Fundamentals of Vector and Tensor Analysis. 9786010426726

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Fundamentals of Vector and Tensor Analysis.
 9786010426726

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AL-FARABI KAZAKH NATIONAL UNIVERSITY

К. А. Zhaksybekova М. А. Zhusupov R. S. Kabatayeva

FUNDAMENTALS OF VECTOR AND TENSOR ANALYSIS Educational manual

Second edition, supplemented

Almaty «Qazaq university» 2017

1

UDC 539.1 (075.8) LBC 22.383 Zh 99 Recommended for publication by the decision of the Academic Council of the Faculty of Physics and Technology and Editorial and Publishing Council of Al-Farabi Kazakh National University (Protocol №4 dated 26.05.2017) Reviewers: doctor of Physics and Mathematics, Professor Yu.V. Arkhipov candidate of Technical Sciences, Associate Professor G.А. Abdraimova

Zhaksybekova К.А. Zh 99 Fundamentals of Vector and Tensor Analysis: educational manual / К.А. Zhaksybekova, М.А. Zhusupov, R.S. Kabatayeva. – 2nd ed., suppl. – Almaty: Qazaq university, 2017. – 148 p. ISBN 978-601-04- 2672-6 The tutorial contains the lectures being given at the Department of Theoretical and Nuclear Physics of al-Farabi Kazakh National University. The tutorial is intended for students in physics and mathematics and can be used for the active mastering of different physical problems solution methods. Publishing in authorial release.

UDC 539.1 (075.8) LBC 22.383 ISBN 978-601-04-2672-6

© Zhaksybekova К.А., Zhusupov М.А., Kabatayeva R.S., 2017 © al-Farabi KazNU, 2017

2

Mathematical methods are used in physics when describing properties, laws of motion and interaction of different physical objects. The use of scalar quantities allows describing the simplest physical properties ob bodies. However, for example, for the quantitative description of bodies’ interaction the scalar quantities are not enough. In this case it is necessary to use the more complicated mathematical quantities − directed segment or vectors. The tensors which have more complicated mathematical nature are used for characterization of deformations, inertia at rotational motion and etc. Since the scalars, vectors and tensors are chosen for the quantitative description of the characteristics of objects of environment, then, from the physical point, they should have the common nature. When considering the physical problems it is convenient to determine the specific vectors and tensors relatively to coordinates systems. At this the coordinates systems themselves can be chosen in arbitrary way since they carry the secondary character. That is, all systems of coordinates should be equivalent. In the mathematical apparatus of tensor calculus, as in its particular case – vector algebra and calculus, the coordinates systems equality are included since the physical characteristics and physical regulations formulation should not be dependent on the choice of coordinates systems. This allowed to express the mathematical formulations of physical laws in opportune and obvious forms and gave the opportunity of its active use in the modern physics. The present tutorial is devoted to statement of the fundamentals of vector and tensor calculus for physicists in a volume necessary for solution of problems of classical mechanics, electrodynamics, quantum mechanics and etc. The treatment of the theory is illustrated with examples. The tutorial contains the problems for self-solution as well. 3

In the first part of the present tutorial there is a consideration of questions connected to the operations over vectors, integral theorems of vector calculus. The other two chapters contain the information about curvilinear orthogonal systems of coordinates and tensor calculus.

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1.1. Fundamental concepts In the science and technique there are often the quantities which are totally determined by only number, the so called absolute value, for example, mass, time, temperature and etc. These quantities are called scalar ones. However many physical quantities are determined by not only the number but by the direction as well, for example, displacement, velocity, force, momentum and angular momentum. These quantities are called vector ones. It is interesting to notice, that the all mentioned vector quantities are taken from mechanics, however at the development of mechanics the vector calculus was not used; moreover it was not created so far. The need in the vector calculus appeared after that Maxwell developed the electromagnetic theory and the vector nature of the electric and magnet fields became clear. Graphically it is opportune any vector quantity (further we will call it as vector) to represent by an arrow, length of which is proportional to the quantity of vector, and the direction determines the vector’s direction. The direction of the arrow stands for the opposite direction. At such a determination the sum of vectors

   C  A B

(1.1)

means the convergency of the origin of vector B with the end of the vector A . An arrow, joining the origin of the vector A with the end of the vector B , defines the vector C . This procedure of vectors addition by the triangle rule (1.1) is illustrated in the figure 1. 5

Fig. 1. Triangle rule for vectors addition

Supplementing the triangle obtained to a parallelogram, one can see (fig. 2), that

     C  A B  B  A.

(1.2)

Fig. 2. Parallelogram rule for vectors addition

Let’s note that vectors are the geometrical objects not dependent on coordinates systems. For example, the vector A (fig. 3), directed from the origin of reference system, end in the point (х1,у1,z1).

Fig. 3. Components of vector in the Cartesian coordinate system

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Any vector quantity (momentum, electric field strength and etc.) can be denoted by a symbol A , however some vector quantities, for example, a distance from the origin of coordinates to a point  (х1,у1,z1) is denoted by a special symbol r (radius-vector)

r  ( x1 , y1 , z1 ) .

(1.3)

Let’s denote the absolute value of the radius-vector with a symbol

r  r . It is easy to understand (fig. 4), that the coordinates

of the end of the vector is connected with the absolute value of the vector by the relations

x1  r  cos ,

y1  r  cos , z1  r  cos .

(1.4)

Fig. 4. Directing cosines

Here cos , cos 

and

cos  are directing cosines, and

 ,  and   respectively the angles between the vector and the positive direction of axis x, y and z . The quantities x1 , y 1 and z 1 

are called the components (Cartesian) of the radius-vector r or its projections. 7

Any vector A can be decomposed on components (or projected on the coordinate axes):

Ax  A  cos , Ay  A  cos , Az  A  cos .

(1.5)

Let’s introduce the unite vectors in the direction of each of the

  

coordinate axes. Let i , j , k  respectively the vector of unit length directed along the positive semiaxes x, y , z . Then according to the operation of vector addition,

A  i x  j  y  k z .

(1.6)

If A  0 , then  x   y   z  0 . In accordance with the Pythagorean theorem the absolute value of the vector A equals

   2x   2y   2z . An addition and a subtraction of vectors can be done using the component representation. For vector A

  i x  j  y  k z and vector 

  i x  j  y  k z it is true

    i   x   x   j   y   y   k   z   z  . (1.7) Problems 1. Two vectors are given: C1  A  B and C2  A  B . Construct the vectors

A and B . 8

2. Two points are given

M1 ( x1 ; y1 ; z1 ) and M 2 ( x2 ; y2 ; z2 ) . Define the

vector M1M 2 in a coordinate form. 3. Find the unit vector collinear (parallel) to a vector directed by the bisectrix of the angle ВАС of the triangle АВС, if its vertexes are given: А(1; 1; 1), В(3; 0; 1), С(0; 3; 1). 4. What is the condition for three vectors constructed of them?

a , b and c , that a triangle can be

5. Let   5i  3 j  2k ,   2i  2 j  2k . Define the vectors    and

.

6. A plane flies sequentially along the sides АВ, ВС and СА of the triangle

АВС for the given time intervals t1 , t 2 and t3 respectively. At this the plane is always shifted from the direction by wind with the constant velocity u . Define the intrinsic velocity shifted by wind. 7. A vector

v A,

of the plane and the velocity

u,

with which the plane are

length of which equals 10, does the equal angles with the

coordinate axes. Find Ax , Ay and

Az .

8. Define the components of the unit vector which lies in the plane XY and does the equal angles with the positive directions of the axes x and y. 9. Find the sum of three vectors which have the length а and are built: а) from the vertex of a cube by its three edges; b) from the vertex of a regular pyramid by its three edges.

1.2. Rotation of coordinates system A definition of a vector by an assignment of its absolute value and a direction is not quite strict. There exist quantities, such as, for example, coefficient of elasticity, refraction index in anisothropic crystals, which are characterized by both the absolute value and the direction, but though are not the vectors. Moreover this obvious definition of a vector is inconvenient and cannot be generalized for more complicated quantities.  Using the radius-vector r let’s give the new definition of a vector. For an introduction of the new definition there exist important physical reasons. We describe the environmental world by the instrumentality of mathematics, but any physical description 9

should be independent of mathematical apparatus. Further we will suppose that the space is isotropic. In this case the physical system under consideration or the physical law formulated should not be dependent on the choice or the orientation of the coordinates system.

Fig. 5. Rotation of Cartesian coordinate system





Now we refer again to the radius-vector r . Let’s consider r in two different systems, one of which is rotated relatively to the other one. For the simplicity let’s limit to two-dimensional case (fig. 5). If the coordinate axes x and y are rotated anticlockwise for angle  and  at this the position of the radius-vector r is fixed, one can write the following relations connecting the components of the radius-vector in the stationary system with the components of the same vector in the rotated one:

x  x cos   y sin  ,

y   x sin   y cos  .

(1.8)

A vector can be represented with coordinates of its end point (part 1.1), in other words, the coordinates of the point are proportional to the components of the vector. Hence, the components of the vector should be transformed when rotating of the coordinate axes as well as the coordinates of the point (or as the radius-vector  r ). Moreover, if any pair of quantities ( x , Ay ) , given in the 10

Cartesian coordinate system x, y , is transformed into ( x , Ay ) by a rotation of the coordinate system so that

x  x cos    y sin  , y   x sin    y cos  ,

(1.9)

then one can refer  x and  y as the components of vector A . The vector A is defined now by the law of transformation of its component when rotating of the coordinate system. If  x and  y are transformed as the components of two-dimensional radius-vector, they are the components of the vector A . If  x and  y behave themselves other way when rotating of the coordinate system than no vector can be constructed of them. In order to do the definition of a vector more completed, it is necessary to clear out the meaning of the quantities  x and  y in the equations (1.9). Let’s suppose, that the components of the vector A – are the functions of coordinates and moreover of some constant vector C :

 x   x ( x, y, C x , C y ) ,  y   y ( x, y, Cx , C y ) . (1.10) In the rotated coordinate system the vector А has the components  x and  y :

x  x ( x , y , C x , C y ) ,

y  y ( x, y, Cx , Cy ) .

(1.11)

Using the equation (1.8), the coordinates x, y, Cx , Cy can be expressed through the coordinates of the stationary system and rotation angle  . In general there exists some dependence on the rotation angle. However such a dependence on the orientation is not 11

desirable. That is why we limit to functions not dependent on the orientation. Obviously, in a particular case, when  =0,

 x  x ,  y  y . Examples: 1. A pair of quantities (-у, х) are given. Show, that these quantities form the two-dimensional vector. Let’s consider how these quantities are transformed when rotating of the system for angle  . We have

V x   y cos   x sin  ,

V y  y sin   x cos  ,

where Vx   y, Vy  x . Using (1.8), one obtains V x   y  ,

V y  x  , i.е. the pair of quantities satisfies to the equations (1.9), determining two-dimensional vector. Thus, (-у,х) corresponds the components of the vector.





the

pair



2. Let’s consider V  i x  j y  ( x, y) . According to (1.9),

Vx  x cos   y sin   x, Vy  x sin   y cos    y . Substituting Vx  x and V y   y , one obtains

Vx  Vx cos  Vy sin , Vy  Vx sin   Vy cos . These relations do not satisfy to the definition of a vector. Hence, the pair  x,  y  cannot be referred as a vector. For transition to three- and n-dimensional space it is convenient to use more compact notation. Let x  x1 , a11  cos  , a12  sin  ,

(1.12) y  x2 , a21   sin  , a22  cos  . 12

Then the equations (1.9) can be rewritten:

x1  a11 x1  a12 x2 ,

x2  a21 x1  a22 x2 .

(1.13)

Coefficients a ij can be identified with the directing cosines (as the cosine of the angle between xi and x j ), i.e. a12  cos( x1 , x2 )  sin  , a21  cos( x2 , x1 )  cos( 

 2

)   sin  .

Then the equations (1.13) can be written in a form: 2

x i   a ij x j ,

i  1,2.

(1.14)

j 1

Now it is easy to do a generalization for the case of three, four and more dimensions. A set of N quantities V j ( j  1,..., N ) defines



the components of the N-dimensional vector V then and only then when the values of these quantities in the rotated coordinate system are given with the formulas: N

Vi   aijV j ,

i  1, 2,..., N ,

(1.15)

j 1

Here aij  is the cosine of the angle between xi and x j . Issuing from the definition of aij , one can write in the Cartesian coordinates

aij 

xi x j  . x j xi

(1.16)

These are partial derivatives. Substituting (1.16) into (1.15), one obtains 13

N x xi j Vj   Vj . j 1 x j j 1 xi N

Vi  

(1.17)

The directing cosines aij satisfy the orthogonality condition

a a

  jk ,

(1.18)

a   jk .

(1.19)

ij ik

i

or

a

ji ki

i

Here  jk  is a delta-symbol, determined as

1, if 0, if

j  k, j  k.

 jk  

(1.20)

Substituting aij from (1.12) it is easy to convince, that the equations (1.18) and (1.19) are true as well for the two-dimensional case. In the result for the case when j  k one has:

sin2  cos 2  1. In order to convince in the truth of the equation (1.18) in the general case, one can use the expression (1.16):

x j xk

x j xi

 x  x   x  x i

i

i

i

i

k



x j xk

.

(1.21)

Conclusions. In a new definition of a vector through the law of transformation of its components it is worth to notice the two moments: 1) it is convenient for a description of the different physical phenomena; 14

2) it serves as the base for the transition to a new part of mathematics – tensor calculus. Problems

 1. A constant vector V (Vx  1, Vy  0) is given. Show, that the components of the

vector

in

the

rotated

coordinate

system

have

the

form:

Vx  cos , Vy   sin  , what corresponds to the law of transformations of

vectors (introducing the constant vector we chose the specific direction in the space). 2. Define, if the following quantities satisfy the law of vector transformation (1.15): а) (х-у, х+у, 0) when rotating around the axis z; б) (0, 2z+у, z-2у) when rotating around the axis х; в) ( y 2  z 2 ,  xy,  xz) when rotating around each of coordinate axes. 3. Show, that ( xyCx  y 2C y , x2Cx  xyCy ) forms a vector. The quantities

Cx  and C y are the components of the constant vector C . Do the same for ( xyCx  x 2C y , y 2C x  xyCy ) . 4. Considering the rotation around any of coordinate axes answer the question if 2

2

2

2

2

2

three functions Vx  a1 ( x  y  z ), Vy  a2 ( x  y  z ) and Vz  a3 ( x 2  y 2  z 2 )

are the components of the vector ( ai

 constant).

 5. A two-dimensional vector V is given in a form (ах+ b у, сх+dу), where а,  Show, that the vector V is a linear combination of the radial b , с, d – constant.    vector r  i x  j y and the tangential vector t  iy  jx : V  ar  bt . Note. The law of vector transformation should be obeyed for any angles and any points (х,у).

1.3. Scalar product The laws of vectors product should be mathematically consistent. From all the possible definitions of vectors products let’s choose two which are of interest both mathematically and physically. 15

A product in a form A  B  cos (in which А, В – absolute values of two vectors,   angle between them) is met in physics very often. For example, the expression Work = Force  Displacement  cos .





Let’s define the scalar product of the vectors A and  by the following way:

       x  x  Ay B y  Az B z   Ai Bi .

(1.22)

i

From (1.22)  the relations

           . The unit vectors i , j , k satisfy       i  i  j  j  k  k  1,

(1.22a)

i  j  i k  j i  j k  k i  k  j  0 .

(1.22b)

If one re-orientates the axes and directs the new axis х along the



vector A (fig. 6), then  x  A, Ay  Az  0 and Bx  B  cos . From (1.22) 

   B  cos

Fig. 6. Scalar product

   B  cos 16

(1.23)



 



If     0 and it is known, that A  0, B  0 , then from





(1.23)    90 0 , 270 0 and etc. In this case the vectors A and  should be reciprocally perpendicular, in other words, orthogonal.



 

The unit vectors i , j and k are the orthogonal vectors.

Fig. 7.

For the further development of the concept of orthogonality let’s     suppose, that n  is a unit vector, and r  i x  j y – is a non-zero    vector in the plane XY. If n  r  0 at any r , then n is perpendicular (orthogonal) to the plane XY. Now let’s convince of the truth of the word scalar. For this one   should consider the behavior of the product    when rotating of the coordinate system. With (1.15) let’s present the scalar product in a form:

x  x  y  y  z  z  (1.24)

  axi Ai  axj B j   a yi Ai  a yj B j   azi Ai  azj B j i

j

i

j

Using the indices k, l, one gets: 17

i

j

 A B   a A a B k

k

li

k

l

i

i lj

j



(1.25)

j

  (ali alj ) Ai B j    ij Ai B j   Ai Bi (1.26) i

j

l

i

j

i

The equation (1.26) gives the equality

 A B   A B , k

k

i

k

i

(1.27)

i

corresponding to the definition of the scalar product which is invariant relatively to rotation of the coordinate system.    Similarly let’s consider the product of the vector C  A  B by itself using the invariance of the scalar product:

C  C  ( A  B)  ( A  B)  A      2 A  ,

C C  C2 , 1     (C 2  A2  B 2 ) . 2

(1.28) (1.29) (1.30)

     is invariant under the rotation of the coordinate system

since the right part of the equation (1.30) is invariant. The equation (1.28) can be written in other form:

C 2  A2  B2  2 A  B  cos ,

(1.31)

which is called as the law of cosines (fig. 8). Comparing the equations (1.28) and (1.31), one can be convinced of the vector nature of cosines law one more time.

Fig. 8. Law of cosines

18

Problems 1. Expanding the scalar product show that if two vectors have the directing cosines 1 , 1 ,  1 and  2 ,  2 ,  2 respectively, then cos  1 2  12   1 2 , where



– angle between two vectors.

