1,770 178 13MB
English Pages 368
Theory of Special Relativity Solved Problems Michael Tsamparlis
email: [email protected]
Contents
Mathematical Background: Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1 Mathematical Background: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
2 Classic Experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
3 The Position FourVector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
Relativistic Kinematics: Theory for Chapters 4 and 5 . . . . . . . . . . . . . . . . . . 21 4 FourVelocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 5 FourAcceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 6 ThreeMomentum–Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 Dynamics: Theory of Chapters 7–11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 7 Relativistic Optics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 8 FourForce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 9 Plane Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 10 Relativistic Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 Electromagnetic Field: Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 11 Electromagnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 iii
iv
Contents
Solutions 1 Mathematical Background: Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 2 Classic Experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 3 The Position FourVector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 4 Four Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 5 FourAcceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 6 ThreeMomentum – Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 7 Relativistic Optics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 8 FourForce . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 9 Plane Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 10 Relativistic Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 11 Electromagnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315
Mathematical Background: Chapter 1
Spacetime or Minkowski space is a fourdimensional vector space endowed with the Lorentz metric. The elements of the space are called fourvectors. A fourvector is denoted by a kernel letter and a Latin index which takes the values 0, 1, 2, 3. There are two types of fourvectors: the contravariant which we write as Ai and the 1 , x 2 ,⎞ x 3 ) the contravariant vectors are covariant Ai . In a coordinate system Σ(x 0 , x ⎛ 0 A ⎜ A1 ⎟ ⎟ represented by column matrices, e.g., Ai = ⎜ ⎝ A2 ⎠ and the covariant vectors by A3 Σ row matrices, e.g., Ai = (A0 , A1 , A2 , A3 ). The qualifying characteristic of a fourvector is that under the action of a Lorentz transformation it transforms linearly and homogeneously. That is, if [L ii ] is the matrix representing a Lorentz transformation between the coordinate frames Σ, Σ⎛ , and⎞if in the two frames ⎛ 0 ⎞the components of A A0 ⎜ A1 ⎟ ⎜ A1 ⎟ i i i ⎟ ⎟ ⎜ a contravariant vector A are A = ⎜ ⎝ A2 ⎠ and A = ⎝ A2 ⎠ , respectively, it A3 Σ A3 Σ holds ⎛
⎛ 0⎞ ⎞ A0 A 1 ⎜ A1 ⎟ ⎜ A ⎟ ⎜ ⎟ = [L i ] ⎜ 2 ⎟ . i 2 ⎝A ⎠ ⎝A ⎠ A3 Σ A3 Σ The components of the covariant fourvectors are transformed as follows:
(A0 , A1 , A2 , A3 )Σ = (A0 , A1 , A2 , A3 )Σ [L ii ]−1 . The length of a fourvector is defined by the invariant ηi j Ai A j , where η is the Lorentz metric. The length of a fourvector can be > 0, < 0, or 0. We call the fourvectors with negative length as timelike, with positive length spacelike, and with zero length as null. 1
2
Mathematical Background: Chapter 1
In spacetime there are infinitely many coordinate systems. From all of them we distinguish the ones in which the Lorentz metric has the reduced or canonical form ηi j = diag(−1, 1, 1, 1). We call these coordinate frames Lorentz Cartesian frames (LCF). In physics these frames are identified with the relativistic coordinate frames and will be called relativistic inertial frames (RIO). If Σ is an LCF, i then (and ⎞ then) the length of the fourvector A , which in Σ has components ⎛ 0only A ⎜ A1 ⎟ i ⎟ A =⎜ ⎝ A2 ⎠ , is given by the relation A3 Σ Ai Ai = −(A0 )2 + (A1 )2 + (A2 )2 + (A3 )2 . ⎞ ⎛ 0⎞ A0 B ⎜ A1 ⎟ ⎜ B1 ⎟ i i ⎟ ⎜ ⎟ More generally if A = ⎜ ⎝ A2 ⎠ and B = ⎝ B 2 ⎠ are two fourvectors, then the A3 Σ B3 Σ inner product (in Σ!) ⎛
