- ax 2 =
f (x, t) is given by [Useful integral -oo
a [NET D ec 2016] 2
1 x (b) 1 + 2Dt exp{ - 1 + 2Dt}
1
x
2
x2
(d) exp{---} 1 + Dt
(c) 1 + 4Dt exp{- I + 4Dt} j [email protected]
@Sk J ahiruddin, 2020
Physicsguide
24
Diff. eq and Sp. functions
Prob 1.61. Consider the differential equation
dG(x) + kG(x) = b(x) dx where k is a constant. Which of the following statements is t rue? [JEST 2013] (a) Both G(x) and G' (x) are continuous at x = 0 (b) G(x) is continuous at x = 0 but G'(x) is not. (c) G (x) is discont inuous at x = 0 I
, \
f"'""T"'I ,
,
,.--.,/ 1
\
11 oo :01 : oa true? [JEST 2013] (a) Both G(x) and G' (x) are continuous at x = 0 (b) G(x) is continuous at x = 0 but G' (x) is not. (c) G (x) is discont inuous at x = 0 (d) The cont inuity propert ies of G (x) and G' (x) at x = 0 depend on the value of k
Prob 1.62. Given that
t he value of H4 (0) is (a) 12 (b) 6
[NET June 2013]
(c) 24
(d) -6
Prob 1.63. Consider t he differential equation
!; + ay =
with t he initial condition y(O) = 0. T hen the Laplace t ransform Y (s) of t he solut ion y(t ) is [NET Dec 2017] 1 1 e-bt
(c) a(s + b)
(a) (s + a)(s + b) [email protected]
@Sk J ahiruddin , 2020
25
Physicsguide
Diff. eq and Sp. functions
e-a, - e-b
(d)
b- a
Prob 1.64. T he generating function G(t , x) for t he Legendre polynomial is
11 oo : 01
: 11
Prob 1.64. The generating function G(t,x) for t he Legendre polynomial is
G(t,x) for xl < 1. If the function
=
f (x)
J
1
00
1 - 2xt + x 2
=
L xn Pn(t) 0
is defined by the integral equation
fox f(x')dx' = xG(l , x), it can be expressed as [NET Dec 2017] 00
(a)
L
00
(b)
xn+m Pn(l )Pm(l /2)
xn+m Pn( l )Pm(l )
n,m=O
n,m=O
00
00
(c)
L
L
Xn-mPn(l)Pm(l ) n,m=O
n,m=O
Prob 1.65. For which of the following conditions does the 1
Pm(x)Pn(x)dx vanish for m =/- n where Pm(x)
integral 0
and Pn(x) are t he Legendre polynomials of order m and order n respectively? [JEST 2018] (a) all m , m =/- n (b) m - n is an odd integer
26
jahir@physicsguide. in
Physicsguide
@Sk J ahiruddin , 2020
Diff. eq and Sp. functions
(c) m - n is a nonzero even integer
(d)n= m± l Prob 1.66. If y(x) satisfies
dy = ,,,fl
-4-
/ 1110- ,, ,\ 21
11 oo : 01
: 13
(c) m - n is a nonzero even integer
(d)n=m± l Prob 1.66. If y(x) satisfies
:~ = and y(O)
(a) 0
y [1 + (log y)
2
]
= 1 for x > 0, t hen y( 1r / 2) is (b) 1
[JEST 2018] (d) oo
(c) 1r/2
Prob 1.67. Which one of the following curves correctly represents (schematically) the solut ion for t he equation
df dx
+ 2f
=
3;
f (0) = 07 [JAM 2018]
j [email protected]
27
P hysicsguide
@Sk J ahiruddin, 2020
f(x)
Diff. eq and Sp. ftmctions
f(x) (a)
(b)
1/2 ----------------------
3/2 ----------------------
11 oo : 01
: 16 Diff. eq and Sp. functions
@Sk J ahiruddin , 2020
f(x)
f(x) (a)
(b)
1/2 ----------------------
3/2 ----------------------
X
X
f(x)
f(x) (c)
(d)
1/2 ----------------------
3/2 ---------------- ------
X
X
Prob 1.68. MSQ Let J(x) = 3x6 - 2x 2 - 8. Which of the following statements is (are) true? [JAM 2018] (a) The sum of all its roots is zero. (b) The products of its roots is -8/3 (c) The sum of all its roots is 2/3. (d) Complex roots are conjugates of each other. Prob 1.69. Consider the following ordinary differential equat ion dx
2
dt [email protected]
28
@Sk J ahiruddin, 2020
wit h the boundary condit ions
_ dx= O
dt Physicsguide
Diff. eq and Sp. functions
11 oo :01
: 1a
Diff. eq and Sp. functions
@Sk J ahiruddin , 2020
wit h the boundary condit ions x (t = 0) = 0 and x (t = 1) = 1
The value of x(t) at t
= 2 is
[NET June 2018]
(a) ✓e - 1
(c)
Je + 1 2
Prob 1. 70. In the function Pn(x) e- x of a real variable x, Pn(x) is a polynomial of degree n . The m aximum number of ext rema that t his function can have is [NET June 2018]
(a) n
+2
(b) n - I
(c) n
+I
(d) n
Prob 1.71. The polynomial f(x) = 1 +5x+3x2 is written as a linear combination of the Legendre polynomials
as f( x) 2018]
(a) 1/ 4
= I:n en Pn(x) . The value of co is (b) 1/ 2
(c) 2
[NET Dec
(d) 4
Prob 1. 72. In terms of arbitrary constants A and B , the general solution to the differential equation j [email protected]
@Sk J ahiruddin, 2020
29
Physicsguide
Diff. eq and Sp. functions
11 oo : 01
: 21
[email protected]
P hysicsguide
29
@Sk J ahiruddin, 2020
Diff. eq and Sp. functions
dy x dx2 + 5x dx 2d2y
+ 3y = 0 [NET Dec 2018]
is
A
(a) y = -
X
+ B x3
(c) y= A x+
(b) y =Ax+~ A B (d)y=-+ 3
3 Bx
X
X
Prob 1. 73. A set of polynomials of order n are given by t he formula x2
2 The polynomial P7 ( x) of order n
= 7 is
[TIFR 2019]
(a) x - 21x + 105x - 105x 4 2 6 5 3 (b) x - 21x + 105x - 105x + 21x + x 7 (c) x - 21x 5 + 105x 3 - 105x + 21 (d) x
7
5
7
5
-
21x
3
+ 105x + 35x 3 4
105x
Prob 1. 7 4. For the differential equation
d2y dx~9
y
-
n( n + 1) X 2 = 0
where n is a constant, t he product of its two indep endent solut ions is [GATE 2019]
(b) [email protected]
X 30
Physicsguide
11 oo : 01
(b)
: 23
X
j [email protected]
P hysicsguide
30
Diff. eq and Sp. ftmctions
@Sk J ahiruddin, 2020
Prob 1. 75. The differential equation
dy x dx - xy = exp (x) where y = e2 at x = l , has t he solution y = [TIFR 2019] (a) exp (1 + x) (1 + ln x) (b) exp (x) ln x + exp (1 + x) (c) (1 - x) exp (x) + exp (1 + x) (d) exp ( x 2 + x ) Prob 1. 76. The Euler polynomials are defined by
What is t he value of E 5 (2) + E s(3)?
[JEST 2019]
Prob 1. 77. The solution of t he different ial equation
dy x dx + (l + x)y = e-x wit h t he boundary condition y(x = 1) = 0, is [NET June 2019]
(a) (x - 1) e-x (b) (x ~ 1) e-x (c) (1 - x) e-x
(d)(x- l )2e-x
x2 Prob 1. 78. The equation of motion of a forced simple harX
X
monic oscillator is x+w 2 x = A cos Dt where A is a const ant. At resonance n = w the amplit ude of oscillations at large
11 oo : 01
: 26
(a) x - l e- x (b) x ~ I e- x (c) 1 ~ x e- x (d )(x- 1)2e- x X
X
X
Prob 1. 78. The equation of motion of a forced simple har2 monic oscillator is x+w x = A cos Dt where A is a constant. At resonance n = w t he amplit ude of oscillations at large [email protected]
31
Physicsguide
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D iff. eq and Sp. functions
[NET June 2019]
t imes (a) sat urates to a finit e value (b) increases wit h time as v't (c) increases linearly wit h time (d) increases exporent ially wit h t ime
Prob 1. 79. Which one of t he following is a solut ion of
for k real?
(A) e- kx
(B) sin kx
(C ) cos kx
[GATE 2020] (D ) sinh x
Prob 1.80. T he solut ion of the different ial equat ion y'' 2y' - 3y = e2t is given as C 1 e-t + C 2 e2 t + C3e 3t T he values of the coefficieients C1, C2 and C3 are: [JEST 2020] (A) C1, C2 and C3 are arbitrary (B) C1 , C3 are arbit rary and C2 = - 1/ 3 (C) C 2 , C3 are arbit rary and C 1 = -1/ 3 (D) C1, C2 are arbitrary and C3 = -1/ 3 Pro b 1. 81. Some bact eria are added to a bucket at t ime 10 am . The number of bacteria doubles every minute and reaches a number 16 x 1015 at 10: 18 am. How many seconds
11 oo : 01
: 29
(D) C 1, C2 are arbitrary and C3
==
-1/3
Prob 1. 81. Som e bacteria are added to a bucket at t ime 10 am. The number of bacteria doubles every minute and 15 reaches a number 16 x 10 at 10: 18 am. How m any seconds after 10 am were there 25 x 10 13 bacteria? [JEST 2020]
32
j [email protected]
P hysicsguide
Diff. eq and Sp. funct ions
@Sk J ahiruddin , 2020
Prob 1.82. Which one of the following funct ions has a discontinuity in the second derivative at x == 0, where x is a real variable?
(A) f (x) == x 3 2 (D) f (x) == x
(B) f( x) == x x
Prob 1.83. lim x x is equal to
[JAM 2020]
x ➔O+
(A) 0
(B)oo
(C) e
[JAM 2020] (C) f (X) == cos (IX )
(D) 1
Prob 1.84. If a function y(x) is d escribed by t h e initial2
d y 1 bl va ue pro em , dx 2 y(O) == 2, and
dy
+
5 dx dy
+
6y == 0 , wit . h 1n1t1a . . . 1 con d·t· 1 ions
== 0, t h en the value of y at x == 1 is
dx x=O - - - - - - - - ? (R ound off to 2 decimal places) [JAM
2020]
11 oo : 01
[email protected]
: 32
Physicsguide
33
Diff. eq and Sp. functions
@Sk J ahiruddin , 2020
1.1
Ans keys Ans keys
1.1. d
1.17. a
1.2. b
1.18. d
1.3. b
1.19. b
1.4.
-½ ln 6
1.20. a
1.5. d
1.21. b
1.6. 27.0
1.22. d
1. 7.
C
1.23. d
1.8. d
1.24 . d
1.9. a
1.25. b
1.10. d
1.26. a
1.11. a
1.27.
C
00 : 01 : 34
1.9. a
1.25. b
1.10. d
1.26. a
1.11. a
1.27.
1.12.
1.28. b
C
C
1.13. d
1.29. 0.51 to 0.53
1.14. b
1.30. d
1.15.
1.31. -2 to + 2
C
1.32. 0.25 to 0. 29
1.16. d [email protected]
Physicsguide
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@Sk J ahiruddin, 2020
Diff. eq and Sp. ftmctions
1.33. b
1.52.
1.34. A n
1.53. d
1.35. e- 2t - 2e-t + ~
1.54. d
1.36. 2
1.55.
1.37. a
1.56. a
1.38. d
1.57. d
1.39.
1.58. d
C
C
C
1.40. d
1.59.
C
1.41. b
1.60.
C
1.42.
1.61. b
C
1.4 3. 81
1.62. a
1.44. d
1.63. a
1.45. d
1.64. b
00 : 01 : 37 1.42.
1.61. b
C
1.43. 81
1.62. a
1.44. d
1.63. a
1.45. d
1.64. b
1.46.
C
1.65.
1.47.
C
1.66. d
1.48. a
1.67. b
1.49. d
1.68. a,b,d
1.50. a
1.69.
C
1.51. a
1. 70.
C
j [email protected]
C
35
P hysicsguide
@Sk J ahiruddin , 2020
1. 71.
C
Diff. eq and Sp. functions
1. 78.
C
1. 72. d
1. 79. a
1. 73. a
1.80. b
1. 7 4. b
1.81. 720
1. 75. b
1.82. b
1. 76. 64
1.83. d
1. 77. a
1.84. 0.60 to 0.62
11 oo : 01
[email protected]
: 39
Physicsguide
36
Diff. eq and Sp. func tions
@Sk J ahirudd in, 2020
1.2
Solutions
Sol 1.1. Let us consider a hexagon , t hat can be divided in 6 triangles of equal arm length. Now, we try to cover up t he area with circles of radius ; as shown in the figure . . . Now, t he area of a triangle 1s,
=
1 x ax 2
v'3a 2
2
3v'3a 2
11 oo : 01
: 42
a
l
Again we can see that , in each triangle t he shaded region is covered by the circle which is same as half of the area of a circle. So, the shaded region in any t riangle is I
-7r
2
a
2
1ra2
-
2
8
So, t he fraction of area covered , [email protected]
~1r_il 4
= v'33,ii 3
2
2\1'3 Physicsguide
37
@Sk J ahiruddin , 2020
Diff. eq and Sp. functions
Sol 1.2.
implies f = f (x, t ) aJ aJ df = ax dx + at dt df = j = af ax + af dt ax at at aJ. aJ = ax x + &t Hence
.
7r
•
aJ at
aJ ax
Sol 1.3. Let us consider a t rial function x .
11 oo : 01
: 44
Hence
•
of
of
ot
ox
Sol 1.3. Let us consider a t rial function x .
d - - x dx
d dx +x
f = =
2
. d y Sol 1.4. Given that, dx 2
d df - - x dx +xf dx d2 f d 2 dx2 - x dx - x f + f d2 dx2 - x2 + 1 f
dy + 5 dx
j [email protected]
=0
Physicsguide
38
Diff. eq and Sp. ftmctions
@Sk J ahiruddin, 2020
Now we consider that,
d + x dx
!~ = v dv dx
+ 5v = 0
dv = - 5dx V
In v= -5x
+c
11 oo : 01
: 47 dv = -5dx V
In v= -5x + c dy Now at x= O v = - = 2 dx V ln - = - 5x
c
v
2
dy dx
=
=
Now at x=O
y = 2
y
2e- 5x
5
12
2
+d
= - - -e
d = 12 5 - Sx
5 Now a t y= O e-5x = 6
[email protected]
=
2e- 5x
2 --e-5x
y
= ln 2
5
=}
x =
1 -- ln 6 5
39
P hysicsguide
Diff. eq and Sp. functions
@Sk J ahiruddin, 2020
Sol 1.5. dy
y2
dx dy
x dx
y2
X
1
- - = ln x+c y
11 oo : 01
: 49
dx dy
x dx
y2
X
1 - -= ln x+c y
⇒
Now, at x = l , y= l
c= - 1
1 - - = ln x- 1 y
1 y= 1 - ln x
So, t o have y value real and finit e, we need to have a range of x that doesn 't include 1 or 0. So, the only option will be
3< -
< 00 .
X -
. . df 8f 8f df Sol 1.6. Using Chain Rule, - = a + a x d dx x y x 2 3 3 Now, here f (x, y) = x + y and y = x + 1. J _ 2 . J _ 2 df _ So, ax - 3x , ay - 3y , dx - 2x .
a
a
• • •
• • •
j [email protected]
!~ = 3x df dx
2
=
2
2
+ 3(x + 1) x 2x 3 + 6 x (1 + 1)2 = 27
x=l
40
@Sk J ahiruddin, 2020
Physicsguide
Diff. eq and Sp. functions
Sol 1. 7. We can write t he equation as, 2x dy 3 3 1 d2 y - 2 - 2x - - -2 + - ( - + 1) - - y =O 2 dx (1 - x ) dx 2 2 (1 - x ) 2x
11 oo : 01 : s2 Sol 1. 7. We can write the equation as, 2x d2 y dy 3 3 1 - 2 - 2x - - -2 + - ( - + 1) - - y2= O dx (1 - x ) dx 2 2 (1 - x ) 2x P (x) = - 2x ( ) 2 1 -x 3 3 1 Q(x) = 2(2 + 1) (1 - x2).
So, and
Both P(x) and Q(x) are non-analytic at t he point x 2 That means at x
=
1.
= ±1
Sol 1.8. Given t hat,
dy x dx
+ y = x4
dy 1 - + -y= dx x
⇒
X
3
which is a linear first-order differential equation of t he form
dy dx + P y= Q who's solution is given by where Now, here in this problem , P
l =
Pdx
1
= -. X
I =
Pdx =
dx
eI = eln x = X
- = ln x X
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41
Physicsguide
Diff. eq and Sp. functions
@Sk J ahiruddin , 2020
and
e
-J
1 =X
So, we get,
11 oo : 01 : s4 Diff. eq and Sp. ftmctions
@Sk J ahiruddin, 2020
e
and
l =-
- I
X
So, we get, •
yx
• •
dx
=
X
4
+C
x5
=-+ c 5
x
4
C
y=-+-
⇒
5
Now at x=l , y= l
⇒
x4
⇒
y
X
c = 1 - 1/5 = 4/5 4
= 5 + 5x
So l 1.9.
dx
k1 dt = -k2x + k3 dx dt
k3 - k2x k1 dx dt k3 - k2x k1 1 t - - ln(k3 - k2x) = k2 k1
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42
+c
P hysicsguide
Diff. eq and Sp. ftmctions
11 oo :01 : s1
D iff. eq and Sp. functions
@Sk J ahiruddin , 2020
at t=O x=O
'
In
k3 - k2x k3 - k2 x - I2t - - - =e k 1 k3
So, as t -+ oo
k3 x = k , which means it approaching ,.2
t owards fixed value. So, option a satisfies t he funct ion perfect ly. Sol 1.10. You don 't need to do explicit calculations. Just use
Ot herwise use that the Legendre P olynomials are orthogonal. Then x Pn -+ Pn+l
j [email protected]
(c)Sk J a,hir11ddin. 2020
+ Pn-1 ·
So in t he first option
43
P hysicsguide
D iff. ea and So. ft1nctions
11 oo : 02 : oo
43
j ahir@physicsguide. in
@Sk J ahiruddin , 2020
➔ (Pn+2
P hysicsguide
Diff. eq and Sp. functions
+ Pn + Pn + Pn- 2) Pn+l
The integral will be zero according to t he ort hogonality. Again in option (b)
The integral is also zero. In opt ion (c)
This again goes to zero. In opt ion (d)
➔ (Pn+2
+ Pn + Pn + Pn- 2) Pn+2
In t his int egral t here will be a integral P n+2 P n+2 which will give non zero value.
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11 oo : 02 : 03
44
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Diff. eq and Sp. functions
@Sk J ahiruddin , 2020
Sol 1.11. dy dx
y
(y
+ l)dy = -
y2
x2
X
+1 xdx
2 +y = - 2 +c
1 2 x2 1 - [y + 2y + 1] + - - - = 2 2 2 2 2 x + (y + 1) = (1 + c)
C
So, the equation represents circles having different radii depending on the value of the c. f [n + l ] = n f [n] . Now,
Sol 1.12. We know,
r
1 1 n+- = f n--+ 1 2
2
1
n- 2
1
n - -
2
r
1
n- 2 3 5 n - - n - - ··· 2 2
n - (2n -
1
1
1)2 r 2
(2n - 1)(2n - 3) (2n - 5) · · · l =---------~ 2n = 2n(2n - 1)(2n - 2) (2n - 2)(2n - 4) · · · 2.1 ~ 2n(2n - 2)(2n - 4) • • • 22n
11 oo : 02 : os -
-
2 2 (2n - 1) (2n - 3) (2n - 5) · · · 1 =---------~ 2n 2n(2n - 1)(2n - 2) (2n - 2)(2n - 4) · · · 2.1 = - - - - - - - - - - - - -~ 2n(2n - 2)(2n - 4) • • • 22n [email protected]
P hysicsguide
45
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Diff. eq and Sp. ftmctions
(2n) !fo 2n_n!.2n
(2n) !fo 22nn!
Sol 1.13. Let us consider a trial solut ion, y = em>-. . P utting in the equation we get 2
m - 1= 0
m=±l
So, t he general solution will become, y
= A et + B e- t.
Now,
y(O) = 1 and
y(oo) = 0 ⇒
Sol 1.14. g(x) =
A+B = l
⇒
A=O
⇒
y = e- t = cosh t
xe-x
2
-xe-x
- sinh t
x >0 2
x O
=
-e-x
g'(x)
x=O-
m =
±(-2 + i)
Now t he solution can be written as e- 2x ( A cos x
+ B sin x)
or
e2 x ( A cos x - B sin x)
11 oo : 02 : s2 m
2
=
(3 - 4i)
~► m = ±(-2
+ i)
Now t he solution can be written as 2
e- x(A cos x
+ B sin x)
or
2
e x(A cos x - B sin x)
Now according to the boundary condition t he solut ion converges to zero as x solut ion. And from
-e
➔
f
oo. So we will take only t he first (O) = 1 we get A= - 1 Hence f (1r) =
- 27r
Sol 1.48. This is a plot of Bessel funct ion J0 (x) . For other options, 62
j [email protected]
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@Sk J ahiruddin, 2020
Diff. eq and Sp. funct ions
(b) we know the plot of cos x . The amplitude doesn 't decrease t here. (c) This function doesn 't have a minima at x property of given plot.
= 3, which
is
(d) This function blows up at x = 0. Sol 1.49. Given the init ial condition is, u(x, 0)
= sin(1rx/ L )
We can check that , only option (c) and (d) satisfies t his boundary condition. Now, we will check t he second boundary condition, In between t his two options (c) doesn 't satisfy the different ial equation. and option (d) satisfy t he boundary condit ion. We can say t his from symmetry of the function. Sol 1.50. Given that,
dx
-=x dt
2
⇒
dx -x2 = dt
⇒
1
-- = t + C X
11 oo : 02 : 55 t ion . We can say t his from symmetry of t he function. Sol 1.50. Given that,
dx 2 -= x => dt
dx -=dt=> x2
1
- - = t +C X
Now given t hat
x(O) = 1 =>
1
c= - 1 =>
x=-1 -t
So, x will blow up if t tends to 1. Sol 1 .51. Just do the part ial derivatives. Sol 1 .52. (Same qs has been asked before .. Check t he previous problems) [email protected]
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Sol 1 .53. The main work in t his problem is to do part ial derivatives of the solutions and mat ch with the given bound-
for option (d) and # 0 for all other options. Sol 1 .54. y
=
1 t anh x ex + e- x =y ex - e- x
ex e- X
y+ l y- 1 x = In
1 tanh x
⇒
y+l y- 1
=y
y+ l e2x = y- 1
11 oo :02 : s1 ⇒
ex e- x
y +l y - l
x
e
= In
Y + l =--
2x
y- l
y+l y- l
Sol 1.55.
= 2J1 -
dx
dt
dx
Jl -x2
x2
= 2t
sin - l x = 2t + c From boundary condition we get c So, x = sin 2t.
= 0.
For option (a) and (d) the differential equation is not satisfied. 64
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Again from the given equation, we get that :
= 0
x= l •
1.e.
x = sin 2t
= 1
⇒
7r
2t = -
⇒
2 So the correct option will be option (c) .
G(t ,x) =
Again,
4
1
1
n=O
So,
= -
L n. Hn(x)tn 00
Sol 1.56. Given that,
t
7r
11 oo : 03 : oo _ _ ~ -tn n 8x L-t n ! dx n=O
So,
a c ~ -tn 1 d Hn ___ 8x 2 L-t n ! dx 2 n=O 00 8G _ ~ 1 ndHn t t - L-t n n.1t dX 8 n=O 2
Again, And, Now,
a2 c ax9
-
00
2
ac ac 2x aX + 2t at
d oo d2 _ ~ 1 n Hn _ ~ l n Hn - L-t n.I t dx 2x L-t n.I t dx 2 oo
n=O
d ~ l n Hn + 2 L.-; n n.I t dx oo
n=O n= O 00 tn d2 H dH d n2xdri, + 2nHn(x) 2 I n. X X n= O
=L
=0
Sol 1.57. Again, don't be afraid just seeing the name of '' Gauss hypergeometric function'' . This problem is just a partial derivative calculation. j [email protected]
@Sk J ahiruddin, 2020
dF dz
Physicsguide
65
Diff. eq and Sp. functions
=f
a(a + 1) ... (a+ n - l )b(b + l ) ... (b n=O c(c+ l ) ... (c+n-l)n!
+ n - l ) nzn-l
Now write n ! as n(n - 1)! in t he denominator
=
f
a(a + 1) ... (a+ n - l)b(b + 1) .. . (b + n - l ) zn -l n=O c( c + 1) . .. (c + n - l ) (n - l ) ! ab , , (a + l ) . . . (a + n - l ) (b + l ) . . . (b + n - l )
zn- l
11 oo : 03 : 02 =
f
a(a + l ) ... (a +n -l )b(b + l ) .. . (b+n-1 ) zn-l n= O c( c + 1) ... (c + n - l ) (n - l )!
= ab "" (a+ 1) ... (a + n - l )(b + 1) . .. (b + n - 1) c ~
(c+ l ) . .. (c+n- 1)
zn- l
(n- 1)!
Now write the derivative in this way
= ab c
f (a + 1) . . . [a + (n - 1) - 1](b + 1) . .. [b + (n - 1) - 1] (c
n=O
+ 1) . . . [c + (n -
1) - 1]
z n-I
(n - 1) !
dF ab Hence we get d = -F (a + 1, b + 1, c + 1, z ) Z
C
Sol 1.58. Take Fourier transformation bot h side of the equat ion as done in a example problem in Fourier transformation chapter ?? Sol 1.59. Write the equation in terms of
2 + En+ 1
3 = 1 + l + (2 + En)
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Hence
=>~>
66
En
and En+l
3 -En En+l = - - - 1 = - 3 + En 3 + En P hysicsguide
Diff. eq and Sp. functions
- 1 - 1 +E-+ as En-+ 0 3 3 En Sol 1.60. You can solve the partial differential equation directly. But t hat process is little lengthy. I recommend to directly check t he options by substit ut ing them in t he orig-
En+l
inal equation. Just a few partial derivative calculations are enough. Verify the ans.
11 oo : 03 : os rectly. But t hat process is little lengthy. I recommend to directly check t he opt ions by substit u t ing them in the original equation. Just a few partial derivative calculations are enough. Verify the ans .
Sol 1.61. Here, the equation suggests t hat, G(x) is Green 's Function. And we know t hat, at x
= 0 G(x)
is cont inuous,
but G' (x) will have a discontinuity because of t he presence of o(x).
Sol 1.62.
tn
L Hn(X) n . = oo
I
e-t2+2tx
n=O
~ Hn(O) tn = e - t2 ~
n!
oo
tn
n=O
-t2
( - t2)2
H4(0)
1 4! 2! 4! H4(0) = - = 12 2! 67
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Diff. eq and Sp. functions
!~
Sol 1.63. Given that, + ay = e-bt . If we take Laplace Transformation of both side, we get
L [y'] + aL[y] = L[e-bt] sY - y 0 + aY
1
= --
s+b
1
[Where Y = L[y]]
11 oo : 03 : 01 If we take Laplace Transformation of both side, we get
L [y'] + aL[y] = L[e-bt] 1
sY - y 0 + aY = - -
[Where Y = L[y]]
s+ b
⇒
Y ( s + a) =
⇒
y =
1
s+
b
[as Yo = O]
1
(s + a)(s + b)
Sol 1.64. Let's calculate G(l, x) first
hence we see that 1
---;:==~ -
00
1
xnP 1
-- -
Now 00
xG(l , x)
x 1 -x
=
=
X
'"""" xn+l Pn( l ) L.t n=O
We can easily see that J(x) j [email protected]
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@Sk J ahiruddin , 2020
00
=
d d
=
o
J(x')dx' 1
X
x l -x
(1 -
X) 2
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Diff. eq and Sp. ftmctions
11 oo : 03 : 1o
00
=
L
Xn+m pn(l )Pm(l )
n.m=O , Sol 1.65. Let us consider first, m = n = 1 T hen P 1 (x) = x, 1
1
2
[P1( x)] dx
=
0
1
2
x dx = -
3
0
So, it does not vanishes when m = n or m - n = 0 Now we consider m = 1, n = 2 1
x[(3x
2
) -
1] =
0
4
1
5
3
2
6
- - - = -
So, it doesn 't vanishes when m - n = ± 1 or an odd int eger. So, t he only option left is (c). Let us check. Now, we consider m = 0, n = 2. 1
1
1
Po(x) P1 (x)dx = -
[(3x
2
0
) -
l ]dx = 0
0
69
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Sol 1.66.
dy
r1
,
1,
11 oo : 03 : 12 Diff. eq and Sp. functions
@Sk J ahiruddin, 2020
Sol 1.66.
:~ = y[ l + (log
2 y) ]
dy y[l + (log y ) 2 ] Now, we t ake logy ⇒ ⇒
⇒
⇒
p
=
1 - dy y
=
dp
dp --=x+c 1 + p2 1 tan- = x + c
= tan(x + c) log y = t an (x + c)
p
From the boundary condit ion we get c = 0
logy So, at
x
7r
=
2
= tan x
y = oo
d2 f Sol 1.67. Here we can see dx 2 = - 2. So, the curvature of
function doesn 't change . Now, taking 1= 2 Jx we get the
j [email protected]
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Diff. eq and Sp. functions
11 oo : 03 : 1 s [email protected]
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Diff. eq and Sp. functions
solut ion , ye 2x
=3 =
⇒
3 - e 2x 2
e2xdx
+C
+c
3 y = - + ce-2x 2
Putting the boundary condit ion we get, c = ~. 3 So, y = (1 - e 2x). 2 3 So, at x ➔ oo y = . 2 Sol 1.68. For a general polynomial,
f (X) =
axn
b Sum of t he roots =-a
+ bx n - l + · · · + z z
and product of the roots =
a
Even degree polynomial
z Odd degree Polynomial a Here, the polynomial is a even degree polynomial, where
a = 3, b = 0, z = - 8.
8 So, sum= O and product=-= --. a 3 z
And , as t he product is real and sum is zero, t h at means complex roots are conj ugate of each other. [email protected]
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11 oo : 03 : 11 complex roots are conjugate of each other. j [email protected]
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Sol 1.69.
dx dt put dx dt Then
2
_dx=O dt
=P d2x dt 2
=
dp dt
=
dp P dx
So t he equation becomes
dp 1 2 p-+-p -p = 0 dx x or
dp dx
+P-
x The integrating factor is exp
1=0
(J ~ ) =
x
The equation now
dp xdx +p-x=O d dx (xp) x2
Xp =
2
=x
+ C1
dx p= - = (x/2+c1/x) dt dx = d t X + C1 2
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2
=
x + 2c1 -2x
X
72
P hysicsguide
11 oo : 03 : 20
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Diff. eq and Sp. functions
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Now t he integration is easy
dt
= 2x dx x
put x 2 + 2c1 = z
~>
2xdx
=
2
+ 2c1
dz
So t he integration
dz = dt z or
x
2
+ 2c1 = X2
=
z = c2et t
C2e -
2C1
Now we will apply t he boundary conditions x(t
0, x(t
= 1) = 1
And
Hence at t - 2 2 - 1 e 2 x = - - ~~> x = ✓e + 1 e- 1
0)
11 oo : 03 : 23 e1 -
1
Hence at t - 2
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Diff. eq and Sp. functions
Sol 1. 70. When a function is extremum (maximum or minimum) we know dy
dx
= 0 Hence
P~(x)e-x
2
2
+ Pn(x)(- 2x)e-x = 0
or
So we get
P~(x) - 2x Pn(x) = 0 Pn(x) is an algebric polynomial of order n. The x multiplied with it make is order n
+l
From the theory of differential equation an equation of order n can have maximum n distinct root. As our equation is of order n + l , so the n + l roots will be able to create maximum (n + l ) extrem as.
Sol 1. 71. Given that ,
Now,
11 oo : 03 : 26
Now, P2 ( x)
=
~ (3x
2 3x = 2P2(x)
jahir@physicsguide. in
@Sk J ahiruddin,
2
+1=
1)
-
2P2(x) + Po(x)
74
Physicsguide
2020
Diff. eq and Sp. functions
Now,
J(x) = l + 5x + 3x
2
+ 5P 1(x) + 2P2 (x) + Po(x) =2Po(x) + 5P1 (x) + 2P2(x)
= Po(x)
Now, cornparing with
we get
co = 2 Sol 1. 72. Let us first consider here , z
=
In x
So,
dy dy dz 1 dy dx dz dx x dz dy dy . . . [i] X-=dx dz d IT. dy l = d dy = d
I l
Idy l dz
11 oo : 03 : 2a
⇒
dy dydz 1 dy dx dz dx xdz dy dy .. . [i] X = dx dz dy - d dy d d dy dz dz dz dx dx x dx dx dz d2 y dy 1 d2y x dx 2 + dx xdz 2 d2y 2d2y dy · ·· [ii] x dx 2 + x dx dz 2
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Diff. eq and Sp. ftmctions
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Now,
dy x dx 2 + 5x dx 2d2y 2
y dy dy x dx2 + x dx + 4x dx + 3y = 0 d2 y dy [iii] dz 2 + 4 dz + 3y = 0 2d
⇒
+ 3y = 0
Now, let u s con sider the trial solution is, y
[iii] we get that, 2
m +4m + 3 = 0
(m + l )(m + 3) = 0 m
=
-1 , m
= -3
So, t he general solution is ,
+ B e- 3z = Ae-ln x + B e- 3ln x
y = A e-z
= em z
So, from
11 oo : 03 : 31 So, the general solution is, y =Ae-z + B e- 3z
= A e-ln x + Be-3ln x
A B =-+-3 x x Sol 1. 73. Don 't do the explicit calculation. It will be huge. Think of t he properties of Legendre Polynomials 2
Sol 1.74. Write the equation as x 2 d solution is obvious y 1 j [email protected]
@Sk J ahiruddin,
;
dx
= n(n + l)y. One
= xn+l within a constant factor. As 76
Physicsguide
2020
Diff. eq and Sp. functions
we see from t he general form of 2nd order equation y'' + P (x )y' + Q(x )y = 0 t he P (x) = 0 in our case. So the second solution is X [Y1 (x2)]
2
(Refer to Arfken , 7th ed Section 7.6) So
dx x2n+2 1 2nl X ---2n- l x-n
2n+ 1 So the product of the two solutions is y 1y2
=
x
2n + 1 So t he ans would be x within a constant factor.
11 oo : 03 : 33 x-n 2n + 1 So t he product of t he two solutions is y 1y 2
=
x
2n + 1 So t he ans would be x within a constant factor.
Sol 1. 75. Write the equation as
dy exp(x) --y=--dx x P(x)
=
- 1. So the integrating factor is exp{f P(x )dx}
77
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Diff. eq and Sp. functions
e- x. Now t he solution is 1
e- xf racdydx - e- xy = -
X
d -ye-x dx
ye
-x
1 X
=
dx - = ln x+c X
Now use to the boundary condition y = e 2 at x = l to get c = e . Now t he solution is
ye- x = ln x Sol 1. 76.
+e
:=::-~>
y
= ex ln x + exp(x + 1)
11 oo : 03 : 36 ye- x
= ln x + e
Sol 1. 76.
For s = 2
Fors = 3
2e3x X
e
00
+1 =
xn
L E 1i ( 3) n . n=O I
Adding the two equations
78
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@Sk J ahiruddin , 2020
Physicsguide
Diff. eq and Sp. functions
Hence
Now expand e2x 2 l
+
2x 1!
+
(2x ) 2!
2
+
(2 x ) 3
3!
+ .. .
Now the first term in the series is for n = 0, t he second t erm is for n = 1, the t hird term for n = 2 and so on. 25x5 • Equating So the sixth t erm for n = 5 in LHS is 2 5.1 5 with x term in RHS we get
11 oo : 03 : 38 n=O 78
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Diff. eq and Sp. ftmctions
@Sk J ahiruddin, 2020
Hence
Now expand e2x
2x 2 l + 1!
+
(2x ) 2!
2
+
(2x ) 3!
3
+ .. .
Now the first t erm in the series is for n = 0, the second t erm is for n = l , the t hird t erm for n = 2 and so on. 25 5 So t he sixth t erm for n = 5 in LHS is 2 ~ . Equating 5. wit h x 5 term in RHS we get
Hence
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11
00:03:41
n=O 78
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Diff. eq and Sp. ftmctions
@Sk J ahiruddin, 2020
Hence
Now expand e2x
2x 2 l + 1!
+
(2x ) 2!
2
+
(2x ) 3!
3
+ .. .
Now the first t erm in the series is for n = 0, the second t erm is for n = l , the t hird t erm for n = 2 and so on. 25 5 So t he sixth t erm for n = 5 in LHS is 2 ~ . Equating 5. wit h x 5 term in RHS we get
Hence
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11 oo : 03 : 44 n=O 78
j [email protected]
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Diff. eq and Sp. ftmctions
@Sk J ahiruddin, 2020
Hence
Now expand e2x
2x 2 l + 1!
+
(2x ) 2!
2
+
(2x ) 3!
3
+ .. .
Now the first t erm in the series is for n = 0, the second t erm is for n = l , the t hird t erm for n = 2 and so on. 25 5 So t he sixth t erm for n = 5 in LHS is 2 ~ . Equating 5. wit h x 5 term in RHS we get
Hence
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11 oo : 03 : 4s n=O 78
j [email protected]
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Diff. eq and Sp. ftmctions
@Sk J ahiruddin, 2020
Hence
Now expand e2x
2x 2 l + 1!
+
(2x ) 2!
2
+
(2x ) 3!
3
+ .. .
Now the first t erm in the series is for n = 0, the second t erm is for n = l , the t hird t erm for n = 2 and so on. 25 5 So t he sixth t erm for n = 5 in LHS is 2 ~ . Equating 5. wit h x 5 term in RHS we get
Hence
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P hysicsguide
11 oo : oo : 01
Problellls and Solutions in Integral Transforlllations Sk J ahiruddin Arnab Choudhury Assistant Professor Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay 1\11.Sc Pl1ysics 2009-2011 batch He ranked 007 in IIT J Al\11 2009 and 008 (JRF) in CSIR NET J une 2011 He has been teaching CSIR NET aspirants since 2012
1
@Sk J ahiruddin, 2020
Integral transformations
11 oo : oo : 03 1
@Sk J ahiruddin, 2020
Integral transformations
Contents 1 Problems from NET , GATE, JEST, TIFR & JAM papers
3
Ans keys .
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
23
1.2 Solutions .
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
24
1.1
j [email protected]
2
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11 oo : oo : 06
j ahir@physicsguide. in
2
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1
Integral transformations
Problems from NET, GATE, JEST, TIFR & JAM papers
Prob 1.1:
f (x) is a
periodic function of x with a p eriod of
21r. In t he int erval - 1r
0. The ratio (:), where X
f (x) = x
x is the
exp(-f) where
most probable value
and (x) is t he mean value of the variable x, is 1 (a) 2 (b) + A (c) ~ (d) 1 [TIFR 2014]
l-A
A
Prob 1.21. If x is a continuous variable which is uniformly
distributed over t he real line from x = 0 to x -+ oo according to t he distribution f (x) = exp( - 4x), t hen the expectation vaalue of cos 4x is [TIFR 2016] (a) 0 (b) 1/ 2 (c) 1/ 4 (d) 1/ 16 Prob 1.22. A bag contains many balls, each with a number j [email protected]
@Sk J ahiruddin, 2020
8
Physicsguide
Probability
painted on it . There are exactly n balls which have the number n (namely one ball wit h 1, 2 balls with 2, and so on until N on them). An experiment consists of choosing a ball at random, noting t he number on it and returning it to t he bag. If the experiment is repeated a large number of [NET t imes, t he average value t he number will tend to June 12]
(a) 2N + 1
(d) N(N + 1)
2 3 Prob 1.23. Consider three particles A , B and C, each with
an attribute S t hat can take two values: ± 1. Let SA = l , SB = 1 and Sc = - 1 at a given instant. In the next instant, each S value can change to -S with probability 1/ 3. The
11 oo : oo : 24 Prob 1.23. Consider three particles A , B and C , each with an attribute S that can take two values: ± 1. Let SA = l ,
SB = land Sc= - 1 at a given instant. In the next instant, each S value can change to -S with probability 1/ 3. The [NET probability t hat SA+ SB+ Sc reains constant is; June 13]
(a) 2/ 3
(b) 1/ 3
(c) 2/ 9
(d) 4/ 9
Prob 1.24. Let u be a random variable uniformly distributed
in the interval [O , 1] and V = -c ln(u), where c is a real const ant . If V is t o be exponent ially distributed in the interval (0, oo) with unit standard deviation, then t he value of c should b e: (a) ln (2) (b) 1/ 2 (c) 1 (d) - 1 [NET June 13]
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P robability
Prob 1.25. Let y = i(x1 + x2 - µ) where x1 and x2 are indep endent and identically distributed Gaussian random
variables of mean µ and st andard deviation O". Then (y4 ) / 0" 4 is [NET June 14] (a) 1
(b) 3/ 4
(c) 1/2
(d ) 1/ 4
Prob 1.26. Three real variables a , b and c are each randomly chosen from a uniform probability distribution in the
interval [O, 1]. The probability that a + b > 2c is June 15] (a) 3/ 4
(b) 2/ 3
(c) 1/ 2
(d) 1/ 4
[NET
11 oo : oo : 21 domly chosen from a uniform probability distribution in the interval [O, 1]. The probability that a+ b > 2c is [NET June 15]
(a) 3/ 4
(b) 2/ 3
(c) 1/ 2
(d) 1/ 4
Prob 1.27. Let X and Y be two independent random variables, each of which follow a normal distribution with t he
same standard deviation a, but with means spect ively. Then t he sum X + Y follows a 16]
+µ
and - µ re[NET June
(a) dist ribution with two peaks at ±2 and mean O and standard deviation aJ2 (b) normal distribution with mean O and standard deviation 2a (c) dist ribut ion with two peaks at ±2 and mean O and standard deviation 2a (d) normal distribution with mean Oand standard deviation
aJ2 j [email protected]
@Sk J ahiruddin , 2020
10
Physicsguide
P robability
Prob 1.28. T wo independent variables m and n which can take t he integer values 0, 1, 2 ... oo , follow t he Poisson dis-
t ribution, with dist inct mean values µ and v respectively. Then [NET Dec 14] (a) The probability distribution of the random variable l = m + n is a binomial distribut ion (b) The probability distribution of t he random variable r = m - n is also a Poisson distribut ion (c) The variance of t he random variable l = m + n is equal
11 oo : oo : 29 (a) The probability distribut ion of t he random vai·iable l = m + n is a binomial distribution (b) The probability distribut ion of t he random variable r m - n is also a Poisson distribut ion (c) The variance of t he random variable l
=
= m + n is equal
toµ + V (d) The mean value of t he random variable r equal to zero
= m - n is
Prob 1.29. There are on average 20 buses per hour at a point, but at random t imes. The probability that t here are
no buses in five minut es is closest to
[JEST 2013]
(a) 0.07
(b) 0.60
(c) 0.36
(d) 0.19
Prob 1.30. You receive on average 5 emails p er day during a 365-days year . The number of days on average on which
you do not receive any em ails in t hat year are: 2016] (a) more th an 5 of the above [email protected]
(b) More t han 2
(c) 1
11
[JEST (d) None
Physicsguide
@Sk J ahiruddin , 2020
P robability
Prob 1.31. Consider two 1·adioactive atoms, each of which has a decay rate of 1 per year. The probability that at least
one of t hem decays in t he first two years is
(a)
!4
Dec 16]
(b)
3 4
(c) 1 - e- 4
(d) (1 - e- 2 ) 2 [NET
11 oo : oo : 31 one of t hem decays in the first two years is
1 3 (a) (b) 4 4 D ec 16]
(c) 1 - e- 4
(d) (1 - e- 2 ) 2 [NET
Prob 1.32. The random variable x(-oo < x < oo) is dist ributed according to the normal distribution
1 2 2 P (x) =-----;::::::= exp (-x /2a ) 21ra 2 The probability density of the random variable y = x 2
[NET June 1 7]
is (a) ~ exp (-y/2a
2
y
2 11"0"
(b) ~ exp (-y/2a 2 2
27ro- y
), 2
0
- 1, a > 0)
n+l 2
Hence
Hence 1 -00
00
1
4
-(x -µ / 2)2
x e
21r (O" I ~)2
2
dx
-oo
using t he result 4
(]"4
Solution 1.26. Let a+ b = x and 2c = y So x>y
e is ess than
a+ b
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@Sk J ahiruddin, 2020
2
c
2c is 1/ 2 Solution 1.27. If X and Y are two independent random variables, each of which follow a normal distribution, such
t hat, X has standard deviation a x means µ x Y has standard deviation ay means µy
Z = X +Y
J
Then Z has standard deviation a z = al + a~ means µz = µx + µ y So, here X + Y is a normal distribution with ax +Y = v'2a and mean µ - µ = 0. Solution 1.28. Sum of two independent variable obeying Poission Distribution is also a poission distribution with
mean= µ+ v. 2 Variance of a poission distribution = a =mean. So, variance of sum of those random variables is = µ + v. Solution 1.29. Probability in Poission Distribution is given by,
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29
Physicsguide
P robability
11 oo : 01
: 21
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P robability
Where µ is the mean rate in the span asked, and n is the amount asked. Here, 20 buses per hour means ~ buses per five minutes. So, µ = 5/3, And n = 0. So, t he probability is P = e-~ = 0.188
Solution 1.30. Here µ = 5, n = 0. So, probability is P = e- 5 .
So, number of days are = 365 x e- 5 more t han 2.
2.49, that means
Solution 1.31. One atom per year means two atoms p er two year. So, µ = 2.
If an atom does not decays in two days, t hen the probability 2 of that occurrence will be given by, P1 = e- . 4 2 SO, for two atoms it will be P2 = P1 = eSo, Probability that at least one atom decays will be given by, 4 P = l - P2 = 1 - eSolution 1.32. As P(x) is even function then we can write 00
P (x) dx = -oo
00
1
21ra 2 1
= 2 --;::::=====:2 21ra
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2
30
2
exp ( -x / 2a )dx
----;:=~
- oo 00
2
o
2
exp (-x / 2a )dx
Physicsguide
11 oo : 01
: 24
= 2 ----;::::=====:2 21ra
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exp (-x 2 / 2a 2 )dx
o
30
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P robability
Now if we put x 2 = y ⇒ 2xdx = dy , then we get, 00
1
00
2
2
exp ( -x / 2a )dx = 0
P(y)
=
2x
1
J exp (-y/2u 2 21ra y
2
2
1
) =
J 21ra y exp (-y/2u 2
2
)
Solution 1.33. Suppose we toss one fair coins hundred t imes, t hen probability of n number of head occurs at t he end of
100 toss is (using binomial distribut ion formula) n
lOOC n
1
2
l
100- n
2
So when you toss both coins t hen the probability of occurring same n number of head at t he end of 100 toss is just mult iply t he probabilit ies .
lOOCn
l -
2
n
~
100- n
l x lOO C n
2
n
-
2
l
100- n
-
2
Now just sum over n from Oto 100 to get t he probability to get same numbers. The ans is (d)
! n=l
100 100
L cooen) n =l
2
11 oo : 01
: 26
ow JUS sum over n rom to get same numbers. The ans is (d) 100
L
1 (l00Cn )2
n =l
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200
!
-
2
100 100
L
eoocn)2
n =l
31
Physicsguide
P robability
Solution 1.34. The standard deviation of binomial distribution = y0Vpq_ step size to right or left is 2a so mean square displacement = 2ay0Vfig_ Solution 1.35.
11 oo : 01
: 29
n =l j [email protected]
n =l
31
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Physicsguide
P robability
Solution 1.34. The standard deviation of binomial distribution = ,JNpg_ step size to right or left is 2a so mean square displacement = 2a,JF[pq_ Solution 1.35.
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: 31
n =l j [email protected]
n =l
31
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P robability
Solution 1.34. The standard deviation of binomial distribution = ,JNpg_ step size to right or left is 2a so mean square displacement = 2a,JF[pq_ Solution 1.35.
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11 oo : oo : oo
Problellls and Solutions in N ulllerical Methods, Group Theory, Tensors and Greens Functions Sk J ahiruddin Assistant Professor Sister Nibedita Govt. College, Kolkata Aut hor was the topper of IIT Bombay lVI.Sc Physics 2009-201 1 batch He ranked 007 in IIT JAlVI 2009 and 008 (JRF) in CSIR NET J une 2011 He has been teaching CSIR NET aspirants since 2012
1
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Advanced topics in Mathematical Physics
11 oo : oo : 03 1
Advanced topics in Mathematical P hysics
@Sk J ahiruddin, 2020
Contents 1
Numerical Methods
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1.1
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Group Theory
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2. 1 Ans keys and solutions
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2.2 Solutions . . . . . . . .
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Tensors
3. 1
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3.2
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Green's functions
4.1
Ans keys . . ..
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38 •
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Physicsguide
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2
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Advanced topics in Mathematical P hysics
Physicsguide
4. 2 Solutions . . . . . . . . . . . . . . . . . . . .
41
11 oo :oo :oa
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Advanced t opics in 1\ll athem atical P hysics
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Problems from NET, GATE, JEST, TIFR & JAM papers
1
Numerical Methods 1
#dx is to be evaluated up to
Prob 1.1. The int egral 0
3 decimal places using Simpson's 3-point rule. If the interval [O , 1] is divided into 4 equal parts, t he correct result is [NET June 14] (a) 0.683 (b ) 0.667 (c) 0.657 (d) 0.638 Prob 1.2. Given t he following xy data. X
1.02
2.0
3.0
4.0
5.0
f(x) 0.002 0.601 0.948 1. 21 1.42 Which of t he following would be t he best curve, wit h constant positive parameters a and b, to fit this data? [TIFR 2018]
(a) y = ax - b (c) y = a log 10 bx
(b) y = a + ebx
(d)y = a -
e(- bx)
Prob 1.3. Consider the differential equation ~~
= x2
-
y
11 oo : oo : 11 2018]
(a) y
=
ax - b
(b) y
(c) y = a log 10 bx
=
a + ebx
(d)y = a -
e(- bx)
Prob 1. 3. Consider the differential equation ~~ with the initial condition y = 2 at x = 0. Let
= x2
Y(i )
and
-
y
Y(~)
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Advanced topics in Mathematical Physics
Physicsguide
be the solutions at x = 1 obtained using Euler 's forward algorithm with step size 1 and ½ resp ectively. The value of Y(1) -Y(½)
Y(½)
(a)
-½
•
[NET June 15]
IS
(b) - 1
(c) ½
(d) 1 8
Prob 1.4. The value of t he integral
=
- - evaluated o x2 + 5 2 is
(c) 0.698
(d) 0.736 [NET
using Simpson's 1/ 3rd rule with h (a) 0.565
(b) 0.620
dx
Dec 15] Prob 1. 5. In finding the roots of the polynomial f (x) = 3x 3 - 4x - 5 using t he iterative Newton-Raphson m ethod, t he init ial guess is taken t o be x = 2. In the next iteration, [NET June 16] its value is nearest to: (a) l.671
(b) 1.656
(c) 1.959
(d) 1.551
Prob 1.6. Given the values sin45° 0.7071 , sin 50° 0. 7660, sin 55° = 0.8192 and sin 60° = 0.8660, the approximate value of sin 52°, computed by Newton's forward difference method, is
[NET Dec 16]
11 oo : oo : 14 Prob 1.6. Given the values sin45° = 0.7071 , sin50° = 0. 7660, sin 55° = 0.8192 and sin 60° = 0.8660, the approximate value of sin 52°, computed by Newton's forward dif-
[NET Dec 16]
ference method, is (a)0.804
(b )0. 776
(c)O. 788
(d )O. 798
Prob 1.7. The interval [0,1] is divided into 2n parts of equal j [email protected]
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Advanced topics in Mathematical Physics
Physicsguide
1
2
ei 1rxdx using Simpson 's
length to calculate t he integral 0
1/ 3 rule. What is t he minimum value of n for t he result to
[NET June 1 7]
be exact ?
(a) oo
(d) 4
(c) 3
(b) 2
2.2
Prob 1.8. The value of
xexdx by using t he one-segment 0.2
t rap ezoidal rule is close to (a) 11.672 (b) 11.807
[JEST 2014] (c) 20 .099
(d ) 24.119
Prob 1.9. The approximation cos 0 l is valid up to 3 decimal places as long as I0 is less t han: (take 180° / 1r 57.29°) [NET June 13] f"'-.J
f"'-.J
(a) 1.28°
(b) 1.81°
(c) 3.28°
(d ) 4.01°
Prob 1.10. The differential equation dy( x ) = a X 2 with dx t he init ial condit ion y(O) = 0, is solved using Euler 's method. If YE(x) is the exact solution and y 1v (x) the numerical solut ion obtained using n steps of equal length, then the relative error
YN(x) - YE(x ) . l) , ....
r
ry, \
.
1s proportional to
[NET Dec 17]
11 oo : oo : 16 dx
t he initial condit ion y(O) = 0, is solved using Euler's method. If YE(x) is the exact solut ion and YN(x) the numerical solut ion obtained using n steps of equal length, then the relative YN(x - YE x .
(a) 1/n
2
YE
.
[NET Dec 17]
X
(b) 1/n
3
(c) 1/n
4
(d) 1/n
Prob 1.11. The interval [O, 1] is divided into n parts of 1
equal length to calculate the integral
exp (i21rx )dx uso
ing the trapezoidal rule. The minimum value of n for which [email protected]
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Advanced t opics in :Nlathematical P hysics
t he result is exact is (a) 2 (b) 3 (c) 4
Physicsguide
[NET Dec 17] (d) oo
Prob 1.12. The fractional error in estimating the integral 1 f0 xdx using Simpson's ½rule, using a step size 0.1 , is nearest to [NET June 18]
(a) 10-4
(b) 0
(c) 10-2
3
X
10-4
Prob 1.13. The value of the integral
0
evaluated using t he trapezoidal rule with step size of 0.2, is [NET Dec 18] (a) 0.30 (b) 0.39 (c) 0.34 (d) 0.27
Prob 1.14. If the Newton-Raphson method is used to find t he positive root of the equation x = 2 sin x t he iteration
11 oo :oo :1a (a) 0.30
(b) 0.39
(c) 0.34
[NET D ec 18] (d) 0. 27
Prob 1.14. If the Newton-Raphson method is used to find t he positive root of the equation x = 2 sin x t he iteration equation is [NET June 2019] 2 x n - 2 (sin X n + Xri cos Xn) ( a) Xn+ l = - - - - - - - - - 1 - 2cosxn _ 2 (sin Xn - Xn COS Xn) (b) Xn+ l 1 - 2cosxn x~ - 1 + 2 (cosxn - Xn sin xn) ( C) Xn+ l = . Xn - 2 Sln Xn [email protected]
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Advanced topics in l\llathem a tical Physics
P hysicsguide
_ x~ - 1 - 2 (cos Xn + sin Xn) (d) Xn+ l . Xn - 2 s1n xn Prob 1.15. The positive zero of the polynomial f (x) = x 2 - 4 is determined using N ewton-Raphson method, using an initial guess x = l . Let the estimate, after two iterations, x( 2) - 2 x 100% is [NET be x( 2) . The p ercentage error 2 Dec 2019] (a) 7.5% (b) 5.0% (c) 1.0% (d) 2.5%
11 oo : oo : 21
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8
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Advanced topics in :Nl athematical Physics
1.1
Physicsguide
Ans keys
1.1. d
1.6.
C
1.11. a
1.2.
1.7. b
1.12. b
1.3. b
1.8.
1.13.
1.4. a
1.9. b
1.14. b
1.5. b
1.10. d
1.15. d
1.2
C
C
C
Solutions
Solution 1.1. The interval is divided into 4 equal parts.
11 oo : oo : 23 1.2
Solutions
Solution 1.1. The interval is divided into 4 equal parts. So, h = 0.25 X
f( x)
Xo
=0
X1
Yl
Yo= 0
= 0.25
S2
Now
=
= 0.5
v'xdx =
0.707 y3
Y2 h
=
= 0.75 =
X4 =
0.866 Y4
l
=1
0.707
[s1 + 4s2
3 0.25 [ = 1+4 3
0
X3
= Yo + Y4 = 1 + 0 = 0 = Y1 + Y3 = 0.5 + 0.866 = 1.366
S3 =
i
Y2
0.5
=
s1
X2
+ 2s3] X
1.366 + 2
X
] 0.707 = 0.657
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Solution 1.2. As first look f (x) is increasing with x, but t he rate of change is slowing down. So option (a) and (b) eliminated. Then ind the slopes of t he function. Also find t he double derivative. Then with t he values find the values of a and b. Then check what option matches. Remember in T IFR you are allowed to use calculator. Solution 1.3. After first step , . 2
Y(i )
=y +h x (x -y) =
A
1
2 + 1(0 - 2) n
,
1I A
r. \
=
0
11 oo : oo : 26 Solution 1.3. After first step, Y(l )
2
=y+ h x (x -y)
2 + 1(0 - 2)
=
1
Y(l /2) =
2 + (0 - 2) 2
Y(l ) - Y( ½) =
0- 1= - 1
And
=
0
=
1
So, 1
Y(~)
Solution 1.4. Const ruct t he table X
Xo = 0
f(x)
Yo = 0.2
2 Yi = 0. 111 X1
=
X2
= 4
Y2 = 0.048
j [email protected]
S2
= Y1
S3 =
0
X4 = 8 6 0.024 Y4 = 0.014
=
Physicsguide
Advanced t opics in Mathem atical P hysics
s1 = Yo + Y4 =
Now
y3 =
10
@Sk J ahiruddin , 2020
i
X3
.jxdx =
Y2 h
+ Y3 =
=
0.2 + 0.014 = 0.214 0.111
+ 0.024 = 0.135
0.048
[s1 + 4s2
+ 2s3]
3 2 = [0.214 + 4 3
X
0.135 + 2 X 0.048] = 0.565
11 oo : oo : 29 S3 = Y2 =
i
Now
.jxdx =
h
0.048
[s1 + 4s2
+ 2s3]
3 2 = [0.214 + 4 3
0
X
0.135
+2X
0.048] = 0.565
Solution 1.5. The algorithm of Newton Raphson method •
1S
Now here xo
=2
So, f (xo) = 3x - 4x - 5l xo = 11 2 And, f'( xo) = 9x - 4 xo = 32 So, x 1 = 2 - 1~ = 1.656 3
Solution 1. 6. construct t he table XQ
45° 50° 55° 60°
f (xo) =
l:iy1
Yo
0.7071 0.7660 0.8192 0.8660
!:iy2
!:iy2
0.0589 0.0532 -0.0057 0.0468 -0.0064 -0.0007
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Advanced topics in 1\ll athematical Physics
Now, we apply Newton's forward difference method ph
2 p=5
=2
p(p + 1) , l:iy2 sin 52 = Yo + pl:iy1 + 2 •
0
2
= 0. 7660 + ::-
X
0.0532 +
~(l + l) 5 5 ,,.. ,
( -0.0064)
11 oo : oo : 31 5 •
Sln 52
0
= Yo + pfl. y1 +
p(p + 1) I
2.
fl. y 2
~ ~
2
+1
= _
0.1064 7 X 0.0064 0 7660 + 5 25 0.7660 + 0.0213 - 0.0018
= = 0.788
Solution 1.7. T he exact value of t he int egral is zero. Let's divide the integral in two equal interval. T hen h = 0.5, x 0 = 0, x1 = 0.5, x2 = 1, and so f (xo ) = 1, f (x1) = - 1, f (x2) = 1 05 . Simpson 's l/3rd met hod give the ans 3 (1- 4 + 1) = - 1/ 3. hence t he ans is not exact. If we divide the int egral into four equal int ervals h ¼, XO= 0, X 1 = 0.25 , X2 = 0.5, X3 = 0.75, X4 = 1.
The value of t he int egral would be
1 (1+ 12
4(i- i) + 2(- 1)+
1) = 0 [email protected]
@Sk J ahiruddin , 2020
12
P hysicsguide
Advanced topics in Mathematical P hysics
Hence if we divide t he int erval in four equal length t hen t he integral will be exact. Hence according to t he question
n = 2.
Solution 1.8. Here f (x) = xex Now. f (2.2) = 19.855 and f(0 .2)
=
0.244
11 oo : oo : 34 ence 1 we
e 1nterva in our equa
engt
t en
t he integral will b e exact. Hence according to t he question
n = 2. Solution 1.8. Here f (x) = xex Now, f (2.2) = 19.855 and f (0.2) = 0.244 2.2 2 So, xexdx = -[19.855 + 0.244] = 20.099 0.2 2 Solution 1.9. Expand cos x in taylor series about taylor series about x = 0. The highest value provider is cos x 2 . Now if t he value has to be correct upto th1·ee decimal point t hen x 2 / 2 < 0.0005. Hence x ::; 57.29 x J0.001 in degrees. Solution 1.10. In Euler m ethod, Relative Error = Local error x no of step. Again, local error rv h 2 l where h = - where l is t he range and n is no of step. n 2 z2 z 1 So, Relative error = 2 xn=-~n n n
Solution 1.11. If we do t he integration analytically, we get 1
e(21rix)dx
e(21rix)
1
21ri
0
e21ri -
1
= --
o
21ri j [email protected]
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=O Physicsguide
Advanced topics in Mathematical Physics
Now, let us consider t hat the interval is d ivided into two 1- 0 1 equal parts. So, h = = . 2 2
00: 00: 37 Advanced topics in Mathematical Physics
Now, let us consider that the interval is divided into two 1- 0 1 . equal parts. So, h = = 2 2 1
e(i
• • •
0
1rxldx = h [f (O) + f (l ) + 2f (1/2)] 2
2
= ~ [1 + e21ri + 2ei1r] 4
1 = -[1 + 1 - 2] = 0 4
So, for n
= 2 analytical and numerical solutions are same.
Solution 1.12. The error in Simpson's 1/ 3 rule is
where t is some point in b etween the interval. Here, h = 0.1, (b- a)= l , f (x) = x, so f''( x) = 0, so the term f 4 ( t ) term in t he error formula will give you ZERO. Hence the error is ZERO.
f(x) = x
Solution 1.13. Here,
2
5 0.2 0.4 0.6 0.8 1 0 Xn f( xn) 0.00 0.04 0. 16 0.36 0.64 1.00
n=
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0
1
2
3
4
14
Physicsguide
Advanced topics in :Nlathematical Physics
Now, integration in trapezoidal rule is given by,
11 oo : oo : 39 @Sk J ahiruddin, 2020
Advanced t opics in Nl athematical P hysics
Now, integration in trapezoidal rule is given by,
So, here, 1
2
x dx
02
= ;
[{0.00
+ 1.00} + 2{0.04 + 0. 16 + 0.36 + 0.64}]
0
= 0.1 [1.00 + 2 = 0.1
X
[email protected]
@Sk J ahiruddin , 2020
3.40
X
1.20]
= 0.34
15
Physicsguide
Advanced topics in Nlathematical P hysics
11 oo : oo : 42 15
jahir@physicsguide. in
@Sk J ahiruddin , 2020
Physicsguide
Advanced topics in Mathem atical P hysics
Group Theory
2
Prob 2.1. Which of the following matrices is an element of t he group SU(2)? [NET June 11]
(a) (d)
I I 0 I v'3 -1 2
v'3 2
(b)
l +i
1
v'3
v'3
1
1- i
v'3
v'3
(c)
•
2 +i
'l
3
I+ i
2
-1 2
Prob 2.2. Let A and B b e two vectors in t hree-dimensional Eucledian space. Under rotation, t he tensor product: T i j =
[NET Dec 13]
A i BJ
(a) reduces t o a direct sum of three 3-dimensional representations (b) is an irreducible 9-dimensional representation (c) 1~educes to a direct sum of a I-dimensional, a 3-dimensional and a 5-dimensional irreducible representat ions (d) reduces to a direct sum of a I-dimensional and an 8dimensional irreducible representations Prob 2.3. Let a and f3 be complex numbers. Which of
the following set s of matrices forms a group under matrix multiplicat ion? [NET Dec 14]
j ahir@physicsguide. in
16
Physicsguide
11 oo : oo : 4s
j a [email protected]
16
@Sk J a hiruddin , 2020
(b)
0 0
a a* /3 /3*
(c) a
Advanced topics in Mathematical Physics
/3
(a)
2
+ 1/32 =
Physicsguide
where
1
/3
where
1
a/3* is real
(d)
a/3 -/=
1
/3 -/3* a*
where
1
Prob 2.4. A p art of t h e group multiplication table for a six elem ent group G = e, a, b, c, d, f is shown below (e is the identity element of G). e
a b
C
d
f
e
e
a b
C
d
f
a
a b
b b C
C
d
d
f
f
e
e d X
f
The entries x, y and z should b e: (a) x = a, y = d and z = c (c )x = c, y = d and z = a
y z
[NET June 16] (b) x = c, y = a and z = d
(d)x = a, y = c and z = d
11 oo:oo:47 (c) x = c, y = d and z = a
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(d) x = a, y = c and z = d
P hysicsguide
17
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Advanced topics in :Nlathematical Physics
Prob 2.5. The 2 x 2 ident ity matrix I and the P auli mat rices ax, aY, az do not form a group under matrix mult iplication . The minimum number of 2 x 2 matrices, which includes t hese four matrices, and form a group (under ma-
(NET Dec 16]
trix mult iplication) is:
(a)20
(b) 8
(c) 12
(d) 16
Prob 2.6. Which of t he following sets of 3 x 3 matrices (in which a and b are real numbers ) form a group under matrix multiplication? [a, b E ~ ]
[NET June 1 7]
(a)
1 0 a 0 1 0 b O1
(b)
1 a 0 0 1 b 0 0 1
(c)
1 0 a 0 1 b
:
0 0 1
1 a 0 (d)
b 1 0 0 0 1
Prob 2.7. Consider an element U(cp) of the group SU(2) , where is any one of the parameters of t he group. Under an infinitesimal change ➔
U( ) + 6U() =
+ 6,
it changes as U () ➔ (1+ X (6))U(cp) . To order 6, t he matrix rl\.TV ry, n .,.,. .,..
1 ,.,.l
11 oo : oo : so Prob 2.7. Consider an element U(cp) of the group SU(2), where
. which is
•
2000 2.426 x 10- 2
~A
=
82440
. he Sol 1.13. The Energy(E) of the photon 1s hv or A The incident wavelength is 0.2 nm. T he expression for Compton (0\ r1
' '
•
•
,\
\
r,. \
•
')
11 oo : 01 : oo ~A
=
2.426 x 10- 2
=
82440
Sol 1.13. The Energy(E) of the photon is hv or : c The incident wavelength is 0.2 nm. The expression for Compton
Sacttering is ~;\ = 2;\c sin
2
~2
Now for maximum wavelength shift we must have 0 = 180°. So the wavelength of the scattered photon
~;\ =
2;\c
;\' - ;\ =
2;\c
,:\' =
2;\c
+ ;\
,:\' = 4.86 pm + 0.2 nm
;\' = 0. 20486 nm [email protected]
@Sk J ahiruddin , 2020
22
Physicsguide
Old Quantum Theory
he Energy( E ) of the initial photon 0.2 nm he Energy( E' ) of t he scattered photon 0.20486 nm Sol 1.14. The expression for Compton Scattering is ~;\ = 2-Acsin
2
~
Here Ac = Compton wavelength of the electron = 0.00243 nm ;\ = Wavelength of incident t he X-rays= 0.24 nm 0 = Angle of the scattered beam is relative to t he incident beam = 60° ;\' = Wavelength of t he deflected X-rays Thus we get
11 oo : 01
: 03
ave eng e -rays = . 0 = Angle of the scattered beam is relative to t he incident beam = 60° A' = Wavelength of t he deflected X-rays Thus we get
9
A = 2 x 0.00243 x 10- x sin 1
2
60°
2
+ 0.24 X
10-
9
A = 0.241215 nm 1
23
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Physicsguide
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Old Quantum Theory
According to the law of conservation of energy K e = E - E' he he A A' 6.626 X 10- 34 X 3 0.24 x - 9 = 4.172 X 10- 18 J
= 26.04 eV Sol 1.15. That is
X
108
6.626
10- 34 X 3 X 108 0.241215 x - 9 X
11 oo : 01
: 06
= 4.172 X 10- 18 J = 26.04 eV Sol 1.15. That is h mpe
= 1836 t h of>..c 1 2.426 _ 12 A 1836 X l O
= 1.32 fm Sol 1.16.