2. Find the cosine of the angle between vectors

     B  i  j  k ( solution: cos  0,   ). 2 3. Two unit vectors

 ai

and

 aj

    A  3i  4 j  k and

are either parallel or perpendicular. Show that

the condition of orthogonality of the directing cosines (1.18) follows from the scalar product of these vectors. 4. Find the angle subtended by the vectors

a  im cos  jm sin  and

b  in cos  jn sin  . 5. Two vectors are given: l1  3a  2b  c and l2  3b  2c  a . Find the third vector which does a triangle with these two vectors. 6. The vector is given: a  i  cos   j sin  . Find vector

a and the unit

a0 .

7. In the triangle АВС the following vectors are given: a  AB and b  BC . Find vectors coinciding with the medians Am1 , Bm2 and Cm3 of the triangle ( m1 , m2 , m3 – middles of the sides of the triangle). 8. Find the sum (a  F )  (b  F )  (c  F ) if vectors a , b and c subtend a triangle. 9. The vertexes of the triangle are given: А(-1; 1), В(-5; 4), С(7; 2). Find the scalar product AB  AC and the square of the triangle. 10. The vectors are given:

a (6; -8; 5 2 ) and b (2; -4;

subtended by the vector a  b with the axis Оz. 11. The unit vectors

e1 , e2 , e3 satisfy the condition: e1 + e2 + e3 =0.

Find ( e1  e2 + e2  e3 +

e3  e1 ).

19

2 ). Find the angle

1.4. Vector product This form of vectors product is connected with the use of the sine of the angle between two vectors

   C   ,

(1.32)

where C  A  sin , but in difference from the scalar product in



this case the vector C – is a vector and we suppose by the definition





that this vector is perpendicular to the plane of vectors A and  ,    and its direction is that the set of the vectors A ,  and С do the right-handed coordinate system. At the given choice of the direction one has:     (1.32a)        (anticommutation).

i i  j  j  k k  0 ,

i  j  k,

j  k  i , k i  j,

j  i   k , k  j  i , i  k   j .

(1.32b) (1.32c)

The vector product has an important geometric interpretation (fig. 9).    A  B  sin – is a square of parallelogram. So, the







vector C     is perpendicular to the plane of the parallelogram and by magnitude is equal to its square.

Fig. 9. Representation of vector product in a form of parallelogram

20



   connected with the notation of the component of the vector C :

The another definition of the vector product C     is

Cx  Ay Bz  Az By , C y   Ax Bz  Az Bx , (1.33)

Cz  Ax By  Ay Bx , or

Ci  Aj Bk  Ak B j ,

i, j, k  different,

(1.34)

with the cyclic permutation of the indices і, j, k. It is convenient to write the vector product in a form of the determinant:

   i j k  C  x y z . x

y

(1.35)

z

Let’s show the equivalence of the definitions of vector product

 

(1.32) and (1.33). For this let’s consider the scalar products A  C

 

and B  C :

  C  A  ( A  B)  Ax ( Ay Bz  Az By )  (1.36)

 Ay ( Az Bx  Ax Bz )  Az ( Ax By  Ay Bx )  0, B  C  B  ( A  B)  0 ,

(1.37)

  , the equations (1.36) and (1.37) show that the vector C is   perpendicular both to the vector A and to the vector B , and, hence,

is perpendicular to the vectors plane. 21

Let’s further consider the product

( A  )  ( A  )  A2 B 2  ( A  ) 2  (1.38)

 A B  A B cos   A B sin  , 2

2

2

2

2

2

2

2

 C  A  B  sin .

(1.39)

 

In the equation (1.38) we expanded the vector product A   on components in a form (1.33) and then used the formulas of the scalar product (1.22). From the equations (1.36), (1.37) and (1.39) it follows that two definitions of the vector product (1.32) and (1.33) are equivalent.    Now let’s prove that C     is a vector actually, i.e. obeys the law of vectors transformation (1.15). In the rotated coordinate system

Ci  Aj Bk  Ak Bj    a jl Al   akm Bm   akl Al  a jm Bm  l

m

l

 ( a jl akm  akl a jm )Al Bm ,

m

(1.40)

l ,m

where і, j, k are taken in the cyclic order. The expression in the brackets disappears when m= l . That is why the indices j, k take the definite values in dependence on the choice of і and 6 combinations of m and l . If і=3, then j=1, k=2 (cyclic order), and we get the set of the directing cosines:

a11a22  a21a12  a33 , a13a21  a13a11  a32 , a12 a23  a22 a13  a31. Substituting (1.41) into the equation (1.40) one gets: 22

(1.41)

C3  a33 A1 B2  a32 A3 B1  a31 A2 B3  a33 A2 B1  a32 A1 B3  a31 A3 B2 

(1.42)

 a31C1  a32C2  a33C3   a3nCn . n

Permuting the indices one gets C1 and C 2 , and after this it is easy



to show that the condition (1.15) is true, and C – is actually a vector. Problems 1. The vectors are given

        A  2i  4 j  6k and B  3i  3 j  5k .     A  B and A  B .

Define the scalar and vector products 2. Show that

( A  B)  ( A  B)  A2  B2 ,

( A  B)  ( A  B)  2 A  B,

A  ( B  C )  A  B  A  C,

A  ( B  C )  A  B  A  C.

3. Find the scalar and vector products of two vectors between them

  150

0

a

and

b

, if the angle

, and their lengths equal m and 2m respectively.

4. Find the scalar and vector products of the vectors a   b and a   b , if

a  b and b  2 a   . 5. Show that the vectors

a, b

and

c

are coplanar if

a  (b  c ) .

6. Find the vector product of the vectors l1  i   k  and l2  i   k  . 7. The vertex coordinates of a triangle are given by points (2, 1, 5), (5, 2, 8) and (4, 8, 2). Find the triangle square using vector calculus. 8. Three vectors are given

          3i  2 j  k , Q  6i  4 j  2k and

    R  i  2 j  k . Define which two of them are reciprocally perpendicular and which two are antiparallel to each other. 9.

Using

the

vectors

  icos  jsin , Q  icos  jsin ,

R  icos  jsin , prove the known trigonometric formulas: 23

sin(   )  sin cos  cos sin , cos(   )  cos cos  sin sin . 

10. Define the vector  perpendicular to the vectors

    U  2i  j  k and

     V  i  j  k . What the vector  should be if its magnitude equals unit? 11. Four vectors

a, b , с

and

 d

are in one plane. Show that

(a  b )  (с  d )  0 . 12. Find the sides and angles of the spherical triangle АВС (fig. 10) subtended by the vectors A =(1,0,0), B  ( 1 , 0, 1 ), C  (0, 1 , 1 ). The origin of each 2 2 2 2 vector coincide with the coordinate origin.

Fig. 10. Spherical triangle

 B is defined by Lorentz equation     F  q(V  B), where V – is a velocity of the electric charge q and F – is a 13.

The

magnetic

induction

force acting on the charge. When conducting three experiments it is true that: 1)

  V i,

   F  2k  4 j ; q

2)

V  j,

F  4i  k ; q 24

F  j  2i . q

V  k,

Using the results of these experiments find the magnetic induction.







Solution: B  i  2 j 14. Prove the equality:

  4k .

[ab ]  [bc]  [ca]  [(a  b )(b  c )] . 15. Prove:

[ac[ac ] []db[db ] ] and [bc ]

[da ] ,

if (a( ab )b ) (c(c  d )d ) and (a  b ) (c  d ) . 16. Prove:

(ia )i  ( ja ) j  (ka )k  a . 17. Calculate:

a  [i ( j )][(i )k ] ; s  [( j )(i ))][k (i )] . 18. Prove:

[a[bc ]  [b[ca ]]  [c[ab ]]  0 .

1.5. Scalar triple and vector triple products       Let one has three vectors combinations A  ( B  C ) and A  ( B  C ) ,







which are met very often. The combination of vectors A  ( B  C ) is known as scalar triple product.   The product B  C gives a vector which is then multiplied by  the vector A , in the result one has a scalar. Let’s note, that

   ( A  B)  C is a product of a scalar and a vector, but such an

operation is not defined yet. That is why let’s agree in advance not to consider this operation. 25

Using formulas (1.33), (1.22) one gets:

A  B  C  Ax ( By C z  Bz C y )  Ay ( Bz C x  BxC z )   Az ( Bx C y  By Cx )  Bx ( Az C y  Ay Cz )   By ( Ax Cz  Az Cx )  Bz ( Ay C x  AxC y ) 

(1.43)

 B C  A  C  A B   AC  B   C  B  A   B  A  C , or

      A B  C  A B  C .

Ax    A  B  C  Bx Cx

Ay By Cy

Az Bz . Cz

(1.44)

(1.45)

From the rules of interchanging of rows and columns in the determinant the permutation relations (1.43) follow at once, whereas    the symmetry of the vectors A, B , C in such a notation provides the truth of the condition (1.44). The scalar triple product has the obvious geometric

   C subtend a A, B and    parallelepiped (fig. 11), then A  ( B  C ) equals the volume of this interpretation: if three vectors parallelepiped.

26

Fig. 11.

   Now let’s consider the vector triple product A  ( B  C ) . In this case the brackets are necessary, what is obvious if one considers the special case

         i  (i  j )  i  k   j , ((i  i )  k  0) .

(1.48)

The given product of three vectors is itself a vector. In  addition, the resultant vector j is perpendicular to the vector

   A and ( B  C ) . The plane defined by the vectors B and    C is perpendicular to ( B  C ) , and, hence, the vector    A  ( B  C ) is in the this plane. This means that the vector    A  ( B  C ) should be a linear combination of vectors B and

 C .

         A  ( B  C )  B( A  C )  C ( A  B) .

(1.49)

Using the scalar triple and vector triple products one can simplify the other vector products. 27

Example: The scalar triple product finds an interesting application when    constructing the reciprocal crystal lattice. Let a , b and c (it is not necessary that they are reciprocally perpendicular) – vectors defining the crystal lattice. The distance between two points of the lattice

    r  na a  nb b  nc c ,

where na , nb and n c – some integer numbers. Using the given vectors let’s write the relations

         a b ca b c   ; b     ; c      (1.50) a  a b c a b c a b c  From (1.50)  a  is perpendicular to the plane of vectors b  and c and by magnitude is proportional to a 1 . It is easy to show that

a '  a  b '  b  c '  c  1, a'  b  a'  c  b '  a  b '  c  c '  a  c '  b  0 . The last equations define the so-called reciprocal lattice. The reciprocal lattice is connected with the problems on scattering of waves on different planes of the crystal. Problems



   

 

 

1. Prove the formula   a  b  b  a  a  b .

 



 



 

2. Show that a  (b  c )  b  (c  a )  c  (a  b )  0 . 3. The vector

 A

is expanded on the radial

 Ar

and tangential vectors

– unit vector in the radial direction. Show that

    At  r0  (r0  A) .

  At , r0

    Ar  r0 ( A  r0 ) and

4. Prove that the necessary and sufficient condition of complanarity of three

 

(non-zero) vectors A, B and 5. The vectors are given

 C

is the equality of the scalar triple product to zero.

28

A  i  2 j  3k ,

B  4i  5 j  6k , C  3i  2 j  k

,

D  6i  5 j  4k . Find: a) sum and difference of the vectors:

A  B  C  D; A  B  C  D; A  B  C  D;

 A  B  C  D; b) angles which are subtended by the vectors A, B, C , D and the coordinate axes; c) magnitudes of vectors A, B, C , D ; d) scalar and vector products of sum of first two vectors and the sum of last two vectors; e) angles which are subtended by the vector f) projection of the vector g) vector products

A

and angles which are subtended by

D;

h) square of the parallelograms subtended by the vectors

D;

B , C , D;

A on the direction of vectors B, C , D;

A  B, A  C , B  C

the products and the vector

and the vectors

A

and

find the length of diagonals of these parallelograms; j) show that all vectors k) find products:

   A B C

A, B, C

and

D

are in one plane;

D  B  C; A  D  C and D  B  A;    A  ( B  C ), C  ( A  B) and B  (C  A) ; and

D  ( B  C ), C  ( D  B) 29

and

B  (C  D) ;

B, C

ad

A  ( D  C ), C  ( A  D)

and

D  (C  A);

A  ( B  D), D  ( A  B)

and

B  ( D  A) .

given A  3i  2 j  2k , B  6i  4 j  2k ,    C  3i  2 j  4k . Find A  B  C and A  ( B  C ), C  ( A  B) and    B  (C  A) . 6.

Three

vectors

are

7. The vectors are given

A  i  2 j  3k , B  4i  5 j , C  3i  2 j  k . Which system (right- or left-handed) do these vectors form? Define: a) volume of the parallelepiped subtended by these vectors; b) vectors figuring two (issuing from the end of the vector A ) diagonals of the parallelepiped subtended by these two vectors and find the length of these diagonals; c) square of the diagonal section of the parallelepiped subtended through the vector

A

and

A  ( B  C ), C  ( A  B)



and

   B  (C  A) .

8. The force F acts on a body, placed in point r . Show that the resultant momentum relatively to any of axes issued through the coordinate origin equals

 L  r  F  a , where a – unit vector in the direction of the axis. 9. Given:        ca  b c  a b a     b      c      a b c a b c a b c    and a  b  c  0 Show that

x  y   xy ( x, y  a, b , c ); b' c' a  b  c  (a  b  c ) ; a  . a b c' '

'

1

30

10. Using the Euler formula

V   r,

define the linear velocity of the

center of a rectangular rotating around one of the vertexes and having sides a  2 cm and b  4 cm at the moments when the instantaneous angular velocity has the magnitude 5

1 and is directed: sec

a) by little, b) by large side of the rectangular. 11. Define the moment of force with a magnitude of 5 N, directed by one of the edge of a cube, relatively to all its vertexes and axes subtending through the edges (the length of the cube edges equals а cm). 12. The kinetic moment relatively to the center О (angular momentum) of a system of n material points is called the vector sum n

L0   ri  mi vi , i 1

where

ri – radius-vector of i

point with mass

mi and velocity vi . Define:

a) kinetic moment of two particles with masses rotate with angular velocity

 5

m1  1 g and m2  2

g which

1 around the axis (z) and describing sec

circumferences with radii of 3 and 6 cm; b) kinetic moments of particles with masses 1 g and 2 g, moving in opposite sides with speed of 3 cm/sec by two opposite edges of the cube relatively to all its vertexes (the length of the cube edges equals а cm). 13. Let

a and b – two vectors, defining the sides of the parallelogram with a  b and a  b ; show that:

diagonals a) the sum of diagonals squared equals the sum of its sides squared; b) parallelogram’s diagonals are perpendicular then and only then if this parallelogram is a rhomb; c) the square of a parallelogram (А), subtended by the diagonals of another parallelogram (В), is two times larger than the square of this parallelogram (В).

Vector calculus (differentiation of vectors) 1.6. Gradient Suppose that  ( x, y, z ) – is a scalar function of a point in a space, i.e. such a function that the magnitude of it depends on the 31

coordinates values (x,y,z). As a scalar it should have the same value for the fixed point in the space not depending on the rotation of the coordinate system, i.e. (1.51)  ( x1, x2 , x3 )   ( x1 , x2 , x3 ). Differentiating with respect to x i and using the equation (1.16), one gets:

 ( x1, x2 , x3 )  ( x1 , x2 , x3 )   xi xi

(1.52)

 j

 x j x j   '  aij   aij  . ' x j xi xi x j j

A comparison of the expression (1.52) with the vector transformation law (1.17) shows at once that we constructed a vector with components

 . We call this vector as a gradient of  . It is x j

convenient to move to a symbolic notation

         i j k x y z or i

    j k . x y z



(1.53)

(1.54)

Here  (nabla) – is a vector differential operator. This operator possesses the properties of vectors and obeys the law of partial differentiation. Example: Calculate the gradient of the function

f (r )  f ( x 2  y 2  z 2 ) 32

f f f j k , x y z f (r ) f (r ) r df x    . x r x dr r

f (r )  i

The other components are found similarly. Then:

f (r )  (ix  jy  kz )  where r0 

1 df r df df    r0 , r dr r dr dr

r – is a unit vector in the positive direction of the radiusr

vector.  One of the direct application of  is connected with the length increment calculation

    dr  i dx  j dy  k dz

(1.55)

Considering the previous notation one gets:

( )dr 

   dx  dy  dz  d , x y z

(1.56)

a change of the scalar function  , corresponding to the change of  the position dr . Consider two particles P and Q on the surface  ( x, y, z )  C . The distance between these two particles is dr. Then when displacing from P to Q the change of the function on the surface  ( x, y, z)  C equals

  d  ( )dr  0 ,

(1.57)

since the displacement occurs along the surface  ( x, y, z )  C . It



follows that ( )  dr . Since dr can be drawn in any direction  from point P to Q on the surface, and it means that dr is always on 33

the surface,  should be perpendicular to the surface   сonst in its any point.  If one supposes now that the dr is directed from one surface   c1 to the neighboring one   c2 (fig. 12), then d  c2  c1  c  ( )  dr .

(1.58)



For the given d the absolute value of dr is minimal, if

   (cos=1), or, vice versa, at the given dr the change of   the scalar function  is maximal for dr  .  dr

Fig. 12. Gradient



This defines the  as a vector pointing the direction of the maximal velocity of a function change. The gradient of a scalar quantity plays an important role in physics when determining the relation between the forces field and the potential field:  Force = -  (Potential) (1.59) This is true both for the gravitational and electric fields. 34

Problems 1.

Show

that

   (uv)  vu  uv , where u  u( x, y, z) and

v  v( x, y, z )

– differentiable scalar functions. 2. Find the gradient of scalar field

u  x  2 y  3z.

3. Find the greatest slope (velocity) of the surface rise u  x y in the point М(2,2,4). 4. Find the unit vector of the normal to the surface of the level of scalar field

u  x 2  y2  z2. 5. Find the gradient of the field

u  (a, b , r ), where a and b – constant

vectors, r – radius-vector of a particle. 6. Find the gradient of the distance

r  ( x  x0 )2  ( y  y0 )2  ( z  z0 )2 , where P( x, y, z ) – is the field point under consideration, and the point

P0 ( x0 , y0 , z0 ) – is some fixed point. 7. Find the angle



between the gradients of the functions

u  x2  y 2

and v  x  y  2 xy in point М0 (1, 1). 8. Find the derivative in the direction of the radius-vector r for the function

u  sin r , where r  r field

.