Ai B i = −A0 B 0 + A1 B 1 + A2 B 2 + A3 B 3 . In spacetime there are infinitely many types of coordinate transformations. Lorentz transformations are special transformations which satisfy (among others) two basic conditions: (a) They are linear. (b) They preserve the inner product of the fourvectors. The geometric characteristic of Lorentz transformations is that they preserve the canonical form diag(−1, 1, 1, 1) of the metric. Therefore, they relate an LCF with an LCF or a RIO with a RIO. The same holds for the Galilean transformation in Newtonian Physics. Lorentz transformations constitute a group which we call the Poincar´e group. The Poincar´e group has 10 parameters of which 4 correspond to translations in spacetime and 6 to rotations. The rotations form a closed subgroup of the Poincar´e group which we call the Lorentz group. This latter has four components (parts) one of which is a group because it is the only one which contains the identity. The other three parts are not groups. We call the elements of the first part as the proper Lorentz transformations. In the following we shall use these transformations only. Consider two LCF Σ, Σ which are moving so that their spatial axes are parallel and the relative velocity of Σ in Σ is u = βc. Then the proper Lorentz transformation relating Σ, Σ has the form L(u) =
γ −γ u μ μ γ u μ δνμ + γu−1 2 u uν
,
Mathematical Background: Chapter 1
3
⎞ ux where u μ = (u x , u y , u z ) , u μ = ⎝ u y ⎠ , and δνμ + uz matrix: ⎛
⎛ ⎜ ⎜ ⎜ ⎜ ⎝
1+
γ −1 2 ux u2
γ −1 ux uz u2
γ −1 ux u y u2
1+
γ −1 μ u uν u2
γ −1 2 uy u2
γ −1 u y uz u2
1+
is the symmetric 3 × 3
⎞ ⎟ ⎟ ⎟. ⎟ ⎠
γ −1 2 uz u2
In the special case, Σ, Σ are moving so that the axes y, y and z, z are kept parallel whereas the x axis slides along the xaxis we say that Σ, Σ are moving in the standard configuration with relative speed u = βc. The Lorentz transformation for this special motion is called boost and has the following representation: ⎡
γ ⎢ −βγ L=⎢ ⎣ 0 0
−βγ γ 0 0
⎡ ⎤ 0 γ ⎢ βγ 0 ⎥ −1 ⎥, L = ⎢ ⎣ 0 0 ⎦ 1 0
0 0 1 0
βγ γ 0 0
0 0 1 0
⎤ 0 0 ⎥ ⎥. 0 ⎦ 1
Consider a RIO Σ and another RIO Σ whose space axes are parallel to those of Σ and moves with velocity u (wrt Σ). Suppose a fourvector Ai has components ⎞ A0 ⎜ Ax ⎟ ⎟ Ai = ⎜ ⎝ Ay ⎠ , Az Σ ⎛
⎞ A0 ⎜ Ax ⎟ ⎟ Ai = ⎜ ⎝ A y ⎠ , A z Σ
⎛
then the (proper) Lorentz transformation gives A·u , A =γ A − c u A·u A = A−γ A0 + (γ − 1) 2 u. c u 0
0
In the special case of a boost the transformation equations read
A0 = γ (A0 − β A x ), Ax = γ (A x − β A0 ), Ay = A y , Az = A z . These relations hold for an arbitrary fourvector!
4
Mathematical Background: Chapter 1
The basic fourvector in Special Relativity is the position fourvector x i of a relativistic mass point (ReMaP) say P. During the course of its existence the “motion” of P traces a curve in Minkowski space which we call the world line of P. If in a RIO Σ a ReMaP P at the time moment ct (of Σ!) has position vector r (in Σ!), then the fourposition vector of P in Σ has components: xi =
ct r
Σ
.
The length of the position fourvector x i x i = −τ 2 c2 defines the proper time of P which is an invariant characteristic of P. Physically the proper time is identified with the clock time in the proper frame Σ+P of P. This frame need not be a RIO. The proper timeis related with the coordinate time t in a RIO Σ, in which P has γ factor γ = 1/ 1 − β 2 by the relation t = γ τ > τ. This relation is called time dilation formula. Consider two points in the threespace of a ReMaP P which the proper moment τ of P have spatial distance d P . Let dΣ be the difference of the spatial coordinate distance of the two points in the RIO Σ in which P has factor γ . Then dΣ =
dP < dP . γ
This relation is called the length contraction formula. During the course of time there have been proposed many thought experiments with the view to prove the inconsistency, hence the validity, of the Theory of Special Relativity. Most of them have to do with the length contraction and the time dilation because the concepts of space and time are fundamental to our thinking. These “experiments” have been called paradoxes and, contrary to their intension, have established the validity and the consistency of the theory.