~ ).. =
h (1 - cos 0) m ee
cos 0 = 1 - ~ Am ee h 3.25 X 10- 13 X 9.1 X 10- 31 X 3 X 108 = 1 - -----------6.626 X 10- 34 cos 0 = 0.866
0 = 30 j [email protected]
P hysicsguide
24
@Sk J ahiruddin , 2020
Old Quantum Theory
Sol 1.1 7. Equation for phot oelectric effect is hv = W
or ~c = W
+K
+K
Now we have t wo equat ions
he = W + K )..
he
).. = W + 5K
Solving these two equations we get ½ = 0.775V and in
p
second case we get ½ = 3.875V Now Current (I ) = E
11 oo :01 : oa he -A=
W +5K
Solving these two equat ions we get second case we get
½ = 0.775V and in p
½ = 3.875V Now Current (I ) = E I = p
E I = p he
or,
-A
= 50 mA
Putting all t he values we get I
Sol 1.18. Here we have two types of incident electric field. W1
W2
= =
6.28
10 15 s -
X
12.56
X
l
10 15 s-l
The corresponding wavelengths 21r x 3 x 1015 1- w 6.28 x 1015 = 300. 15 nm 21r X 3 X 1015 -A2 = 12.56 x 1015 = 150.07 nm
A _ 21re
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Physicsguide
Old Q t1antum Theory
Incident Energy on the metal surface is E1
=
E2 =
10-34 X 6.28 X 10 15 hv = nw = - - - - - - - - - J 21r 6.622 x 10- 19 eV = 4.13 eV 1.602 X 10- 19 15 34 6.626 X 10- X 12.56 X 10 J 21r 6.626
X
11 oo : 01 E1 = hv =
liw
: 1o
= - ·- - - - - - - - - J
21r
19 6.622 X 10V 3 V ----e = 4.1 e 1.602 X 10- 19 34 15 E2 = 6.626 X 10- X 12.56 X 10 J
21r 83.22 X 10- 19 1.602 x 10- 19 eV ~ 52 eV T hus Kinetic Energy
Ek= E1 - Wo = 4.13-2 eV = 2.13 eV
Sol 1.19. T he E lectric F ield
+ sin wt cos w0 t) E = a cosw0t + ~ sin(w + w 0 )t + ~ sin(w E = a (cos w0t
Now t he m aximum angular frequency 6 X 10 15.
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Wmax
w 0 )t
= w + wo
26
Physicsguide
@Sk J ahiruddin , 2020
Old Quantum Theory
Maximum energy incident on the metal surface
10- 34 X 5 X 10 15 = liwm ax = 1T J 2 5.272 x 10- 19 e V = 3.29 e V 6.626
E max
1
nn?.
x 1n- 19
X
11 oo : 01
: 13
Maximum energy incident on the metal surface
10- 34 X 5 X 1015 = rlwmax = 7r J 2 5.272 x 10- 19 eV = 3.29 eV 1.602 X 10- 19 6.626
E max
X
The maximum kinetic energy of the photons
(3.29 - 2) eV = 1.29 eV
E k= E max - Wniax =
Sol 1.20. The stopping potential is given by
½ = he
W
e;\
e
6.626 X 10-34 X 3 X 108 320 X 10- 9 X 9.109 X 10- 31
or,
2.1
1.602 X 10- 19 9.109 X 10- 31 X
½ = 1.77 V
Sol 1.21. For t ungsten
e½ = hv -W hv = e½
+W
l.8
=
+ 4.5 =
6.3 eV
For sodium
e½
j [email protected]
=
hv - W = 6.3 - 2.3
= 4
eV
27
Physicsguide
@Sk J ahiruddin , 2020
Old Qt1antum Theory
Sol 1.22. Intensity(! ) ex T 4 (Temperature) Now
I2
Tf
T, -
T .4
11 oo : 01
: 1s
@Sk J ahiruddin, 2020
Old Quantum Theory
4
Sol 1.22. Intensity(! ) ex T (Temperature) Now
rt
I2 4 11 Tl 12 (2T) 4 11 T4
or,
12 = 16 11
or, Sol 1.23. Sol 1.24. Einstein relation
de Broglie wavelength
A= ~ p
From t his relation we will find the expression of moment um for free electron and photon. For photon We know that photon has no rest mass. Thus the expression for momentum becomes
E
Pph
= -
C
Thus the expression fo De-broglie wavelength is A ph
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For free electron
= 28
he E Physicsguide
Old Quantum Theory
11 oo : 01
: 1a
@Sk Jahiruddin, 2020
Old Qt1antum Theory
For free electron Thus the expression for momentum becomes
Pe =
JE2 - (moe2)2 e
Thus the expression fo De-broglie wavelength is
Ae =
he (moc2)2
JE2 -
Sol 1.25. 1
A - A = 2AcSin
,\' - ,\ =
2
0 2
h
me ,\' ,\ l he he me2 1 1 1 -E' E me2 1 1 1 -=-+--2 E' E me 1 1 1 E' = 20 + Sll = 19.23 KeV According to the law of consevart ion of energy K e= E - E'
= (20 - 19.23) KeV = 0.77 KeV j [email protected]
@Sk J ahiruddin, 2020
29
Physicsguide
Old Quant11m Theory
11 oo : 01 [email protected]
: 21
29
P hysicsguide
@Sk J ahiruddin , 2020
Old Quantum Theory
Sol 1.26. This discussion is available in any book of Quant um mechanics. My suggestion is to read t he old quantum
t heory from Art hur Beiser's "Concept of modern physics" first. Sol 1.27. At first we have need t o have a simple knowledge of a charged part icle moving in a circle in a magnetic field.
At first we will equate Lorentz force with Centripetal force and derive the expression of radius of the circular field. 2 mv
qv B = - r
~ ~>
mv r=qB
Angular Momentum m2v2 mvr=--
qB
According to Bohr's model, Angular Momentum is given by nh - By equatinng we get 21r
qB
21r n hqB 2 mv = - 21rm 1 hqB 2 -mv = n 2 41rm [email protected]
30
P hysicsguide
11 oo : 01 1 2 -mv 2
: 23
=n
j [email protected]
hqB 41rm
Physicsguide
30
@Sk J ahiruddin, 2020
Old Qt1antum Theory
Sol 1.28. Bohr radius is propotional to n2
rn ex: Zµ
Here Z = Atomic number µ = Reduced mass of the atom n = Principle quantum number The reduced mass of positron is
(me- ) (me+) µ= (me- )+ (me+ ) (me- )(me+) 2(me+) (me- ) 2 In case of Hydrogen , electron and proton are moving around t he center of mass. As the mass of proton is very large compared to mass of electron, therfore centre of mass shifts nearer to proton. Infact proton is at t he position of centre of mass and may be considered stationary, while electron revolving around it. Now centre of mass of Hydrogen atom
(m e) (mp) µ= (me)+ (mp) (me) (mp) (mp) ,-...J ,-...J
~
_ _ __
me
11 oo : 01
: 26
31
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Old Quantum Theory
Here mp > > m e so of Positron Tn
~~> mp
+ m e ~ mp.
The Bohr radius
==
n2 P-
[P
IS T HE PROPORT IO NALITY CONSTANT]
Zµ
2n 2 == P - Zme-
0
== 2
x 0.53A
0
==
1.06A
Sol 1.29. Refer to the previous problem. Energy is propotional to Zµ E n ex 2 n
Earlier we have shown t hat reduced mass of Positron is half of the reduced mass of Hydrogen atom. For this problem keeping all parameters constant in the expression for energy, t he ground state energy of the Positron will be half of the ground state energy of Hydrogen atom. Thus energy of ground state of Positron E ==
3 5 - l · 2
==
- 6.8 eV
11 oo : 01
: 29
will be half of the ground st ate energy of Hydrogen atom. Thus energy of ground state of Posit ron
3 6 E =- l2 · =-68 eV . Sol 1.30.
En= - 13.6
X
Z
2
n2 j [email protected]
32
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Physicsguide
Old Quant11m Theory
For ground state n = l and for H e+ ion the atomic number is Z = 2 Thus we have -13.6 X 4 Eri = l - 54.4 eV Sol 1.31. Read the first chapter of Sakurai's "Modern Quantum Mechanics'' book or read the Spin section from notes
of angular momentum. Sol 1.32. Refer to previous problem.
This is an example of sequential experiments. As we know the, when the beam passes t hrough a Stern-Gerlach apparatus the beam is divided into two seperate beams depending on the orientation of spin along the magnetic field. Here the beam oriented along the magnetic field is allowed to pass through. So when t he unpolaried bean is incident on the first Stern-Gerlach apparatus, electrons whose spin is oriented along z-axis is allowed to pass. As discussed above, the intensity of this beam(I 1 ) is half of the incident beam(I 0 ). Now the incident beam (11 ) is passed through second SternGerlach apparatus, whose magnetic field is oriented along
11 oo : 01
: 31
along z-axis is allowed to pass. As discussed above, t he intensity of this beam(/ 1 ) is half of the incident beam(/ 0 ). Now t he incident beam (11 ) is passed through second SternGerlach apparatus, whose magnetic field is oriented along y-axis. As usual electrons whose spin is oriented along yaxis is allowed to pass. The intensity of t his final beam(I 1 ) is half of t he incident beam(/ 1 ). So we can easily say t hat intensity of I 1 is one-fourt h of 10 . [email protected]
33
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P hysicsguide
Old Quantum Theory
Sol 1.33. This is Davisson-Germer experiment because t here's a maximum at ~ 50 deg in number of electrons detected
which supports electron interference
00 : 01 : 34 intensity of I 1 is one-fourt h of I 0 . j [email protected]
33
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P hysicsguide
Old Quantum Theory
Sol 1.33. This is Davisson-Germer experiment because t here's a maximum at ~ 50 deg in number of electrons detect ed
which supports electron interference
j [email protected]
34
P hysicsguide
00 : 01 : 36 intensity of I 1 is one-fourt h of I 0 . j [email protected]
33
@Sk J ahiruddin, 2020
P hysicsguide
Old Quantum Theory
Sol 1.33. This is Davisson-Germer experiment because t here's a maximum at ~ 50 deg in number of electrons detect ed
which supports electron interference
j [email protected]
34
P hysicsguide
00 : 01 : 39 intensity of I 1 is one-fourt h of I 0 . j [email protected]
33
@Sk J ahiruddin, 2020
P hysicsguide
Old Quantum Theory
Sol 1.33. This is Davisson-Germer experiment because t here's a maximum at ~ 50 deg in number of electrons detect ed
which supports electron interference
j [email protected]
34
P hysicsguide
00:01 :41 intensity of I 1 is one-fourt h of I 0 . j [email protected]
33
@Sk J ahiruddin, 2020
P hysicsguide
Old Quantum Theory
Sol 1.33. This is Davisson-Germer experiment because t here's a maximum at ~ 50 deg in number of electrons detect ed
which supports electron interference
j [email protected]
34
P hysicsguide
00: 01 : 44 intensity of I 1 is one-fourt h of I 0 . j [email protected]
33
@Sk J ahiruddin, 2020
P hysicsguide
Old Quantum Theory
Sol 1.33. This is Davisson-Germer experiment because t here's a maximum at ~ 50 deg in number of electrons detect ed
which supports electron interference
j [email protected]
34
P hysicsguide
11 oo : oo : oo
Problems and Solutions in Basic formalism of Quantum Mechanics: Part-1 Sk J ahiruddin * Souradeep Mondal
*Assistant Professor Sister Nibedita Govt. College, Kolkata Author was t he topper of IIT Bombay M.Sc P hysics 2009-2011 batch He ranked 007 in IIT JAM 2009 and 008 (JRF) in CSIR NET J une 2011 He has been teaching CSIR NET aspirants since 2012
1
@Sk J ahiruddin , 2020
Basic Formalism: Part-1
11 oo : oo : 03 1
@Sk J ahiruddin, 2020
Basic Formalism: Part-1
Contents 1
Problems from NET, GATE, JEST, TIFR & JAM papers 1.1 Probability 1.2 Normalization of Wave Function . 1.3 Orthogonal Stat es . 1.4 Schrodinger Equation . • 1.5 Hysenberg Uncertainty Principle . 1.6 Commutators 1.7 Probability Current Density 1.8 Expectation Values 1.9 Ans Keys • • • 1.10 Solutions . • •
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3 4 5 6 9 13 16 16 20 21
Physicsguide
11 oo : oo : 06
j [email protected]
2
P hysicsguide
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1
1.1
Basic Formalism: Part-1
Problems from NET, GATE, JEST, TIFR & JAM papers Probability
Prob 1.1. It is necessary to apply quantum statistics to a [GATE 2007] system of particles if
(a) t here is substantial overlap b etween the wave functions of the particles (b) t he mean free path of the particles is comparable to the inter-particle separation (c) the particles have identical mass and charge (d) t he particles are interacting Pro b 1. 2. Which of the following is an allowed wave funct ions for a particle in a bound state? N is constant and
a, /3 > 0.
[GATE 2010]
- ar
(a) 'lj; = Ne 3 (b) 'lj; = N(l r (c) 'ljJ = N e - a x e -f3(x2+y2 +z2 ) non-zero constant
0
e - ar)
if r < R if r > R
Prob 1.3. The state of a syst em is given by
11 oo :oo :oa if r > R
0
Prob 1.3. The state of a system is given by
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Physicsguide
Basic Formalism: Part-1
Where c/> 1 ), c/>2 ) and c/>3 ) form an orthonormal set . The probability of finding the system in t he state lc/>2 ) is __ . (Give your answer upto two decimal places) [GATE 2016] Prob 1.4. A two-state quantum state has two observables A and B. It is known that t he observable A has eigenstates a1 ) and la2 ) wit h eigenvalues a 1 and a 2 respectively, while
B has eigenstates /3 1 ) and /32 ) with eigenvalues b1 and b2 respect ively, and that t hese eigenstates are related by, [TIFR 2015]
Suppose a measurement is made of t he observable A and a value a 1 is obtained. If the observable B is now measured, t he probability of obtaining t he value b1 will be (a) 0.80 (b) 0.64 (c) 0.60 (d) 0.36
1.2
Normalization of Wave Function
A normalised wave-function means t hat
Prob 1. 5. A particle in one dimension is in a potential
11 oo : oo : 11 A normalised wave-function means t hat
Prob 1. 5. A particle in one dimension is in a potential V (x) = Ab (x - a) . Its wavefunction \JI ( x) is continuous [email protected]
Physicsguide
4
@Sk J ahiruddin, 2020
Basic Formalism: Part-1
. t· ·t . d\J! . everywh ere. T h e d 1scon 1nu1 y 1n dx at x = a 1s
[NET
Dec 2016] 2m (a) /i2 A \JI (a)
Ii (b) A ('11 (a) - \JI (-a)) (c) m A (d) 0 2 Prob 1.6. The normalized wave function of a particle can be written as oc
1
where n(x) are t he normalized energy eigenfunctions of a
[TIFR 2017]
given Hamiltonian. The value of N is
(a)
1.3
1/ 7
(b)
6/7
(c)
3/7
(d)
(6 - 2v1) / 7
Orthogonal States
Prob 1.7. The quantum state sin xl t) + exp (itI
Here K is a constant, and a > d. The position uncertainty
[GATE 2019]
(~ x) of the particle is
12
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(a)
(c)
a
2
+ 3d
2
(b)
12
d2
Basic Formalism: Part-1
(d)
3a
2
+d
2
12
d2
24 6 Prob 1.29. Consider the m otion of a p article along t he x-
11 oo : oo : 34 (a)
(c)
(b)
12 d2
6
(d)
12
d2 24
Prob 1.29. Consider t he motion of a particle along t he xaxis in a potential V (x) = F x . Its ground state energy E 0 is estimated using the uncertainty principle. Then E 0 is proportional to [GATE 2019] (a) F 1 (b) F ~ (c) F ~ (d) F ~
1.6
Commutators
Now you know very well how commutators and uncertainty principles are relat ed. So we have discussed t his topic just after t he uncertainty principle. As you have been int roduced with commutators from your undergraduate days and hopefully you are well accustomed with commutators, here we will state the answer without explicitly solving. For your benefit we have listed different commutator relations at the beginning.
Prob 1.30. If [x, p] = i n, t he value of [x 3 ,p] is [GATE 2007] 2 2 ( c) 3ilix ( d) -3ilix (a) 2ilix 2 (b) - 2ilix 2
j [email protected]
@Sk J ahiruddin, 2020
13
Physicsguide
Basic Formalism: Part-1
Prob 1.31. If x and p are t he x components of t he posit ion and the momentum operators of a particle respectively, the 2 2 commutator [x ,p ] is [GATE 2016]
00: 00: 37 Basic Formalism: Part-1
Prob 1.31. If x and pare t he x components of t he posit ion and the moment um operators of a particle resp ectively, t he 2
commut ator [x , p
(a) i n(xp - px) (d) 2i!i(xp + px)
2
]
[GATE 2016] (c) i!i(xp + px)
is
(b) 2i!i(xp - px )
Prob 1.32. For operat ors P and Q, t he commutator [P, Q-
(b) - Q- 1 [P, Q]Q- 1
Prob 1.33. Consider t he operator a
= x+
!
acting on
smooth funct ions of x . The commutator [a, cos x] is [NET Dec 2016] (a) - sin x (b) cos x (c) cos x (d) 0 Prob 1.34. Given t he usual canonical commutation rela-
t ions, the commutator [A, B ] of A = i (xpy - YPx) and B =
(YPz + zpy) is (a) !i(xpz -PxZ ) (d) - !i(xpz + PxZ)
[NET Dec 2012] (b) -!i(xpz -PxZ) (c) !i(xpz+ PxZ )
Prob 1.35. If the operators A and B sat isfy t he commutation relation [A, B)= l , where I is t he identity op erat or ,
[NET June 2013]
t hen
(a) [eA, B] =
eA
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@Sk J ahirudd in, 2020
(b) [eA, B] = [eB , A] 14
]
[JEST 2016] (c) Q- 1 [P, Q]Q
is
(a) Q- 1 [P, Q]Q- 1 (d) - Q[P, Q]Q- 1
1
(c) [eA, B] = P hysicsguide
Basic Formalism: Part-1
11 oo : oo : 39
@Sk J ahiruddin, 2020
Basic Formalism: Part-1
Prob 1.36. Let x and p denote, respectively, t he coordinate and momentum operators satisfying the canonical commu-
tation relation [x, p] = i in natural units (Ii = 1). Then the commutator [x, pe-P] is [NET Dec 2014] (a) i( l - p) e- P (b) i(l - p 2)e-P (c) i( l - e-P)
(d) ipe- P Prob 1.37. Three operators X , Y and Z satisfy t he com-
mutation relations
[X , Y ] = iliZ , [Y, Z]
and [Z, X ] = iliY The set of all possible eigenvalues of the operator Z, in units [GATE 2007] of Ii, is = iliX
1
3
5
(a) {0, ± 1,±2,±3, ..... } (b) { 2 ' 1, 2 ' 2' 2 ' ..... } 1 3 5 1 1 (c) {0,±-,±l ,±-,±2,±-, ..... } (d) {-2, + 2} 2 2 2 Prob 1.38. Let x denote t he position operator and p the canonically conjugate momentum operator of a particle. The commutator [2~p stants, is zero if
(a) , = f3
2
2
2
2
+ {3x , ¾iP + 1 x ] where f3 and, are con[NET Dec 2017]
(b) , = 2/3
(c) , =
v2/3
(d) 2, = {3
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15
Physicsguide
Basic Formalism: Part-1
11 oo : oo : 42 15
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1. 7
Basic Formalism: Part-1
Probability Current Density
Prob 1.39. A particle of mass m is represented by the wavefunction w(x) = A eikx, where k is the wave-vector and A is a constant. The magnitude of the probability current density of the particle is [GATE 2006]
(a) IA 2 nk
(b) A 2 nk 2m
m
(c)
IA 2 2w 2w(nk )
2
( d)
A 2 (nk )
2
2m Prob 1.40. The wavefunction of particle moving in free space is given by W = eikx + 2e- ikx m
The probability current density for the real part of the wavefunction is [GATE 2012]
(a) 1
(b) nk
(c) nk
(d) 0
2m Prob 1.41. Consider t he wave function Aeikr (r 0 /r), where m
A is normalization constant . For r = 2r0 , t he magnitude of probability current density up to two decimal places, in units of (A 2 nk/m) is__ . [GATE 2013]
1.8
Expectation Values
Prob 1.42. A free particle of mass m moves along t he x direction. At t=O, t he normalized wave function of t he parj [email protected]
16
Physicsguide
11 oo : oo : 44 Prob 1.42. A free particle of mass m moves along t he x direction. At t=O, t he normalized wave function of t he parj [email protected]
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Physicsguide
Basic Formalism: Part-1
1 t icle is given by w(x , 0) = ( 1ra)l/ 4 exp 2 a is a real constant
x2
--
4a 2
+ ix
, where
[GATE 2009]
i) The expectation value of the moment um, in this state is
(c) a ii) The expectation value of the particle energy is 2 2 2 n l n 2 n 4a + 1 ( c) 2m 4a 3/ 2 (a) 2m 2a3/ 2 (b) 2m a fi2
(d) 8ma3/2 Prob 1.43. If t he expectation value of t he momentum is (p) for the wave function W(x), then the expectation value of momentum for t he wave function eikx/nw(x) is [JEST 2013] (a) k
(b) (p )-k
(c) (p)
+k
(d) (p)
Prob 1.44. A particle moving in one dimension has the un2
normalised wave function w(x) = xexp - : 2
where A is
a real constant. The expectation value of its momentum is (p) = [TIFR 2014] n x2 n2 n n (a) A exp A2 (b) A2 - 2 A (c) A exp ( - 1) (d) zero Prob 1.45. Consider the two lowest normalized energy eigen-
11 oo:oo:47 ri
ri
(c) ,\ exp ( - 1)
(a) ,\ exp (d) zero
Prob 1.45. Consider t he two lowest normalized energy [email protected]
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Basic Formalism: Part-1
functions \J! 0 (x) and \Jf 1 (x) of a one dimensional syst em. . d\J! o They satisfy \J! 0 (x) = \J!0(x) and \J! 1 (x) = a - , where a dx is a real constant. The expectation value of the moment um operator in t he state \JI 1 is ri ri
(a) -
a2
(b) 0
(c)
a2
Prob 1.46. The wave function of a quantum mechanical
particle is given by
where 1 (x) and 2 (x) are eigenfunctions with corresponding energy eigenvalues -1 eV and -2 eV, respectively. The energy of the particle in t he state \JI is [JAM 2011] -41 - 11 36 -7 (a) eV (b) eV (c) eV (d) eV 25 25 5 5 Prob 1.47. A particle is in the normalized state \JI) which is a superposit ion of t he energy eignestates E 0 = lOeV) and E 1 = 30eV). The average value of energy of t he particle in t he state \JI) is 20 eV. The state \JI) is given by [GATE 2009]
1 v'3 (a) Ea= l OeV) + IE 1 = 30eV) 4 2 1 2 (b) ~ E 0 = l OeV) + ~ E 1 = 30eV)
11 oo : oo : 49 is 20 eV. T he state '11 is given by t he state 2009] 1 v'3 (a) Ea= lOeV) + IE 1 = 30eV) 2 4 (b)
1
v'3 E 0 =
lOeV )
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2
+ v'3 E 1 =
GATE
30eV)
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Physicsguide
Basic Formalism: Part-1
v'3 1 (c) 2.IEo = lOeV ) - - 4 E1 = 30eV) 1 1 (d) y'2 E 0 = l OeV ) - y'2 E 1 = 30eV) Prob 1.48. The wave funct ion of a particle is given by 1 y'2 o + 1 , where o and 1 are the normalized eigen-
functions with energies E 0 and E 1 corresponding to the ground state and first excited state, respectively. The expectation value of the Hamiltonian in the state '11 is [NET June 2011] Ea (a) + E1 2
11 oo :oo : s2
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1.9
Physicsguide
19
Basic Formalism: Part-1
Ans Keys
1.1. a
1.17. b
1.33. a
1.2.
1.18.
1.34.
C
C
C
1.3. 0.28
1.19. b
1.35. a
1.4. d
1.20. nJ(a/ 2m)
1.36. a
1.5. a
1.21. b
1.37.
1.6. b
1.22. d
1.38. b
1.7. d
1.23.
C
1.39. a
1.8. a
1.24. b
1.40. d
1.9. d
1.25. b,d
1.41. 0.25
1.26. b
1.42. b,c
1.11. b
1.27. 0.3 to 3.97
1.43.
1.12. b
1.28. b
1.44. d
1.13.
1.29. d
1.45. b
1.10.
C
C
C
C
11 oo : oo : 55 1.11. b
1.27. 0.3 to 3.97
1.43.
1.12. b
1.28. b
1.44. d
1.13.
C
1.29. d
1.45. b
1.14.
C
1.30.
C
1.46. a
1.15. b
1.31. d
1.47. d
1.16. b
1.32. b
1.48. d
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1.10
C
Basic Formalism: Part-1
Solutions
Sol 1.1. This is standard t heory. Whenever there is a overlap of wave functions you need to apply quantum mechanics. To be more precise when the de Broglie wavelength becomes
comparable to mean t hermal wavelength , you must apply quantum mechanics and quantum stat mech to deal wit h t he problem. Sol 1.2. Think of the graphs of t he solut ions.
The first option diverges at r = 0, the second becomes constant as r increases, the fourth is a constant inside a definite region and discontinuous at r = R which is not allowed. Only option remains is option (c) . Sol 1.3. We need to find the probability that the system which is now in state \JI) will be in state 2 ) subsequent to A
a measurement of any observable Q. 'T1l, ~ h
rQ+ + ".:IQ
1- i Q
+"
n f"'\rTn ".:I
1i Q ~ I \TI \
'T1l, ~
n
T"\rf"'\
h ".:I hi 1i +"'( r " f h n rl _
11 oo :oo : s1 option remains is option (c). Sol 1.3. We need to find the probability that the system which is now in state '11) will be in state 2 ) subsequent to "' a measurement of any observable Q. The first task is to normalise '11). Then probability of finding the system in state 2 is given by
12 2 Probability = ( '11) (\JI W)
22 2 2 2 1 +2 +3 = 0.28 j [email protected]
2 7
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Physicsguide
Basic Formalism: Part-I
Sol 1.4. The eigenstate is a 1 ) at the time of first measurement. Now write the state a 1) in terms of t he /31) and
/32)
3 4 la1) = /31) + /32) 5 5 Now if B is measured, the probability to get b1 will be 3
5
2
=
9 = 0.36
25
Sol 1.5. As we know to find the discontinuity we must approach the function from both sides. So the limit will be a - E to a + E. Now the Schrodinger Equation is d2 w - m dx 2 + Ab(x - a)W = E\JJ 2
n2
11 oo : 01 : oo a-
to a+ E. Now the Schrodinger Equation is E
d2 w - m dx 2 + A8(x - a) w = Ew 2
n2
Now integrating both sides within the discussed limit is n2
2m
a+E
d2w dx2 dx
a+E
a+E
b(x - a)w dx
+A
a-E
=
E
w dx
a-E
/i2
d\J!2 dx
2m
Thus the discontinuity is
a-E
d\J! 1 + A\ll(a) dx d'112 dw 1 dx dx
2m /i
2
=0 2
= ~;i w(a)
AW (a)
22
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Basic Formalism: Part-1
Sol 1 .6 . q>n(x) are t he normalized energy eigenfunctions of a given Hamiltonian. Now we will expand the w (x). CX)
n
1
vl7 = N
q>o +
1
1
1
vl7
vl7
vl7
Now 1 7 2
~> l =N (
1
l\
1- = )
+
1 7
1
2
+
3 • • •
7
11 oo : 01
: 03
1
7 1=
====;:~>
1
+ 7
2
1
3
+ 7
• • •
N2 _ _1__
1 1-7 1= N
====;:~>
2
X
1
6 -
7
6 7
Sol 1. 7. Orthogonal states mean that the inner product between the states will be zero
There is no short process to solve t his problem other than plucking the states from the options and putting in t he def23
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Physicsguide
Basic Formalism: Part-1
inition and wait to see which option holds. We found t hat
option d satifies the definition of orthonormal states. Now W1)
= sin x
t)
+ exp(icp) cosx l -!-)
Thus
('112 W1) = ( - exp (icp) cos x(t
=-
+ sin x(-!- I) (sin x t) + exp(icp ) cos x -!-) )
exp( icp) cos x sin x(t I t) - exp (2icp) cos x(t I -!-) 2
+ sin
+ exp (icp) cosxsinx(-!- -!-) = - exp(icp) cos x sin x + exp (icp) cos x sin x =0
2
x(-!- t)
11 oo : 01 : os =-
exp (icp) cos x sin x(t
I t)
2
- exp (2icp) cos x(t .J,)
2
+ sin x(.J, t) + exp (icp) cosxsin x(.J, .J,) = - exp (icp) cos x sin x + exp (icp) cos x sin x =0 Sol 1.8. At first we will list out wh at is known t o us. "
0 1) = 111) (u v) = 0
"'
"'
0 12) = 2 2)
(v O v) = 7/4
Now
(v cos
2
6 v) = 7 / 4 2
cp + 2 sin cp = 7 / 4 2 sin cp = 7 / 4 - 1 cp =
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7r
3
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Basic Formalism: Part-1
Simila rly
(u v) = cos 0 cos cp (1I 1) + sin 0sin cp (2 I2) = 0 =;►
cos(0 - cp) = 0 7r
0- cp = 2 0-
7r
7r -
- 2+ 3-
0=
51r Jr- -
6
51r
6
7r
= --
6
11 oo :01 : oa
Sol 1. 9. The Schrodinger Equation is
-n,2 d2\J! + V\J! = E\J! 2m dx 2 It is given that n = 2m = 1 and E = 0. Putting these values we have d2 \J! - V\J! = 0 dx 2 d2
dx 2 [A exp( -x
4
)] -
d2
dx 2 [exp(-x
4
[-12x exp(-x
4
)
+ 16x 4
6
exp(-x
exp (-x [16x )
6
-
4
12x
V[exp(-x
4
V[exp (-x
4
V[exp (-x
4
V[exp (-x
4
)] -
d 4 3 dx [-4x exp(-x )] 2
V [A exp (-x
4
-
)] 2
] -
V == l6x 25
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6
-
)]
=0
)]
==
)]
== O
)] )]
12x
0
=0 =0
2
Physicsguide
Basic Formalism: Part-1
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Sol 1.10. We will solve following the procedure illustrated in t he previous problem but we will add some insights.
Now the Schrodinger Equation is
n,2 d2\J! d ?. 2m x.,
+ V\J!
The wave function is \J! ( x) = ,...
==
e - iax+b
E\J!
11 oo : 01
: 1o
Now the Schrodinger Equation is
The wave function is \ll ( x)
=
e - iax+b
2
Now
d
dx 2
(e - iax+b ) =
(-ia)2 X e - iax+b =
-a2e- iax+b
Thus we have
n2a2 \ll 2m
+ V \ll = E\ll
Now we know that energy (E) is t he eigenvalue of t he Schrodinger Equation which is a constant number. In the solut ion of Schrodinger Equation we will equate the constant terms with energy (E) and all the t erms containing t he independent variable with potent ial (V). Thus from the solution we can easily say t hat the potential • 1s zero.
Sol 1.11. Needless to say we will follow t he steps depicted
26
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. 1n
Physicsguide
Basic Formalism: Part-1
pro blem ?? ... \JI
2
= exp( -x /2) cosh (v'2x)
!! = -x exp(-x / 2) cosh(v'2x) + v'2 exp (-x / 2) sinh(v'2x) 2
..J2 ,T,
2
11 oo : 01
: 13
. pro 61em ?? 1n ...
w = exp( -x
2
/ 2) cosh ( h x)
!! = -x exp (-x / 2) cosh ( hx) + V2 exp(-x / 2) sinh( hx) 2
2
d w 2 2 2 dx 2 = (x + 1) exp( -x / 2) cosh( h x) - 2V2 exp(-x / 2) sinh( v 2
Now the Schrodinger Equation is
d2 w 2m dx2 + Vw = Ew
n2
ld2 w - 2 dx2 + Vw
-x 2 2
w + V2 exp (- x / 2) sinh( h x) 2
1
= Ew
w + Vw = 2
Ew
Now 2
V'V V
= ~ IV - V2 exp( -x / 2) sinh( h x) 2
=
x2 2
- v'2tanh(v'2x)
Sol 1.12. Same as previous problem . Sol 1.13. This is a t heory. Read Delta function well discussion from t he notes of one dimensional potentials. Sol 1.14. Bot h the ground and first excited states need to follow the Schrodinger equation. Using n = l and m = 1/ 2 j a hir@physicsguide. in
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Basic Formalism: Part-1
we get 2
8 7/Jo
- -ox2 + V 'l/Jo =
-4'l/Jo
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: 16
@Sk Jahiruddin, 2020
Basic Formalism: Part-1
we get 2
8 7/Jo
- -ox2 + V 1/Jo = -41/Jo and
a - ox2 + V 'l/J1 = 2 1/J1
l 1P1
Now put '1µ 1 = 1/Jo sinh x the second equation. Keep in mind t hat
dsinh x h dcosh x - - - = cos x · dx dx ' Calculate t he derivatives and obtain . h 02'1f'o
Sln
=
X OX 2 ·
+
2
COS
h o'l/Jo X OX
=
. h ,,,,
+ Sln
X 'f'O
. h sin x
+ V 'ljJ0 sinh x
- 'ljJ0 sinh x
Now use the Schrodinger equation for ground state wave function 82 '1/Jo - ax2
_
+ V '1f'o -
-4'1po here. Doing some algebra you can easily get
0'1f'o
- 4'1µ 0 - 2 coth x -
OX
- '1f'o = -1Po
Now doing integration or solving first order linear equat ion you can get 2
'1f'o = sech x
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Sol 1.15. The double derivative gives
Physicsguide
Basic Formalism: Part-1
11 oo :01
:1
a
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Basic Formalism: Part-1
Sol 1.15. The double derivative gives
d2 dx2A exp
b2x2 -2
b2x2 = A exp - - 2
(-b + b x 2
4 2
)
Substitute the wave function in the Schrodinger equation.
n,2 2
m Aexp
= EAexp Hence
n,2
--
Now at x
=
2
4 2
(-b + b x
2m 0, b = 0. So
)
+V = E
n2b2
E =2m
n,2
4 2
V (x) = - b x 2m Sol 1.16. This is a theory. Read Delta function well discussion from the notes of one dimensional potentials. Sol 1.17. Think about t he boundary conditions. At t he boundary x = 0 t he potential diverges to infinity. So t he wave function must be ZERO there. And when x becomes
large, i.e x -+ oo the wave function again must converge to ZERO. The only option satisfies those is option (b). j [email protected]
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Physicsguide
Basic Formalism: Part-1
11 oo : 01 [email protected]
: 21
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Basic Formalism: Part-1
Sol 1.18. The original Hysenberg uncertainty principle is given by axay 2::
n
2
where ax and ay are the standard deviations in the measurement of any dynamical variables. Refer to Introduction of Quantum Mechanics(2 Ed.) by David J. Griffith (page 112 - 120) We have an "uncertainty principle'' for every pair of observables whose operators do not commute. The most important and commonly used pairs include 1) Position - Momentum 2) Energy- Time 3) Angular Position - Angular Momentum For the sake of completeness we must mention some important concept about Energy - Time uncertainty principle. Position, momentum, energy t hese are all dynamical variables but we cannot measure ''time'' of a particle. In particular flt in t he Energy - Time uncertainty principle is not the standard deviation of a collection of time measurements but rather the t ime the system takes to change substantially. Hopefully we have intrigued your interest in this context . Refer to Introduction of Quantum Mechanics(2 Ed.) by David J. Griffith (page 116) to know more. [email protected]
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: 23
Refer to Introduction of Quantum Mechanics(2 Ed.) by David J. Griffith (page 116) to know more. j [email protected]
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Basic Formalism: Part-1
The position-momentum uncertainty principle is written in t he form
n
i:l.x !.:l.p > - 2
where i:l.x is the standard deviation in the measurement of posit ion and i:l.p is t he standard deviation in t he measurement of momentum. Now !.:l.x ~ 21rr = 2 x 3.14 x 10-15 m Thus we have !.:l.p
=-
n
21.:l.x 6.63 X 10- 34 4 X 3.14 X 10- 15
= 5.279 ~ 100
x 10-
20
x
1
_ x _ 28 eV / c 5 36 10
MeV/c
Even t hough the actual uncertainty principle states that t he product of standard deviations must be greater t han or equal to ~' there are some problems where we assume the product to be greater than n only. Whether we take ~ or n t he order of the answers will be same. Here if we assume t he uncertainty principle as
11 oo : 01
: 26
t he order of the answers will be same. Here if we assume the uncertainty principle as
31
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Basic Formalism: Part-1
Then the expected value of momentum comes to be around ~P ~ 200 M eV /c
We notice that using both t he equations t he order of the •
answer 1s same.
Sol 1.19. Refer t o previous problem Prob 1.18 .. The posit ionmomentum uncertainty principle is written in t he form
Ii ~ x ~p > - 2 Now ~ x ~ L 3 = (10- 12) 3 m 3 = 10- 36 m 3 Thus we have
Ii ~ p=2~x 6.63 X 10- 34 2 X 10- 36 = 331.5 units Now expression for kinetic energy is given by E ~p2
E= 2m
(331.5)
2
=
~ 2
p
2m
11 oo :01 : 2a . . energy is . given . by E Now expression for k.inetic
6.p=-
2m
6_p2
E= 2m 2
=
(331.5) 2 X 1.67 X 10- 27 31 3.3 X 10 J 32
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Basic Formalism: Part-i
Sol 1.20. In the last example we have seen how to estimate t he energy from the Position - Momentum uncertaint y
relation. This example is a perfect elaboration. The general expression of Energy is given by 2
E= P + V 2m p2
E = 2m
+ ax2
Now we replace x by 6. x and p by 6.p. Thus expression for Energy becomes
E =
6_p2
2 - - + a6. x 2m
Now according to the Uncertainty Principle (as given in the quest ion)
n
6.p= 26.x The relation for Energy becomes t;,2
E =
8m
6.
X
2
+ a6. x
2
11 oo : 01
: 31
---
2~x The relation for Energy becomes /i2
E =
8m
~
X
2
+ a~x
2
To get the minimum energy we need to take t he derivative
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Basic Formalism: Part-1
of energy dE
d(~ x) = O 2'/i2 8m~ x3
+ 2a~x =
0 1
/i2 ~ X=
4
8am
Now t he minimum energy is 1
/i2
2
E = - - - - -1 + a /i2
8m
2
8am
8am
liva, =-----+-2
li xv'Sam,
v]m
8mli
liva, = --+-riy'a
v]m 2
X
v]m
riVo,
2 x v'2ffi .
,--
11 oo : 01
: 34
8mli nva + _li_ va_a ~
~
/iyia 2 x v2ffi 2
X
nva y2m Sol 1.21. The uncertainty in Energy or more accurately the standard deviation in the measurement of energy is given by
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Basic Formalism: Part-1
Now t he considered state is \JI =
W1
1
\JI
+ W2
NOT NORMALISED
1
= y'2 \JI 1 + y'2 \Jf 2
The expectation value of Energy of t he oscillator in state \JI •
IS
1
v'2 Similarly the expectation value of square of Energy of the oscillator in state \JI is
11 oo : 01
: 36
Similarly the expectation value of square of Energy of the oscillator in state \JJ is
2
Ef+
1
v'2
35
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Basic Formalism: Part-1
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Now t he standard deviation of the energy of t he state \JI bt.E =
J (E
2
) -
(E) 2
1 2 1 2 1 ( - E 1 + -E2 - - E1 2 2 4 2Ef
+ 2E? -
+ E2
)2
Ef - E? - 2E1E2 4
Ef
+ E? -
2E1E2
4 E1 -E2 2
Sol 1.22. This is little bit lengthy. You need to calculate the average value of the momentum operator
11 oo : 01
: 39
Sol 1.22. This is little bit lengthy. You need to calculate
t he average value of t he momentum operator
(p) =
a
v'f5 (a2 -
-a
4a5/ 2
x2) (-iii)
v'f5 a (a2- x2) dx 4a 5/ 2
ax
This integration will be ZERO as you will be integrating over an odd function with symmetric limit both side of t he or1g1n. T he average value of p 2 over the wave function is •
•
\P2) = -n2
15
a
x 16a5
( a2 -
x2)
82 8x2 (a2 -
x2) dx
- a
I leave the integration on you. The result will be
sn
2
2a 2 [email protected]
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Basic Formalism: Part-1
Hence
sn
2
2a2
_
= 0
v'5n v"ia
Sol 1.23.
Get t he posit ion space wave function 1
1/J(x ) = y2n -
CX)
2
Then calculate t he average values of x and x over 'lj; (x). Sol 1.24 . C;nm n::1.t .ih lP( rnmm11 t.inP-l nnP.r ::1.t nr ~ h::1., rP. r.n m nl P.t .P.
11 oo : 01 'lj; (x) =
: 42 +oo
1
J21r
cp(k )eikx dk
- oo
Then calculate the average values of x and x 2 over 'ljJ (x) . Sol 1.24. Compatible( commuting) operators have complet e sets of simultaneous eigenfunctions. Here operat ors A and B share all t he eigenstates i.e. t hey are commuting. So the product of uncertaint ies ~A~B is 0. Refer to Introduction of Quantum Mechanics(2 Ed.) by David J. Griffith (page 112). Sol 1.25. Standard Theory In problem Prob 1.24. we have ment ioned that compatible operators have complet e sets of simultaneous eigenfunctions. Here we will elaborate this point a bit. If two operators are commutating then every non-degenerate eigenfunction of one operator must be the [email protected]
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Basic Formalism: P art-1
eigenfunction of the other operator. Let us explain this with an example. Let the two observables 2
be 01 = J and 0 always commute.
2
2
= l z. We know that [J , Jz] = 0 They
Now let the states be denoted by J, l z) For J = 2 we have Jz = - 2, - 1, 0, 1, 2 Now we consider a state '11)
1
1
= v'2 2, 1) + v'2 12, - 1)
Now I'll) is an eigenfunction of J 2 but not of l z. We see t hat for a degenerate eigenst at e of J 2 we do not have a simultaneous eigenstate of l z.
11 oo : 01
: 44 J
Now we consider a state '11)
1
1
= v'2 2, 1) + v'212, - 1) 2
Now '11) is an eigenfunction of J but not of l z. We see t hat for a degenerate eigenstate of J 2 we do not have a simultaneous eigenstate of Jz. Note : To know how J 2 and Jz act on a state refer to t he chapter on Angular Momentum. Sol 1.26. This is easiest. The width of the potential well a is t he posit ion uncertainty.
Momentum uncertainty is fj_p= -
Ii
2/j_x
2a
The actual moment um will at least of the order of moment um uncertainty. We take
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Basic Formalism: Part-1
Hence t he energy p2 E = 2m
2
2a
1
fi2
2m
8ma 2
Now substitute t he values. Sol 1.27. Proceeding similarly as previous problem fj_p
=
Ii 2a;
11 oo : 01
: 47
Now st1bstitute the values. Sol 1.27. Proceeding similarly as previous problem
n
l:ip
= 2a;
Sol 1.28. Calculation is lengthy. Better to skip Sol 1.29. From uncertainty principle. (Ignored t he factor 1/ 2 as t hat won't change the result) . !:ip
=-
n
!:ix
The potential is a modulus potential. Let t he part icle is confined between t he length !:ix . As in t he discussion in t he previous problems, t he moment um is of the order of uncertainty of t he momentum. So t he energy 2
E = (l!.p) 2m
or
+ F (l!.x)
fi2 E
[email protected]
=
2m(!:ix)2
+ F!:ix Physicsguide
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Basic Formalism: Part-1
Now to find t he best estimate of the energy, we need to minimize it with respect t o variable !:ix (we did it when we derived t he energy of hydrogen atom or harmonic oscillator using uncertainty principle) dE d!:i x
fi2 m(!:ix)3
+F =
0
11 oo : 01
: 49
derived t he energy of hydrogen atom or harmonic oscillator using uncertainty principle)
dE d~x
n,2 m (~ x)3 + F = 0
You get n2
1; 3
mF
Substitute the value of ~ x in the expression of energy. Hence you get the energy proportional to F 213 Sol 1.30. Simple
Sol 1.31. The commutator relations we will be using are list ed at t he b eginning.
[x2, p2]
=
[x2, pp]
=
p[x2,p]
+ [x2, p]p
= 2i n px + 2i n xp
=
2i n (xp + px)
Sol 1.32. Commutator [P, Q- 1] = PQ- 1
j [email protected]
Q- 1p
40
@Sk J ahiruddin , 2020
Now
-
P hysicsguide
Basic Formalism: Part-1
_ Q-l [P, Q]Q-1 = -Q- l [PQ _ QP]Q-l
= -Q-l [PQQ-l - QPQ-l ] = - Q- l [P - QPQ- l]
11 oo : 01 : s2 Now
- Q-l [P, Q]Q-l = - Q-l [PQ - QP]Q-l
= = = =
- Q-l [PQQ- l - QP Q-l] - Q- l [P - QPQ- l] - Q- lP
+ Q- lQPQ - l
PQ- l - Q-lp d
Sol 1.33. The operator is a= x + dx . Now we will calculate t he commutator
[a, cos x]
=
x
d
+ dx , cos x
= [x, cos x]
d
+ dx ' cosx
Let the commutator act on a state W
d
[x, cos x] W + dx , cos x W = [x cos(x) - cos(x )x]w + =0+
d dw dx cos(x )w - cos(x) dx
dw dw cos(x) dx - sin(x )w - cos(x) dx
= - sin(x) w Thus we have [a, cos x]
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Sol 1.34.
= - sin(x)
41
Physicsguide
Basic Formalism: Part-1
11 oo : 01 : s4 @Sk J ahiruddin, 2020
Basic Formalism: Part-1
Sol 1.34.
[A, B] = [i(xpy - YPx), (YPz + zpy)]
= i [xpy, YPz] + i [xpy, zpy] - i[YPx, YPz] - i[YPx, zpy] = i [xpy, YPz] + 0 - 0 - i[YPx, zpy] = ix[py, YPz] + i[x, YPz]Py - iy [px, zpy] - i[y, zpy]Px
= ix[py, YPz] + 0 - 0 - i [y, zpy]Px = ix X ( -i fi )pz - iz X ( i fi )Px = fi[ XPz + ZPx] Sol 1.35. To solve this problem we will use the list of formulae mentioned above. Now 2 A
A A3 eA= l+ -+-+-+ · ·· 1! 2! 3! Now [
A
e '
B] = [1 B] '
2
[A, B] +
_ O
-
1!
3
[A , B] +
2!
[A , B] ... +
3!
A[A, B] + [A, B]A
2
+ l + 2! + 2 2A 3A =1 + - + + ··· 2! 3! A A2 A3 = l +-+-+-+ ··· =eA 1! 2! 3!
2
A [A , B] + [A , B]A . . . 3!
Thus we have [eA, B] = eA [email protected]
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Basic For malism: Part-1
11 oo :01 : s1
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Basic Formalism: Part-1
Sol 1.36. You can solve the problem just as we have solved the previous problem. Here we have used the relation from
t he previous problem.
[x,pe-P] = [x, p]e-P + p[x, e-P] = ie-P - ipe-P
= i( l
- p)e- P
Sol 1.37. These operators follow the Angular momentum formalism. So their eigenvalues will be integers or half inte-
gers. Both positive and negative, including zero. Sol 1.38. The commutator b ecomes
/3 m
ry 2m
We have calculated b efore
which is non zero. Hence the commutator will be zero only if ry /3 = 0 ⇒ ry = 2/3 m 2m j [email protected]
(c)Sk J a,hir11ddin. 2020
43
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Basic Forma.lism: Part-1
11 oo : 01 : sg
43
j [email protected]
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Basic Formalism: Part-1
Sol 1.39. The expression for probability current density in 1-d is given by
.
iii J = 2m Now we know that
w(x) = A eikx w*(x) = Ae-ikx Putting these values in t he expression we get
j = i ii [A eikx
X ( -ik )A e- ikx
2m iii . = - x( - 2ik)x 2m = A 2lik
- Ae- ikx
X ( ik )Aeikx ]
Al
2
m
Sol 1.40. This problem is somewhat similiar to problem Prob 1.39. but here we have to consider t he real part of the wavefunction The expression for probability current density in 1-d is given by
.
i ii J= 2m [email protected]
44
Physicsguide
11 oo : 02 : 02 .
J=
in 2m 44
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Basic Formalism: Part-1
Now we know that \ll (x)
=
eikx
+ 2 e - i kx
Now the real part of t he wave-function is \ll (x)
real =
cos(kx)
+ 2 cos(kx) = 3 cos(kx)
Since W = \II* the probability current density is zero.
j =
in [-9cos(kx) sin(kx) + 9cos(kx) sin(kx)] 2m
=0 Sol 1.41. Refer to problem Prob 1.39. The expression for probability current density in 1-d is given
by
.
J=
in 2m
Now we know that
w(x) = w(x)* =
Aeikr
ro r
A e - ikr
ro r
11 oo : 02 : os w(x) = w(x)* =
[email protected]
A e ikr
-
r
ro
A e - i kr
r
45
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Basic Formalism: Part-1
Putting these values in the expression we get
. in
J=
To -
ikr
Ae
2m
in
Ae
2m
r
X
m
nk
-
X
m
(ik) A e
r
2
-
r
ro -
-ikr
in - x Al X 2m in 2 - x Al X 2m nk
( - i·k)A e - i kr -ro
(-ik)
ro ro
r
ro -
r
·k
- Aei r
r 2
r5 r3
r
(-2ik) r
ro A l2
2·k
- A e- i kr ro2
ro r2
ro (ik) -:;:
-
2
+
r2
0 r3
2
2
2r0
l
A l2 -
4
= o. 25 nk x IA 2 m
Sol 1.42. (i) The definition of expectation value of any her-
mitian operator is given by A
(A) =
A
A
w*A w dx = (w A w)
The first definition involving the integral over the whole space follows from the wave-function representation of Quant um Mechanics. The second definition follows from the m a-
11 oo : 02 : 01 (A)
=
= (\JI A \JI)
\JI* A \JI dx
The first definit ion involving t he integral over the whole space follows from the wave-function representation of Quant um Mechanics. The second definit ion follows from t he mat rix representation of Quantum Mechanics. The expect ation value of momentum is calculated as fol46
j [email protected]
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Basic Formalism: Part-1
lows A
A
(P) =
'l!* P 'l! dx
1
00
-oo 00
-oo
-
x2
.
x2
1
A
-ix p (2wa)l/4 exp - 4a2 + ix dx (21ra)l/4 exp 4a 1 x2 . d l x2 - ix (- in dx) (2na)l/4 exp - 4a2 + ix 1/4 exp 2 (2?ra ) 4a x2 x2 exp - - 2 -ix- - 2 +ix [- : 4a 4a 2 oo
00
1
= (-in) (21ra)l/2
-
-
2
dx
+ i] dx
2
00
(-in) (21ra)1/2
2
x exp - 2a2 oo
[
X
+i
- 2a2
00
(in) x exp 1 2 2 (2na) / x (2a ) _00
]
dx
n
x2 2a
dx
2
+ (21ra)1/2
x2
00
-oo
exp
2a
2
The first part of the integral is an odd int egral. We know t he integ1nation of odd integral over a complete interval is zero. So we neglect the first part.
,. ,
n
(P ) X
Let
v122a = t
= (2na )1 / 2
=~► dx
,. ,
(P ) =
x2
00
- oo
exp
2a2
= v12a dt
nv12a I~
roe exp r-t l dt 2
dx
dx
11 oo : 02 : 1o exp
(P) = (21ra)l/2 X
Let
=~► dx
v1220'. = t
"
2a2
- oo
= y12a
nv12a
OO
(P) = v'2iro,_ 21ra
ny12a
dt
exp [-t
2
]
dt
- oo X
ft
v'2ira = nv/a 47
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Basic Formalism: Part-1
(ii) The expect ation value of Energy is given by "
(E ) =
\J!*E \J! dx 1
00
=
-oo
x2
n2 d2
•
-i x (21I"O'. )1/4 exp -4a 2
2mdx 2
1 (21ra)l/4 exp -:
Now d\J! 1 x2 -=---e xp -+ix - 4a 2 dx (21ra) 1/ 4 d 2 \J! dx 2
=
1 x2 (21ra)l/4 exp - 4a2
+ ix
x2 - (21ra)l/4 2a2 exp - 4a2 1
1
X
.
X
--+i 2a 2
X
2
- 20'.2
+i
+ ix
Thus we have
" - n,2 (E ) = 2m(21ra) 112
..
r
x2 . exp - -2 -ix 4a 00
CX)
')
77
x2
exp --+ix 4a 2
2
11 oo : 02 : 12 Thus we have - fi2
A
x2
00
(E ) = 2m(21ra)l/2 1
-oo
x
exp - -2 + ix 4a
- - -ix
exp
4a 2
2
- 2a2 exp - 4a2 - n2
x2
.
+ ix
dx
x2 exp ix + ix 2m(21ra) 1/ 2 _00 4a2 4a2 x2 ix 1 - 4- -2- 1 - -2 dx 4a a 2a x2
oo
48
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Basic Formalism: Part- I
x2
oo
2m(21ra) 1/ 2 •
00
4a4
00
'l
a
_
2
x2
exp
2a 2
dx-
x2
xexp
2a
-oo
t;,2
+2m(21ra)1/2
dx
2
1+
1 2a2
00
x2
exp - oo
2a
2
dx
We will neglect t he integration with imaginary part which is actually an odd integral and per£orm the rest of the integration. Let
X
M
v2a
=t
-ri2 (E) = 2m(21ra) 1/ 2 A
1+
1
2a
2
x2 x2 4 4 exp 2 2a -oo a 00 x2 exp dx 2a 2 oo
dx
2
11 oo : 02 : 1 s "'
oo
-/i2
(E)
=
2m(21ra) 112
-
1 1 +2a2
- oo oo
4
a
4
x2 exp
-oo
dx
t2
v'2a exp [-t2] 00
1 2 2a
dx -oo
49
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Basic Formalism: Part-1
yW 2m(21ra) 112 -fi2
2y'2a
2m(21ra) 1/ 2
2v'2a
1+ l 2a 2
Vr.;--2 ~7rQ
yW - ~a - yW
v'2a
-/i2
yW - 4a 2 yW - 2yW
2m(21ra) 112
2y'2a
2 2 /i (4a
2
dx
2a2
- oo
2m(21ra) 1/ 2
2a
x2
exp 00
1+
x2
+ 1)
8ma3/ 2
Sol 1.43. We know how to find t he expectation value of any operator. The expectation value of moment um for t he wave funct ion w(x) is 00
(p) = - 00
w*
-iii d
dx
'11 dx
11 oo :02 : 1a The expectation value of moment um for t he wave funct ion w (x) is 00
(p) =
. d w* -in dx -oo
'11 dx
00
= (-in)
w*w' dx - oo
Now the new wave function is e·ikx/nw (x). The expectation
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Basic Formalism: Part-1
value of momentum for this wave function is 00
-ikx * exp n w
. d - in dx
ikx exp n w dx
-00 00
= (-in)
exp
-ikx ,T,*
n
-oo 00
+ (-in)
exp
'±'
n
-ikx ,T,*
n
ikx exp n wdx
ik
'±'
exp
ikx ,T,, d
n
'±'
X
- oo 00
= (k)
00
w*w dx - oo
+ (-in)
w*w' dx - oo
= (p) + k Sol 1.44. In this problem before computing t he integral to
11 oo : 02 : 20 = (k)
\JJ*\JJ dx
+ (-iii)
\JJ*\JJ' dx
- oo
- oo
= {p) + k Sol 1.44. In this problem before computing t he int egral to find t he expectation value of momentum we must normalise t he wave function . As we have solved the previous problems following t he footsteps we will get opt ion d as t he answer.
d\Jlo Sol 1.45. The st at e is '111(x) = a dx The expectation value of moment um for t he wave funct ion \JJ (x) is oo -
(X)
\JI* -iii d 1 dx
j [email protected]
\JI 1 dx
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P hysicsguide
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Basic Formalism: Part-1
Integrat ing by parts
d\JJ o d2\JJ o d I = (- i fia*a) 2 X _ 00 dx dx 00 d'1lo dwo dx dx _00 oo
_ 00
I = -
\JI* x a la*
- oo
00
. d -iii dx
00
I = - {
X
'117 ( -iii~
00
- oo
2
d \JI o dw o dx dx 2 dx
X
d\Jf 0 a dx dx
l \JI , dx
11 oo : 02 : 23 1 = 0 - ( -in a
2
) _
1= -
oo
Wf
X O'.
- oo 00
1= -
dx2 dx dx 00 . d d\llo d -indx a dx x
w*1 -in d
'111 dx
dx
- oo
0
0
1= - 1
21 = 0
1= 0 Thus we have \Px) = 0 Sol 1.46. We had already shown how to find the expecta-
tion (average) value of any operator in the wave-function representaion of Quant um Mechanics. Here in this sum we will learn how to find t he expectation (average) value of any operator in t he matrix representaion of Quantum Meehan•
lCS .
We know t hat in Hilbert space any wave function can be written as a linear combination of ort honormalized eigen52
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Basic Formalism: Part-1
functions of an observable operator provided t he eigenfunct ions are complete.
w (x, t) =
L Cn(t)fn(x) n
Here f n, ( x) = Normalized eigenfunction of t he operator which is complete en(t ) = It is t he co-effecient offn which basically t ells ll l
1
•
1
•
J
1
•
,, T , , ,
11 oo : 02 : 2s n
Here f n(x) = Normalized eigenfunction of the operator which is complete en ( t ) = It is the co-effecient of f n which basically tells us "how much of f n is contained within \JI" The expect ation value of any operator will be given by
Here Qn = Eigenvalue of the eigenfunction f n when operator Q acts Cn = Probability of the eigenfunction f n to occur To understand more about this and to study the complete derivation please refer to Introduction of Quantum Mechanics(2 Ed.) by David J. Griffith (page 108 - 110) The expression for average Energy is A
Here En = Energy of t he n -t h state Cn = Probability of the n-th state to occur [email protected]
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Basic Formalism: Part-1
The average Energy is
(E ) =
L 3 5 - 9
2
Cn En 2
= 25 +
4 E1 + 5 - 16 x2 25
11
oo : 02 : 2a
(E) =
L 3
2
Cn En 2
-
E1
5 - 9
4
+ -
5 - 16x2
= 25 +
25
41
= -25 eV Sol 1.47. In problem number Prob 1.46. we learnt how to find the expectation value of an operator if the operator and t he probability is given. In t his problem we have to find the state which will give us the average energy 20 eV. Now if we look at t he given solutions we will see that t he wavefunction in option d will give us average energy of 20
eV . The average Energy is
2
1
J2 = 5 + 15
1
Eo+
J2
= 20 eV Sol 1.48. The wave function is w) j [email protected]
=
54
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Basic Formalism: Part-1
expectation value of t he Hamiltonian
H
The
"
= (w H w)
(w w) 1,T, 1
Eo
1 ,/..
,
I
Dl ,J.. \
11 oo : 02 : 30
j [email protected]
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Basic Formalism: Part-1
expectation value of t he Hamiltonian
-
2
j [email protected]
55
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j [email protected]
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Basic Formalism: Part-1
expectation value of t he Hamiltonian
-
2
j [email protected]
55
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11 oo : 02 : 36
j [email protected]
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Basic Formalism: Part-1
expectation value of t he Hamiltonian
-
2
j [email protected]
55
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11 oo : 02 : 39
j [email protected]
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Basic Formalism: Part-1
expectation value of t he Hamiltonian
-
2
j [email protected]
55
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j [email protected]
00:02:41
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Basic Formalism: Part-1
expectation value of t he Hamiltonian
-
2
j [email protected]
55
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j [email protected]
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Basic Formalism: Part-1
expectation value of t he Hamiltonian
-
2
j [email protected]
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Problems and Solutions in Basic formalism of Quantum Mechanics: Part-2 Sk J ahiruddin * Souradeep Mondal
*Assistant Professor Sister Nibedita Govt. College, Kolkata Author was t he topper of IIT Bombay M.Sc P hysics 2009-2011 batch He ranked 007 in IIT JAM 2009 and 008 (JRF) in CSIR NET J une 2011 He has been teaching CSIR NET aspirants since 2012
1
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Basic Formalism: Part-2
11 oo : oo : 04 1
@Sk J ahiruddin, 2020
Basic Formalism: Part-2
Contents 1
Problems from NET, GATE, JEST , TIFR & JAM pap ers 1.1 Hermitian Operators 1.2 Operators • • • 1.3 Time Evolution 1.4 Mixed Problems . 1.5 Ans Keys • 1.6 Solutions . •
•
j [email protected]
•
•
•
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3
•
14 21 22
Physicsguide
11 oo : oo : 06
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1
P hysicsguide
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Basic Formalism: Part-2
Problems from NET, GATE, JEST, TIFR & JAM papers
1.1
Hermitian Operators
Prob 1.1. Which of t he following operators is Hermitian? [GATE 2016] d2 d3 d
(a) dx
(b) dx2
(d) dx3
Prob 1.2. Which one of the following operator is Hermit ian?
[GATE 2017] (c) exp (i Pxa)
Prob 1.3. The hermitian conjugat e of the operator
(a)
a
ax
a
--
8x
[JEST 2013]
(b) -
a
ax
(d) -i
a
ax
Prob 1.4. The adj oint of a differential operator d acting
dx
on a wavefunction W (x) for a quantum m echanical system is: [JEST 2016]
11 oo : oo : 09 Prob 1.4. The adjoint of a differential operator - acting dx on a wavefunction \JI (x) for a quantum mechanical system is: [JEST 2016]
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Physicsguide
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d (a) dx
Basic Formalism: Part-2
(b) -in d
dx
,...
d (c) - -d x
,...
Prob 1.5. A and B represent two physical characteristics of ,... a quantum sysyt em. If A is Hermitian, t hen for the product ,... ,... AB to be Hermit ian , it is sufficient t hat [GATE 2009] (a) B is Hermitian (b)
B is anti-Hermitian
(c)
B is Hermitian and A and B commute
(d) Bis Hermit ian and
A and B anti-commute
Prob 1.6. If A, B and C are non-zero Hermitian operators, which of the following relations must be false? [NET Dec 2013] (a)[A, B]=C (b)AB + BA= C (c)ABA = C
(d)A
+B=
C
Prob 1. 7. Consider a quantum mechanical syst em with three linear operators A , B and C , which are related by AB - C = d I where I is the unit operator. If A = - and B = x, then dx ,... C must be [TIFR 2013] d d (b) d (a) zero (c) -x dx (d) x dx dx A
A
A
A
/"',
A
A
A
A
A
11 oo : oo : 11 1near opera ors
, w 1c are re a e d I where I is the unit operator. If A= - and B = x, then dx C must be [TIFR 2013] A
,
an
A
(a) zero
A
A
d
(b) d dx
d
(d) x dx
(c) -x dx
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1.2
Basic Formalism: Part-2
Operators
Prob 1. 8. In a quantum mechanical system, an observable A is represented by an operator A. If 1/;) is a state of the system, but not an eigenstate of A , then t he quantity A
satisfies the relation (a) r < 0 (b) r = 0
[TIFR 2013] (d) r > 0
(c) r > 0
Prob 1. 9. The quantum mechanical operator for t he moment um of a particle moving in one dimension is given by
[GATE 2011] n2 d2 (d) 2mdx 2
d (b) -in dx
Prob 1.10. If a Hamiltonian H is given as
H =
IO)(0
- 1) (1I + i 0) (1 - i i1) (0
, where IO ) and I1) are orthonormal states, t he eigenvalues [JEST 2015] of H are (a)
±1
(b)
±i
(c)
±v'2
(d)
±iv'2
11 oo : oo : 14 H = IO) (0 - 1) (1I + i 0) (1 - i i1) (0 , where IO) and 11) are orthonormal states, t he eigenvalues of H are [JEST 2015]
(a) ±1
(b) ±i
(c) ±v'2
(d) ±iv'2
Prob 1.11. The Hamiltonian operator for a two-state system is given by j [email protected]
Physicsguide
5
@Sk J ahiruddin , 2020
Basic Formalism: Part-2
H = a( 1) (1 - 2) (2 + 1) (2 + 2) (1 ) where a is a positive number with t he dimension of energy. The energy eigenstates corresponding to t he larger and smaller eigenvalues respectively are: [JEST 2014] (a) 11) - (v'2 + 1) 2), 1) + (v'2- 1)12)
(b) 1) + (v'2 - 1) 2) , 1) - (v'2 + 1) 2) (c) 11) + (v'2-1) 2) , (J2 + 1) 1) -12) (d) 1)-(v'2 + 1) 2),(v'2-1) 1)+ 2) Prob 1.12. The state 'lj;) of a quantum mechanical syst em, in a certain basis, is represented by the column vector
'lj;) =
1/v'2 0 1/ v'2
A
The operator A corresponding to a dynamical variable A , is given, in the same basis, by t he matrix
11 oo : oo : 11
"
The operator A corresponding to a dynamical variable A , is given, in t he same basis, by t he matrix
A
A=
1 1 1 1 2 1
1 1 2 If, now, a measurement of t he variable A is made on the [email protected]
Physicsguide
6
@Sk J ahiruddin, 2020
Basic Formalism: Part-2
system in t he state 'l/;), the probability that t he result will [TIFR 2013] be + 1 is (a) 1/ v'2 (b) 1 (c) 1/ 2 (d) 1/4 Prob 1.13. Two different sets of orthogonal basis vectors
1 0 0 ' 1
and
1
1
1
1
y'2
1
' y'2
- 1
are given for a two dimensional real vector space. The mat rix representation of linear operator A in these bases are related by a unitary t ransformation. The unitary matrix may be chosen to be [NET June 2015] 0 -1 0 1 1 1 1 (a) 1 0 (b) 1 0 (c) v'2 1 - 1 A
(d) 1
v'2
1 0 1 1
11 oo : oo : 20 (a) 1
0
(b)
(c)
1 0
v'2
1 -1
1 0
(d) 1
v'2
1 1
Prob 1.14. The wavefunction of a particle in one-dimension is denoted by 'lj; (x) in the coordinate representation and ipx •
by cp(p) = J 'lj) (x)e n dx in t he momentum representat ion. If t he action of an operator T on 'ljJ (x) is given by T'lj)(x) = 'lj)(x + a) , where a is a constant t hen T 'lj) (p) is [NET June 2015] given by A
A
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7
@Sk J ahiruddin , 2020
Basic Formalism: Part-2
•
•
iap
•
(a) -
'l
napcp(p)
(b) e-
n (P)
i ap
(C) e +
n (p)
. 'l
(d) (1 + nap )cp(p) Prob 1.15. Let 'lj;1 )
1
0
represent
1 0 two possible states of a two-level quantum system . The state obtained by t he incoherent superposition of 'lj;1 ) and 'lj;2 ) is given by a density matrix t hat is defined as p c1 'l/J1) {'l/J1 I+c2 'l/J2) {'l/J2 . If c1 = 0.4 and c2 = 0. 6, the matrix element p 22 (rounded off to one decimal place) is [GATE 2019]
11 oo : oo : 23 element p22 ( rounded off to one decimal place) is 2019]
1.3
[GATE
Time Evolution
Prob 1.16. The wave function W of a quantum mechanical system described by a Hamiltonian H can be written as a linear combination of 'lj;1 and 'lj;2 which are the eigenfunctions of H with eigenvalues E 1 and E 2 respectively. At t =0, A
A
!
t he system is prepared in the state \¥ 0 = : 'lj;1 + 'lj;2 and t hen allowed to evolve wit h time. The wavefunction at t ime h T = (Ei _ E ) will be (accurate to within a phase) [TIFR 2 2 j [email protected]
Physicsguide
8
@Sk J ahiruddin, 2020
Basic Formalism: Part-2
(d) 'l/J2
Prob 1.17. Consider a spin-1/2 particle in the presence of homogeneous magnetic field of magnitude B along z-axis 1 which is prepared initially in a state 'lj;) = (I t) + --1-)) at 2 time t = 0. At what time t will t he particles be in the state - 'lj;) (µB is Bohr magnetron)? [JEST 2012]
(a) t
= -
1rn
µB E (d) Never
(b) t
21rn
= -
µBE
( c)
t
1rn
= --
2µBE
11 oo : oo : 24 - ?jJ) (µB is Bohr magnetron)? 2 (a) t = 1r n (b) t = 1rn µB E
µB E
[JEST 2012]
1rn
(c) t = 2µBE
(d) Never Prob 1.18. A spin-½ particle in a uniform external magnetic field has energy eigenstat es 1) and 2). The system
minimum energy difference between two levels is 2016] h
(b) h 4T
(a) 6T
[JEST
h (c) 2T
(d) h T [email protected]
@Sk J ahiruddin , 2020
9
Physicsguide
Basic Formalism: Part-2
Prob 1.19. The wave function of a part icle at time t = 0 1 is given by 1'11 (0) ) = v'2 (lu1) + u2)), where lu1) and u2) are the normalized eigenstates with eigenvalues E 1 and E 2 respectively., (E2 > El). The shortest t ime after which ?jJ (t)) will become orthogonal to '11 (0)) is [NET Dec 2011]
- n1r (a) 2(E2 - E1) (d)
2n1r E2 - E1
11 oo :oo :2a -li1r
(a) 2(E2 - E1 ) (d)
2/i1r E2 -E1
Prob 1.20. Let 7f;1 and 7f;2 denote the normalized eigenstates of a particle with energy eigenvalues E 1 and E 2 re-
spectively, with E2 > E 1. At time t prepared in a state
The shortest time T at which \JI (t to w(t = 0) is 2/i1r (a) (E2 - E1) 2/i1r
=
0 the particle is
= T ) will be orthogonal [NET Dec 2014] li1r
(c) 2(E2 - E 1)
(d)
Prob 1.21. A two-state quantum system has energy eigenj [email protected]
10
Physicsguide
@Sk J ahiruddin , 2020
Basic Formalism: Part-2
values ±c corresponding to t he normalized st ates 7/J±). At 1 t ime t = 0, t he system is in quantum state v'2 [ 7/J+) + 7/J- )]. The probability that the system will be in the same st ate at
t=
:E is ...... (up to two decimal places) .