9. Find in the point М0 (1,1,1) the direction of the greatest change of the scalar u  xy  yz  xz and the value of this greatest change in this point. 10. The function is given

S ( x, y, z )  ( x 2  y 2  z 2 ) 3 / 2 . Define in the point (1,2,3) the



 S , its absolute value and the directing cosines

of the S . 11. The vector is given

    r12  i ( x1  x2 )  j ( y1  y 2 )  k ( z1  z 2 ) . 35

 1r12 (the gradient of the absolute value of the vector r12  respect to variables x1 , y1 , z1 ) is a unit vector directed along the vector r12 . Show that

12. Find the 13. Find the

2 2 2 grad r , where r  x  y  z grad (rc ), where c  const. .

with

.

  (u)  (v)  0 is necessary and sufficient one for two functions u ( x, y, z ) and v( x, y, z ) to be related in a form f (u, v)  0 . Show that in a case of u  u( x, y) and v  v( x, y) the condition   (u)  (v)  0 gives the two-dimensional Jacobian: 14. Prove that the condition

u  u , v  x   I   x, y  v x

u y 0 . v y

The functions u and v are supposed to be differentiable. 15. Prove that the condition

   (u)  (v)  ( )  0 – is necessary and

u( x, y, z ) , v( x, y, z ) and  ( x, y, z ) to be F (u, v, )  0 . Show also that the scalar triple

sufficient one for three functions

related through some function product of the gradients is equivalent to three-dimensional Jacobian

u x  u , v,   v I   x, y, z  x  x

u y v y  y

It is supposed the existence of the derivatives. 16. Prove that if the vector function coordinates and time, then

u z v  0. z  z

  F  F ( x, y, z, t ) depends on the spatial

36

     F dF  (dr  ) F  dt . t 17. Find the gradient of the scalar field

u  ln( x2  y 2  z 2 ) in point М0 (1,1,-1). 18. Let

v  v( x, y, z ), u  ln( x2  y 2  z 2 ) – are differentiable functions in point М0 (x,y,z). Show that a)

grad u   grad u,   const;

b)

grad (u  v)  grad u  grad v ;

c)

grad (uv)  v grad u  u grad v ,   const ;

d)

grad

19. Calculate

u v grad u  u grad v , v  0.  v v2 grad ( x m y n ).

1.7. Divergence The differentiation of the vector function is a generalization of  the differentiation of the scalar quantities. Let’s suppose that r (t ) describes the position of some body in a space at the moment t (fig. 13). Then the differentiation with respect to time gives

   dr (t ) r (t  t )  r (t )   v,  lim t 0 dt t



where v – is a linear velocity, and the operator  was defined in the paragraph 1.6 as a vector operator. 37

Now, keeping in mind its vector and differential properties, let’s consider the action of the  on a vector.

Fig. 13. Differentiation of a vector

The scalar product of this vector on a vector gives the expression

  V V y Vz ,   V  x  z y x

(1.60)



which is called a divergence of the vector V . The divergence is a scalar in a sense how it was defined in the part 1.3. For example:

  r  (i

   x y z  j  k )(ix  jy  kz )     3; x y z x y z

     f   r f (r )  ( xf (r ))  ( yf (r ))  ( zf (r ))  3 f (r )  r . x y z r In particular, if f (r )  r n1 , then

      r r n1    r0 r n  3r n1  (n  1)r n1 38

The divergence of this quantity equals zero when n=-2. For more clear representation of the physical essence of the    divergence let’s consider   (   v ) , where v ( x, y, z ) – is a velocity of flow of compressible fluid;  ( x, y, z ) – is a density of this fluid in the point (x,y,z). If one considers some element of volume dxdydz (fig. 14), then the quantity of liquid coming into this volume in a unit time through the surface EFGH, will be expressed in a form: (inflow) EFGH =  vxdydz. The quantity of liquid escaped from the volume through

the

surface

ABCD

equals:

(outflow)

ABCD

=

   =   vx  (   v x )dx  dydz ; the derivative takes into account the x  

possibility of dependence of nonuniform density or velocity, or at once both these quantities on the variable х. The total fluid flow through these two surfaces equals the difference of two flows or to a flow in the direction of axis х:

 (   v x )dxdydz . x

The additional fluid flow occurs through the other four surfaces of this element of volume, the total flow (in a unit of time) equals

     x (   vx )  y (   v y )  z (   vz )  dxdydz    (1.61)

   (   v )dxdydz. Hence, the total quantity of the compressible fluid passed   through the volume unit in a unit of time equals   ( v ) . Hence the denomination divergence follows. One of the examples of divergence use is an continuity equation.

      (  v)  0 t

39

(1.62)

Fig. 14. Differential rectangular parallelepiped

Hence

  ( fv ) 

   ( fvx )  ( fv y )  ( fvz )  x y z

v v v f f f  vx   v y   vz  f x  f y  f z  x y z x y z

(1.63)

 

 f  v  f   v , 

where f – scalar, and v – vector function.



 

In a particular case when   B  0 the vector B is called solenoidal. Problems



  

 

 

1. Prove the formula   a  b  b  a  a  b . Note. Consider the left part of the formula as a scalar triple product.

 

 

2. Rotating the coordinate system, show that   v     v , and hence, by definition, the divergence of the vector – is a scalar (it is sufficient to consider twodimensional case).

40

div r . 4. Calculate the div ( a ), 3. Calculate the

where



– scalar function,

a

– vector function

of the field.

div (rc ), div(r 2c ) where c – constant vector. 6. Calculate div ( r ), where  – constant scalar. r 7. Calculate the div . r 8. Calculate the div b (r  a ), div r (r  a ), where a and b – 5. Calculate the

constant

vectors.

div (r 4 r ). 10. Find the div (r  (  r )),

where



– constant vector.

div (a  (r  b )),

where

a

and

9. Find the

11. Find the

b

– constant vectors.

12. A solid body rotates with a constant angular velocity linear velocity

 v



 . Show that the

is solenoidal.

 13. The electrostatic field of the point charge q equals E 

 

 r0 .  4 0 r 2 q

Calculate the   E . 14. Show that:

div (r n c )  nr n2 (rc ), (n  2); b) div (c  r )  0; a)

c).

div r n r  (n  3)r n .

15. Find the flux of the radius-vector

r

through the surface of sphere.

16. Find the divergence of the vector

a

 (r ) r

r , where r  r – is a

distance from the coordinate origin to the variable point M ( x, y, z ) .

div [ grad  (r )]. x yz 18. Find the div  r. xyz

17. Find the

41

1.8. Rotor The operation of vector product of the vector  and a vector one can define by the formula:

      v  i  vz  v y   z   y

   j  vx  vz   x   z (1.64)

i       k  v y  vx   y  x  x vx

j  y vy

k  . z vz



The expression obtained is called the rotor on the vector v . When opening the determinant or at any other operations with  it is necessary to consider its differential nature. Let’s specify that the   product v   is defined as a new vector differential operator. In a     general case v      v . If one multiplies the vector  in a vector way by the product of a scalar and a vector then one has:

     fv  x  i  ( fvz )  ( fv y )   z  y 

v y f   v f  i  f z  vz  f  vy   z z   y y

 

(1.65)

 f  v x  f  v . x

Doing the cyclic permutation of the coordinates it is easy to obtain the у- and z-components. One can convince of 42

 

  fv   f  v  f  v .

(1.66)

The expression obtained is an analogue of the expression (1.63). Example:

 rf (r )  f (r ) r  (f (r ))  r . 1)

i j k    r   0. x y z x y z   df 2) Using the equality f (r )  r0   , dr   df     r f (r )   r0  r  0 , dr     since r  r0  r , r0  r0  0 . The nomination rotor appeared in   connection with that the product   v describes a rotation of the  vector field v in a point where the rotor is calculated. Let one considers a solid body in the plane ху, rotating around



the axis z with the angular velocity  . The linear velocity V in the point defined by the radius-vector r equals

V  r . 

(1.67)



In order to define the product   r let’s consider

V   (  r ) ,

(1.68)

 (  r )  ( r )  r  ( )   r       r . (1.69) 43

Here the vector  is multiplied in a scalar way by the first vector but as a differential operator it works on both the vectors:

 (  r )   r  r   r    r.

(1.70)

When   const then the second and the third terms in the equation (1.70) are equal to zero. As it is known

  r  3,

that is why

 r  x

(1.71)

  (ix  jy  kz )   y (ix  jy  kz )  x y (1.72)

z

 (ix  jy  kz )  i x  j y  k  z  . z

Substituting (1.71) and (1.72) into (1.70) one gets

V   (  r )  2 ,

(1.73)

i.e. the rotor of the linear velocity of the solid body equals the angular velocity doubled. If

    v  0,

 then the vector V is called irrotational.

(1.74)

The most important physical examples of the irrotational vectors are presented by the gravitational and electrostatic forces

V C

r0 r C 3 , 2 r r



(1.75)

where С – constant, r0  unit vector directed in the radius-vector. 44

Problems 1. Show that the vector

   U  V is solenoidal if U , V – irrotational vectors.

2. Show that the vector

A r

3. Find the rotor of the vector

 – irrotational vector. a  ( x  z )i  ( y  z ) j  ( x 2  z )k is solenoidal if 2

2

.

2

4. Find the rotor of the field F  y zi  z xj  x yk . 5. Rotating the coordinate system show that the components of rotor obey the law of vector transformation. Note. Use the directing cosines from the equation (1.41). 6. Show that the rotor

   V  i V x ( x, y)  j V y ( x, y)

   V

is perpendicular to the vector

  and   V  0 .

V

if

7. In quantum mechanics the operators of angular momentum are defined by the relations:

L x  i( y

L y  i( z Show that

   z ), y z

   x ), x z

Lz  i ( x

   y ). y x

[ Lx, Ly ]  i Lz , and hence, L  L  i L .

8. Prove the vector equalities:

                 ( A  B)  ( B  ) A  ( A  ) B  B  (  A)  A  (  B),  ( A  B)  ( B ) A  ( A ) B  B( A)  A( B).

9. Show that: a) rot r b) c)

 0; rot (c  r )  2c ; rot c (ar )  a  c.

.

10. Find: a) rot[(c  r )  a ], b) where

a, c

rot[(c  r )  r ], – constant vectors.

11. Find the divergence and the rotor of field of velocities

V

and field of

accelerations W of the solid body rotating around the stationary point knowing that

45

V    r ; W    r    (  r ), where the field of accelerations  ,  – constant vectors. 12. Find the rotor of the following vectors: a)

a  ( x2  y 2 ) i  ( y 2  z 2 ) j  ( z 2  x2 ) k ;

a  z 3 i  y 3 j  x3 k ; 1 2 2 c) a  ( y i  x j ). 2 13. Represent the vector a  grad  , where a b)

– constant vector, in a form of a rotor of a certain vector. 14. The vector r (  r ) , where  – constant vector, is a solenoidal one. Represent it in a form of a rotor of a certain vector.

1.9. Sequential application of the operator  Using the introduced notations of gradient, divergence and rotor one can get a vector, scalar and combinations of vectors. Working on each of the introduced quantities by the operator  , one gets

  ,

 ,

V , V ,  (V ) .

These expressions are often used in the second order differential equations in mathematical physics. First of them,   , divergence of the gradient, is called the Laplacian of  :

   (i (

       j  k )(i j k ) x y z x y z

2 2 2   )     2 . x 2 y 2 z 2

The second operation can be written in a form 46

i    

 x  x

j  y  y

k  , z

(1.76)

 z

 2  2     i (  ) yz zy  2  2  2  2  j(  ) k(  )  0. zx xz xy yx

(1.77)

It is supposed here that one can change the order of differentiation. One can do it only if the first partial derivatives are continuous. Further from the equation (1.77) it follows that the rotor of the gradient equals identically zero: rot grad  0 . Hence, the gradient – is irrotational vector. The fourth expression is a scalar triple product:

   x y z       V   x y z Vx Vy Vz



2 2  2Vz  2Vx  Vy  2Vx  Vy  2Vz       0, xy yz zx zy xz yx

47

(1.78)

That is

V  0.

(1.79)

Thus, the divergence of the rotor equals zero div (rot )  0 , i.e. the rotor – is always solenoidal vector. Let’s consider the last expression:

 (V )  V  V .

(1.80)

If the vector V is expanded on components in the Cartesian coordinate system then the vector  V – vector Laplacian  is reduced to a vector sum of usual scalar Laplacians:

  V  i   Vx  j   Vy  k   Vz . The important use of the equality (1.80) is connected with the wave equation in the electromagnetic theory. In the vacuum Maxwell’s equations take the form:

a)

c)   B   0 0

  B  0,

 , t (1.81)

b)

    0,

d)   

 . t





Here  – electric field; В – magnetic induction;  0 ,  0 



electric and magnetic permeabilities. Suppose that  is defined from the equations (1.81c) and (1.81d). This can be done taking the rotor from both the parts of the equation (1.81d):

 ( )  

      . t t 48

(1.82)

Further let’s take the derivative with respect to time of both the parts of the equation (1.81c):   2 (  )   0 0 2    (  ), t t

 2 , t 2

 ( )   0 0

  (  )       0 0

    0 0

(1.83)  2 , t 2

 2 . t 2

(1.84)

The expression (1.84) is a vector wave equation of the electromagnetic field.



If one expands the vector E in the Cartesian coordinate system then the expression (1.84) disintegrates on three scalar wave equations containing the scalar Laplacian. Problems

    ( )  0 .   2. Prove that the vector (U )  (V ) is solenoidal if the quantities U and V 1. Prove that

are differentiable scalar functions. 3. The scalar

 



satisfy the Laplace equation

 2  0 . Show that the vector

is solenoidal and irrotational.

4.

Prove

that

C3   ( a )

the

expressions

C1   , C2   a

and

– are solutions of the vector wave equation:

2C  k 2C   C   C  k 2C  0 . Here

a



satisfies

the

scalar

wave

equation

constant vector. Prove also that the quantities

49

 С1

2  k 2  0 and and

 С2

are orthogonal,

the vector

 С1

is irrotational, and the vectors

 С2

and

 С3



are solenoidal

vectors. 5. Prove that the equality  (V )   V   V comes from the rule

BAC  CAB

for vector triple product. Explain the arbitrary position of the

multipliers in the terms

and

BAC

CAB .

       k 2  0     2   k 2   0 and the

6. Show that any solution of the equation automatically satisfies the Helmholtz vector equation

 

condition of solenoidity     0 . 7. Prove if the following scalar fields are harmonic: a)

u  x 2  2 xy  y 2 ;

b)

u  x2 y  y 2 z  z 2 x;

c)

u  x2  y 2 .

8. Show that the scalar field harmonic one. 9. The scalar potential

1 u  ln , r

x2  y 2

where r 

U LM  j L (ka)YLM ( ,  )

(r  0)

is a

satisfies the Helmholtz

2 2 scalar equation  U LM  k U LM  0 . Using the operator of angular    momentum L  i (r  ) one can construct the vector potentials:

1 E ALM    LU LM , k

M ALM  lLU LM .

Show that both the potentials satisfy the equation

   k 2   0. 10. Time-dependent Schrodinger equation has a form:

 2     2m    V (r )   (r , t )  i t .   Suppose

that

representation of imaginary parts):



   (r , t )  A(r , t )  e iS ( r ,t ) / 

.

Show

that

such

a

gives the following two equations (separately for the real and

50

2 S (S ) 2 2 A  V   , t 2m 2m A

m

A A  (A)(  S )   2 S  0. t 2

In quantum mechanics the density of the probability of discovering a particle in the given point of a space current

J



is defined by the quantity

– by the quantity

 A2   S  . Show that the second equation is m 

equivalent to the equation of continuity:

 J  11. Let the function



 2 , and the density of the

  0. t

– is a scalar one, show that it satisfies the equation:

2 2      2 2   ( r  )  ( r  )   r    r  2r . 2 r t

1.10. Integration of vectors Following the vectors’ differentiation let’s consider the vectors’ integration. Let’s start from the linear integration, and then move to surface and volume integrals. In each of these cases the integral of vector will be reduced to an integral of scalar functions. Using the increment of the length dr , one can define the linear integrals: a)

  dr ,

b)

  V   dr , C

C

c)

  V   dr .

(1.85)

C

Here the integration is by a certain cicuit С, open or closed ones.

a)

  dr  i   ( x, y, z)dx 

C

C

51

 j   ( x, y, z )dy  k   ( x, y, z )dz. C

(1.86)

C

Such a disintegration of the primary integral is possible due to the equation

 i  dx  i   dx ,

(1.87)

which is written down with account of the property of the unit



vectors i , j , and k , remaining in the rectangular coordinate system to be constant by the magnitude and the direction. Three integrals in the right part of the equation (1.86) correspond the usual scalar integrals and are the Riemann integrals. However the integral with respect to variable х cannot be calculated without the knowledge of dependence of у and z on variable х (the same should be noted also to integrals of another variables). This means the necessity of exact determination of the integration circuit C. If only the integrand does not possess the special properties (in the result of which the integral will depend on only the position of the final points of the circuit), the value of the integral is defined by the peculiarities of the choice of the circuit C. For example, for the particular case   1 the integral (1.85а) will be the exact vector distance from the origin of circuit C to its final point. In this case the value of the integral does not depend on the choice of integration way between









the fixed ends. When dr  i dx  j dy  k dz than the second and the third integrals considered above are also reduced to the integrals of scalar quantities. The integral (1.85b) is exactly equal to the integral which defines the work done by the force on a given segment of way:

W   F  dr .

(1.88)

Example: Let’s integrate the scalar function r 2  x 2  y 2 from the origin of coordinates until the point (1,1), using the increment of the length dr (fig. 15). 52

(1,1)



( 0, 0 )

   (1,1)  (1,1) ( x 2  y 2 )(i dx  j dy )  i  ( x 2  y 2 )dx  j  ( x 2  y 2 )dy  ( 0, 0 )

( 0, 0 )

(the integration is done by the circuit imaged in fig. 15) x 1



 i

( x 2  y 2 )dx  j

(0, y  0)

at the circuit of integration

 i

1

 x

( x  0, 0 )

2



y 1

1 4 ( x 2  y 2 )dy  i  j , 3 3 ( x 1,0)



0,0  0,1  1,1 :

  y 2 dx  j

1



( x 2  y 2 )dy 

( 0, y 1)

4 1  i  j, 3 3

the integration by the circuit x=y gives the value: (1,1)

i

4 4 ( x 2  y 2 )dx  j  ( x 2  y 2 )dy  i  j . 3 3 (0,0)



Fig. 15. Integration circuit

Thus, the value of the integral depends on the choice of the circuit С, along which the integration is done. The surface integrals are written down in the same way as the  linear ones, but the dr is changed for the vector d . It is often that 53



this element of surface is written down in a form ndA , where  n – unit (normal) vector of positive direction (fig. 16). There are two ways of positive direction choice. If the surface is closed, then let’s agree to call the direction from the volume subtended by the surface to be the positive one. For the open surfaces let’s agree that the positive direction depends on the direction of bypass of the surface perimeter. If one places the fingers of the right hand in the direction of bypass by the edge of the surface then the direction of thumb coincides with the positive direction.

Fig. 16. Right hand rule when choosing the positive direction

The surface integral





 V  d

can be interpreted as a flux

through the given surface. The volume integrals are rather simpler since the volume element d – is a scalar.

 Vd  i  V

V

V

X

d  j  Vy d  k  Vz d . V

(1.89)

V

Using the surface and volume integrals one can define the differential relations in a different way: 54

  lim  d 0



In this equations the

  d ,  d

(1.90)

 V  lim  d 0

 V  d ,  d

(1.91)

  V  lim  d 0

 d  V .  d

(1.92)

 d

 is a certain small volume of a space,

d  is a vector element of a surface of the volume. Let’s show now that the expression (1.90) corresponds indeed to the quantity  earlier introduced by the equation (1.53). For the simplicity let’s change the

 d

by the differential volume dxdydz and place the

coordinate origin into the geometric center of this element of the volume (fig. 17).

Fig. 17. Differential rectangular parallelepiped

55

The surface integral is reduced to six integrals by each of the six  edges of the parallelepiped. The vector d is directed outside, 



then one has d  i   d for the surface EFGH and d for the surface ABCD, that is why

   i 

( 

EFGH



j

( 

AEHD

k



( 

AEFB

 (i

 dx  dx  )dydz  i  (   )dydz  x 2 x 2 ABCD

 dy  dy  )dxdz  j  (   )dxdz  y 2 y 2 BFGC

(1.93)

 z  dz )dxdy  k  (  )dxdy  z 2 z 2 DCGH

   )dxdydz. j k x y z

Dividing the expression obtained by the

 d  dxdydz , ove

can prove the truth of the expression (1.90). When proving one neglected the correction terms containing the derivatives of higher orders. The additional terms in connection with the consideration of the Taylor series disappear in the limit

 d  0

(dx  0, dy  0, dz  0) .

For more strict verification of the equations (1.90), (1.91) and (1.92) it is necessary to do the limiting process mentioned. Problems 1. Find the indefinite integral of the vector-function

a (t )  i cos t  jet  k . 56

2. Find the integrals of the following vector-functions: а)

a (t )  tet i  sin 2 t j 

1 k; 1 t2

2 t i  t et j  cos t k ; 2 1 t sin t c) a (t )  cos t e i  t cos t 2 j  k ; 1 2 t d) a (t )  t i  t sin t j  2 k . 2

b)

a (t ) 

3. Calculate the following integrals: 

а)

 a (t ) dt , где a  sin

2

t cos t i  cos2 t sin t j  k ;

0

1

b)

 a (t ) dt ,

где

a

0 1

c)

ei / 2 ei / 2 i j  et k ; 2 2 1

 a (t ) dt , где a  3 cos  t i  1  t j  2t k ; 0 1

d)

 a (t ) dt , где a  (2t   ) i  t sin t

j   k.

0

4. The field of forces acting on two-dimensional linear oscillator can be written







down in a form F  i kx  j ky . Compare the work, which is done when moving from the point (1,1) until the point (4,4) in the field of these forces in the case of three different ways of displacement: а) (1,1)  (4,1)

 (4,4);  b) (1,1) (1,4)  (4,4); c) x=y.

(4,4)

For this estimate the integral





F  dr .

(1,1)

5. The field of forces is given

F 

iy jx .  2 2 x y x  y2 2

57

Define the work done when moving around the circumference of unit radius: а) counterclockwise from 0 to  ; б) clockwise from 0 to  .

1 r  d taken by the surface of unit cube which is 3 S

6. Calculate the integral

defined by the point (0,0) and the unit segments in the positive directions of axes х, у and z. For three faces r  d  0 , and each of the remained faces contributes into the integral equally.

 d  V

7. Prove that

lim S  d  0

 d

  V

.

S

Note. When proving use the elementary volume dxdydz. 8. Find the work which is done when displacing from the point (1,1) to the point (3,3). The force applied equals

   F  i ( x  y)  j ( x  y) . Define exactly

the way of displacement. Note that this force is non-conservative.

1.11. Gauss’s theorem



 

Let the quantities V and   V are given, continuous in all area interesting for us. The Gauss’s theorem says that

 V  d    Vd .  

S

(1.94)

V

The   V is a quantity of liquid escaped from the unit volume. Hence, the right part of the equation (1.94) equals the total amount of liquid escaped from the volume V, by which the integration is done. Proving that the left part of the equation describes the flow of the liquid through the surface S, which subtends the given volume, one thereby proves the Gauss’s theorem. The more detailed and mathematically strict prove of the Gauss’s theorem can be found in the literature recommended in the end of the present tutorial. 58

From the Gauss’s theorem the useful consequence known as the Green’s theorem follows. If u and v – two scalar functions then one has:

 (uv)  uv  (u)  (v),

(1.95)

 (vu)  vu  (v)  (u).

(1.96)

Subtracting (1.96) from (1.95), integrating by the volume (u, v and their derivatives are supposed to be continuous) and applying the formula (1.94) (the Gauss’s theorem), one gets the Green’s theorem:





( (u v  v u )d  (uv  vu )  d . V

(1.97)

S

The equation (1.95) admits the different notation:

 uv  d   u vd   u vd . S

V

(1.98)

V

Despite the expression (1.94), containing the divergence, is an important form of the Gauss’s theorem notation, one can also meet such a form of the theorem when the volume integrals contains the gradient and the rotor. Suppose that

V ( x, y, z )  V ( x, y, z )  a,

(1.99)

where a – vector constant by magnitude and direction (the direction is chosen arbitrary, but the direction chosen then always remains to be fixed). Using the relation (1.62а) the equation (1.94) in this case is rewritten in a form:

a  Vd     aVd  a  Vd ,

(1.100)

  a    Vd   Vd   0. V S 

(1.101)

V

V

V

59

a  0,



 Vd    Vd . S

(1.102)

V

Similarly, taking that V  a  P ( a – constant vector) it is easy to prove that

 d  P    Pd . S

(1.103)

V

1.12. Stoke’s theorem The Gauss’s theorem connects the volume integral of the divergence of a certain function with the integral of the same function over closed surface subtending the volume. Now let’s consider the similar relation between the surface integral of the divergence of a certain function and the linear integral of the same function and the linear integration is done by the perimeter of the surface given. With this purpose let’s transform the surface integral of the rotor, and for this let’s apply the formula of scalar triple product to the integrand:

Vx

  Vd   ( z d S



y



S

Vy z

Vy Vx d z  d z  y x (1.104)

d x 

Vz V d x  z d y ). y x

The surface integral given is taken by a certain surface S given. Let’s orientate the axes of the Cartesian coordinate system so that the surface crosses the plane x=c along the line АВ (fig. 18). The positive direction on this line corresponds to the direction from А to В, the  direction of the d is pointed in the fig. 18. In particular, as it is shown in the fig. 19,

d y  dxdy, d z  dxdy. 60

(1.105)

Fig. 18.

The increment dx corresponds to the surface subtended between the planes x=c and x=c+dx. Integrating the derivatives of Vx by the pointed increment of the surface one gets:

dVx

 ( z

d y 

S

dVx V V d z )   ( x dy  x dz )dx . (1.106) y y z S

Since х remains to be constant when integrating from А to В, then

Vx V dy  x dz  dVx y z

(1.107)

or B

 dx  dV   V ( x, y x

x

A

B

, zB )dx   Vx ( x, y A , z A )dx .

(1.107а)

The specified choice of the direction when bypassing the edges of the area means that dx  d x – in the direction to the point В and

dx   d x – in the direction to the point А, where d  – is a vector of length increment along the perimeter.

Vx

 ( z S

d y 

Vx d z )   Vx  d x . y 61

(1.108)

Fig. 19.



denotes the integration by the closed way, in this The symbol case by the perimeter of the surface given. For another coordinates one gets the same expressions, that is why finally

 V  d   (V d  x

x

 Vy d y  Vz d z )   V  d  .

(1.109)

This is the Stoke’s theorem. Using the Stoke’s theorem one can find the additional relations between the surface and linear integrals:

 d     d  ,

(1.110)

S

 (d )     d   .

(1.111)

S

In truth of the expression (1.110) one can easy be sure by   substituting of the expression V  a   into the formula (1.109),  where a  is a vector constant by magnitude and direction: 62

 ( a )  d   (a  )d  a    d . S

S

(1.112)

S

For the linear integral

so

 a  d   a    d 

(1.113)

a (    d    d  )  0.

(1.114)

S

Since the direction of the vector a is taken arbitrarily, the expression in round brackets equals zero. Similarly one can prove the relation (1.111), where one should take the V  a   . Let’s again consider the equation (1.109), where the term

  V   d can be considered as a flow of the fluid circulating by the

closed circuit. If in the capacity of the surface one chooses the circle







of square k d , then the   V d  equals the circulation of the

  allows measuring the rotor of the vector V by rotating of the small

vector along the closed circuit of square of d in the plane ху. This screw. If the screw does not rotate, the circulation equals zero and



hence on the base of the Stokes theorem the vector V – irrotational. Problems 1. Calculate the flux of the vector field the sphere surface

F (M )  x3 i  y3 j  z 3k through

x2  y 2  z 2  R2 .

2. Calculate the flux of the intensity field E  through the closed surface

qr of the point charge q r3

S , not containing the charge q

inside itself . 2

2

3. Calculate the flux of the vector field F ( M )  xy i  x y j  zk through the closed surface subtended by the coordinate planes x  0, y  0, z  0 and the part of the surface of paraboloid

4  z  x2  y 2 ,

which lies in the first octant .

63

4. The points are given:

A(1;0;0), B(0;2;0), C (0;0;3)

F  ( x  y 2 ) i  ( y  z 2 ) j  ( z  x 2 )k .

and the field

Find the flux of the field

F

through the squares of triangles ОВС, ОАС and ОАВ. Find the flux of the field through the total surface of the pyramid ОАВС. 5. Prove the Stokes theorem in a form (1.111).

F







6. Let t  i y  j x . Use the Stokes theorem to prove that the integral along the continuous closed curve in the plane ху equals

1   1  t  d   ( xdy  ydx)  A , 2 2 where А – is a square of the surface subtended by this curve. 7. Integrating by the perimeter of the surface situated in the plane ху show that the absolute value of the integral itself. 8. Show that

 r  dr

is tow times larger than the surface

      V  d  0 , if S – is a closed surface. S

9. Prove the relations

 U V  d     V U  d  ,  U V  d    (U )  (V )  d . S

Test questions to chapter 1 1.

Find the center of gravity of a system of three material points M1 (r1 ) ,

M 2 (r2 ) , M3 (r3 ) where masses m1 , m2 , m3 are focused knowing that the

center of gravity of two masses is on the line connecting these masses and divides it in a relation inversely proportional to the masses. 2. Prove that the vector x = b(a  c) - a(b  c) is perpendicular to the vector 3.

What is the angle between the vector

a

b , if the vector a + 3b is a - 4b is perpendicular to the

and

perpendicular to the vector 7a - 5b , and the vector vector

7a - 2b ?

4. A solid body rotates with the constant angular velocity linear velocity

v

is solenoidal.

64

c.

 . Show that the

5. Find the (c ) r , where

r is a radius-vector.

rot (b(r  a)), where a and b – constant vectors. 7. Prove the Gauss’s theorem in a form V d  V d .   6. Calculate the

S

8. Prove the Gauss’s theorem in a form

 d  P     Pd . S

9. Prove the Stoke’s theorem in a form

V

 d      d . S

65

V

In the chapter 1 we entirely limited to the Cartesian coordinate system in which it is supposed that the unit vectors are constant. We introduced the radius-vector. However not all the physical problems are successfully solved in the Cartesian coordinate system. For example, for the central forces (such as the gravitational or electrostatic) the Cartesian coordinates can occur to be very inconvenient that is why one uses such a system in which one of the coordinates is the distance in the radial direction. The coordinate system should be chosen on the condition of the best correspondence to the problem stated, using the different conditions and symmetry typical for the problem under consideration. The correct choice of the coordinate system allows getting the solution sooner, i.e. the partial derivative differential equations in the new system can be reduced to the first order differential equations by variables separation method. Let’s first consider coordinates in which the equation

2   k 2   0

(2.1)

admits the variables separation. The equation (2.1) has rather more general meaning than it can seem from the first view: 1) when k 2  0 it represents the Laplace equation; 2) when k 2     const – Helmholtz equation; 3) when k 2    const – equation of diffusion (spatial part);

4) when k 2  const  kinetic energy – Schrodinger wave equation. 66

2.1. Curvilinear coordinates The Cartesian coordinates are formed by three classes of reciprocally perpendicular planes: х=const, y=const, and z=const. Let’s imagine that we put on this system the three other classes of surfaces. The surfaces of any of these classes are not parallel to each other and in addition they should not be the planes. All three new classes of surfaces should not be reciprocally perpendicular however for the simplicity we omit this condition. The position of any point can be given by the crossing of three planes in the Cartesian system or by the crossing of three surfaces which form the new system of curvilinear coordinates. Admitting the surfaces of the curvilinear coordinates to be q1  const , q2  const , q3  const , we thereby fix the position of the given point by the coordinates

 q1 , q2 , q3  as well as the coordinates (х1,х2,х3). This means thst in

principle one can write:

x  x  q1 , q2 , q3  , y  y  q1 , q2 q3  , z  z  q1 , q2 , q3  .

(2.2)

The opposite dependence is possible:

q1  q1  x, y.z  , q2  q2  x, y, z  , q3  q3  x, y, z  .

(2.3)

To each of the surfaces class qi  const one can corresponds the



unit vector ai , normal to the surface qi  const and directed to increase of the qi . The square of the distance between two points is calculated by the formula:

ds 2  dx 2  dy 2  dz 2   hij 2 dqi dq j .

(2.4)

ij

The coefficients hij are called Lame coefficients; They can be considered as certain parameters characterizing the given coordinate 67

system q1 , q2 , q3 . The set of Lame coefficients defines the metrics of the coordinate system. In order to determine the hij2 , let’s differentiate the equations (2.2):

dx 

x x x dq1  dq2  dq3 , q1 q2 q3

dy 

y y y dq1  dq2  dq3 , q1 q2 q3

dz 

z z z dq1  dq2  dq3 ; q1 q2 q3

dx 2  

x x  dqi dq j , qi q j

dy 2  

y y  dqi dq j , qi q j

dz 2  

z z  dqi dq j . qi q j

ij

ij

(2.5)

Substituting the obtained expressions into the (2.4), one gets:  x x y y z z  ds 2      dqi dq j ,  qi q j qi q j  ij  qi q j

(2.6) hij2 

x x y y z z   . qi q j qi q j qi q j 68

Let’s limit to the orthogonal coordinate system (reciprocally perpendicular surfaces). Mathematically it means that

hij  0, i  j .

(2.7)

In order to simplify the denotations let’s put hii  hi , then

ds 2   h1dq1    h2 dq2    h3dq3  . 2

2

2

(2.8)

In the subsequent parts each coordinate system will be defined by the Lame coefficients assignment. And vice versa for any given dqi , admitting the other q to be constant, it is convenient to determine these quantities using the relation

dsi  hi dqi .

(2.9)

The curvilinear coordinates q1 , q2 , q3 are dimensionless. The Lame coefficients can depend on q and can have the dimension. The product hi dqi can have the dimension of length. From the relation (2.9) the elements of surface and volume follow immediately

d ij  dsi ds j  h i h j dqi dq j ,

d  ds1ds2 ds3  h1h2 h3dq1dq2 dq3 .

(2.10) (2.11)

The expressions (2.10), (2.11) are entirely consistent with the law of transformation (2.2). Problems 1. Show that the condition (2.7) satisfies the requirement of coordinate system’s orthogonality. 2. Show that the Jacobian element

 x, y , z  J   h1h2 h3 and hence the volume  q1 , q2 , q3 

69

 x, y , z  J  dq1dq2 dq3  h1h2 h3dq1dq2 dq3  q1 , q2 , q3  are in agreement with the (2.11).

2.2. Differential vector operators In the base of consideration of operators of gradient, divergence and rotor in the curvilinear coordinates we put the definition of the gradient of specified function as a vector with magnitude and the direction of the maximal velocity of change of this function in a space (see part 1.6).