[GATE 2018]
Prob 1.22. The Hamiltonian operator for a two-level quan-
t um system is H =
E1 0 0 E2
. If the state of the syst em 1
(
1 \
11 oo : oo : 30 Prob 1.22. T he Hamiltonian operator for a two-level quan-
E1 0 0 E2
t um system is H = at t
=
0 is given by 'lj;(O))
. If t he state of t he syst em
f2
=
1
1
t hen
at a later time t is 1 (a) ( + e - (E1 - E2)t/ li) 2
(c)
[GATE 2019]
1
(b) ½( 1 _ 1
I('lj;(O) l'l/J(t))
2
e-(E1- E2)t/li)
(1 + cos [(E 1
-
E 2 ) t/n])
2 1 (d) (1 - cos [(E1 - E 2 ) t / n]) 2 Prob 1.23. Consider a three-state system wit h energies E , E and E - 3g (where g is a constant) and respective eigenstates 1
1
1P2)
=
J6
1 -2
1
'lj;3)
=
J3
1 1
11
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@Sk J ahiruddin , 2020
Basic Formalism: Part-2
If the system is initially(at t
=
0 ) , in st ate 1Pi)
=
1 0
0 what is t he probability t hat at a later time t , system will 0
be in st ate 'l/J1) =
[JEST 2014]
0 1 A
11 oo : oo : 33 what is the probability that at a later t ime t , system will 0 be in state 'l/J1) = 0 [JEST 2014] 1 4 3gt 3gt 2 (a) 0 (b) sin 2n 2n 9
4 .
(d) 9 Sln
2
E - 3gt
2n
Prob 1.24. The Hamiltonian of a two-level quantum sys1 1 1 . A possible initial state in tem is H = 2 1 - 1 which t he probability of the system being in t hat quant um stat e does not change wit h t ime, is [NET Dec 2017] cos 4 cos 8 (a) (b) sin 1r4 sin 1r
nw
1[
1[
8
(c)
cos !!:2 sin rr2
(d)
cos!!:6 sin 1r 6
Prob 1.25. The wavefunction of a particle in a one-dimensional
potential at time t = O is
[email protected]
12
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P hysicsguide
Basic Formalism: Part-2
where w0 (x) and '11 1(x) are the ground and t he first excited states of t he particle wit h corresponding energies E 0 and E 1 . The wavefunction of t he particle at a time t is [GATE 2006]
(a) lr.::-e-i(Eo+E1)t/2h [2'llo(x)- '111 (x))
(b) ~ e-iEot/li,[2 '11o(x)-
11 oo : oo : 3s x an 1 x are t e groun an t e rst excite states of the particle with corresponding energies E 0 and E 1 . The wavefunction of t he particle at a time t is [GATE 2006] 0
W1 (x)] W1( X )e- iE1 t/2l'i] Prob 1.26. At t = 0, t he wavefunction of an otherwise free particle confined between two infinite walls at x = 0 and
x = Lis
w(x, t
=
0)
2 L
=
. 7r X . 31rx Sln L - sin L
Its wavefunction at a lat er time t
m L2
= - - is a1rn
2018]
(a) (b)
(c)
2
L 2
L
2 L
. 7rX . 31rx sin L - sin L
. 7rX sin L
.
+ sin
31rx
L
•
i1r
exp
-1,Jr
exp
L
j [email protected]
6 •
. 7rX . 31rx Sln - - Sln - - exp L
6 •
-1,Jr
8
13
Physicsguide
@Sk J ahiruddin, 2020
(d)
2
L
. 7rX sin L
[NET June
Basic Formalism: Part-2
.
+ sin
31rx
L
•
-1,Jr
exp
8
00: 00: 38 Basic Formalism: Part-2
2
L
(d)
1.4
. 7rX sin L
.
+ sin
31rx
L
•
-i1r
exp
8
Mixed Problems
Before you start attempting these problems I request you to read the Identical Particles notes first Prob 1.27. The wavefunctions of two identical particles in
states n and s are given by n(r1) and cp8 (r2), respectively. The particles obey Maxwell-Boltzmann statistics. The state of the combined two-particle system is expressed as [GATE 2006]
(a) n(r1)
+ s (r2)
[1i(r1)s(r2)
+ n(r2) s(r1)]
(c) J2[n(r1)s(r2)
n (r2) s(r1)]
(b)
12
(d) n(r1) s(r2) Prob 1.28. Two electrons are confined in a one dimensional box of length L. The one-electron state are given by
Wn ( x) =
2 L sin
n1rx
L
. What would be the ground state
wave function w(x 1 , x 2 ) if both electrons are arranged to have the same spin state? [JEST 2013]
[email protected]
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14
Physicsguide
Basic Formalism: Part-2
11 oo : oo : 40
@Sk J ahiruddin , 2020
Basic Formalism: Part-2
1
2 . y'2 L sin
2 .
+ L sin
•
Sln
•
Sln
1rx2 L
(b) w(x1, X2) = 2 . 7rX1 y'2 L Sln L 1
(c) w(x1, X2) (d)
•
= L Sln
w (x1 , X2) =
-
Sln
2 .
2 .
7rX1
L
2 . 21rx1 L sin L
-
L
Sln
•
Sln
21rx2
•
Sln
L •
Sln
7rX2
L
Prob 1.29. Consider the wave function 1/; = 1/;(7 1 , 7 2)Xs for a fermionic system consisting of two spin-half particles. The spatial part of the wave function is given by
'l/;(71, 72) = }i[1(71)2(72) + 2(71) 1(72)] where 1 and 2 are single particle states. The spin part Xs of the wave function with spin states a(+ 1/2) and /3(- 1/ 2) should be [GATE 2012] (a) ~(a/3 + f3a) (b) }i(a/3- f3a) (c) aa (d) /3/3 Prob 1.30. The ground state and first excited state wave function of a one dimensional infinite potential well are 1/;1 [email protected]
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15
Physicsguide
Basic Formalism: P art-2
11 oo : oo : 43 15
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Basic Formalism: Part-2
and 'lj;2 respectively. When two spin-up electrons are placed in t his potential which one of t he following wit h x 1 and x 2 denoting t he posit ion of t he two electrons correctly represents the space part of the ground state wave function of t he system? [GATE 2014]
(a) (b) (c) (d)
~['l/J1(x1)'l/J2(x1) - 'l/J1(x2)'l/J2(x2)] ~ ['l/J1(X1 )'l/J2(X2) + 'lp1(X2 )'l/J2(X1)] ~['lf;1(x1)'l/J2(x1) + 'l/J1(x2)'l/J2(x2)] ~ ['l/J1(x1)'l/J2(x2) - 'l/J1(x2)'l/J2(x1)]
Prob 1.31. Consider a system of two non-interacting ident ical fermions, each of mass m in an infinite square well potential of width a . (Take the potential inside the well to be zero and ignore spin) The composite wavefunction for t he system with total energy E =
2014]
2
(a) -
a
(b)
2 a
2
.
Sln
1I"X1
-
a
.
Sln
21rX2
.
(c) a sin
1rx1
a
sin
2a
2 (d) sin 1rx1 cos
1rx 2
a
a
a
j [email protected]
•
Sln
+ sin
31r x2
.
- sin -
sin
a
a
•
Sln
a
31rx1 2a
1rx 2
a 16
[NET June
is
- Sln
a
.
2ma
2
•
2 2 1 sin 1rx sin 1rx a
51r2n2
.
a 1rx2
Sln
a
cos 1rx1 a Physicsguide
11 oo : oo : 4s (d)
~ a
•
cos
Sln
j [email protected]
•
-
cos
Sln
Physicsguide
16
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Basic Formalism: Part-2
Prob 1.32. The wave function of a free part icle in one dimension is given by w(x) = A sin x + B sin 3x. Then w(x) is an eigenstate of [JEST 2012] (a) t he position operator (b) the Hamiltonian (c) t he momentum operator (d) the parity operator Prob 1.33. For the parity operator P, which of t he following statements is NOT t rue? [GATE 2016] (a) p t = P (b) P 2 = - P (c) p t = I (d) p t= p-1
i
X
wavefunction in two dimensions? [JEST 2017] (a) kg m- 1 s- 2 (b) kg s- 2 (c) kg m 2 s- 2 (d) kg s- 1
Prob 1.35. It is required to construct t he quantum t heory of a particle of mass m moving in one-dimension x under the influence of a constant force F . The ch aract eristic lengthscale in this problem is [TIFR 2015] 1
(a)
mF h,2
(b)
n
mF
(c)
n m 2F
3
1 2
(d)
ri mF
3
Prob 1.36. Which one of the functions given below repred2 sents t he bound state eigenfunction of t he operator - dx 2
11 oo : oo : 49 m 2F
Prob 1.36. Which one of the functions given below repre-
sents the bound state eigenfunction of the operator [email protected]
2
P hysicsguide
17
@Sk J ahiruddin, 2020
!
2
Basic Formalism: Part-2
in t he region, 0 < x < oo , with t he eigenvalue -4? [GATE 2009] 2 2 (a) A0 e x (b)Aoe- x (c) Ao cosh 2x (d)Aosinh 2x Prob 1. 3 7. The wavefunction of particle moving in free space is given by W = e i kx + 2 e-ikx
The energy of the part icle is
5n2 k2 (a) 2m
3n2 k2 (b) 4m
n2 k2 (c) 2m
[GATE 2012] (d) ri,2k2 m
Prob 1.38. Suppose Hamiltonian of a conservative system in classical mechanics is H = wxp , where w is a constant
and x and p are the posit ion and moment um respectively. The corresponding Hamiltonian in quantum mechanics, in t he coordinate representation, is [NET D ec 2014] b 1 b 1 b (a)-inw(x x - ) (b)-inw(x x + ) (c)-i nwx x 2 2 6 6 6
(d)
_i':x8~
Prob 1.39. Consider a hypothetical world in which t he elec-
t r·on has spin ~ instead of ~. What will be the electronic configuration for an element wit h atomic number Z = 5? [JEST 2019] 4 1 2 2 1 1 1 5 3 (a) 1s , 2s (b) 1s , 2s , 2p (c) 1s (d) 1s , 2s ,2p
11 oo : oo : so tron has spin ~ instead of ~. What will be the electronic configuration for an element with atomic number Z = 5? [JEST 2019] (a) 1s 4 ,2s 1 (b) 1s 2 , 2s 2 , 2p1 (c) 1s 5 (d) 1s 3 , 2s 1 , 2p1 Prob 1.40. The wave function of which orbital is spherijahir@physicsguide. in
18
Physicsguide
@Sk J ahiruddin , 2020
Basic Formalism: P art-2
cally symmetric: (a) Px (b) Py
[GATE 2017]
(c) s
(d) Pxy
-
Prob 1.41. Consider the operator if= jJ- qA , where jJ is
t he momentum operator, A = (Ax, Ay, Az is the vector potential and q denotes the electric charge. If B = (Bx, By, B z)
denotes the magnetic field , the z-component of t he vector operator if x if is [NET Dec 2016] (a) iqnBz+ q(AxPy-AyPx) (b) -iqnBz- q(AxPy-AyPx) (c) -iqnBz (d) iqnBz Prob 1.42. 1000 neutral spinless particles are confined in
a one-dimensional box of length 100nm. At a given instant of time, if 100 of these particle have energy 4E and the remaining 900 have energy 225E t hen the number of part icles in t he left half of t he box will be approximately [TIFR 2015] (a) 441 (b) 100 (c) 500 (d) 625 Prob 1.43. If the root-mean-squared momentum of a part icle in the ground state of a one-dimensional simple har-
11 oo : oo : s4 (a) 441
(b) 100
(c) 500
(d) 625
Prob 1.43. If the root-mean-squared momentum of a part icle in the ground state of a one-dimensional simple har-
monic potential is p 0 , then its excited state is June 2017]
(a) Pov'2
(b) Pov'3
j [email protected]
(c) Po 19
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2/3
(d) Po
[NET 3/2
Physicsguide
Basic Formalism: Part-2
Prob 1.44. Suppose the spin degree of freedom of two part icles (nonzero rest mass and nonzero spin) is described com-
pletely by a Hilbert Space of dimension twenty one. Which of t he following could be the spin of one of t he particles? [JEST 2018]
(a) 2
(b) 3 2
(c) 1
11 oo : oo : s6
[email protected]
20
@Sk J ahiruddin , 2020
1.5
Physicsguide
Basic Formalism: Part-2
Ans Keys
1.1. b
1.16.
C
1.31. a
1.2. a
1.17. a
1.32. d
1.3. a
1.18.
1.33. b
1.4.
C
1.19. b
1.5.
C
1.20. b
C
1.6. a
1.21. 0.25
1. 7. d
1.22.
1.34. d 1.35. d 1.36. b
C
1.37.
C
1.23. b
1.8. a
1.38. b 1.9. b 1.10.
1.24. d C
1.11. b 1.12. d
1.25. d 1.26. b 1.27. d
1 .39. a 1.40.
C
1.41. d
11 oo : oo : sg 1.9. b 1.10.
1.24. d 1.39. a
1.25. d
C
1.11. b
1.40.
1.26. b
C
1.12. d
1.27. d
1.41. d
1.13.
C
1.28. b
1.42.
1.14.
C
1.29. b
1.43. b
1.30. d
1.44. a
1.15. 0.6
21
j [email protected]
P hysicsguide
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1.6
C
Basic Formalism: Part-2
Solutions
Sol 1.1. The definit ion of Hermit ian operator is 00
00
w;Aw1 dT =
(Aw2)*w1 dT
-oo
-oo
Now let us check for opt ion a 00
oc
- oo
w*d'111 d 2
- oo
dx
= [w*w ]00 2 1 -
(X)
x
00 dw; w d d l X
- oo
X
00
- oo
As Quantum mechanical wave function must be well behaved so the first part of integration is equal to zero.
11 oo : 01
: 01
00
- oo
As Quantum mechanical wave function must be well behaved so the first part of integration is equal to zero. 00
- 00
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22
Physicsguide
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So t he operator option b.
Basic Formalism: Part-2
d! is not Hermitian. Now we will check for
00
w*d2'111 dx 2 dx2 -oo oo
-oo
w*d'111 2 dx
oo - (X)
00
_ 00
d'11 2d'll 1 dx dx dx
00
w*dw 1 2 dx
- (X)
00
w*d'111 2 dx
+
- oo
00
- oo
d2 . H .. Thus t he operator dx 2 1s erm1t1an .
11 oo : 01
: 04
X
X
- oo
- oo
CX)
- (X)
.. Thus t h e operator dx 2 1s erm1t1an. There is a second and fast method to check if an operator is hermitian. An operator is hermitian if d2
H
.
We must be careful while using this definition as always t his definit ion will not give us a definit ive answer. For t his particular sum this definit ion will not be beneficial to us. Sol 1.2. Refer to problem Prob 1.1. ,.. ,.
PxX2 - X2Px
,.
23
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Basic Formalism: Part-2
@Sk J ahiruddin, 2020
A_t = - i((PxX2)t - (x2px)t) 2 2 2 . (X Px - PxX ) = -i- - - - 2 2 2 . (PxX - X Px) =i----2 A
=A A
Sol 1.3. The operator A is
a
ax
11 oo : 01
: 06 2
"
=A
a
"
Sol 1.3. The operator A is
ax
00 00
-00
w* - aw1 dx 2
ax
- 00
00
=-
,T,* ,Tr ] 00
[ 'J:' 2 'J:' 1 -oo
Thus the hermit ian conjugate is
+
aw5 d 8x '±' 1 X ,Tr
-oo
a
ax
Sol 1.4. This sum will be solved exactly as t he previous
sum is solved.
j [email protected]
@Sk J ahiruddin, 2020
"
The operator A is 00
Physicsguide
24
Basic Formalism: Part-2
d
dx 00
dx -oo
-oo
00
- oo
11 oo : 01
: 09
CX)
-
CX)
d dx
Thus the hermitian conjugate is Sol 1.5. Operator
(A.B)t
=
(A.B)T =
A is Hermitian i.e. (.A)t = A _BTAT = _BT AT = _Bt At = BA = AB A
A
Thus we see that for the product of AB to be Hermitian B must be Hermitian and A and .B must commute (i.e.
AB = BA) Sol 1.6. We have shown in the previous problem that AB
operator is hermitian then the two operators must commute. That is [A, BJ = 0. But C i- 0. So option a is incorrect.
25
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Physicsguide
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Basic Formalism: Part-2
A
A
A
Sol 1. 7. Let the operator AB - C act on any state f.
d -dxx - C
(AB - C)f =
A
f
d dx (x f) - C(f) A
A
A
A
A
d = (f + x dx (f) - C (f)) A
A
For this AB - C = I equation to hold true C must be equal ,J
11 oo : 01 AB-C
f =
-dxx - C
: 11 f
d dx (xf) - C(f) A
A
A
A
A
A
For this AB - C = I equation to hold true C must be equal d to x dx. Sol 1.8. Basic theory Sol 1.9. The form of the Hamiltonian is given but if we need to find t he eigenvalues of t he Hamiltonian then we need to form the matrix form of the operator . Here we would learn how to form the matrix form of any operator. We will use the eigenstate of the operator as a base ket and will represent the operator as a square matrix. (a(l )
A a (1))
(a(l )
• • •
A a(2))
·••
• • •
•
•
•
Please refer to Modern Quantum Mechanics by J .J. Sakurai (page 20) Now we return to the problem at hand. 26
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H O)
=
Basic Formalism: Part-2
0) - i 1) and H I)
=
i IO) - 1)
Now the matrix form of t he Hamiltonian
(0 H 0) (0 H 1)
H=
( 11HIO\
(11H l1\
1
•
1,
-i - 1
11 oo : 01
: 1s
0) - i I1) and H l) = i IO) - 1) Now t he matrix form of t he Hamiltonian H O)
=
(0 H 0) (0 H 1)
H=
•
1
(1 H 0) (1 H 1)
1,
-i - 1
The characteristic equation •
1,
-i
=0
- 1- A
[-(1 - A)(l + A) - (ix -i)] = 0 [- 1 + A
2
2
A
1] = 0
-
-
2= 0
A=
±v12
Sol 1.10. Here in this problem we know t he form of t he
hamiltonian but we have no information about its eigenstates. Now at first we must know how the hamiltonian acts on a state H 1) = a( 1)(1 - 12)(21+ 1)(21+ 2)(1 ) 1) = a( 1) + 2)) H 2) = a(l l )( l j [email protected]
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2)(2 + 1)(2 + 2)(1 )12) = a( 1) - 2)) 27
Physicsguide
Basic Formalism: Part-2
For this sum we have to individually check which of t he option is an eigenstate of t he Hamiltonian. We have found t hat option b is t he answer. Here we will work out t he
11 oo : 01
: 16
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Basic Formalism: Part-2
For this sum we have to individually check which of t he opt ion is an eigenstate of t he Hamiltonian. We have found t hat option b is the answer. Here we will work out t he sum with option b and show you how to solve t his type of problems. H [ 1 )+( ✓2 - 1) 2)]
=a( ll ) + 2)) + a( ✓2 - 1)(11) - 12)) =a( l + ✓2 - 1) 1) + a( l - ✓2 + 1)12) =a ✓2 [ 1) + ( ✓2 - 1) 2)]
Again
H [ 1 )-(✓2+ 1) 2)] =a(l)+ 2))-a( ✓2 + 1) ( 1 )- 2))
V2 - 1) 1) + a(l + V2 + 1)12) ah [11) - (h + 1) 2)]
= a( l -
=Sol 1.11. Sol 1.12.
Sol 1.13. One set of basis is the unit vectors. Hence t he problem becomes quite easy. It operator will be t he second
set of basis written each vector column-wise.
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28
Physicsguide
Basic Formalism: Part-2
Please see the Linear Algebra notes under Mathematical
11 oo : 01
: 20
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Basic Formalism: Part-2
Please see the Linear Algebra notes under Mathematical Physics. The change of basis and similarity transformation are discussed well. Sol 1.14. This is translation operator. Discussed in Generators of Transformation in Basic postulates of quantum mechanics notes. Sol 1.15. From the definition of the matrix element of Op-
erator Aij
= ('l/Ji A I'lp_j )
P22
=
('l/J2 P lrf2)
Now substituite
And calculate
Sol 1.16. 'lj;1 and 'l/J2 are the eigenfunctions of Hamiltonian H. This means rf 1 and 'lj;2 are stationary states. Now t he A
time developement of a stationary state is given by
n
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29
Physicsguide
Basic Formalism: Part-2
11 oo : 01
: 22
29
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P hysicsguide
Basic Formalism: Part-2
Now
= exp
-iE 1t ri
4 1P1 5
+
3 1P2 exp 5
-iE1 h = exp - - x - - - ri 2(E1 - E2) 4n /, 3n/, -i(E2-E1) h - > ml 2 ) is approximately 2 Dec 2015]
n, 2Vo ( a ) vo+-TT
L m
(b)
Vo+~ Vo Lm
(c)
Vo+
[NET
n Vo
4Lm
n Vo
(d)Vo+ 2L m
Prob 1.4. A part icle of mass m moves in one dimension
under the influence of t he potential V (x) = -ab (x), where a is a posit ive constant. The uncertainty in the product (~ x)(~p) in its ground state is [NET June 2016] (a) 2n (b) n/2 (c) n/-/2 (d) -/2n Prob 1.5. In a one-dimensional system, the boundary con-
dition that t he derivative of the wavefunction 'l/J' (x) should be continuous at every point is applicable whenever [TIFR 2019] (a) t he wavefunction 'ljJ (x) is itself continuous everywhere. (b) t here is a bound state and t he potential is piecewise continuous. (c) t here is a bound state and t he potential has no singularity anywhere. (d) there are bound or scattering states with definite momentum.
11 oo : oo : 11 t inuous. (c) t here is a bound st ate and the potent ial has no singularity anywhere. (d) there are bound or scattering states with definit e momentum.
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1.1
lD Potentials: Part-1
Free Particle in 1D Box
Prob 1.6. A free particle is moving in +x direction with a linear momentum p. The wavefunction of the particle normalized in a length Lis [GATE 2006] •
(b)
l
p
vfL cos nx
(c)
1 -
ipx
v!Le n
•
i px
Prob 1. 7. The wave function \JI n ( x) of a particle confined t o a one-dimensional box of length L with rigid walls is given by Wn(x)=
j; sin
n~x , n= l , 2, 3, ...
[JAM 2008]
i) Determine t he energy eigenvalues. Also, determine t he eigenvalues and the eigenfunctions of t he momentum operator. ii) Show that the energy eigenfunctions are not t he eigenfunctions of the momentum op erator. [This Qs contains many important concepts. Read infi-
11 oo : oo : 13 ator. ii) Show that t he energy eigenfunctions are not t he eigenfunctions of the momentum operator. [This Qs contains many important concept s. Read infinite p otent ial well from Arthur Beiser first , then from David Griffiths, then think of the problem again]
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1D Potent ials: Part-1
Prob 1. 8. A particle of mass m is confined in a one-dimensional box of unit length. At time t = 0 t he wavefunction of the
particle is Wx,o = A sin(21rx) cos(1rx), where A is t he nor[JAM malization constant . 2009]
i) Write the wavefunction W(x ,t ) at a later t ime t. ii) Find t he expectation values of momentum and energy at t = 0. Prob 1. 9. A particle of mass m is confined in the ground
state of a one-dimensional box, extending from x = -21 to x = +21. The wavefunction of t he part icle in t his state is 7r X \ll (x) = Wo cos L where \ll 0 is a constant. [GATE 4 2007] i) The normalization factor WO of t his wavefunction is
(a)
(b)
1
4£
(c)
1 2£
(d)
11 oo : oo : 16 2007]
i) The normalization factor WO of t his wavefunction is
(a)
(b)
1 4L
1 2L
(c)
(d)
ii) The energy eigenvalue corresponding to this state is n,27r2
(a) 2mL2
n,2 7r2
n,27r2
(b) 4mL 2
(c) 16mL 2
n,27r2
(d) 32mL 2
iii) The expect ation value of p 2 (p is the momentum opera-
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P hysicsguide
lD Potentials: Part-1
tor) in this state is n,27r2
(a) 0
(b) 32£2
n,27r2
(c) 16£2
Prob 1.10. A part icle is confined inside a one-dimensional
box of length l and left undisturbed for a long time. In the most general case, its wave-function MUST be, [TIFR 2018] (a) t he ground st at e of energy. (b) p eriodic, where l equals an integer number of periods. (c) a linear superposition of the energy eigenfunctions. (d) any one of the energy eigenfunctions. Prob 1.11. Consider a syst em of eight non-interacting, iden-
t ical quantum particles of spin - ; in a one dimensional box of length L. The minimum excitation energy of the system , 7r2n,2
in units of m L 2 is __ [GATE 2015] 2 Here before solving sums on this topic we must bring to
11 oo :oo :1a •
•
v .1..1. u l
Y
uv.1..1..1.
lh.._....
-1
u~i u.v
1--.., l 1
t ical quantum particles of spin -~ in a one dimensional box of length L. The minimum excitation energy of the system, 7r2fi2
in units of m L 2 is __ [GATE 2015] 2 Here before solving sums on this topic we must bring to your attention one simple mistake which we have observed many students make. What is t he difference between probability density and probability? To find probability we have to integrat e t he square of a wave function within a range.
Prob 1.12. A particle is placed in a one dimensional box [email protected]
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7
P hysicsguide
ID Potent ials: Part-1
of size L along t he x-axis (0 < x < L). Which one of the following is t rue? [GATE 2008] (a) In t he ground state, the probability of finding the L 3L particle in the interval is half 4' 4 (b) In t he first excited state, the probability of finding the L 3L particle in the interval is half. This also holds for 4' 4 states with n=4, 6, 8, ..... (c) For any arbitrary st at e Iw), the probability of finding t he particle in the left of the well is half (d) In the ground state, the particle has a definite moment um
Prob 1.13. Given t he wave function of a particle 1/J(x) 1 : sin ( :
x) for O < x < L and Oelsewhere t he probability
11 oo : oo : 21 (d) In the ground state, the particle has a definite moment um Prob 1.13. Given t he wave function of a part icle 'l/J(x) = 2 7r L sin L x for O < x < L and O elsewhere t he probability
of finding the particle b etween x = 0 and x = L / 2 is __ (Round off t o 1 decimal places) [JAM 2019] Prob 1.14. A part icle is in t he ground st ate of an infinite square well potent ial given by, 0 for -a< x < a V (x ) = oo otherwise The probability t o find the particle in the interval between
[NET Dec 2013]
- ; and ; is j [email protected]
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1 (a) 2
1
(b) 2
1D Potent ials: Part-1
1
+-
1
1
2
7r
(c) - - -
7r
Prob 1.15. A part icle moving in one dimension is confined
inside a rigid box located between x
~ a and x
=
= ;.
If
2 1rX t he particle is in its ground st at e wO( x) = - cos - the a a quant um mechanical probability of its having a moment um
p is given by
,
-- ....
~
[TIFR 2016]
rri,
1•
1
•
("
I
•
,
.
,
11 oo : oo : 24
Prob 1.16. The normalised eigenfunctions and eigenvalues of t he Hamiltonian of a p article confined to move between 0
< x < a is one dimension are 2
n1rx
a
a
'l/Jri(x) = - sin - and n21r2 n 2
En = - - 2ma2 respectively. Here 1, 2, 3, · · · . Suppose the state of the p art icle is 7rX 7rX 'lf;(x) = A sin 1 + cos a a [email protected]
Physicsguide
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lD Potentials: Part-1
where A is the normalisation constant. If the energy of the particle is measured , the probability to get as 7r2fi2 X • - - lS 2ma2 100
What is t he value of x?
[JEST 2018]
Prob 1.1 7. A particle of mass m is in a one dimension al potential 0 for O < x < L V (x ) = 00 otherwise At some inst ant its wave function is given by
11 oo : oo : 21 Prob 1.1 7. A particle of m ass m is in a one dimensional potential 0 for O < x < L V (x) = 00 ot herwise At some inst ant its wave function is given by
'lp (X)
1
= J3 'lp1(X) + i
where 'lj;1 ( x) and 'lj;2 are t he ground and t he first excited states, respectively. Identify the correct statement. [JAM 2018] L n2 31r2 (a) < x >= 2; < E >= 2m £2
(b) < x >=
2L
3
; < E >=
L
n,2 w2
2m L 2 n 2 81r2
(c) < x >= -· < E >= - 2
2' 2m L 2 2 2L n 4 1r (d) < x >= 3 ; < E >= 2m3L 2
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1D Potent ials: Part-1
Prob 1.18. A particle in t he infinite square well V (x) 0 O 0
as shown in the figure
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Physicsguide
11 oo : oo : 44
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1D Potentials: Part-1
V(x)
Vo i - - - - - -
E 0
X
For E < Vo , the space part of the wavefunction for x > 0 is of the form. Where a is a real positive quantity. [GATE 2011]
(b)
e - ax
(d)
e - iax
Prob 1.31. A particle of mass m and energy E moving in t he positive x direction, encounters a one dimensional po-
tent ial barrier at x = 0. The barrier is defined by 2005]
[JAM
= 0 for x < 0 V = Vo for x > 0 (Vo is positive and E > V0 ) V
If t he wave function of t he particle in t he region x < 0 is given as Aeikx + B e - ikx (a) Find the ratio B/ A (b) If B / A = 0.4, find E /Vo , and t he transmission coefficients.
11 oo:oo:47 given as
A eikx
+ B e - ikx
(a) Find the ratio B/ A (b) If B / A = 0 .4, find E / Vo , and the transmission coefficients. [email protected]
P hysicsguide
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lD Potentials: Part-1
Prob 1.32. A particle of mass m and energy E, moving in t he posit ion x direction, is incident on a step potential at
x= O, as indicated in the figure. The height of the potential is V0 , where V0 > E . At x = x 0 , where x 0 > 0, the probability of finding the electron is ¼ times t he probability of finding 2m V0 -E it at x= O. If a = 2016] V(x) '
E -~
X=0
2
(a) -
O'.
1
(c) 2a
X = Xo
(d) 1 40'.
Prob 1.33. A two-dimensional square rigid box of side L
contains six non-interacting electrons at T = OJ( . The mass of the electron is m. What is the ground state energy of the n,27r2
system of electrons, in units of m L 2 ? 2
[GATE 2016]
11 oo : oo : so contains six non-interacting electrons at T = OK. The mass of t he electron is m. What is the ground state energy of the fi27r2
system of electrons, in units of m L 2 ? 2
[GATE 2016]
Prob 1.34. A particle of energy E moves in one dimension under t he influence of a potential V(x). If E > V(x) for [email protected]
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Physicsguide
lD Potentials: P art-1
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some range of x, which of the following graphs can represent a bound state wave function of the particle? [TIFR 2013] 'I'(x) 'lfx)
X
'¥(x)
X
X
Prob 1.35. A beam of identical particles of mass m and energy E is incident from left on a potential barrier of width L (between O < x < L ) and height Vo as shown in the figure (E < Vo) For x > L , t here is tunneling with a transmission coefficient T > 0. Let Ao , AR and AT denote the amplit udes for the incident, reflected and t he t ransmitted were, respectively. [GATE 2008]
11 oo : oo : s3 0
L , t here is t unneling wit h a transmission
coefficient T > 0. Let Ao, A R and A r denote t he amplit udes for the incident, reflected and t he t r ansmitted were, respectively. [GATE 2008] i) Throughout O < x < L , the wave-function (a) Can be chosen to be real (b) is exponent ially de-
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P hysicsguide
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1D Potentials: Part-1
"
V(x)
·-----------
------·
E .
(c) is generally complex
(d) is zero
ii) Let the probability current associat ed with the incident wave be S0 . Let R be the reflection coefficient. Then (a) t he probability current vanished in the classically forbidden region (b) t he probability current is T S 0 for x > L (c) for , x < 0, t he probability current is S 0 (1 + R) (d) for ,x > L , the probability current is complex iii) The ratio of the reflected to t he incident amplitude A R/ A 0 ,• c
11 oo : oo : 55 0
(c) for , x < 0, the probability current is S 0 (1 + R) (d) for ,x > L , the probability current is complex iii) The ratio of the reflected to the incident amplitude AR/ A 0 •
18
(b) ✓( 1 - T ) in mag-
(a) 1 - Ar/Ao nitude (c) a real negative number
(d)
IAT )2 E Ao Vo- E
(1
Prob 1.36. Consider a potential barrier V (x) of t he form: [email protected]
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lD Potentials: Part-1
V(x)
V(x)
= oo
t-------.Vo
x=O
x=a
where Vo is a constant. For particles of energy E < Vo incident on this barrier from the left , which of the following schematic diagrams best represents t he probability density 'l/J (x) 2 as a function of x. [GATE 2019]
(B)
(A)
x= O
x=a
x= b
X
x= O
I tJJ(x)l 2
x=a
x= b
X
11 oo : oo : sa (B)
(A)
x=O
1I/JCx)l
" '
"
x=a
2
x=O
X
lt/J(x)l 2
(C)
x=a
X
(D)
A
1""x=O
x=a
x= b
x=O
X
x= a
x= b
X
Prob 1.37. A free particle of energy E collides with a one-
dimensional square potential barrier of height V and width W. Which one of the following statement (s) is / are correct? j [email protected]
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21
P hysicsguide
1D Potent ials: Part-I
[JAM 2016] (a) For E > V , the t ransmission coefficient for the part icle across the barrier will always be unity (b) For E < V , the transmission coefficient changes more rapidly with W t han with V (c) For E < V , if V is doubled, t he transmission coefficient will also be doubled (d) Sum of t he reflection and the t ransmission coefficients is always one Prob 1. 38. Consider a potential barrier A of height
Vo
and width b, and another potential barrier B of height 2V0 and t he same width b. The ratio TA/T B of tunneling probabilit ies TA and T B, t hrough barriers A and B respectively, for a particle of energy Vo/ l 00, is best approximated by [NET
11 oo : 01 : oo Prob 1.38. Consider a potential barrier A of height
Vo
and width b, and another potential barrier B of height 2V0 and t he same width b. The ratio TA / T B of tunneling probabilities T A and TB, t hrough barriers A and B respectively, for a particle of energy Vo/100, is best approximated by [NET June 2017] (a) exp [(vT.99 - Jo.§9) ✓8mV0 b 2 /n2 ]
(b) exp [(J"f98 -vif§8) ✓8mVob 2 /n2 ] (c) exp[(~ - Jo.§9) ✓8mV0 b2 /n2 ] (d) exp [(v'2.§8 - vif§B) ✓8mV0 b 2 / n2 ] Prob 1.39. A free electron of energy l eV is incident upon a
one-dimensional finite potential step of height 0.75eV. The [email protected]
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lD Potent ials: Part-1
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probaility of its reflection from t he barrier is __ (up to two [GATE 2017] decimal places)
1.4
Finite Potential Well
Prob 1.40. There are only t hree bound states for a particle of mass m in a one- dimensional potential well of t he
form shown in the figure. The dept h Vo of the potential satisfies [GATE 2007]
-a / 2
V
+a / 2 y
11 oo : 01
: 03 [GATE 2007]
satisfies
•
-a / 2
V
+a / 2
-----+- ~ - -; - - - + .-
--.....---·····-·........... _
X
~
0
Prob 1.41. A particle moving in one dimension, is placed in an asymmetric square well pote11t ial V (x) as sketched below.
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lD Potentials: Part-1
-a
V(x) +a
X
The probability density p( x) in t he ground state will most closely resemble [TIFR 2019]
(a) p
(c)
p
11 oo : 01 : os
p
(a)
(c)
0'====:...___,!..._ _ ____..::...X
Ob=c------=~--"------=x -a +a
+a
-a
(d)
(b) p
0 1::::::::::...._ _L___
p
p
----=::::::~X
+a
-a
+a
-a
Prob 1.42. Consider a square well of depth
a with Voa fixed. Let
Vo -+
well has (a) No bound states (c) 2 bound states
and width
oo and O -+ 0. This potential [JEST 2014]
(b) 1 bound state (d) In finitely rnany bound states
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1.5
-Vo
1D Potentials: Part-1
Ans Keys
1.1. b
1.5. c
1.2. a 1.6. d 1.3. d 1.4.
1. 7. do yourself
C
1.8. sin(1rx) exp (-iE 1t/n)+sin(31rx) exp (-i E 3t/n) wh ere
n / 2m,
2 2 2
E,,., = n 1r
0 and 51r
n / 2m
2 2
11 oo :01 : oa 1.3. d 1.4.
1. 7. do yourself
C
1.8. sin(1rx) exp (-i E1t/ ti)+sin(31rx) exp ( -iE3t/ti) where 2 2 2
2 2
En = n 1r ti / 2m, 0 and 51r ti / 2m gn,27r2
1.9. c,d ,c
1 • 20 •
1.10.
C
1.21. a
1.11. 5
1.22. d
1.12. a
1.23. b
1.13. 0.50
1.24. a
1.14. b
1.25. b
1.15. a
1.26. a or b
1.16. 80
1.27. b
1.17. a
1.28. b,c
1.18.
C
1.29. d
1.19. d
1.30. b
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25
Physicsguide
@Sk J ahiruddin, 2020
1.32.
C
1.33. 24
lD Potentials: Part-1
1.38. a
1. 3 9. 0 .10 to O.12
1.34. a 1.40. a 1.35. b,b,b
00 : 01 : 10 1.32.
1.38. a
C
1.33. 24
1.39. 0.10 to 0.12
1.34. a 1.40. a 1.35. b ,b ,b 1.36. a
1.41. a
1.37. b ,d
1.42. b
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lD Potentials: P art-1
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1.6
P hysicsguide
Solutions
1.1. Here the potential is symmetrical. We know that for
a symmetric (even) potential the eigenfunctions of t he one dimensional Schrodinger equation can always be chosen to ,
,
rl
• '
'
,
.
,
,
11 oo : 01 1.6
: 13
Solutions
1.1. Here the potential is symmetrical. We know that for
a symmetric (even) potential the eigenfunctions of t he one dimensional Schrodinger equation can always be chosen to have definite parity i.e. t he eigenfunctions can be chosen to be odd or even. We also know t hat int egration of odd function over a complete interval is equal to zero. Thus we have
('1/Jo x 'l/Jo) = 0
and
The expectation value (x) of t he position operator x in the stat e 'ljJ) is
(x)
=
+ a1 ('lj; )x(ao 'l/Jo) + a 1 'lj;)) = aoa1[('1/Jo x '1/J1) + ('1/J1 x 'l/Jo)]
(\Jll x \JI ) = (ao('l/Jo
11
111
-VO - - -
1.2. To solve this problem we have to closely follow t he
solutions of a particle in a one-dimensional square well. j [email protected]
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27
P hysicsguide
1D Potentials: Part-1
For bound stat e the range of energy must be - Vo < E < 0 The wavefunctions of the different regions are as follows.
11 oo : 01
: 16
@Sk J ahiruddin , 2020
lD Potentials: Part-1
For bound stat e the range of energy must be - Vo < E < 0 The wavefunctions of the different regions are as follows. and and
'11 2 = A sin kx + B cos kx
W3=Cexp(-ax)
Here a=
2ml E I n,2
and
2m(Vo + E )
k=
n,2
Please go through example 4.13 in page 270 in t he book Quant um Mechanics written by Nouredine Zettili(Second Edition) . The lowest value Vo giving n bound states is given by
Vo =
7r2ri2
8ma 2
2
(2n - 1)
where n is t eh number of bound st at es. Now for n = 1, we get
Vo=
7r2ri2
8ma2
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1.3. We have to expand the potential
Physicsguide
lD Potentials: Part-1
11 oo : 01
: 1a
@Sk J ahiruddin , 2020
lD Potentials: Part-1
1.3. We have to expand t he potential X
Vo cosh L = Vo 2
Vo 2
Vo 2
x
[exp(x/ L ) + exp(-x/ L)]
+
Vo 2 1
= Vo + 2
+ Vo
Vo x 2 L
£2
Vo
2
x l + L + 2! L l
+ ... + 2
x
l -
l
L + 2!
X
L
2
+ .. .
2
x2
This potential looks like t hat of Simple harmonic motion. Now we can say k
= Vo
£2
and
w=
Vo m£ 2
So,the ground state energy is
Vo - Vr Ii m £ 2 - o + 2£
Vo m
1.4. At first you need to study Delta Function Well from t he 1D Potential notes or any books like Griffit hs or Zettili The X < ( foexp(ax) wave funt ion is written as W(x) = X > ( foexp(-ax) As the wavefunction is even about x = 0 so we can easily depict t hat j [email protected]
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Physicsguide
lD Potentials: Part-1
11 oo : 01 [email protected]
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29
P hysicsguide
lD Potent ials: Part-1
@Sk J ahiruddin , 2020
X
(x) = 0
(p) = 0
and
Now 00
2
x exp(-2ax) dx =
0
1 2a 2
Thus we have
~x =
✓ (x2) - (x)2 =
_ 1_
v12a
Before calculating the uncertainty of momentum please go t hrough the remark section in page 510 in t he book Quant um Mechanics written by Nouredine Zettili(Second Edit ion). In t he problems where the first derivative of the wave funct ion is discontinuous at a given value of x, one has to be careful while calculating momentum or one would get neg-
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30
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11 oo : 01
: 24
careful while calculating momentum or one would get neg-
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30
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1D Potentials: Part-1
ative momentum. We have to use t he following expression
- 00
Thus the uncertainty in momentun
Thus we have !J.x/J.p
l
= -- X
v'2a
ria
= -
Ii
v'2
1.5. There are many discussions on when the derivative of wave function becomes discont inuous. Physically, when the wave function has a boundary condition involving an infinite potential, there is a discontinuity in t he derivative because t he t he wave function must immediately vanish at
t he boundary (because you will never find t he particle in t he potent ial), and it may not do so smoothly, usually creating a corner. For a finite potent ial, the particle may be found inside it, even if it's not very likely - so t he wave funct ion is not required to immediately drop to zero and instead smoothly transit ions to an exponential. For mathematical argument, please follow t he derivation of Dirac Delta potential well. At that problem the derivative of 1/J(x) is discont inuous.
11 oo : 01
: 26
smoothly t ransit ions t o an exponential. For mat hematical argument, please follow the derivat ion of Dirac Delta potent ial well. At that problem the derivative of 'ljJ (x) is discontinuous. 31
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lD Potentials: Part-1
discussion in Arthur Beiser Concepts of Modern P hysics, Chapter 4. 1. 7. This is basic t heory. P lease read t he Discussions from
Beiser and Griffiths and One dimension pot ent ial not es form t he Quantum mechanics not es t hen come in problem solv1ng. •
1.8. Now the wave function is Wx,o = A sin 21rx cos 1rx A 'Y x,o = (sin 31rx + sin 1rx)
2
Now we have t o find t he Normalising Const ant A.
(w w) = 1 A 2
2
A
+ -2
2
=~► A =
=1
v'2
Now t he actual wave function at t ime t = 0 is 1 'Y x,o = -J2 (sin 31rx + sin 1rx) The Wave Funct ion w(x, t) at a later time (i.e. t
= t) is
11 oo : 01 : 2a Now the actual wave function at time t = 0 is 1 Wx,o = v'2 (sin 31Tx + sin 1TX) The Wave Function w(x, t) at a later time (i.e. t
1
= t) is
1
J2 sin (7rx) exp (-iE 1t/n) + J2 sin (37rx) exp (-iE3t/ n) 32
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1D Potentials: Part-1
where E1 =
?T
2 2
n / 2m 2 2
E3 = 9?T n / 2m Now t he expectation value of Momentum is
1 (PA) -_ 2
1 ( sin 31TX 0
. 31TX -1 i(Sln 2 O ( -in) 2 (- i n) 2
+ sin 1TX)P" (sin 31TX + sin 1TX) dx
. 1TX ) + Sin
·t:;
-'l, 11,
a (Sln . 31TX + Sin . 1TX ) d X
8X
1
(sin 31Tx
+ sin 1TX) (3 cos 31Tx + cos ?TX)
0 1
( 3 sin 31Tx cos 31TX
+ sin 31Tx cos 1TX
0
+ 3 sin 1TX cos 31Tx + sin 1TX cos 1TX)
=0
dx
1
-in 4
dx
(3 sin 61Tx + sin 21Tx 0
+ 4 sin 41Tx -
2 sin 21Tx) dx
11 oo : 01 4
(3 sin 61rx
: 31
+ sin 21rx + 4 sin 41rx -
2 sin 21rx) dx
0
=0
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33
lD Potent ials: Part-1
@Sk J ahiruddin, 2020
1.9. (i) 2L
- 2L
w5 w 5 [ 2 X
X
w02
2L
2
- 2L
w 5 - 2L + 2 X
]2L
2L
cos
2
-2L
1fX
-
4L
dx
=l
dx
=
1 + cos
1fX
2£
1fX
2L
2£
- 2L
1f
•
Sln
2£
w2
l
= 1
° X 4£ = 1 2 1
2£ (ii) Here the length of the one-dimensional box is 4£.
11 oo : 01
: 34
Here the length of the one-dimensional box is 4£.
n21r2 n,2 En=--2ma2
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lD Potentials: Part-1
(iii)
2L
1rX
cos
n,2
-
2£ n,2
4£
2L
cos,.
-2L
1rX
1 + cos
x - -2 16£
- 2L
7r2
X
16£2
[x]2~L
7r2
X
16£2
1rX
+ 4L
dx
2L
n,2
X
4L
dx
dx
4L
2L
n,2
4£
?
x
n,2
4£
cos
4L
- 2L
1rX
7r2
X
16£2
7rX
X
sin 2£
2L - 2L
ri27r2
X
4£
= 16£2
1.10. To answer t his question we must be well versed with
t he postulates of Quantum Mechanics. Please read properly t he Mathematical Background and Basic postulates notes. When a system is left undisturbed or no measurement is done on the syst em t hen the wave function is a linear
11 oo : 01
: 37
1.10. To answer this question we must be well versed with t he postulates of Quantum Mechanics. Please read properly t he Mathematical Background and Basic postulat es notes. When a system is left undisturbed or no measurement
is done on the system t hen the wave function is a linear superposition of t he energy eigenfunctions. If the system is disturbed or a measurement is made then after a long time the system tends to attain a wave function which is a linear superposition of t he energy eigenfunctions 1.11. The spin is - ~. So the degeneracy will is (2s + 1)
= 4.
The sysytem conists of 8 particles. The ground state of the syst em will be formed of 4 particles in the ground state and 35
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1D Potent ials: Part-1
4 particles will be in the first excit ed st ate. 2 2 4n n n n Ea = 4 x 2m L2 + 4 x 2mL2 2 2
2 2 20n n 2mL2
The first excited state of t he system will be formed of 4 particles in the ground state and 3 particles will be in t he first excited state and 1 particle in the second excited state. n 2 n2 E1 = 4 x 2mL2
4n 2n2 9n 2n2 + 3 x 2mL2 + 1 x 2mL2
25n 2n2 2mL2
Now t he minimum excitation energy is 25n 2n2 E1 - Ea = 2mL2
20n 2 n2 2mL2
5n2 n2 2mL2
1.12. The wave function in posit ion space is given by
11 oo : 01
: 39
251r2fi 2 EI - E o = 2m£2
201r 2!i2 2m £ 2
57r2fi2 2m L 2
1.12. The wave function in posit ion space is given by
2 .
-
Slil
L
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n1rx
L
Physicsguide
36
@Sk J ahiruddin, 2020
-3L4
ID Potentials: Part- I
w*w dx
L 4
-3L 4
2 L
Sln 4
SIIl
3L
1 - cos
L
4
3L
L
---
4
L
4
X
1
1
--
L 21r
=---= 0.1817 2 7r
L
dx
L
L
1
dx
21rx
4
-
1TX
•
L
L
1
1
1TX
•
L
•
SIIl
3L L x 4
21r
-
•
SIIl
21r L -XL 4
11 oo : 01
: 42
L
X
21r
1 1 = - - - = 0.181 7
2
7r
-3L 4
w*w dx
L 4
3L
2 L
21rx
4 •
Slll
Slll
L
L 4
3L
1 L
L
1
3L
4
1 - cos
4
-
L
L
-
-
4 -
1 L
-
4
41r L
•
Slll
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21rx
•
L
41rx
dx
L
41r
•
Slll
L
X -
3L
L x 4 L
X
4
dx
1 2
41r
P hysicsguide
37
@Sk J ahiruddin , 2020
lD Potentials: Part-1
1.13. -L
2
\¥ 2=
w*w dx
0
2
-
-L2 •
Slll
1I"X
L 0 L L 2 1 1 - cos L 0 1 L 1 - - 0 -L 2 L 1
-
1I"X
•
Slll
L
21rx
dx
L •
Slll
dx
21r
L
xL 2
-
•
Slll
21r L
xO
L X
21r
11 oo : 01 1 - cos
L - - 0 L 2
1 -L
1
dx
L
0
: 44
21r L xL 2
•
Sln
-
•
Sl n
211" L
X
Q
X
L 21r
1
-
2
1 .14. Be careful and see that the infinite square well ranges from -a ::; x ::; a and not our usual r ange that extends from
0 < x < a. Basically when we choose t he range O < x < a t he origin always contains ant i-nodes of the wave funtion. When t he range extends b etween -a < x < a t he origin contains bot h t he nodes and ant i-nodes of the wave function. The wave function of a particle within an infinite square
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ID Potentials: Part-1
potential well extending from -a < x < a
2 n1rx -cos L L 2 n1rx - Sln L L
W=
for odd value of n
•
for even value of n
For the ground state n=odd= l . Now to the probability to ("'
1
I
1
I
•
1
• , 1
•
11
•
1
11 oo : 01
: 47
2 . n1rx - Sln L L
for even value of n
For the ground state n= odd= l . Now to the probability to find the particle wit hin the said range a 2 - a 2 a 2 -a 2
P=
w*w dx 2 cos 2a
a 1 2 cos2 -a
a
2 cos 2a
1fX
2a
1fX
2a
dx =
1fX
2a
dx
a 2 2 cos 2 -a
1
2a
2
1fX
2a
dx
2
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lD Potentials: Part-1
a 1 2a 1 x 2a 1 ra
2 - a 2
1 + cos
a + -sin 7f
a
21rx
2a
a 1fX
a
a i ..
2
-a
2 .. \ l
dx
11 oo : 01
: 49
2 l 2a
a .
X
+ -1r Sl n
a 2 -a 2
1rx a
1.15. Find t h e momentum space wave function
1 ~
cp(p) =
CX)
'l/Jo(x) e- ipx/fidx - (X)
1
~
2 a
a/2
cos - a/2
7rX fi
e-ipx/!idx
a
Do t he integration and t ake m odulus.
40
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1D Potentials: Part-1
1.16. At first we have to normalise t he wave-function . 'lj; ( x)
= A sin aA 2
7rX
a
1 + cos 2 • - Sl n a
7rX
a
7rX
a
+
2 • - Sl n a
7rX
a
cos
7rX
a
11 oo : 01 : s2 1.16. At first we have to normalise t he wave-function. 1rX
'lj; (x) = A sin
1 + cos
a
2 - Sl n a
~A
•
2
1rX
a
1rX
a
a
•
+
2 . 1rX -sin -
~A 2
2 1rX 1rX - Sln cos a a a
2 1rX - x 2 sin cos a a
l +2
a
1rX
a
aA 2 Now to normalise
a 2 -A (1 1) 2
==>~►
1 + 4 (2 2)
==!l~>
A
2
5a
X -
8
=1 =1 8 5a
Finally the wave function is
'l/J (x) =
a 2
Now the prob ability to get t he ground state energy
·1·t
P ro b a b 1 1 y [email protected]
=
I(1
w) 2 ('11 '11)
41
4 5 Physicsguide
@Sk J ahiruddin , 2020
ID Potentials: Part-1
Finally we n eed to find x 4
X
==>~► X
= 80
11 oo : 01 : s4 @Sk J ahiruddin, 2020
1D Potent ials: Part-1
Finally we need to find x 4
X
5
100
-
=~► X
== 80
1.17. In this problem just by finding the expectation value of energy we will be able to find t he correct answer. Nev-
ertheless we will provide you wit h t he idea how to find the expectation value of position. The given wave function is normalised. So the expectation value of Energy can be written as
Thus we get answer a as t he correct option. Now if we wanted to compute t he expectation value of post ion we had to follow t he next few steps \ X)
1 == 3 (Wl IX Wl)
2
+ 3('l/;2 X 'l/;2)
i +3
3 • • •
1.18. To solve t his problem we need the following t [email protected]
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P hysicsguide
1D Potent ials: Part-1
11 oo : 01 : s1
@Sk J ahiruddin , 2020
lD Potentials: Part-1
metric ident ity. sin(3A)
~~►
3
= 3 sin A - 4 sin A 3
3
sin A =
sin A -
4
~ sin(3A) 4
Now we will try to rearrange t he wave function to a well known form.
\J! (x) = A sin
7rX
3
a~ a 2 2 a
3 . = A- Sln
1 . - A - sin
7rX
4
a
4
a
~
. 3 Sln
X
2 . - Sln a
a
7rX
2
a
31rx
a
a - cp3 2
Normalise
(\JI \JI) =
A2
A=
9a2 32
2
a =1 + 32
32 l Oa
F inally, t he form of the wave function stands as a 2
32 lOa T A oc exp [email protected]
Vo
and E
=
~~ . Thus 1
Vo Vo - 100
- J2m (0.99½ ) 52
Physicsguide
lD Potentials: Part-1
@Sk J ahiruddin , 2020
Similarly for barrier B T B oc exp
-
~~> T B oc exp
2m
2Vr0 -
Vo 100
- J2m (1.99V0 )
Only opt ion a has an expression consist ing of the calcut at ed t ransmission probabilit ies.
11 oo : 02 : 2s ===>~► TB ex exp - ✓2m (1.99V0 ) Only opt ion a has an expression consisting of the calcutated t ransmission probabilities. 1.39. Just use the expression of reflection coefficient v'E - ✓E - Vo
R=
2
v'E + ✓E - Vo
E = 1, o = 0.75 here. 1.40. You can prove t hat if t he number of energy levels are
(N + 1), then N1r
1r
- 2 - 0 if X < 0
where w is a const ant. Which one of the following represents t he possible ground state wave function of the p art icle?
[JEST 2019]
11 oo : oo : 09
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lD Potent ials: Part-2
@Sk J ahiruddin , 2020
q,
+I
(8)
(A)
+oo
+oo :t
X •
-1
II/
+l
- 1
.
(D)
(C)
+oo
+oo -1
-1
Prob 1.3. Consider t he 1-D asymmetric double-well potent ial V(x) as sketched below. The probability distribution
p(x) of a part icle in t he ground state of this potential is best represented by [TIFR 201 7]
V(x)
11 oo : oo : 11
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lD Potentials: P art-2
p(x)
p(x)
X
X
p(x)
p(x)
X
----=:::::___
_,____-=----:x
Prob 1.4. A part icle is confined inside a one-dimensional
potential well V (x), as shown below.
v(.r)
One of t he possible probability distributions 'ljJ (x)
2
for
11 oo : oo : 14
One of t he possible probability distributions 'ljJ (x) 2 for [TIFR 2016] an energy eigenstate for this particle
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1D Potent ials: Part-2
IV' (x)l2
IV'( X )!2 (a)
IV'(x)lz (d)
(b) X
0
(c)
X
0
X
0
X
Prob 1. 5. A part icle is constrained to move in a truncated
harmonic potential well (x > 0) as shown in the figure. Which one of the following st atements is CORRECT? [GATE 2012]
V(.Y)
X
(a) The parity of t he first excited state is even (b) The parity of the grounded state is even (c) the ground state energy is ½liw
11 oo : oo : 16 X
(a) The parity of the first excited state is even (b) The parity of the grounded state is even (c) the ground state energy is ½nw (d) The first excit ed state energy is ~ nw Prob 1.6. The energy levels of a particle of mass m in a 00
potential of the form V (x) =
X
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lD Potentials: Part-2
@Sk J ahiruddin, 2020
are given in terms of quant um number n=0,1,2,3 .... by [GATE 2007] (a) (n + ~)nw (b) (2n + ~)nw (c) (2n + ~)nw (d) (n + ~)nw Prob 1. 7. T wo harmonic oscillator A and B are in excited eigenstates wit h t he same excitation energy E , as measured
from their respective ground state energies. The natural frequency of A is twice that of B.
- 10
-5
0
5
10
If the wavefunction of B is as sketched in t he above pict ure, which of the following would best rep1~esent t he wave-
11 oo : oo : 19 -10
0
-5
5
10
If t he wavefunction of B is as sketched in t he above pict ure, which of the following would best represent t he wave[TIFR 2016] fucntion of A?
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lD Potentials: Part-2
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(a)
- 10
(b)
-5
5
10
-10 -5
1.2
10
5
10
(d)
(c)
-10
5
-5
5
10 -10
Probability
Prob 1. 8. A one-dimensional harmonic oscillator is in t he
state
11 oo : oo : 22 Prob 1.8. A one-dimensional harmonic oscillator is in the
state
where w0(x) \lf 1(x) and w2(x) are the ground, first excited and second excited states, respectively. The probability of finding the oscillator in the ground state is [GATE 2006] 3 9 (a) 0 (b) JM (c) 14 (d) 1 Prob 1.9. Consider a harmonic oscillator in t he state
l"P) =
a2
e
2
e a at
0) , where 0) is the ground state, at is the raising
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1D Potentials: Part-2
operator and a is a complex number. What is t he probability t hat t he harmonic oscillator is in the n-th eigenstate n) ? [JEST 2015] a2 2n
la2J _a_ (a ) e n 1.
(b)
2
e
la n Vnf
a2 a12n
(d)e
1.3
2
n!
Uncertainty Principle
Prob 1.10. Let 0) and 1) denote the normalized eigen-
states corresponding to t he ground and t he first excited
11 oo : oo : 24 1.3
Uncertainty Principle
Prob 1.10. Let 0) and 1) denote the normalized eigenstates corresponding to t he ground and the first excited states of a one-dimensional harmonic oscillator. The uncertainty ~ x in t he state 0) + II)) is [NET D ec 2012] (a) ~ x = J!i/2mw (b) ~ x = J!i/mw
f2(
(c) ~ x = J2!i/mw
(d) ~ x = J4!i/mw
Prob 1.11. The product ~ x ~ p of uncertainties in t he position and momentum of a simple harmonic oscillator of mass m and angular frequency w in the ground state IO) , is !i/ 2. The value of the product ~ x ~p in the state e- ipf/h 0) (where R, is a constant and p is t he moment um operator) is [NET [email protected]
P hysicsguide
9
@Sk J ahiruddin, 2020
lD Potentials: Part-2
Dec 2018]
(a) fi 2
mwf
2
(b) Ii
h
(c) !i/ 2
h2 (d) mw£2
Prob 1.12. Let 0) and 1) denote t he normalized eigenstates corresponding to t he ground and first excited states of a one dimensional harmonic oscillator. The uncertainty ~ Pin the state 0) + 1) ), is [NET Dec 2011]
f2 (
(a) ~p = ~ / 2
(c) ~ p =
!imw
(b) ~p = J!imw / 2 (d) ~ p = ✓2!imw
11 oo : oo : 21 ~Pin t he state
f2 (0) + 1) ), is
(a) ~P = ~
/2
(c)
(b) ~P (d ) ~P
~ p=~
1.4
=
[NET Dec 2011]
Jnmw/2
= ✓2nmw
Expectation Values
Prob 1.13. A particle is in a state which is a superposition of t he ground state 0 and t he first excited state 1 of a one-dimensuonal quantum harmonic oscillator. The state is
.
given by
1 = ✓5 0 5
+
2 . ✓5 1 . The expectation value of the 5
energy of t h e particle in t his state (in units of nw, w being [JAM 2015] t he frequency of the oscillator) is _ _
Prob 1.14. In a one-dimensional harmonic oscillator , 0 , 1 and 2 are respectively the ground, first and the second excited states. These tree states are normalized and are ort hogonal to one another. W1 and W2 are two states defined j [email protected]
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1D Potent ials: Part-2
by
where a is a constant
[GATE 2011]
i) The value of a which '11 2 is orthogonal to '11 1 is
(a) 2
(b ) 1
(c) -1
(d ) -2
ii) For the value of a det ermined in t h e above question, the PYn Prt.:::itinn , ,.:::i l11 P nf Pn Pr o-, , nf t.h P n~rill.:::it.nr in t h P ~t.:::.tP \lJ "
11 oo : oo : 29 i) The value of a which '11 2 is orthogonal to '11 1 is (a) 2 (b) 1 (c) -1 (d) -2 ii) For the value of a determined in the above question, the expectation value of energy of the oscillator in the state w2 •
IS
(a)
nw
(b) 3nw/ 2
(c)
3nw
(d) 9nw/ 2
Prob 1.15. A one dimensional harmonic oscillator is in t he superposition of number state n) given by
The average energy of t he oscillator in the given state is _ nw [GATE 2014] Prob 1.16. The ground state energy of 5 identical spin 1/ 2 particles which are subject to a one dimensional simple [JEST harmonic oscillator potential of frequency w is 2012]
(b) (13/2)nw
(a) (15/2)nw [email protected]
(c) (1/2)nw
11
(d)
5nw
Physicsguide
@Sk J ahiruddin , 2020
ID Potent ials: Part-2
Prob 1.17. A harmonic oscillator has the wave function,
w(x, t) =
1
5
[3~0 (x, t) - 2J2~ 1 (x, t) + 2J2~ 2 (x, t)]
where ~n(x, t) is the eigenfunction belonging to t he n-th 1 energy eigenvalue (n + )nw. The expectation value (E) of
2
energy for the state w(x, t) is (a)
1.58nw
(b)
0.46nw
[TIFR 2013] (c)
nw
(d )
1.46nw
11 oo : oo : 32 where o - 21>1 + 31>2 1>o - 1>1 + a 1>2) = 0 (1>o 1>o)
+ 2(1>1 1>1) + 3a(1>2l 1>2) = 0 3a = -3 a = - 1
Here vve have used t he property
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P hysicsguide
ID Potentials: Part-2
Thus t he state \JI 2 is given by
The state \JI 2 is orthogonal to \JI 1 but is not normalised. As t he state is simple it can be normalised easily by inspection. The normalised state is
11 oo : 01
: 11
The state \JI 2 is orthogonal to \JI 1 but is not normalised. As t he state is simple it can be normalised easily by inspection. The normalised stat e is
(ii) The expectation value of Energy of the oscillator in state \JI 2 is
1.15. This type solved previously 1 1.16. The spin of t he particle is . The degeneracy due to 2 1 spin is (2s + 1) = (2 x + 1) = 2. This means that in
2
a particular state two particles can accommodate wit hout 26
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lD Potentials: Part-2
@Sk J ahiruddin , 2020
breaking t he P auli's Exclusion Principle. Now the ground state energy of t he system E
= 2 x o+
1 -2
'fu»+2x
1 2+ 2
11 oo : 01
: 13
breaking t he P auli 's Exclusion Principle. Now t he ground state energy of t he system
E =2x
1 0+ 2
1 2+2
liw+2x
= 13 liw 2 1.17.
9
8
(E) = 25Eo + 25E1
8
+ 25E2
where
1 n+- liw 2 1.18. We solved previously that for half harmonic oscillator
3 7 Eo = nw, E1 = nw 2 2 So, t he expectation value of energy
(E) =
1 3
4 7
-5 · -+· 2 5 2
1.19. The expression for the position operator of a linear harmonic oscillator is X"' = •
1,1[
The state is w)
j [email protected]
@Sk J ahiruddin, 2020
Now
= 1
J2 27
Physicsguide
1D Potent ials: Part-2
11 oo : 01
: 16
@Sk J ahiruddin , 2020
lD Potentials: Part-2
Now !i (a+at) w)
2mw
_n_{w (a+ at) w)
-
2mw
2 1 2
1
2
2mw •
•
1,Jr
'l7r
•
1,7r
Ii
1 2
2mw
e 2 {'llo 'llo) •
-
2
e2
2mw 1
2v'2
mw
1
Ii
2v'2 =0
mw
cos
+-
2
•
1,Jr
1
l
-1,Jr
+e 7r
2
2
+ isin
7r
2
+ cos
7r
2
-
•
'l
•
Sln
7r
2
xO
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28
Physicsguide
lD Potentials: Part-2
1.20. The expression for the position operator of a linear
11 oo :01
:1
a
@Sk J ahiruddin , 2020
lD Potentials: Part-2
1.20. The expression for the position operator of a linear harmonic oscillator is i;
=
2 li (a+ al) mw
i
1
i
The state is w) = Now
v'3IO) + v'3 1) + v'3 2)
(w x w) = (w
fi
2mw
(a + at) w)
!3
2mw
~
;w (W l(a 0) + a1IO)) + (ai jl ) + a1i 1)) + (ai 2) + ali j2) 2
1 3
!3
fi (w (a+at) (0) +il l ) +i 2))
n
2mw fi
2mw
(w lv'l 1) + v'li o) + v'2il 2) + v'2i 1) + v'3i 3)
(0 - i( l l - i(2 i v'l 0) + (v'l
+ i v'2) 1) + iv'2 2) + iv'3 3) l 3 l 3 2 3
ri ivl(O 0) - (i 2mw fi [i - i +
2mw
v'2 +
V1 + i
X
X
i v'2)(1 l ) - i
X
iv'2(2 2)
v'2]
fi mw
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@Sk J ahiruddin , 2020
29
Physicsguide
lD Potentials: Part-2
11 oo : 01 3
: 21
mw 29
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P hysicsguide
lD Potentials: Part-2
@Sk J ahiruddin , 2020
1.21. We can writ e t he position operator in terms of ladder operator
or n ' X" n
2mw
\fn8n',n- l
+ Jn + l 6n' ,n+l
or
n
"
X IO) =
2mw
10)
It will be b etter to use Heisenb erg picture here. Still we need to do lit t le bit calculation. The operator will evolve in t ime as
or
(0 X (t)X (O)IO) = 0 eiHt/nx (O)e-iHt/nx (o) O
2mw
O eiHt/nx (O)e-iHt/n l
Now e- iHt/n is t he t ime transnat ional operator which make t he state evolve in t ime
e- iHt/n 0) = 0(t)) [email protected]
30
Physicsguide
11 oo : 01
: 23
make the state evolve in time e- iHt/n j [email protected]
0) = O(t)) Physicsguide
30
@Sk J ahiruddin, 2020
1D Potentials: Part-2
and e- iHt/ n 1)
= l (t))
You can do explicit integration from here. But better to A
say that t he action of X on 1 (t)) will give and
n - - 1 (t)) . 2mw
. There 1s
another
?
n 2mw
IO(t))
n at t he outside.
-rru,,;
Hence t he factor outside will be h . 2mw Please complete t he explicit integration also. 1.22. You can solve t hese types easily if you have solved t he coherent states of Harmonic oscillator. I'm t rying to give a
solution here. See t hat
where f is a complex number. We define a state f ) as quantum state
If) = e-lfl2/2efat 0) t he action of a a f) =ff)
and at
11 oo : 01
: 26
a J) =
f J)
and at
31
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Physicsguide
@Sk J ahiruddin, 2020
lD Potentials: Part-2
Aft er t his my calculation becoming little complicat ed . If you can do in simpler way, please inform m e.
1.23. You need to perform the integral. 00
1 X
-
7r[2
CX)
1/4
1+
v'2x l
2
x2
exp
2z 2
you need to use
oo
xn exp [-axm ] dx =
1 f 2x - x
(n+l)
xnexp [- axm] dx =
1 f 2x - x
(n+l)
oo
m
m
n +l a
1.24. Use
oo 00
m
m ·n +l a
1. 2 5. The normalized wave function is w(x ) = Co cf>o (x) + C 1 c/> 1 ( x) . As t he wave function is normalised it means t hat
C5 +Cl== 1. Now we need to find the values of C0 and C 1 for which t he expaectation value of displacement for the state be maximized.
w will
For a harmonic oscillator we know that (c/>i x Ic/>j ) = Now
Oij
11 oo : 01
: 29
Now we need to find the values of C0 and C 1 for which the expaectation value of displacement for the state '11 will be maximized. For a harmonic oscillator we know that (i x lj) = bij Now
32
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Physicsguide
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1D Potent ials: Part-2
(w(x)lx l'll (x))
+ C1(1 (x)) x (Co o (x)) + C1 1 (x))) = CJ(o x o) + cf (1 x 1) + CoC1 (ox 1) + CoC1 (1 x o) = (Co (o (x)
= 2CoC1 (o x 1) We want to m aximize (x) so 2C0 C1(o x 1) must be maximum. •
(x)max = [( C5
2
+ cf) - (Co - C1 ) ] (o X 1) 2
= [1 - (Co - C1) ](o x 1) For the expectation value to be maximized we need to have
(Co-C1) 2 =0
~► Co=C1
1.26. We will solve this problem in t he footsteps of t he previous problem number Prob 1.25. We want to minimize
(x) so 2b1b2(0lx ll) must be minimum. 2 (X) min = [(bf + b~) - (b1 - b2) ](0 X I1)
= r1 - (b1 - b2) l (Olx 1) 2
11 oo : 01
: 31
vious problem number Prob 1.25. We want to minimize (x) so 2b1b2(0lx ll ) must be minimum.
(X) min
2 = [(bi + b~) - (b1 - b2) ] (0 XI1) 2 = [1 - (b1 - b2) ](Olx 1)
For the expectation value to be minimised we need to maximise (b 0 - b1 ) 2
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P hysicsguide
33
lD Potentials: Part-2
@Sk J ahiruddin , 2020
1.27. From problem Prob 1.25. we will get t he following results.
Applying t he normalised condit ion we will get t he value of 1 Co= C1 =
v'2
Now t he expression of x in t erms of raising and lowering operator is
x=
n (at + a) 2mw 1
Now 1
n
(x) max = 2coc1 (0 X 1) = 2 v'2 v'2
2mw
(o (at + a) 1)
n 2mw 1.28. T his is coherent st at e for
00
1
f =
l
00
1
11 oo : 01
: 34
2mw 1. 28. This is coherent st at e for
In t his coherent stat e for
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f =
f =
1
1, the expectation values
P hysicsguide
34
@Sk J ahiruddin, 2020
lD Potentials: Part-2
you can prove
(f x
2
f)
n, =2mw
=
(f p
2
f (a + a )
2
f
n (! + !* + (f*f + 1) + f *!) 2mw 2
mnw f) = -2
=-
t
2
t
2
(a - a )
f
mnw (! + ! * 2 2
2
-
f
(f *f
+ 1) -
f*f)
Now put f = 1 and use t he expression of Hamiltonian The det ailed discussion of coherent states are found in problem in Griffiths 3.35 (2nd edition), 3.42 (3rd edit ion) and in Sakurai page no 96 and problem 2.19 (second edit ion). I'll include t he coherent states in our notes as soon as possible.
11 oo : 01
: 36
problem in Griffiths 3.35 (2nd edition), 3.42 (3rd edition) and in Sakurai page no 96 and problem 2.19 (second edit ion). I'll include t he coherent states in our notes as soon as p ossible. 1.29.
1.30. In the Heisenberg picture we know that t he observables of the syst em evolve wit h t ime.
da = dt
Now
a does
[a iI] aa in ' + at 1
not depend on time explicitly.
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35
Physicsguide
@Sk J ahiruddin, 2020
1D Potent ials: Part-2
da
1
in [a, iIJ
dt
2
1
"a -+-mw-x Px l ? 2 ' 2m 2
in 1 in
+
+
0,
+o
2
" Px a , 2m
!mw 2 x 2 '2
1
mw
in
2n
mw 2n
•
A
ip
x+ mw . "'
ip
x+ mw
p; '2m
1 2 2 -mw x '2
11 oo : 01 1
mw
in
2n
mw
in
2n
mw
+
2n mw
+
2n
A
ip
p; ' 2m
x+ mw A
ip
1 2 2 -mw x '2
x+ mw
2n
1
•
•
mw
+
: 39
2
, ., Px x, 2m
+
mw
2n
. ,...
ip
p~ mw' 2m
x' ~2 mw 2 x 2 ip •
A
1
- mw 2 x 2 mw' 2
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36
@Sk J ahiruddin, 2020
lD Potentials: Part-2
2
1
mw
,. . Px
in
2n
x, 2m
1
mw
in
2n
1
mw
in 2n
+
2n
2
,. . Px x, 2m
ip , -l mw 2 x 2
+
mw 2
in X 2p iw ( - - - + - - 2X
2n
mw
mw
2m
p
2
X
.t;_ )
'l 1i
•
- - iwx m
= -iwa •
A
1.31. This type of problem has been solved in the chapter
''Basic Formulation'' . Still here we will give an outline of how to solve the problem. The stat e at time t = 0 is
11
00:01 :41
= -iwa •
A
1.31. This type of problem has been solved in the chapter
''Basic Formulation''. Still here we will give an outline of how to solve the problem. The stat e at time t = 0 is
The evolved state at time t
l'l/J (t)) =
1
v'2
exp
-
= t is
i Eot
2)
0) + exp
Ii
Now we need to find t he time at which t hese two states will
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37
lD Potentials: P art-2
@Sk J ahiruddin, 2020
be orthogonal
~►
1
2
iEot
exp
+ exp
Ii
~~►
exp
{'lf;(O) 'l/; (t)) = 0 iE2t
exp
D
\ -1- \
=0
Ii
iEot Ii
i(E2 - Eo )t Ii
Equating the real part we get If D
-
= -exp
= -1
11 oo : 01 : 4s =-1
exp Equating t he real part we get
=-1
cos
~► t =:::::,~►
==:,~►
= (E
t=
2
Ii _
Ea) arccos( -1)
Ii --X7r 2/iw 7r
t=-
2w
1.32. This problem has been solved in t he chapter ''Basic
Formulation'' . Here if you closely follow the sum you will not ice t hat t he ground state and the first excit ed has flipp ed . We alredy know t he minimum time required for flipping of states which is
T =
h
2(E zower j [email protected]
-
E higher ) Physicsguide
38
@Sk J ahiruddin , 2020
1D Potentials: Part-2
Putting all t he necessary values we will get option d as t he answer. 1.33. Energy Level of two Dimensional harmonic potential
well is
E (nx, ny) = (nx + ny
+ 1)/iw
Now the total energy of t he state is 4/iw. Thus we have n x + ny
=3
11 oo : 01
: 47
E (nx, ny) = (nx + ny
+ l )nw
Now the total energy of t he state is 4nw. Thus we have
nx + ny = 3
1
2
2
1
II Total 3 3
So the st ate is 2 fold degenerate. 1.34. The trick to solve this type of problem is to express
t he 'x' or 'y' part of the equat ion as whole square terms. We will explain this by solving the following examples. But t his problem just requires some re-arranging. 1 1 1 2 2 2 2 2 2 2 V (x , y) = mw (x + 4y ) = mw x + m4w y
2
2
2
Thus the Energy expression will be En =
nx +
1
2
nw +
1 ny + 2nw 2 For the first excited state we have n x = l ; ny = 0
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39
@Sk J ahiruddin , 2020
lD Potential s: Part-2
Thus 1 1 +2 1.35. This is not a perturbation problem. By rotating t he
axis t he problem becomes two simple harmonic oscillator problem. What are the frequencies? The PE matrix is /
'\
11 oo : 01
j [email protected]
: 49
Physicsguide
39
@Sk J ahiruddin, 2020
lD Potentials: Part-2
T hus 1 1 +2
1 o+-2
1uu+
2ru.u
=
5 luu 2
1.35. This is not a perturbation problem. By rotating the
axis the problem becomes two simple harmonic oscillator problem. What are the frequencies? T he P E matrix is mw
2
5
3
2
2
3 2
5 2
The KE matrix m
1 0 0 1
Let the eigenfrequencies are w' 5 2 3 2
-32 5 2
- mw'
2
1 0 0 1
=0
Solving we get eigen frequencies 2w,w
Hence t he energy for ground state is 3 = - luu 2 2 2 1.36. The x state is in first excited state and the y state is in ground state. 1
1 - n2w + - luu
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40
Physicsguide
11 oo : 01 : s2
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Physicsguide
39
@Sk J ahiruddin, 2020
lD Potentials: Part-2
T hus 1 1 +2
1 o+-2
1uu+
2ru.u
=
5 luu 2
1.35. This is not a perturbation problem. By rotating the
axis the problem becomes two simple harmonic oscillator problem. What are the frequencies? T he P E matrix is mw
2
5
3
2
2
3 2
5 2
The KE matrix m
1 0 0 1
Let the eigenfrequencies are w' 5 2 3 2
-32 5 2
- mw'
2
1 0 0 1
=0
Solving we get eigen frequencies 2w,w
Hence t he energy for ground state is 3 = - luu 2 2 2 1.36. The x state is in first excited state and the y state is in ground state. 1
1 - n2w + - luu
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40
Physicsguide
11 oo : 01 : s4
j [email protected]
Physicsguide
39
@Sk J ahiruddin, 2020
lD Potentials: Part-2
T hus 1 1 +2
1 o+-2
1uu+
2ru.u
=
5 luu 2
1.35. This is not a perturbation problem. By rotating the
axis the problem becomes two simple harmonic oscillator problem. What are the frequencies? T he P E matrix is mw
2
5
3
2
2
3 2
5 2
The KE matrix m
1 0 0 1
Let the eigenfrequencies are w' 5 2 3 2
-32 5 2
- mw'
2
1 0 0 1
=0
Solving we get eigen frequencies 2w,w
Hence t he energy for ground state is 3 = - luu 2 2 2 1.36. The x state is in first excited state and the y state is in ground state. 1
1 - n2w + - luu
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40
Physicsguide
11 oo : 01 : sa
j [email protected]
Physicsguide
39
@Sk J ahiruddin, 2020
lD Potentials: Part-2
T hus 1 1 +2
1 o+-2
1uu+
2ru.u
=
5 luu 2
1.35. This is not a perturbation problem. By rotating the
axis the problem becomes two simple harmonic oscillator problem. What are the frequencies? T he P E matrix is mw
2
5
3
2
2
3 2
5 2
The KE matrix m
1 0 0 1
Let the eigenfrequencies are w' 5 2 3 2
-32 5 2
- mw'
2
1 0 0 1
=0
Solving we get eigen frequencies 2w,w
Hence t he energy for ground state is 3 = - luu 2 2 2 1.36. The x state is in first excited state and the y state is in ground state. 1
1 - n2w + - luu
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40
Physicsguide
11 oo : 02 : oo
j [email protected]
Physicsguide
39
@Sk J ahiruddin, 2020
lD Potentials: Part-2
T hus 1 1 +2
1 o+-2
1uu+
2ru.u
=
5 luu 2
1.35. This is not a perturbation problem. By rotating the
axis the problem becomes two simple harmonic oscillator problem. What are the frequencies? T he P E matrix is mw
2
5
3
2
2
3 2
5 2
The KE matrix m
1 0 0 1
Let the eigenfrequencies are w' 5 2 3 2
-32 5 2
- mw'
2
1 0 0 1
=0
Solving we get eigen frequencies 2w,w
Hence t he energy for ground state is 3 = - luu 2 2 2 1.36. The x state is in first excited state and the y state is in ground state. 1
1 - n2w + - luu
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40
Physicsguide
11 oo : 02 : 02
j [email protected]
Physicsguide
39
@Sk J ahiruddin, 2020
lD Potentials: Part-2
T hus 1 1 +2
1 o+-2
1uu+
2ru.u
=
5 luu 2
1.35. This is not a perturbation problem. By rotating the
axis the problem becomes two simple harmonic oscillator problem. What are the frequencies? T he P E matrix is mw
2
5
3
2
2
3 2
5 2
The KE matrix m
1 0 0 1
Let the eigenfrequencies are w' 5 2 3 2
-32 5 2
- mw'
2
1 0 0 1
=0
Solving we get eigen frequencies 2w,w
Hence t he energy for ground state is 3 = - luu 2 2 2 1.36. The x state is in first excited state and the y state is in ground state. 1
1 - n2w + - luu
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40
Physicsguide
11 oo : 02 : os
j [email protected]
Physicsguide
39
@Sk J ahiruddin, 2020
lD Potentials: Part-2
T hus 1 1 +2
1 o+-2
1uu+
2ru.u
=
5 luu 2
1.35. This is not a perturbation problem. By rotating the
axis the problem becomes two simple harmonic oscillator problem. What are the frequencies? T he P E matrix is mw
2
5
3
2
2
3 2
5 2
The KE matrix m
1 0 0 1
Let the eigenfrequencies are w' 5 2 3 2
-32 5 2
- mw'
2
1 0 0 1
=0
Solving we get eigen frequencies 2w,w
Hence t he energy for ground state is 3 = - luu 2 2 2 1.36. The x state is in first excited state and the y state is in ground state. 1
1 - n2w + - luu
j [email protected]
40
Physicsguide
11 oo :02 : oa
j [email protected]
Physicsguide
39
@Sk J ahiruddin, 2020
lD Potentials: Part-2
T hus 1 1 +2
1 o+-2
1uu+
2ru.u
=
5 luu 2
1.35. This is not a perturbation problem. By rotating the
axis the problem becomes two simple harmonic oscillator problem. What are the frequencies? T he P E matrix is mw
2
5
3
2
2
3 2
5 2
The KE matrix m
1 0 0 1
Let the eigenfrequencies are w' 5 2 3 2
-32 5 2
- mw'
2
1 0 0 1
=0
Solving we get eigen frequencies 2w,w
Hence t he energy for ground state is 3 = - luu 2 2 2 1.36. The x state is in first excited state and the y state is in ground state. 1
1 - n2w + - luu
j [email protected]
40
Physicsguide
11 oo : oo : 01
Problems and Solutions in 3D Systems in Quantum Mechanics Sk J ahiruddin * Souradeep Mondal
*Assistant Professor Sister Nibedita Govt. College, Kolkata Author was t he topper of IIT Bombay M.Sc P hysics 2009-2011 batch He ranked 007 in IIT J AM 2009 and 008 (J RF) in CSIR NET J une 2011 He has been teaching CSIR NET aspirants since 2012
1
@Sk J ahiruddin , 2020
3D Systems
11 oo : oo : 04 1
@Sk J ahiruddin, 2020
3D Systems
Contents 1 Problems from NET, GATE, JEST, TIFR & JAM papers 1.1 Free Particle in a 3D Box 1.2 Free Particle in 3D Harmonic Well Potential 1.3 Hydrogen Atom Problems 1.4 Ans Keys 1.5 Solutions . • •
j [email protected]
•
•
•
•
•
•
•
•
•
•
•
•
•
•
3 6 7
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
2
3
17 18
Physicsguide
11 oo : oo : 06
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2
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@Sk J ahiruddin , 2020
1
1.1
3D Systems
Problems from NET, GATE, JEST, TIFR & JAM papers Free Particle in a 3D Box
Prob 1.1. A particle of mass m is confined in a potentialbox of sides Lx, Ly, Lz, as shown in the figure. By solving schrodinger equation of the part icle, find its eigenfunctions and energy eigenvalues. [JAM 2012]
z
•• ••• •
j(0,0,0)
••···-····--··········· ..........
(O,l ,0)
y
••••••••••••
••
.x
Pro b 1. 2. The ground st ate energy of a particle of mass m in a t hree-dimensional cubical box of side l is not zero 2 3h but Bml . This is because [TIFR 2015]
11 oo : oo : 09 Pro b 1. 2. The ground state energy of a particle of mass m in a three-dimensional cubical box of side l is not zero 2 3h but Bml . This is because [TIFR 2015] [email protected]
Physicsguide
3
@Sk J ahiruddin , 2020
3D Systems
(a) the potential at the boundaries is not really infinite, but just very large. (b) this is the most convenient choice of the zero level of potential energy. (c) potential and momentum cannot be exactly determined simultaneously. (d) t he ground state has no nodes in t he interior of t he box. Prob 1.3. A particle of mass m is in a cubic box of size a. The potent ial inside the box (0 < x < a, 0 < y < a, 0
H ~rni ltoni ~n
h11t not o f T.2 or T. _
11 oo : oo : 19 ro
e
.
nlm
eno e VAAe e1gens a es o a y rogen
atom in the usual notation. The state IV = ~[2 W200-3W211 + is an eigenstate of [NET Dec 2015] but not of the Hamiltonian or Lz
v'7'11 210 -
(a) L 2 ,
y'5'11 21 _ 1]
(b) the Hamiltonian , but not of L 2 or Lz 2
(c) the Hamiltonian L and Lz (d) L 2 and Lz, but not of the Hamiltonian Prob 1.13. The ground state wavefunction for the hydro-
gen atom is given by
'11 100
1
1
J4rr
ao
= --(-) 312 e- r/ao , where ao is
t he Bohr radius. The plot of t he radial probability density, [email protected]
Physicsguide
7
@Sk J ahiruddin , 2020
3D Systems
P (x) for the hydrogen atom in the ground state is [GATE 2012]
(a)
(b)
P(r)
P(r)
(d)
(c) P(r)
P(r)
11 oo : oo : 21 P(rj
P(rj
Prob 1.14. A part icle in the 2s state of hydrogen has t he wave function 1
1P2s(r) = y'21r 4
1 a0
3 2
_ r 2 a0
exp
-
r
2a0
where r is t he radial coordinate w.r.t. t he nucleus as origin and a0 is t he Bohr radius. The probability P of finding the electron somewhere inside a sphere of radius Aao centered at the nucleus, is best described by the graph [TIFR 2014]
j [email protected]
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8
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3D Syst ems
p
p
(a) (b )
).
).
p
p (c)
(d)
/ J..
11 oo : oo : 24 (c)
(d)
Pro b 1.15. An electron is in the 2s level of the hydrogen atom, with the radial wave-function
qJ(r) =
1
v'2 3/ 2 2 2a
2- r
ao
0
exp
r 2ao
•
The probability P (r) of finding this electron between distances r to r + dr from t he centre is best represented by the [TIFR 2018] sketch
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9
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(a)
3D Systems
"(b)
P(r )
~
r)
. _ _---1 r (c)
(r )
_
(d)
P(r )
~
r
11 oo : oo : 26
---==-._
Prob 1.16. Let drogen atom. given below
\lf 0)
___.J r
denoted the ground state of the hy-
Choose the correct statement from those
(a) [Lx, Ly] Wo) = 0 (d) [Sx, Sy] Wo) = 0
(b) J
2
IwO) =
0
[GATE 2008] (C) L.§ -/- 0
Prob 1 .17. The radial wave function of t he electrons in t he state of n 2 r exp 312
a0
= 1 and l = 0 in a hydrogen atom is R 10 =
ao
, ao is the Bohr radius. The most probable
[GATE 2008]
value of r for and electron is
(a) ao
(b) 2ao
(d) 8ao
(c) 4ao
Prob 1.18. In t he ground state of hydrogen atom , t he j [email protected]
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10
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3D Systems
most probable distance of the electron from t he nucleus, in unit s of Bohr radius a0 is: [JEST 2016] 1 (d) 3 (b) 1 (c) 2 (a) 2 2 Prob 1.19. An electron in the ground state of t he hyr 1 drogen atom has t he wave function \lJ ( r) = -----;::===e ao na~
where a0 is constant. The expectation value of t he operator
Q=
z
2
/
r
2
,
where z = rcos0 is I
[GATE 2014]
11 oo : oo : 29 r 1 drogen atom has t he wave function w(r) = -----;:===e
ao
na~
where a0 is constant . The expectation value of t he operator
Q=
z2
Hint:
r 2 , where z = rcos0 is
-
--
---
a n+l
0
[GATE 2014] -
an+l
(b)
-a5
Prob 1.20. The normalized ground state wavefunction of 41ra
is the Bohr radius and r is the distance of t he electron from t he nucleus, located at t he origin. The expectation value 1 r 2 is [GATE 2011] 4 (d) 2 (a) ? ( c) 2 a2 a2 a= a Prob 1.21. The electronic ground state energy of the Hy1T"
(b) 41r
drogen atom is -13.6 eV. The highest possible electronic en-
[GATE 2017] ergy eigenstate has an energy equal to (a) 0 (b) 1 eV (c) +13.6 eV (d) oo [email protected]
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11
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3D Systems
Prob 1.22. The wavefunction of a hydrogen atom is given by the following superposit ion of energy eigen functions Wnzm(r)(n , l, mare the usual quantum numbers) :
The ratio of expectation value of t he energy to the ground
11 oo : oo : 32 Wnlm r n , l, m are the usual quantum numbers :
The ratio of expectation value of t he energy to the ground state energy and t he expectation value of £ 2 are, respect ively: [JEST 2016] 2 2 101 229 d 12n 101 nd 121i (b) 2 c and n (a ) 504 an ( ) 7 504 a 7 504 (d)
~~: and Ji
2
Prob 1.23. The normalized wave functions of a Hydrogen atom are denoted by wn,l,rri (x)' where n, l and m are, respectively, the principal, azimuthal and magnetic quant um numbers respectively. Now consider an electron in the mixed st ate
W(X) = ~ W1,0,0 ( X) + ~ '¥2,1,0( X) + ~ W3,2,-2 ( X) The expectation value (E) of the energy of this electron, in [TIFR 2012] electron-Volts (eV) will be approximately (a) -1.5 (b) -3.7 (c) -13.6 (d) -80.1
Prob 1.24. Let the wavefunction of the electron in a [email protected]
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12
Physicsguide
3D Systems
drogen atom be
where nzm(r) are the eigenstates of the Hamiltonian in t he
11 oo : oo : 34 rogen atom
e
where ef>nlm(r) are the eigenstates of the Hamiltonian in t he standard notation. The expectation value. of the energy in t his stat e is [NET Dec 2018] (a) - 10.8 eV
(b) - 6.2 eV
(c) - 9.5 eV
(d) -5.1 eV
Prob 1.25. An energy eigenstate of the Hydrogen atom has t he wave function
312
r Wnzm(r, 0,ef>) = ~ sin0 cos 0 exp[-(+ ief>)] 81 1r ao 3ao where a0 is t he Bohr radius. The principal (n), azimuthal (l ) and magnetic (m) quantum numbers corresponding to t his wave function are [TIFR 2013] (a) n = 3, l = 2, m = 1 (b) n = 2, l = l , m = 1 (c) n = 3, l = 2, m = -1 (d) n = 2, l = l , m = ±1 1
1
Prob 1.26. A rigid rotator is in a quantum state described 3 by t he wave function 'l! (0, q>) = sin 0 sin q> where 0 and 41r q> are t he usual polar angles. If two successive measurements of L 2 are made on this , the probability that the second measurement will yield t he value + Ii is [TIFR 2014] j [email protected]
13
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(a) 0.25
(b) 0.33
3D Systems
(c) 0.5
(d ) negligible
Prob 1.27. An electron is in the ground state of a hydro£len atom. The nroba,bilitv that it is within the Bohr radius
00: 00: 37
(a) 0.25
(b) 0.33
(c) 0.5
(d) negligible
Prob 1.27. An electron is in the ground state of a hydro-
gen atom. The probability that it is within the Bohr radius [NET June 2014] is approximately equal to
(a) 0.60
(b) 0.90
(c) 0.16
(d) 0.32
Prob 1.28. If an electron is in t he ground state of the hydrogen atom , t he probability that its distance from the
proton is more t han one Bohr radius is approximately [NET June 2011]
(a) 0.68
(b) 0.48
(c) 0.28
(d) 0.91
Prob 1.29. If t he position of t he electron in the ground state of a Hydrogen atom is measu1~ed, the probability that it will be found at a distance r > a0 ( a0 being Bohr radius) is nearest to [NET Dec 2018]
(a) 0.91
(b) 0.66
(c) 0.32
(d) 0.13
Pro b 1. 30. The normalized wavefunction of a particle in 1 t hree dimensions is given by 1jJ (r, 0, cp) = e-r / 2a, where 81ra3
a> 0 is a constant . The ratio of the most probable distance from the origin to the mean distance from t he origin, is, [NET Dec 2017]
(a) 1/3
(b) 1/ 2
(c) 3/ 2
(d) 2/3
Prob 1.31. A simple model of a helium like atom with [email protected]
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14
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3D Systems
electron-electron interaction is replaced by Hooke's law force
11 oo : oo : 39
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3D Systems
electron-electron interaction is replaced by Hooke's law force is described by Hamiltonian
-n 1 - (\72 + \72) + -mw2(r2 + r2) 2
m
2
1
2
2
1
2
A -mw2 4
What is the exact ground state energy?
(a) E (b) E (c) E (d) E
71- 722 [JEST 2013]
= ~nw( l + ✓1 + A) = ~nw( l + ~ ) = ~ nw ( ✓l - A) = ~nw(l + ✓,--1-A)
Prob 1.32. The energy levels of the non-relativistic elect ron in a hydrogen atom (i.e. in a Coulomb potent ial V (r) ex 2 -1 /r ) are given by Enzm ex -1/n , whe1~e n is the principal quantum number, and the corresponding wave functions are given by Wnzm , where l is t he orbital angular moment um quant um number and m is the magnetic quantum number. The spin of t he electron is not considered . Which of the [NET June 2011] following is a correct stat ement? (a) There are exactly (2l + 1) different wave functions Wnzm, for each Enlm· (b) There are l ( l + l ) different wave functions wnlm, for each Enlm · (c) Enzm does not depend on l and m for t he Coulomb potent ial. [email protected]
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15
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3D Systems
11 oo : oo : 43 15
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3D Systems
(d) There is a unique wave funct ion
for each
W nzm
E nlm ·
Prob 1.33. Suppose t hat the Coulomb potential of t he
hydrogen atom is changed by adding an inverse-square term 2
Ze
g
such t hat t he total potential is V (7) = - - + 2 , where g r r is a constant. The energy eigenvalues Er,,lm in the modified potential [NET June 2016] (a) depend on n,l but not on m (b) dep end on n but not on l and m (c) dep end on n and m, but not on l (d) depend explicitly on all t hree quantum numbers n,l and m
Prob 1.34. Given that
tainty
t)..p1•
c5 - in c5r
p
1•
+ - , the r
uncer-
in the ground state
[NET June 2014]
of the hydrogen atom is
(a) rt ao
1
(b) ,/'in a0
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2/i
()-
c 2a0
16
(d) ao
Physicsguide
11 oo : oo : 4s
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1.4
3D Systems
Ans Keys
1.1. Standard
1.12. b
1.24. d
Theory
1.13. d
1.25.
C
1.2.
1.14. d
1.26.
C
C
1.3. d 1.4.
1.15. b
C
1.16. a
1.5. a
1.27. d 1.28. a
1.17. a 1.29. b
1.6. a
1.18. b
1.7. b
1.19. d
1.8. a
1.20. d
1.30. d 1.31. b
1.9. a
1.21. d
1.32.
1.10. 2.5
1.22. a
1.33. b
1.11.
1.23. b
1.34. a
C
C
11 oo : oo : 48
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1.5
3D Systems
Solutions
Sol 1.1. This is theory. Can be found in any Quantum
book like Griffiths and Zettili Sol 1.2. This is also the basic concepts . No quantum system has ZERO ground stat e energ)'· Kinetic energy, and
t hus velocity of a quantum particle can't be zero as the zero velocity will precisely determine the momentum which will violate uncertainty principle. Thus every quantum particle will have some velocity at every quantum st ate. Sol 1.3. For the cubic box potential, energy expression is 2 2
E nxnynz
n 1r 2 2) ( 2 = 2m £ 2 nx + ny + n z
Wave function 8 • Sln a3
n x 7r --X
a
. Sln
n y1r
a y
•
Sln
a
To make 14 in numerator of the energy, the n' s should be 3, 2, 1 not necessarily in the same order. Only option (d) matches t his requirement . Sol 1.4. Following the same argument from previous problem , then' swill be 2, 2, 1, not necessarily in the same order.
11 oo : oo : so 3, 2, 1 not necessarily in the same order. Only option (d) matches this requirement . Sol 1.4. Following the same argument from previous problem, then' swill be 2, 2, 1, not necessarily in the same order.
There could be three combination of the eigenvalues 2, 2, 1, 2, 1, 2 and 1, 2, 2. Thus t here would be three eigenvalues. [email protected]
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3D Systems
Sol 1.5. Ground state eigenvalue is for 1, 1, 1 31r2 n2
E111 = - -2 = Eo 2m L
First excited state
61r2n2
E 211 = - -2 = 2Eo 2mL Sol 1.6. Solving this from scratch is difficult in t he exami-
nation hall. I have tried to solve by using uncertainty principle but could to get the exact answer. If you can solve then please inform me. Sol 1. 7. When the potential is
1
w
2 2
x
,
the energy is
1
nw.
2 2 2 2 So when potent ial 2w y , you can think t hat w is trans1 formed into 2w. So t he energy is !i2w . And when potential 2 2
8w z
2
you can think t hat w is t ransformed into 4w. So the 1 energy is n4w. So the total energy will be ,
2
Sol 1.8. The EnerQ'v exoression for 3-dimensional isotrooic
11 oo :oo : s2 energy is
2
fi4w. So the total energy will be 1
n(w
2
1
+ 2w + 4w) = 2n7w
Sol 1. 8. The Energy expression for 3-dimensional isotropic quantum harmonic oscillator is fiw= j [email protected]
3 n +2
19
Physicsguide
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3D Systems
The expression to find t he degeneracy for 3-d isotropic quantum harmonic oscillator is
+ l)(n + 2) Degeneracy = - - - - - (n
2
For the third energy level we have n = 2. (2 + 1)(2 + 2) Thus Degeneracy = - - - - - = 6 2
1
1
0
2
1
0
1
0 2
1
1
2 2
0 2
0 0 2
0 0
0
2 2 2
Sol 1.9. The degeneracy of n th level of 3D isotropic har-
monic oscillator is
11 oo : oo : 55 2
0
0
2
Sol 1.9. The degeneracy of n th level of 3D isotropic har-
monic oscillator is
1 9n = (n 2
+ l ) (n + 2)
better to remeber t his result. Proved in Zettili. Another result which can b e useful for furt her problems is t he degeneracy of Hydrogen atom which is n 2
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Sol 1.10. There are many ways to solve this problem. One is direct integration. Another is breaking the r 2 into
r2
= x2 + y2 + z 2
and then use op erator relations a and at . Finally t he easiest way is to use virial theorem If there is a cent ral force rn so V(r) ex rn+l t hen
n+l T = - -V 2 Here n
=
l. So -
-
- E V =-
-
T = V· ' Energy of 3D isotropic oscillator is E=
3 n+2
Energy in first excited state is
2
f (r)
ex
11 oo : oo : sa Energy of 3D isotropic oscillator is
Energy in first excited state is
E= V
=
~nw 2
1 ') 2 -mw--(r ) 2
=
5 -nw 2
(r2) = 5 Ii 2mw Ii Hence in the unit of - , the ans is ~ mw j [email protected]
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3D Systems
Sol 1.11. We need to do a scaling in the integration to t he expectation value of kinetic energy 00
(T) =
'lj;* (r,t)
2m When we change t he wave function, t he integration becomes 0
n,2
00
(T ) =
'lj;* (ar, t )
2m
0
Now put ar
= r'
The Laplacian changes to
Hence t he Kinetic energy integration
11 oo : 01 : oo The Laplacian changes to
Hence t he Kinetic energy integration 2
00
(T) = a a3 1 O'.
n2
'lj)* (r' ,t )
2
0
2m
00
fi2
'lj)* (r' ,t)
"\1 'ljJ (r', t ) r' dr' 2
2
V 'lj; (r' , t) r' dr'
2m
0
2
(T )
Sol 1.12. In the chapter of Angular Momentum we have solved t hese type of problems. But for the sake of complet eness we will solve it here also. We will use the process 22
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3D Systems
of elimination to solve the problem. We know that
LW = ...
Jl (l + l)nw
Lz l, m) =
mn
l, m)
We will use these relations to check which operator.
w is
an eigenstate of
l
11 oo : 01
: 03
-
5
1
2
= n, [(0 5
X
1)2'lf 200 - (1
X
2)3'lf 211
+(1 X 2)V7'lf210 - (1 X 2)V5\Jl21-1
= ~ n [-3'11211 + v'7'11210 # alw) 2
Lz'Y =
=
1 Lz[2'11200 - 3'11211 + \/7'11 210 - V5'1121- 1] 5 1 - n, 0 X 2\Jf 200 - 1 X 3'11211 + 0 X V7\Jl210 5
-(- 1)
X
V5\Jl21-1
2 2 = n, [-3\JI 211 5 # /3 \JI)
~
+ V 5 \JI 21- 1]
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@Sk J ahiruddin,
V5'1121-1]
23
2020
Thus we see that
Physicsguide
3D Systems
w is neither the eigenstat e of L 2 nor of Lz.
This eliminates all t he options except option b.
Sol 1.13. Follows from t he wave function. Given in every book. Sol 1.14. This is not easy. You are asked to find the probability within a radius. Not t he probability density. So the probability P vvhich was plotted is
1 r.:::
1
3/ 2 (
r 2 - - exp ( - Ar
2
dr
11 oo : 01
: 06
a i ity wit in a ra ius. Not t e pro probability P which was plotted is
1
2
3/ 2
2- r
ao
0
ns1 y.
ao
exp
r -2a0
dr
This will increase as A increases. Between the option (b) and (d) t he option (d) is easily chosen seeing t he Gaussian nature of t he function. Sol 1.15. This problem is different than t he previous one.
Here probability yo find within a range is asked. Here P is
r+dr
1
r 2- -
ao
r
exp
r 2a0
2
dr
Now you need to look carefully at the graphs. There is no t rick, I guess! j [email protected]
24
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3D Systems
Sol 1.16.
. The value Lz is zero on t he ground state where L = 0. Spin is not zero as S = 1/ 2. All the other t hree operators involves spin. Sol 1.17. The probability density 1
11 oo :01 : oa involves spin .
Sol 1.17. T he prob ability density
1P =
e- r/a
1
~ 2 2 P = 'ljJ 41rr dr =
4
- e- 2rfar 2 dr
p(r)dr
=
a3
4
p(r) = -r2e-2r/a a3
Take t he derivative and m ake it to ZERO
dp = 4 2re-2r/a dr a3 =
+ r2
a
8r e-2r I a 1 - r a3
Hence t he ans is r
- 2 e-2r/a =
0
a
=a
Sol 1.18. Sam e problem as previous.
25
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3D Systems
Sol 1.19. CX)
Q
-0
21r
7f
2 'lj}* Q'lj}r dr
A
sin 0 d0 0
0
CX)
21r
7f
'lj}* (z 2 0
dcp
-
r 2 )'lj; r 2 dr
sin 0 d0 0
CX)
2 2 2 2 'lj}*(r cos 0 - r )'lj} r dr A
d, < O"y > and < O"z > [NET June 2018] (a) only < O" x > changes with time. (b) only < a y > changes with time. (c) only < a z > changes with t ime. (d) all three changes with time.
B
1.5
Expectation Values
Prob 1.35. A hydrogen atom is in t he state 8 -
21
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'l'200 -
12
Physicsguide
Angular Momentum
where n , l , m in 'Ynzm denote the principal, orbit and magnetic quantum numbers, respectively. If L is the angular momentum operator, the average value of £ 2 is __ /i2 [GATE 2014] -+
11 oo : oo : 37 w ere n, , m 1n nlm enote t e pr1nc1pa , or it an mag-+ netic quantum numbers, respectively. If L is the angular momentum operator , the average value of £ 2 is __ /i2 [GATE 2014]
Prob 1.36. If L+ and L_ are the angular momentum ladder operators then the expectation value of (L+L- + L_L+) in t he state ll = 1, m = 1) of an atom is __li [GATE 2014] Prob 1.37. Let ll, m) be the simultaneous eigestates of £ 2 and Lz. Here L is the angular momentum operator with Cartesian components (Lx, Ly, Lz), l is the angular momentum quantum number and m is t he azimuthal quant um number .The value of (1, OI(Lx + iLy) 1, - 1) is [GATE 2016] (a) 0 (b) n (c) v2li (d) ~n -+
Prob 1.38. A spin-I particle is in a stat e w) described 2
J2
in the Bz basis. What is
2i t he probability t hat a measurement of operator S z will yield t he result n for the state Bx w)? [JEST 2016] 1
(a) -
2
1
(b) 3
j [email protected]
@Sk J ahiruddin , 2020
1
(c) -
4
1
(d) 6
13
Physicsguide
Angular Niomentum
where t) and t) are the eigenstates of Sz operator. The P.xn P.) =
1
v3 I1' -
1) + I1' 0) ei"!
3
+ I1'
2 3 1) e i" /
where, S, M 8 ) denote t he spin eigenst ates wit h eigenvalues n2 S(S + l ) and nMs respectively. Find (Sx), i.e. t he expectation value of x component of t he spin. [TIFR 2018] Prob 1.41. If s1 and s2 are the spin operators of the two elect rons of a He atom , t he value of (s1 .s2) for the ground state is [GATE 2016]
(a) - 3 !i,2 2
(b) - 3 !i,2 4
(
c) 0
(d )
! !i,2 4
Prob 1.42. The wa,re function of a st ate of t he hydrogen atom is given by \ll = W200 + 2 W 211 + 3 W210 + v'2W21- 1 where Wnlm is t he normalized eigen funct ion of the st ate wit h quant um number n , l and m in t he usual notation. T he expectation value of L z in t he state W is [NET Dec 2012] (a)15n/ 16 (b)lln/ 16 (c)3n/8 (d)n/8 [email protected]
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Prob 1.43. Let
Physicsguide
Angular Momentum
W nzm
denote t he eigenfunct ions of a Hamil-
11 oo : oo : 42
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Angular JVIomentum
Prob 1.43. Let W nzm denote the eigenfunctions of a Hamiltonian for a spherically symmetric potential V (r). The expectation value of Lz in the state [NET Dec 2013] 1 w = 6 [w200
+ -Jsw210 + v'law21- 1 + v'20w211]
•
IS
(a) - 5!i 18
(b)
5n 6
(c)
(d) 5!i 18
n
Prob 1.44. Electron in a given system of hydrogen atoms are described by t he wave function •
'l7r
\J!(r, 0, cp) = 0.8\J!100 + 0.6e 3 W311 where t he \JI nlm denoted normalized energy eigenstates. angular moIf (Lx, Ly, Lz) are the components of t he orbital ,... mentum operator, t he expectation value of L~in t his system is [TIFR 2017] (a) l.5!i2 (b)0.36!i2 (c) 0.18!i2 (d)zero Prob 1.45. If \JI nlm denotes t he eigenfunction of t he Hamiltonian with a potential V = V(r) t hen the expectation value of the operator L ; + L~ in t he state 1 \JI = [3\Jf 211 + W210 - v115w21-1] 5
[NET June 2013] 2 2 (c) 2!i ( d) 26!i / 25
is 2 (a) 39!i / 25 [email protected]
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2
(b) 13!i / 25 15
Physicsguide
Angular Momentum
11 oo : oo : 4s 15
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A ngular Moment um
Prob 1.46. An electron in a hydrogen atom is in a state described by t he wavefunct ion:
where 'lfJnPm (x) denotes a normalized wavefunction of the hydrogen atom with the principal quantum number n, angular quantum number £ and magnetic quantum number m. Neglect ing the spin-orbit interaction, t he expectation values of Lz and £2 for this state are [TIFR 2019]
(a) 10 ' 5 2 3n 9n (c) 5' 10
-
25
8n 3n (d) 10 '
2
5
Prob 1.47. (Data fo1· next two questions)
i) In a system consisting of two spin 1/ 2 particles labeled 1 and 2 let §( 1) = ~5(1 ) and § (2) = ~5(2) denote t he corre' 2 2 sponding spin operators. Here 5 (ax, ay, az) and ax, ay, az are t he three Pauli matrices. In the standard basis the ma1 2 1 2 ) ) trices for the operators ) and ) are respectively, [NET June 2011] 2 t;,2 1 0 n -1 0 (a) 4 0 1 0 - 1 ' 4
si si
(b) n,2 4
i
O
0 -i
j [email protected]
2
ri ' 4
si si
-i 0
0
•
1,
16
Physicsguide
11 oo:oo:47 2
(b) li
2
O
i
0 -i
4
-i 0 0 i
li ' 4
j [email protected]
16
P hysicsguide
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Angular Niomentum
•
0 0 0 -1, 0 0 1, 0 . 0 -1, 0 0 . 1, 0 0 0
fi2
(c) 4
0 1 0 0 1 0 0 0 0 0 0 - 1, 0 0 'l 0
(d) n,2
•
4
0 0 0 0 0 - 1, . 0 1, 0 . 1, 0 0
n,2
•
0
' 4
•
-1,
•
' 4
n,2
•
0 0 0
•
0 0 'l 0 0 0 0 0 0 1 0 0 1 0 -1,
•
ii) These two op erat ors satisfy t he relation
(a) {s1) st 1
2
),
st ) si 1
2
)} =
si ) si ) 1
2
(b) { s1) st
2
1
),
s~)s1) } = 1
2
0
(c)
[si )si ), st )sl 1
2
1
2
)J =
Prob 1.48. If Y xy
iSi )si ) 1
2
(d)
[sl )st ),st )si 1
2
1
2
)J =
o
1
= v'2 ( Y 2,2- Y 2,-2) where Yt ,m are spher-
ical harmonics then which of t he following is t ru e? [JEST
2016] (a) Y x y is an eigenfunction of both L 2 and Lz (b ) Y x y is an eigen function of L 2 b ut not Lz (c) Yxy is an eigenfunction of both L z but not L 2 (d) Y x y is not an eigenfunction of eit her L 2 and Lz
P rob 1.49 . Let W nlm denote t he eigenfunct ion of a Hamilt onian for a sph erically symmet ric p otent ial V (r). The wave1 function W = - ['11 210 + v'5'11 21 _ 1 + ,Jiow 211 ] is an eigenfunc4
t ion on ly of
[N ET June 2012]
11 oo : oo : so Prob 1.49. Let Wnzrri denote t he eigenfunction of a Hamiltonian for a spherically symmetric p otential V (r). The wave1 function W = -['11210 + \ibW21- 1 + vtiOw211] is an eigenfunc4 t ion only of [NET June 2012] [email protected]
Physicsguide
17
@Sk J ahiruddin, 2020
Angular Momentum
2
(a) H , L and L z (c) H and L 2
(b) H and L z (d) L 2 and L z
Pro b 1. 50. Consider the normalized wavefunction (3k - w
2
m)
2
-
2
T =0 k
2
=0
Solving this equation we will have Wx
=
4k
2k
m
m
- and Wy
Now the expression for energy will be
1 n Y +2 The ground state energy will be
n
Eo = 2
n 2k -m + -2 -m + VO 4k
TT
11 oo : 02 : 16 The ground state energy will be
n
Eo = 2
n 2k -m + -2 -+Vo m 4k
The first excited state energy will be 4k
3n
m+2
2k
l
-+ vo m ,T
Note :Be careful to choose the first excited state. Thus the difference between the two energies will be
2k m j ahir@physicsguide. in
49
@Sk J ahiruddin , 2020
Physicsguide
Approximation Methods: part-1
Sol 1.36. This is a problem of Stark Effect. This problem has been explicitly solved and explained in Quantum Mechanics Concepts and Applications(2 Ed.) by Nouredine Zettili (page 494, Example 9.2) Sol 1.37. Sol 1.38. The wave function is given in spherical coordinate system. So we have to convert the perturbing potential into spherical co-ordinates. Put x = r sin 0 cos cp 2 2 2 We have H' = br sin 0 cos cp The first order correction to the ground state energy is 1
E6) = ('l/Jo IH' 'l/Jo) =
r1/,!
T-l 11/,,.,r 2 Q1n {}rl{}rlrf,
11 oo : 02 : 19 We have H'
2
2
2
= br sin 0 cos cp
The first order correction to the ground state energy is
Ea = (wo lH' wo) 1
)
2
'i/;0H''i/Jor sin 0d0dcp 00
b
4
r exp
7f
- 2r
cos 2 cpd
sin3 0d0
dr
1ra~ 0 ao b 3a 5 4 -X OX - X 7r 1ra5 4 3
27f
0
0
= ba5 We have not solved t he integration explicitly hoping that you will be able to solve t his simple integration on your own. [email protected]
Physicsguide
50
@Sk J ahiruddin, 2020
Approximation Methods: part-1
Sol 1.39. Int egration 00
o
1 / 2r 1 ; - - e- r ao E cos - - - - e - r a 0 dr
v,ia,3
ao v,ia,3
Better to write the cos as exp and tal a
1/ 3
3
(c) -
4
n
2g-?
1/ 3
m
1/3
I I,
m
Prob 1.5. T he ground state energy of a particle of m ass m
ized trial wavefunction 'ljJ (x)
=
a
1/ 4
2
e-ax /2 IS
'
7r
June 2016] Use and
(a) 3
2m
n,2131 / 3
a
00
1r
- oo
a
00
7r
-oo
2 2 x e - ax
dx
1
=-
2a
2 4 x e-ax
(b) 8 n2131/3 3m
•
dx
=-
3
4a2
[NET
11 oo : oo : 13 and 3 (a) n,2(31/3 2m
2 (c) n,2(31/3 3m
Q
00
7r
-oo
4
2
x e-ax dx
=-
3
4a2
8 (b) n2 (31/3 3m 3 (d) n2 (31/3 8m
Prob 1.6. Using t he t rial function j [email protected]
Physicsguide
5
Approximation Methods: part-2
@Sk J ahiruddin, 2020
-a< x < a
\J!(x) =
otherwise the ground state energy of a one-dimensional harmonic oscillator is ~ [NET June 2017]
(a)
nw
(b)
5
nw 14
(c)
!tuu 2
5
nw 7
(d)
Prob 1. 7. Given a part icle confined in one-dimensional box between x = -a and x = +a, a student attempts to find
t he ground state by assuming a wave-function
(x) =
A(a 0
2
-
2 3 2 x ) 1
for Ix < a for Ix > a
The ground state energy Erri is estimated by calculating t he exp ectation value of energy wit h this trial wave-function. If E 0 is the true ground state energy, what is the ratio
Em/ Ea?
[TIFR 2018]
11 oo : oo : 16 The ground state energy Em is estimated by calculating t he expectation value of energy with this trial wave-function. If E 0 is the true ground state energy, what is t he ratio
Em/ Eo?
1.2
[TIFR 2018]
WKB Approximation
Prob 1. 8. Consider a particle confined by a potential V (x) = klx , where k is a positive constant . The spectrum E n of the [email protected]
Physicsguide
6
@Sk J ahiruddin, 2020
Approximat ion Methods: part-2
system , within the WKB approximation is proportional to [JEST 2017] (a) (n + ;)3/2 (b) (n + ;)2/3
(c) (n + ½)1/2
(d) (n
+ ½) 4/3
Prob 1.9. Consider a part icle of mass m in t he potent ial V (x ) = alx, a> 0. The energy eigenvalues E n(n = 0, 1, 2, .. ) in t he WKB approximation, are, [NET Dec 2014] [NET June 2016] ~ 1/3 ~ 2/3 3a,"1r ( 3a,"1r ( 1) 1) (b) 4-J2rn n + 2 (a) 4-J2m n + 2 4/ 3
3an1r ( 1.) (C) ~ n+ 2 4v 2m
Prob 1.10. A one-dimensional system is described by t he 2
;m x
Hamiltonian H = + >. (where >. > O) The ground stat e energy varies as a funct ion of ). as [NET Dec 2018]
11 oo : oo : 19
Prob 1.10. A one-dimensional system is described by t he 2
Hamiltonian H =
;m + >. x (where >. >
0) The ground
stat e energy varies as a function of A as [NET Dec 2018]
(a)
,\5/ 3
(c)
,\4/ 3
(b)
,\2/3
Prob 1.11. A particle in one dimension moves under t he influence of a potential V ( x) = ax 6 , where a is a real constant. For large n the quantized energy level En depends on
n as:
[NET June 2011] [NET Dec 201 7]
[email protected]
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7
@Sk J ahiruddin , 2020
Approximation Methods: part-2
(b) En rv n 413
(a) En, rv n 3
(c) En rv n 615
(d) En rv n 312
Prob 1.12. The n-th energy eigenvalue En of a one-dimensional 2
Hamiltonian H = p + ;\x 4 ( where ,\ > 0 is a constant) in 2m t he WKB approximation, is proport ional to [NET June
2018]
(a)
(c)
1.3
1
4/3 ,\ 1/ 3
n+-
2
1
n+ -
2
(b)
1 n+2
4/ 3
1 2
4/3
5/ 3 ,\ 1/3
(d)
n+
_\2/ 3
,\ 1/ 3
Time D ependent Perturbation Theory
Prob 1.13. A particle of charge q in one dimension is in a
11 oo : oo : 21 C
1.3
n+ -
n+ -
2
2
Time Dependent Perturbation Theory
Prob 1.13. A particle of charge q in one dimension is in a
simple harmonic potential with angular frequency w . It is 2 7 subjected to a time-dependent electric field E (t) = A e- (t/ ) where A and T are positive constants and WT > > l. If in the distant past t -+ -oo the particle was in its ground state, t he probability t hat it will be in the first excited state as t -+ +oo is proportional to [NET Dec 2016]
2 (a) exp - ~(wT)
2 (b) exp ~(wT)
j [email protected]
(c) 0
(d)
Physicsguide
8
@Sk J ahiruddin , 2020
Approximation Methods: part-2
Prob 1.14. Consider t he Hamiltonian
1 0 0
H (t) = a
O 2 0
0 0 3
+ {3t
0 0 1 O O 0 1 0 -2
The time dependent function f3 (t) = a for t < 0 and zero fort > 0. Find ( (w(t < 0) w(t > 0)) )2 , where w(t < 0)) is the normalised ground state of t he syst em at a t ime t < 0 and l'll (t < 0)) is t he state of the syst em at t > 0. [JEST 2017] 1 (~1 -( l +rn~(?.rvf11
11 oo : oo : 24 fort> 0. Find ( (w(t < 0) w(t > 0))1) 2, where w(t < 0)) is t he normalised ground state of t he syst em at a t ime t < 0 and l'11 (t < 0)) is t he stat e of the syst em at t > 0. 2017] 1 1 (a) (1 +cos(2at)) (b) (1+cos( at)) 2 2 1 . (d) (1 + sin (at))
[JEST
2
Prob 1.15. In t he usual notation n l m) for the states of a hydrogen like atom, consider the spontaneous t ransitions 2 1 O) ➔ 1 0 0) and 3 1 0) ➔ 1 0 0). If t 1 and t2 are t he lifetimes of the first and the second decaying st at es
!~
[NET
respectively, then t he ratio is proport ional to June 2017] 3 3 2 3 3 3 32 27 (d) (a) 27 (b) 32 (c) 2 3 [email protected]
@Sk J ahiruddin,
1.4
Physicsguide
9
2020
Approximation Methods: part-2
Sudden Expansion
Prob 1.16. A particle of mass m is contained in a onedimensional infinite well extending from x == -L/ 2 to x = L/ 2. The particle is in its ground state given by o(x) == 2/ L cos(1rx / L ). The walls of the box are moved suddenly t o from a box extending from x==-L to x=L. What is the probability that t he particle will be in the ground state after t his sudden expansion? [JEST 2013]
(a) (8/ 31r)
2
(b) 0
(c) (16 / 31r)
2
( d)
(4/ 31r )
2
11 oo : oo : 21 probability that t he particle will be in the ground state after t his sudden expansion? [JEST 2013] 2
(a) (8/31r)
(b) 0
(c) (16/31r)
2
(d) (4/ 31r)
2
Prob 1.1 7. A particle is confined to a one-dimensional box of length L . If a vanishingly thin but strongly repulsive partition is introduced in the exact center of the box, and the particle is allowed to come to its ground state, then the probability density for finding the particle will appear as [TIFR 2014]
j [email protected]
10
Physicsguide
@Sk J ahiruddin , 2020
Ap proximation Methods: part-2
N
8
(b)
;l>-
X
(c)
X
11 oo : oo : 29 X
X
X
X
(c)
Prob 1.18. A particle is in t he ground state of a cubical box of side l . Suddenly one side of t he box changes from l
to 4l. If p is t he probability of finding t he particle in t he [TIFR ground state of the new box, what is l OOOp ? 2018] Prob 1.19. Consider a quant um part icle in a one-dimensional
box of length L. The coordinates of the leftmost wall of t he box is at x = 0 and that of the rightmost wall is at x = L. The particle is in t he ground state at t = 0. At t = 0, we suddenly change the lengt h of t he box to 3L by moving the [email protected]
@Sk J ahiruddin , 2020
11
P hysicsguide
Approximation 1\/fethods: part-2
right wall. What is the probability that the part icle is in t he ground state of the new system immediately after the change? [JEST 2019] 81 9 05 (a) 0.36 (b) (c) (d) · 8n 64n 2 n
1.5
Scattering
11 oo : oo : 32 JEST 2019
change?
(a) 0.36
1.5
(b) 9
81r
81 (c) 641r2
(d) 0.5 7r
Scattering
Prob 1.20. Consider an elastic scatt ering of particles in l = 0 states. If t he corresponding phase shift 80 is 90° and the magnitude of the incident wave vector is equal to v12n" fm- 1 t hen the total scattering cross section in units of fm 2 is __ .
[GATE 2016] Prob 1.21. A phase shift of 30° is observed when a beam of
particles of energy O. l MeV is scattered by a target. When t he beam energy is changed , t he observed phase shift is 60° Assuming that only s-wave scatt ering is relevant and that t he cross-section does not change with energy, t he beam en[NET Dec 2017] ergy is (a) 0.4MeV (b) 0.3MeV (c) 0.2MeV (d) 0.15MeV Prob 1.22. The scattering of particles by a potential can be analyzed by Born approximation. In particular , if the scatt ered wave is replaced by an appropriate plane wave, [email protected]
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12
Physicsguide
Approximation Methods: part-2
t he corresponding Born approximation is known as the first Born approximation. Such an approximation is valid for [GATE 2016] (a) large incident energies and weak scattering potentials. (b) large incident energies and strong scattering p otentials.
11 oo : oo : 34 t e correspon 1ng orn approx1mat1on 1s nown as t e rst Born approximation. Such an approximation is valid for [GATE 2016] (a) large incident energies and weak scattering potentials.
(b) large incident energies and strong scattering potentials. (c) small incident energies and weak scattering potentials. (d) small incident energies and strong scattering potentials.
Prob 1.23. A free particle is described by a plane wave and moving in t he positive z-direction undergoes scattering by a potential
V(r ) =
Vo ,
if r < R
0,
if r > R
(1.1)
If Vo is changed to 2Vo, keeping R fixed, then the different ial scatt ering cross-section , in t he born approximation [NET June 2012] (a) increases to four t imes t he original value (b) increases to twice the original value (c) decreases to half t he original value (d) decreases to one fourth t he original value
Prob 1.24. In the Born approximation, the scattering amplitude f (0) for the Yukawa potential
j [email protected]
13
Physicsguide
@Sk J ahiruddin, 2020
Approximation Methods: part-2
V (r) = f3e - µr r
00: 00: 36 pproximation Methods: part-2
V (r) = f3e - µr r
is given by :(in the following b
= 2k sin ~, E = n2 k 2 /2m) [NET June 2013]
2m/3 (b) -
n2(µ2 + b2)
2m/3 (d) -
n2(µ2 + b2)3
Prob 1.25. The scattering amplit ude f (0) for the potential V(r) = f3e - µr, where /3 and µ are positive constants, is given , in the Born approximation , by [NET June 2014] 0 n2 k 2 (in the following b = 2k sin and E = m) 2 2 4mf3 µ 4mf3 µ (a) - n 2(b2 + µ2)2 (b) -n2b2(b2+µ2) µ 4m/3 () c - n2J(b2 + µ2)
(d)
4m/3 µ - n 2( b2 + µ2)3
Prob 1.26. A particle of energy E scatters off a repulsive spherical potent ial
V (r) =
Vo for r < a 0
where
Vo
for r > a
and a are positive constants. In the low energy
[email protected]
@Sk J ahiruddin, 2020
Physicsguide
14
Approximation Methods: part-2
limit , the total scattering cross-section is O" = 41ra
2
1 (k a
tanh ka -
11 oo : oo : 42 15
[email protected]
Physicsguide
@Sk J ahiruddin , 2020
(a) 16a
2
64a
2
(c)
mV0 a
Approximation Methods: part-2
2
2
mVoa
(b) 16a
fi2
fi2
7r2
mVoa
2
2
(d) 64a
fi2
2
2
2
m V0 a
2
fi2
7r2
Prob 1.29. The differential scattering cross-section : ; for
t he cent ral potential V (r) = (3 e - µr, where (3 and µ are posir tive constants, is calculated in t he first Born approximation. Its dependence on the scattering angle 0 is proportional to (A is a constant below. ) [NET June 2018] A
(a)
(c)
A
2
2
+ sin
+ sin
2
0 -
2
j [email protected]
2
0 -
2
(b)
A
- 2
A
(d)
16
2
2
+ sin
2
0
- l
-
2
+ sin ~
2
2
2
Physicsguide
11 oo : oo : 44
j [email protected]
@Sk J ahiruddin , 2020
1.6
Physicsguide
16
Approximation Methods: part-2
Ans Keys
1.1. d
1.11. d
1.21. b
1.2.
C
1.12. a
1.22. a
1.3.
C
1.13. a
1.23. a
1.4. a
1.14. a
1.5. d
1.15. a
1.6. b 1. 7. Do yourself
1.16. a 1.17. a
1.24. b 1.25. d 1.26. a
1.8. b
1.18. 58
1.27. a
1.9. b
1.19.
C
1.28.
C
1.10. b
1.20. 2
1.29.
C
11 oo:oo:47
[email protected]
@Sk J ahiruddin, 2020
1. 7
Physicsguide
17
Approximation Methods: par t-2
Solutions
Sol 1.1. Almost all the types of Variational approximation problems are solved in the notes. Please go through t hem first .
The Hamiltonian is given by
-n2
d2 H = 2m dx 2
+V
Now we know t hat the potential of the well is V = 0 ranging from x = -a to x = a. As the wave function is normalised then the ground state energy can be calculated as follows
=
-n,2
a
15 (a2 - x2) X 16a5 2m - a a 2 15n 2 2 x ) dx (a 16ma5 -a 4a 3 15n2 5n2 X l6ma5 4ma2 3
d2 dx 2
(a2 - x2)
dx
Sol 1.2. In problem number Prob 1.1. , we did not find the minimum energy t he ground state can attain. We just learnt
11 oo : oo : so 2 2 3 15!i 4a 5!i =---X-=-l6m a5 3 4ma 2
Sol 1.2. In problem number Prob 1.1. , we did not find the minimum energy t he ground state can attain. We just learnt how to find t he epression of energy of t he ground state. In t his problem we will minimise t he expression of energy for a [email protected]
18
P hysicsguide
@Sk J ahiruddin , 2020
Approximation Methods: part-2
given parameter and will try to derive the minimum energy of the ground state. At first we have to normalise t he wave function 00
exp(-2bx
2
)
dx = 1
-oo
2b Now t he expression for ground state energy E
= ('11 H '11) =
A
- !i2
2
X
2m
d2
CX)
- 2 exp(-bx dx
- oo
2
)
dx
00
2
exp(-2bx )6(x) dx -oo 2 2
-4b !i =A x 2m
00
2
2
_00
x exp(-2bx
2
)
2
dx - aA x 1
-4b2!i2 (2b)- 3/ 2 ~ -x x---- - a 2m 2 7r b!i2 2b = -- a 2m 7r 2b
., ,
. . .
.,
.
2b
.
.
,
11 oo : oo : s2 2b
-
X
1r
2
bn
= --a
-4b2 n2 --2m 2b
X
(2b)- 3/ 2 ~ ----- - a 2
2b
2m
Now we will minimise the expression against the parameter
j [email protected]
Physicsguide
19
@Sk J ahiruddin , 2020
Approximation Methods: part-2
"b,, dE db
n2
a 2 - - --X-= 0 2m 2vb 1r
Then, the minimum energy the ground stat e can att ain can be calculated by putting t he value of parameter ''b'' n,2 2a2m2 E =-x -a 2m nn4
2 7r X
am fi2
2 X
7r
a 2m
E =-1rn2
Sol 1.3. Same as previous problem Sol 1.4. We will solve t he problem exactly as in problem number Prob 1.2 ..
The normalising const ant c2 n5
11 oo : oo : 55 Sol 1.4 . We will solve t he problem exactly as in problem number Prob 1.2 .. The normalising constant c2 a5
-a
c=
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15 16
20
@Sk J ahiruddin , 2020
Physicsguide
Approximation Methods: part-2
Now t he expression for ground state energy E
= ('11
H '11) c2 -fi2
dx
=-X--
a5
+
2gc a5
2m
-a
2 0
= _15 X _2_n2_
a (a2
- x2) dx
16 2ma 5 - a 2g 15 a 4 • 2 3 5 2 + -5 x (a x - 2a x + x ) dx a 16 0 2 15 n 4a3 5ga -x-x + 16 ma5 3 16 2 5n 5ga 4ma2 + 16
Now we will minimise the expression against the parameter ''a''
11 oo : oo : sa 2
5n 4ma2
+
5ga 16
Now we will minimise the expression against t he parameter ''a'' 2
+ 5g = O
- 5n 2ma3
dE da
16
n,2
a= 2
===::,~>
j [email protected]
1/3
mg
21
Physicsguide
@Sk J ahiruddin , 2020
Approximation Methods: part-2
Then , t he minimum energy t he ground state can attain 5 n,2 5 n,2 E = --+ g x 2 4ma 2 16 mg
15
n2g2
16
m
1/ 3
1; 3
Sol 1. 5. The expression for ground stat e energy
E = ('1! H '1!)
-n2 2m
X
7r
n2/3
+ 6m
a a
00
exp - oo
d2
dx 2 exp
2
00
4
x exp( -ax
X 7r
-ax 2
- 00
2
)
dx
-ax2 2
dx
00 : 01 : 00 - oo
7r
00
4
x exp( -ax 7r
~2
- ,i a
2
)
dx
- 00
2
00
2
x exp( -ax
2m
2
)
- oo
7r
a
00
1r
- oo
exp(-ax
2
)
dx
2 n /3
3 dx + x 2 6m 4a
n a n a n /3 = --+-+-2
4m n2a
2
2
2m
8ma2
n,2/3
=4m - +8ma - -2 Now we will minimise the expression against the parameter 22
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Physicsguide
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Approximation 1\/fethods: part-2
''a'' dE
n -
2
da 4m ~~► a = 131/3
n /3 ---= 0 2
-
4ma 3
Then, t he minimum energy the ground state can attain n,2131/3
n,2/3
4m
8mf3 1
E = -- +-2 3 = 3 n,2131/3 8m
Sol 1.6. The normalising constant
11 oo : 01
: 03
= 3 ri2131/3 8m
Sol 1. 6. The normalising constant a
A
2
(a
2
2 2 x )
-
dx
=l
-a
15
A=
l6a5
23
jahir@physicsguide. in
Physicsguide
@Sk J a hiruddin , 2020
Approximation Methods: part-2
Now t he expression for ground state energy
E = ('11 H IW)
= A2
X -n,2
2m 2mw 2A2 +--2
dx -a a
x2 ( a 2
-
x 2 ) 2 dx
0
2
5n
15
2
- - +mw x - -5 4ma2 l6a 5n2 mw 2 a 2 =--+-4ma2 14 l\ T
• 11
•
•
•
a
I
1
4 2
(a x
-
2 4
2a x
+
0
•
•
I
I
1
6 2 x )
dx
11 oo : 01 : os 5n
2
4ma 2
+ mw
2
x
a
15
4 2
(a x
16a 5
+
2 4
2a x
-
6 2 x )
dx
0
5n2
mw 2a 2 =--+--4ma2 14
Now we will minimise the expression against t he parameter ''a" 2
dE = 5(-2)n
+ mw
4ma3
da ~~> a2
=
a = O
7
112
35
2
n
2
mw
Then, t he minimum energy tl1e ground state can attain
E=
5n2 4m
mw
X
nw
n 2 35
5 2
2
2 35
mw 2
+
1
+7
14
35 2
X
mw
35 2
5 14 j [email protected]
Physicsguide
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@Sk J ahiruddin , 2020
Approximation Methods: part-2
Sol 1. 7. You can do it yourself. We will just provide you with t he result. The normalising constant a
A
2
(a
2
-
2 3 x )
dx = 1
-a
A2 = 35 32a7
The expression for ground state energy
11 oo :01 : oa A2 = 35 32a7 The expression for ground st at e energy
- fi2
=_
x
35
2m 32a7 2 21fi 16ma2
a
(a2 _ x2)3/2
dx
-a
Now t he ratio 21fi2 16ma 2
21
fi27r
2 1r2
8ma2
Sol 1.8. Please go t hrougth t he WKB notes first. All problems can be done by the the following formula. If t he potent ial varies as
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P hysicsguide
Approximation Methods: part-2
Then the energy varies as
En= Ak (n - 1/ 2)fi where n = 1, 2, 3 .. . In t his problem use v = l
2mk
11 oo : 01
: 1o
En = Ak (n - 1/ 2)fi
2mk
where n = 1, 2, 3 ... In t his problem use v = l
Sol 1.9. Use v = l in the general formula as given in the solution of the previous problem Sol 1.10. Use v
= 1 in the general formula as given in the
solution of the previous problem
Sol 1.11. Use v = 6 in the general formula as given in t he solution of the previous problem. Also solved in the notes. Sol 1.12. Use v = 4 in the general formula as given in the solution of the previous problem. Also solved in the notes.
Sol 1.13. Here we have t o calculate t he probability of transition of the particle from the ground state to first excited state. The probability of transition(first order) is given by
l P1; = n,2
t
2
('¥1 V(t') l'¥;)exp(iw1;t
1 )
dt'
0
In the problem we have been provided with the time depen26
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@Sk J ahiruddin , 2020
Approximation Methods: part-2
dent potent ial along with the time for which the particle was subj ected to the potential. Also we are fairly acquainted with the expression of states of a h armonic oscillat or. Now we will calculate the expression of w f i . 1
1
11 oo : 01
: 13
dent potential along with the t ime for which the p article was subjected to the potential. Also we are fairly acquainted with the expression of stat es of a h armonic oscillator. Now we will calculate the expression of w Ji .
3
1
2 =
2
- ru.u - - ru.u
Wji
n
= W
Now t he transition probability Pfi 2
CX)
exp (iwt') dt'
exp - CX)
Now if you look at the options given, you will know t h at we m ust fo cus only on the integration. We will try to arrange the integration as p er our convenience. CX)
I =
- t'2
exp
72
- CX) CX)
exp - 00 CX)
- 1 72
2
t' -
CX)
4
dt'
iwt' 7 + 2
-W2 7 2
exp -
+ iwt'
exp
- 1 , 2
iw7
2
2
iw7
2
t' -
2
2
dt'
2 iW7
2
2
dt'
2
Now t he transition probability CX)
2 j [email protected]
exp - oo
- 1 T
2
27
@Sk J ahiruddin, 2020
.
'lW7
2
t' - - 2
Physicsguide
Approximation Metho ds: part-2
Thus we can say t h at the transit ion probability -W272
Pf i ex exp
2
dt
11 oo : 01
: 16
@Sk J ahiruddin, 2020
Approximation Methods: part-2
Thus we can say t hat t he transition probability -W2T2
Pfi ex exp
2
Sol 1.14. Sol 1.15. Sol 1.16. We have thoroughly discussed the sudden approximation problems in t he notes with examples. Please go t hrough these first. A brief review is as follows. 1) In Sudden Expansion the initial wavefunction would not resp ond to the sudden change and the system will evolve in such a way that t he expression of t he wavefunction will remain same (only the normalising constant will change). Eg : A potential well of length l is changed to length 2l suddenly. Init ially if the syst em was in ground state then aft er expansion t he system will b e in first excited state. Thus to keep the expression of t he wavefunction same (except the normalising constant) t he system changes t he state. 2) In Adiabatic Expansion, t he wavefunction will change. Eg : A potential well of length l is changed to length 2l very slowly. Initially if t he system was in ground state then after expansion t he syst em will be in ground state also. Thus to remain in ground state the expression of the wavefunction [email protected]
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will seem to have changed.
28
Physicsguide
Approximation Methods: part-2
11 oo :01
:1
a
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Approximation Methods: part-2
will seem to have changed. The wavefunction of t he ground state i:I.
:--
,., ~ ! (c)
l
tl,
0 V'X
(d)
0 Vx
Prob 1. 3. Electrons of mass m in a thin long wire at a temperature T follow a one-dimensional Maxwellian velocity didstribution. The most probable speed of these electrons [JEST 2015] is,
(c)O Prob 1.4. A spherical closed container wit h smooth inner wall contains a monatomic ideal gas. If the collisions between the wall and the atos are elastic, then the Maxwell speed-distribution funct ion ( d:Vv) for t he atoms is best rep-
resented by
[JAM 2016]
11 oo : oo : 11 tween the wall and the atos are elastic, then the Maxwell speed-distribution funct ion ( d~v) for the atoms is best represented by [JAM 2016] Prob 1.5. In low density oxygen gas at low temperature, [email protected]
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I(inetic Theory
only the translational and rotational modes of t he molecules are excited. The specific heat per molecule of the gas is [NET Dec 2014] 3 5 (c) -kB (d)-kB (b)kB 2 2 Prob 1.6. The specific heat per molecule of a gas of diatomic molecules at high temperatures is [NET June 2016] (a) 8kB (b) 3.5kB (c) 4.5kB (d) 3kB ,
.