Fig. 20. Curvilinear volume element

Then the component of the (q1 , q2 , q3 ) in the direction normal to the class of surfaces (figс. 20) is given in a form:

 1 

   S1 h1dq1

(2.12)

( q2 and q3 are fixed). The quantity dS1 – is an increment of the length in the direction of increase of the q1 . In the part 2.1 the unit 70



vector a1 was introduced for pointing of this direction. Obtaining the expression (2.12) for other components and summing them in a vector way let’s represent the gradient in a form:

 (q1 , q2 , q3 )  a1

 a1

    a2  a3  S1 S2 S3

(2.13)

    a2  a3 . h1q1 h2 q2 h3q3

The operator of divergence can be obtained using the equation (1.91) (the Gauss’s theorem):

  V (q1 , q2 , q3 )  lim  d 0

 Vd ,  d

(2.14)

where the product

h1h2 h3dq1dq2 dq3 is taken in capacity of the volume element. The positive directions are chosen so that q1q2 q3 or a1 , a2 , a3 form the fight-handed system. As in the parts 1.7 and 1.10, the integration over two surfaces

q1  const1 and (q1  dq1 )  const2 gives:

   V1h2 h3  dq1  dq2 dq3  V1h2 h3dq2 dq3  V1h2 h3  q1   (2.15)



 V1h2 h3  dq1dq2 dq3 . q1

Here we limit to the differential of the first order since if one considers the limit when dq1dq2 dq3  0 , then 71

   2 V h h  V h h dq  V h h dq12  ... dq2 dq3    1 2 3 1  1 2 3 2  1 2 3 q1 q1      2 V1h2 h3dq2 dq3   V1h2 h3   2 V1h2 h3  dq1  ... dq1dq2 dq3 . q1  q1  Dividing this expression by the volume element and moving to the limit one can see that the coefficients at the differentials of the second and higher orders  dq1 

n 1

reduce to zero.

Adding the similar results for other two pairs of surfaces one gets:

 V  q , q , q d  1

2

3

(2.16)

      V1h2 h3   V2 h1h3   V3h1h2  dq1dq2dq3 . q2 q3  q1  After dividing by the elementary volume one has:

  V  q1 , q2 , q3   

1 h1h2 h3

   (2.17)   V h h  V h h  V h h        1 2 3 2 1 3 3 1 2 , q2 q3  q1 



where Vi – projection of the V on the direction of the

 аi ,

i.e.

Vi  ai V . Combining the equations (2.13) and (2.17), one gets the Laplacian:

72

   h2 h3          q1  h1 q1     1    h3h1       q1 , q2 , q3        . h1h2 h3  q2  h2 q2        h1h2      q  h  q   3  3 3   

(2.18a)

Using the Stoke’s theorem let’s write in an explicit form the

 V and move to the limit approaching the surface square to zero. Let’s consider the differential surface element on the curvilinear surface q1  const . From the expression

 Vd  V

1

h2 h3dq2 dq3

(2.18b)

S

according to the Stoke’s theorem

V 1 h2 h3dq2 dq3   V  d  .

(2.19)

Here the linear integral is taken over the circuit on the surface q1  const .

 V (q , q , q )  d   V h dq 1

2

3

2 2

2



    V3 h3  V3h3  dq2  dq3  q2       V2 h2  V2 h2  dq3  dq2  V3h3dq3  q3   (2.20)

      h3V3    h2V2  dq2 dq3 . q3  q2  73

On the parts of way 1 and 2 the plus sign is taken while on the parts 3 and 4 the minus sign is taken (fig. 21), since in the second case the bypass is done in the negative direction.

Fig. 21.

From the formula (2.19) one gets:

 V

1



1 h2 h3

     h3V3    h2V2  .  q3  q2 

(2.21)

The other two components can be obtained by the cyclic permutation of the indices:

 a1h1  1    V  h1h2 h3  q1   hV 1 1

a2 h2  q2 h2V2

a3h3     . q3   h3V3 

(2.22)

Problems 1. Let

a1

– is a unit vector in the direction of an increase of the

  a1 

q1 . Show that

  h2 h3  h1 h1  . 1 1  ,  a1   a2  a3  h1h2 h3 q1 h1  h3q 3 h2q2  74

2. Show that the orthogonal unit vectors

ai 

ai

can be determined on a form

1 r . In particular prove that the condition ai  ai  1 leads to an  hi qi

expression for the hi , which is consistent with the equation (2.6). 3. Substantiate the statement that the simple scalar and vector products (not containing the operator  ) in the orthogonal curvilinear coordinates are expanded in the same way as in the Cartesian coordinates and do not contain the Lame coefficients. 4. Using the vector equality

 V   V   (V ) , find in the curvilinear coordinates the vector Laplacian  V .

2.3. Cartesian (rectangular) coordinates In the Cartesian coordinate system

h1  hx  1, h2  hy  1, h3  hz  1 . The Cartesian system – is the only in which all Lame coefficients are constant. This circumstance will be especially important when considering the tensor in the chapter 3. Starting from the equations (2.13), (2.17), (2.18) and (2.22), one can get the results considered in the chapter 1:

   j k , x y z V V V  V  x  y  z , x y z

  i

  

 2  2  2   , x 2 y 2 z 2 75

(2.24) (2.25) (2.26)

i    V    x   Vx

j  y Vy

k    . z   Vz 

(2.27)

2.4. Spherical coordinates Fundamental classes of the surfaces of the spherical coordinate system: 1. Concentric spheres with common center in the coordinate origin:

r  x 2  y 2  z 2  const. 2. Concentric surfaces of right round cones with the polar axis z and the vertexes in the coordinate origin:

  arc cos

z x2  y 2  z 2

 const.

3. Semiplanes through the axis z:

  arctg

y  const . x

Due to the arbitrary choice of the polar angle  and the azimuth angle  all the affixments are done relatively to axis z. The connection with the Cartesian coordinate system:

x  r sin  cos  , y  r sin  sin  , z  r cos . 0  r,0    2 ,0     . From the equation (2.6) 76

(2.28)

2

2

2

 x   y   z  h12  h112            r   r   r  2  sin  cos 2   sin 2  sin 2   cos 2   1 2

2

2

 x   y   z  h22  h222                   r 2 cos 2  cos 2   r 2 cos 2  sin 2   r 2 sin 2   r 2 (2.29)

h32  h332  r 2 sin 2  sin 2   r 2 sin 2  cos 2   r 2 sin 2   h1  hr  1, h2  h  r , h3  h  r sin  . The unit vectors r0 ,0 , 0 change the direction which is defined by the angles  and  (fig. 22).

Fig. 22. Spherical coordinates

77

These unit vectors are expressed through the fixed unit vectors in the Cartesian coordinate system i , j , k :

r0  i sin  cos   j sin  sin   k cos  ,

 0  i cos  cos   k cos  sin   k sin  , 0  i sin   j cos  . Putting in the part 2.2

a1  r0 , a2  0 , a3  0 , one gets the basic relations

 1  1   0   0 , r r  r sin      sin   r 2Vr     r 1 ,   2.17    V  2 V   r sin     r   sin V   r    

 2.13    r0

   2    sin   r    r  r        1       2.18a      2  sin  ,    r sin       1 2  2    sin   

78

(2.30)

(2.31)

(2.32)

 r0  1   V  2   r sin  r   Vr

 2.22  

r 0   rV

r sin    0    .    r sin  V 

(2.33)

Sometimes it is required to write the vector Laplacian  2V in the spherical coordinates. This can be done using the vector equality (1.80):





          V   V     V :

 2V

r

 2 2  2 cos     2 2   2  r r r r r sin       Vr   1 2  1 2   r 2  2  r 2 sin 2   2   

 2 cos   2    2   2  2    V    2   V   r  r sin    r sin    (2.34)

  2Vr 

2V



V 2 2 V 2 cos  2 V    V   , r  r2 r 2  r 2 sin  r 2 sin  

  2V 

 2V



1 2 V 2 cos V V  2 r  2 , (2.35) 2 r sin  r  r sin   2

  2V 

V 1 2 V  2  r 2 r sin  r sin   2 cos  V  2 2  . r sin   2

79

(2.36)

Problems 1. Express the unit vectors of the spherical coordinate system trough the Cartesian ones. 2. Get the formulas of inverse transformation:

i  r0 sin  cos    0 cos  cos   0 sin  , j  r0 sin  sin    0 cos  sin   0 cos  , k  r0 cos    0 sin  . 3. A particle moves in a space. Find the components of its velocity and acceleration in the spherical coordinate system:

r  r  r0  0  r   0 r sin  , dr v  r0  r   0 r   0 r sin    , dt vr  r , v  r , v  r sin    , ar  r  r 2  r sin    2 ,

a  r  2r  r sin  cos    2 , a  r sin     2r sin     2r cos    . 4. A motion of a particle with mass m under the action of the central forces is defined

by

the

second

r  r  c  const

second Kepler’s law. 5. Express the

Newton’s

law

mr  r0 f  r 

.

Show

that

and the geometric interpretation of this fact leads to the

   in the spherical coordinates , , x y z

80

 r        x  r x   x   x       sin  cos    cos  cos  1   sin   , r r  r sin        1  cos    ,   sin  sin   cos  sin      y r r r sin        1    cos   sin  .  z r r         4.

Using the results of the task 5 get the formula:

     i  x  y   i x    y That is the quantum operator corresponding to the z-component of the angular momentum. 7. Prove the equivalence of three forms of the

2  r 

(in the spherical

coordinates): 2 1 d  r d  r   1 d 2 d 2 (r ) 2 d (r ) ; r  r ,  .        r 2 dr  dr  r dr 2  dr 2 r dr

The second form is especially convenient for verification of the correspondence between the statement of the problem in the spherical and the Cartesian coordinates. 8. Let





2  0 . Show that 22 r 2  0 .

81

2.5. Separation of variables In the Cartesian coordinates the Helmholtz equation (2.1) takes the form:

 2  2  2  2  2  k 2  0. 2 x y z

(2.42)

Let’s limit to the case k 2  const . The simplest way to solve the partial derivative differential equation in a form (2.42) consists in the reducing of the equation to a system of ordinary differential equations. For this let’s put

  x, y, z   X  x   Y  y   Z  z  ,

(2.43)

and substitute into the (2.42). We in general do not know if the representation of the required solution if a form (2.43) is true or not. But if our attempt is successful then the representation in the form (2.43) will be proved. In the opposite case the equation (2.42) should be solved using the Green’s functions, integral transformations or numerical methods. Let’s put that the representation (2.43) is true, then substitute it into the (2.42):

d2X d 2Y d 2Z YZ  XZ 2  XY 2  k 2 XYZ  0. 2 dx dy dz

(2.44)

Dividing by the   XYZ and rearranging the terms one gets:

1 d2X 1 d 2Y 1 d 2 Z 2   k   . X dx 2 Y dy 2 Z dz 2

(2.45)

Thus, the variables are separated: the left part of the equation depends on only х, while the right one – only on у and z. Since х, у and z – independent variables, the behavior of the х can not be determined by the one of the у and z. Hence, it leaves to set each part of the equation equal to a certain constant, separation constant. 82

1 d2X  l 2 , x dx 2 1 d 2Y 1 d 2 Z k    l 2 , 2 2 Y dy Z dz 2

1 d 2Y 1 d 2Z 2 2   k  l  . Y dy 2 Z dz 2

(2.46)

(2.47)

(2.48)

Let’s set each part of the equation (2.48) equal to the constant:

1 d 2Y  m2 , 2 Y dy 1 d 2Z  k 2  l 2  m2  n 2 . 2 Z dz

(2.49)

(2.50)

The constant n 2 introduced allows getting the symmetric set of three ordinary equations (2.46), (2.49) and (2.50). In the result, the original assumption (2.43) turned out to be justified. Thus, let’s write the solution in a form:

 lmn  x, y, z   X l  x   Ym  y   Z n  z  ,

(2.50а)

where l , m, n – arbitrary numbers, satisfying the condition

k 2  l 2  m2  n2 . The function (2.50а) should be the solution of the equation (2.1), if X l  x  – solution of the equation (2.46), Ym  y  –

solution of the equation (2.49), Z n  z  – solution of the (2.50). The general form of the solution of the equation (2.1) can be presented as a linear combination of the solutions  lmn :

   almn lmn . lmn

83

(2.50b)

The constant coefficients a lmn are chosen so that the boundary conditions of the problem are true. A representation of the solution in a form (2.50b) is based on that



the  2  k 2

 – linear differential operator. By the definition the

linear operator L possess two properties:

L  a   aL, L  1  2   L 1  L 2 , where a – constant. The variables separation method considered works also in the case when k 2  f  x   g  y   h  z   k 2 , (2.50c) where k  – new constant. The equation (2.46) now takes the form: 2

1 d2X  f  x   l 2 . 2 X dx

(2.50d)

The solutions X,Y,Z will be different yet however the transformation of the differential equation and construction of the linear combination of the solutions remain to be the earlier ones. The method of variables separation of the partial derivative differential equation is shown here to illustrate the use of the different coordinate systems. Now let’s consider the equation (2.1) in the spherical coordinate system and try to separate the variables. Using the formula (2.32), one gets:

   2  sin  r  r r 1   2 r sin   1  2  2  sin  

      sin     

84

       k 2 . (2.51)   

Now put

  r , ,    R  r        .

(2.52)

Substitute  2.52    2.51 :

   2 R       sin      r  r r   R   sin      1        2 2  (2.53) r sin   R    2    sin     k 2 R. Divide the equation (2.53) by the R :

1 d  2 dR  1 d  d  r  2  sin   2 R  r dr  dr  r sin  d  d  (2.53а)



1 d 2   k 2. 2 2 2 r sin  d

Note that instead of the partial derivatives in the equation there appeared the simple ones.  2.53a   r 2 sin 2  



 2 1 d  2 dR    k  R  r 2 dr  r dr    1 d      r 2 sin 2   . (2.54) 2   d 1 d  d    2  sin   d    r sin  d  2

The equation (2.54) connects the function Ф, dependent on only the , with the function which is dependent on r and θ. Since the variables r , θ and  are independent then one can set both parts of 85

this equation equal to a certain constant. It is worth to note that almost in all the physical problems the  plays a role of the azimuth angle that is why it is more probable that the solution Ф will have the periodic character, but not the exponential one. Accounting this let’s put the constant of separation to be equal to m2 , then

1 d 2   m2 ,  d 2 1 d  2 dR  1 d  d  r  2  sin   2 r R dr  dr  r sin  d  d  m2  2 2  k 2 . r sin 

(2.55)

(2.56)

Multiplying the (2.56) by the r 2 and rearranging the terms one gets:

1 d  2 dR  2 2 1 d  d  r k r    sin   R dr  dr   sin  d  d  m2  2 . sin 

(2.57)

The variables are again separated. Setting each part of the equation equal to  2 , then one finally gets:

1 d  d  m2 sin      2  0,   2 sin  d  d  sin  1 d  2 dR  2 2R r  k R   0.   r 2 dr  dr  r2

(2.58) (2.59)

We could again reduce the partial derivative differential equation to a system of three ordinary differential equations. The total solution has a form: 86

 lm  r , ,     Rl  r   lm     m   .

(2.60а)

l ,m

The quantity is possible if the

k 2 can also be a variable. The variables separation k 2 is expressed by the formula:

k2  f r  

1 1 g    2 2 h    k 2 . 2 r r sin 

(2.60b)

Problems



1. Act by the operator  2  k 2

 on the sum

a1 1  x, y, z   a2 2  x, y, z 



2

and prove the linearity of this operator, i.e.



 k 2  a1 1  x, y, z   a2 2  x, y, z   









 a1  2  k 2  1  a2  2  k 2  2  x, y, z  2. Prove that the equation

1  2  k  f  r   2 g      r  2  r , ,        r , ,    0 1   h    r 2 sin 2   admits the variables separation (in the spherical coordinates).

k 2  const.

2.6. Cylindrical coordinates The relations which define the connection between the Cartesian and the round cylindrical coordinates (fig. 23) has the form:

x   cos  , y   sin  , z  z. 87

(2.61)

Fig. 23. Cylindrical coordinates

The Lame coefficients:

h1  h  1, h2  h   , h3  hz  1,

(2.62)

Thus coordinate system is formed by the following classes of coordinate surfaces: 1) regular round cylinders with axis z as the common axis:

  x 2  y 2  const. 2) semiplanes going through the axis z:

  arctg

y  Const. x

3) planes parallel to the plane ху: z = сonst.

0    , 0    2 ,    z  .  1    0 k (2.13)  (  ,  , z )  0 ,    z 1  1 V Vz ( V )   (2.17)   V   ,     z

(2.63) (2.64)

  1  2  2 ( ) 2  (2.18)       , (2.65)      2 z 2 2

1

88

0 (2.22)    V 

 0 k

1      V V

 . z Vz

(2.66)

The vector Laplacian

 2V



  2V 

1



2

V 

2 V ,  2 

   1 2 V  2V    2V  2 V  2 ,        2V z   2Vz .

(2.67)

The form of the z- component of the Laplacian is defined that the axes z in the Cartesian and cylindrical systems coincide, i.e. z Cart  z cyl:

 2 ( 0V  0V  kVz )   2 ( 0V  0V )  k  2Vz   0 f (V ,V )  0 g (V ,V )  k  2Vz . Problems 1. Expand the unit vectors of the cylindrical system on the components in the Cartesian coordinate system:

0  icos  jsin ,

0  isin  jcos ,

k0  k .

2. Expand the unit vectors of the Cartesian coordinate system on the components in the cylindrical coordinate system:

i  0cos  0 sin ,

j  0 sin  0cos , 89

k  k0 .

3. A particle moves in a space. Find the components of its velocity and acceleration in the cylindrical coordinate system:

V   , V   , Vz  z; a      2 ,

a     2 , az  z. 4. A conductor by which the current І flows, is situated along the axis z. The vector magnetic potential equals

  I 1 Ak ln( ).  2   I Show that the magnetic induction equals B   0 . 2 2 5. Solve the Laplace equation    0 in the cylindrical coordinates for the

case:



  (  ) .    k ln(



  ) . 0 

cylindrical coordinates the  6. In  the  V (  ,  )   0V (  ,  )   0V (  ,  ) is given. Show

vector that the

function  (  V )

has only the z-component. Test questions to chapter 2 1. 2. 3. 4.

Give the definition of the Lame coefficients. Operators of gradient, divergence and rotor in the curvilinear coordinates. The Helmholtz equation in the Cartesian coordinates. The Lame coefficients for the parabolic coordinates.