,
.,
• r"'
,
,
I
~
I
~
\
11 oo : oo : 14 Prob 1.6. T he specific heat per molecule of a gas of di-
atomic molecules at high temperatures is 2016] (a) 8kB (b) 3.5kB (c) 4.5kB (d) 3kB
[NET June
Prob 1. 7. For which gas t he ratio of specific heats (Gp/ Gv)
will be t he largest? (a) mono-atomic (b)di- atomic j [email protected]
5
[JEST 2014] (c)tri-atomic (d)hexaphysicsguide CSIR NET, GATE
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Kinetic Theory
atomic Prob 1. 8. Consider t he following linear model of a molecule of hydrogen cyanide (H CN) depicted b elow.
0 --H
--C
N
It follows that t he molar specific heat of hydrogen cyanide gas at constant pressure must be [TIFR 2019]
(a) 6R
(b) 4.5R
(c) 5R
(d) 5.5R
Prob 1.9. The molar specific heat of a gas as given from
t he Kinetic t heory is ~R . If it is not sp ecified whet her it is Gp or Gp , one could conclude that t he molecules of t he gas [JAM 2005] (b) are definitely rigid (a) are definit ely mono atomic diatomic
11 oo : oo : 16 ""'"'ry 1 2 . pec1 Gp or Gp , one could conclude that t he molecules of t he gas [JAM 2005] (b) are definitely rigid
(a) are definitely mono atomic diatomic (c) are definitely non-r·igid diatomic
(d) can be monoatomic
or rigid diatomic
Prob 1.10. A rigid triangular molecule consists of t hree non-colinear atoms joined by rigid rods. The constant pressure molar sp ecific heat (Gp) of an ideal gas consisting of [email protected]
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Kinetic Theory
such molecules is
(a) 6R
(b)5R
[JAM 2015]
(c)4R
(d)3R
Pro b 1.11. The speed v of the molecules of mass mof an ideal gas obeys Mazwell's velocity distribution law at an equilibrium temperature T. Let (Vx, Vy, Vz ) denote the components of t he velocity and kB t he Boltzmann constant. The 2 average value of (avx - /3vy ) , where a and j3 are constants, is [NET Dec 2013]
(a)(a
2
-
2
/3 )kBT /m
(c)(a + /3) kBT /m 2
(b) (a
2
+ /3
2
)kBT/m]
2
(d)(a - j3 ) kBT /m
Prob 1.12. At a given t emperature T , t he average energy per particle of a non-interacting gas of two-dimensional classical harmonic oscillator is ______kBT [GATE 2014] (kB is t he Boltzmann constant) Prob 1.13. A classical gas of molecules each having mass m, is in therm:1.l eo11ili hri11m : -1.t the ahsol11te temn er::i.t 11re T.
11 oo : oo : 19 per part icle of a non-interact ing gas of two-dimensional clas[GATE 2014] sical harmonic oscillator is ______kBT (kB is t he Boltzmann constant)
Prob 1.13. A classical gas of molecules each having mass m is in thermal equilibrium at t he absolute temperature T . The velocity components of the molecules along the Carte2 sian axes are Vx, Vy and V z · The mean value of (V x + Vy) is [GATE 2012] (a) kBT (b) 3k BT (c) kBT (d) 2kBT m 2m 2m m Prob 1.14. A syst em of N non-interacting classical point particles is constrained to move on the two-dimensional [email protected]
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face of a sphere. T he internal energy of t he system is [GATE 2010]
(a) ~NkBT
(b) ~NkBT
(c) NkBT
(d) ~NkBT
Prob 1.1 5 . The mean internal energy of a one-dimensional classical harmonic oscillator in equilibrium wit h a heat bat h [GATE 2006] of t emperature T is (a) ~kBT (b) kBT (c) ~kBT (d) 3k BT Prob 1.16. Two boxes A and B contain an equal number of molecules of the same gas. If t he volumes are VA and VB and AA and AB denote respective mean free paths, t hen [JAM 2018]
11 oo : oo : 21 [JAM 2018]
AA
AB
A
B
(c) vl/2 = vl/2
(d) AAVA = ABVB
Prob 1.17. one mole of an ideal gas with average molecular speed vo is kept in a container of fixed volume. If the temperature of the gas is increased such t hat the average speed gets doubled , then [JAM 2016] (a) the mean free path of t he gas molecule will increase (b )the mean free path of the gas molecule will not change (c)the mean free path of the gas molecule will decrease (d) the collision frequency of the gas molecule with wall of j [email protected]
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t he container remains unchanged.
Prob 1.18. Consider a particle diffusing in a liquid contained in a large box. The diffusion constant of the particle 2 2 in t he liquid is 1.0 x 10- cm /s. The minimum t ime after which t he root-mean-squared displacement becomes more [NET June 2018] t han 6 cm is (a) 10 min
(b) 6 min
(c) 30 min
(d)
v'6 min
Prob 1.19. Consider a uniform distribution of particles with volume density n in a box. The particles have an isotropic velocity distribution with constant magnitude v . The rate at which t he particles will be emitted from a hole of area A ·1
r ,,.
1
•
r Ty-,nrr,
n. n -. nl
11 oo : oo : 24 Prob 1.19. Consider a uniform dist ribut ion of part icles with
volume density n in a box. The part icles have an isotropic velocity dist ribut ion with constant magnit ude v . T he rate at which t he part icles will be emitted from a hole of area A on one side of this box is [JEST 2013] (a)nv A (b)nvA / 2 (c)nv A /4 (d)none of the above Prob 1.20. For a diatomic ideal gas near room t empera-
t ure, what fract ion of t he heat supplied is available for external work if t he gas is expanded at const ant pressure? [JEST 2013] (a) l / 7 (b)5/7 (c) 3/ 4 (d)2/ 7 Prob 1.21. T wo containers are maint ained at t he same t emperature and are filled wit h ideal gases whose molecules
have mass m 1 and m 2 respectively. The mean sp eed of molecules of t he second gas is 10 t imes t he r .m.s. speed [email protected]
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the of t he molecules of t he first gas. Find t he ratio of mi m2 [TIFR 2016] nearest int eger Prob 1.22. A gas of molecules each having mass m is in
t hermal equilibrium at a t emperature T . Let Vx, Vy, V z be t he Cart esian component s of velocity, iJ , of a molucule. T he mean value of (vx - a vy + f3vz) 2 is [JAM 2010] ( ) 2 kBT 2 2 kBT 2 ( a) 1 + a + /3 b 1 - a + /3 m
m
11 oo : oo : 21
m
11
Prob 1.23. A t iny dust p article of m ass 1.4 x 10- kg is floating in air at 300K. Ign oring gravity, its rms sp eed (in µm/ s ) due to random collisions wit h air molecules will b e closest t o [JAM 2012] (a) 0.3 (b) 3 (c) 30 (d) 300
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1.1
Kinetic Theory
Ans Keys
1. 1.
C
1. 9. d
1. 17. b
1. 2.
C
1. 10.
1. 3.
C
1. 11. b
1. 4.
C
1. 12. 2
C
1. 18. a 1. 19.
C
1. 20. d 1. 5. d 1. 6. b
1. 13. d 1. 14.
C
1. 21. 118
11 oo : oo : 29 1. 3.
C
1. 11. b
1. 4.
C
1. 12. 2
1. 19.
C
1. 20. d 1. 5. d
1. 13. d 1. 21. 118
1. 6. b
1. 14.
1. 7. a
1. 15. a
1. 22. a
1. 8. d
1. 16. b
1. 23.
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C
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1.2
C
Kinetic Theory
Solutions
Sol 1.1. From t he maxwell boltzman dist ribut ion we came to know that t h e number of particles wit h in t h e velocit y v to v+dv is given by m
3
dn = 41rN(- -) 2 e 21rkT
(-mv 2 ) 2 2 kT
v dv
11 oo : oo : 32 t o know that t he number of particles wit h in t he velocity v to v+dv is given by ffi
3
(-mv 2 )
?
dn = 41rN(- - ) 2 e 2 kT v~dv 21rkT now the rootmeansquared velocity is defined as the square
root of the average of t he square of the velocity.mathematically speaking 00 2 f0 v dn 3RT Vrms = fooo dn = m in similar way we can calculate most probable velocity which is given by dfv = O dv for t he most probable velocity where f v is given by
= 41rN( m )~e(fv 21rkT
2
2~ ;
)v 2
. from t his we will get t hat Vmp
2kT
=
m
hence the ratio of most probable velocity to rms velocity is given by ( Vmp ) = Vrms
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I(inetic Theory
Sol 1.2. From the Maxwell Boltzmann distribut ion we can calculate that most probable velocity is proportional to v'T
so if temperature increases then the peak of the curve will also be held in higher temperature. The velocity distribut ion curve calculates t he number of part icles wit h in t he velocity
11 oo : oo : 34 rom t e axwe o tzmann istr1 ut1on we can calculate that most probable velocity is proportional to vT so if temperature increases then the peak of the curve will also be held in higher temperature. The velocity distribution curve calculates t he number of particles wit h in t he velocity range v to v+dv so integration of t he velocity distribut ion curve over the the full range will give you the total number of particles which is not changing, so integration over the full range in any temperature will get you same value and is given by o 1.2.
00
0
so we have to consider t hat in higher temperature the area under the curve must constant that is why t he curve will be more stipper where as in the lower temperature t he curve will be tending to flatten.so at T1 curve will be flatter than t hat of T 2 . so option c is correct.
Sol 1.3. The Maxwell one dimensional velocity distribution is given by
again i can calculate t he most probable velocity from the definition of t he most probable velcoty but here we have to j [email protected]
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Kinetic Theory
take dgvx dvx
=Q
fnr thP mnc:::t. nrnh~hlP , rPlnri+., , ~n rl it. a-i,rPC::: th~+. (1, .. 1....
=n
00: 00: 37 Kinetic Theory
take dgvx = 0 dvx for the most probable velocity and it gives that (vx)rrip = 0 so option c is correct . Sol 1.4. It is the basic Maxwell Boltzmann distribut ion curve
which t ells that t he option c is correct. Sol 1.5. The
of the gas is defined as heat needed to increase t he temperature of the gas by 1 unit at costant volume and t he cp is defined as t he heat needed to the gas to increase t he temperature by i unit at constant pressure and we know that for an ideal gas t hey are related by the following equat ion Cp - Cv = R where R is the universal gas constant.To calculate the Cv of the gas it is most important to know the internal energy of t he gas because we know that Cv
Cv
=
au f)T
we know from the equipartition t heorem that each of the digrees of freedom will get an equal amount of energy and at t emperature T it is given by k~T SO if t he gas has f numbers of digrees of freedom t hen that gas will have total internal energy is given by U
=
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14
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so the gas wll have
fKBT 2 physicsguide CSIR NET, GATE
Kinetic Theory
Cv
=
f~B
now we know t hat f is related
11 oo : oo : 40
@Sk J ahiruddin, 2020
Kinetic Theory
so the gas wll have Cv = 1~n now we know that f is related by the total numbers of particles in the following way f = 3N - k where N is the total numbers of the particle and k is t he obstacles. for this problem we have oxygen gas which is diatomic so we have number of digrees of freedom is 5 so we have t he specific heat per molecule is ~kB
Sol 1. 6. Whenever you are asking to calculate the specific heat at higher temperature then you have to take all the degrees of freedom in to account. Total degrees of freedom =transnational and rotational digresses of freedom( contribut all the values of T)+ vibrational degrees of freedom(gets effective in only higher temperature) it is found that for linear molecules the vibrational degrees of freedom is f v = 3N - 5 and for the nor linear molecules this degrees of freedom is f v = 3N - 6
the diatomic molecule is always linear so the vibrational degrees of freedom is 1 and each vibrational digress of freedom takes an amount of energy kB T so total specific heat is given
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Kinet ic Theory
11 oo : oo : 42 15
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I(inetic Theory
by 5 7 Cv = -kB + kB = -kB = 3.5kB 2 2 Sol 1. 7. We define
2 1=-= l +Cp
f
cv
in order to do so we have to calculate t he Cp and Cv of different gas and take ratio for mono atomic its value is 1.67 for di-atomic its value is 1.40 for linear triatomic value is 1.28 and non linear 1.33 for hexa atomic 1.14 (molecules 6 and obstacles 6) so t he highest values is for mono atomic molecule.
Sol 1.8. Here the all degrees of freedom is accessible as the dotted curve represents so the possible degrees of freedom is given by rotational degrees of freedom 9 and t he vibrational degrees of freedom is 1 so total Cv
=
9 R+R 2
=
5.5R
Sol 1. 9. It is surely be monoatomic or rigid diatomic as we know t hat t he Cv for diatomic=cp for monoatomic as it is not mentioned if the solution is cporcv so both possibility •
remains.
Sol 1.10. As molecule is non linear so obstacles 3 so we will have f (t he degrees of freedom) is 6 so cp = Cv + R = 4R j [email protected]
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11 oo : oo : 4s Sol 1.10. As molecule is non linear so obstacles 3 so we will have f (the degrees of freedom) is 6 so Cp = Cv + R = 4R j [email protected]
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Sol 1.11. We have
now a and /3 are constanst so they come of the integration. now to evaluate the
and this is equatl to
and in similar way if you calculate
, < Vy > v.re will get
zero because Vx, Vy these are the components so logically if we take the velocity distribution curve then t he components are equally distributed from -oo to oo so it's average is obviously zero. then we are getting putting all the values 2
a2
+ 132
< (avx - /3vy) >= ---kBT m
Sol 1.12. Here the system is in two dimension so contribut ion comes from 2 co ordinates and each co ordinate will have potential energy contribution and the kinetic energy
contribution and from the generalized equipartition t heorem will give us the potential energy contribution from the
11 oo : oo : 47 t ion comes from 2 co ordinates and each co ordinate will have potential energy cont ribution and the kinetic energy cont ribution and from the generalized equipart ition t heorem will give us t he potential energy contribution from the [email protected]
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Kinetic Theory
each co ordineat e is k~T and similarly t he contribution for kinetic energy from each coordinat e also k~T so total energy
=4(k~T)=2kBT hence the answer is 2. Sol 1.13. We have
and we know from t he previous discussions t hat
=
= 0 and the rest of the two t erms will give you equal contribution which is kBT ( remind here it is only < (vx)2 > 2 m
Vy
but not
(v
2
so we get only k1:nT) . so total value k::,_T.
Sol 1.14. Here t he particle is constr ained to move on t he
surface of t he sphere so it is two dimensional surface hence contribution comes from only kinetic energy in two dimension and hense we get t he energy for N numbers of particle is NkBT. Sol 1.15. For one dimensional classical harmonic oscillator
we have total cont ribution to internal energy is coming from 2 2 1) kinetic energy = (~~ and 2) potential energy k; and each having average value k~T so total energy contribution kBT.so option b is correct . Sol 1.16. We know that mean free path of a gas is given by
11 oo : oo : so 2
1) kinetic energy = (~~ and 2) potential energy each having average value k~T kBT.so option b is correct .
2
and so total energy contribution k~
Sol 1.16. We know that mean free path of a gas is given by
A=
1
--/21rnd2 [email protected]
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Kinet ic Theory
where we haven= ~ puut ing t his back in t his equation we get t hat
A=
V
~1rNd2 so from this i can conclude that .A is proportional to V so we will get
so option B is correct . Sol 1.17. We know t hat mean free path is defined as the average distance t raveled by a gas molecule or other particle
between collisions with t he other particles.from the correct ion made by Maxwell to t he fo1~mulae made by the Clausius we get t he mean free path of the gas molecule is given by v7 =
1
--/21rnd2 where n is the number of molecules per unit volume and d 2 is t he diameter so 1rd effective cross sectional area and we see that this is independent of the temperature so mean free path of t he molecules will not change. Sol 1.18. The root mean square displacement is propor-
11 oo : oo : s3 w ere n 1s e num er o mo ecu es per uni vo ume an is t he diamet er so 1rd2 effect ive cross sectional area and we see that t his is independent of t he t emperature so mean free path of the molecules will not change. Sol 1.18. The root mean square displacement is proport ional to t ime so with t he increment of t ime t he displace-
ment also increases so we get
= 6Dt
19
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Kinetic Theory
where D is the diffusion constant . Here we get D = 10- 2 cm 2 / s and r is given by which is 6 cm so < r rms 2 >= 36cm 2 so putting all values in t he equation we get t = 10 min Sol 1.19. We know from t he first law of t hermodynamics dQ = dU + dw where dU is called the change in t he internal
energy of t he system , and dw is the workdone on t he syst em so first law of thermodynamics just tells you t he fact t hat if you give some amount of energy dQ to t he syst em then one of it's part will be given to internal energy and t he other part will be expended to get in the work of t he system.now in isobaric process the work done is given by P dV = R dT and we know t hat dU = cvdT .now we know t hat Cp - Cv = R ep = 1 so putting all the values in t he equation and we define Cu we get dU = ~~1; now i can say t hat dw
=, - l dQ r
and for diatomic molecule we have , = ; and hence we get dw = ; dQ so here we get only ; of total energy will be
00: 00: 55
dw
ry
and for diatomic molecule we have ry = ; and hence we get dw = ~dQ so here we get only ~ of total energy will be converted to workdone so opt ion d is correct. Sol 1.20. The ans is (c) . I'm writing the proof in two ways below.
Consider t he process by which molecules escape t hrough a hole in a vessel and into a vacuum [email protected]
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Kinetic Theory
Hole of Area "A"
We assume t hat: (1) A is so small t hat t he pressure in the vessel is unchanged ; (2) The effusion does not perturb the velocity of t he gas in t he vessel; (3) There are no collisions when the molecules pass through t he slit . Molecules t hat would have been incident on t he portion of t he wall where the hole is, now pass through the hole. n
r
,
r-
1
1
r
11 oo :oo : s1 (3) There are no collisions when the molecules pass through t he slit . Molecules t hat would have been incident on t he portion of t he wall where the hole is, now pass through the hole. This creates a flux of particles defined as t he number of particles per unit area per unit t ime that leave the vessel. Consider a square hole of area dA . A particle that is a distance vdt from the hole moves with speed v and at angle 0 from the surface normal toward the hole.
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0
vdt
Draw a parallelepiped around the hole wit h lengt h equal to v dt , at angle 0 from the normal. All molecules within t his volume moving toward t he hole (i.e. , with the correct
11 oo : 01 : oo
Draw a parallelepiped around the hole with length equal to vdt, at angle 0 from the normal. All molecules within t his volume moving toward the hole (i.e., with the correct
0, cp angle) with speed v will pass t hrough the hole in t ime interval dt.
n
=
molecular density of gas.
Volume of parallelepiped = v cos 0 dA dt (Note that , at grazing angles, 0 r-v 1r /2, t he volume is small) No of molecules crossing t hrough da in dt is
nvcos0dadt
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Kinetic Theory
(number density t imes volume) No of molecules - - - - - - - = nv cos 0 = FLUX. dAdt We must integrate t his expression over the distribution of velocities of the gas to obtain the average flux J. The Maxwell-Boltzm ann distribution is F(V)dv
=
F(v, 0, cp)v dvsin0d0dcp
m 21rkT Thus for average flux T
2
3 2
e-
mv 2 kT
2
v dv sin 0d0dcp
11 oo : 01
: 03
m 21rkT Thus for average flux
J=n
m 21rkT
3 2
3 2
e-
mv 2
kT
v 2 dv sin 0d0d(j)
7r
00
e-
2 mv
kT
21r
2
v 3dv
cos 0 sin 0d0 0
0
def> 0
The integration limit in 0 is taken to
1r /
2 instead of
1r
to take t he forward direction only. In the integration of v, 2 contribution of v comes from volume element and v comes from flux. The result is 3/2 1 m J =n 21rkT 2
n
8kT
4
1rm
11 2
2kT m
2
1 - [21r] 2
nv 4 23
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I(inetic Theory
The alternative way to obtain t he int egral is Consider a volume of gas behind the hole t hat contains n dV molecules. Then J simply is:
0
m 21rkT
3 2
e
mv
2
2kT
Only t he molecules with v z > 0 can exit t he hole. Integration give you same result.
Sol 1.21. The mean speed of gas molecules can be obtained from the maxwell boltzman velocity distribution and we get
11 oo : 01
: 06
Only the molecules with Vz > 0 can exit t he hole. Int egration give you same result . Sol 1.21. The mean speed of gas molecules can be obtained from the maxwell boltzman velocity distribution and we get t hat
and doing this we get
=
BkBT m1r
and we know 3 k::,_T
from t he previous discussion we know Vrms = so now putting condition 10(Vrms)1 = (Va,ug)2 will give as ratio of m 1 and m 2 is 118. Sol 1.22. Using t he generalized equipartition t heorem as
discussion before avg of the square of the velocity only cont ributes but any other t erms or only average of t he velocity terms contributes zero so taking that into account
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which gives us
•
Sol 1.23. Use the formulae of rms velocity.
Kinetic Theory
11 oo :01 : oa j [email protected]
24
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I(inetic Theory
which gives us
•
Sol 1.23. Use the formul ae of rms velocity.
j [email protected]
25
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11 oo : 01 j [email protected]
: 1o
24
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@Sk J ahiruddin , 2020
I(inetic Theory
which gives us
•
Sol 1.23. Use the formul ae of rms velocity.
j [email protected]
25
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11 oo : 01 j [email protected]
: 13
24
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@Sk J ahiruddin , 2020
I(inetic Theory
which gives us
•
Sol 1.23. Use the formul ae of rms velocity.
j [email protected]
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11 oo : 01 j [email protected]
: 16
24
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@Sk J ahiruddin , 2020
I(inetic Theory
which gives us
•
Sol 1.23. Use the formul ae of rms velocity.
j [email protected]
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11 oo :01 j [email protected]
: 1a
24
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@Sk J ahiruddin , 2020
I(inetic Theory
which gives us
•
Sol 1.23. Use the formul ae of rms velocity.
j [email protected]
25
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11 oo : 01 j [email protected]
: 21
24
physicsguide CSIR NET, GATE
@Sk J ahiruddin , 2020
I(inetic Theory
which gives us
•
Sol 1.23. Use the formul ae of rms velocity.
j [email protected]
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11 oo : 01 j [email protected]
: 24
24
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@Sk J ahiruddin , 2020
I(inetic Theory
which gives us
•
Sol 1.23. Use the formul ae of rms velocity.
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11 oo : 01 j [email protected]
: 26
24
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@Sk J ahiruddin , 2020
I(inetic Theory
which gives us
•
Sol 1.23. Use the formul ae of rms velocity.
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11 oo : oo : 01
Problems and Solutions
•
Ill
Second Law- of Thermodynamics Sk J ahiruddin * Suchismito Chattopadhyay
*Assistant Professor Sister Nibedita Govt. College, Kolkata Aut hor was t he topper of IIT Bombay M.Sc Physics 2009-2011 batch He ranked 007 in IIT J AM 2009 and 008 (J RF) in CSIR NET J une 2011 He has been teaching CSIR NET aspirants since 2012
1
@Sk J ahiruddin , 2020
Second Law of Thermodynaics
11 oo : oo : 03 1
Second Law of Thermodynaics
@Sk J ahiruddin, 2020
Contents 1
Problems from NET, GATE, JEST , TIFR & JAM pap ers 1.1 Ans keys . 1.2 Solut ions . • • •
j [email protected]
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
2
•
•
3 •
•
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•
•
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•
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physicsguide CSIR NET, GATE
11 oo : oo : 06
j ahir@physicsguide. in
2
Second Law of Thermodynaics
@Sk J ahiruddin , 2020
1
physicsguide CSIR NET I GATE
Problems from NET, GATE, JEST, TIFR & JAM papers
Prob 1.1. A Carnot cycle operates as a heat engine be-
tween two bodies of equal heat capacity unt il their t emperat ures become equal. If t he init ial temperatures of the bodies are T1 and T 2 respectively and T1 > T 2 t hen t heir common fin al temperature is [NET Dec 2013]
2 (a)T1 /T2
2 (b)T2 /T1
(c)~
(d)½(T1 +T2 )
Prob 1.2. Efficiency of a perfectly reversible (Carnot) heat
engine operating between absolute t emperature T and zero [JEST 2012] is equal to (a) 0 (b) 0. 5 (c) 0.75 (d)l Prob 1.3. The ent ropy-temperature diagram of two Carnot
engines, A and B, are shown in t he figure. The efficiencies of the engines are T/A and T/B respectively. Which one of the following equalities is correct? [JEST 2015]
(a) T/A == 11,f T/A
(b) T/A == T/B
(c) T/A == 3TJB
(d)
== 2TJB
Prob 1.4. The pV diagram for a Carnot cycle executed by
an ideal gas wit h Cp / Cv = , > l is shown below. Note t hat 1,2,3 and 4 label t he change-over points in t he cycle.
11 oo :oo :oa Prob 1.4. The p V diagram for a Carnot cycle executed by
an ideal gas with Cp / Cv = , > 1 is shown below. Note that 1,2,3 and 4 label t he change-over points in the cycle.
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3
@Sk J ahiruddin ) 2020
Second Law of Thermodynaics
,II
/
"
,t
s \
v A ,,
'
'
/.13 ✓• ,...
' I
'r
•
-
•
-
If, for t his cycle, P2 p3
X
[TIFR 2013]
t hen X =
p
1 2
4 3
~------~ V
ft... \ (\
(;J\
1 /_.
11 oo : oo : 11 4 3
.____ _ _ _ _ ___, V
(a)
1 - 1/,
(b) O
(c) l
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Second Law of Thermodynaics
@Sk J ahiruddin , 2020
Prob 1.5. Consider a Carnot engine operating between t emperature of 600K and 400K . T he engine performs 1000 J
of work per cycle. T he heat (in Joules) extracted per cycle from the high temperature reservoir is. . . . . . . . . . . . .. (Specify your answer to two digits after t he decimal point) [JAM 2017] Prob 1.6. Which one of t he figures correctly represents t he
T - S diagram of a Carnot engine?
T
[JAM 2018]
T
(a)
(b)
s T (c)
s T
(d)
11 oo : oo : 14 T
T
s j [email protected]
s 5
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Second Law of Therrnodynaics
@Sk J ahiruddin, 2020
Prob 1. 7. In a heat engine based on the Carnot cycle, heat is added to the working substance at const ant [JAM 2019]
(A) Entropy (B) Pressure (C) Temperature (D) Volume Prob 1.8. A heat pump working on t he carnot cycle maintains t he inside temperature of a house at 22°C by supplying 450k J s- 1 . If t he outside temperature is 0°C , the heat taken, [GATE in kJ s- 1 , from the outside air is approximately
2007] (a) 487
(b) 470
(c) 467
(d) 417
Prob 1.9. A Carnot cycle operates on a working substance
between two reservoir at temperatures T 1 and T2 with T1 > T2 . During each cycle an amount of heat Q 1 is extracted from t he reservoir at T1 and an amount Q 2 is delivered in t he reservoir T2 which of t he following statements is INCORRECT? [GATE 2011] ( a) work done in one cycle is Q1 - Q2
11 oo : oo : 16 1
\.Al .I. .I.
2
l,,A/Ol .I. .I.
t he reservoir T 2 which of t he following statements is IN-
CORRECT?
[GATE 2011]
(a) work done in one cycle is Qi - Q2
(b)
Q1
T1
=
Q2 T2
(c) entropy of the hotter reservoir decreases (d) ent ropy of t he universe (consisting of t he working substance and t he two reservoirs) increases [email protected]
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6
2020
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Second Law of Thermodynaics
Prob 1.10. A reversible Carnot engine is operated between temperatures T1 and T 2 (T 2 > T 1) wit h a photon gas as the working substance. The efficiency of the engine is [GATE 2017] (a) 1 - 3T1 / 4T2 (b) 1 - T1 / T2 (c) 1 - (T1/T2)¾
4
(d) 1 - (T1/ T2) 3
Prob 1.11. When a gas expands adiabatically from V1 t o V2 by a quasi-static reversible process, it cools from t emperature T 1 to T 2 . If now t h e same process is carried out adiabatically and irreversibly, and T~ is the temperature of the gas when it has equilibrated, then (a)T~ = T2
(b)T~ > T 2
(c)T~
[NET Dec 2014]
= T2( ½ -
½
Vi)
(d)T~ = T2 V1
½ Prob 1.12. A h eat engine is operated b etween two bodies t h at are kept at constant pressure. The constant - pressure
11 oo : oo : 19 2 -
2
Prob 1.12. A heat engine is operated between two bodies t hat are kept at constant pressure. The const ant - pressure heat capacity Gp of the reservoirs is independent of temperature. Initially the reservoirs are at temperatures 300 Kand 402 K . If, after some t ime, they come to a common final
temperature T1, the process remaining adiabatic, what is [TIFR 2018] t he value of T1 (in Kelvin ?)
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Prob 1.13. Consider a monoatomic ideal gas operating in a closed cycle as shown in t he P - V diagram given below. The ratio ; ~ is ____ [JAM 2018]
(Specific your answer upto two digits after the decimal point)
P-2
adiabatic
V
11 oo : oo : 21 Pi l---~~--4-----=:::==--,. V
Prob 1 .14. The heat capacity of t he interior of a refrigerator is 4.2 kJ / K The minimum work t hat must be done to lower t he internal temperature from 18°C to 17°C when the outside temperature is 27°C will be [NET Dec 2015] (a)2.20kJ (b)0.80kJ (c)0.30kJ (d)0 .14kJ Prob 1 .15. When an ideal monatomic gas is expanded adiabatically from an initial volume Vo to 3V0 , its temperature
[NET
changes from To to T . Then t he ratio T / To is June 2016] j [email protected]
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Second Law of Thermodynaics
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(a) 1/ 3
(b)
(½)
2
3
(c)
(½)
1
3
(d) 3
Prob 1 .16. An ideal gas engine is run according to t he cycle shown in the s -T diagram below, where the process from D to A is known to be isochoric (i.e. maintaining V = constant s
D
.
A
•
B
-
C
T
The corresponding cycle in t he p - V diagram will most l
l
l
l
rm
TT'."\T""-
""r,.
-t r,. 1
00: 00: 24 B..___
___._ _ ____,(
T
The corresponding cycle in the p - V diagram will most
[TIFR 2019]
closely resemble
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@Sk J ahiruddin , 2020
A
p
C
p
...___o.
(a)
A
p
(b) V
V
D p
B
(d)
(c)
V
V
11 oo : oo : 21 p
B
(d)
(c) A
V
D
V
Prob 1.17. An ideal gas reversible engine operates in a [JAM closed cycle. The P-V diagram is shown below. 2011]
p
V j [email protected]
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Second Law of Thermodynaics
(a)Find t he efficiency of the reversible engine assuming both specific heat , Cp and Cv as constants. (b) Identify the t hermodynamic processes and draw the corresponding T-S diagram schematically. Prob 1.18. One mole of monoatomic ideal gas is init ially at pressure P0 and volume Vo. The gas t hen undergoes a
t hree-stage cycle consisting of the following processes: (i) An isothermal expansion till it reaches volume 2V0 , and hP.::i.t O -Rows int.n t.hP. !T::J.s.
11 oo : oo : 29 Prob 1.18. One mole of monoatomic ideal gas is initially at pressure P0 and volume Vo. The gas t hen undergoes a
t hree-stage cycle consisting of the following processes: (i) An isothermal expansion till it reaches volume 2Vo , and heat Q flows into the gas. (ii) An isobaric compression back to the original volume V0 (iii)An isochoric increase in pressure till t he original pressure P0 is regained [TIFR 2017] The efficiency of this cycle can be expressed as p To -----------
Po
I I I I I I I I I I I
I I I I I I I
' I I I
-----------?-----, I I
I I I
1 2
: To Po -------, ---~: - - ~:------2 To : : I
I I I
I I I
I
I
I I
I I
.___. . ._,______, - -- v Vo
(a)
E=
4Q + 2RTo (b) 4Q + RTo
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E
4Q + 2RT0 (c) 4Q - 3RTo 11
@Sk J ahiruddin ) 2020
(d)
E=
2Vo E=
4Q-2RT0 4Q + RTo
physicsguide CSIR NET ) GATE
Second Law of Thermodynaics
= 4Q - 2RT0 4Q + 3RTo
Prob 1.19. 1 m 3 of an ideal gas with ry = Cp/Cv = 1.5 is at a pressure of 100 kPa and atemperature of 300 K. initially the state of the gas is at the point a of the PV diagram shown. The gas is taken through a reversible cycle
a
➔
b
➔
c
➔
a. The pressure at a point b is 200 kPa
11 oo : oo : 31 p
V
= .
is at a pressure of 100 kPa and atemperature of 300 K. initially the state of t he gas is at t he point a of t he PV diagram shown. The gas is taken t hrough a reversible cycle a ➔ b ➔ c ➔ a. The pressure at a point b is 200 kPa and the line ba, when extended, passes through the origin. [JAM 2007]
b
/
/
V (a)Calculate the work done by the gas in each of t he steps a ➔ b ➔ c and c ➔ a (b) Calculate the change in entropy of t he gas in each of the t hree steps above [email protected]
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Second Law of Thermodynaics
Prob 1.20. One mole of an ideal monatomic gas in an initial state a wit h pressure, Pi and volume ¼ is to be t aken 2 to a fin al st ate d wit h PJ = B pi and VJ = ¼/ B t hrough t he path a ➔ b ➔ c ➔ d as shown in t he figure below for a particular value of B(> 1). Here a ➔ b and c ➔ d are
11 oo : oo : 34 1.20. ne mo e o an 1 ea monatom1c gas in an init ial state a wit h pressure, Pi and volume ¼ is to be taken
to a fin al st ate d with PJ = B 2pi and VJ = ¼/ B t hrough t he path a ➔ b ➔ c -+ d as shown in t he figure below for a particular value of B(> 1). Here a ➔ b and c -+ d are adiabatic paths while b ➔ c is an isot herm with temperat ure T 0 . States b and c correspond to (p 1, ½) and (p 2 , V2 ) , respectively. [JAM 2010]
1)
b(Pi, V.)
V Find the ratio ~ and t he total work done by t he gas in terms of Pi , ¼ , T 0 and B. Prob 1.21. The P-V diagram below represents an ideal j [email protected]
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Second Law of Therrnodynaics
monatomic gas cycle for 1 mole of a gas. In terms of the gas constant R , calculate the temperatures at the points J ,K,L and M. Also calculate the heat rejected and heat absorbed
00: 00: 37 econd Law of Thermodynaics
monatomic gas cycle for 1 mole of a gas. In terms of the gas constant R , calculate the temperatures at the points J ,K,L and M. Also calculate the heat rejected and heat absorbed during t he cycle, and t he efficiency of t he cycle. [JAM 2012]
80 - - ~-------' 'I
I
I
I ,
~,
0•10 I
I
Pro b 1. 2 2. A hollow cylinder (closed at both ends) with adiabatic walls is divided into n equal cells (C 1 , C2 , . .. C,n)
using discs D1, D2, ... D n -1 (see figure) . The discs can slide freely wit hout f1·iction. The first (D 1 ) is adiabatic and the remaining discs are diathermal (t hermally conducting) . Each cell contains one mole of ideal monoatomic gas. Let t he init ial pressure , volume and temperat ure of each cell be P0 ,V0 [email protected]
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Second Law of Thermodynaics
and T0 , respect ively. The gas in cell C 1 (first cell) is heated
11 oo : oo : 39
@Sk J ahiruddin, 2020
Second Law of Thermodynaics
and To, respectively. The gas in cell C1 ( first cell) is heated slowly until t he temperature of t he gas in cell Cn (last cell) reaches final equilibrium temperature 4T0 . Find the volume of t he first cell interms of the number of cells (n) and the initial volume (Vo) . [JAM 2013]
•
c,
cl-
s
• •
•
.... . - ..
•
en
Prob 1.23. In the t hermodynamic cycle shown in t he figure, one mole of a monatomic ideal gas is taken t hrough a cycle.AB is a reversible isothermal expansion at a temperat ure of 800K in which the volume of the gas is doubled. BC is an isobaric contraction to t he original volume in which t he temperature is reduced to 300K. CA is a constant volume process in which the pressure and temperature return to t heir initial values. The net amount of heat (in joules) absorbed by the gas in one complete cycle is [JAM 2015]
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Second Law of Thermodynaics
11 oo : oo : 42 15
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Second Law of Thermodynaics
@Sk J ahiruddin , 2020
I
.
'
'\
PJ
'
• V
Prob 1.24. For an ideal gas , which one of t he following
T-S diagram is valid?
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[JAM 2016]
16
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11 oo : oo : 44
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Second Law of Therrnodynaics
@Sk J ahiruddin , 2020
s
s
1.1
5
Ans keys Ans keys
1.1.
C
1. 7.
C
1.13. 0.16
1.2. d
1.8 . d
1.14 . d
1.3. d
1.9. a
1.15. b
1.4. a
1.10. b
1.16. a
1.5. 3000
1.11. b
1.17. Do yourself
1.6. b
1.12. 347
1.18. d
11 oo : oo : 48 1.4. a
1.10. b
1.16. a
1.5. 3000
1.11. b
1.17. Do yourself
1.6. b
1.12. 347
1.18. d
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1.19. 100 - ln2 KJ/ K
KJ , ~PiVi ( l
1.20. w
1.2
-
B)
2RT0 ln B 1.21. rJ
= 0.736
+
1.22.
V
=
Vo
3
8(n- l ) o
1.23. 453 1.24. a
Solutions
Sol 1.1. Heat engines are special kind of engine that works between two different t emperature source one is higher t emperature and t he other is lower t emperature.an ideal heat engine takes heat energy from the hot source and convert all the heat in to t he work that is why the efficiency of an ideal heat engine is I .let heat engine takes Q 1 amount of heat from the hot source and leave Q 2 ( < Q 1 ) amount of heat in t he cold source converting Q 1 - Q2 amount of heat in to the work so the efficiency of t he heat engine is defined as follows . e ffi c1ency
=
work output ----heat input
hence for ideal gas engine no heat is left at t he cold source so the total efficiency is 1, but total efficiency of value 1 is impossible practically. Carnot cycle is a reversible cy-
11 oo : oo : so heat input hence for ideal gas engine no heat is left at t he cold source so t he total efficiency is 1, but total efficiency of value 1 is impossible practically. Carnot cycle is a reversible cycle containing an ideal gas.t he gas is first taken through an isothermal expansion at temperature T1 from t he volume v1 [email protected]
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Second Law of Thermodynaics
to volume v2 (½ < V2 ) t hen t he gas is been t hrough an adiabatic expansion to volume v 3 and the temperature of t he gas becomes T2 ( < T1 ) and t hen t he gas is taken t hrough an isothermal compression to volume ¼ and then the gas is been taken through an adiabatic compression to temperature T1 and volume V1 so t he gas is been taken back to t he init ial state. if we perform t his process still then we get efficiency of the system is given by
but you see t hat this is less than one and only equal to 1 if t he lower temperatu1·e is zero kelvin but zero kelvin is un achievable ( we will discuss it in the later problem that why it is un achievable) . now er are comparing t his result to the efficiency of the heat engine and we get t he following
that is we can say t hat t he factor ;J_; remains unchanged during t he reversible process. this quantity is defined as
11 oo : oo : s2
t hat is we can say t hat the factor f remains unchanged during t he reversible process. t his quantity is defined as t he entropy of the system t hat is entropy of the system is defined as Q S(entropy) = T j [email protected]
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Second Law of Thermodynaics
its unit is joule/kelvin. It 's change in the reversible process is zero as we see from t he previous example. (reversible process is t hat kind of process where a system is brought back to it's initial stage in with out accompanying any kind of change in the surroundings) t he entropy of t he system is also a state dependent function that is it only depend on t he final st ate of t he system and t he initial state of t he system and it does not depend on the factor t hat how the state has been achieved and we only can measure the change of t he entropy but the exact value of entropy in a particular state can not be determined. we are going to use t hese fact to solve t his problem.as the system is working between two different temperature so t he syst em will keep working unt il the two different temperature are coming in t he equilibrium. say t hat the equilibrium temperature is T so t he hot source will loss it's entropy and t he cold source will gain t he entropy until t he change in t he entropy of t he whole system comes to zero so we get the change in the entropy of the hot source i c rri "l ron
h"',
11 oo : oo : 55 t hat t he equilibrium temperature is T so the hot source will loss it 's entropy and t he cold source will gain t he entropy until t he change in t he ent ropy of t he whole system comes to zero so we get the change in the entropy of the hot source is given by dQ
dS=
T
( the change in the ent ropy is counted as t he ratio of the heat taken or given and at the particular temperature it is
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Second Law of Thermodynaics
given or taken)so the entropy change of t he hot body is
entropy change of the hot body
=
T
epdT
T
T1
and t he change of the entropy of the colder one is
entropy change of the cold body
=
T T2
CpdT
T
so the total change of the system is given by
change of the entropy of the hot body + change of the cold body entropy ==total change of the entropy of the system==O using t his principle we get that
T1 T Cp loge T = Cp loge T
2
which gives us that final temperature is Im
rn
11 oo : oo : sa T1
Cp
loge T
=
T
Cp
loge T2
which gives us that final t emper ature is
Sol 1. 2. It is an ideal Carnot cycle so t he efficiency is 1 as no heat ha been left to t he cold source and the total heat taken from the hot source by t he system is converted to the work output. Sol 1.3. The most important fact about t he systems are t hat t hey are operating in the different t emperature. The j [email protected]
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Second Law of Thermodynaics
@Sk J ahiruddin, 2020
Carnot cycle in the t-s diagram looks like a rectangle(why?) . Carnot cycle is consisting of t he two isothermal process and two adiabatic process. adiabatic process in t he T-S diagram is 1·epresented by the straight line parallel to t he T (temperature) because the fact t hat ds = 0 in t he adiabatic process as the dq=O in t he adiabatic process so change in t he ent ropy is zero so t he entropy is const ant where as the isothermal process is represented as t he straight line parallel to s curve. so as a whole the Carnot cycle is represented by a rectangle. so this two is representing nothing but two different Carnot cycle op erate in b etween 4 different t emperature.let each of the vertical line represent 1 kelvin and each of the pa1·allel line represent 1 joule/kelvin so the a cycle is operating between 1 kelvin and 2 kelvin temperature ~nn thP R r,,r lP i ~ nnPr~tinO"
h P t .,~TPP TI
thP
~
k Pl,,in ~nn thP 4
11 oo : 01 : oo different Carnot cycle operate in b etween 4 different t emperature.let each of the vertical line represent 1 kelvin and each of the parallel line represent 1 joule/kelvin so the a cycle is operating between 1 kelvin and 2 kelvin temperature and t he B cycle is operating between the 3 kelvin and t he 4 kelvin.so the efficiency of A Cycle and B cycle is given by rJ A
T2
= (1 - Ti ) = 1 - ( 1/ 2) = 1/ 2
T2 represents t he lower t emp and the T1 is representing the higher t emp. now in similar way 'r/B
= 1/ 4
'r/A
=
so we get that
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2r;B
22
physicsguide CSIR NET ) GATE
Second Law of Thermodynaics
@Sk J ahiruddin ) 2020
Sol 1.4. It is actually representing a Carnot cycle but now in the p-v diagram. I ➔ 2 represent isothermal process so
temperature T2 is constant t hrough out the process and we have V2 Pl P2 V1 but t he second one is representing (2 ➔ 3) an adiabatic process so we have
p3 P2
where we have use the concept t hat as t he gas is going t hrough t he adiabatic expansion t he temperature also changes
11 oo : 01
: 03 p3 P2
where we have use the concept t hat as t he gas is going t hrough the adiabatic expansion t he temperature also changes and goes t o T 3 . again t he 3 ➔ 4 represents isothermal compression hence we get p3
V4
p4
V3
and t he final process we get again
so from the second condit ion we get t hat option 1 is correct. Sol 1.5. Efficiency of t he Carnot's cycle = [email protected]
@Sk J ahiruddin, 2020
23
4
1- -
6
=
1/ 3
physicsguide CSIR NET, GAT E
Second Law of Thermodynaics
we know that
work output rJ = heat input so work out put is lOOOjoule so we get heat input per cycle is 3000 Joule. Sol 1. 6. The Carnot cycle consisting of t he two isothermal and two adiabatic process and isotherm process are repre-
sents in t h T-S diagram through a straight line parallel t o s and adiabatic process are represents by the parallel to T line. the first process in the Carnot cycle is represented by t he isot hermal expansion so t he volume of the system will
11 oo : 01 : os and two adiabatic process and isotherm process are represents in t h T-S diagram through a straight line parallel to s and adiabatic process are represents by the parallel to T line. the first process in the Carnot cycle is represented by t he isothermal expansion so the volume of the syst em will increase and hence the entropy of the system must increase and heat will be taken inside the system hence the option b is correct because t he first process represents t he entropy increase. •
Sol 1. 7. A refrigerator is working on the basic principle of t he fact that a heat engine is now taking heat from the cold source and rejecting heat in the hot source by amount of some work doing on it externally. so let a heat engine is taking is t aking Q 2 amount of heat from the cold source and by doing Q1 - Q2 amount of work on it externally rejecting Q1 ( > Q2 ) amount of the heat to the hot source.so hot source is getting hotter and cold sink is getting colder. in this case we define a paramet er called coefficient of p erformance and j [email protected]
24
physicsguide CSIR NET, GATE
Second Law of Therrnodynaics
@Sk J ahiruddin , 2020
is denoted by/3 and defined as
/3 = heat t aken from the cold source v.rrokdone perfromed so we can say th at where w denotes the workdone on the syst em and w = Q 1 Q 2 so the heat reject ed in t he hot source can be defined as
11 oo :01 : oa so we can say where w denotes the workdone on the syst em and w = Q 1 Q 2 so the heat rejected in the hot source can be defined as Q1 = w( l
+ /3)
, on the basis of t hese we can say that here
where we use the fact that Q1
T1
Q2
T2
and hence we have t he relation between
/3
and 'T/ is given by
1 rJ=
l +/3
so here i can calculate the eta and that is
/3 =
22 295
and hence
~;~ and we know that heat t aken from t he cold source
is Q 2 = W /3 = 4l 7kj / s [email protected]
@Sk J ahiruddin, 2020
25
physicsguide CSIR NET, GATE
Second Law of Thermodynaics
Sol 1.8 . Entropy of the universe increases only in t he case of irreversible process and for the reversible process it is
equal to zero so here for the cycle being consisting of the configuration of the Carnot cycle so the entropy change must be equal to zero. Sol 1 .9. In a Carnot cycle t here is only one process where
11 oo : 01
: 11
equal to zero so here for the cycle being consisting of the configuration of the Carnot cycle so t he ent ropy change must be equal to zero. Sol 1.9. In a Carnot cycle there is only one process where heat is entering the system and that is isothermal expansion
of the working substance and in the isothermal process the temperature of the system is constant so only temperature is constant when heat is added to the system hence the option c is correct. Sol 1.10. From the previous discussion we came to know t hat the carnoot cycle's efficiency is not depending upon t he substance you are using in t he system it is depending
on the temperature of the hot and sink temperature hence T1 77 = l - T2 as T1< T2 Sol 1.11. Irreversible process is always accompanied with positive change in th entropy so we must have dS > 0 for the irreversible process and for the reve1~sible one we must
have dS = 0 . This summerise the second law of the thermodynamics t hat entropy must be greater or equal to zero. [email protected]
@Sk J ahiruddin ,
2020
26
physicsguide CSIR NET, GATE
Second Law of Thermodynaics
The first case t ell you t he process is reversible as well adiabatic where as the case of second one the process in irreversible and adiabatic, the first one is isoentropic process but the second is definitely not an isoent ropic process. so t he t emperature in t he second one would be greater than
11 oo : 01
: 13
The first case tell you t he process is reversible as well adiabatic where as the case of second one the process in irreversible and adiabatic, t he first one is isoent ropic process but the second is definitely not an isoentropic process. so t he temperature in t he second one would be greater t han t han previous one. Sol 1.12. As the system is working between two different temperature so t he system will keep working until the two
different temperature are coming in the equilibrium. say t hat the equilibrium temperature is T so the hot source will loss it 's entropy and t he cold source will gain t he entropy until t he change in t he entropy of t he whole syst em comes to zero so we get the change in t he entropy of the hot source is given by dQ dS= T ( the change in the ent ropy is counted as t he ratio of the heat taken or given and at the particular temperature it is given or taken)so the entropy change of the hot body is entropy change of the hot body
j [email protected]
@Sk J ahiruddin, 2020
27
=
physicsguide CSIR NET 1 GATE
Second Law of Thermodynaics
and t he change of the entropy of the colder one is entropy change of the cold body
=
Ti
CpdT
r
11 oo : 01
: 16
@Sk J ahiruddin, 2020
Second Law of Therrnodynaics
and t he change of the entropy of the colder one is
entropy change of the cold body
=
so the total change of the system is given by Change of t he ent ropy of t he hot body + change of t he cold body entropy = total change of the ent ropy of the system= O. Using t his principle we get t hat
T1 Cp loge T1 =
T1 Cp loge T2
which gives us that final t emperature is
T1 =
T2T2
so now putting all the values of T1 and T2 i get Ti = 347.2K
Sol 1.13. From the figure we can say t hat in the adiabatic process P2V1'
= Pl 3v1 'Y
but for the monoatomic gas the ry = Pl P2
=
i now we will have
3-5/3 = 0.16.
Sol 1.14. We know that the relation between the coefficient of performance and efficiency is given by 1 rJ= [email protected]
l + ,8 28
@Sk J ahiruddin, 2020
and we can easily calculate the
physicsguide CSIR NET, GATE
Second Law of Thermodynaics
11 oo :01
: 1a
@Sk J ahiruddin, 2020
Second Law of Thermodynaics
and we can easily calculate the
+ 18 = 273 + 27
= l _ 273
rJ
0.03
so the coefficient of performance
/3
1 = 0.03 - 1 = 32.33
now the heat must be taken out to decrease t he temperature is given by CpdT = 4.2 x (1) = 4.2kJ and from the principles of refrigerator we know t hat heat has been taken out from the cold source is q2
= w/3
so t he workdone must be 0.14KJ. Sol 1.15. From t he adiabetic process we come to know t hat
Tova, - l = T3v0, - 1 so we get that 2 3
1 T 3 To Sol 1.16. First the engine goes from say A -+ B it is an isothermal process but as t he ent ropy decreases so we can j [email protected]
@Sk J ahiruddin, 2020
29
physicsguide CSIR NET, GATE
Second Law of Thermodynaics
11 oo : 01 [email protected]
: 21
29
physicsguide CSIR NET, GATE
Second Law of Thermodynaics
@Sk J ahiruddin , 2020
say that this co responds to release of heat t hat means it is an isothermal compression so in pv diagram volume must decreases.then B -+ c is an adiabatic process but again it corresponds to the adiabatic compression so volume again decreases. now c -+ D corresponds to an isothermal expansion as t he ent ropy increasing so in pv diagram volume must increase now finally D -+ A is an isochoric process so in pv diagram is a straight line parallel to P as the volume is unchanged so the best represented curve is option a. Sol 1 .17. (a) according to the given graph
W AB
= 0 (IS-
COCHORIC PROCESS) so we have that
we know that Pv
= n RT T = pv
' nR so t he workdone in t he AB process can be written down
and during BC process heat input is zero as it is a adiabatic process and we have dT
[email protected]
=
30
-cv(P2V1 - p1v2) R physicsguide CSIR NET, GATE
11 oo : 01
: 24
dT = -cv(P2V1 - p1v2) R j [email protected]
physicsguide CSIR NET 1 GATE
30
Second Law of Thermodynaics
@Sk J ahiruddin , 2020
and lastly
~QcA=ncp
Cp
dT = R (p1v1- p1 v2
)
and t he work done
we get the efficiency total wor kdone rJ = heat absorped WAB + WBc + WcA ~QAB + ~QBc + ~QcA hence we have all the elements calculated before putting all we get t he exact result. (b)
- 0
A ..........____,. : - ~ C
Cp
11 oo : 01
: 26
Cp
31
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physicsguide CSIR NET, GATE
Second Law of Thermodynaics
@Sk J ahiruddin , 2020
change in the entropy of system is given as !:lS
= !:lQ T
now for A to B process i:lSAB =
fraccvdTT
so we have from this
similarly sor t he C to A process we will have
T =
kl e(s/ cp)
but we know that Cp > Cv so we have the slope of one is different from the other and curt of AB is more steeper t han t hat of t he AC. So t he figure is given above in the sT diagram is justified. Sol 1.18. The gas has been expanded to two times of it 's initial volume at const ant temperature T0 so heat inserted Q
=
dQin
=
RToln 2
t hen t he gas has been compressed by t he isobaric process so t he heat rejected
11 oo : 01
: 29
initial volume at const ant temperature T0 so heat inserted Q = dQin = RToln 2
t hen the gas has been compressed by the isobaric process so the heat rejected dqout
=-
3 To Po R + (-Vo ) 2 2 2 32
j [email protected]
physicsguide CSIR NET I GATE
Second Law of Thermodynaics
@Sk Jahiruddin, 2020
here the negative sign denotes that heat is going out . Again for isochoric process t he gas is gaini11g heat
as t he temperature of the gas is increasing.so efficiency 1] = 1 -
which can be resolved 3RTo+2Povo
'l]
= 1-
4
4Q+3RTo 4
so we further simplify and get 4Q- 2RTo 'lJ = 4Q + 3RTo
where we use the ideal gas equation PoVo = RT0 Sol 1.19. (a) As for t he path Pa
a ➔
Pb
b p ex: v so we have
11 oo : 01
: 31
where we use the ideal gas equation PoVo = RT0 Sol 1.19. (a) As for t he path a~ b p ex: v so we have
so we have
l OOkP a l m3
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200kPa vb 33
@Sk J ahiruddin ) 2020
physicsguide CSIR NET ) GATE
Second Law of Thermodynaics
and hence we get
now let us take the process AB and analyses it. for this we take p = a v and we need to know that t he a is to evaluate t he a we take init ial condition that when pressure is 100 Kpa volume 1 cubic meter and t his give us a = lOOK P a/m3 and t he work done in the process is 2
2
avdv = 50K J
pdv = 1
1
but during t he BC the work done is zero as the process is isochoric so no volume change so work done is zero. during t he CA process it is the isobaric process so work done WcA =
Pa(V a
-
vb) = 100(1 - 2)kJ = -100 K J
(b) For the process AB t he change in the entropy is given by
11 oo : 01 WcA =
Pa(Va - vb)
=
: 34 100(1 - 2)k J
=
- 100 K J
(b) For t he process AB t he change in the entropy is given by ~QAB
T
= ncv
now from the ideal gas equation we get
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Second Law of Thermodynaics
@Sk J ahiruddin, 2020
sow ecan use t his formulae to solve this chnageinentropy
=
so applying all the values we get change is entropy during AB process is given by
100 x 103 x ln(4 ) x 3
l
+
1 1.5 - 1
= 2ln2kj /k
in similar way for t he second means BC process change in t he ent ropy is ncv
T as we know that in the isochoric process no work is done so total heat is been used up to enhance the internal energy of t he system. and hence we get T2
rn R
?1 A
')
11 oo : 01
: 36
as we know t hat in t he isochoric process no work is done so total heat is been used up to enhance t he internal energy of t he system. and hence we get
T3 nR Pc nevln - = - -ln T2 ry - 1 Pb
= -
2 - x ln2kj / k 3
now we come to t he last process that is CA process ao the entropy change in t he CA process is given by
llQcA
=
ncp
dT T3T1 T = n R
1 1 + -1' - 1
In Va Vb
= - ln2 KJ/ k 35
j [email protected]
physicsguide CSIR NET, GATE
Second Law of Therrnodynaics
@Sk J ahiruddin , 2020
Sol 1.20. According to the PV curve we have a to b is an adiabatic process so we will have 5/ 3
V1
Pi Pl
Vi
for t he b to c process as it is a iso thermal process we have V2 V1
Pl P2
and for t he cd process being adiabatic again we have Vi
5/3
P2 1 B 3pi
now multiplying t he relation of a to b process by the c to d process we get
11 oo : 01
: 39 1
B 3pi
now multiplying the relation of a to b process by the c to d process we get 1 V2 B2 now we are coming to t he second part of calculating t he t otal workdone
W=w(ab)+w(bc)+w(cd) Now we have
w(ab) = p;v;2/:1v1 t he work done in the be process
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physicsguide CSIR NET , GATE
@Sk J a hiruddin, 2020
Second Law of Thermodynaics
and t he work done is the cd process is given by
so total work done can be written as
here we use the fact that p 1v1 = p 2 v2 as we get in t he isothermal process.but we know t hat PJVJ = Bpivi putting this in t he equation we get
11 oo : 01
: 42
mal process.but we know that PJVJ = Bpivi putting this in t he equation we get
Sol 1. 21. Using t he ideal gas equation ,pv one mole gas .so
= n RT for the
and t he Tk
=
PkVk
= 580K
R and the other two temperatures are using same technique is given by TL = 193.33K, TM = 97 K now we come to know t he fact that area under the curve is going to given you the total wor kdone in the cycle as t hey both have same
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37
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Second Law of Thermodynaics
@Sk J ahiruddin , 2020
dimension but only in pv curve the area under curve is equal to the work dome so the total work done areaaunderJKLM = 0.07 x (80 x 103 ) = 5600Joule
change in the internal energy during t he cycle ~Q
=
0 + w = 5.65
as t he system returns to the init ial state so t he total internal change in the energy is zero.heat associated with each path can be calculated as follows
11 oo : 01 ~Q
: 44
= 0
+w =
5.65
as t he syst em returns to the init ial state so the total internal change in the energy is zero .heat associated with each pat h can be calculated as follows Cv OTJK
3R = 2
X
(580 - 290)
again for KL process
and we again have
and for ~ Q MJCpTJ - T M=
4.03kJ
so heat absorbed in the cycle
~Qa = (3.6 + 4.03)K J j [email protected]
=
7.63K J
physicsguide CSIR NET 1 GATE
38
Second Law of Thermodynaics
@Sk J ahiruddin, 2020
and heat reject ed
~Qr = 1.2 + 8.02
=
9.22kJ
=
0.736
so efficiency of the cycle rJ =
W ~Qa
=
5.6 _ 76
•
,
.
...
..
...
..
..,,
...
11 oo : 01
: 47
W
5.6 _ 76
rJ = !:::,.Qa =
0.736
=
•
Sol 1.22. Let us consider a hollow cylinder which is devided in to n equal compartments.final equilibrium temperature of t he gas in the compartment 2 to n is 4T0 , for the adiabetic expansion of the first compartment is given by
PoVo 1 = pv 1 for the adiabet ic expansion of t he remaining (n-l)cells we can write 1 1 (n - l )Po - T01 = p 1- -r 4T 01 and here we know t hat 5 ry = 3
and we get that p
= 32Po(n - 1)
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39
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@Sk J ahiruddin , 2020
Second Law of Thermodynaics
now we are using this pressure in the first equation we get
Pavo1 = 32P0 (n - l )v1 and hence we get the volume V
Vo = 3 8(n - l ) s
11 oo : 01 ovo
: 49
= 32Pa
n - l v
and hence we get the volume
v=
Vo 3
8(n - 1)5
Sol 1.23. AB is the isothermal expansion. work done from A to B process W AB =
V2
n RTln -
=
l x 8.31 x 800ln2
=
4608j
V1
so t his is t he heat absorbed in the AB process , now let us calculate t he heat absorbed in the BC processes QAB
= CpdT = - 10387.5]
as the process makes t he compression so t he heal is being rejected so there is a negative sign. Now during t he AB process t he total work done in t he process is zero because it is an isochoric process so heat absorbed is equal to increase in t he internal energy. so
dU
= qisochoric =
Cv(6T ) =
3
R (800 - 300)
2
=
6232.5]
so net heat absorbed
Qr= 4608 + (- 10387.5) j [email protected]
@Sk J ahiruddin , 2020
40
+ 6232.5 = 453J physicsguide CSIR NET I GATE
Second Law of Thermodynaics
Sol 1.24. Change in t he ent ropy of system is given as
now for A to B process
11 oo : 01 : s2 Qr
=
4608 + (- 10387.5)
j ahir@physicsguide. in
40
+ 6232.5 =
453J
physicsguide CSIR NET, GATE
Second Law of Thermodynaics
@Sk J ahiruddin , 2020
Sol 1.24. Change in t he ent ropy of system is given as
now for A to B process CvdT
T so we have from this
similarly for the C to A process we will have
but we know t hat Cp > Cv so we have the slope of one is different from t he other and curt of AB is more steeper t han t hat of t he AC. So the figure a is given above in t he TS diagram is justified.
j ahir@physicsguide. in
41
physicsguide CSIR NET, GATE
11 oo : 01 Qr
=
: 55
4608 + (- 10387.5)
j ahir@physicsguide. in
40
+ 6232.5 =
453J
physicsguide CSIR NET, GATE
Second Law of Thermodynaics
@Sk J ahiruddin , 2020
Sol 1.24. Change in t he ent ropy of system is given as
now for A to B process CvdT
T so we have from this
similarly for the C to A process we will have
but we know t hat Cp > Cv so we have the slope of one is different from t he other and curt of AB is more steeper t han t hat of t he AC. So the figure a is given above in t he TS diagram is justified.
j ahir@physicsguide. in
41
physicsguide CSIR NET, GATE
11 oo :01 : s1 Qr
=
4608 + (- 10387.5)
j ahir@physicsguide. in
40
+ 6232.5 =
453J
physicsguide CSIR NET, GATE
Second Law of Thermodynaics
@Sk J ahiruddin , 2020
Sol 1.24. Change in t he ent ropy of system is given as
now for A to B process CvdT
T so we have from this
similarly for the C to A process we will have
but we know t hat Cp > Cv so we have the slope of one is different from t he other and curt of AB is more steeper t han t hat of t he AC. So the figure a is given above in t he TS diagram is justified.
j ahir@physicsguide. in
41
physicsguide CSIR NET, GATE
11 oo : oo : 01
Problems and Solutions in Max-well's relation and Thermodynamic Potentials Sk J ahiruddin * Suchismito Chattopadhyay
*Assistant Professor Sister Nibedita Govt. College, Kolkata Aut hor was t he topper of IIT Bombay M.Sc Physics 2009-2011 batch He ranked 007 in IIT J AM 2009 and 008 (J RF) in CSIR NET J une 2011 He has been teaching CSIR NET aspirants since 2012
1
@Sk J ahiruddin , 2020
Nlaxwell's relation and Thermodynamic Potent ials
11 oo : oo : 03 1
@Sk J ahiruddin, 2020
Maxwell's relation and Thermodynamic Potentials
Contents 1
Problems from NET, GATE, JEST , TIFR & JAM pap ers 1.1 Free expansion of Gas 1.2 Ans keys . • • • 1.3 Solut ions . •
j [email protected]
•
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•
•
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•
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12 13
physicsguide CSIR NET, GATE
11 oo : oo : 06
j ahir@physicsguide. in
@Sk J ahiruddin , 2020
1
physicsguide CSIR NET I GATE
2
Maxwell's relation and Thermodynamic Potent ials
Problems from NET, GATE, JEST, TIFR & JAM papers
Prob 1.1. A thermodynamic system is maintained at a con-
stant t emperature and pressure. In thermodynamic equilib(JAM 2008] rium, its (a) Gibbs free energy is minimum (b) ent halpy is maximum (c) Helmholtz free energy is minimum (d) Int ernal energy is zero Prob 1.2. If U,F ,H and G represent internal energy, Helmholtz free energy, ent halpy and Gibbs free energy respectively,then which of t he following is a correct thermodynamic relation? [JAM 2016]
(a) dU
=
P dV - T dS
(c)dF = - P dV
(b )dH
+ SdT
=
V dP
(d)dG = V dP
+ T dS + sdT
Prob 1. 3. W hich among t he following sets of Maxwell relat ions is correct? (U-internal energy,H -ent halpy,A-Helmholt z
free energy and G-Gibbs free energy) (a) T
=
(Zi)8 and P =
(i~ )v
(b) V
=
(i ~) 8
(~~)P
and T
=
[GATE 2010]
11 oo :oo :oa
(a) T
=
(b) V =
(ii)8 and P =
(i~)v
(i ~) 8
(~~)P
and T =
[email protected]
3
@Sk J ahiruddin, 2020
(c) P = (d) P
=
physicsguide CSIR NET, GATE
l\/[axwell's relation and Thermodynamic Potentials
-(ii )T and V = (ii)s - (~1)T and S = (it )v
Prob 1.4. The internal energy of n moles of a gas s given by E = ~nRT where V is te volume of t he gas at temperature T and a is a positive constant. One mole of t he gas in state (T1 , Vi ) is allowed to expand adiabatically into vaccum to a final state (T2 , ½). The t emperature T2 is [GATE 2006]
t,
(a) T1 + Ra( J2 +
(b) T1 - ~ Ra (
J 1
)
J J 2
-
1
)
..l..) (c) T1 + ~3 Ra(..l.. V2 V1
Prob 1.5. The equat ion of state of one mole of a van der Walls gas is P
+;
2
(V - b)
=
RT and its internal energy U (T , V )
;2 ,
is given by U(T, V) = Uo + CvT where Uo and Cv can be t aken as constants. [JAM 2009] (a) prove that in a reversible adiabatic process t he t emperature and volume satisfy the equation T (V - b)Rf Cv =
11 oo : oo : 11 is given by U (T , V ) = Uo + CvT - {:2 , where Uo and Cv can be taken as constants. [JAM 2009] (a) prove t hat in a reversible adiabatic process t he t emperature and volume satisfy t he equation T (V - b)Rf Cv = constant. [email protected]
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@Sk J ahiruddin , 2020
Nlaxwell's relation and Thermodynamic Potentials
(b) Calculat e t he change in entropy of the gas when it undergoes a reversible isot hermal expansion from volume V0 to 2Vo. Prob 1.6. The relation bet ween the internal energy U, ent ropy S, temperature T , pressure p , volume V, chemical potential µ and number of particles N of a t hermodynamic system is dU = T dS - pdV + µdN. T hat Uis an exact different ial implies t hat [NET Dec 2017] ap (a) - as
V,N
au
(c) Par
s,N
=
aT av l s ,N
=
l au - r av
s ,µ
au
(b) par
au
l s,J\T
ap (d) - as
= s av
V,N
=
aT av
S,µ
S,N
Prob 1. 7. A t hermodynamic function G(T , P, N) = U T S + PV is given in t erms of t he internal energy U, Temperature T , ent ropy S, pressure P, volume V, and t he number of particles N. Which of t he following relations is t rue? (µ is chemical potential) [NET June 2017] aT
(c) V = -
N,P
ac
N,P
(d) µ = -
aP
N .T
ac aN
P.r
11 oo : oo : 14 is chemical potential) fJG (a) S = fJT N,P (c )V= -
[NET June 2017] fJG fJT
(b) S =
fJG fJP
(d) µ = N,T
N ,P
fJG f)N
PT
'
Prob 1.8. Starting with the equation T DS
= dU + pdV
and using t he appropriate Maxwell's relation along with the expression for heat capacity Gp (see useful information), the j [email protected]
@Sk J ahiruddin , 2020
derivat ive
5
physicsguide CSIR NET, GATE
Maxwell's relation and Thermodynamic Potentials
op
for a substance can be expressed in terms fJT s of its specific heat cp, density p , coefficient of volume expansion /3 and temperatu1~e T. 3 3 For ice cp = 2010 J/ Kg - K , p = 10 km/m and /3 = 1.6 x 10-4 /° K. If t he value of
at 270 K is N x 107 Pa/ K ,
: s
t hen t he value of N is ..... . Specify your answer to two digits after the decimal point [JAM 2017] Prob 1.9. For temperat ure T1 > T 2
the qualitative temperature dependence of t he probability dist ribution F( v) of t he speed v of a molecule in t hree dimensions is correctly represented by t he following figures [NET June 2013] ,
00 : 00 : 16
[email protected]
@Sk J ahiruddin, 2020
6
physicsguide CSIR NET, GATE
Nlaxwell's relation and Thermodynamic Potentials
Prob 1.10. For a system at constant temperature and volume, which of the following statements is correct at equilib[GATE 2016] rium? The Helmholtz free energy attains a local minimum The Helmholtz free energy attains a local maximum
11 oo : oo : 19 ume, which of the following statements is correct at equilib[GATE 2016] rium? The Helmholtz free energy attains a local minimum T he Helmholtz free energy attains a local maximum The Gibbs free energy attains a local minimum The Gibbs free energy attains a local maximum
Prob 1.11. The free energy of a photon gas enclosed in a volume V is given by F = -½aT4 where a is a constant and T is the temperature of the gas. The chemical potent ial of t he photon gas is [GATE 2006] (a) 0 (b) ;avT 3 (c) ! aT 4 (d) aVT3 [email protected]
physicsguide CSIR NET, GATE
7
Maxwell's relation and Thermodynamic Potentials
@Sk J ahiruddin , 2020
Prob 1.12. MSQ: A thermodynamic system is described by t he P, V, T coordinates. Choose t he valid expression(s) for the system. [JAM 2019]
(A)
(gC)r (i~ )p = - (g; )v
(gC)r (ii) P = (i; )v
(B)
Prob 1.13. MSQ: Which of the following relations is (are) t rue for t hermodynamic variables? [JAM 2018]
(a) T dS = CvdT
8P
+ T BT
dV V
(b) T dS = CpdT -T
(c) dF = -SdT + PdV (d) dG = - SdT + VdP .. - . ., .....
~
av 8T
dP p
,.,
"
_,
,
11 oo : oo : 21 av oT
(b) TdS = CpdT - T
dP p
(c) dF = -SdT + PdV (d) dG = -SdT + VdP Prob 1.14. Consider an ensemble of t hermodynamic sys-
tems each of which is characterized by the same number of particles, pressure and temperature. The t hermodynamic [JAM 2018] function describing the ensemble is (a) Enthalpy (b) Helmholtz free energy (d) Entropy (c) Gibbs free energy Prob 1.15. A real gas has specific volume v at temperat ure T. Its coefficient of volume expansion and isothermal
compressibility are a and kr, respectively. Its molar spej [email protected]
@Sk J ahiruddin, 2020
8
physicsguide CSIR NET, GATE
:Niaxwell's relation and Thermodynamic Potentials
cific heat at constant pressure C p and molar specific heat at constant volume Cv are related as [JAM 2014] T va (a)Cp = Cv + R (b )Cp = Cv + kr
Tva 2 (c)Cp = Cv + kr
(d)Cp = Cv
Prob 1.16. A thermally insulated ideal gas of volume ½
and temperature T expands to another enclosure of volume ½through a prous plug. What is t he change in the temperature of the gas? [JEST 2012] (a) 0 (b) Tln(½ /½ ) (c) Tln(V2 /½)
(d) T In [(½ - V1) / V2]
11 oo : oo : 24 ature of t he gas? (a) 0 (b) Tln(½ / V2 )
[JEST 2012]
(c) Tln(½/½)
(d) T ln [(½ - V1) / V2]
1.1
Free expansion of Gas
Prob 1.17. During free expansion of an ideal gas under adiabatic condition, the internal energy of t he gas [JAM 2019] (A) Decreases (B) Initially decreases and then increases (C) Increases (D) Remains constant
Prob 1.18. MSQAn isolated box is divided into two equal compartments by a partition (see figure) . One compartment [email protected]
@Sk J ahiruddin, 2020
9
physicsguide CSIR NET, GATE
Maxwell's relation and Thermodynamic Potentials
contains a van der Waals gas while the other compartment is empty. The partition between the two compartments is now removed. After the gas has filled the entire box and equilibrium has been achieved, which of the following statement (s) is (are) correct? [JAM 2017] (a) Internal energy of t he gas has not changed (b) Internal energy of the gas has decreased (c) Temperature of t he gas has increased (d) Temperature of the gas has decreased
Prob 1.19. A frictionless, heat conducting piston of neglio-ihlP m::-i.~~ ::in n hP::-it. r::-in::-irit,r ni,rin P~ ::-i '\TPrtir::-il in ~,, 1:::-it.Pn
11 oo : oo : 21 (b) Internal energy of the gas has decreased (c) Temperature of t he gas has increased (d) Temperature of the gas has decreased Prob 1.19. A frictionless, heat conducting piston of neg-
ligible mass and heat capacity divides a vertical, insulat ed cylinder of height 2H and cross sectional area A into two halves. Each half contains one mole of an ideal gas at temperature T0 and presure P0 corresponding to STP. The heat capacity ratio , = Gp/ Cv is given. A load of weight W is tied to t he piston and suddenly released. After t he system comes to equilibrium, t he piston is at rest and the temperat ures of the gases in t he two compartments are equal. What is t he final displacement y of t he piston from its initial position, assuming yW > ToCv? [JEST 2018] (A) 2H (D) 2H (C) H (B) H , ~ ~ Prob 1.20. A thermally-insulated container of volumeV0 is
'
j [email protected]
10
@Sk J ahiruddin , 2020
physicsguide CSIR NET I GATE
Maxwell's relation and Thermodynamic Potentials
divided into two equal halves by a non-permeable partition. A real gas with equation of st ate 3
b
a2 p+ V 3
= n RT
where a and b are constants,is confined to one of these halves at a t emperature T0 . The partition is now removed suddenly and t he gas is allowed to expand to fill the entire container. The final t emperature of the gas, in t erms of its specific
11 oo : oo : 29 where a and bare constants,is confined to one of these halves at a temperature T0 . T he part it ion is now removed suddenly and t he gas is allowed to expand to fill t he ent ire container. T he final t emperature of the gas, in terms of its sp ecific heatCv , will be [TIFR 2014] 3a2 2a2
(a)To - 2Cv¼2 3a
(b)To - 3Cv¼2
2
(c) To + 2CvV02
2a
(d)To + 3Cv½2
[email protected]
@Sk J ahiruddin, 2020
1.2
2
11
physicsguide CSIR NET, GATE
l\/[axwell's relation and Thermodynamic Potent ials
Ans keys Ans keys
1.1. a
1.8. 4.65
1.15.
C
1.2. b
1.9. a
1.16.
C
1.3. b
1.10. a
1.17. d
11 oo : oo : 32 1.1. a
1.8. 4.65
1.15.
C
1.2. b
1.9. a
1.16.
C
1.3. b
1.10. a
1.4.
1.11. a
C
1.17. d 1.18. a,d
1. 5. Do yourself
1.12. a .c
1.6. a
1.13. b ,d
1.19.
1.7. a
1.14.
1.20. a
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1.3
J
C
12
C
physicsguide CSIR NET, GATE
Nlaxwell's relation and Thermodynamic Potentials
Solutions
Sol 1.1. The Gibbs free energy is defined as
G == U +PV-TS
11 oo : oo : 34 Sol 1.1. The Gibbs free energy is defined as
G == U + PV - TS so we can say
dG == dU
+ PdV + VdP -
TdS - SdT
but we actually know that from t he first law of thermodynamics that dU + PdV == dQ == TdS so putt ing t his in the equation we get
dG == VdP - SdT but at constant pressure and temperature dP and dT is zero so Gibbs free energy should have t he minimum. because at t he constant t emperature and pressure the change in the Gibbs energy is zero that is it remains unchanged. Sol 1.2. The enthalpy function is defined as
H == U + PV so
dH = dU + PdV + VdP j [email protected]
@Sk J ahiruddin , 2020
13
physicsguide CSIR NET, GATE
Maxwell's relation and Thermodynamic Potentials
but first law of t hermodynamics is telling us that
dQ = dU
+ PdV == TdS
00: 00: 37 Maxwell's relation and Thermodynamic Potentials
but first law of t hermodynamics is telling us that dQ = dU
+ P dV =
T dS
and putting this in the equation we get dH = V dP
+ TdS
Sol 1.3. We get to know t hat dH
= VdP+TdS
So from this we can say that is p is constant then
and if we say t hat s is constant t hen ds=O so we are getting
so option b is correct . Sol 1.4. For the adiabatic system we always have that the heat received or heat rejected is zero that is no heat can enter t he system and no heat can go away from the system
so we have dU
= -dw
where dU is t he change in t he internal energy and U is the internal energy here U is equal to E as E is defined as t he [email protected]
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14
physicsguide CSIR NET, GATE
Nlaxwell's relation and Thermodynamic Potentials
internal energy and hence dE is t he change i n the internal
11 oo : oo : 40
@Sk J ahiruddin, 2020
l\/[axwell's relation and Thermodynamic Potentials
internal energy and hence dE is t he change i n the internal energy.so we have 3
dw == -dE ==
a nRdT + v 2 dv
2
now to get the total work done in the free expansion is zero so we have T2 3 -nRdT + T1 2 and t his gives us that 2
1
1
T2 == Ti + - Ra(- - -) 3 ½ V1 . (in t he free expansion there is no work done because t here is no external pressure. a free expansion is an irreversible process and as t he vacuum has no pressure so a gas is allowed to expand in vacuum is essentially a free expansion as t he gas has to do no work against t he ext ernal pressure as p ==
0). Sol 1.5. (a) during an isoentropic process t here are no dissipative effects and the system neither absorbs nor give off t he heat. for t his reason the isoentropic process is called t he reversible adiabatic process. now t he given equation of state is a P + V 2 (V - b) == RT [email protected]
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15
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Maxwell's relation and Thermodynamic Potentials
11 oo : oo : 42 15
[email protected]
@Sk J ahiruddin, 2020
physicsguide CSIR NET, GATE
Nlaxwell's relation and T hermodynamic Potent ials
from t his we can say that
P=
RT V-b
a
v2
now we come to know t hat from the first law of thermodynamics dQ = dU + dW and it is given by that a U(T, V) = Uo + CvT - V 2
so the change in t he internal energy is given by
and t hat is why we can say t hat
dQ = evdT +
RT V _ bdV
but due to isoentropic process dQ= O and hence we get
dT Cv T
+
R dV Cv
= O
V- b
so we get
so we get
T (V - b)Rf cv = constant j [email protected]
16
physicsguide CSIR NET, GATE
11 oo : oo : 4s so we ge
T(V - b)Rf cv = constant j [email protected]
16
@Sk J ahiruddin , 2020
physicsguide CSIR NET, GATE
Maxwell's relation and Thermody namic Potentials
(b) from the first law of thermodynamics
dQ
=
dU + dW
=
RT cvdT + V _ bdV
and we know that
dQ = TdS so using this we get
dS = CvdT T
RdV + V-b
but now we are considering the isothermal process so we must get dT=O and this condition will give us change in the entropy 2 0 V RdV 2½ - b Vo V - b = Rloge Vo - b Sol 1.6. From t he relation i can conclude t hat
au = T OS
'
au
av
=- P
'
now we take the first one and we take
BT
av but it also can be written as
au f)N
=µ
11 oo : oo : 47 avas
av
but it also can be written as 2
a u 8s8V [email protected]
aP 8S physicsguide CSIR NET, GATE
17
@Sk J ahiruddin, 2020
Nlaxwell's relation and Thermodynamic Potentials
and we know that
so the option a is correct. Sol 1. 7. Given that
G(T, P, N)
= U - TS + PV
we can say that dG
aG aG 8G aTdT + aPdP + aNdN
= = dU - TdS - SdT + PdV + VdP
and we know from t he first law of the t hermodynamics dQ Tds = dU + PdV and we are using this and get dG = -SdT
=
+ VdP
now we compare the co efficient of dT on both side and get
S =-
8G
BT
N, P
Sol 1.8. So we have if entropy as a function of pressure and
temperature then
11 oo : oo : so ac aT
N ,P
Sol 1.8. So we have if ent ropy as a function of pressure and temperature t hen
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18
Maxwell's relation and Thermodynamic Potentials
@Sk J ahiruddin , 2020
now multiply both side by T we get T dS = T
as aP
dP + T T
as aT
dT p
so we get the fact dS = TV f3dP
+ epdT
so here we put t he fact
/3 =
! aV v aT now we are to calculate in const ant entropy t he term value
ap aT
Cp
TV/3
s
= 4.65
X
107
so N = 4.65 putting all the values of the given quantities. Sol 1. 9. T he area under t he curve is always denoting t he total number of the part icles and t he most probable velocity of the particle is proport ional to JT so more temperature gets the curve more steeper. So option a is correct. Sol 1.10. T he Helmholtz free function is defined as V
-
TT
rrc
11 oo : oo : s2 of the particle is proportional to JT so more temperature gets the curve more steeper. So option a is correct. Sol 1.10. T he Helmholtz free function is defined as
F = U-TS So the change in the Helmholtz free function is given by
dF = dU - TdS - SdT j [email protected]
19
@Sk J ahiruddin , 2020
physicsguide CSIR NET, GATE
:Niaxwell's relation and Thermodynamic Potentials
but we know that from t he first law of t hermodynamics
dQ
=
dU + P dV
=
T dS
and putting this in the equation we
dF = - P dV - SdT so keeping volume and temperature is constant will give us no change in the Helmholtz free energy and hence Helmholtz free function attains a local minimum. Sol 1.11. Chemical potential is t he change in t he internal
energy due to add or subtract of particle but in a system t he total number of particles is not conserved so chemical potential is zero. Sol 1.12. We have t hat for an ideal gas we get
av
8P
8T
8P r
8T v
av
p
=-1
so from this relation we get to know that A and C is correct
11 oo : oo : 55 Sol 1.12. We have t hat for an ideal gas we get
av
8P
8T
8P r
8T v
av
p
=
- 1
so from this relation we get to know that A and C is correct option. equation of state f(x,y,z)=O [equation of the state f(p,V,T )=O)
df =
8f
dx+
ax
y,z
8f 8y
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dy+
az
x,z 20
@Sk J ahiruddin , 2020
8f
dz
=
0
y,x
physicsguide CSIR NET, GATE
Maxwell's relation and Thermodynamic Potentials
setting dx,dy,dz equal to zero, successively we get a set of equation like
8y
az
X
and again we have
az ax
y
fx fz
hence we get
av
8P
8T
8P r
8T v
av
=
- 1
p
Sol 1.13. From t he previous discussion we have seen t hat dG = -SdT + V dP and most importantly we have also see t hat one of the T dS relation is given by
TdS = CpdT - T
av
BT
r,
dP
11 oo : oo : sa e previous 1scuss1on we ave seen dG = -SdT + V dP and most importantly we have also see t hat one of the T dS relation is given by ~
av ar
TdS=CpdT- T
ii
dP p
one of the other TdS relation can be obtained by using the fact that s can be a function of volume and the temperat ure.so option b and d is correct . Sol 1.14 . From the previous discussion we get t his is Gibbs free energy.
j [email protected]
21
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Maxwell's relation and Thermodynamic Potentials
Sol 1.15. We have
aQ
ar
aQ
ar
p
v
and we know that dQ= TdS and we are going to use this
as ar
- T p
as ar
v
and we use the
dS=
as av
dv + T
as ar
dT V
now we have so we
as ar
p
as av
T
av as + ar ar p
v
11 oo : 01 : oo T
V
now we have so we
as aT t hen
av as as + av raT P aT v
P
as as aT P aT v
as av avT ar
p
and we know from the maxwell relation
as av
aP ar
V
V
AND WE USE THIS RELATION IN THE ABOVE EQUATION Cp-C-u = T
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av aP aTpaT v
22
@Sk J ahiruddin , 2020
physicsguide CSIR NET, GATE
l\/[axwell's relation and Thermodynamic Potent ials
for ideal gas we know t hat
av aP
r
aP ar
v
ar av
we use t his relation to replace the t ion and we get
p
=-1
(if)v in t he above equa-
T va 2 Gp= Cv+ kr where a and kr carries as usual meaning.
Sol 1.16. The porous plug experiment is essentially giving you a more important condition which is given by the change inn the enthalpy is zero and t he dQ=O so we have the ent halpy function is defined as
11 oo : 01
: 03
Sol 1.16. The porous plug experiment is essentially giving you a more important condition which is given by the change
inn the enthalpy is zero and the dQ=O so we have the ent halpy function is defined as
H = U + PV hence the change in t he enthalpy is given by
dH
=
dU + PdV + VdP
=
T dS + VdP
and we have dH= O and T dS = 0 so we have fin ally V dP = O now we know PV = nRT So, we can say
VdP + PdV = n RdT bu we know that vdP = 0 so we are getting
PdV = RdT [email protected]
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23
physicsguide CSIR NET, GATE
Nlaxwell's relation and Thermodynamic Potentials
and from the ideal gas equation p = n RT/ V so putting this we get V2 dv = Tln V2 dT = T V1 V V1 Sol 1.1 7. In free expansion t he gas is allowed to expand in to a vacuum.t his happens quickly , so there is no heat t ransferred. as the gas expands in vacuum so there is no work done against the external pressure.so according to the first law t his means that internal energy of t he system is not changing so t he temperature is constant . The temperature remains the same. (some times it is also called irreversible
11 oo : 01
: 06
t ransferred. as the gas expands in vacuum so there is no work done against the external pressure.so according to the first law this means that internal energy of t he system is not changing so the t emperature is constant . The temperature remains the same. (some times it is also called irreversible isothermal expansion) Sol 1.18. As we know t hat t he internal energy is a funct ion of t mperature and volume as we are considering t he
vanderwalls gas and in the vanderwalls gas t he internal energy depends on t he volume so we have dU =
au
ar
dT + V
au
dV
av
r
and the change in the t otal internal energy is zero as the gas's internal energy is not changing so according to t his condiation we can say
au
ar
au
dT +
av
V
j [email protected]
24
@Sk J ahiruddin , 2020
r
dV =
o
physicsguide CSIR NET, GATE
Maxwell's relation and Thermodynamic Potentials
and we have TdS = dU +pdV
So, we have
T
as av
au
av
+p
and now we are going to use the maxwell's second equation
as av
TP '
aP ar
VS
'
11 oo :01 : oa and now we are going to use t he maxwell,s second equation
as av
8P 8T
TP )
VS
'
using t his in t he above equation will gives us
T aP -p= au
aT
av
from t he 2nd equation we can say t hat
au
0=
dV
av
T
from t his we can say that l
TJ
=~- -
Cv
VJ Vi
au
av
dV T
so temperature is decreasing. so option a and d is correct. Sol 1.19. At equilibrium say finally t he to sector comes to some equilibrium temperature say Tp 1 so for the first compart ment gas we can write
PoVo To [email protected]
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25
physicsguide CSIR NET, GATE
Nlaxwell's relation and Thermodynamic Potentials
now Vo= H A and Vp = (H - y)A where we have consider t he fact t hat aft er t he weight has been released t he piston comes down y dist ance so t he first or say one of the compart ment will have increased lengt h by y amount where as ot her corn part ment will have decreased length by t he same amount of length and t he first corn partment has final pressure P P, and final temperature T P,. now we can sav t hat
11 oo : 01
: 11
comes down y dist ance so the first or say one of the compartment will have increased length by y amount where as other compartment will have decreased length by t he same amount of length and the first compartment has final pressure Pp1 and final temperature Tp1. now we can say t hat
PF l
_ Tp1 x P0 H To(H - y)
in similar way the final pressure in the second compartment is given by p _ Tp1 x P0H 2 F - To( H + y) in this case we say t hat the equilibrium temperature would be t he same for the both gas.now according to the given condition is pressure difference must be equal to ~ so mathematically
w
Pp1 - Pp2 = A
now again we are considering the fact t hat the process is adiabatic process so we will have dU = ev(TF1 - To) = Wy = dw
but we have t he condition(yW > ToCv so we get Tp1 = ~: now we will apply t his in t he previous equation of Pp1 and [email protected]
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26
physicsguide CSIR NET, GATE
Maxwell's relation and Thermodynamic Potentials
get the following
which is giving us
1
1
H- y
H +y
1 A
11 oo : 01
: 13
get the following 1
1
H- y
H +y
1 A
which is giving us H
y=
Sol 1.20. From t he given equation of state 2
p = nRT _ a b3 v3
This sudden change of wall leads to free expansion of gas. For free expansion Change of internal energy is ZERO. From internal energy equation
au av
T
aP - P =T aT V TNR nRT b3 = a 2/ V 3
j [email protected]
@Sk J a hiruddin, 2020
27
b3
a2
+ V3
physicsguide CSIR NET , GATE
:Niaxwell's relation and Thermodynamic Potentials
Now write the internal energy in the differential form
dU = nCvdT+
au
dV
11 oo : 01
: 16
j [email protected]
27
@Sk J ahiruddin , 2020
physicsguide CSIR NET, GATE
Nlaxwell's relation and Thermodynamic Potentials
Now writ e the internal energy in the differential form dU
au
= nCvdT+ -
av
r
dV
a2
= n,CvdT + V 3 dV T1
=
nCv
dT
Vo
+
Vo/2
To
a2
V 3 dV = 0
as dU = 0
Hence The answer a2
Vo
=0
nCv(T1 - ~ ) - 2v2 Vo/ 2
=
n Cv(T1 - ~)
a2
1
2
(Vo/2) 2
2
hence
T.t = ~ -
j [email protected]
2n
3a C v, 2 V
0
28
physicsguide CSIR NET, GATE
11 oo : 01
: 1a
j [email protected]
27
@Sk J ahiruddin , 2020
physicsguide CSIR NET, GATE
Nlaxwell's relation and Thermodynamic Potentials
Now writ e the internal energy in the differential form dU
au
= nCvdT+ -
av
r
dV
a2
= n,CvdT + V 3 dV T1
=
nCv
dT
Vo
+
Vo/2
To
a2
V 3 dV = 0
as dU = 0
Hence The answer a2
Vo
=0
nCv(T1 - ~ ) - 2v2 Vo/ 2
=
n Cv(T1 - ~)
a2
1
2
(Vo/2) 2
2
hence
T.t = ~ -
j [email protected]
2n
3a C v, 2 V
0
28
physicsguide CSIR NET, GATE
11 oo : 01
: 21
j [email protected]
27
@Sk J ahiruddin , 2020
physicsguide CSIR NET, GATE
Nlaxwell's relation and Thermodynamic Potentials
Now writ e the internal energy in the differential form dU
au
= nCvdT+ -
av
r
dV
a2
= n,CvdT + V 3 dV T1
=
nCv
dT
Vo
+
Vo/2
To
a2
V 3 dV = 0
as dU = 0
Hence The answer a2
Vo
=0
nCv(T1 - ~ ) - 2v2 Vo/ 2
=
n Cv(T1 - ~)
a2
1
2
(Vo/2) 2
2
hence
T.t = ~ -
j [email protected]
2n
3a C v, 2 V
0
28
physicsguide CSIR NET, GATE
11 oo : 01
: 23
j [email protected]
27
@Sk J ahiruddin , 2020
physicsguide CSIR NET, GATE
Nlaxwell's relation and Thermodynamic Potentials
Now writ e the internal energy in the differential form dU
au
= nCvdT+ -
av
r
dV
a2
= n,CvdT + V 3 dV T1
=
nCv
dT
Vo
+
Vo/2
To
a2
V 3 dV = 0
as dU = 0
Hence The answer a2
Vo
=0
nCv(T1 - ~ ) - 2v2 Vo/ 2
=
n Cv(T1 - ~)
a2
1
2
(Vo/2) 2
2
hence
T.t = ~ -
j [email protected]
2n
3a C v, 2 V
0
28
physicsguide CSIR NET, GATE
11 oo : oo : 01
Problems and Solutions in Thermodynamic equation of State Sk J ahiruddin * Suchismito Chattopadhyay
*Assistant Professor Sister Nibedita Govt. College, Kolkata Aut hor was t he topper of IIT Bombay M.Sc Physics 2009-2011 batch He ranked 007 in IIT J AM 2009 and 008 (J RF) in CSIR NET J une 2011 He has been teaching CSIR NET aspirants since 2012
1
@Sk J ahiruddin, 2020
Thermodynamic equation of State
11 oo : oo : 03 1
Thermodynamic equation of State
@Sk J ahiruddin, 2020
Contents 1
Problems from NET, GATE, JEST , TIFR & JAM pap ers 1.1 Ans keys . 1.2 Solut ions . • • •
j [email protected]
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11 12
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11 oo : oo : 06
j [email protected]
physicsguide CSIR NET I GATE
2
@Sk J ahiruddin , 2020
1
Thermodynamic equation of State
Problems from NET, GATE, JEST, TIFR & JAM papers
Prob 1.1. The internal energy of a system is given by E
=
tt,
where b is a constant and other symbols have t heir usual meaning. The temperature of this system is equal to [NET Dec 2011] 2 3 bS (a) bS2 (b) 3bS (d)( S )2 (c) v2N N VN VN Prob 1.2. For a part icular t hermodynamic system t he ent ropy S is related to t he internal energy U and volume V by S = cU3/4vl/4 where c is a constant. The Gibbs potential G = U - TS+pV for this system is [N ET June 2014]
(a) 3pU 4T
(b) cU
( )
3
c zero
(d) US 4V
Prob 1.3. A theoretical model for a real (non-ideal) gas gives the following expressions for the internal energy ( U)
and t he pressure ( P) U (T, V)
= av-
2 3 /
+ bV
2 3
1T
2
and
P (T , V)
= ~av- ! + ~bv- 1 T 5 3
1 3
2
11 oo :oo :og U (T , V )
= a v-
2 3 /
+ bV
2 3
1T
2
and
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3
Thermodynamic equation of State
@Sk J ahiruddin ) 2020
where a and b are constants. Let Vo and To be t he init ial volume and initial temperature respectively. If t he gas expands adiabatically, t he volume of the gas is proport ional to [JEST 2018]
(a)T
(b)T 3/2
(d)T- 2
(c)T -3/2
Prob 1.4. Suppose that the number of microstat es available to a syst em of N part icle depends on N and t he com2
bined variable UV , where U is the internal energy and V is t he volume of the syst em. The syst em init ially has volume 2m 3 and energy 200J. It undergoes an isent1~opic expansion t o volume 4m 3 . What is the final pressure of t he system in
[JEST 2017]
SI units?
Prob 1.5. The equation of state of a gas is given by
V == RT _ !2_
p
T
where R is the gas constant and b is another constant paramet er. The specific heat at constant pressure C p and the sp ecific heat at constant volume Cv for t his gas is related by Gp - Cv = [TIFR 2015]
11 oo : oo : 11 where R is t he gas constant and b is anot her constant parameter. The sp ecific heat at constant pressure C p and the specific heat at constant volume Cv for this gas is related by Gp - Cv = [TIFR 2015] RT2 2 (a) R (b) R 1 + bP [email protected]
Thermodynamic equation of St ate
@Sk J ahiruddin, 2020
(c) R 1 +
physicsguide CSIR NET, GATE
4
2
bP RT2
Prob 1.6. An ideal gas consists of t hree dimensional polyatomic molecules. The temperature is such that only one vibrational mode is excited. If R denotes t he gas constant, t hen the specific heat at constant volume of one mole of t he [JAM 2018] gas at this t emperature is
(b) ;R
(a) 3R
(c) 4R
(d) ~R
Prob 1. 7. The free energy for a photon gas is given by F = -(~)VT4 , where a is a constant. The entropy Sand t he pressure P of t he photon gas are [GATE 2007]
4 c) S = a VT 4 P ( 3 '
(d) S
=
=
_aT 3 3
1aVT3 p = _ 4aT 4 3
'
3
Prob 1.8. The energy density and pressure of a photon gas are given by u = aT4 and P = u/3, where T is t he t emperature and a is the radiation const ant. The entropy per unit
11 oo : oo : 13 (d) S
=
laVT3 p = _ 4aT 4 3
'
3
Prob 1. 8. The energy density and pressure of a phot on gas 4 are given by u = aT and P = u/ 3, where T is t he t emper-
ature and a is t he radiation const ant. The ent ropy per unit volume is given by aaT 3. The value of a is ______ (up t o two
[GATE 2017]
decimal places)
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Thermodynamic equation of State
@Sk J ahiruddin , 2020
Prob 1 .9 . The ent ropy of a gas containing N particles en-
closed in a volume V is given by S = NkB ln
3 2 ;
a~f1
,
where E is t he total energy, a is a constant and kB is the Bolt zmann const ant. T he chemical potent ial µ of t he system at a t emperature T is given by [GATE 2015]
(a) µ = -kBT ln
(b) µ = - kBT In (c) µ = - kBT In
aV E 312 N 5/ 2
5 2
aVE 312
3 2
aVE 312
5 2
N 5/ 2
N 3/ 2
aV E 312 N 3/2
3 2
Prob 1 .10. The ent ropy function of a syst em is given by
S (E ) = aE (E o - E ) where a and E 0 are posit ive constant s. [GATE 2013] T he temp erature of t he system is
11 oo : oo : 16 Prob 1.10. The entropy function of a syst em is given by
S(E) = aE(Eo - E ) where a and E 0 are posit ive constants. The t emperature of the system is [GATE 2013] (a) negative for some energies (b) increases monotonically with energy (c) decreases monotonically with energy (d) zero [email protected]
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6
Thermodynamic equation of State
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Prob 1.11. The total energy, E of an ideal non-relativistic where Fermi gas in three dimensions is given by E ex
f:;:
N is the number of part icles and V is t he volume of the gas
Identify the CORRECT equation of state (P being the pressure) . [GATE 2012] (a) PV = ½E (b) PV = j E (c) PV = E (d) PV = i E Prob 1.12. A red star having radius rR at a temperature TR and a white star having radius rw at a temperature Tw ,
radiate the same total power. If these stars radiate as per[JAM 2019] fect black bodies, then (A) rR > rw and TR > Tw (C) rR > rw and TR < T1,v
(B) rR < rw and TR > Tw (D) rR < rw and TR < Tw
Prob 1.13. The equation of st ate for one mole of a non-
i ),
where t he coefficient ideal gas is given by PV = A (1 + A and B are temperature dependent. If the volume changes fr()rn 1/4 f() 1/n 1n ~ n l~flt.h Prrn~ l
nr()('P~ ~
t.h P 1xr()rk rl ()n P h1r
11 oo : oo : 19 Prob 1.13. The equation of state for one mole of a non-
i) ,
ideal gas is given by PV = A (1 + where t he coefficient A and B are temperature dependent . If the volume changes from V1 to ½ in an isothermal process, the work done by the gas is [JAM 2018] 1 1 ½ (a) AB -½ ½ (b) ABln Vi
(c) Aln
V2
V1
1 -1
+ AB
½
V2
[email protected]
7
(d) Aln
½-½ Vi
+B
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T hermodynamic equation of Stat e
@Sk J ahiruddin , 2020
Prob 1.14. The equation of state for a gas is given by aN
p+
2
V
where P,V,T,N and kB represent pressure, volume,temperature, number of atoms and t he Boltzmann constant respectively, while a and /3 are constants specific to the gas. If the critical point C corresponding to a point of inflexion of t he p-V curve, then t he critical volume Ve and critical pressure Pc for t his gas are given by [TIFR 2016] 2 2 (a) Vc = 3/3 N , p0 = a /3/3
a/ 27/3 2 2 (c)Vc = 3(3 N , pc = 8a / 27/3 (d)Vc = 3(3 N , p0 = a 2 / 27/3 2 (b)Vc
=
3/3N, p0
=
11 oo : oo : 22 (a) Vc (b)Vc
= =
3f3 N ,Pc 3f3N,pc
= a =
/3/3
a / 27/3
2
2
(c)Vc = 3/3 N ,pc = 8a / 27(3 (d )Vc
=
3f3 N ,pc
=
a 2 /27/3 2
Prob 1 .15 . The molar equation of state of a gas at tem-
perature T , pressure P and volume V is given by
RT a p = V - b - TV 2 where, a and b are two constants and R is the gas constant. The critical temperature and pressure for this gas will be [TIFR 20 19]
j [email protected]
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T hermodynamic equation of Stat e
@ Sk J ahiruddin, 2020
a RTc (a) Tc = Pc = 27Rb' b 8a RTc (b) Tc= Pc= 27Rb' 8b 4a RTc (c) Tc = Pc = 4b 27Rb ' 8a RTc (d) Tc = Pc= 3R ' 8b Prob 1 .16. The van der Wall's equation of stat e for a gas is given by (P
a
+ V 2 )(V -
b) = RT
where P,V and T represent the pressure, volume and t emperature respectively, and a and b are constant parameters.
11 oo : oo : 24 is given by
(P
a
+ V 2 )(V -
b) = RT
where P,V and T represent the pressure, volume and t emperature respectively, and a and b are constant parameters. At the critical point, where all the roots of the above cubic [NET equation are degenerate , the volume is given by June 2014] a a 8a (d) 3b (a) 9b (b) 27b2 (c) 27bR Prob 1.17. The van der Waals equation for one mole of a gas is (p + ; ) (V - b) = RT . The corresponding equation of state for n moles of t his gas at pressure p, volume V and temperature T , is [NET June 2018]
p +~
(a) (b)
2
2
(P
+~
2)
(V-nb)=n RT
(V - nb)
[email protected]
= n RT 9
@Sk J ahiruddin , 2020
(c) (d)
p +~: (V -nb)= RT (p+ ~2 ) (V-nb )= RT
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Thermodynamic equation of State
11 oo : oo : 21
j [email protected]
10
@Sk J ahiruddin , 2020
1.1
physicsguide CSIR NET I GATE
Thermodynamic equation of State
Ans keys
1.1. b
1.6. c
1.11. b
1.16. d
1.2.
C
1.7. a
1.12.
C
1.17. a
1.3.
C
1.8. 1.33
1.13.
C
1.4. 25
1.9. a
1.14. d
1.5.
1.10. a
1.15. b
C
11 oo : oo : 29 1.8. 1.33
1.13.
1.4. 25
1.9. a
1.14. d
1.5. c
1.10. a
1.15. b
1.3.
C
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1.2
C
Thermodynamic equation of State
Solutions
Sol 1.1. We know that dq
= TdS =
dE
+ pdV
but here we have fixed volume and hence dv = 0 so we easily can say that
11 oo : oo : 32 dq = TdS = dE
+ pdV
but here we have fixed volume and hence dv = 0 so we easily can say that aE =T
as
v
now we are going to use t he upper formulae and we are getting 2 3bs T = VN . so option b is correct.
Sol 1.2. We know from t he Maxwell's equations of t hermodynamics that
as
1
au
T
using this we get
T = 4 u1J4 v - 1J4 3c
and we have TdS=dU+ PdV
12
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Thermodynamic equation of State
@Sk J ahiruddin , 2020
so we can easily say
as av and from this we get
P
r
r
11 oo : oo : 34 so we can eas1 y say
as av
P
T
T
and from t his we get
G = u - 4 u l J4v-l /4
cu 3J4vl /4+
X
3c 4 u1J4v - 114 x c u 3J4v - 3J4 x v = o 3c 4 Sol 1.3. We know for t he ad iabat ic process dq= O so t he first law of thermodynamics says that dU = -dw and we know t h at dw = pdV so we get
au dT 8T
au dV = [~ v-
+ 8V
3a
5/ 3
2bv- 113T 2 ]dV
+3
hence we get
from t h is we can easily integrate to get t he dependence of v on T and is given by
dT
T
2dV
---
3V
and we integrate both side and get Vex
2 3 T- /
SO opt ion c
is correct. j [email protected]
@Sk J ahiruddin , 2020
13
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Thermodynamic equation of State
Sol 1.4. solution t he t h ermodynamic prob ability is given by
00: 00: 37 Thermodynamic equation of tate
Sol 1.4. solution the thermodynamic probability is given by
where b is merely a proportionality constant. now we know that
and as the process is isoentropic we get dS = 0 = bkB
+ 2 dV
dU U
V
which essentially gives me dU U
= _ 2 dV V
and we know t hat from the first law of thermodynamics we get dU
=
-pdV
so we get
p=
2U
V so init ially U =200J and V =2 cubic meter and hence we get
p=initially =200 pascal.now we know that TdS = 0 = dU [email protected]
@Sk J ahiruddin, 2020
which essentially gives us
14
+ pdV physicsguide CSIR NET, GATE
Thermodynamic equation of State
11 oo : oo : 40
@Sk J ahiruddin, 2020
Thermodynamic equation of State
which essentially gives us
au
av
s
=
- p
and we get pV= 2U
so we get
so we get pV 3
=K
and we we use this to get the final pressure 25 atm.
Sol 1.5. We have known before that
8P ov 8T v 8Tp Now we are going to derive the each term first let us calculate v . in order to do that we use
(if)
RT P=--
v +;
so we get
8P 8T v [email protected]
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15
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Thermodynamic equation of State
11 oo : oo : 42 V
15
[email protected]
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Thermodynamic equation of St ate
@Sk J ahiruddin , 2020
and we get the
av 8T
R
b
p + r2
P
putt ing all this values we get bP
2
1 + RT2
Sol 1.6. We know that for the poly-atomic molecules the specific heat at constant volume is given by in 3 dimension c,;
= (3 + f)R
where f gives vibrational mode here Cv
f = l so we have
= 4R
Sol 1. 7. we know that F is defined as follows F = U-TS
so dF
= dU - T dS - S dT
and we have from the first law of the thermodynamics dQ = TdS = dU j [email protected]
16
+ P dV physicsguide CSIR NET, GATE
11 oo : oo : 4s dQ
=
TdS
j [email protected]
=
dU + P dV
16
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Thermodynamic equation of State
@Sk J ahiruddin , 2020
so we replace this in the equation and we obtain
= -S
8F
8T
v
8F
8V
= -P T
using this we get from the first relation S
= 4aVT3 P = aT4 3
3
Sol 1.8. From t he previous several discussion we come to know that
dS= dQ T and we have from the first law of the thermodynamics dQ = dU +PdV now from the given relation we get 4
3
dU = 4aT dT
aT P =3
u 3
now using this relation in t he above equation
Now we know for t he photon gas
PV = ~U 3
11 oo : oo : 48 + -T Now we know for t he photon gas
PV= ~U 3 [email protected]
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17
Thermodynamic equation of State
@Sk J ahiruddin, 2020
from t his we can say that
P dV
= ~dU - VdP
to simplify this we use V = 3~ and putting instead of V in the above equation and we get
P dV =
1 -dU -dP 3
and we know that P = U/ 3 so workdone is actually zero. and we are left with
dS = 4a
2
T dT = iaT 3 3
so a= 1.33 Sol 1.9. We know t hat
µ =-
8G
8N
S =-
8G 8T
from t he second relation we get to know t hat G that is Gibbs free energy can be written as follows /
... ?",........") / ' ) ,
11 oo : oo : so µ=-
S=-
8N
ar
from t he second relation we get to know t hat G that is Gibbs free energy can be written as follows
G [email protected]
dT
N 5/2
physicsguide CSIR NET, GATE
18
Thermodynamic equation of Stat e
@Sk J ahiruddin, 2020
so we can write aVE 312 Ns/ 2
T
+A
where A ia just a mere integration constant . now from the first relation we get
aNk B ln µ=
aVE3/ 2 N s/2
8N
and hence we get
[ actually µ calculates the change in the Gibbs energy due to adding or subtracting particle from the system t hat is we are considering here t hat number of particle also can change so in general t he change in the Gibbs energy is defined as dG = V dP - SdT - µdN
]
11 oo : oo : s2 so in general t he change in the Gibbs energy is defined as dG = V dP - SdT - µdN
] Sol 1.10. We use t he relation
as
1 T
8E v j [email protected]
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Thermodynamic equation of State
@Sk J ahiruddin , 2020
t his gives us 1 T = a(Eo - E)
+ (-a E)
a(Eo - 2E)
=
now when E < Ea Then temperature would be negative so for some case temperature is negative. Sol 1.11. We know from t he first law of thermodynamics
dQ = dE+pdV
t he total energy is given for t he system so we have dQ= O and t his giving us p= -
8E
av
N
2E -3V
hence t he equation of t he state is given by 2 PV = E 3
11 oo : oo : 55 hence t he equation of t he state is given by
PV =
2
3
E
Sol 1.12. Black body radiation is a remarkable discovery of t he modern science. when heated a solid glows and emits
t hermal radiation but it is different from the radiation emitted by a gas. when a solid radiates t hen it's sharp contrast [email protected]
@Sk J ahiruddin , 2020
20
physicsguide CSIR NET, GATE
Thermodynamic equation of State
of continuous radiat ion pattern with that of t he discrete pattern structure of a gas really bound us to investigate the problem. a perfect black body is that body which absorbs all t he radiation failing in it so appears to be black when subjected to some external radiation. when t hese perfect black bodies are heated t hen it radiates energy and the radiation process follows a particular continuous pattern. A perfect black body is a good absorber as well as a good radiator. (which is actually the core idea or core t hought of the Kirchhoff 's law of black body radiation) Max Plank was t he first person to give the satisfactory theoretical justification to t he black body radiation considering t he fact t he bodies radiates in small packets of energy (which is known as quanta) . he first showed t hat energy density per unit wave length is can be written as 1
11 oo : oo : s1 body radiation considering the fact t he bodies radiates in small packets of energy (which is known as quanta) . he first showed t hat energy density per unit wave length is can be written as 81rhc 1 u(A, T ) = A5 ehc/ >.kT _ 1 and hence the energy radiated by the blackbody at temperature T per unit area per unit time can be obtained 00
00
uv, Tdv = 0
0
81r hv3 c3(ehv/ kT _
1)
dv = 4 a-T4 c
and this is the Stephan Boltzman law and the maximum of j [email protected]
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physicsguide CSIR NET I GATE
21
Thermodynamic equation of St ate
t he energy density curve satisfies the following relation
this is known as the Wien's displacement law. This summarizes the whole black body radiation. now coming to this particular problem. the red star has maximum wave length say Ar and the white star has Aw . according to the Wien's displacement law
and this enables us to conclude that
11 oo : 01 : oo and t his enables us to conclude t hat
so we get
now t hey are emitting same power so we have 4
4
E = - aTw x 41rrw
2
C
t his is t he energy radiated by the white star. the energy radiated by the red star
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22
@Sk J ahiruddin ) 2020
Thermodynamic equation of State
but according to t he given condition t hey are both emitting same power so we get
TR
4
TR
2
=
'77 4 2 .L W rw
which enables us to conclude t hat 2
> 1
so we must have rR > rw Sol 1.13. The work done in t he isothermal process is given by V2
pdv
=
A+
AB
_ _v_dV 1/
11 oo : 01
: 03
so we must have rR > rw Sol 1.13. The work done in t he isothermal process is given by V2
pdv
=
A+
AB
_ _v_dV V
integrating we get
Sol 1.14. There exists a critical temperature for any gas and p-V curve at t his temperature is called the critical temperature Tc. We know that in van der Walls t heoretical
curve , minima and maxima appear and they approach each other as we consider higher and higher isothermal process [email protected]
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23
Thermodynamic equation of State
@Sk J ahiruddin, 2020
.At a particular temperature t he maxima and minima coalesce to a single point called t he critical point.the pressure and volume corresponding to t hat critical point are called critical pressure and critical volume. Here we have the van der Walls equation of state of the real gas and at crit ical point we have
8P
av a2 p av2
=0 =0
11 oo : 01
: 06
8P
av a2p
av
2
=0 = 0
from t he given equation of state we can say that aN V
2
so according to the first relation we have
in similar way the second relation giving us
from third and fourth equation we can easily conclude that t he critical volume Ve = 3f3 Nnow we can put t his value of j [email protected]
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24
Thermodynamic equation of State
@Sk J ahiruddin, 2020
Ve
in the 3rd equation and we get 2
8a Tc= 27f3 K B and putting t his both value in expression of Pressure obtained from the equation of state will give us a2
p
option d is correct.
= 27/3 2
11 oo :01 : oa tained from t he equation of state will give us 0'.2
p= 27/3 2
opt ion d is correct. Sol 1.15.
8P
=0
av f)2 p
= 0
av 2 t he 1st equation gives us 2a TV 3
RT (V - b)2
and t he 2nd equation giving us
RT (V - b)3
3a TV 4
now we get by doing 7/ 8 Ve = 3b and putt ing t his in t he 3rd equation will give us Tc =
2~~b
and putting all in this in
t he equation of state will give us Pc = correct. [email protected]
@Sk J ahiruddin, 2020
25
~fc. so option b is
physicsguide CSIR NET, GATE
Thermodynamic equation of State
Sol 1.16. From t he discussion of the problem 115 we get t o see that the crit ical volume is Ve = 3b. so option d is
correct . Sol 1.1 7. For n mole real gas t he opt ion A is t he equation of state as t he correction terms to be evaluated for n mole gas.
11 oo : 01
: 11
correct. j [email protected]
25
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Thermodynamic equation of St ate
@Sk J ahiruddin, 2020
Sol 1.16. From t he discussion of the problem 115 we get to see that t he crit ical volume is Ve = 3b. so option d is
correct. Sol 1.17. For n mole real gas the option A is t he equation of state as t he correction terms to be evaluated for n mole gas.
j [email protected]
26
physicsguide CSIR NET, GATE
11 oo : 01
: 13
correct. j [email protected]
25
physicsguide CSIR NET, GATE
Thermodynamic equation of St ate
@Sk J ahiruddin, 2020
Sol 1.16. From t he discussion of the problem 115 we get to see that t he crit ical volume is Ve = 3b. so option d is
correct. Sol 1.17. For n mole real gas the option A is t he equation of state as t he correction terms to be evaluated for n mole gas.
j [email protected]
26
physicsguide CSIR NET, GATE
11 oo : 01
: 1s
correct. j [email protected]
25
physicsguide CSIR NET, GATE
Thermodynamic equation of St ate
@Sk J ahiruddin, 2020
Sol 1.16. From t he discussion of the problem 115 we get to see that t he crit ical volume is Ve = 3b. so option d is
correct. Sol 1.17. For n mole real gas the option A is t he equation of state as t he correction terms to be evaluated for n mole gas.
j [email protected]
26
physicsguide CSIR NET, GATE
11 oo : oo : oo
Problems and Solutions
•
Ill
Entropy Calculation Sk J ahiruddin * Suchismito Chattopadhyay
*Assistant P rofessor Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay 1\11.Sc Physics 2009-2011 batch He ranked 007 in IIT J Al\lI 2009 and 008 (JRF) in CSIR NET J une 2011 He has been teaching CSIR NET aspirants since 2012
1
@Sk J ahiruddin, 2020
Ent ropy Calculation
11 oo : oo : 03 1
Entropy Calculation
@Sk J ahiruddin, 2020
Contents 1
Problems from NET, GATE, JEST , TIFR & JAM pap ers 1.1 Ans keys . 1.2 Solut ions . • • •
j [email protected]
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11 oo : oo : os
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@Sk J ahiruddin , 2020
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Entropy Calculation
Problems from NET, GATE, JEST, TIFR & JAM papers
Prob 1.1. The internal energy E )T ) of a system at a fixed volume is found to depend on the temperature T as E (T ) = aT 2 + bT 4 . T hen t he ent ropy S (T ), as a function of t emperature is [NET June 2016] 2 4 2 4 3 (a) ~aT +!bT (b) 2aT +4bT (c) 2aT +ibT (d) 2aT + 2bT 3 Prob 1.2. The heat capacity Cv at constant volume of a met al, as a function of t emperature, is aT + f3T 3 , where a and f3 are const ants. The temperature dependence of t he entropy at constant volume is [NET D e c 2018]
(a) aT + 1f3T 3 3 (c) ~aT + 1f3T
(b) a T + f3T 3 3 (d) ~aT + ! f3T
Prob 1.3. In a syst em comprising of approximately 1023 distinguishable particles, each particle may occupy any of 20 distinct states. The maximum value of the entropy per particle is nearest to [NET June 2019] (a) 20kB (b) 3kB (c) 10(ln2)kB (d) 20(ln2)kB Prob 1.4. Consider an ideal gas of mass m at temperature
11 oo : oo : os [NET June 2019] (c) lO (ln 2)kB (d) 20(ln 2)kB
particle is nearest to (a) 20kB (b) 3kB
Prob 1.4. Consider an ideal gas of mass mat temperature [email protected]
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3
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EntrOP)' Calculation
T1 which is mixed isobarically (i.e. at const ant pressure) wit h an equal mass of same gas at t emperature T2 in a thermally insulat ed container. What is the change of entropy of t he universe? [JEST 2012] (a) 2mCp ln( Ti+ T2 ) (b ) 2mCp ln( Ti - T2) 2~ 2~ 1 1 2 (c) 2m Cp ln(T + T ) (d) 2m Cp ln(T + T2 ) 2T1T2 2T1T2 Prob 1.5. A 100 ohms resistor carrying current of 1 amp is maintained at a constant t emperature of 30°C by a heat bath. What is the rat e of entropy increase of the resistor? [JEST 2014] (a) 3.3 Joules/K/sec
(b)6.6 Joules/K/sec
(c)0.33 Joules/K/sec
(d) None of t he above
Prob 1.6. A certain amount of fluid with heat capacity Gp joules/°C is initially at a t emperature 0°C. It is then brought into contact with a heat bath at a t emperature of 100°C , and the system is allowed t o come into equilibrium. In t his process, t he ent ropy (in J oules/°C) of t he Universe changes by [TIFR 2013] (a) lOOCp (b) 0 (c) 0.055Cp (d) 0.044Cp Prob 1. 7. Consider a sealed but t hermally conducting con('I
•
•
,
,
TT
,
•
,
•
•
., ., .
. , '
11 oo : oo : 11 In this process, the entropy (in J oules/°C) of the Universe [TIFR 2013] changes by (a) l OOCp (b) 0 (c) 0.055Cp (d) 0.044Cp Prob 1. 7. Consider a sealed but t hermally conducting container of total volume V, which is in equilibrium with a [email protected]
4
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Entropy Calculation
@Sk J ahiruddin , 2020
t hermal bath at t emperature T. The container is divided into two equal chambers by a thin but impermeable partit ion. One of these chambers contains an ideal gas , while [TIFR 2012] t he other half is vaccum (see figure)
gas
vacuu
If the partition is removed and ideal gas is allowed to expand and fill the entire container , t hen the entropy per molecule of the system will increase by an amount
Prob 1.8. Experimental measurements of heat capacity per mole of Aluminum at low temperatures show t hat the data can be fitted to t he formula Cv = aT + bT3 , where a =
11 oo : oo : 13 Prob 1.8. Experimental measurements of heat capacity p er
mole of Aluminum at low temperatures show t hat t he data can be fitted to t he formula Cv = aT + bT 3 , where a = 0.00135J K- 2moze- 1 , b = 2.48 x 10-5 J K- 4 moze- 1 and T is t he temperature in Kelvin. THe ent ropy of a mole of Aluminum at such temperat ures is given by t he formula [JAM j [email protected]
5
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Ent ropy Calculation
2007]
(a)aT + i T 3 + c , where c > 0 is a const ant (b) + 3 + c , where c > 0 is a constant 3 (c)aT + t T + C
ai :r
(d)ai +
1T3 +
C
Prob 1.9. 2 kg of a liquid ( specific heat= 2000JK - 1 kg- 1 ,indepc
of temperature ) is heated from 200K to 400K by eit her of t he following two processes P 1 and P 2 : P 1 :bringing it first in contact wit h a reservoir at 400K. P 2 :bringing it first in contact wit h a reservoir at 300K t ill equilibrium is reached , and t hen bringing it in contact with another reservoir at 400K. Calculate the change in the entropy of the liquid and t hat of the universe in processes p 1 and P2 . Neglect any change in volume of the liquid. [JAM 2012] Prob 1.10. Consider the free expansion of one mole of an
11 oo : oo : 16 of t he universe in processes p 1 and P2 . Neglect any change in volume of the liquid. [JAM 2012] Prob 1.10. Consider t he free expansion of one mole of an ideal gas in an adiabatic container from volue V1 to ½ . The ent ropy change of the gas, calculated by considering a reversible process between t he original state (Vi, T ) to the final state (½, T ) where T is t he temperature of the system , is denoted by ~ S 1. The corresponding change in the [email protected]
@Sk J ahiruddin, 2020
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Ent ropy Calculation
entropy of t he surrounding is ~ S 2 . Which of t he following [JAM 2011] combinations is correct ? (a) ~ S1 = R ln (V1/½), ~ S2 = -R ln (Vi/V2) (b) ~ S1 = -Rln(V1/V2), ~ S2 = R ln(Vi/V2) (c) ~ S1 = Rln(V2/V1), ~ S2 = 0 (d)~ S1 = - Rln(V2/V1), ~ S2 = 0 Prob 1.11. A solid met allic cube of heat capacity S is at temperature 300K. It is bought in contact wit h a reservoir at 600K. If t he heat t ransfer takes place only between the reservoir and the cube, the entropy change of t he universe [JAM 2014] after reaching t he t hermal equilibrium is (a)0.69S (b)0.54S (c)0.27S (d)0. 19S Prob 1.12. One gram of ice at 0°G is melted and heated to water at 39°G. Assume t hat t he specific heat remains constant over t he ent ire process. The latent heat of fusion
11 oo : oo : 1s (a)0.69S
(b )0.54S
(c)0.27S
(d)0. 19S
Prob 1.12. One gram of ice at 0°G is melted and heated
to water at 39°G. Assume t hat the specific heat remains constant over the ent ire process. The latent heat of fusion of ice is 80 Calories/ gm . The entropy change in the process [JAM 2015] (in Calories per degree) is Prob 1.13. A rigid and thermally isolated tank is divided into two compartments of equal volume V , separated by a t hin membrane. One compartment contains one mole of an
ideal gas A and the other compartment contains one mole [email protected]
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Entrop)' Calculation
of a different ideal gas B. The two gases are in t hermal equilibrium at a temperature T. If t he membrane raptures, t he two gases mix. Assume that the gases are chemically inert. The change in t he total entropy of t he gases on mixing is [JAM 2015] (a) 0 (b) R In 2 (c) ~ R In 2 (d) 2R ln 2 Prob 1.14. Consider two identical, finite, isolated systems of constant heat capacity C at temperature T1 and T 2 (T1 >
T 2 ). An engine works between them until t heir temperatures become equal. Taking into account that the work performed by the engine will be maximum (= W max). if t he process is reversible (equivalently, the entropy change of the entire system is zero), the value of Wmax is: [JAM 2017]
11 oo : oo : 21 by the engine will be maximum (= Wmax). if t he process is reversible (equivalently, the ent ropy change of the ent ire system is zero) , t he value of Wm a,x is: [JAM 201 7]
(d) C(PJ - P2) 2 j [email protected]
@Sk J ahiruddin, 2020
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Ent ropy Calculation
Prob 1.15. Each of t he two isolated vessels, A and B of fixed volumes, contains N molecules of a perfect monatomic gas at a pressure P. The temperatures of A and B are T1 and T2 respectively. The two vessels are brought into t hermal contact . At equilibrium, t he change in entropy is [GATE 2006]
Prob 1.16. In a thermally insulated container , 0.01kg of ice at 273K is mixed with 0.1kg of v\rater at 300K. Neglecting t he specific heat of the container , the change in t he
11 oo : oo : 24 (c) ~NkB ln
(~~i) 2
(d ) 2NkB
Prob 1.16. In a thermally insulated container , 0.01kg of ice at 273K is mixed with 0.1kg of V\rater at 300K. Neglect-
ing t he specific heat of the container, the change in the entropy of t he system in J / K on attaining thermal equilibrium (rounded off to two decimal places) is _____ [GATE 2019] (Specific heat of water is 4.2kJ /kg - K and t he latent heat of ice is 335kJ / kg ) . Prob 1.17. The ent ropy S of a system of N spins, which may align either in t he upward or in the downward direc-
t ion, is given by S = - kB N[p lnp + (1 - p) ln(l - p)]. Here kB is t he Boltzmann constant. The probability of alignment
[email protected]
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Entropy Calculation
in the upward direction is p . The value of p , at which the entropy is maximum, is _____ . (Give your answer upto one decimal place)
[GATE 2016]
Prob 1.18. N atoms of an ideal gas are enclosed in a container of volume V . The volume of t he container 4V , while
keeping the total energy const ant. The change in t he ent ropy of the gas, in units of NkB In 2, is ---, where kB is the Boltzmann constant. [GATE 2016]
11 oo : oo : 26 keeping t he total energy constant. The change in t he ent ropy of t he gas, in units of NkB In 2, is ---, where kB is the Boltzmann constant . [GATE 2016]
Prob 1.19. A system of N non-interacting distinguishible particle of spin 1 is in t hermodynamic equilibrium. The en[GATE 2011] t ropy of t he system is (a) 2kBlnN (b) 3kB lnN (c) NkB ln2 (d) NkB ln3 Prob 1.20. Consider a syst em of N atoms of an ideal gas of type A at t emperature T and volume V. It is kept in diffusive contact with another syst em of N atoms of another ideal gas of type B at the same temperature T and volume V. Once t he combined syst em reaches equilibrium , [GATE 2008]
(a) t he total entropy of the final system is t he same as the sum of the ent ropy of the individual system always j [email protected]
10
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Entropy Calculation
(b) t he entropy of mixing is 2N kB In 2 (c) the entropy of the final system is less t han t hat of the sum of the init ial entropies of the two gases. (d) the entropy of mixing is non-zero when the atoms A and B are of t he same type.
Prob 1.21. An ideal gas with adiabatic exponent , undergoes a process in which its pressure P is related to its volume V by t he relation P = P0 - a V , where P0 and a are posi-
11 oo : oo : 29 B are of t he same type. Prob 1.21. An ideal gas wit h adiabatic exponent , under-
goes a process in which its pressure P is related to its volume V by the relation P = P0 - a V , where P0 and a are posit ive constants. The volume st arts from being very close to zero and increase monotonically t o ~. At what value of t he volume during t he process does t he gas have maximum entropy? [JEST 2016] Po (b) , Po , Po d Po
(a)a(l+,1)
a( l -,)
(c)a(l+,)
( )a(l -,)
Prob 1.22. The value of entropy at absolute zero of t em-
perature would be [JAM 2005] (a) zero for all t he materials (b) finite for all the materials (c)zero for some materials and non-zero for others (d)unpredictable for any material
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11
@Sk J ahiruddin ) 2020
1.1
EntrOP)' Calculation
Ans keys Ans keys
1.1.
C
1. 7.
C
1.12. 0.39
1.18. 2
1.13. d
1.19. d
1.2. a
1.8. a
1.3. b
1.9. Do your- 1.14. d
1 _20_ h
11 oo : oo : 31 1.1.
1. 7.
C
C
1.12. 0.39
1.18. 2
1.2. a
1.8. a
1.13. d
1.19. d
1.3. b
1.20. b
1.4. a
1.9. Do your- 1.14. d self 1.15. C
1.5.
1.10.
1.16. 1.03
1.21.
1.17. 0.5
1.22. a
C
1.6. d
C
1.11. d
12
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Entropy Calculation
@Sk J ahiruddin, 2020
1.2
C
Solutions
Sol 1.1. We know t hat dq
== TdS == dE + pdV
11 oo : oo : 34 ons Sol 1.1. We know t hat
dq = TdS = dE
+ pdV
but here we have fixed volume and hence dv=O so we easily can say that 1 8E v T now we are going to use the upper formulae as Eis given in t he problem. dE dS= T and we have 3 dE = 2aTdT + 4bT dT
as
we will put this in t he integration and we get S(T )
=
. So option c is correct. Sol 1. 2. Assuming t hat t he volume is not changing we will
get
dU dS = T which gives us the right option is option a where c is an integration const ant. j [email protected]
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Entropy Calculation
Sol 1.3. We know that if there is N number distinct of sates in a system and n umber of particles in that syst em t hen
t here is accessible st ates in the syst em is given by
11 oo : oo : 36 , 2020
Entropy
alculation
Sol 1.3. We know that if there is N number distinct of sates in a system and n umber of particles in t hat system t hen
t here is accessible st ates in the syst em is given by
we use t his formulae to evaluate t he entropy of the system as we ha,,e known that ent ropy is related to the microst ates of t he system and as t he microstates increases the entropy of the syst em also increases in t hat sense we can say that t he entropy measures t he disorderness in t he syst em and the relation is given by
s = kB lnn so the ent ropy is given by
S
= kB
ln(20
1023
=
)
23
23
10 kBln20 = 10 kB3
so t he ent ropy per molecule is given by
Sol 1.4. First of all t he fact that we have to calculate the
final temperature TJ
= T1 + T2
2 t he ent ropy gain by t he gas at temperature T1 is given by dSgain =
mCp
T1
T1 [email protected]
@Sk J ahiruddin, 2020
and t he entropy loose
14
dT
T
T1
=
mcplnT1
physicsguide CSIR NET, GATE
Ent ropy Calculation
11 oo : oo : 39 @Sk J ahiruddin, 2020
EntrOP)' Calculation
and t he ent ropy loose
dSzose = mep non total change is
+ dszoss =
dsgain
2mcp ln
Sol 1. 5. To calculate t he increase of ent ropy you have to understand first why the heat is increasing? as t he heat is
increasing the disorderness in t he system also will increase and hence t he entropy must increase. now t he resistor is maintained at constant t emperature (273+30)Kelvin t emperature. and 1 amp current is flowing t hrough it so it must consists of joule heat energy and it is given by
H
=
2
i R
=
2
1 x 100 = lOOJ
so t he increase rate of entropy is given by 100 ds = T = = 0.33J oules / k / sec 303 dQ
Sol 1.6. The heat bath will be at constant temperature as we are assuming that it is an endless source of heat energy
comparing to t he body so at equilibrium the body temperature become equal to the heat bath temperature. So the [email protected]
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Entrop)' Calculation
11 oo : oo :42 15
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Ent ropy Calculation
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loss of the entropy of the heat bath is given by 373 - 273 100 dszoss = Gp = Gp 373 373
and t he gain of the entropy is given by 373
ds 9 ain =
dT
Gp T = Gp ln
273
373 273
so change in t he ent ropy of the universe= 100 dSroTAL = -Gp 373
373 + Gp In 273 = 0.044Gp
Sol 1. 7. The gas has been allowed to expand in the vacuum so it is t he free expansion and the temperature of t he gas is remaining unchanged so the change in entropy is given by V
dS
=
pdvT
= Rln2
V/ 2
it is t he total change of the gas system so t he change in entropy per molecule is
j [email protected]
16
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11 oo : oo : 4s
j [email protected]
16
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@Sk J ahiruddin , 2020
Entropy Calculation
Sol 1.8. Assuming that the volume is not changing we will get dU dS =
T
which gives us the right option is option a where c is an integration constant.
Sol 1.9. For an ideal liquid change in the volume , dV using t he entropy ds = dU
T
and specific heat at constant volume. we have 2
ds
= 1
evdT
T
so from this we know
for an isother·mal process change in entropy S
1
Qr = 300
2
Qr = 400
and S
=0
11 oo:oo:47 and S
2
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17
Entropy Calculation
@Sk J ahirudd in, 2020
•
using ~U + ~KE + ~ pE
=
Q- w
so we get 1 1 bS = - 2000 = 5 400 200 Sol 1.10. Free expansion of one mole of an ideal gas in adiabatic container from v1 ,T to v 2 ,T and T is not changed. during isothermal process ds1
v2
=
V1
pdV
=
Rln v2
T
V1
and t he change of entropy for the surroundings we get ds2
=
dQ
=0
T
AS the adiabatic container no heat was coming out side t he container and no heat goes inside t he container so the entropy of the surrounding remains same. Sol 1.11. Entropy loss for t he reservoir 600k where we t ake t hat the reservoir becomes at constant temperature .
~s _ s(600 loss -
300)
600
and t he entropy gain from _ann
__
11 oo : oo : so t hat the reservoir becomes at constant temperature . l::l.S
_ loss -
s(600 - 300) 600
and t he entropy gain from
dT
600 l::l.Sgain
=C
T
300 [email protected]
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18
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Entrop)' Calculation
so total change 6S = s(-0.5
+ ln 2) =
0.19s
[here we take reservoir temperature is constant as it is endless source of heat energy so just the body temperature increase and in equilibrium it 's temperature would be equal to reservoir temperature] Sol 1.12. T he gain of entropy
dS . - (0.001
X
335)
273
gain -
39 273
+
+
0.001
X
4.2dT
T
273
So, We get 0.0012 + 0.0005 = 0.0017 J/ K. SO the change in the entropy = gain in entropy + loss in entropy = 0.39 cal per degree Celsius Sol 1.13. Change of entropy of the gas 1
dS=
CvdT
T +
2v
R
=
p
-dV V
T
as the gas has initial volume v and after mixing the total vol-
11 oo :oo : s2 dS
=
CvdT T
+
2v
R V
p
-dV T
as the gas has init ial volume v and after mixing t he total volume is available to the gas but as the temperature remains unchanged so the the first term is zero but the second term dS1 j [email protected]
=
Rln2
19
physicsguide CSIR NET 1 GATE
Ent ropy Calculation
@Sk J ahiruddin, 2020
and same for t he second gas also so we are getting again dS2
=
Rln2
so t he mixture has entropy more t hat previous configuration 2Rln2.
Sol 1.14. From the previous discussions on the Carnot cycle gave us the in sight that during this reversible process t he total change in the entropy is ds = 0 and that leads us to t he the fact that t he final t emperature of t he hot source or cold source becomes ,JT;T;, so the heat that has been taken from the hot source is
and t he heat rejected at t he cold source is
SO total workdone
11 oo : oo : 55 and t he heat rejected at t he cold source is
SO total workdone
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Entropy Calculation
Sol 1.15. First of all the fact that we have to calculate the final t emperature Tf = T1 + T2 2 t he entropy gain by t he gas at temperature T1 is given by
dSgain = NCv and t he entropy loose dSzose = NCv non total change is
11 oo :oo : s1
Sol 1.16. Let us first calculate the final temperature after t he mixing. it is just the use of the calorimetry.
(0.01 x 335)
+ 0.01
x 4.2(T1 - 273) = 0.1 x (300 - T1)
from this we get that t he final temperature is 290.29kelvin. now let us calculate the loss of the entropy for the water at j [email protected]
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Entropy Calculation
@Sk J ahiruddin , 2020
300k and we get 290
dSloss
= 300
0.1 X 4.2dT T
and the gain of entropy dS . _ (0.01 x 335) gain 273
290
+
273
0.01 x 4.2dT T
SO the change in the ent ropy =gain in entropy entropy= 1.03 Joule/K.
+
loss in
Sol 1.17. For a particular value of the entropy is maximum
so we have ds = O dp
and t his condition will be giving us
00 : 01 : 00 so we have ds = O dp and t his condit ion will be giving us lnp + l - ln( l - p) - 1 = 0 and t his gives us that p = 1/ 2. Sol 1.18. As t he total energy of the system remains constant so t he temperature of t he gas remains constant and prom the previous discussion we can say t hat t he change in
t he entropy is just due t he enhancement of the volume of t he gas and is given by 4v
4v
pdV = NKB V
V
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@Sk J ahiruddin ) 2020
22
dv R - = 2N K Bln 2. V
physicsguide CSIR NET ) GAT E
EntrOP)' Calculation
so t he correct ans is 2. Sol 1.19. We know that the ent ropy of the system is depending on the thermodynamic probability of a system . What is t hermodynamic probability? it is t he number of microstates accessible t o t he system. t he more is t he number of microst at es the more will be t he entropy so in a way
you can say that this entropy is a measure of disorderedness in a system as the more t hermodynamic probability t he more is t he disorderedness. one can show that t hat the t he entropy is related to t he thermodynamic process probability through t he given relation
11 oo : 01
: 02
t he more is t he disorderedness. one can show that t hat the t he entropy is related to t he thermodynamic process pro bability t hrough t he given relation
where D is t he thermodynamic probability of t he system. for a spin one system the number of accessible state of the syst em is given by
n = (2J + 1) = 3 as J=l . Now we can say that
as the system consists of N particles. [email protected]
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@Sk J ahiruddin, 2020
Ent ropy Calculation
Sol 1.20. Change of entropy of the gas A is
dS=
CvdT
T +
2v
NK
p
-dV
B V
T
as the gas has initial volume v and after mixing the total volume is available to t he gas but as the temperature remains unchanged so t he the first term is zero but t he second term
and same for t he second gas also so we are getting again
11 oo : 01 : os unchanged so t he the first term is zero but t he second term •
•
and same for t he second gas also so we are getting again
so t he mixture has entropy more t hat previous configuration 2NKBln2 . Sol 1. 21. The process is related as follows
P = Po - aV the entropy of the system can be obtained ds
= dU + P dV T
T
now we know t hat P V= RT j [email protected]
24
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Entropy Calculation
@Sk J ahiruddin , 2020
so we get P dV
+ Vd P =
R dT
and we al so know t hat R Cv = - '°Y - 1 putting all t his in t he equation we get T dS
= P dV + Vd P + P dV '°Y - 1
11 oo : 01 : os Cv = - ' - 1 putting all t his in t he equation we get TdS = PdV + VdP
+ P dV
,- 1
SO solving this we get
T dS=
'
,- 1
P dV+ VdP
,- 1
now we know that
dP = -adV from the process it is given. Putting it back in the equation we get , VdV TdS = - - (Po - aV)dV - a - 1- l , -1 but for maximum of S dS= O so we get
0=
' ( Po -aV )dV-aVdV
,- 1
,- 1
from t his we can see
, Po = V a (l + , ) [email protected]
@Sk J ahiruddin, 2020
25
physicsguide CSIR NET, GATE
Entropy Calculation
Sol 1.22. Zero for all the materials as t he t hird law of t hermodynamics says that as the temperature goes to zero en-
t ropy also tends to zero.
00 : 01 : 10 a(l j [email protected]
+ ,) 25
physicsguide CSIR NET, GATE
Ent ropy Calculation
@Sk J ahiruddin, 2020
Sol 1.22. Zero for all the materials as t he third law of t hermodynamics says that as the temperature goes to zero en-
tropy also tends to zero.
j [email protected]
26
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11 oo : oo : oo
Problems and Solutions in Phase Transition Sk J ahiruddin * Suchismito Chattopadhyay
*Assistant P rofessor Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay 1\11.Sc Physics 2009-2011 batch He ranked 007 in IIT J Al\lI 2009 and 008 (JRF) in CSIR NET J une 2011 He has been teaching CSIR NET aspirants since 2012
1
@Sk J ahiruddin, 2020
Phase Transition
11 oo : oo : 03 1
@Sk J ahiruddin, 2020
Phase Transition
Contents 1 Problems from NET, GATE, JEST , TIFR & JAM pap ers 1.1 Ans keys . 1.2 Solut ions . • • •
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2 12 13
Problems from NET, GATE, JEST, TIFR & JAM papers
Prob 1.1. For a liquid to vapour phase t ransit ion at T tr which of the following plots between specific Gibbs free energy g and t emperature T is correct? [JAM 2012]
j [email protected]
2
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11 oo : oo : 06
j ahir@physicsguide. in
2
physicsguide CSIR NET I GATE
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Phase 'Iransition
I
r
.