90

3.1. Introduction. Fundamental concepts Tensors play an important role in many fields of physics in particular in the general relativity and magnetic theory, are widely used when studying the anisotropic properties of the solid body. Let’s consider as an example the Ohm’s law :

j    ,

(3.1)



where j – density of current; E  electric field;   conductance. If the medium under consideration is isotropic, then the   scalar and for example, for the х-component of the current the equality is true:

j1   1.

(3.2)

However, if the medium is anisotropic, as for example, in many crystals, then the density of the current in the х-direction can be dependent on the electric fields in the у- and z-directions. Supposing the linear dependence, one can rewrite the equation (3.2) in a form:

j1  111  122  133

(3.3)

ji    ik k .

(3.4)

or in the general form:

k

91

The scalar conductivity is given by the set of 9 elements  ik :

  11  12  13        21  22  23  .    31  32  33 

(3.5)

The quantities which do not change when rotating the coordinate system, i.e. invariant, are called scalar. The quantities, components of which are transformed by the same law that the components of the radius-vector do, are called vectors:

i   aij Aj ,

(3.6)

j

where aij – is a set of cosines of angles between the axes xi and x j . But such a definition of the vector contains some uncertainty.  Let’s take the radius-vector r , then

xi   j

xi  xj. x j

(3.7)

If one defines the derivatives as

aij 

xi , x j

(3.8)

then the equations (3.6) and (3.7) turned out to be identical. An arbitrary set of quantities A j , transforming by the law

Ai   j

xi Aj x j

(3.9)

defines the contravariant vector. Let’s consider the gradient of the scalar

  i

   j k , x1 x2 x3 92

(3.10)

which is transformed by the law

    x j  x j     ,. xi j x j xi j x j xi

(3.11)

where    ( x, y, z)   ( x , y , z )     scalar. The equation (3.11) differs from the (3.9). The equation (3.11) defines the covariant vector. In the Cartesian coordinates

x j xi   aij xi x j

(3.12)

and hence, the contravariant and the covariant transformations coincide. In other coordinate systems the relation (3.12), in general, does not exist. We will note the components of the contravariant vector with the sup-index A i , and the components of the covariant vector with the sub-index Ai . The scalar is called the zero-order tensor, and the vector – firstorder tensor. Now let’s define the contravariant, mixed and covariant tensors of second order using the relations:

xi xj kl        xk xl  kl   xi xl k    l  ji   xk xj . kl    xk xl Cij    ' Ckl  kl xi x j   ij

93

(3.13)

It is seen that the A kl is contravariant with respect to both the indices, Cij is covariant with respect to both indices, and the lk is transformed contravariantly with respect to the first index k , but covariantly with respect to the second index l . In the Cartesian coordinates all three types of second-order tensors – the contravariant, mixed and covariant  coincide. The second-order tensor A (with components  kl ) is convenient to represent writing its components in a form of the square table (3  3 in the case of three-dimensional space):

 11 12 13        21  22  23  .  31 32 33   

(3.14)

This does not mean however that any square table of numbers or functions forms the tensor. The significant condition imposed on the tensor components is that they are transformed by the law (3.13). For example, let’s consider the two-dimensional tensor:

  xy  y 2     2  . x xy   

In the rotated coordinate system the component 11 should be  equal to  x y  . Let’s verify if the 11 is transformed by the law (3.13):

11   x y    kl

x1 x1 kl     a1k a1l  kl , x k x l kl

where i, j  1. Substituting instead of the a1k , a1l and  kl their real values one gets:

94

( xcos  ysin )( xsin  ycos ) 

cos sin =  sin cos

cos 2 11  cos sin  12  sin cos  21  sin 2 22    xy  cos 2  y 2cos sin  x 2 sin cos  xysin 2 . The equality appeared which shows that the condition (3.13) is  true for the 11 (and other components). Thus, the Т – is a really second-order tensor. The adding of the tensors is defined similarly to the one for the vectors: А+В=С,

(3.15)

if  ij   ij  C ij . At that the tensors А and В should have the same order and both should be given in a space of equal dimension. The equation (3.13) can be written in a more compact form. If two identical indices meet in one part of an expression while one index is sup and the other one is sub then the summation is with respect to these indices. That is why the second expression from the (3.13) can be rewritten in a form:

 j 'i 

xi ' xl k  l , xk x j '

(3.16)

where the summation with respect to k and l is implied. This defines the rule of summation. For illustration of this rule let’s show that the  – symbol corresponds to a mixed second-order tensor  lk . Firstly, it is necessary to know if the  lk is transformed in accordance with the (3.13), i.e. if it is a tensor. With account of the summation rule one has: 95

 lk

xi xl xi xk    , xk xj xk xj

(3.17)

while,

xi xk xi   . xk xj xj

(3.18)

But x j and x i – independent coordinates, 

xi   j ik ,  x j



(3.19)

so

 lk

x i x l i    j , x k x j

i.e.  lk – is really the second-order mixed tensor. The Kronecker symbol possesses one more interesting property: it has the identical components in all rotating coordinate systems and that is why it can be called isotropic. In general, the  mn does not depend on the  nm , so the order in which the tensor indices are represented is important. However there are several interesting special case; so if

mn  nm ,

(3.20)

Then the tensor is called symmetric. If

mn  nm ,

96

(3.21)

then the tensor is called antisymmetric. It is obvious, that any tensor (of second order) can be expended on the symmetric and antisymmetric parts:

mn 

1 mn 1   nm    mn  nm  .  2 2 symmetric tensor

(3.22)

antisymmetric tensor

Problems 1. Prove that

are tensors, and

 y 2  xy  ,    xy x 2   

  xy    y2 

x2

 y2 C    xy

 xy D   2 x

y2    xy 

xy  , x 2 

 ,  xy 

are not tensors. 2. In the general relativity the four-dimensional tensor of curvature is a fourthorder one (Riemann-Christofel) and satisfies the symmetry conditions

Riklm   Rikml   Rkilm : a) show that the number of components at the condition

Riklm   Rikml   Rkilm

decreases from 256 until 36; b) show that the condition

Riklm  Rlmik additionally decreases the number of

independent components until 21; c) show that if the equality

Riklm  Rilmk  Rimkl  0 is true, that the number of

independent components equal 20. Note. The last relation can be considered as the additional condition only in a case when all four indices are different.

  xy   y2 

3. Expand the tensor 

x2   on the symmetric and antisymmetric parts. xy 

4. Prove that if an arbitrary order tensor components equal zero in the given coordinate system then they equal zero also in other coordinate systems. 5. The components of the tensor А equal the corresponding components of the tensor В in a certain coordinate system, i.e.

97

A0ji  Bij0 .

Show that the tensor А equal the tensor В in all coordinate systems, i.e.

Aij  Bij .

3.2. Direct product We defined the scalar product (part 1.3) in a form of a sum of the corresponding components products:

   i i  .

(3.23)

A generalization of this expression in the tensor calculus is the operation of contraction. Two indices, one of them is covariant and the other one is contravariant, are supposed to be equal to each other and then (in accordance with the summation rule) a summation is done by this circulating index. For example,

ji  i i 

xi xl k xl k  l  l . xk xi xk

(3.24)

With account of the equations (3.18) and (3.19)

i i 

xl k Bl   kl lk  kk . xk

(3.25)

Thus, the contracted mixed second-order tensor is invariant and hence is a scalar. This exactly corresponds to what we obtained in the part 1.3 for the scalar product of two vectors and in the part 1.7 for the divergence of a vector. In general, the operation of contraction decreases the order of a tensor by two. The components of the covariant and contravariant vectors (firs order tensors) can be multiplied by each other, in the result one gets the term ai b j . According to the (3.13), the obtained product is the second-order tensor: ' xk x j l xk xj aib  ak b   ak bl . xi xl xi xl 'j

98

(3.26)

Contracting one gets the simple scalar product:

aib'i 

xk xi ' x  ak bl  k ak bl   lk ak bl  ak b k . (3.27) xi xl xl

The operation shown is called the direct product. In the case of two vectors the direct product corresponds to the second-order



tensor. Just in this sense one can understand the quantity  , which was not defined in the framework of the vector calculus. The direct product of two tensors is a tensor of the order which equals the sum of orders of two primary tensors, i.e.

ij kl  C ikl j ,

(3.28)

where C ikl j  is a fourth-order tensor. Up till now we kept the difference between the covariant and contravariant transformations since it exists in the non-euclidean space and plays an important role in the general relativity. Further we will not differ the covariant and contravariant tensor that is why let’s take the system of sub-indices. In addition, we will use the summation rule and the contraction operation. Summation rule. If an index (a letter, but not a number) is met two times on the same side of the equation then the summation is implied by this index. Contraction. Contraction consists in a setting of two different indices to be equal to each other and henceforth in an application of the summation rule. Problems 1. The

n -order tensor T...i

is given. Prove that

...i / x j

order tensor (in the Cartesian coordinates). 2. The

n -order ijk ...

tensor is given, prove that

(n  1) -order tensor (in the Cartesian coordinates).

99

– is an

 

ijk ...

(n  1) -

/ x j  is an

3. A quantity L – is a scalar function of non-cartesian variables derivatives

q i

and in addition it depends on time t

qi ,

their time-

explicitly, i.e.

L(qi, qi, t )  L(qi , qi , t ) . Show that the d  L   L corresponds the dt  qi  qi

vector components. Note. It is counted that the But

qi

and

q i

are independent variables.

q j / q j  0.

3.3. Quotient rule If Ai and B j – are vectors, then it is easy to prove that  i  j – is a second-order tensor. Let’s consider the series of inverse dependencies.

Ki Ai  B ,

(3.29a)

Kij Aj  Bi ,

(3.29b)

Kij Ajk  Bik ,

(3.29c)

Kijkl Aij  Bkl ,

(3.29d)

Kij Ak  Bijk .

(3.29e)

In each of these equations the А and В – are known tensors, order of which is determined by the number of indices and in addition the А is arbitrary. In each case the К – is unknown quantity. It is necessary to find out the behavior of the quantity К when it is transformed. According to the quotient rule if the equation of interest is true in any rotating (rotated) Cartesian coordinate system, then the К – is a tensor of the pointed order. As an illustration let’s consider the equation (3.29b). Accounting the vector properties of transformation of the В, one can write down that in the stationary coordinate system 100

Kij Aj  Bi  aik Bk .

(3.30)

The equation (3.29b) is true in any rotating Cartesian coordinate system, so

aik Bk  aik ( Kkl Al ) .

(3.31)

In the last equation let’s again write the А in the rotating coordinate system:

l   j



x l  j   a jl Aj x j j

ij j  aik  kl a jl j

(3.32)

   a

(3.33)

ij

ik

a jl  kl  j  0 .

The last equality is true at any і and in any rotating system. Since the j is arbitrary then

 ij  aik a jl  kl ,

(3.34)

what coincides with the definition of a second-order tensor. Similarly one can consider the other equations (3.29). In the conclusion it is worth to aware of incorrect application of the quotient rule. It cannot be true if В=0. In this case the properties of transformation are not determined. 3.4. Pseudotensors Up till now all coordinate system transformations were limited to a pure rotation. Let’s now consider the operation of portrayal or inversion. If the coefficients of transformation aij   ij are given, then 101

xi   xi .

(3.35)

It is essentially that this transformation replaces the primary right-handed coordinate system by the left-handed one. The radius vector r  x1 , x 2 , x3    x1 ,  x 2 ,  x3 . This new vector has the negative components relatively to the new axis transformed. The simultaneous change of the signs both of the axes and the components does not change the vector (the direction in a space, fig. 24). The radius-vector r and all other vectors, which behave in the similar way when inversion of the coordinate system, are called the polar vectors. Absolutely in other way the vector which is equal to the vector     product of two polar vectors behaves. Let the C    , , where 



and  – are polar vectors. The equation (1.33) defines the components of the vector C : С1=А2В3-А3В2.

(3.36)

Fig. 24. Inversion of Cartesian coordinates. Polar vector

The vector C when inversion behaves not as the polar one. In order to differ them let’s call the vector C as pseudovector or axial vector (fig. 25). 102

Fig. 25. Axial vector

     momentum L  r  , rotation moment L  r  f , magnetic field          . In the right-handed coordinate system  , for which t the vector C characterizes a rotation which is connected to the right

For example, the angular velocity   r  , angular

hand rule. In the left-handed, inverted system the rotation is changed for the left one. In general, the pseudovectors and pseudotensors are transformed by the formulas:

Ci  a aij C j ,

'ij  a aik a jl kl ,

(3.37)

where a  is a determinant consisting of the elements of tables for the coefficients amn . In the case of inversion the determinant has a form:

1 0 0 a  0  1 0  1 . 0 0 1 103

(3.38)

When inversion of only axis х

1 a  0 0

0 1 0

0 0  1 . 1

(3.39)

For any pure rotation it is always true that a  1 . The quantities, which are transformed in accordance with the (3.37), are often called the tensor densities.



 

The mixed product S      C behaves similarly to a scalar (when rotating). But when inversion of coordinates (3.35) S  S , i.e. S – is a pseudoscalar. This property of the mixed product disappears due to its geometric interpretation as the volume of a parallelepiped. Indeed, if all three parameters – length, width and height – are changed for the negative ones, then the product of these three quantities will be negative. The electric charge is also a pseudoscalar. Let’s introduce for convenience the three-dimensional symbol of Levi-Civita  ijk :

123   231   312  1,  132   213

 ,   321  1, 

(3.40)

all other  ijk  0. Let the third-order pseudotensor  ijk in a certain coordinate system equals the  ijk . Then, by the definition of the pseudotensor,

 'ijk  a aip a jq akr  pqr . 123

(3.41)

 a11a22 a33 1  a12 a23a31 1      a  a1 p a2 q a3r  a    a13a21a32 1  a11a23a32     a a a  a a a   12 21 33 13 22 31  2

 a  1  a1 p a2 q a3r  pqr  a . 104

(3.42)

For other components one gets similarly:

 ijk   ijk .

(3.43)

It follows hence, that  ijk – is an isotropic pseudotensor with identical components in any rotating coordinate system. To any antisymmetric second-order tensor C ij (in the threedimensional space) one can corresponds the dual pseudovector C i , defined as

1 Ci   ijk  C jk , 2 C12  C31  0   С jk   C12 0 C23  . C   31  C23 0 

(3.44)

(3.45)

The double contraction (pseudo-) of fifth-order tensor  ijk C jk shows that when rotating of the coordinate system the quantity C i should behave as a vector but the existence of the pseudotensor  ijk leads to that the C i is an pseudovector. The components of the



pseudovector C are given as

 C1,

C2 , C3    C23 , C31 , C12  .

(3.46)

Note, that the cyclic order of indices appeared because of the circularity of the components of  ijk . This duality means that the three-dimensional vector product can be written in a form of pseudovector or in a form of the antisymmetric second-order tensor.

  

If the (polar) vectors , , C are given, then one can determine:

i Vijk   j k

i

Ci

 j C j  i  j Ck  i k C j  . . . k Ck 105

(3.47)

Each term  p  q C r should be a third-order tensor. Since the determinant (3.47) is entirely antisymmetric, then when permuting of any two indices the change of sign occurs. The dual quantity

V

1  ijkVijk 3!

(3.48)

is a pseudoscalar. Opening the determinant in an explicit form

1 1 C1 V  2 3

2

C2 ,

(3.49)

3 C3

it is easy to show that this is a mixed product. In order to prove the covariance of Maxwell’s equations it is necessary to develop this result for the four-dimensional space and in particular show that the four-dimensional volume element dx1 dx 2 dx3 dx 4  is a pseudoscalar. Let’s introduce the four-dimensional Levi-Civita symbol  ijkl , the analogue of the three-dimensional one  ijk . By the definition,

 ijkl is entirely antisymmetric with respect to all four indices.  ijkl  1, if the even number of indices permutation,  ijkl  1, if the odd number of indices permutation. Introducing the fourth-order tensor Н

 ijkl 

i

i

j

 j C j Dj

k

k Ck Dk

l

l

Ci

Di

Cl Dl

106

,

(3.50)

elements of which are the components f the polar vectors

, , C, D , one can define the dual quantity:

H

1  ijkl H ijkl . 4!

(3.51)

Since the  ijkl is a pseudotensor, then the H ijkl is also a pseudotensor. Suppose now, that the , , C, D have infinitely small extension along the four coordinate axes (Minkowski space):

   dx1 , 0, 0, 0  ,    0, dx2 , 0, 0  , ... (3.52)

and the four-dimensional volume element

H  dx1dx2 dx3dx4

(3.53)

is a pseudoscalar. We moved to the four-dimensional space by simple mathematical generalization of the three-dimensional space. One can similarly consider also the N-dimensional spaces. Problems 1.

The

antisymmetric

table

(C1 , C 2 , C3 ) form a pseudovector

is

given,

 0 C3  C2   0     C3 0 C1    C12     C2  C1 0   C13

the

C12 0  C23

element

of

the

table

C13   C23  . 0 

Supposing that the relation

Ci  is true in all coordinate systems prove that

1  ijk C jk 2!

C jk 

is formulated in other form).

107

is a tensor (here the quotient rule

2. The operator which

x 4  ic  t

2 

4 2 1 2 can be written in a form of sum  2 , in i 1 xi c 2  2

. This four-dimensional Laplacian, usually called the

D’Alembertian, is denoted by the symbol 2. Show that the 2 – is a scalar operator. 3. Show that

 ii  3,  ij ijk  0,  ipq jpq  2 ij ,  ijk  ijk  6 . 4. Show that

 ijk   pqk   ip jq   iq jp .

5. Prove that every of the following fourth-order tensor:

ij kl ,  ik jl   il jk ,  ik jl   il jk is isotropic, i.e. the form of every one does not depend on the coordinate system rotation. 6. Applying the inversion, prove, that the isotropic tensor really has the pseudotensor nature.