.
d I
Prob 1.2. In the given phase diagram for a pure substance, regions I, II, III, IV, respectively represent [JAM 2019]
p Critical •••••••••••••••• Point
III
•
• ••
II~• I T
11 oo : oo : 09 •
II;• I T
[email protected]
3
@Sk J ahiruddin ) 2020
physicsguide CSIR NET ) GATE
Phase Transition
(A) Vapor, Gas, Solid, Liquid (B) Gas, Vapor, Liquid, Solid (C) Gas, Liquid, Vapor, Solid (D) Vapor, Gas, Liquid, Solid Prob 1.3. Consider the transition of liquid water to st eam as water boils at a temperature of 100°C under a pressure of 1 atmosphere. Which of the following quant ities does not [NET June change discontinuously at the t ransition? 2011] (a)The Gibbs free energy (b) The Internal Energy (c) The Entropy (d) The Specific Volume Prob 1.4. Consider the melting transition of ice into water at a constant pressure. Which of the following thermodynamic quant ities does not exhibit a discont inuous change across the phase transition? [NET Dec 2013] (a) Internal energy (b) Helmholtz free energy (c) Gibbs free energy (d) entropy Prob 1. 5. The entropy of a system , S , is related to the accessible phase space volume r by S = kB In r (E , N, V ) where E ,N and V are t he energy, number of particles and volume respectively. From this one can conclude that r [NET Dec 2012]
11 oo : oo : 12 Pro b 1. 5. The entropy of a system , S , is related to the accessible phase space volume r by S = kB In r (E, N, V )
where E,N and V are t he energy, number of particles and volume respectively. From t his one can conclude that r [NET Dec 2012] (a)does not change during evolut ion to equilibrium [email protected]
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Phase Transition
(b )oscillates during evolution to equilibrium (d) is a minimum at (c)is a maximum at equilibrium equilibrium Prob 1. 6. The entropy S of a thermodynamic system as a
function of energy E is given by the following graph
t
s
The temperature of the phases A,B and C , denoted by TA, TB and Tc , respectively, satisfy the following inequalit ies: [NET Dec 2013] (a)Tc > TB > TA (b)TA > Tc > TB
(c) TB > Tc> TA
(d)TB > TA > Tc
Prob 1. 7. The condition for the liquid and vapour phass
11 oo : oo : 14 [NET Dec 2013]
t ies:
(a)Tc > TB > TA
(b )TA > Tc > TB
(c) TB > Tc> TA
(d )TB > TA > Tc
Prob 1. 7. The condit ion for the liquid and vapour phass of a fluid to be in equilibrium is given by t he approximate
~f
equat ion ~ T Vvap Qt (Clausius- Clayperon equation), where Vvap is t he volume per part icle in t he vapour phase, and Qz j [email protected]
5
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Phase Transition
is t he latent heat , which may be taken to be a const ant. If t he vapour obeys ideal gas law, which of the following plots is correct? [NET June 2015]
0
T
Qnf 0
t--+-----
-r
Qt-+---- - -
-r
Prob 1.8. The phase diagram of a pure substance is given in t he figure below. Where T denotes the triple point and
11 oo : oo : 11
Prob 1.8. The phase diagram of a pure substance is given in the figure below. Where T denotes the t riple point and C denotes the critical point.
The phase transitions occurring along the lines marked a, /3 and , are [TIFR 2011]
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6
Phase 'I'ransition
@Sk Jahiruddin, 2020
C
~
--
::,
~
Q)
a. E
~
T
Pressure - -
(a)a=melting: /3=condensation: ,=sublimation (b )a=sublimation: /3=vapourisation: ,=melting (c) a=melting: /3 = vapourisation: ,=condensation (d)a=sublimation: /3= melting: ,=vapourisation Prob 1.9. The vapour pressure p (in mm of Hg) of a solid, at a temperature T , is expressed by lnp = 23 - 3863/T and
t hat of its liquid phase by ln p = 19 - 3063/ T. The triple
11 oo : oo : 20 (d)a=sublimation: ,B= melting: ,1=vapourisation Prob 1.9. The vapour pressure p (in mm of Hg) of a solid, at a temperature T , is expressed by In p = 23 - 3863 /T and
t hat of its liquid phase by ln p = 19 - 3063/ T. The triple [GATE 2007] point (in Kelvin) of the material is (a) 185 (b) 190 (c) 195 (d) 200 Prob 1.10. The pressure versus temperature diagram of a given system at certain low temperature range found to be parallel to the temperature axis in t he liquid-to-solid tran-
sition region. The change in t he specific volume remains constant in this region. The conclusion one can get from [email protected]
7
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Phase 'Iransition
@Sk J ahiruddin, 2020
[GATE 2008]
t he above is
(a) t he entropy solid is zero in this temperature region (b) t he entropy increases when the system goes from liquid to solid phase in t his temperature region (c) t he ent ropy decreases when the system transforms from liquid to solid phase in t his temperature region (d) t he change in entropy is zero in the liquid-to-solid transition region Prob 1.11. Identify which one is a first order phase tran-
sition? [GATE 2009] (a) A liquid to gas transition at its critical temperature (b) A liquid to gas transition close to its triple point (c) A paramagnetic to ferromagnetic transition in the abQt:)nf"P
f'lf
Q rn~O'nPtif"'
h Plrl
11 oo : oo : 22 sition? [GATE 2009] (a) A liquid to gas transition at its critical temperature (b) A liquid to gas transition close to its triple point (c) A paramagnetic to ferromagnetic transition in the absence of s magnetic field (d) A metal to superconductor t ransition in the absence of magnetic field Prob 1.12. In a first order phase t ransition , at the transition temperature, specific heat of the system [GATE 2011]
(a) diverges and its entropy remains the same (b) diverges and its entropy has finite discontinuity (c) remains unchanged and its entropy has finite discontij [email protected]
physicsguide CSIR NET 1 GATE
8
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Phase Ttansition
nuity (d) has finite discontinuity and its entropy diverges Prob 1.13. Across a first order phase transition , the free energy is [GATE 2013] (a) proportional to the temperature (b) a discontinuous function of the temperature (c) a continuous function of t he t emperature but its first derivative is discontinuous (d) such that the first derivative with respect to temperature is continuous Prob 1.14. Ice of density p1 melts at pressure P and absolute t emperature T to form water of density r ho 2 . The 1
J
J
,
r
1
I
•
('
•
•
T
,. T Tl
,
•
, 1
11 oo : oo : 24 (d) such t hat the first derivative wit h respect to temperature is cont inuous Prob 1.14. Ice of density p1 melts at pressure P and absolut e t emperature T to form water of density r ho2 . The latent heat of melting one gram of ice is L. What is t he change in the internal energy ~ U resulting from t he melt-
ing of 1 gram of ice? (a) L + P ( l - l) (b) L - P ( l - l ) P2
Pl
P2
(c) L- P (l l) P1 P2
[JEST 2014]
Pl
(d)L + P (l - l) P1
P2
Prob 1.15. The free energy F of a system depends on a t hermodynamic variable ~ as F
2
= - a~ + b~
6
with a,b > 0. T he value of ~ , when t he syst em is in t hermodynamic equilibrium, is [NET June 2014] [email protected]
9
physicsguide CSIR NET, GATE
@Sk J ahiruddin , 2020
(a) 0
Phase Transition
1 4
(b) ±(a/ 6b) !
1 4
(c) ±(a/ 3b) 1
1 4
(d) ± (a / b) 1
Prob 1.16. Water freezes at o° C at atmospheric pressure (1.01 x 105 P a) . The densities of water and ice at t his
temperature and pressure are 1000 kg /m 3 and 934 kg/m3 respectively. The lat ent heat of fusion is 3.34 x 105 J / Kg . T he pressure required for depressing t he melting temperat ure of ice by 10°c is _____GPa (up to two decimal places) [GATE 2017] Prob 1.17. At atmospheric pressure (=105 Pa), aluminium r
11 oo : oo : 21 t ure of ice by 10°c is _____GPa (up to two decimal places) [GATE 2017]
Prob 1.17. At atmospheric pressure (=105 Pa), aluminium melts at 550K. As it melts, its density decreases from 3 x l0 3 kg/m 3 to 2.9 x l0 3 kg/ m 3 . Latent heat od fusion of aluminum is 24 x l03 J /kg . The melting point of aluminum at a pressure of 107 Pa is closest to [JAM 2014] (a)551.3K (b)552.6K (c)558.7K (d)547.4 Prob 1.18. A many-body system undergoes a phase transition between two phases A and B at a temperature Tc. The temperature-dependent specific heat at constant volume Cv of t he two phases are given by c~) = aT 3 + bT and ciB) = cT 3. Assuming negligible volume change of the system, and no latent heat generated in the phase transition, Tc is [TIFR 2018]
j [email protected]
10
physicsguide CSIR NET I GATE
@Sk J ahiruddin , 2020
(a)
4b
c- a
Phase 'Iransition
(b)
3b
c- a
(c)
-2b C
(d)
b c- a
11 oo : oo : 30
[email protected]
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Phase Transition
@Sk J ahiruddin ) 2020
1.1
Ans keys Ans keys
1.1.
C
1. 7.
C
1.13.
C
1.2. b
1.8. b
1.14. d
1.3. a
1.9. d
1.15.
C
11 oo : oo : 32 1.1.
1. 7.
C
1.13.
C
C
1.2. b
1.8. b
1.14. d
1.3. a
1.9. d
1.15.
1.4.
C
1.10.
1.5.
C
1.11. b
1.17. a
1.6.
C
1.12. b
1.18. b
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1.16. 0.17
C
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physicsguide CSIR NET, GATE
@Sk J ahiruddin, 2020
1.2
C
Phase Transition
Solutions
Sol 1.1. Standard theory. The phase t ransition is not dis-
cussed in details in the notes. It will be updated quickly. P lease refer to Garg, Banshal and Ghosh. .
.
11 oo : oo : 3s Sol 1.1. Standard theory. The phase transition is not dis-
cussed in details in the notes. It will be updated quickly. Please refer to Garg, Banshal and Ghosh. Sol 1.2. The triple point refers to that temperature and pressure where three states of substance namely solid , liquid
and vapour co exist. 1) the equilibrium pressures and temperatures of a solid in contact with it's vapour up to triple point when plotted in a p-T diagram gives rise to t he sublimat ion curve . (the curve between II and III region denotes to the vapourization and t he curve between IV and II is sublimation curve and the curve between III and IV is called fusion curve and the point where t hey meet is called triple point .) 2)the equilibrium pressures and temperatures of a liquid in contact with it 's vapour up to the crit ical point when plotted gives rise to t he vapourisation curve is a continuation of the sublimation curve. 3) t he fusion curve is t he locus of equilibrium pressures and temperatures when a solid in contact with it 's liquid starting point being the t riple point. if
8P 8T
8P
8P 8T
= U1 and aT I
j [email protected]
@Sk J ahiruddin, 2020
Then we must have
I
13
II = U3 1
physicsguide CSIR NET, GATE
Phase Transition
00: 00: 37 Phase TI.·ansition
Then we must have
must not equal to zero. The III denotes liquid state IV denotes solid state I state denotes gas and the II denotes vapour. Sol 1.3. From t he previous discussion the Gibbs energy is
t hat quantity which changes cont inuously across the phase t ransition temperature as at phase transition point pressure and the temperature remains constant so change in the Gibbs energy is zero. Sol 1.4. The entropy is the quantity which cahnges abruptlty or discontinuously at phase transition temperature as the melting of ice into water is a first order phase t ransition. Sol 1.5. Entropy of the syst em always tend to maximize which is known to us so if we are associated to some ir-
reversible process then we must accompany the increase of t he entropy. and as we know s is related to t he accessible microstates of t he system so we must have that microstates will be maximum at equilibrium. Sol 1.6. We know t hat
[email protected]
@Sk J ahirudd in, 2020
as
1
8E
T 14
physicsguide CSIR NET, GATE
Phase TI.· ansition
(g~) this is called t he slope of the curve. so for t he b case the
11 oo : oo : 40 @Sk J ahiruddin, 2020
Phase Transition
(g~) t his is called t he slope of the curve. so for t he b case t he slope is zero so t emperature at b process is much larger t han c and b process but for c process t he internal energy E is greater than t hat of A process so definitely t he temperature at A process is less t han t hat of the c process so we have
•
~f
Sol 1. 7. ~ T ~Lap and we knovv that PVvap=RT and we will put t hat in the equation we will get t hat
dT
r2
so from this we get
C lnP = - - +a T so t his satisfies t he plot c. Sol 1.8. From t he previous discussion we can tell t hat t he
opt ion b is correct . Sol 1.9. The P - T diagram is continuous so at crit ical temperature
23 _ 3863 Tc [email protected]
@Sk J ahiruddin , 2020
= 19 _ 3063 Tc 15
physicsguide CSIR NET, GATE
Phase Transition
11 oo : oo : 42 [email protected]
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physicsguide CSIR NET, GATE
Phase Transition
Hence or
Tc= 200 Sol 1.10. When a solid transforms to liquid t hen it requires energy and the solid has less disorderness that liquid so when
liquid converts to solid t hen it rejects heat and it's disorderness reduces so ent ropy decreases. Sol 1.11. The option 2 is correct because the last two de-
notes 1) lambda phase transition where the second order derivative at the t ransit ion t emperature has an infinite discont inuity and t he 2) the last one is 2nd order phase transit ion as the 2nd order derivative at t he transit ion temperature has t he finite discontinuity. so option 3 and 4 can never be the answer. only 2 denotes t he 1st order phase transition •
Sol 1.12. In a first order phase transition the first order
derivative of t he Gibbs free energy has a finite discont inuity but it's second order derivative diverges at t he crit ical temperature. we know that entropy of as system is the first order derivative of the gibbs free energy with respect totemperature so it has the finite discontinuity at the transit ion j [email protected]
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11 oo : oo : 4s order derivative of the gibbs free energy with respect to temperature so it has the finite discontinuity at the transition j [email protected]
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@Sk J ahiruddin , 2020
Phase Transition
temperature but t he second order derivative is t he specific heat so it diverges. dG = - SdT + V dP so we can say 8G 8T
ac 2
p
=
- S ,T
ar2
8Q
p
=
=
-T
as
8T
p
-Cp
8T p hence we see 2nd order derivative is actually related to the specific heat. Sol 1.13. The first order phase t ransition is a special kind of phase transition where the first order derivatives of gibbs free energy becomes discontinuous accross the critial or phase changing temperature. so only the option 3 is correct. Sol 1.14. From t he previous discussion we have seen t hat t he internal energy changes wit h volume in the following
manner
au
=T 8P - P 8V 8T and t he Clausius Clapeyron equation gives us 8P 8T
L
11 oo : oo : 48 av
ar
and t he Clausius Clapeyron equation gives us
8P
L
ar [email protected]
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physicsguide CSIR NET, GATE
@Sk J ahirudd in, 2020
Phase 'I'ransition
so putting t his in t he equation we get
L
dU= - -
dV -
P dV
and we know that v = 1/ p so punting all t his in t he equation we get ~U
=
L
+P( l PI
-
l ) P2
[as at t he phase changing temperature t he pressure remains constant so it comes simply out of the integration .]
Sol 1.15. So at equilibrium for a particular value of 1/; F would be maximized so we have 8F 82 F = 0 81/; = 0' 8'lj) 2 2
so we have from t he first condit ion -2a'l/; + b'l/J 6 = 0 from t his 1 4 we get 'l/;=±(a/3b) 1
Sol 1.16. From t he Clausius Clapeyron equation we came to know t hat 8P L 8T . the meting point of ice is 273 kelvin , and t he depression of temperature 10°c here v 2 = (1/ 1000) = 1/ p1 and the v 1 = (1/ 934) = 1/ p2 and hence
11 oo : oo : so 8T . the meting point of ice is 273 kelvin , and the depression of temperature 10°c here v2 = (1/1000) = 1/ p 1 and the v 1 = (1/ 934) = 1/ p2 and hence L 263 - - - dT 273 T (V2 - v1) 18
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Phase 'Iransition
SO integrating we get P2 = 1.01 x 10
5
+(
3.34
1 - 934
X
105
+
1 ) 1000
263 5 ln = 0 .1 7 x 10 pa 273
Sol 1.17. From the Clausius Clapeyron equation we came to know that
8P 8T
L
now we have T =550K L= latent heat of phase change = 24 x 10 3 J / kg densities are given so volume can be easily deter3 3 mined P1 = 3 x 10 = lV1 similarly p2 = 2.9 x 10 = lV2 so we have 7 10 - 105 24 X 103 1 ( - 3x 10 3
55 0 ~T + 2_9;103 ) from this we get the change in the melting point in creases by 2.607 kelvin. so the final melting point at t he given pressure would be 552.6K. Sol 1.18. The Entropy is continuous in the second order phase transition where there is no lat ent heat. Entropy in state A is
CvdT = aT3
r
q
+
bT
11 oo : oo : s3 Sol 1.18. The Entropy is continuous in t he second order phase t ransition where t here is no latent heat. Entropy in state A is CvdT = aT3 bT T 3 +
Entropy in state B is CvdT = cT T 3 j [email protected]
@Sk J ahiruddin , 2020
19
3
physicsguide CSIR NET 1 GATE
Phase Ttansition
At crit ical temperature t hese two entropy become same
Hence the ans
00: 00: 55 T j [email protected]
19
@Sk J ahiruddin, 2020
3 physicsguide CSIR NET, GAT E
Phase Transition
At critical temperature these two entropy become same SA= SB
Hence t he ans
j [email protected]
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11 oo : oo : oo
Problems and Solutions in Microcanonical Ensembles Sk J ahiruddin * Suchismito Chatterjee
*Assistant P rofessor Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay 1\11.Sc Physics 2009-2011 batch He ranked 007 in IIT JAl\lI 2009 and 008 (JRF) in CSIR NET June 2011 He has been teaching CSIR NET aspirants since 2012
1
@Sk J ahiruddin, 2020
Microcanonical Ensembles
11 oo : oo : 03 1
@Sk J ahiruddin, 2020
Microcanonical Ensembles
Contents 1
Problems from NET, GATE, JEST , TIFR & JAM pap ers 1.1 Ans Keys 1.2 Solutions . • • •
j [email protected]
•
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physicsguide CSIR NET, GATE
11 oo : oo : os
j [email protected]
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1
physicsguide CSIR NET I GATE
Microcanonica l Ensembles
Problems from NET, GATE, JEST, TIFR & JAM papers
Prob 1.1. Thermodynamic variables of a syst em can be
volume V , pressure P , t emperature T , number of part icles N , internal energy E, chemical potent ial µ etc. For a syst em to be microcanonical (MC) , canonical (CE) and grand canonical ( GC) ensembles, the parameters required [GATE 2008] are (a) JVIC:(N, V, T ); CE:(E, N, V ); GC:(V, T , µ) (b) MC:(E, N, V); CE:(N, V, T ); GC: (V, T , µ) (c) MC:(V, T , µ); CE:(N, V, T) ; GC:(E, N, V ) (d) MC:(E , N , V ); CE:(V, T, µ); GC: (N , V, T ) Prob 1.2. A paramagnetic syst em consist ing of N spin- ~
particles is kept in an external magnetic field. It is found t hat N / 2 spins are aligned parallel and t he remaining N / 2 spins are aligned anti-parallel to the magnetic field. The stat istical entropy of the syst em is [GATE 2012]
(a) 2NkBln2
(b) (N/2)kBln2
(c) (3N/ 2)kBln2
(d) NkBln2
Prob 1.3. Consider a linear collection of N independent
spin-~ particles, each at a fixed location. The ent ropy of the
11 oo : oo : os
Pro b 1. 3. Consider a linear collection of JV independent spin- ~ particles, each at a fixed location. The ent ropy of the [email protected]
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3
Microcanonical Ensembles
@Sk J ahiruddin ) 2020
syst em is
(a) 0
(b) NkB
(c) (l / 2)NkB
[GATE 2013] (d) NkBln2
Prob 1.4. A system of N distinguishable particles, each of which can be in one of t he two energy levels O and c, has a total energy nc, where n is an integer. T he entropy of t he [NET June 2015] system is proportional to NI
(a) Nlnn
(b) nlnN
(c) ln
n !.
(d) ln
NI n !(N ~ n)!
Pro b 1. 5. Consider a gas of N classical part icles in a twodimensional square box of side L. If t he total energy of the gas is E, t he entropy (apart from an additive constant) is [NET Dec 2016] 2 (a) NkB ln(L E / N ) (b) NkB ln (LE/ N)
Prob 1.6. In a t hermodynamics system in equilibrium each molecule can exist in t hree possible states with probabilities 1/ 2, 1/3 and 1/ 6 resp ectively. The entropy per molecule is [NET June 2017]
1
(d) kB ln 2 + ~kB ln 3
Prob 1. 7. The number of microstates of a gas of N par-
11 oo : oo : 11 [NET June 2017]
molecule is (a) kB ln 3
(b) ~kB ln 2 + ~kB ln 3 (d) ~kB ln 2 + ~kB ln 3
Prob 1. 7. The number of microstates of a gas of N par-
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4
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physicsguide CSIR NET, GATE
Microcanonical Ensembles
t icles in a volume V and of internal energy U, is given by
n (u, V, N) = (V - Nb)N
a;:
3N/ 2
(where a and b are positive constants) . Its pressure P , volume V and temperature T , are related by [NET Dec 2017]
(a) (P + at)(V - nb) = NkBT (b) (P - at)(V - nb) = NkBT (c) PV = NkBT (d) P (V - nb) = NkBT Prob 1.8. In a system comprising of approximately 10 23 distinguishable particles, each particle may occupy any of 20 distinct states. The maximum value of the entropy p er [NET June 2019] particle is nearest to (a) 20kB (b) 3kB (c) 10(ln 2)kB (d) 20(ln2)kB Prob 1.9. Consider a system maintained at temperature T with available energy states E 1 and E 2 each with degen-
11 oo : oo : 14 particle is nearest to (a) 20kB (b) 3kB (c) lO (ln 2)kB
[NET June 2019] (d) 20(ln 2)kB
Prob 1.9. Consider a system maintained at temperature T with available energy states E 1 and E 2 each with degeneracies g1 and g2. If P1 and p2 are the probabilities of occupancy of these ststes, what is t he entropy of the system? [JEST 2012] j [email protected]
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(a) S = - kB[P1ln(p1/ g1 ) + p2ln(p2 / g2) ]
+ p2ln(p2g2) ] (c) S = - kB [p1 l n (Pi + P2 l n (p~ (d) S = - kB [(I / p1)ln(p1 / 91) + (l / p2)ln(p2/ 92)] (b) S = -kB[P1ln(p1g1) 1
2
)
)]
Prob 1.10. Consider a particle with 3 possible spin states: s = 0, ± 1. There is a magnetic field h present and t he energy for a spin state s is -hs. The system is at temperature T . Which of t he following is true about t he entropy S(T )? [JEST 2013] (a) ln3 at T = 0, and 3 at high T (b) ln 3 at T = 0, and O at high T (c) 0 at T = 0, and 3 at high T (d) 0 at T = 0, and ln3 at high T
Prob 1.11. Consider a system of 2N non-interacting spin~ particles fixed in position and carrying magnetic moment µ . The system is immersed in a uniform magnetic field B.
11 oo : oo : 16 (d) 0 at T = 0, and ln3 at high T Prob 1.11. Consider a system of 2N non-interacting spin~ particles fixed in position and carrying magnetic moment
µ. The system is immersed in a uniform magnetic field B.
The number of spin-up particles for which the entropy of [JEST 2014] t he system will be maximum is (a) 0 (b) N (c) 2N (d) N/2 Prob 1.12. A system having N non-degenerate energy eigenstates is populated by N identical spin-zero particles [email protected]
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and 2N identical spin-half particles. There are no interactions between any of these particles. If N = 1000, the entropy of the system is equal to [TIFR 2011] (a) 13.82kB (b) 0 (c) 693.lkB (d) lOOOkB (e)5909.693kB
(f)6909kB
Prob 1.13. The entropy S of a black hole is known to be of the form S = akBA, where A is the surface area of
the black hole and a is a const ant , which can be written in terms of c (velocity of light in vacuum) , Ii (reduced Planck's constant) and GN (Newton's constant of gravitation). Tak2 ing t he radius of the black hole as R = GJM, it follows that t he entropy is
__
___
A(nc) 4 AGNM (Hint: Use dimensional analysis)
[TIFR 2013]
Alie
Prob 1.14. A system at temperature T has 3 energy states: ,. . , . , , .
11 oo : oo : 1s 2
kB G7vM ( ) a A(lic) 4
2
(b) lickB AGNM
( ) G'JvM kB C
Alic4
(Hint: Use dimensional analysis) Prob 1.14. A system at temperature T has 3 energy states:
0,±c. The entropy of the system in t he low t emperature (T ➔ 0) a nd high temperature (T ➔ oo) are respectively [TIFR 2013]
(a)
sT➔O =
0, sT➔co
(b) sT➔O
= sT➔oo =
(c) sT➔O
=
=
kB exp (-3)
kB ln 3
0, sT➔oo = kB In 3
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Prob 1.15. Consider the CO molecule as a syst em of 2
point particles which has both t ranslational and rotat ional degrees of freedom. Using classical statistical mechanics, t he molar specific heat Cv of CO gas is given in t erms of [TIFR 2014] Boltzmann constant kB by
(a) ~kB
(b) 2kB
(c) ~kB
(d) ~kB
Prob 1.16. Cosmic r ay muons, which decay spontaneously with proper lifetime 2.2µs are produced in t he atmosphere,
at a height of 5 km above sea level. These move straight downwards at 98% of t he sp eed of light. Find the percent rat io 100 x (NA/NB) of the number of muons measured at t he top of two mountains A and B , which are at heights 4, 848m and 2, 682m respectively above mean sea level. [TIFR
11 oo : oo : 21 downwards at 98% of t he speed of light. Find the percent ratio 100 x (N A/NB) of the number of muons measured at the top of two mountains A and B, which are at heights 4, 848m and 2, 682m respectively above mean sea level. [TIFR 2017]
Prob 1.17. Consider N non-interacting distinguishable particles in equilibrium at an absolute temperature T. Each particle can only occupy one of two possible states of energy 0 and E respectively (E > 0). The entropy of the system, in terms of /3 = c/kBT is [TIFR 2019] (a) NkB In (1 + e- 13 ) - 1:~~13 j [email protected]
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(b)
(c) (d)
NkB In
(1 + e-f3) -
13 el -e-/3
Prob 1.18. A monatomic crystalline solid comprises of N atoms, out of which n are in interstitial positions. If t he available interstitial sites are N' , then t he number of possible microstates is [GATE 2006] (N' + n) ! b N! N'! (a) n!N! ( ) n!(N + n)! n!(N' + n )! I N ( ) . c n !(N' - n) !
(d)
NI
.
N'I
.
n !(N - n)! n !(N' - n) !
00: 00: 24 (a) (N' + n) ! n !N !
NI
(b) n !(N ~ n) ! n !(N'
NI
(c) n!(N'
~ n)!
N' I
NI
(d) n !(N
~
~ n) !
N' 1 n) ! n !(N' ~ n) !
Prob 1.19 . Consider a system of N non-interacting spin~ particles, each having a magnetic moment mu in a magnetic field B = B z. If E be t he total energy of t he system, t he number of accessible microstates D is given by [GATE 2015] E N N! µB (a) n = (b) D = 1 E I~ E E I• NN+ N +
-
2
µB
•
µB
2
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µB
' •
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1
I
•
Microcanonical Ensembles
E
,~ N - - ·2 (c) D = 2 µB
E
N + µB !
N! (d) D= - - N+ µB. E '
Prob 1.20. Consider N non-int eracting, distinguishable particles in a two-level system at temperature T . The energies of the levels are O and c. In t he high temperature limit (kBT >> c), what is t he populat ion of part icles in t he level [GATE 2017] wit h energy c? N N 3N ( b) N (d) (a) (c) 2 4 4 Prob 1.21. A microcanonical ensemble consists of 12 atoms
11 oo : oo : 26 par IC es In
[GATE 2017]
with energy c?
N N (d) 3N (a) (b) N (c) 2 4 4 Prob 1.21. A microcanonical ensemble consists of 12 atoms with each taking either energy O state, or en ergy
E
state.
Both states are non-degenerate. If t h e total energy of t his ensemble is 4E, its entropy will be ____ kB (up to one decimal place), where kB is the Boltzmann constant. [GATE 2018]
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1.1
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Microcanonical Ensembles
Ans Keys
1. 1. b
1. 8. b
1. 15. a
1. 2. d
1. 9. a
1. 16. 052
1. 3. d
1. 10. d
1. 17. b
1. 4. d
1. 11. b
1. 18. d
1. 5.
C
1. 12. f
1. 19. a
1. 6.
C
1. 13. d
1. 20. a
11 oo : oo : 29 1. 3 . d
1. 10. d
1. 17. b
1. 4 . d
1. 11. b
1. 18. d
1. 5.
C
1. 12. f
1. 19. a
1. 6.
C
1. 13. d
1. 20. a
1. 14 .
1. 21. 6.2
1. 7. d
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1.2
C
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Solutions
Sol 1.1. t he concept of ensemble in the statistical mechanics is very important concept . each microstate of a N part icle gas is represented by a point in the 6N dimensional phase space .sO if one considers all possible microstat es which t he gas can have , one will have a huge collection of points in t he
11 oo : oo : 31 ics is very important concept. each microstate of a N particle gas is represented by a point in the 6N dimensional phase space .sO if one considers all possible microstates which t he gas can have , one will have a huge collection of points in t he phase space of all possible microstates of t he syst em, this is called ensemble.one just can think of t hat each point in t he phase space is just a copy of t he possible configuration of t he syst em and such a haze collection of the points is called t he ensemble. there are three kind of possible ensemble in t he statistical mechanics . these are 1) microcanonical ensemble 2) canonical ensemble 3) grand canonical ensemble. in the micro canonical ensemble we take the volume of t he ensemble and energy of the ensemble and the number of t he particle in t he ensemble remains constant. in the canonical ensemble we take the energy number of particle and t he temperature remains constant. and in t he grand canonical ensemble we take the volume temperature and the chemical potential of the syst em is constant. so option b is correct.
Sol 1.2. t he important concept of the statistical entropy is t hat t he entropy is the measure of the dis order of the sytem and the n1ore microstates available to a system the more will [email protected]
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be the entropy of the system. the relation of the entropy and t he available microstates of t he system is given by
whP-r P- ko is thP- Rolt,7,m::1,n n r.onst.::1,nt ::t.nrl t,hP- w is thP- m i-
11 oo : oo : 34 e entropy o t e system. t e re ation o t e entropy and the available microstates of t he system is given by
where kB is the Bolt zmann constant and the w is t he microstates available to the syst em. here only two microstates are available to the system of N particles , so we have
hence t he option d is correct. Sol 1. 3. if the spin of a syst em is ½ then the number of
available microst at es of the syst em is given by
hence the entropy of the system in the case of the N number of the particle is given by
hence the option d is correct. Sol 1.4. this one again is required to calculate the number
of ways the arrangements can be done such that the total energy is remaining constant . this can be dome if n number j [email protected]
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Microcanonical Ensembles
of particle to be in t he higher energy levels and all the other particles remain in O energy level. that is now the question is how much way we can choose n number of t he particle
00: 00: 36 Microcanonical Ensembles
of particle to be in t he higher energy levels and all the other particles remain in O energy level. that is now the question is how much way we can choose n number of t he particle from the N number of the particle so this can be done in the following number of t he ways and is given by
D=
N! (N - n) !(n) !
hence we have the entropy of the system
N! S = kn ln (N _ n) !(n) ! . so option d is correct.
Sol 1. 5. we know t hat the number of shells in the phase space 1s _ dxdpxdydpy dnn,2 •
( here we consider that if a particle has momentum change say dp and position change say dx t hen if there implication is less than Ii then we will be unable to detect that if t he particle makes t he change of its position ) so we can say that is the phase space is n dimensional then number of shells we have in the phase space is
_ dx1 ...... dxNdPxl·····dPxN dn n,N [email protected]
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from this concept we write down the upper equation. so in
11 oo : oo : 39 @Sk J ahiruddin , 2020
Microcanonical Ensembles
from t his concept we writ e down t he upper equation. so in t his case t he available space is
and hence we have
ftEfN
L2
n = r,,2 o
L2
21rpdp =
E r,,2 21r N
so t he microstates available
D= n N
L221r E r,,2 N2
=
now entropy would be S=NkB lnD = 2NkBln L,/E/N
Sol 1.6. the entropy of the system is given by
hence we have
1
1
S = -kB( ln(l/2) + ln(l /3) 2 3 2 1 = - kB ln 2 + - kB ln 3 3 2
1
+ 6ln(l/6))
Sol 1. 7. we know that
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11 oo : oo : 42 15
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Microcanonical Ensembles
so we can write down that
aU
3N/ 2
N
which can be simplified to write down as
but we know that from the first law of thermodynamics dQ= TdS= dU+ PdV so from this we can say that
as
1
au v
-
r
now from the above relation we can write that 1 T
3N k N a 2 Bau N
now we want to evaluate t he pressure and from the first law of thermodynamics we again see
as av
P
u
r
this gives us t hat
p
1 T =NkBV -Nb j [email protected]
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11 oo : oo : 4s p
1 T =NkBV -Nb j [email protected]
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Microcanonical Ensembles
so t he above relation can be writ ten as
ao t he last option is correct. Sol 1.8. we know t hat if there is N number distinct of sates in a system and n umber of particles in t hat syst em t hen t here is accessible st ates in the system is given by
we use t his formulae to evaluate the entropy of the system as we ha,,e known that ent ropy is related to the microstates of t he system and as t he microstates increases the entropy of the system also increases in t hat sense we can say that t he ent ropy measures t he disorderness in t he syst em and the relation is given by so t he ent ropy is given by
so the ent ropy per molecule is given by S / N = 3kB
11 oo:oo:47 so t he ent ropy per molecule is given by
S / N = 3kB [email protected]
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Sol 1. 9. t he entropy of t he system is defined also as S
= - kB
L Piln pi .
'l
where p represents t he probability of each st ate. so we have
as t he first state has probability p 1 and t he second state probability p 2 Sol 1.10. to solve t his kind of problem we need to know what is part ition funct ion . by mathematical expression the
partit ion function is given by for discrete syst em
and in terms of t he partit ion function we can define t he Helmholtz free energy function
and we know t hat
8F
S =entr opy = - BT
for this part icular system we can define
11 oo : oo : 49 and we know that
8F
s =entropy= - ar
for t his particular system we can define
Z =l +
h
ekBT
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+ e-
18
h
kBr
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Microcanonical Ensembles
now we can define In Z
h
= ln(l + e kBT + e -
h
kBr)
so we can say that h
F = - kB T In ( 1 + e kBT
h
+ e- kBr )
so t he entropy can be calculated from the previous relation S
=
kB ln (l
+ kBT
h
+ ekBr + e -
h
kBT )
h
l h ekB r
+
hence we see t hat as T
➔
(1 +
h e - kBT )
h
2 (-ekBT
+ e-
h
kBr )
kBT
oo the S = kB In 3. so option d is
correct.
Sol 1.11. let there be n numbers of spin up and the spin down are then (2N-n) now we have to calculate in how many way t hat spins can be arranged because we need to calculate t he thermodynamic probability and we have n _ ~
t. -
n n _ ~ l,1 ~ l,2 -
2N! 2N! (2N - n) !n ! n !(2N - n) !
as the thermodynamic probability is multiplicative quantity.
11 oo :oo : s2 t he thermodynamic probability and we have n _ ~
G -
2N! 2N! (2N - n)!n! n !(2N - n)!
n
n _ ~ Gl~ l2 -
as the thermodynamic probability is multiplicative quantity. so t he entropy of the system we have S = kBlnO t hen the entropy of the system 2N ! S = 2kB ln (2 N _ n)!n! j [email protected]
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Microcano11ical Ensembles
now we will use t he star ling approximation and that tells lnN! = NlnN - N hence we will have S = 2kB [2Nln2N - 2N - (2N - n) ln(2N - n)
+(2N - n) - n ln n + n] for maximum we must have
as= 0 an which gives n= N hence option B is correct . Sol 1.12. for classical one t he number of orientation is given by w = NN hence the entropy S=6908KB but for t he spin half ones we have 2N number of particle will be accommodate in N states so only one arrangement is possible so we
have total entropy S=6909KB Sol 1.13. from the concept of t he dimension analysis we takes= cxnYGNzkBA So we have r
1
rr
1
rrn - l 1 r .ry
1
r
11
,r - l
r
2rn-21
r-1--1
r
11
,r2rn- l 7
11 oo : oo : 55 have total entropy S=6909KB Sol 1.13. from t he concept of the dimension analysis we take S = cxfiYGNzkBA So we have
so applying this in the above equation we have
[S] = [M L 2r - 2 0- 1 ] = [[L][T-l]] x[M-1 L2T-2] z[M2T-l]Y [M LT-20- 1] [email protected]
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Microcanonical Ensembles
now comparing both side we get S
2
= !iGNM kB Arie
Sol 1.14. see the problem 15. Sol 1.15. let us calculate the degrees of freedom t hat is
given by f=3N-k= 3x 2-1=5 and from t he equipartition t heorem we have ea h degrees freedom will have 0. 5kBT hence we internal energy U = ~kBT hence Cv =
j~
= ~kB
Sol 1.16. Sol 1.17. let us first calculate t he part ition function and that is given by z
= (1 + e-c/kB T)N
hence t he Helmholtz free function
11 oo :oo : s1 z
= (1 + e-c/ kBT)N
hence t he Helmholtz free function
now we have to calculate t he entropy
S= _ 8F 8T so doing this we get
S
=
E
NkB ln(l
which gives S
=
e - c/ kBT
+ e - c/kBT) + - - - - kBT 1 + e - c/ kBT
NkB In (1
+ e- /3c ) + l+ef3e - f1 .B
21
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Microcanonical Ensembles
Sol 1.18. first let us see t hat in how much way t he n is arranged in t he N number of t he atoms and it is given by D1
=
n!ff~n!
now we are given N' interstitial sites so in how
much way n is arranged in that sites we have D2 so t he total probability will be n n _ ~ll~ l2 -
=
(N'~~)!n!
N' ·I N I· n !N - n ! (N' - n) !n !
hence opt ion d is correct.
Sol 1.19. we have let n 1 part icle has spin 1/ 2 and n 2 has spin down (-1/ 2) so we have n1 + n2 = N and also-n1µ B + n 2 µ B = E hence we can say n 1 = ~ N - µ~ and we have n 2½ N
+ µ~
so t he number of states are given by
N! R \
1 /
R \
so option a is correct .
11 oo : 01 : oo spin down (-1/ 2) so we have n1 + n2
=N
and also-n1µ B
n 2 µB = E hence we can say n 1 = ~ N h ave n 2~ N
+ µ~
2
N- -
E
µB
and we
so t he number of states are given by
N!
O= 1
µ~
+
so option a is correct.
,!·2
N+ E I µB.
Sol 1.20. t he part ition function Z
= 1 + e c:/kBT so the frac-
t ion of population in t he higher energy state is given by e-c:/ kBT N2
= N 1 + e c:/kBT
so as t is going to higher t emperature N 2 =
1
Sol 1.21. we have to see t hat how many ways we can arrange 12 atoms in t he 2 energy state such that t he energy [email protected]
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is 4E t he number of the ways is given by 12!
0 = =495 4!8! hence t he entropy S = kBlnO = 6.204kB so upto one decim al we have 6.2 .
11 oo : 01
: 03
range 12 atoms in t he 2 energy state such that t he energy j [email protected]
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Microcanonical Ensembles
is 4E the number of the ways is given by 12! D= =495 4!8! hence t he entropy S = kBlnO = 6.204kB so upto one decimal we have 6.2 .
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Problems and Solutions in Canonical Ensembles: Part-I Sk J ahiruddin * Suchismito Chatterjee
*Assistant P rofessor Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay 1\11.Sc Physics 2009-2011 batch He ranked 007 in IIT JAl\lI 2009 and 008 (JRF) in CSIR NET June 2011 He has been teaching CSIR NET aspirants since 2012
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Car1onical Ensembles: Part-1
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Canonical Ensembles: Part-1
Contents 1
Problems from NET, GATE, JEST , TIFR & JAM pap ers 1.1 Ans Keys 1.2 Solutions . • • •
j [email protected]
•
•
•
•
•
•
•
•
•
•
•
•
•
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•
2
•
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3 •
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1
2
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Canonical Ensembles: Part-1
Problems from NET, GATE, JEST, TIFR & JAM papers
Prob 1.1. The partition function of a single gas molecule
is Za . T he partition funct ion of N such non-interacting gas [GATE 2007] molecules is t hen given by (a) (Za)N / N ! (b) (Za)N (c) N(Za) (d) (Za)N / N Prob 1.2. C0 2 molecule has t he first few energy levels separated by approximately 2.5 meV. At a temperature of
300K, the ratio of number of molecules in t he 4 th' excited state to t hat in t he 2n d excited state is [GATE 2010] (a) 0. 5 (b) 0. 6 (c) 0.8 (d) 0. 9 Prob 1.3. In I-dimension , an ensemble of N classical parp2 1 2 t icles has energy of the form E = x + -kx . T he average 2m 2 internal energy of the system at temperature T is [JAM 2014]
(a) ~NkBT
(b) ½NkBT
(c) 3NkBT
(d) NkBT
(Hint: you can directly use equipartit ion theorem) Prob 1.4. An ensemble of quantum harmonic oscillators is kept at finite temperature T = l /kB /3 - [GATE 2007]
(i) T he partition function of a single oscillator wit h energy levels (n + ½) nw is given by
11 oo : oo : os is kept at finite temperature T = 1/ kB/3 - [GATE 2007] (i) T he partition function of a single oscillator wit h energy levels (n + ½) nw is given by [email protected]
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(a) Z =
Canonical Ensembles: Part-1
e-f3nw/ 2 1 - e- /3nw
1
-(c) Z -- 1 - e- /3nw
e-f3nw/ 2
(b) Z -- 1 + e- /3 nw 1 (d) Z -- 1 + e-f3nw
(ii) The average number of energy quanta of t he oscillators is given by
(a) (n) = (c) (n) =
1 ef3nw - 1
1 ef3nw
(b) (n) =
+ 1 (d) (n) =
e- f3nw ef3nw - 1 e- f3nw e- f3nw + 1
Prob 1.5. Consider a system whose 3 energy levels are 0, E and 2c. T he energy level c is 2-fold degenerate and t he other energy levels are non-degenerate. T he partition
function of the system with /3 = 1/kBT is given by [GATE 2012] (a) 1 + 2e-/3c (b) 2e- /3c + e- 2/3c
(c) (1 + e- 13€)2
( d)
1 + e- /3c + e- 2/3c
Prob 1.6. A gas of N non-interacting particles is in thermal equilibrium at temperature T . Each part icle can be in any one of the possible non-degenerate stat es of energy 0, 2c and 4c. T he average energy per part icle of t he gas, when
/3E > L kept vertically in t he Earth's gravitational field. The average energy of t he gas at low temperatures such that mgL >> kBT is given by [NET Dec 2011] (a) NkBT/2 (b) 3NkBT/ 2 (c) 2NkBT (d) 5NkBT/2 Pro b 1. 9. The free energy of a gas of N part icles in a volume V and at a temperature T is
11 oo : oo : 13 Prob 1.9. The free energy of a gas of N particles in a volume V and at a temperature T is
where a0 is a constant. The internal energy of the gas is
[NET June 2012] (b) (5/ 2)NkBT
(a) (3/ 2)NkBT j [email protected]
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5 2
(c) NkBTln[aoV(kBT) 1 /N] - (3/ 2)NkBT 5 2
(d) NkBTln[a0 V /(kBT ) 1 ] Prob 1.10. A system has two normal modes of vibration, with frequencies w1 and w2 = 2w1 . What is the probability t hat at temperature T , t he system has energy less than 4nw1 ? [In t he following x = e-f3nwi and Z is the partition function.] [NET June 2012]
(a) x 312 (x + 2x 2 )/ Z (c) x 312 ( l
+ 2x 2 )/Z
+ x + x 2 )/Z (d) x 312 (1 + x + 2x 2 )/Z (b) x 312 (l
Prob 1.11. A system can have 3 energy levels: E = 0, ±E. The energy level E = 0 is doubly degenerate, while others are non-degenerate. The average energy at inverse temperature {3 is [NET June 2014]
(c) 0
Prob 1.12. A system of N non-interacting classical parti-
11 oo : oo : 16
Prob 1.12. A system of N non-interacting classical particles each of mass m is in a 2-dimensional harmonic oscillator
potential of the form V(r) = a(x 2 +y 2 ), where a is a posit ive constant. The canonical part ition function of the syst em at temperature T is [,B = l / kBT] [NET June 2015]
[email protected]
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Canonical Ensembles: Part-1
@Sk J ahiruddin, 2020
(a)
(c)
a 2m
an
2 7r
N
(b)
,B N
(d)
2m,B
2mn
2N
a,B
2mn
2
N
a,82
Prob 1.13. For a system of independent, non-interacting
I-dimensional oscillators, the value of free energy per oscil[NET Dec 2015] lator in the limit T ➔ 0 is
(a) (U;J/2
(b)
(c) 3nw/2
(U;J
(d) 0
Prob 1.14. A gas of non-relativistic classical particles in
one dimension is subj ected to a potential V (x) = a x (where a is a constant). The partition function is (,B = 1/ K = kBT) [NET June 2016]
(a)
(b)
11 oo : oo : 19 [NET June 2016]
(a)
(c)
41rm f32a,2 n,2
81rm 132a2 n,2
(b) (d)
21rm f32a,2n,2
31rm f32a,2 n,2
Prob 1.15. The partition function of a two-level system governed by the Hamiltonian ,' - > ~ )varies with temperature [NET Dec 2018] as j [email protected]
10
Canonical Ensembles: Part-1
@Sk J ahiruddin, 2020
3 2
(a) 1/ T 1
(b) 1/ T
physicsguide CSIR NET I GATE
3
(
c) 1/T
(d) 1/ T
2
Prob 1.24. The Hamiltonian of a classical nonlinear one
dimensional oscillator is H = 2~p2 + Ax 4 where A > 0 is a constant. The specific heat of a collection of N independent such oscillators is [NET June 2019] (a) 3NkB/2 (b) 3NkB / 4 (c) NkB (d) NkB/2 (Hint: Generalized equlipartition theorem) Prob 1.25. For a system in t hermal equilibrium with a
11 oo : oo : 29 [NET June 2019] (a) 3NkB/2 (b) 3NkB / 4 (c) NkB (d) NkB / 2 (Hint: Generalized equlipartition theorem )
such oscillators is
Prob 1.25. For a syst em in t hermal equilibrium with a heat bath at t emperature T , which of the following equali-
==
t ies is correct?
[JEST 2015]
Prob 1.26. A particle in t hermal equilibrium has 3 possible states with energies -E, 0, c . If the system is maintained at a temp erature T > > c / kB, the average energy of the part icle can be approximated to [JEST 2015] 2 2 2 2c 2c c
(a) 3kBT
(b) - 3kBT
(d) O
(c) - kBT
Prob 1.27. A gas of N molecules of mass m is confined 3 in a cube of volume V = L at t emperature T. The box is in a uni£orm gravitational field - g z. Assume that the potential energy of a molecule is U = mgz where z E [O, L ] jahir@physicsguide. in
11
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@Sk J ahiruddin ) 2020
Canonical Ensembles: Part-1
is the vertical co-ordinate inside the box. The pressure P (z)
[JEST 2016]
at height h is
mg
exp N
(z -
½)
kBT
L
(a) P (z) = V m2g - - - - - mgL sinh (
2kBT ma ( z - ~) \
11 oo : oo : 31 mg z - 2 exp kBT N L (a) P(z ) = V m2g - - - - - mgL sinh 2kBT exp
(b) P (z) =
mg
(z - ~)
kBT N L Vm2g - - - - - m gL cosh 2kBT
(c) P (z ) = kBT N V
N (d) P( z ) = Vmg z Prob 1.28. For a quantum mechanical harmonic oscilla-
tor wit h energies E n = ( n
+ t) rii.,J,
where n = 0, l , 2, · · · , [JEST 2016]
Prob 1.29. If t he men square fluctuations in energy of a [email protected]
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Car1onical Ensembles: Part-1
system in equilibrium at temperature T is proportional to ra, then t he energy is proportional to [JEST 2017] (a) y a-2 (b) y a/2 (c) y a-l (d) y a
11 oo : oo : 34 system in equi 1 r1um at temperature T a, then the energy is proportional to (a) r a-2 (b) r a/2 (c) r a -l
j [email protected]
@Sk J ahiruddin ,
1.1
13
2020
is proport1ona to [JEST 2017]
(d)
Ta
physicsguide CSIR NET, GATE
Canonical Ensembles: Part-1
Ans Keys
1. 1. a
1. 11. d
1. 21. d
00: 00: 36 anonical Ensembles: Part-1
1.1
Ans Keys
1. 1. a
1. 11. d
1. 21. d
1. 2.
C
1. 12. d
1. 22. a
1. 3. d
1. 13. a
1. 23. d
1. 4. (i) a, (ii) a
1. 14.
1. 5.
1. 15. b
C
1. 6. a 1. 7.
1. 16. 1. 17.
C
C
1. 24. b 1. 25. a
C C
1. 26. b 1. 27.
C
1. 8. d
1. 18.
1. 9. b
1. 19. a
1. 28. d
1. 10. d
1. 20.
1. 29.
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1.2
Solutions
C
C
14
C
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Canonical Ensembles: Part-1
11 oo : oo : 39 @Sk J ahiruddin, 2020
1.2
Canonical Ensembles: Part-1
Solutions
Sol 1.1. we have t he single particle partition function Za so for the N particle we will have Za N but as we are t aking non interacting particle so we will have counted N! number of
states extra so we must have to devide them by N! so actual partit ion function [2
=
Z aN
N! Sol 1.2. t he ground state energy is E 0 then t he first excited energy (Eo + 2.5)meV t hen the 2nd excited energy (Eo + 5)meV and t he 3rd excited state (E 0 + 7.5 )meV and the fourth excited state has the energy (E0 + lO )meV. so the ratio is N4 Ni0 e-(Eo+l0)/300ks = 0 8 N Na e-(Eo+5)/300kB · 2
Sol 1.3. each degrees of t he freedom has t he energy 0.5knT
at temperature Thence t he total energy 1s < E >=< fm > + < ~kx 2 >= 2 x ~knT = knT so for the N number of particle we have •
E avg
2
= NknT
( here we have directly use t he equipartition theorem for details of equipartition see thermodynamics part first )
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11 oo : oo : 42 15
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Sol 1.4. t he partition function
so we have t here is a constant terms in the partition and it will not have any contribution in any macroscopic contribution.so to calculate first we subside the constant terms and calculate the other term. so we will have the fact that
z=
L e- nnw(3 =
l
+ e- nw(3 + e- 2nw(3 + .....
so this can be written as like as the following z
=
e- f3nw/2 _ _1__ 1 - e- nw(3
so option a is correct. Sol 1.5. t he partition function is defined as
Now here we have that t he first level is non degenerate and t he 2nd level is doubly degenerate and t he t he third level is non degenerate so we have the partition function for t his system Z = l + 2e- /3c + e- 2/3c = (1 + e- (3€ ) 2 so the c option is correct. j [email protected]
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11 oo : oo : 4s Z = 1 + 2e- /3c + e- 2/3c = (1 + e - /3c) 2
so the c opt ion is correct. j [email protected]
@Sk J ahiruddin ,
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Canonical Ensembles: Part-1
Sol 1.6. t he average energy is defined as
< E >= -
a 813
in z
where z is the partition function . now we have single part icle partition funct ion for this case is z
= 1 + e- 2/3c + e- 4/3c
so form this we can define N particle partition function as follows Z
= (1 + e- 2/3c: + e - 4f3c: ) N
now we can define the average energy of t he syst em
8 N (2ce- 2/3c: + 4ce- 4f3c:) < E >= - - In z = - - - - - - - 8(3 1 + e-2/3c + e -4/3c now as
/3E = 2c N so energy par particle
= - - - = 2c N
so option a is correct.
11 oo:oo:47 so energy par particle
= - - - = 2c N
so option a is correct. [email protected]
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Sol 1. 7. here rod can be arranged parallel or perpendicular to the length of the chain. let n 1 be the numbers of the rods which are parallel and n 2 be the number of the rods that are perpendicular to t he length. hence t he partition function of t he system where we take parallel states has the energy E and the perpendicular states has the energy -E now average length we have
= -e --/3c-+-e-f3c- n1 ae- c/3
it should be the average length . but we have that if we make T is very large energy then probability of the rods having parallel situation gets lowered and tensed to zero and the probability of having length perpendicular gets maximized or one can say all the rod take perpendicular configuration so n 2 -+ N hence applying t his in the equation we get N aec/3
< L >=---e-/3c + e/3c which can be simplified to write down as
< L >= Na/ (1 + e-2c/kBT)
11 oo : oo : so < L >= - - - e-f3c:
+ ef3c:
which can be simplified to write down as
< L >= Na/ (1 + e- 2c:/kBT) so option c is correct. [email protected]
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Sol 1.8. t he energy of the particle has two contribution one for the kinetic energy and t he other is due to t he potential
energy . so the energy 2
2
Px+Py +Pz + E = - - - - - mgz 2m where we take the single part icle is at a distance z from t he bottom of the cylinder and we have defined before t he partition function for t he cont inuous system hence we will have
z=
CX)
1
CX)
e- Px2 /3 / 2m dpx
fi3
e- Py 2 /3 / 2m dpy
- oo
-
CX)
L
CX)
e- Pz2f3/2mdpz
e-mgzf3 dz
dxdy
0
-00
so calculating we get Z
= 1rR
2
mkBT 21rn2
312
1 _ e-mgL/kBT
mg/ kBT
so for the n particle system we will have
Zn
=ZN
11 oo :oo : s2 so for the n particle system we will have Zn =ZN
and we have
- B 8T 2
as we take the L is very very large compare to R . j [email protected]
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Sol 1.9. we know that the Helmholtz free energy is related to t he partit ion function t hrough t he relation is given by
so comparing t his to the given quant ity of t he F and we get t hat
which is t he N particle partit ion function . we know that the average energy is related to t he part it ion function is given by _8lnz_ U
8/3 -
so we will now use the expression of ln z in this case we get
so option a is correct. Sol 1.10. t he total energy associated to the system is given
11 oo : oo : 55
so option a is correct. Sol 1.10. t he total energy associated to t he system is given by
E = (n1
1
1
+ 2 )nw1 + (n2 + 2 )n2w1
so t he ground state energy of the syst em is when n 1 = n 2 = 3 1 hence E 1 = ~w and the first excited state has the energy 5 E 2 = ~w and t he 2nd excited state is doubly degenerate and [email protected]
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Canonical Ensembles: Part-1
7
has the energy E 3 = ~w and the other states have energy ..., greater than the given energy so we have probability p
e-3(3nw/2 + e-5(3nw/2 + 2e-7f3nw/ 2
= _ _ _ _ _ _ _ _ _ _ = x 312 (l + x + 2x2 )/ Z z
Sol 1.11. t he average energy of the system is given by
I: E ie - Ei/3
= - - - -
z
where Z is the partition function of the syst em hence we have £e- c/3 - sec/3
= - - - - 2 + ec/3 + e- c/3
so we will have
< E >=
e/3c/2 _ e - /3c/ 2 -s e/3c/2+ef3e12
/3£ =
- £
tanh 2
Sol 1.12. t he energy of the harmonic oscillator in two di-
mension is given by
11 oo :oo : s1 e/3c/2 _ e-/3c/2
< E >=
/3£ = -£ tanh
e/3c/2+ef3e12
-£
2
Sol 1.12. t he energy of t he harmonic oscillator in two di-
mension is given by
E
PX 2+pY 2 +
=----
2m
ax2+ y 2
t hen the single part icle part ition function is given by 1 Z = h2 CX)
00
00
e- Py 2 /3 /2m dpy
e- Px2f3/2mdPx - oo
- oo
2 e- o:f3x dx
- (X)
00
2
e- o:f3y dy - (X)
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Canonical Ensembles: Part-I
@Sk J ahiruddin , 2020
so we have
41rmkBT1rkBT z=------Ji2 O'.
so we will have Zn
=
2m7r2 a f32
N
Sol 1.13. t he part ition function
z=
e - nw/32
now we know t hat t he Helmholtz free function we have
so we put t he value of t he partition function in t he part !iw
F • 11
1
= kBT 2kBT
11 oo : 01 : oo so we put t he value of t he part it ion function in t he part F
so as T
➔
=
nw kBT 2kBT
0 we will have F = nw/ 2
Sol 1.14. t he part ition function is defined as 00
Z = 2_ h
00
p2 e2mkBT
-
dp
alxl
e ksT dx -oo
-00
1/2
21rm kBT
0
aJxJ
eksT dx
h2
- oo
00
+
_ aJxJ
e ksT dx 0
1/ 2
so opt ion c is correct. [email protected]
@Sk J ahiruddin ) 2020
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Canonical Ensembles: Part-1
Sol 1.15. let us first calculate the eigen value of t he system solving eigen value equation we get
now the eigen value has two possible values A = ✓rr2 + 62 now we will have the second value A = - ✓ry 2 + 62 so the partition function
Sol 1.16. Sol 1.17. t he part ition function of the system
11 oo : 01
: 02
Sol 1.16. Sol 1.17. t he part ition function of t he syst em
Z =l
+ 2e- 136
so definitely the energy would b e
8 ln z 2ce- /3c = - - - = - - 8/3 l + 2e-/3c and we know that
- au -
2 - /3€ 1 + 2e-/3c cv - 8T - 2kB(/3c) e (1 + 2e- /3c)2 so we t ake the limit ing condition t hat is /3c > > 1 so we haveCv = 2kB(/3c ) 2 e- /3c 23
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Car1onical Ensembles: Part-1
Sol 1.18. here the energy is proportional to the cube of t he moment um but we first try to think that if energy is proportional to some power of the momentum then we get
E ex: PS , but we know that pressure is related to the energy is given by
-
sE P =-3V where P is the pressure. hence we have
p
-
s
()(
-
11 oo : 01 : os by
-
sE P =-3V where P is the pressure. hence we have
p
-
E
s
ex -
3
here s=3 given so we have pressure P = E V
at fixed T. so option is c. Sol 1.19. t he partition function of the syst em is given by
z = 1 + e-f./kBT + ef./kBT SO t he probability wit h energy E=O and we have
eo
eo
P =-=---z 1 + 2 cosh2
E
= 2kBT hence
1 1 + 2 cosh2
so option a is correct. j [email protected]
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Canonical Ensembles: Part-1
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Sol 1.20. from t he kinetic theory of the gases we have that force is given by 2
m < v > ML - 2 < F >=---= t L we have n mole ideal gas so we have nIVI=m where M is t he molecular mass. hence we have
,
In
11 oo : 01 : os we have n mole ideal gas so we have nJVI=m where M is t he molecular mass. hence we have
which gives that t=LT- 112 Sol 1.21. t he Helmholtz free energy is related to t he par-
t ition function is given by F = -kBT In Z now we are to evaluate t he partition function and is given by
z
=
L (2l + 1)e- Et/kBT = L (2l + l
2 1 )e- h l(l+l )/2I0 k 8 T
l
now t his can be simplified F
=
-ksTinZ
=
-ks Tln
1 + L (2l + l )e-h2l(l+l) / 2Ioks T l
but we know that ln( l F = - K ET L (2l
+ x)
2
= x - x / 2... so
+ l )e- h2l (l+l )/2IokBT =
- 3kBTeh2 / IokBT
l
where we neglect t he higher order terms [email protected]
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Canonical Ensembles: Part- I
Sol 1.22. t he energy of diatomic molecule considering t he molecule as the simple harmonic oscillator we get En =
(n + ~) nw now we nee to calculate t he part ition function of t he system and to do this we would like to refer to the problem to the previous one problem 30 from which we get t he partition function which is given by
11 oo : 01 (n
: 11
+ ~) fiw
now we nee to calculate t he part ition function of t he system and to do this we would like to refer to the problem to the previous one problem 30 from which we get t he partit ion funct ion which is given by e- f3nw/2
Z =--1 - e- f3nw
so the probability we have for t he part icle being in t he lowest possible energy state
so the option a is correct. Sol 1.23. t he part ition function of t he system
z = 1 + e- 6./ kBT + e- 26. /kBT hence we can say that the average energy is given by
8 ln 1 + e-6./3 < E >= 8{3
+ e- 26./3
2~ e- 26./3 + ~ e- 6./3 1 + e-6./3 + e- 26./3
now t he specific heat
au
cv = 8T jahir@physicsguide. in
@Sk J ahiruddin , 2020
which essentially gives us
26
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Canonical Ensembles: Part-1
11 oo : 01
: 13
which essentially gives us
now we take
/3~ =
= _kB_T_
ax
4
and we have
j [email protected]
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Car1onical Ensembles: Part-1
@Sk J ahiruddin , 2020
hence we have
< E >= _3N_kB_T_ 4
00 : 01 : 16 Canonical Ensembles: Part-1
@Sk J ahiruddin , 2020
hence we have
< E >= _3N_kn_T_ 4
for N particle and as we have {JU 3Nkn Cv ={JT= 4
Sol 1.25. we know that the
-
8/3
now we are to evaluate
8 8(3
8
8 In Z
8(3
8/3
l8Z 8/3 z 8/3 {)
now t his can be written as
8 8(3
1
az
z2
8(3
2
1 a2 z
-
-
z
8/3 2
now let us calculate what t his quant ity gives you if we define
t hen t he 2nd t erm of the above equation is giving you 1 82 Z -
z
~ E 2e- f3E
z
a13 2 28
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Canonical Ensembles: Part-1
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which is basically < E
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2
> and the first term is basically
11 oo : 01
: 1s
@Sk J ahiruddin, 2020
Canonical Ensembles: Part-1
2
which is basically < E > and the first term is basically 2 giving you < E > so t he option a is correct. Sol 1.26. t he average energy is given by ce-c/3 - c ec/3
=----l
+ ec/3 + e-c/3
which can be written as function to get t he following
[here we take Taylor expansion and t ake only first order contribution] Sol 1.27. to solve t his problem we have to first find out t he partition function . this partit ion function has been
calculated in t he problem 34 please see that we will take t he result here. the partition function 21rmkBT
3N/ 2
h2 so Helmholtz free energy function is given by
so t he pressure is given by
j [email protected]
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Canonical Ensembles: Part-1
11 oo : 01
: 21
P = -
= --
8V
j [email protected]
V
29
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Car1onical Ensembles: Part-1
@Sk J ahiruddin , 2020
Sol 1.28. See solution of problem Prob 1.4. Sol 1.29. t he mean square fluctuation is defined as
which gives us
au 8T
r a- 2
ex
so by integrating we get U ex
r a -l
option c is correct.
j [email protected]
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11 oo : oo : oo
Problems and Solutions in Canonical Ensembles: Part-2 Sk J ahiruddin * Suchismito Chatterjee
*Assistant P rofessor Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay 1\11.Sc Physics 2009-2011 batch He ranked 007 in IIT JAl\lI 2009 and 008 (JRF) in CSIR NET June 2011 He has been teaching CSIR NET aspirants since 2012
1
@Sk J ahiruddin , 2020
Car1onical Ensembles: Part-2
11 oo : oo : 03 1
Canonical Ensembles: Part-2
@Sk J ahiruddin, 2020
Contents 1
Problems from NET, GATE, JEST , TIFR & JAM pap ers 1.1 Ans Keys 1.2 Solutions . • • •
j [email protected]
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
2
•
•
3 •
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
13 14
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11 oo : oo : 06
j a hir@physicsguide. in
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Canonical Ensembles: Part-2
@Sk J a hiruddin , 2020
1
Problems from NET, GATE, JEST, TIFR & JAM papers
Prob 1.1. If t he Hamiltonian of a particle is 2
H =
t hen (x + xy + y 2017] 2
(a)
kBT
(b)
2
)
2
Px + Py+ - - - xy
2m
at temperature T is equal to 1
kBT
2
(c)
2kBT
[JEST
(d)
(Hint: Generalized equlipartition t heorem) Prob 1.2. Let a particle of mass 1 x 10- 9 Kg, constrained
to have one dimensional motion, be initially at t he origin (x = Om) . T he particle is in equilibrium wit h a t hermal bath (KBT = 10- 8 J . What is (x 2 ) of the particle atfer time t = 5s? [JEST 2017] Prob 1.3. For a classical system of non-interacting par-
ticles in t he presence of a spherically symmetric potential V (r) = , r 3 , what is t he mean energy per particle? , is a [JEST 2018] constant.
11 oo : oo : os
,r
V (r) = constant.
3,
what is the mean energy per particle? , is a [JEST 2018]
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(Hint: Generalized equlipartition theorem) Prob 1.4. A large cylinder of radius R filled with particles
of mass m . The cylinder spins about its axis at an angular speed w radians per second , providing an acceleration g for t he particles at the rim. If t he temperature T is constant inside the cylinder , what is t he ratio of air pressure P0 at [JEST 2018] t he axis to the pressure Pc at the rim? (a) exp
mgR
(b) exp
2kbT
(d) 2kbT mgR
Prob 1.5. A quantum system has 3 energy levels -0.12 eV, -0.2 eV and -0.44 eV respectively. 3 electrons are dis-
t ributed among these levels. At a temperature 1727°C, the system has total energy -0.68 eV. The free energy of the [TIFR 2010] system is approximately (a) +1.5 eV (b) + 0.3 eV (c) -0.1 eV (d) -0.3 eV (e) - 1.0 eV (f)- 1.5 eV Hint : Calculate Helmholtz Gibbs both Prob 1.6. N particles are distributed among 3 states having energies E = 0, kBT and 2kBT respectively. If t he total
11 oo : oo : 11 (e)-1 .0 eV (f) - 1.5 eV Hint : Calculate Helmholtz Gibbs both
Prob 1.6. N particles are distributed among 3 states having energies E = 0, kBT and 2kBT respectively. If t he total equilibrium energy of t he system is 138.06 kBT, find the [email protected]
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number N of particles.
4
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Car1onical Ensembles: Part-2
[TIFR 2016]
Prob 1. 7. In a monatomic gas, t he first excited state is only l .5eV above the ground state, while t he other excited states are much higher up. The ground state is doublydegenerate, while the first excited state has a 4-fold degeneracy. If now t he gas is heat ed to a temperature of 7000 K, t he fraction of atoms in t he excit ed state will be approximately [TIFR 2015] (a) 0.42 (b) 0.3 (c) 0.14 (d) 0.07 Prob 1.8. Consider a system of non-interacting part icles with integer angular momentum J at a t emperature T. This system is placed in a magnetic field B in t he z direction. The energy of a state wit h Jz = m ri is Em = mµB B with µB > 0. The fr actional magnetization of the particles as a function of µBB / kBT can be represented as [TIFR 2017]
00 : 00 : 14
j [email protected]
physicsguide CSIR NET, GATE
5
@Sk J ahiruddin , 2020
Canonical Ensembles: Part-2
--0.3
0.0
--0.4
-0.2
--0.5
-
-, -0.4 E -0.6
~
E
--0.6 --0.7 --0.8 --0.9
-0.8
· 1.0
0.0
0.5
1 .0
1.5
2.0
1.=::::...__ _ __ 0.0
0.5
1.0
1.5
2.0
µ8 B/K 8 T
IJeBIKeT
0.8
1.0
--....._
0.9
0.6
0.8 ~
E
0.4
-,
0.7
e o.6
0.3
0.5
0.4
0.0
0.3
0.0
0.5
1.0
1.5
2.0
0.0
1Je8/K8 T
0.5
1.0 µ9 B/K 9 T
1.5
2.0
Prob 1.9. A one-dimensional quantum harmonic oscillator of natural frequency w is in t hermal equilibrium with a
heat bath at temperature T . The mean value (E ) of the energy of t he oscillator can be written as [TIFR 2017]
nw
nw
nw
(a) 2 sech 2kBT
(b)
,, nw
, _, nw
_fnw\
2
csch
nw
11 oo : oo : 16 heat bath at t emperature T . The mean value (E ) of t he [TIFR 2017] energy of the oscillator can be writ t en as
nw
nw
nw
(a) 2 sech 2kBT
nw
(c) 2 coth
(b ) 2 csch
nw
nw
(d) 2 tanh
2kBT
nw
2kBT
nw 2kBT
Prob 1.10. A classical ideal gas of atoms wit h masses m is confined in a three-dimensional pot ential
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physicsguide CSIR NET, GATE
Canonical Ensembles: Part-2
at a temperature T . If kB is the Boltzmann const ant, the root mean square (r.m.s.) distance of the at oms from the or1g1n 1s [TIFR 2018 •
•
•
(b)
(a)
(c)
(d)
Prob 1.11. N particles are distributed among three en-
ergy levels having energies: 0, kBT and 2 kBT resp ectively. If the total equilibrium energy of t he syst em is approximately 42 .5 kBT then find the value of N (to t he closest [TIFR 2018 integer).
Prob 1.12. A statist ical system , kept at a t emperature T , has n discret e energy levels wit h equal level- spacing E,
11 oo : oo : 19 integer).