3.5. Affinors The affinor is introduced with the purpose of development of the rule of simple vector calculus for the second-order tensors. Let’s take   two vectors i and j and form a combination i j . This combination is called the affinor. A multiplication from the left consists in multiplication of the left multiplier by the first multiplier from the pair written from the right:





 ij   i  x  j  y  k  z  i  j   x  j .  

(3.60)

A multiplication form the right supposes the inverse order, i.e.





ij    i  j i  x  j  y  k  z   i  y .   108

(3.61)

It is seen from here that the operation of multiplication is not   commutative. It is necessary to represent exactly that the i and j , forming the affinor i j , do not interact with each other. If they have scalar coefficients, then these coefficients are multiplied, but the unit vectors themselves do not form either scalar or vector product. i j  ji .





Now let’s form a combination of two vectors А and В :



    i  x  j  y  k  z

 i 

x



 j y  k z 

 ii  x  x  ij  x  y  ik  x  z  ji  y  x  jj  y  y  (3.62)  jk  y  z  ki  z  x  kj  z  y  kk  z  z . The quantity    corresponds to the affinor formed of affinors’ combinations. It was found out that the product of two vectors  – is a second-order tensor. Hence, the affinors are also the second-order tensors, written in a form which emphasizes their vector origin. It was yet noted, that the operation of multiplication of a vector and an affinor is not commutative. However there exists one important particular case when this operation possesses the properties of commutativity :

 where   affinor,   If a  i , then  x

a     a ,  a  arbitrary vector.       x , i.e.

(3.63)

i xx  j  x y  k  xz  (3.64)

 i xx  j  yx  k  z x . 109

Setting the separate components to be equal to each other one gets:

x x  x x ,  x  y   y x ,  x z   z x ,

(3.65)

i.e.   c , where c  const . Otherwise, if the multiplication by an arbitrary vector is commutative then the affinor should be symmetric. One of the most important properties of the symmetric affinor is in that by the special choice of the coordinate axes it can always be represented in a normal or diagonal form:

  ii xx  jj  yy  kk zz .

(3.66)

It is interesting and useful to give the geometric interpretation of the symmetric affinor. For the simplicity let’s suppose that the  symmetric affinor  is given in a diagonal form. Then using the  radius-vector r let’s write the equation:

r  r  1 ,

(3.67)



which imposes a limitation to the magnitude of the r in dependence on its orientation.

 ix  jy  kz  ii 

xx

 jj  yy  kk  zz

 ix  jy  kz   1

  . (3.68)  

x 2  xx  y 2  yy  z 2 zz  1 The last equality defines the ellipsoid with the semiaxes

a  xx

1

2

, b   yy

1

2

, c  zz

1

2

.

(3.69)

Hence, a diagonalization of the affinor corresponds to an orientation of the affinor ellipsoid in such a way that its axes coincide with the coordinate axes. 110

If

the

 is given, i.e. U (i  j, i, j  x, y, z ) , then for any

symmetric

Uii  0, Uij  U ji  vector a

affinor

    a  U  U  a .

Otherwise, a multiplication of a vector by the antisymmetric affinor obeys the rule of anticommutation. Problems

 U and V U V  0 .

1. An antisymmetric affinor

a vector

 V

are given. Prove that

V U  U V ,  2. Let U – is an antisymmetric affinor, a – is a unit vector in the direction of the radius-vector r . Show that the end of the radius-vector slides along the    ellipsoid surface when r  U  a . 3. The two-dimensional vectors r  ix  jy, t  iy  jx can be  connected by the equation rU  t . Define a tensor U , using for this purpose the  simple tensor representation. Find the U and give its definition from the point of view of affinors.

4. Show that vector

 V

I i i  j j k k

is a unit affinor in the sense that for any

   I V  V .

Test questions for chapter 3 1. Contravariant and covariant vectors. Give the definitions. 2. Contravariant, mixed and covariant second-order tensors.

111

1. Find the derivative of the scalar field u ( x, y, z ) in the point М in the direction of the normal passing through this point to the surface S , subtending the acute angle with positive direction of the axis Oz . Solution:

u  ln(1  x 2  y 2 )  x 2  z 2 , S : x 2  6 x  9 y 2  z 2  4 z  23, M (3, 0, 4). N  grad F ( x, y, z ). F  x 2  6 x  9 y 2  z 2  4 z  23, grad F 

F F F i j k  (2 x  6)i  18 yj  (2 z  4)k . x y z

N |M  0i  0 j  12k , N  02  02  (12) 2  12. cos   0;cos   0, cos   1. 112

U U U U  cos   cos   cos  . N x y z 2x 2y U U U ; ;    0. 2 2 2 2 x 1  x  y y 1  x  y z U U U |M  0, 6; |M  0; |M  0; x y z U  0,6  0  0  0  0  (1)  0. N 2. Find the angle between the gradients of the scalar fields u( x, y, z ) and v( x, y, z ) in the point М . Solution:

v  x 2  y 2  3z 2 , u 

 1 1 1  x , M  , , . 2 yz  2 2 3

 – required angle. cos  

( grad  gradU ) ; grad  gradU

grad  2 xi  2 yj  6 zk ,

grad |M  2i  2 j  2 3k , grad  ( 2 ) 2  ( 2 ) 2  (2 3 ) 2  4, 113

gradU 

1 x 2x i  2 2 j  3 k, 2 yz y z yz

gradU |M  3 2i  3 2 j  6 3k , grad  (3 2 ) 2  (3 2 ) 2  (6 3 ) 2  12, cos  

2 3 2  2 3 2  2 3 6 3  1    . 4  12

3. Find the vector lines in the vector field a . Solution:

a  3xi  6 zk . The differential equations of the vector lines of the field a :

dy  0  y  C 0 , dx dy dz      2dx dz 3x 0 6 z   z  x

2 ln x  ln z  ln C , x 2  Cz. 4. Find the flux of the vector field а through the surface S , cut by the plane P (the normal is outward to the closed surface subtended by the given surfaces).

114

Solution:

a  xzi  yzj  ( z 2  1)k ,

S : x2  y 2  z 2 ( z  0), P : z  4.

Fig. 26.

F   a  n  dS   ax dydz  a y dxdz  az dxdy , S

S

4

z

0

z

2 2 2 2  ax dydz   xzdydz   z z  y dydz   zdz  z  y dy  S

S

S

4 y 2 z2 y z   z z  y 2  arcsin  | dz  2 2 z  z 0 

4

1 1 1 4    z 3dz    z 4 |  32 . 2 0 2 4 0

115

4

z

0

z

2 2 2 2  ay dxdz   yzdxdz   z z  x dxdz   zdz  z  x dx  S

S

S

4 4 x 2 z2 x z 1 1 1 44 3   z   z |  32 . z  x 2  arcsin  | dz  z dz 2 2 z  z 2 0 2 4 0 0 

2 2 2  az dxdy   ( z  1)dxdz   ( x  y  1)dxdy  S



S

2

4

0

0

3  d  (    )d  

S

2

2

z

0

z

2  d   (   1)d  

2

2  1 4 1 2 4 0  4   2   0|d  56 0 d  56   0|  112 ,

F  32  32  112  176 .

5. Find the flux of the vector field a through the part of the plane Р, situated in the first octant (the normal does the acute angle with the axis Oz ). Solution:

Fig. 27.

116

a  2 xi  3 yj  4 zk , P : 2 x  3 y  z  1.

F   a  n  d   (ax cos   a y cos   az cos  )d  

S

  (2 x cos   3 y cos   4 z cos  ) d . 

n  {3, 2,1}  n  4  9  1  14. 2 3 1 , cos   , cos   , 14 14 14

cos  

d  1  ( z x' )2  ( z 'y ) 2 dxdy  1  4  9dxdy  14dxdy,

F





 2  2x 3 3y 4z     14dxdy  14 14 14  Dxy

 

1/2

1 2  x 3 3

0

0

 dx 

1/ 2

1 2  x 3 3

0

0

 dx 

(4 x  9 y  4(1  2 x  3 y ))dy 

1 (4  4 x  3 y )dy  3

1/ 2

7

  2  10 x  6 x 0

17  1/ 2 1   x  5 x 2  2 x3  |  . 3 2 0 4 117

2

  dx  

6. Find the flux of the vector field a through the part of the plane Р, situated in the first octant (the normal does the acute angle with the axis Oz). Solution:

Fig. 28.

a  9 yj  (7 z  1)k , P : x  y  z  1.

F   a  n  d   (ax cos   a y cos   az cos  )d  

S

  (2 x cos   3 y cos   4 z cos  ) d . 

n  {1,1,1}  n  3. cos  

1 1 1 , cos   , cos   , 3 3 3

d  1  ( z x' )2  ( z 'y )2 dxdy  1  1  1dxdy  3dxdy, 1

1ч

1 1 1   F   dx   0   9 y   7(1  x  y )  1)  3dy  3 3 3 0 0  1

1 x

1

7 9 2  1 x    dx  (8  7 x  7 y  9 y )dy    8 y  7 xy  y 2  y  | dx  2 2  0 0 0 0 118

7  9 2   9  9    (8  9 ) x  x  dx  2  2  8  9 2 7  9 3  1 10  9  9  9  x x  x |  . 2 23 6  2 0 7. Find the flux of the vector field а through the closed surface S (the normal is outward). Solution:

a  (e z  x)i  ( xz  3 y ) j  ( z  x 2 )k , S : 2 x  y  z  2, x  0, y  0, z  0. J  J1  J 2  J 3 ;

J1 

 (e

Dyoz

z

2

2 y

0

0

 x)dydz   dy



  e

z

2

1 2

1 1  y  z dz  2 2 

1 1  2 y 1       e  z  z  yz  z 2  | dy    e y  2  1  y  y 2  dy  2 4  0 4  0 0 1 1 2 1    e y  2  y  y 2  y 3  |   . 2 12  0 3 

Fig. 29.

119

J2 

1

2 2 x

0

0

 ( xz  3 y)dzdx   dx

Dxoz

  xz  6  3z  6 x dz 

1

3 1  2 2 x    xz 2  6 z  z 2  6 xz  | dx  2 2  0 0 1

2 1 1     6  10 x  2 x 2  2 x3  dx   6 x  5 x 2  x3  x 4  |  3 2 0  0 2 1 7 13  6  5    1  . 3 2 6 6 J3 

1

2 2 x

0

0

2  ( z  x )dxdy   dx

Dyox

  2  2 x  y  x dy  2

1

1   2 2 x    2 y  2 xy  y 2  x 2 y  | dx  2  0 0

1

4 1 1     2  4 x  4 x 2  2 x3  dx   2 x  2 x 2  x 3  x 4  |  3 2 0  0 4 1 5  22   . 3 2 6

1 13 5 8 J  J1  J 2  J 3      . 3 6 6 3 8. Find the flux of the vector field а through the closed surface S (the normal is outward). 120

Solution:

a  3xi  zj , 2 2  z  6  x  y , S : 2 2 2   z  x  y ( z  0). F   an d   div a dxdydz ,



div a 

V

ax a y az    3  0  3. x y z

Let’s move to the cylindrical coordinates system:

 x  r cos    y  r sin  z  z 



6

F   3dxdydz  3  2 d  rdr 0

V



6

0

6r 2

 0



6

dz  6 d  r (6  r 2  r )dr  0



0



 6 d  (6r  r 3  r 2 )dr  6 (9  2 6) d  6(9  2 6) |  6(9  2 6) . 0

0

0

0

Fig. 30.

121

9. Find the flux of the vector field а through the closed surface S (the normal is outward). Solution: 2

a  ( zx  y)i  ( xy  z ) j  ( x  yz)k ,  x 2  y 2  2, S :  z  0, z  1. Let’s use the Ostrogradky-Gauss’s formula.

F

 an dS   div a dxdydz, S

diva 

V

ax a y az    z  y  z. x y z

F   ( x  y  z )dxdydz.

Fig. 31.

122

 x  r cos  ,  The cylindrical coordinates system:  y  r sin  ,  z  z.  Hence, 2

1

1

2

1

1  2 2 0 d 0 rdv0 (r cos   r sin   z)dz  0 d 0  r cos   r sin   2 dv 



2

1 3 1 1 1 3 0  3 r cos   3 r sin   2 v  0| d 

2

1

1

1

  3 cos   3 sin   2  d  0

1 1  2 1 1 1   sin   cos     |  0       . 3 2 0 3 3 3

10. Find the F force work when displacing along the lone L from the point M to the point N . Solution:

F  ( x2  2 y)i  ( y 2  2 x) j , L : segment MN , M (4,0), N (0,2).

Fig. 32.

1) MO y  0, dy  0,4  x  0. 0

2 2 2  ( x  2 y)dx  ( y  2 x)dy   x dx  4

L

123

1 3 0 64 x |  . 3 4 3

2) ON x  0, dx  0,0  y  2. 2

1 32 8 y | . 3 0 3 0 L 64 8 72 2 2 L ( x  2 y)dx  ( y  2 x)dy  3  3  3  24. 2 2 2  ( x  2 y)dx  ( y  2 x)dy   y dy 

11. Find the circulation of the vector field а along the circuit G (in the direction of increase of the parameter t ). Solution:

a  xi  z 2 j  yk ,

  x  2cos t , y  3sin t , G:   z  4cos t  3sin t  3.

dx  2sin tdt , dy  3cos tdt , dz  4sin t  3cos t. C   ax dx  a y dy  az dz  Г



2

 (2  2cos t sin t  3cos t (4cos t  3sin t  3)  0

3sin t (4sin t  3cos t ))dt 

2

 (2sin 2t  12 cos

2

t  9sin t cos t  9 cos t 

0

12sin 2 t  9sin t cos t )dt 



2

2

 (2sin 2t  12  9cos t )dt  (cos 2t  9sin t 12t ) |  24 . 0

0

12. Find the magnitude of the circulation of the vector field а along the circuit G . 124

Solution:

a  2 yi  5zj  3xk ,

2 x 2  2 y 2  1, G:  x  y  z  3. Let’s use the Stoke’s theorem:

C

 a  n  dS   n  rota  d Г

i  rot a  x 2y

j  y 5z

k   5i  3 j  2k . z 3k

 n  rota  d   (5cos   3cos   2cos  )d . n  {1,1,1}, n  3  cos   cos   cos   d  1  ( z x' ) 2  ( z 'y ) 2 dxdy   1  (1) 2  (1) 2 dxdy  3dxdy.

125

1 . 3

 x  r cos   3 2   5     dxdy  10  dxdy   y  r sin    C  3   3 3 3   Dxy Dxy

2

1/ 2

0

0

 10  d



2 1 1/ 2 1 1 rdr  10 |  r 2 |  10  2    5 . 0 2 0 2 2

Projection on XOY

126

1. Find the derivative of the scalar field u  u  x, y, z  in the direction of the gradient of the scalar field     x, y, z  . 2. Find the gradient of the scalar field u  Cr , where C – is a constant vector, and the r – radius-vector. Which are the surfaces of the level of this field and how are they situated relatively to the vector C ? 3. Prove that if the S – closed sectionally smooth surface and the C – non-zero constant vector, then

 cos  n,C  dS  0 , S

where n – is a vector, normal to the surface S . 4. Prove the formula

  an dS    diva  agrad  dV , 0

S

V

where     x, y, z  ; S – surface, subtending the volume V ; n 0 – unit vector of the outward normal to the surface S . Find out the conditions of the formula applicability. 5. Prove that if the function u  x, y, z  satisfies the Laplace equation:

 2u  2u  2u    0 , then x 2 y 2 z 2 127

u

 n dS  0 , S

u – is a derivative in the direction of the normal to the n sectionally smooth closed surface S . 6. Prove that if the function u  x, y, z  is a second-order where

polynomial and the S – is a sectionally smooth closed surface then the integral

u

 n dS S

is proportional to a volume subtended by the surface S . 7. Let a  Pi  Qj  Rk , where P, Q, R – is linear functions of the x, y, z and let the G – is a closed sectionally-smooth curve situated in a certain plane. Prove that if the circulation

 adr

differs



from zero, then it is proportional to the square of the figure subtended by the circuit G . 8. A solid body rotates with the constant angular velocity around the stationary axis passing through the coordinate origin. The vector of angular velocity   x i   y j  z k . Find the rotor and the

divergence of the field of linear velocities v   r  of the points of

the bodyа (here r – radius-vector). 9. Find the derivative of the scalar field u  x, y, z  in the point

M in the direction of the normal to the surface S , subtending the acute angle with the positive direction of the axis Oz .





9.1. u  4ln 3  x 2  8xyz, S : x 2  2 y 2  2 z 2  1, M 1,1,1 . 9.2. u  x y  y z ; S : 4 z  2 x 2  y 2  8, M  2, 4, 4  .





9.3. u  2ln x 2  5  4 xyz, S : x 2  2 y 2  2 z 2  1, M 1,1,1 . 9.4. u 

1 2 1   x y  x 2  5 z 2 , S : z 2  x 2  4 y 2  4, M  2, ,1 . 4 2   128

9.5. u  xz 2  x 3 y , S : x 2  y 2  3 z  12  0, M  2, 2, 4  . 9.6. u  x y  yz 2 , S : x 2  y 2  4 z  9, M  2,1, 1 . 9.7. u  7 ln 1 13  x 2   4 xyz, S : 7 x 2  4 y 2  4 z 2  7, M 1,1,1 .

9.8. u  arctg  y x   xz , S : x 2  y 2  2 z  10, M  2, 2, 1 . 9.9. u  ln 1  x 2   xy z , S : 4 x 2  y 2  z 2  16, M 1, 2, 4  .

9.10. u 

x 2  y 2  z , S : x 2  y 2  24 z  1, M  3, 4,1 .

10. Find the angle between the gradients of the scalar fields u  x, y, z  and   x, y, z  in the point M .

x2 yz 2 1 1    6 y 3  3 6 z 3 , u  2 , M  2, , . 2 x 2 3   1 3 4 6 6 3 10.2.     , u  x 2 yz 3 , M  2, ,  . x 9y z  3 2 1 y3 4z3 z3 3 3  10.3.   9 2 x  , u  2 , M  , 2, . xy 2  2 2 3 3 3 4 1 z 1   10.4.     , u  3 2 , M 1, 2, . x y x y 6z 6 

10.1.  

x3 x2 1   6 y 3  3 6 z 3 , u  2 , M  2, , 2 yz 2  1 y2 z2 2 3  3 2 z , u  2 , M  , 2, 10.6.   3 2 x  xy 2 3

10.5.  

129

1  . 3 2 . 3 

xz 2  1 1  , M , ,1 . y  6 6  6 6 2 yz 2  1 1 1  10.8.     , u , M , , . 2 x 2 y 3z x  2 2 3 10.7.   6 6 x3  6 6 y 3  2 z 3 , u 

1 y2 xy 2 2 2 10.9.   3 2 x   3 2 z , u  2 , M  , 2, . z 3  2 3 x3 y 2 3 4 1 1   10.10.     , u , M 1, 2, . x y z 6z 6  2

11. Find the vector lines in the vector field a . 11.1. a  4 yi  9 xj . 11.2. a  2 yi  3xj . 11.3. a  2 xi  4 yj . 11.4. a  2 yi  3xj . 11.5. a  xi  4 yj . 11.6. a  3xi  6 zk . 11.7. a  4 zi  9 xk . 11.8. a  2 zi  3xk . 11.9. a  4 yj  8 zk . 11.10. a  yj  3zk . 130

12. Find the flux of the vector field a through the part of the surface S , cut by the planes P1 , P2 (the normal is outward to the closed surface subtended by the given surfaces).

a  xi  yj  zk 12.1. S : x 2  y 2  1,

P1 : z  0, P2 : z  2. a  xi  yj  zk 12.2. S : x 2  y 2  1,

P1 : z  0, P2 : z  4. a  xi  yj  2 zk 12.3. S : x 2  y 2  1,

P1 : z  0, P2 : z  3. a  xi  yj  z 3k 12.4. S : x 2  y 2  1,

P1 : z  0, P2 : z  1. a  xi  yj  xyzk 12.5. S : x 2  y 2  1,

P1 : z  0, P2 : z  5. a   x  y  i   x  y  j  z 2k 12.6. S : x 2  y 2  1,

P1 : z  0, P2 : z  2. 131

a   x  y  i   x  y  j  xyzk 12.7. S : x 2  y 2  1,

P1 : z  0, P2 : z  4.

a   x3  xy 2  i   y 3  x 2 y  j  z 2 k 12.8. S : x 2  y 2  1,

P1 : z  0, P2 : z  3.

a  xi  yj  sin zk 12.9. S : x 2  y 2  1,

P1 : z  0, P2 : z  5. a  xi  yj  k 12.10. S : x 2  y 2  1,

P1 : z  0, P2 : z  2. 13. Find the flux of the vector field a through the part of the plane P , situated in the first octant (the normal does the acute angle with the axis Oz ). 13.1.

a  xi  yj  zk P : x  y  z  1.

13.2.

a  yj  zk P : x  y  z  1.

13.3.

a  2 xi  yj  zk P : x  y  z  1. 132

13.4.

a  xi  3 yj  2 zk P : x  y  z  1.

13.5.

a  2 xi  3 yj P : x  y  z  1.

13.6.

a  xi  yj  zk P : x 2  y  z  1.

13.7.

a  xi  2 yj  zk P : x 2  y  z  1.

13.8.

a  yj  3zk P : x 2  y  z  1.

13.9.

a  xi  2 yj  zk P : x  y 2  z 3  1.

13.10.

a  2 xi  yj  zk P : x  y 2  z 3  1.

14. Find the flux of the vector field a through the part of the plane P , situated in the first octant (the normal does the acute angle with the axis Oz ). 14.1.

14.2.

a  7 xi  (5 y  2) j  4 zk , P : x  y 2  4 z  1.

a  2 xi   7 y  2  j  7 zk , P : x  y 2  z 3  1. 133

14.3.

14.4.

a  9 xi  j  3zk , P : x 3  y  z  1. a   2 x  1 i  yj  3 zk , P : x 3  y  2 z  1.

14.5.

a  7 xi  9 yj  k , P : x  y 3  z  1.

14.6.

a  i  5 yj  11 zk , P : x  y  z 3  1.

14.7.

14.8.

14.9.

a  xi   z  1 k , P : 2 x  y 2  z 3  1. a  5 xi   9 y  1 j  4 zk , P : x 2  y 3  z 2  1.



a  2i  yj  3



zk , 2 P : x 3  y  z 4  1.

14.10.

a  9 xi   5 y  1 j  2 zk , P : 3x  y  z 9  1.

15. Find the flux of the vector field a through the closed surface S (the normal is outward). 15.1.

a  ez  2x  i  ex j  e yk , S : x  y  z  1, x  0, y  0, z  0. 134

15.2.

15.3.

15.4.

a   3z 2  x  i   e x  2 y  j   2 z  xy  k , S : x 2  y 2  z 2 , z  1, z  4. a   ln y  7 x  i   sin z  2 y  j   e y  2 z  k , S : x 2  y 2  z 2  2 x  2 y  2 z  2. a   cos z  3x  i   x  2 y  j   3z  y 2  k , S : z 2  36  x 2  y 2  , z  6.

z 2 15.5. a   e  x  i   xz  3 y  j   z  x  k ,

S : 2 x  y  z  2, x  0, y  0, z  0.

15.6.

a   6 x  cos y  i   e x  z  j   2 y  3z  k ,

15.7.

a   4 x  2 y 2  i   ln z  4 y  j   x  3z 4  k ,

15.8.

S : x 2  y 2  z 2 , z  1, z  2.

S : x 2  y 2  z 2  2 x  3.



 



a  1  z i  4 y  x j  xyk , S : z 2  4  x 2  y 2  , z  3.

15.9. a 





z  x i   x  y  j   y2  z  k ,

S : 3x  2 y  z  6, x  0, y  0, z  0.

15.10.

a   yz  x  i   x 2  y  j   xy 2  z  k , S : x 2  y 2  z 2  2 z.

16. Find the flux of the vector field a through the close surface S (the normal is outward). 135

a   x  zi   z  yk , 16.1.

 x 2  y 2  9, S :  z  x, z  0  z  0  . a  2 xi  zk ,

16.2.

2 2  z  3x  2 y  1, S : 2 2  x  y  4, z  0.

a  2 xi  2 yj  zk ,

16.3.

 y  x 2 , y  4 x 2 , y  1  x  0  S :  z  y, z  0.

a  3xi  zj , 16.4.

 z  6  x 2  y 2 , S : 2 2 2  z  x  y  z  0  . a   z  y  i  yj  xk ,

16.5.

 x 2  y 2  2 y, S :  y  2. a  xi   x  2 y  j  yk ,

16.6.

 x 2  y 2  1, z  0, S :  x  2 y  3z  6.

a  2 z  y j   x  z k , 16.7.

 z  x 2  3 y 2  1, z  0, S : 2 2  x  y  1. 136

a  xi  zj  yk , 16.8.

 z  4  2  x2  y 2  ,  S : 2 2  z  2  x  y  .

a  zi  4 yj  2 xk , 16.9.

 z  x2  y 2 , S :  z  1. a  4 xi  2 yj  zk ,

16.10.

3x  2 y  12, 3x  y  6, y  0, S :  x  y  z  6, z  0.

17. Find the flux of the vector field a through the closed surface S (the normal is outward).

a   x2  y 2  i   y 2  x2  j   y 2  z 2  k , 17.1.

 x 2  y 2  1, S :  z  0, z  1. a  x 2i  y 2 j  z 2 k ,

17.2.

2 2 2  x  y  z  4, S : 2 2 2  x  y  z ,  z  0  . a  x 2 i  yj  zk ,

17.3.

 x 2  y 2  z 2  1, S :  z  0,  z  0  .

137

a  xzi  zj  yk , 17.4.

 x 2  y 2  1  z, S :  z  0.

a  3xzi  2 xj  yk , 17.5.

 x  y  z  2, x  1, S :  x  0, y  0, z  0.

a  x 2i  y 2 j  z 2 k , 17.6.

17.7.

 x 2  y 2  z 2  2, S :  z  0  z  0  .

a  x 3i  y 3 j  z 3 k , S : x 2  y 2  z 2  1. a   zx  y  i   zy  x  j   x 2  y 2  k ,

17.8.

17.9.

 x 2  y 2  z 2  1, S :  z  0,  z  0  .

a  y 2 xi  z 2 yj  x 2 zk , S : x 2  y 2  z  1. a  x 2i  xyj  3zk ,

17.10.

 x2  y 2  z 2 , S :  z  4.

138

18. Find the F force work when displacing along the line L from the point M to the point N .

F   x2  2 y  i   y 2  2x  j , 18.1. L : 2- x 2 8  y,

M  4, 0  , N  0, 2  . F   x  y  i  2 xj ,

 y  0 , M  2, 0  , N  2, 0  .

18.2. L : x 2  y 2  4

F  x 3i  y 3 j ,

 x  0, y  0  . M  2, 0  , N  0, 2  .

18.3. L : x 2  y 2  4,

F   x  yi   x  y j, 18.4. L : y  x 2 ,

M  1,1 , N 1,1 .

F   2 xy  y  i   x 2  x  j ,

 y  0 , M  3, 0  , N  3, 0  .

18.5. L : x 2  y 2  9,

F   x  yi   x  y j, 18.6. L : x 2  y 2 9  1  x  0, y  0  ,

M 1, 0  , N  0,3 . 139

F  yi  xj , 18.7. L : x 2  y 2  1  y  0  ,

M 1, 0  , N  2, 0  .

F   x2  y 2  i   x2  y 2  j ,  x, 0  x  1; 2  x, 1  x  2;

18.8. L : 

M  2, 0  , N  0, 0  . F  yi  xj , 18.9. L : x 2  y 2  2,

M



 

 y  0 ,



2, 0 , N  2, 0 .

F  xyi  2 yj ,

 x  0, y  0  , M 1, 0  , N  0,1 .

18.10. L : x 2  y 2  1,

19. Find a circulation of the vector field a along the circuit G (in the direction of increase of the parameter t ).

a  yi  xj  z 2 k , 19.1.





 x  2 2 cos t , y  G:  z  sin t.



a   x 2 y 3i  j  zk , 19.2.

 x  4 cos t , y  4 sin t , G:  z  3. 3

3

140



2 2 cos t ,

a   y  z i   z  x j   x  y k , 19.3.

 x  cos t , y  sin t , G:  z  2 1  cos t  .

a  x 2 i  yj  zk , 19.4.

 

 x  cos t , y  2 sin t  G:  z  2 cos t 2. 





2,

a   y  z i   z  x j   x  y k , 19.5.

 x  4 cos t , y  4sin t , G:  z  1  cos t. a  2 yi  3xj  xk ,

19.6.

 x  2 cos t , y  2sin t , G:  z  2  2 cos t  2sin t. a  2 zi  xj  yk ,

19.7.

 x  2 cos t , y  2sin t , G:  z  1. a  yi  xj  zk ,

19.8.

 x  cos t , y  sin t , G:  z  3. a  xi  z 2 j  yk ,

19.9.

 x  cos t , y  2sin t , G:  z  2 cos t  2sin t  1. 141

a  3 yi  3xj  xk , 19.10.

 x  3cos t , y  3sin t , G:  z  3  3cos t  3sin t.

20. Find the magnitude of a circulation of the vector field a along the circuit G .

a   x 2  y  i  xj  k , 20.1.

 x 2  y 2  1, G:  z  1. a  xzi  j  yk ,

20.2.

2 2  z  5  x  y   1, G:  z  4.

a  yzi  2 xzj  xyk , 20.3.

 x 2  y 2  z 2  25, G: 2 2  x  y  9  z  0  . a  xi  yzj  xk ,

20.4.

 x 2  y 2  1, G:  x  y  z  1.

a   x  y  i  xj  zk , 20.5.

 x 2  y 2  1, G:  z  1. 142

a  yi  xj  z 2 k , 20.6.

 z  3  x 2  y 2   1, G:  z  4. a  yzi  2 xzj  y 2 k ,

20.7.

 x 2  y 2  z 2  25, G: 2 2  x  y  16  z  0  . a  xyi  yzj  xzk ,

20.8.

 x 2  y 2  9, G:  x  y  z  1.

a  yi  1  x  j  zk , 20.9.

 x 2  y 2  z 2  4, G: 2 2  x  y  1  z  0  . a  yi  xj  z 2 k ,

20.10.

 x 2  y 2  1, G:  z  4.

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Абсолютная величина вектора Аксиальные векторы Определитель Смешанное произведение Аффинор Направляющие косинусы Векторный элемент поверхности Единичный вектор Единичный (нормальный) вектор Вектор Интегрирование векторов Векторное произведение Векторный дифференциальный оператор Векторный лапласиан Векторные величины Теорема Гаусса Уравнение Гельмгольца Градиент Теорема Грина Декартовы координаты Дивергенция Физическая сущность дивергенции Двойное векторное произведение Свертывание Операция свертывания Закон косинусов Символ Кронекера Антисимметричные тензоры Правило суммирования Безвихревые векторы Лапласиан Коэффициенты Ламе Уравнения Максвелла Ортогональные векторы Полярные векторы

144

Magnіtude of vector Axіal vector Determіnant Trіple scalar product Affіnor (dyadіcs) Dіrectіon cosіnes Vector area element Unіt vector Normal vector Vector Іntegratіon of vectors Vector product; vector (cross) product Dіfferentіal vector operator Vector Laplacіan Vector quantіtіes Gauss's theorem Helmholtz equatіon Gradіent Green's theorem Cartesіan coordіnates Dіvergence Physіcal sіgnіfіcance of dіvergence Trіple vector product Convolutіon (faltung); Contractіon The law of cosіnes Kronecker delta Antіsymmetrіc tensors Sum rules Іrrotatіonal vectors Laplacіan Metrіc; Lamer's coeffіcіents Maxwell's equatіons Orthogonal vectors Polar vectors

Псевдовекторы Псевдотензоры Радиус-вектор Ротор Симметричные тензоры Скаляр Скалярное произведение Скалярная функция Теорема Стокса Правило частного Тензор - двумерный Тензорные плотности Приращение длины

Pseudovectors; axіal vectors Pseudotensors Radіus-vector Curl Symmetry tensors Scalar Scalar product; dot product of vectors Scalar functіon Stoke's theorem Quotіent rule Tensor - two-dіmensіonal Tensor densіty Dіfferentіal length

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Basic 1. 2. 3. 4. 5.

Arfken G. Mathematical methods in physics. – М.: Atomizdat, 1970. – 712 p. (in russian). Kochin N.Е. Vector calculus and tensor calculus fundamentals. – М.: Science, 1965. – 426 p. (in russian). Borisenko А.I., Taranov I.Е. Vector calculus and tensor calculus fundamentals. – М.: Higher school, 1963. – 262 p. (in russian). Grechko L.V., Sugakov V.I., Tomasevich О.F. et al. Collection of problems on theoretical physics. – М.: Higher school, 1972. – 335 p. (in russian). Krasnov М.L., Kiselev А.I., Makarenko G.I. Vector calculus. – М.: Book house «LIBROCOM», 2009. – 140 p. (in russian).

Additional 1. Zhaksybekova К.А. Guidance to solution of problems on electrodynamics. Part 1. – Almaty: Kazakh University press, 2003 (in russian). 2. Arfken G.B., Weber H.J. Mathematical Methods for Physicists. – Elsevier, 2005. – 1182 p. 3. Chow T. L. Mathematical Methods for Physicists: a concise introduction. – Cambridge University Press, 2000. – 555 p.

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INTRODUCTION .............................................................................................. 3 Chapter 1. VECTOR CALCULUS ..................................................................... 5 1.1. Fundamental concepts ................................................................................. 5 1.2. Rotation of coordinate system ..................................................................... 9 1.3. Scalar product ............................................................................................. 15 1.4. Vector product ............................................................................................. 20 1.5. Scalar triple and vector triple products ........................................................ 25 Vector calculus (differentiation of vectors) ........................................................ 31 1.6. Gradient ....................................................................................................... 31 1.7. Divergence................................................................................................... 37 1.8. Rotor ............................................................................................................ 42 1.9. Subsequent application of operator  ........................................................ 46 1.10. Integration of vectors ................................................................................. 51 1.11. Gauss’s theorem ........................................................................................ 58 1.12. Stokes theorem .......................................................................................... 60 Chapter 2. SYSTEMS OF COORDINATES ...................................................... 66 2.1. Curvilinear coordinates………………… .................................................... 67 2.2. Differential vector operators … ................................................................... 70 2.3. Cartesian (rectangular) coordinates ............................................................. 75 2.4. Spherical coordinates ................................................................................... 76 2.5. Separation of variables ................................................................................ 82 2.6. Cylindrical coordinates ................................................................................ 87 Chapter 3. TENSOR CALCULUS ..................................................................... 91 3.1. Introduction. Fundamental concepts ............................................................ 91 3.2. Direct product .............................................................................................. 98 3.3. Quotient rule ................................................................................................ 100 3.4. Pseudotensors .............................................................................................. 101 3.5. Affinors ....................................................................................................... 108 Examples of problems solution .......................................................................... 112 Problems for independent work .......................................................................... 127 GLOSSARY ...................................................................................................... 144 REFERENCES ................................................................................................... 146

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Educational edition Zhaksybekova Kulyan Aitmagambetovna Zhusupov Marat Abzhanovich Kabatayeva Raushan Sarsembekovna

FUNDAMENTALS OF VECTOR AND TENSOR ANALYSIS Educational manual Second edition, supplemented Computer page makeup and cover designer N. Bazarbaeva The website used for front-page designing http://hqtexture.com/ IS No.10990 Signed for publishing 09.06.17. Format 60x84 1/16. Offset paper. Digital printing. Volume 9,25 printer’s sheet. Edition 80. Order No.3496 Publishing house «Qazaq university» Al-Farabi Kazakh National University, 71 Al-Farabi, 050040, Almaty Printed in the printing office of the «Qazaq university» publishing house

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