[TIFR 2018
Prob 1.12. A statistical system , kept at a temperature T , has n discrete energy levels wit h equal level- spacing c, starting from energy O. If, now, a single part icle is placed in the system what will be the mean energy of t he system in the limit as n ➔ oo? [The answer should not be left as a [TIFR 2018 summation ] Prob 1.13. Consider a thermal ensemble at temperature T which is composed of identical quant um harmonic [email protected]
7
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@Sk J ahiruddin , 2020
Canonical Ensembles: Part-2
lators of frequency w 0 wit h non-overlapping wavefunctions. The probability that there will be an even number of energy quanta in t he system is [TIFR 2019] (a) 1 exp (- nwo/ kBT) + 1
(C)
i
(d )
tanh ( nw0 I 2kBT )
exp (-nwo/kBT) - 1 Prob 1.14. Consider a system of 2 non-interacting classical particles which can occupy any of t he 3 energy levels: E = 0, E and 2c with degeneracies g(E) = 1, 2, 4 respect ively. T hee mean energy of the system is [GATE 2008] 4 exp(-c/kBT)
+ 8 exp(-2c/kBT )
(a)c
1 + 2 exp(-c/kBT ) + 4 exp(-2c/kBT )
(b)c
2 exp(-c/kBT) + 4 exp(-2c/kBT) 1 + 2exp( -c/kBT) +4exp(-2c/kBT) I
'"' ~--- I
_
I,_
rr \
,
o ~--- I
() _ I 1_
rr \
\
11 oo : oo : 21 (a)c:
4 exp(-c:/kBT ) + 8 exp(-2c:/kBT ) 1 + 2 exp ( -E / kB T ) + 4 exp ( - 2c /kBT )
(b )c:
2 exp(-c:/kBT) + 4 exp(-2c:/ kBT) 1 + 2 exp(-c:/kBT) + 4 exp(- 2c:/kBT)
(c)c:
2 exp(-c:/kBT) + 8 exp(-2c/ kBT) 1 + 2 exp(-c:/kBT ) + 4exp(-2c:/kBT )
(d)c:
ex p(- c:/kBT ) + 2 exp(- 2E/ kBT ) 1 + exp(-c:/kBT) + exp(-2c:/kBT)
Prob 1.15. Consider a gas of atoms obeying MaxwellBoltzmann statist ics. T he average value of eia-f over all the moments
p of each of t he particles (where a is a constant
vector, a is the magnitude, m is t he mass of each atom, T is t he t emperature and kB is Boltzmann constant) is [GATE j [email protected]
8
Car1onical Ensembles: Part-2
@Sk J ahiruddin , 2020
2013] (a) 1
physicsguide CSIR NET 1 GATE
(b) 0
Prob 1.16. At a given t emperature T , the average per particle of a non-interacting gas of two-dimensional classical harmonic oscillators is _____ kBT. [GATE 2014]
Prob 1.17. The average energy U of a I-d imension al qu ant um oscillator of frequency w and in contact wit h a h eat bath at temperature T is given by [GATE 2015] 1 1 1 1 (a) U = 1uv coth (3nw (b) U = 1uv sinh (3!uv 2 2 2 2 1 1 1 1 (c) U = !uv tanh f3 nw (d) U = !uv cosh (3 !uv
2
2
2
2
11 oo : oo : 24 at temperature T is given by (a) U (c) U
1
1
2
2
1
1
2
2
= nw coth = nw tanh
[GATE 2015]
{3nw
(b) U
{3nw
(d) U
1
1
2 1
2
= nw sinh = nw cosh
2
{3nw
1
{3nw
2
Prob 1.18. Consider a triatomic molecule of the shape shown in the figure below in t hree dimensions. T he heat capacity of this molecule at high t emperature (t emp much higher than t he vibrational and rotational energy scales of the molecule but lower than it s bond dissociation energies) is [GATE 2017] (a) 3/2kB (b) 3kB (c) 9/2kB (d) 6kB Prob 1.19. T he partition function of an ensemble at a N
k;T ,
temperature T is Z 2cosh where kB is the Boltzmann constant. The heat capacity of t his ensemble at [email protected]
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9
Canonical Ensembles: Part-2
@Sk J ahiruddin, 2020
•••••• •• ♦ ♦ ♦ ♦
♦
♦
T =
•
♦ ♦ ♦
is X NkB, where t he value of X is ____ (up to two decimal places) . [GATE 2018] k~
Prob 1.20. Consider a one-dimensional gas of N noninteracting part icles of mass m with t he Hamiltonian for a single particle given by,
11 oo : oo : 26 Prob 1.20. Consider a one-dimensional gas of N noninteracting particles of mass m wit h t he Hamiltonian for a single part icle given by,
1
p2
H
+ -mw (x + 2x) 2m 2
=-
2
2
The high temperature sp ecific heat in units of R = NkB (K B is t he Boltzmann constant ) is [GATE 2019]
(A) 1
(C) 2
(B) 1.5
(D) 2.5
(Hint: use generalized equipartition theorem)
Pro b 1. 21. A 2-level syst em has energies O and E . The level with zero energy is non-degenerate, while t he level wit h energy E is t riply-degenerate. T he mean energy of a classical part icle in this system at temperature T is [GATE 2016] E e - E/kBT
(a) 1 + 3e- E/ kBT
E e - E/kBT
(b) 1 + e - E/kBT
j [email protected]
10
(c)
3 E e - E/ kBT 1 e - E / kBT
+
(d)
physicsguide CSIR NET I GATE
Canonical Ensembles: Part-2
@Sk J ahiruddin , 2020
3 E e-E/ kBT
1 + 3e- E/ kBT
Prob 1.22. A syst em has 2 energy levels wit h energies c and 2c . The lower level is 4-fold degenerate while the upper level is doubly degenerate. If there are N non-interacting classical part icles in t he system, which is in t hermodynamic equilibrium at temperature T , t he fraction of part icles in t he upper level is [GATE 2011]
(a)
1
1
:FI Qr k
( b)
1
,
r,
~FI Qr k
11 oo : oo : 29 egenera e. non-in erac 1ng classical part icles in the system, which is in thermodynamic equilibrium at t emperature T , t he fraction of particles in t he upper level is [GATE 2011] 1
1
(a) 1 + eE-/kBT
(c)
2e•fkaT
(b) 1 + 2eE-/kBT
~ 4e2 0). One system is populated by spin-~ fermions and t he other by bosons. What is the value of E p - EB where E p and EB are the ground state energies of the fermionic and bosonic systems respectively? [email protected]
@Sk J a hiruddin , 2020
4
physicsguide CSIR NET , GATE
Quantum Stat Mech
[NET June 2013]
(a) 6co
(b) 2co
(c) 4cO
( d)
co
Prob 1. 7. The minimum ener·gy of a collection of 6 noninteracting electrons of spin-½ and mass m placed in a onedimensional infinite square well potential of widt h L is [NET Dec 2012] 2 2 2 2 2 2 2 2 2 (a) 141r n /mL (b) 911r n /mL (c) 71r n /mL (d) 31r2 n2 / mL 2 Prob 1.8. The number of ways in which N identical bosons can be distributed in two energy levels is (a) N + 1 (b) N(N - 1)/ 2 (c) N (N + 1)/2 (d) N [NET June 2012] Prob 1.9. 3 identical spin-~ fermions are to be distributed in 2 distinct, non-degenerate energy levels. The number of ways t his can be done is [NET Dec 2013] (a) 8 (b) 4 (c) 3 (d) 2 Prob 1.10. Consider a gas of Cs atoms at a number den-
11 oo : oo : 14 in 2 distinct, non-degenerate energy levels. The number of ways this can be done is
(a) 8
(c) 3
(b) 4
[NET Dec 2013]
(d) 2
Prob 1.10. Consider a gas of Cs atoms at a number den12
sity of 10 atoms / cc. When the typical inter-particle distance is equal to the thermal de Broglie wavelength of the particles, the temperature of t he gas is nearest to (Take the j [email protected]
5
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Quantum Stat Mech
@Sk J ahiruddin, 2020
26
mass of a Cs atom to be 22.7 x 10- kg) 2016] (a) lx10- 9 K (b)7x10- 5K
[NET June
(c) 1 x 10- 3 K
Prob 1.11. The number of ways of distributing 11 indist inguishable bosons in 3 different energy levels is. June 2018] (a )
311
(b) 11 3
(13) ! (c) 2! (11 )!
[NET
(11)!
( ) d 3!8!
Prob 1.12. Consider a syst em of two particles A and B. Each can occupy one of three possible quantum states 1), 2) and 3). The ratio of the probability t hat the two particles are in the same state to t he probability that two particles are in different states is calculated for bosons and classical (Maxwell- Boltzmann) particles. They are respectively [JEST 2013] (a) l , O
(b)l/2,1
(c) l , 1/ 2
(d) 0, 1/ 2
11 oo : oo : 16 cles are in different states is calculated for bosons and classical (Maxwell- Boltzmann) particles. They are respectively [JEST 2013] (a) 1, 0 (b) 1/ 2, 1 (c) 1, 1/ 2 (d) 0, 1/ 2
Prob 1.13. Consider 3 situations of 4 particles in a 1dimensional box of width L with hard walls. In case(i) t he particles are fermions, in case(ii) they are bosons and in case(iii) t hey are classical. If the total ground state energy of t he 4 particles in these 3 cases are E p ,EB , Eel respectively, [email protected]
6
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Quantum Stat Mech
@Sk J ahirudd in, 2020
which of the following is true? (a) E p=EB=Ecz (b) Ep > EB=Ecl
(c) E p < EB < Eel
[JEST 2013]
(d) Ep > EB > Eel
Prob 1.14. The lowest quantum mechanical energy of a particle confined in a I-dimensional box of size L is 2 eV . The energy of the quantum mechanical ground state of a [JEST system of 3 non-interacting spin- ~ particles is 2014] (a) 6eV (b) lOeV (c) 12eV (d) 16eV Prob 1.15. A 2-dimensional box in a uniform magnet ic field B contains N/ 2 localized spin-t particles with magnetic moment µ and N / 2 spinless particles which do not interact with each other. The average energy of the system at temperature T is [JEST 2016] 1 µB (a) 3NkBT- NµBsinh kBT 2
11 oo : oo : 19 netic moment µ and N / 2 spinless particles which do not interact with each other . T he average energy of the syst em at temperature T is 1 . (a) 3NkBT - NµBs1nh 2
[JEST 2016] µB kBT
1 (b) NkBT- - NµBtanh
µB 2 kBT 1 1 µB (c) -NkBT- -NµBtanh k T 2 2 ~B
3 1 (d) NkBT+ Nµ B cosh 2 2
µB kBT
Pro b 1.16. The energy spectrum of a particle consists of jahir@physicsguide. in
@Sk J a hiruddin , 2020
7
physicsguide CSIR NET , GATE
Quantum Stat lVIech
four states wit h energies 0, E, 2E, 3E. Let ZB(T), Z p(T ) and
Zc(T ) denote the canonical partit ion functions for four noninteracting particles at temperature T. The subscripts B , F and C corresponds to bosons, fermions and distinguishable classical particles, respectively. Let y == exp - k~T . Which one of t he following statem ent s is true about ZB(T), Zp(T ) and Zc(T )? [JEST 2019] (A) They are polynomials in y of degree 12, 6 and 12, resp ectively. (B) They are polynomials in y of degree 16, 10 and 16, resp ectively. (C) They are polynomials in y of degree 9, 6 and 12, respect ively. (D) They are polynomials in y of degree 12, 10 and 16, re-
11 oo : oo : 21 sp ectively. (C) They are polynomials in y of degree 9, 6 and 12, respect ively. (D) They are polynomials in y of degree 12, 10 and 16, respectively. Pro b 1.1 7. A gas of non-interacting particles, each of rest
mass l Me V , is at a temperature T = 2.0 x 107 K and has an average particle density n = 2.7 x 1034 cm- 3 . We can obtain a reasonably correct treatment of t his system [TIFR 2016] (a) only by using special relativity as well as quantum mechanics. (b) by neglecting quant um mechanics but not special t heory j [email protected]
@Sk J ahiruddin, 2020
8
physicsguide CSIR NET 1 GATE
Quant um Stat Mech
of relativity. (c) by neglecting sp ecial relativity but not quantum mechanics. (d) by neglecting bot h special relativity and quantum mechanics. Prob 1.18. A quant um mechanical system consist s of a
one-dimensional infinite box, as indicated in the figures below.
• • •
• ••• • • •
11 oo : oo : 24
•••• • • •
• • •
--------3 (three) identical non-interacting spin 1/ 2 particles, are first placed in t he box, and t he ground state energy of the system is found to be E 0 = l 8eV. If 7 (seven) such ident ical particles are placed in the box, what will be t he ground state energy, in units of eV? [TIFR 2017]
Prob 1.19. MSQ Consider a one dimensional harmonic oscillator of angular frequency w . If 5(five) identical particles occupy the energy levels of t his oscillator at zero t [email protected]
9
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@Sk J ahiruddin , 2020
Quantum Stat Mech
perature, which of the following statement (s) about t heir [JAM 201 7] ground state energy E 0 is (are) correct? . 13 (a) If the particles are electrons, E 0 = liw. 2 . 25 (b) If t he particles are protons, Ea = liw. 2 2 (c) If t he part icles are spin-less fermions, E 0 =
;nw.
(d) If the particles are bosons, E 0 =
~nw.
Prob 1.20. For a system of 2 bosons, each of which can occupy any of the 2 energy levels: 0 and c, t he mean energy of the system at t emperature T is [GATE 2014] n
no n
n
11 oo : oo : 26 (d) If t he part icles are bosons, E 0 =
~nw.
Prob 1.20. For a system of 2 bosons, each of which can occupy any of t he 2 energy levels: 0 and c, t he mean energy
of the system at t emperature T is c:e- f3c: + 2c:e- 2f3c: l + c:e- f3c: (a) 1 + 2e- f3c: + e- 2f3c: (b) 2e- f3c: + e- 2f3c:
+ c:e- 2f3c: (c) 2 + e-f3c: + e-2f3c: 2c:e-f3c:
[GATE 2014]
+ 2c:e- 2f3c: 2 + e-f3c: + e-2/3c: c:e-f3c:
(d) Prob 1.21. Consider a system of 3 fermions which can occupy any of t he 4 available energy states wit h equal prob[GATE 2014] ability. The ent ropy of t he system is (a) kBln2 (b) 2kBln2 (c) 2kB ln4 (d) 3kBln4 Prob 1.22. Consider a system having 3 energy levels wit h
energies: 0, 2c: and 4c:, wit h respect ive degenerates 2, 2 and 3. Four bosons of spin-0 have to be accommodated in t hese levels such that t he total energy of t he system is 1Oc. T he number of ways in which it can be done is ? [GATE 2016] j [email protected]
@Sk J ahiruddin , 2020
10
physicsguide CSIR NET I GATE
Quant um Stat Mech
Prob 1.23. Consider two particles and two non-degenerate
quant um levels 1 and 2. Level 1 always contains a particle. Hence, what is t he probability t hat level 2 also contain a particle for each of t he two cases. (i) When t he two particles are dist inguishable (ii) when t he two particles are Bosons. [GATE 2017] (a) (i) 1/ 2 and (ii) 1/ 3 (b) (i) 1/ 2 and (ii) 1/ 3
11 oo : oo : 29 particle for each of the two cases. (i) When t he two particles are distinguishable (ii) when t he two particles are Bosons. [GATE 2017] (a) (i) 1/ 2 and (ii) 1/3 (b) (i) 1/ 2 and (ii) 1/ 3 (c) (i) 2/ 3
(d) (i) 1
and (ii) 1/ 2
and (ii) 0
Prob 1.24. Three particles are to be distributed in four non-degenerate energy levels. The possible number of ways of distribution: (i) for distinguishable particles, and (ii) for [GATE 2018] identical Bosons, respectively, is
(a) (i) 24, (ii)4 (c) (i)64, (ii)20
(b) (i) 24, (ii )20 (d) (i)64, (ii)l6
Prob 1.25. The quantum effects in an ideal gas become important below a certain temperature TQ when de Broglie wavelength corresponding to the root mean square thermal speed becomes equal to the inter-atomic separation. For such a gas of atoms of mass 2 x 10726 kg and number density 6.4 x 1025 /m3 , TQ = ___ x 103 K (up to one decimal place) [email protected]
@Sk J a hiruddin ) 2020
11
physicsguide CSIR NET ) GATE
Quantum Stat Mech
[GATE 2018] (kB
= 1.38 x 10- 23 J / K, h = 6.6 x 10-34 J-s)
Prob 1.26. The energy-wave vector (E-k) dispersion relation for a particle in two dimensions is E = Ck, where C is a constant. If its density of states D (E) is proportional
11 oo : oo : 32 (kB = 1.38 x 10- J / K , h = 6.6 x 10- J-s) Prob 1.26. The energy-wave vector (E-k) dispersion relation for a particle in two dimensions is E = Ck, where C
is a const ant . If its density of states D (E) is proportional to E P t hen the value of p is - [GATE 2019]
Prob 1.27. Consider two syst ems A and B each having two dist inguishable particles. In bot h the syst ems, each
part icle can exist in states wit h energies 0, 1, 2 and 3 units wit h equal probability. The total energy of the combined system is 5 units. Assuming t hat t he system A has energy 3 units and t he system B has energy 2 units, the ent ropy of t he combined system is kB ln ..\. The value of ,\ is - - - [GATE 2019] Prob 1.28 . There are 4 energy levels: E, 2E, 3E and 4E (where E > 0). The canonical part it ion function of 2 parti-
[GATE 2013]
cles, if these are (i) 2 ident ical fermions
(a)
e - 2/3E
+ e - 4/3E + e - 6/3E + e - 8/3E
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12
physicsguide CSIR NET, GATE
@Sk J ahiruddin, 2020
+ e-4/3E + e-5/3E + e-6/3E + e-7/3E (C) (e - f3E + e - 2/3E + e - 3/3E + e - 4f3E)2 (d) e-2/3E _ e-4/3E + e-6/3E _ e-8/3E (b) e-3/3E
Quantum Stat Mech
11 oo : oo : 34 (C) (e - f3E + (d)
e - 2(3E
e-2f3E _ e-4f3E
+ e - 3(3E + e - 4f3E) 2
+ e-6f3E _
e-8f3E
(ii) 2 distinguishable particles
a)
e - 2f3E
(b)
+ e - 4f3E + e - 6/3E + e - 8f3E
e - 3(3E
+ e - 4(3E + e - 5(3E + e - 6(3E + e - 7(3E
(C) (e - f3E + (d)
e - 2(3E
e - 2f3E _ e - 4f3E
+ e - 3(3E + e - 4f3E) 2
+ e - 6f3E
_ e - 8f3E
Prob 1.29. Consider a system of non-interacting particles in d dimensional obeying the dispersion relation E = Ak , E is the energy, k is the wave vector, s is an integer and A is a constant . T he density of states N ( E) is proportional to 8
[NET June 2012]
(a)E: -l
(b )E~- l
(c)E~+l
(d)E ~+l
Prob 1.30. Consider a gas of Cs atoms at a number density of 10 12 atoms / cc. When the typical inter-particle is equal to t he t hermal de Broglie wavelength of t he particles, t he temperature of the gas is equal to (Mass of Cs atom j [email protected]
13
Quantum Stat Mech
@Sk J ahiruddin , 2020
= 22.7
X
26
10- kg
(a) l xl0- 9 K 10-8 K
physicsguide CSIR NET, GATE
[NET Dec 2016]
(b)7xl0- 5K
(c) 1 x 10-3 K
(d) 2 x
00: 00: 37
= 22.7
X
10- 26 kg
(a) l x 10- 9 K
[NET Dec 2016] (b )7x 10- 5K
(c) 1 x 10-3 K
(d) 2 x
10-8 K
[email protected]
@Sk J ahiruddin, 2020
1.1
Ans Keys
14
physicsguide CSIR NET, GATE
Quantum Stat Mech
11 oo : oo : 40 @Sk J ahiruddin, 2020
1.1
Quantum Stat Mech
Ans Keys
1.1. a
1.11.
C
1.21. b
1.2.
1.12.
C
1.22. 18
C
1.3. a
1.13. b
1.23.
C
1.4. d
1.14.
C
1.24.
C
1.5. d
1.15.
C
1.25. 84
1.6. b
1.16. a
1.26. 1
1. 7. a
1.17.
1.27. 12
1.8. a
1.18. 132
1.28. (i) b (ii) c
1.9. b
1.19. a,c,d
1.29. b
1.10. d
1.20. a
1.30. d
1.2
C
solution
Sol 1.1. t he dist inguishable particle t hat is the classical particles obey t he Maxwell Boltzmann distribution function where as t he in dist inguishable part icle obey t he quant um statistics. act ually due to t he indist inguishablity we can not say t hat which particle belongs to which stat e and the probability remains unchanged from this condit ion arise two different quant um st at ist ics hence we can conclude t hat [email protected]
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15
physicsguide CSIR NET, GATE
Quantum Stat lVIech
11 oo : oo : 42 [email protected]
15
physicsguide CSIR NET, GATE
Quantum Stat Mech
@Sk J ahiruddin, 2020
t ion a should be the answer b ecause indistinguishable part icle does not obey t he classical statistics. Sol 1. 2. as we are considering t he spin half particle so each
sate can contain two particles so t he lower energy state will have 2 particle where as t he higher energy state will have 1 particle as we have total 3 part icle in hand and hence t he energy is given by n 2 n2 4n 2 n2 3n2 n2 E = 2x - - + - = 2ma2 2ma2 ma 2
hence option c is correct. Sol 1.3. t he number of ways t he we can arrange according to t he MB statistics is given by N = aj nJ = 22 = 4 now we
are going to evaluate the the number of ways we can arrange in t he quantum fermi statistics N = nj 1.aj-nj ai' = 1 and we 1. 1 1 ((ni_-=_a )· -1 _)1' a1 1 .n 1 .
have for t he Bose Einst ein statistics N = hence if we take the ratio we will have NMB 4 : 3 : 1 hence option a is correct.
:
=
3
N BE : N F n
=
Sol 1.4. let t here are 3 states and 3 particle hence we can say t hat t he possible configuration we have 1) t he lower state
have 2 particle and the middle state has only one part icle 2) t he middle takes 2 and t he lowe1~st at e has one 3)the middle has 2 and the upper most has one particle 4) t he upper most j [email protected]
16
physicsguide CSIR NET, GATE
11 oo : oo : 44 t he middle takes 2 and the lower state has one 3)the middle has 2 and the upper most has one particle 4) t he upper most j [email protected]
16
physicsguide CSIR NET, GATE
Quantum Stat Mech
@Sk J ahiruddin , 2020
take 2 and t he middle takes only one 5) t he upper most t ake 2 particle and t he lower most take one 6) t he upper most take only one and lower most take 2 particle 7) each of state takes 1 particle each. so possible configuration is 7. option d is correct. Sol 1.5. two identical particle to be arrange in the 3 states.
now let us see the possible configuration in the fermeonic case and in the fermeonic sate there can never be two part icle in the single particle state hence would be zero. now take t he bosonic state t here could be t hree double particle configuration and there could be 3 single particle configuration so ratio is one . where in the classical state we will have to in corporate the distiguishability and hence we have to divide t he ratio of b statistics by 2! and hence we have t he ratio is given by rB : rp: re = l : 0: 1/ 2 Sol 1.6. let us first calculate the fermeonic case we have t he
for t he lowest energy to have t he most justified configurat ion of the system would be 2 particle would be t he lowest possible state and 1 particle would be in t he state of 3c state hence t he energy of t he configuration would be E p = 2c
+ 3c =
5c
and for t he bosonic case all particle will be at t he lowest
11 oo:oo:47 hence the energy of t he configuration would be E p = 2c + 3c = 5c
and for the bosonic case all part icle will be at t he lowest [email protected]
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energy sate hence we have EB = 3c hence we have the difference in the energy is given by
Sol 1. 7. each state can occupy two electrons hence t he energy of t he system is given by
n n E = 2x + 2 x 2ma 2 2 2
4n n 2ma 2
2 2
2 2
+ 2x
9n !i 2ma2
ma2
hence the opt ion a is correct . Sol 1.8. we have seen from the previous discussion that the number of ways we can arrange in t he bosonic syst em is given by ( N + 2 - 1) ! n=-----=----=N+ l (aj - l )!nj ! N !l! ( nj
+ aj
- l) !
so opt ion a is correct. Sol 1.9. we have the possible configuration given below
[A( 1/ 2, - 1/ 2 > )B( - 1/2 > )] [A( ll /2, - 1/ 2 > ), B ( 1/ 2 > )] [A(ll / 2 > ), B ( 1/ 2, - 1/ 2 > )]
fA(I - 1/ 2 > )B (l l / 2, - 1/ 2 > )l
11 oo : oo : so [A( 1/ 2, - 1/ 2 > )B( - 1/ 2 > )] [A( ll /2, - 1/ 2 > ), B( 1/2 > )] [A( ll /2 > ), B( 1/ 2, - 1/2 > )]
[A ( - 1/ 2 > )B ( 1/ 2, -1/2 > )] so possible configuration is 4. 18
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Quantum Stat lVIech
Sol 1.10. t he thermal debroglie wave lengt h satisfies t he given relation N n3 - - - - - -2 = 1 V (21rmkBT) 31
at any temperature T now we have the atomic density hence we have
12 10
h2(n)2/3 T =-----
(2.61)213 21rmkB now just we are putting t he requi1~ed values and we will get
T =
6.28
X
6 62 10- 7 . x = 1.16 x 10-8 22.7 X 1.38 X (2.61 ) 2/3
hence option d is correct. Sol 1.11 . we have seen from the previous discussion that t he number of ways we can arrange in the bosonic system
is given by
( nj + a_i - l ) ! 13! n=-----=-(aj - l)!n_i ! 2!1 1!
hence we have the correct option is option c. So l 1.12. two particle will be in the same sate for t he bosonic
system will be 3 and the for classical system there will also
11 oo :oo : s2 l)!nj ! hence we have the correct option is option c. (aj -
Sol 1.12. two particle will be in the same sate for t he bosonic syst em will be 3 and t he for classical system there will also be 3 sates where there will be two particle in the same state hence the ratio of the occupancy is given by
p(B) P= j [email protected]
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and the single particle st ates in t he bosonic configuration will be 3 but for t he classical case the configuration will be 6 hence the ration will be 1/ 2 hence the we have t he answer 1, 1/ 2 and the option c is correct. Sol 1.13. in t he classical and t he bosonic system there could be we sate where all the part icles come to the same particle stage. so there is the lowest possible state energy for the boson and the classical case would be equal when the all particle comes to the lowest possible energy state and the energy of the syst em is given by
EB = Ec= 4
7r2t;,2 X --
2ma2
but in the fermeonic state each single part icle sate can occupy only one particle or none because if we want to accommodate t he 2 particle in the same single part icle state t hen t he wave funct ion vanishes t hat is the system prohibits to put two particle in the same single particle state hence the energy in the fermeonic case is given by
11 oo : oo : 55 cupy on y one part1c e or none ecause 1 we want o accommodate t he 2 particle in the same single particle stat e t hen t he wave function vanishes that is the system prohibits to put two particle in the same single particle state hence the energy in the fermeonic case is given by w2n2 4w2n2 9w2n2 16w2n2 Ep = - - + - + + 2ma2 2ma 2 2ma 2 2ma 2 hence we see that E p > EB = E e so option bis correct. Sol 1.14. t he lowest possible energy state has the energy 2eV hence we have to state that the lowest possible energy [email protected]
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of the one D box is given by
w2n2 --= 2eV 2 2ma
from the given condition we have and the system has 3 spin 1/ 2 particle hence t he lowest state comprises t he 2 particle and the higher state has t he only one particle. hence the energy of the system
w2n 2 4w 2 n2 E = 2 x - - + - -2 2ma2
2ma
6w 2 n 2
--=
2ma 2
12eV
hence the option c is correct. Sol 1.15. here is two contribution to t he internal energy 1) the first contribution is coming from the degree of freedom here we are considering only two dimensional system and hence two degrees of the freedom and the each degree of
t he freedom has l / 2kBT amount of t he energy and hence f"'{)ntrih11ti{)n ic:: k- r..T tl--iic:: f"'{)ntrih11ti{)n f()r th P N / ½ ? [NET Dec 2014] (a) ~ N ln 2 (b) 2 ln 2 (c) ½ln 2 (d) N ln 2 Prob 3.6. Consider 3 Ising spins at the vertices of a tri-
11 oo : 01
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1
µB / kBT for which the mean value (Si) > ? [NET Dec 2014] (a) ½N ln 2 (b) 2 ln 2 (c) ½ln 2 (d) N ln 2 Prob 3.6. Consider 3 Ising spins at the vertices of a triangle which interact with each other with a ferromagnetic j [email protected]
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Two state arid other systems
Ising interaction of strength J. The partition function of t he system at temperature T is [,B = 1/ kBT ] [NET June 2015] (a) 2e3f3 1 + 6e-f3 1 (b) 2e- 3f3J + 6ef3J (c) 2e3f3 1 + 6e- 3f3 J + 3e131 + 3e-f3J ( d) (2 cosh ,BJ) 3 Prob 3. 7. The partition function of N Ising spins is Z = Af + Af where A1 , A2 are functions of temperature but in-
dependent of N. If A1 > A2 , the free energy per spin in the limit N ➔ oo is [NET Dec 2015] (a) -kBTln(A1/ A2 ) (b)-kBTlnA2
(c) -kBTln(A1A2) (d) -kBTlnA1 Prob 3.8. The Hamiltonian of a syst em of N non-
31
spin- ½particles is H = - µoB LJiS[ , where S f= ± 1 a:_ _ _ ___. ponents of itli spin along an external magnetic field B . At µoB a temperature T , such that ekBT = 2, the specific heat per particle is [NET Dec 2015]
(a) * kB
(c) kB(ln2)
(b) ~kBln2 2
(d) * kB(ln2)
2
Prob 3.9. The Hamiltonian for three Ising spins S0 , S 1 , S2
,
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: 26
particle is
[NET Dec 2015]
(b) ~kBln2
(a) ~kB
(c) kB(ln2)
(d) ~~ k B (ln2)
2
2
Prob 3.9. The Hamiltonian for t hree Ising spins S0 , S 1 , S 2 taking values ± 1, is H = - J S 0 (S1 + S 2 ). If t he system is in equilibrium at temperature T , t he average energy of t he [NET June 2017] system, in terms of f3 = (kBT )- 1 is 31
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Two state and other systems
a) _ 1 + cosh (2/3 J ) ( sinh(2/3 J ) 213 (c) _ 21/3
(b) - 2J(l
+ cosh (2/3 J )]
(d) _ 2J
sinh(2f3J ) 1 + cosh(2/3 J) Prob 3.10. The Hamiltonian of a one-dimensional Ising model of N spins (N large) is N
=-J
H
L O"ZO"i+l i= l
where t he spin az = ± l and J is a posit ive constant. At inverse temperature /3 = l / kBT t he correlation 32 between t he nearest neighbour spins (azaz+i) is Dec 2018] (a) e-f3J / ( ef3J + e-f3J) (b) e- 2f3J (d) coth(f3 J )
(c) tanh(/3 J )
Prob 3.11. The Hamiltonian of t hree Ising spins S 1 , S 2 and S3 , each taking values ± 1, is TT
T
/
rv
r'"'f
.
rv
rv \
'l
rv
,
11 oo : 01 (c) tanh(,BJ )
: 29
(d) coth(,BJ )
Prob 3.11. The Hamiltonian of t hree Ising spins S 1 , S 2
and S3 , each taking values ± 1, is
where J and h are posit ive constants. The mean value of S3 in equilibrium at a temperature T = 1/ (kB,B), is [NET June 2019] 2 3 (a) tanh (,BJ ) (b) tanh(,Bh) tanh (,BJ ) 32
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Two state and other systems
2
(c) sinh (,8 h) sinh (,8 J )
(d) 0
33
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3.1 3. 1.
: 31
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Tv.ro state arid other systems
Ans Keys C
3. 2. b
3. 5.
C
3. 9. d
3. 6. a
3. 10. 3. 3. d
3. 7. d
3. 4. b
3. 8. d
C
3. 11. b
34
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3.2
34
: 34
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Two state and other systems
solution
Sol 3.1. let us first take S 1 =S2=83 =1 t hen the energy of
t he system would be E = -J(l x l + l x 1+ 1 x 1) = -3Jnow again we take S1 =S 2=8 3=-l so we have E = - J (- 1 x - 1 + - 1 x - 1 + - 1 x - 1) = -3 J so there would be degeneracy 2 hence we have option 2 is correct.
Sol 3.2. let t he magnetic moment of the system . . . .. 34 angle 0 in general wit h the magnetic field hence we H =
e - µB /3 - eµB /3 B/3 B /3 e- µ eµ
+
= - tanh µB /3
so for N number of the particle of t he system we have < Si >= - N tanh µB /3 now according to the given condition we have
/N =l/3 at least. so we have tanh µB /3
= 1/3
t his gives us 1 µB /3 = ln2 2 Sol 3 .6 . we have the Hamiltonian
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Two state arid other systems
and each spin can take either 1 or -1 value . so we have to see t he possible number of configuration and t here corresponding energy . we have taken a simple designation s 37 which gives you t he energy and t he spin configura t his is (S1 , s2, s3 , E ). so possible cases are (1 ,1,1,3J), (1,1,1,-J), (1,-1 ,1,-J), (-1 ,1,1,-J), (-1 ,-1 ,1,-J), (-1 ,1,-1 ,-J), (1,-1 ,-
11 oo : 01
: 44
ing energy . we have taken a simple designation skim here which gives you t he energy and t he spin configuration and t his is (S1, s2, s3, E ). so possible cases are (1,1 ,1,3J), (1,1,1,-J) , (1,-1 ,1,-J), (-1 ,1,1,-J), (-1 ,-1 ,1,-J) , (-1 ,1,-1 ,-J), (1,-1 ,1,-J) , (-1 ,-1 ,-1 ,3J) hence we see t hat 3J energy is doubly degenerate and the -J energy state has degeneracy 6 hence we have the partit ion function
Z = 2e- bf3J
+ 6e131
hence the option b is correct . Sol 3. 7. we know that
and as the z is given by we have that
F = - kBTln(A1 N
+ A2N)
and we have t he A1 > A2 hence we modify the case and get
F = -kBTln
N
A1 (1
+
A 2N
A1
N)
is the N is very large then A2 N / A1 N ➔ 0 hence we only have F = - NkBTlnA 1 so we have F / N = - kBTlnA 1 so option d is correct. j [email protected]
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Two state arid other systems
Sol 3.8. from the previous discussion we have sever ~ ~.Q.Q..---. prove that the energy of the given kind of t he syst 38 -µ 0 B t anh µ 0 B /3 now we have t o calculate t he specr_ _ _ _ of t he system is given by
......a
11 oo : 01
: 47
Sol 3.8. from the previous discussion we have several t imes prove t hat t he energy of t he given kind of t he system E =
-µ 0 B t anh µ 0 B /3 now we have to calculate t he specific heat of the system is given by
dU 2 2 Cv = dT = (µo B /3 ) NkB sec hµ 0 B f3 but we know sec h(µo B /3) = e - µoB/3~ eµoB/3 and we are given t hat eµoB /3 = 2 so putting this in the equation we get Cv / N = ~~ kBln 2 hence option d is correct. 2
Sol 3. 9. here we can say t he Hamiltonian of the system is given by H = - J (s 0 s 1 + s0 s 2 ) and each spin can take either + 1 or -1 hence we are now wanting to calculate the possible configuration and t here corresponding energy and this will be denoted in t he manner given by (s 0 , s 1 , s 2 , E) .T he possi-
ble states are (1,1,1, -2J ),(-1 ,1,1,2J),(1,-1 ,1,0),(-1 ,-1,1 ,0),(1,1,1,0),(-1 ,1,-1 ,0) ,(1,-1 ,-1 ,2J) ,(-1 ,-1 ,-1 ,-2J) hence we see that t he energy of zero state has degeneracy 4 and 2J has degeneracy 2 and dengen of -2j is also 2 .hence t he average energy of the system is given by 0 + 2J e- 2f3J - 2J e2f3 1 E =-----4 +2e- 2f31 + 2e2f3 1
2J sinh 2f3 J 1 + cosh 2f3 J
hence option d is correct. j a [email protected]
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physicsguide CSIR NET, GATE
Two state and other systems
39 Sol 3.10. here O"i = 1 or - 1 and hence we have the combination and t he are designat ed as (O"i O"i+ l , E) an?n:l'IESF_ ___.
11 oo : 01 j [email protected]
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Two state arid other systems
Sol 3.10. here a i = 1 or - 1 and hence we have the different combination and the are designat ed as (a i a i+l, E ) and t hese
are given by (1,1,-J),(1,-1 ,J ),(-1 ,1,J ),(-1 ,-1 ,-J) hence t he energy J has degeneracy 2 and similarly -J has . so we have Z = 2e131 + 2e- 131 hence t he average correlation function we have
= - - - - - - - - - = tanh f3 J z
hence opt ion c is correct . Sol 3 .11. Lengthy calculation
j [email protected]
40
physicsguide CSIR N
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11 oo : oo : oo
Problems and Solutions Ideal Fermi Systems
•
Ill
Sk J ahiruddin * Suchismito Chatterjee
*Assistant P rofessor Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay 1\11.Sc Physics 2009-2011 batch He ranked 007 in IIT JAl\lI 2009 and 008 (JRF) in CSIR NET June 2011 He has been teaching CSIR NET aspirants since 2012
1
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Ideal Fermi Systems
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Ideal Fermi Systems
Contents 1
Problems from NET, GATE, JEST , TIFR & JAM pap ers 1.1 Ans Keys 1.2 Solut ions . • • •
j [email protected]
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Ideal Fermi Systems
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1
Problems from NET, GATE, JEST, TIFR & JAM papers
Prob 1.1. T he Fermi-Dirac distribut ion function [n(c:)] is (kB is t he Boltzmann const ant , T is t he temperature and c p is t he Fermi energy) (JAM 2019] 1 (a) n (c) = € - cp e k BT - l
Prob 1.2. The probability t hat an energy level c at t emperature T is unoccupied by a fermion of chemical potent ial µ is [GATE 2009]
(a)
1 exp
c-µ ka T
(b)
+1
1 exp
c- µ k BT
(c) -
1
1 exp
µ -c
kB T
+1
1 (d) - - exp - 1
Prob 1.3. Which one of t he following is a fermion? [GATE 2014] (a) a p art icle (b) Be~ nucleus (c) hydrogen atom (d) deut eron Prob 1.4. The pressure of a free, non-relativist ic Fermi gas in 3-dimensions dep ends, at T = 0, on the density of
11 oo : oo : os deuteron
Prob 1.4. The pressure of a free, non-relativistic Fermi gas in 3-dimensions depends, at T = 0, on the density of [email protected]
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fermions n as (a) n5/3 (b) n l/3
Ideal Fermi Systems
(c) n 2/3
[NET June 2014] (d) n 4/3
Prob 1.5. The electrons in graphene can be t hought of as a two-dimensional gas wit h a linear energy-moment um relation E = fi1v, where p = (Px, Py) and v is a const ant . If p is t he number of electrons per unit area, the energy per unit area is proport ional t o [NET Dec 2016] (a) p3/2 (b) p (c) pl/3 (d) p2 Pro b 1. 6. The dispersion relation of a gas of spin-½ fermions ➔ ➔ in two dimensions is E = nv k , where Eis t he energy, k is t he wave vector and v is a constant with the dimension of velocity. If the Fermi energy at zero temperature is Ep , t he number of particles per unit area is [NET Dec 2017] €3
(a) (4;~n)
3/2
(b) (G1r2{3n2 )
(c) (;::n3)
€2
(d) (21rJn2)
Prob 1. 7. Consider an ideal Fermi gas in a grand canonical ensemble at a const ant chemical potent ial. The variance of t he occupation number of the single part icle energy level with mean occupation number n is [NET Dec 2018]
(a ) n( l - n)
(b)
vfn
(c) n
(d)
1
vn
11 oo : oo : 11 of the occupation number of the single particle energy level [NET Dec 2018] with mean occupation number n is
(a) n(l - n)
(b)
\ffi
(c) n
(d) 1
vn
Prob 1.8. At low t emperatures, in the Debye [email protected]
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Ideal Fermi Systems
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t ion, the contribut ion of the phonons to t he heat capacity of a two- dimensional solid is proportional to [NET Dec 2018] (a) T 2 (b) T 3 (c) T 112 (d) T 312
Prob 1.9. An ideal gas of non-relativistic fermions in 3dimensions is at OK. When both t he number density and t he mass of the particles are doubled, then t he energy per particle is multiplied by a factor [JEST 2014] (a) 21 / 2 (b) 1 (c) 21/ 3 (d) 2- 1/ 3 Prob 1.10. A cylinder at temperature T = 0 is separated into two compartments A and B by a free sliding piston. Compartments A and B are filled by Fermi gases made of spin 1/2 and 3/2 particles respectively. If particles in both t he compartments have same mass, the ratio of equilibrium density of the gas in compartment A to that of gas in compartment Bis [JEST 2017]
(a) 1
1
( b) 32/5
1
(c) 22/s
1
(d) 22/3
Prob 1.11. Consider a non-relativistic two-dimensional gas of N electrons with t he Fermi energy E p . What is the
11 oo : oo : 14 IS
1
(a) 1
(b)
32/ 5
1
1 (d) 22/3
( C) 22/ 5
Prob 1.11. Consider a non-relativistic two-dimensional gas of N electrons with t he Fermi energy E p . What is the
average energy p er particle at temperature T 2019] (A) ~E p (B) i E p (C) ~Ep (D) E p
j [email protected]
= O? [JEST
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Ideal Fermi Systems
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Prob 1.12. The energy per oscillator of an isolated system of a large number of non-interacting, identical fermions
in a I-dimensional harmonic oscillator potential is 5fU.;J / 4 where w is the angular frequency of t he harmonic oscillator. The ent ropy of t he syst em p er oscillator is given by [TIFR 2016] (a) 0.25 (b) 0.56 (c) 0.63 (d) 0.75 Prob 1.13. In two dimensions, two metals A and B , have t he number density of free electrons in t he ratio nA : nB = 1 : 2. The ratio of t heir Fermi energies is [TIFR 2017]
(a)
2:3
(b)
1:2
(c)
1:4
(d)
1:8
Prob 1.14. A white dwarf star has volume V and contains N electrons so t hat the density of electrons is n =
N/V
.
- - Taking t he temperature of t he star to be OK, the av•
1 m where m is t he mass of t he elect ron. The electronic pressure
in t he st ar is
[JAM 2017] 1
2
11 oo : oo : 16 - - Taking t he temperature of t he star t o b e OK, the av•
1 m where m is t he mass of the electron. The electronic pressure
[JAM 2017] 1 2 (a) n Eo (b) 2nEo (c) -nEo (d) -nEo 3 3 Pro b 1.15. Consider a 2-dimensional electron gas wit h a density of 1019 m - 2 . The Fermi Energy of the syst em is ______eV (up t o two decimal places) (me= 9.31 x 10-31 kg , h = 6.626 x 10- 34 J s, e = 1.602 x 10- 19 C) [GATE 2017] in t he st ar is
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Prob 1.16. At t emperature T Kelvin (K ), the value of the Fermi function at an energy O.5eV above the Fermi energy is 0.01 . Then T , t o the nearest integer , is - - - [GATE 2019] 5 (kB = 8.62 x 10- eV / K) Prob 1.17. For an ideal Fermi gas in 3 dimensions, t he electron velocity Vp at t he Fermi surface is related to electron concent ration n as [GATE 2012] (a)Vp ex n 213 (b)Vp ex n (c)Vp ex n 1/ 2 (d)Vp ex n l/3
Prob 1.18. The pressure of a non relativistic free Fermi gas in t hree-dimensions depends, at T = 0, on the density [NET June 2014] of fermions n as (a) n 5/3
(b) nl/3
( c) n 2/3
( d) n4/3
11 oo : oo : 19 Prob 1.18. The pressure of a non relativistic free Fermi gas in t hree-dimensions depends, at T = 0, on t he density of fermions n as [NET June 2014]
(a) n 5/3
(b) n l/3
(c) n2/ 3
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1.1
(d) n4/ 3
Ans Keys
1.1.
C
1.7. a
1.13. b
1.2.
C
1.8. a
1.14. d
1.3. b
1.9. d
1.15. 2.36
1.4. a
1.10.
C
1.16. 1263
1.5. a
1.11.
C
1.17. d
1.6. d
1.12. b
1.18. a
11 oo : oo : 21
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1.2
Ideal Fermi Systems
Solutions
Sol 1.1. t he fermi statistics applied for only t he fermeonic
system where a single particle state can accommodate only one or none particle. and the fermi energy is defined as t he energy level upto which all states a1·e completely occupied and at temperature higher than O kelvin t he occupancy is half and the for t he fermi systems µ = Ep where E p is the fermi energy level hence we have
so opt ion c is correct. ~
..
-
...,.
.,
,
, .,. ,
,.
•
1
11 oo : oo : 24 n(c) =
1 -E -EF--
e kBT
+l
so option c is correct.
Sol 1.2. the occupancy means the probability of a single particle state gets occupied and for the fermionic system it is given by 1 exp
+1
Sol 1. 3. if we consider the nuclear energy level particle dist ribution then we have even number of protons and neutron hence B e~ can never be fermeonic syst em. only deuterium has t he PROTON+ ELECTRON+ NEUTRON=3 and t hat is odd hence the system is fermeonic and hence option d is correct. [email protected]
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Ideal Fermi Systems
Sol 1.4. t he pressure of a fermeonic system can be defined as Pp = jnEp now we can show t hat Ep ex n 213 hence we can surely say t hat Pp ex n 513 hence the option a is correct. Sol 1.5. now we are considering the momentum space, the number of state in K to K + dK is given by
2 n(k )dk = (L / 21r ) 21rKdK now we h ave E = nKV so we take dE = ndKV now we are calculating the number of the particle Ep
N = "
L2 L2 'lrn.21,2 EdE = ?.7rn.21,2 EP2
11 oo : oo : 26 now we have E = nKV so we take dE = ndKV now we are calculating the number of the particle Ep L2 L2 N = o 7r!i,2v2 EdE = 21rri,2v2 E p2 2
so we can say Ep = 21rn2v 2 p now we are calculating the average energy at zero kelvin foEF
E .g(E)dE
2L 2 J2it,,nv p312
< E >= _.;;..__ _ _ _ = - - - - - N
so we can say E / L
2
3N
2 3 = j E p 2 3 but we know that Ep ex n 1 the pressure of the system is defined 2N
P=3V < E >
now N/V= n hence we have P ex n 513
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11 oo : oo : oo
Problems and Solutions
•
Ill
Ideal Bose Systems Sk J ahiruddin * Suchismito Chatterjee
*Assistant P rofessor Sister Nibedita Govt. College, Kolkata Author was the topper of IIT Bombay 1\11.Sc Physics 2009-2011 batch He ranked 007 in IIT JAl\lI 2009 and 008 (JRF) in CSIR NET June 2011 He has been teaching CSIR NET aspirants since 2012
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@Sk Jahiruddin, 2020
Ideal Bose Systems
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Ideal Bose Systems
Contents 1 Problems from NET, GATE, JEST, TIFR & JAM papers 1.1 Basics, Bose Einstein condensation superconductivity . . . . . . . . . . . . . . . . . . . . 1.2 Photon gas: Blackbody radiation . . . . . . 1. 3 Ans Keys . . . . . . . . . . . . . . . . . . . 1.4 Solutions . . . . . . . . . . . . . . . . . . . .
j [email protected]
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3 3 7 16 17
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Ideal Bose Systems
@Sk J ahiruddin , 2020
1
Problems from NET, GATE, JEST, TIFR & JAM papers
1.1
Basics, Bose Einstein condensation and
superconductivity Prob 1.1. Which of the following atoms cannot exhibit Bose-Einstein condensation even in principle? [GATE 2010]
(a) H t
(b) H e~
(c)
Narr
(d)
Kt8
Prob 1.2. A system has energy levels: E 0 , 2E0 , 3E0 , · · · where t he excited states are triply degenerate. 4 non-interacting bosons are placed in this syst em. If t he total energy of t hese bosons is 5E0 , the number of microstates is [GATE 2007] (a) 2 (b) 3 (c) 4 (d) 5
Prob 1.3. In Bose-Einstein condensates, the particles [GATE 2015]
(a) have strong inter-particle attraction. (b) condense in real space. (c) have over lapping wave functions. (d) have large and positive chemical potential. Prob 1.4. The chemical potential of an ideal Bose gas at
11 oo : oo : os (c) have over lapping wave functions. (d) have large and positive chemical potential.
Prob 1.4. The chemical potential of an ideal Bose gas at [email protected]
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Ideal Bose Systems
any temperature is (a) necessarily negative
[JAM 2008] (b) eit her zero or negative (d) either zero or positive
(c) necessarily positive
Prob 1.5. An ideal Bose gas in cl-dimensions obeys t he -+
dispersion relation: c( k = Ak8 ) where A and s are constants. For Bose-Einst ein condensation to occur, the occupancy of excited states 00
0
e
d- s s
e /3 (c-µ)- l
where c is a constant, should remain finite even for µ = 0. This can happen if [NET June 2015] d l 1 d l d l d (a) s < 4 (b) 4 < s < 2 (c) s > l (d ) 2 < s < l
Prob 1.6. Consider an ideal Bose gas in 3-dimensions with energy-momentum relation c ex p 8 , with s > 0. The range of s for which this syst em may undergo Bose-Einstein condensation at a non-zero temperature is [NET June 2011]
(a) 1 < s < 3
(b) 0 < s < 2
(c) 0 < s < 3
(d) 0 < s < oo 4
Prob 1. 7. Bose condensation occurs in liquid He kept at
11 oo : oo : 11 ~.rrr,v t.L ti'J I lb OJ O OGl.fi i":llOij
densation at a non-zero temperature is [NET June 2011]
(a) 1 < s < 3
(b ) 0 < s < 2
(c) 0 < s < 3
(d) 0 < s < oo 4
Pro b 1. 7. Bose condensation occurs in liquid He kept at an ambient pressure at 2. 17K. At what t emperature will [email protected]
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Ideal Bose Systems
Bose condensation will occur in He4 in t he gaseous state, t he density of which is 1000 times sm aller than that of He4 ? [Assume that it is a p erfect Bose gas.] [NET June 2012]
(a) 2.17mK
(b) 21 .7mK
(c) 21 .7µK
(d ) 2.17µK
Prob 1.8. Non-int eracting bosons undergo Bose-Einstein condensation (BEC) when trapped in a 3-dimensional isotropic simple harmonic p otential. For BEC to occur, the chemical
[NET Dec 2012]
potential must be equal to
(a) nJ.;J/ 2
(b)
nJ.;J
(c) 3nJ.;J/ 2
(d) 0
Pro b 1. 9. T wo identical bosons each of m ass m are placed 2 2 in a I-dimensional potential V (x) = ½mw x . The bosons interact via a weak potent ial ½ 2 = Vo exp[-mD(x 1 -x 2 ) /4n] where x 1 and x 2 denote t he co-ordinates of t he part icles. 2
Given that t he ground-state wave function of t he h armonic oscillator is 'lj; 0 (x)
= (7;~) ! exp[-mwx 2 / 2n].
The ground
stat e energy of the two boson system , to the first order in V0 is [NET June 2013]
(a)
nJ.;J
+ 2Vo
(b)
nJ.;J
+ Vo£
(c)
nJ.;J
+ Vo( l + £)-~
11 oo : oo : 13 2
oscillator is 'lj;0 (x) = (7;~ )! exp[-mwx / 2n]. The ground state energy of t he two boson system , to the first order in V0 is [NET June 2013]
nw + 2Vo (d) nw + Vo~ (a)
(b)
nw + Vo£
(c)
nw + Vo( l + £)-~
Prob 1.10. An ideal Bose gas is confined inside a container that is connect ed to a particle reservoir. Each parj [email protected]
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Ideal Bose Systems
t icle can occupy a discret e set of single-particle quantum states. If the probability that a part icular quantum st ate is unoccupied is 0.1, t hen the average number of bosons in t hat state is [NET Dec 2014] (a) 8 (b) 9 (c) 10 (d) 11
Prob 1.11. Consider a quant um syst em of non-interacting bosons in contact wit h a particle bath. The probability of finding no particle in a given single particle quantum state is 1o- 6 . The average number of part icles in that state is of [NET Dec 2017] t he order of (a) 10 3 (b) 106 (c) 109 (d) 1012 Prob 1.12. T wo ident ical bosons may occupy any of two energy levels 0, E where E > 0 lowest energy state is doublydegenerate and t he excited state is non-degenerate. Assume t hat the two-particle system is in thermal equilibrium at a temperature T. Calculat e the average energy (E). What - - - - at low temperature? exp - E fTTFR. 20171
11 oo : oo : 11 egenera e an v~~e exc1 e1 egenera e. t hat the two-particle system is in t hermal equilibrium at a temperature T. Calculate t he average energy (E). What
- - - - at low temperature? exp - E [TIFR 2017]
Prob 1.13. A large number N of ideal bosons, each of
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Ideal Bose Systems
@Sk J ahiruddin, 2020
mass m, are trapped in a three-dimensional potential 2 2
V(r)
= mw r 2
The bosonic system is kept at temperature T which is much lower than t he Bose-Einstein condensation temperature Tc. The chemical potential (µ) satisfies [GATE 2019]
(A) µ < ~nw (c)
1.2
(B) 2nw > µ > ~nw
3nw > µ > 2nw
(D) µ = 3nw
Photon gas: Blackbody radiation
Prob 1.14. The blackbody spectrum of an object 0 1 is such t hat its radiant intensity (i.e. intensity per unit wavelength interval) is maximum at a wavelength of 200 nm. Another object 0 2 has maximum radiant intensity at 600 nm. The ratio of power emitted per unit area by 0 1 to that
11 oo : oo : 1s ro e o ac o y sp ec rum o an o Jec 1 1s such t hat it s radiant intensity (i.e. intensity per unit wavelength interval) is maximum at a wavelengt h of 200 nm.
Another object 0 2 has maximum radiant int ensity at 600 nm . T he ratio of power emitted per unit area by O1 to that by 0 2 is [JAM 2007] (a) 1/81 (b) 1/ 9 (c) 9 (d) 81 Prob 1.15. When the temperature of a blackbody is doubled , t he maximum value of it s sp ectral energy density, wit h respect to t hat at init ial t emperature would b ecome [JAM 2012] (a) 1/ 16 t imes
(b) 8 t imes
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(c) 16 t imes 7
(d) 32 times
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Ideal Bose Systems
@Sk J ahiruddin , 2020
Prob 1.16. A blackbody at temp erat ure T emit s radiat ion at a p eak wavelength ,\. If it s t emperature becomes
4T , t he new p eak wavelength is
(a)
2;6 ,\
(b)
614 ,\
( c)
[JAM 2013]
f6 ,\ (d) ¼,\
Prob 1.17. In t he radiation emitted by a black body, t he rat io of t he spectral densit ies at frequencies 2v and v will [JAM 2017] vary wit h v as
(a) [ehv/kBT (c)
]-l 1]
_ l
[ehv/ ksT _
+ l ]-l [ehv/ ksT _ 1]
(b) [ehv/ ksT (d)
Prob 1.18. A red star having radius rR at a temperature
TR and a white st ar having radius rw at a temperature T w, radiat e the same t otal power. If these st ars radiate as perfect black bodies, then [JAM 2019] (A) rR > rw and TR > Tw (B) rR < rw and TR > T w
11 oo : oo : 22 Prob 1.18. A red star having radius rR at a temperature TR and a white star having radius rw at a temperature Tw, radiate the same total power. If these stars radiate as perfect black bodies, then [JAM 2019] (A) rR > rw and TR > Tl,v (B) rR < rtt1 and T R > Tw (C) rR > rw and TR < Tw (D) rR < rw and TR < Tw Prob 1.19. The partition function of a gas of photons is given as: •
[GATE 2010] (i) The specific heat of the photon gas varies with temperature as
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Ideal Bose Systems
T
T
(ii) The pressure of the photon gas is
T
T
11 oo : oo : 24 T
T
(ii) The pressure of the photon gas is
(a) n2 (ksT )3 15n3 c3
(c) n2(ksT)4
(b) n2(ksT)4 8n3c3 (d) n2(ksT)3/2
45n3 c3 45n3 c3 Prob 1.20. For a blackbody radiation in a cavity, photons are created and annihilated freely as a result of emission and absorption by the walls of t he cavity. T his is because [GATE 2015] (a) t he chemical potential of the photons is zero. (b) photons obey P auli exclusion principle. (c) photons are spin-1 particles. [email protected]
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Ideal Bose Systems
(d) t he entropy of the photons is very large. Prob 1.21. The total power emitted by a spherical black-
body of radius R at t emperature T is P1 . Let P2 be the power emitted by another spherical blackbody of radius R / 2 at temperature 2T. The ratio (P1 / P2 ) is [GATE 2016] (Give your answer upto two decimal places.) Prob 1.22. A cavity contains blackbody radiation in equi-
librium at temperature T. The specific heat per unit volume of the photon gas in t he cavity is of t he form Cv = ,1T 3 , where 1' is a constant. The cavity is exp anded to twice its
11 oo : oo : 21 Prob 1.22. A cavity contains blackbody radiat ion in equilibriu m at t emperature T . The specific heat per unit volum e of t he photon gas in t he cavit y is of t he form Cv = ryT 3 , where ry is a const ant. The cavity is expanded to twice it s origin al volume and t hen allowed to equilibrium at t emp erature T . T he new internal energy per u nit volume is [NET
June 2011] 4 4 (a) 4ryT (b )2ryT
( c )ryT
4
4
( d )ryT /
4
Prob 1.23. Consider black-body radiation contained in a cavity whose walls are at t emperature T . The radiat ion is in equilibrium wit h the walls of the cavity. If the t emperature of t he walls is increased to 2T and t he radiat ion is allowed t o come to equilibrium at the new t emperature , the ent ropy
[NET June
of the radiation increases by a factor of
2012] (a) 2
(b) 4
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(c) 8
(d) 16
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@Sk J ahiruddin , 2020
Prob 1.24. Which of t he graphs below gives t he correct qualitative behaviour of t he energy densit y Er(>..) of blackbody radiation >.. at two t emperatures T1 and T2,(T1 < T2)?
[NET June 2014]
,,--
c
T2 so the peak of the curve in t he case one will be higher t han t hat of t he peak of the 2nd one hence
t he option d is correct . Sol 1.33. t he volume would be t ripled in t he fut ure times now the energy radiated per unite area is E = aT4 in future ,xrP ,xr-ill h ~ , rP Ti:, = rrT, 4 h,,t tri+~ l PnPr> a from t he loop. T he current loop and the line are all in the same plane.
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EMT: Test-1
@Sk Jahiruddin: 2020
r
I
Figure 1: 2
2
(a) µoa 2I (b) µoa I r
Ans= c Solution:
r
Consider t he line to be part of an infinite loop , with the return part at infinity. By t he reciprocal law (M 12 = M 21 for mutual inductance), flux through t he infinite loop due to . current I in t he small loop is t he same as the flux through the small loop due to the current I in the infinite loop. Flux =
1ra2 µo f
21rr
= µoa2 I
2r
Qs.3: Consider a thin conducting spherical shell with radius
11 oo : oo : os Flux = 1ra2 µo f = µoa2 I 21rr 2r Qs.3: Consider a thin conducting spherical shell wit h radius [email protected]
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EMT: Test-1
R carrying a total charge Q . W hat is t he force per unit area
due to t he electric field acting on t he shell's surface? 1
1
Q2
( a) 41rE 0 81r R 3
(C)
Q2
(b) 41rEo 41r R 4
Q2 (d) 1 Q2 41rEo 41r R 3 41rEo 81r R 4 Ans = d Solution: Capacity of a charged sphere is C = 41rE 0 R. Stored energy is W = ~ = ~ 4,;co Increasing t he radius by dR yields 1
°d
a;
dW l 1 Q2 41r R A dR = dR dR = - 2 41rEo R 2dR 2F
Thus, ~ erywhere.
= 4,;co 8~~4 . The force points radially outwards ev-
Qs.4: In a double-slit int erference experiment, d is t he dis-
tance between t he centers of the slits and w is t he width of each slit , as shown in t he figure above. For incident plane waves, an interference maximum on a distant screen will be '' missing'' when
11 oo : oo : 11 '' missing'' when
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E:NIT : Test-I
@Sk Jahiruddin: 2020
w{ • •
(a) d = v!2w
d
(b) d = \!'3w
(c) 2d = w
(d ) 2d = 3w
Ans= d Based on the given diagram, we know it can't be true that d < w so (C) can be eliminated. Next, recall that the equation for constructive interference in double slit diffraction is dsin(0)
= m 1A
addit ionally, we know t hat we will get a '' missing'' interference maximum when the constructive double slit equation coincides with the single slit diffraction, so we use w sin(0)
= m2 A
get both equations equal to sin (0) and set them equal to one another and solve for d,
11 oo : oo : 14 •
•
co1nc1
S Wl
sin
ac 10n, so we use w sin(0) = m2A
get both equations equal to sin (0) and set them equal to one another and solve for d, w
d
m 2
m 1
d= m 1w m2 [email protected]
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EMT: Test-1
@Sk Jahiruddin: 2020
comparing t his to (A) and (B) , we aren 't going to get an irrat ional number (i.e. v'2 or v'3) with some fraction of integers so we can confidently choose (D).
Qs.5: An electro-magnetic plane wave propagates in free space with angular frequency wand wave number k. The wave is incident normally on an absorbing sheet. The resistance of t he sheet is adjust ed until there is no reflection. Find an expression for t he resistance of a square of the sheet of side l , and estimate t he numerical value in Ohm.
(a) 81
(c) 280
(b) 190
(d) 370
Ans= d
Solution: This is sometimes taught as theory. The Electric potential V = El Amperes' law gives
E l= µof
I =
El mu0
11 oo : oo : 16 Amperes' law gives
I = El mu0 Resistance
V
R
El E Bl/µo = µ 0 B
=I=
For EM wave in free space
E 1 -=c=--B ~ [email protected]
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EMT: Test-1
@Sk J a hiruddin: 2020
So,
µo = 41r R
=
X
10-7 ,
co = 8.85
8~~:~i-=-:2
rv
✓1 .3
X
X
10- 12
10 5 rv 370D
Qs.6: Two parallel infinite line charges are fixed a distance h above a grounded conducting plane. They are a distance 2h apart and have linear charge density - .A. A positive charge
Q is placed a distance h above the plane midway between the two line charges and feels no net force. Find Q in t erms of ,\ and h . (a) 32:h (b) 21:h (c) 13:h (d) 7~h
Ans= a Solution: Charged conductors over a grounded conducting plane Use the method of images: Remove the conducting plane and replace with line charges
11 oo : oo : 19 Ans= a
Solution: Charged conductors over a grounded conducting plane Use the method of images: Remove the conducting plane and replace with line charges of +,\ a distance h b elow the line charges and a charge of -Q below t he point charge. Obviously, the parallel components of t he electrostatic force cancels. The upward force from both the line charges is 2,\Q
Fz = 2 v'5 cos 0 h 5
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@Sk Jahiruddin: 2020
EJVIT: Test-1
and cos 0 = 2/ v'5. The force from the point image charge is Q2
FQ
= - 4h2
Then the total force must sum to zero: F, - T -
Q2
4h 2
+
8,\Q -
5h - 0
-+ Q = 32-\h 5
Qs. 7: A point charge +q is fixed at the point (0, 0, d) above the grounded conducting plane z = 0 What would be the radius of the circle centered at the origin in the xy plane which encloses half of the total charge induced on the plane? (a) d (b) 2d (c) dv0, (d) dv13 Ans= d
Solution: From image problem the potential r
7
11 oo : oo : 23 origin in t he xy plane which encloses half of t he total charge induced on the plane? (a) d (b) 2d (c) dv'2 (d) dv13 Ans= d Solution: From image problem the potential 1 q 47rEO J x 2 + y2 + (z - d)2
V (x, y ,z) =
q
Induced surface charge a = - Eo
av
av
l
az
47rEo
8z
z =O
+ d) ) ------+ - - - - - - -2 2 3 [x2 + y2 + (z _ d)2] / [x2 + y2 + (z + d)2)3/ J - q (z - d)
q(z
- qd a (x' Y) = 2 ( 2 2 d2) 3/2 1f X +y + [email protected]
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2020
ENIT: Test-1
We use polar coordinate
- qd a( r ) - - - - - 21r (r2
+ d2)3/2
Charge bounded wit hin a circle of radius x is 21r
0
- qd qd - - - - - rdr dcp = -----;:::::= = x o 21r (r2 + d2)3/2 J r2 + d2 o X
=
qd
-----;:::::= = - q
J x 2 + d2
T his should be half of t he total charged induced in the plane. The total charge induced in t he plane is - q, as it should be. Then (]
11 oo : oo : 26 This should be half of t he total charged induced in the plane. The total charge induced in the plane is - q, as it should be. Then qd ✓x2
+d2 -
q
q
=-2
Now do the algebra to get X
=
v13d
Qs.8: A point charge q is located at the center of a toroidal coil of rectangular cross section, inner radius a , outer radius a+ w, and height h, which carries a total of N tightly-wound turns and current I.
What will be t he electromagnetic momentum p of this configuration, assuming that w and h are both much less t han [email protected]
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EMT: Test-1
@Sk Jahiruddin: 2020
a (so you can ignore the variation of the fields over t he cross
section). a) µo qNiwh (b) µo qNiw2 h (c) µo qNiwh (d) µo qNiw 2 h ( 41r 2a2 41r 2a3 41r a2 41r a3 Ans= c Solution: For points inside t he toroid the fields are
The electromagnetic moment um density q µoN I A µoqNI A g = Eo(E x B ) = Eo z= z 2 2 47rEoa 21ra 81r a 3
Inside the toroid r
~
as and s
~
a So, the electromagnetic
11 oo : oo : 30 The electromagnetic momentum density q
µoNI A µo qNI A g = Eo (E x B) = Eo z = ? z 2 3 41fEoa 21ra 81r,., a
Inside the toroid r ~ as and s ~ a So, the electromagnetic momentum is t he momentum density multi- plied by the volume µo q N I h) µo q N I wh 1raw = -z 2 p = - 2- -z 81r a3 41r a2 A
(
A
Qs. 9: Consider two statements (a) T wo conducting rings, A and B , have identical dimensions. The self-inductance of ring A will be the same as that of ring B even when t he rings are made of very different metals. (b) In comparing two capacitors which each are l µF but which differ significantly in size, t he largest of the two can probably be charged up to a higher potential difference t han the smaller one. It will thus be able to hold more energy than the smaller one. [email protected]
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EMT: Test-1
Which of the statements are true? (a) a only (b) b only (c) Both (d) None Ans= c Solution: Both are true (a) TRUE: Identical currents in the two rings will generate identical magnetic fields, which will in turn result in identical magnetic fluxes. Self-inductance is simply the ratio of this flux to the current , therefore it will be the same for the two rings regardless of their differing composition.
11 oo : oo : 33 identical magnetic fields, which will in turn result in ident ical magnetic fluxes. Self-inductance is simply the ratio of this flux to t he current , therefore it will be the same for the two rings regardless of their differing composition. (b) TRUE: Consider two parallel-plate capacitors in air: a smaller one with plate area A and plate separation d and a larger one with with plate area 2A and plate separation 2d. Both have t he same capacitance C = EoA/ d. However, if both are charged to the same potential difference V, the electric field between the plates of t he smaller capacitor will be E 1 = V / d while t he electric field between the plates of the larger one will be E 2 = V / 2d. Thus the larger capacitor can be charged to twice as high a potential difference before reaching 2 the breakdown E for air, and can hold more energy ( CV /2) . If the dielectric constant, K,, is t he same for both, t he same reasoning holds. It is t herefore very probable that the capacitor which is significantly larger in size t han t he other, can be charged to a higher potential.
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EMT: Test-1
Qs.10: Consider two statements
(a) Charge particles in a magnetic field can not experience ..... an acceleration due to t he iJ x B force. (b) When the magnetic flux t hrough an open surface attached to a conduct ing loop is zero, there cannot be an induced EMF in t hat loop. Which of the statements are t rue? (a) a only (b) b only ( ("l \
TI "+ h ( rl \ l\Tf"\Yl Cl
11 oo : oo : 3s [email protected]
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E:NIT: Test-I
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Qs.10: Consider two statements
(a) Charge particles in a magnetic field can not experience _, an acceleration due to the v x B force. (b) When the magnetic flux through an open surface attached to a conducting loop is zero, there cannot be an induced EMF in t hat loop. Which of the statements are true? (a) a only (b) b only (c) Both (d) None Ans= d Solution: None are t rue (a) FALSE: Any unbalanced force on a particle causes an acceleration (F = ma) If t he force is perpendicular to the particle's velocity (as is the case for magnetic forces) , then the acceleration causes a change in the direction of velocity. (b) FALSE: It is not the value of the flux itself that gives an induced EMF, but rather the time derivative of the flux. 'PB can be zero at an instant, and d'PB / dt can still be non-zero.
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