Chapter 1
Mathematical Background: Problems
⎛
(1)
(2)
(3) (4)
⎞ e xc ⎜ xe x ⎟ ⎟ In the LCF Σ the vector field Ai has components Ai = ⎜ ⎝ 0 ⎠ . Compute the x Σ i divergence A,i in the LCF Σ which moves wrt Σ in the standard configuration with velocity u = kcxˆ (k < 1). In the Euclidean Cartesian frame (ECF) {x μ , μ = 1, 2, 3} of E 3 a tensor of type (0, 3) is defined as follows: Tμνρ = xμ2 + 2xν2 + xρ2 . Compute the div and the curl of the vector Tμνμ . Consider the Euclidean tensor Aμν = xμ2 + xν2 . Compute Aμν,ν and Aμν,μν . Consider the two Lorentz Cartesian frames Σ(x, y, z) and Σ (x , y , z ) which are related with the transformation
x 1 = − sinh φx 0 + cosh φx 1
x 2 = x 2, x 3 = x 3, x 0 = − sinh φx 1 + cosh φx 0 where φ is a real parameter. (a) Show that the transformation is a Lorentz transformation. (b) A fourvector V i in Σ(x, y, z) has components (0, 1, 0, 1)t . Compute the components of V i in Σ (x , y , z ). The only nonzero components in Σ (x , y , z ) of a tensor Ti j of type (0, 2) are the T0 0 = T3 3 = 1. Compute the components of Ti j in Σ(x, y, z). Then compute in Σ(x, y, z) the covariant vector Ti j V i and the invariant Ti j V i V j . (c) Derive the results of (b) using matrix multiplication (the notation is obvious) as follows: [V ]Σ = [L][V ]Σ , [Ti j ]Σ = [L −1 ]t [Ti j ]Σ [L −1 ], [a]Σ = ([T ]t [V ]Σ )t , ai V i = [a]Σ [V ]Σ .
5
6
1 Mathematical Background: Problems
(5) Consider the twodimensional Lorentzian metric ds 2 , in which the coordinates (t, x) has the form ds 2 =
1 (−dt 2 + d x 2 ). t2
Compute the geodesic equations and solve them. Comment on the type of motion they describe. (6) Determine the covariant Lorentz transformation using reasonable physical assumptions.
Chapter 2
Classic Experiments
2.1. (a) The speed of a monochromatic beam of light propagating in a homogeneous and isotropic medium with refraction index n at rest equals v = nc < c. When the medium is moving this formula is not valid. J. A. Fresnel (in 1817) studied the speed of light in moving isotropic and homogeneous media and suggested that the speed of light in these media is given by the formula v = nc + κ V where κ is a coefficient, which he called drag coefficient, and V is the projection of the speed of the medium in the direction of propagation of light. The physical explanation given by Fresnel was that the light was “dragged” from the moving medium. Years later (in 1891) he proposed the following experiment which justified his explanation. A monochromatic light source is producing a beam which by means of a semitransparent glass tile is split into two synchronous beams, each beam propagating in opposite directions. The two beams interfere in the telescope T due to the difference in to the optical path. The displacement of the interference fringes measures the number of the wavelengths. If N = Δλ/λ is the displacement per wavelength, reduced, is the total distance of traveling of the beam in the medium (water), v is the speed of the medium in the tube, and n the refraction index (n = 1.33) of the medium, compute the drag coefficient. (b) Using the relativistic composition rule for threevelocities justify the Fresnel hypothesis. The explanation of the dragging of light by a moving medium without any extra hypothesis has been an additional experimental verification of Special Relativity.
2.2. (a) A variation of the classical Fizeau experiment is the following. Two identical glass tubes are filled with water, sealed and are placed parallel to each other on a horizontal circular table which can rotate about its center as in Fig. 2.1. A source produces a monochromatic beam which is split by means of a halftransparent glass tile in two synchronous beams. One of the beams is directed directly into the telescope and the other is reaching the telescope after it has being propagated through the glass tubes by means of a system of reflectors. The two beams interfere in the telescope creating interference fringes. When the table is rotating with constant angular velocity about the perpendicular axis 7
8
2 Classic Experiments
Fig. 2.1 The Fizeau experiment
through its center the water in the tubes is moving with the result the difference of the optical paths to change and consequently the interference fringes to be displaced. Compute the change in the optical path when the angular speed of the table is ω. (b) If the dragging coefficient κ = 1 − n12 , = 20 cm, n = 1.33, R = 20 cm, and ˚ compute the angular speed ω so that the maximum difference of λ = 5300 A the optical path is αλ (0 < α < 1).
Chapter 3
The Position FourVector
3.1 The Speed of Light 3.1.1. Consider two inertial Newtonian frames N , N and let v be the velocity of N relative to N . Let A be a Newtonian vector physical quantity which in N , N is given by the expressions A, A , respectively. Then the action of the Galileo transformation on the quantity A implies A = A + vt, where t is time in N (or N , since t = t ). Let B be another Newtonian vector physical quantity of the same nature as A and expression B, B in N , N , respectively. (a) Show that under the action of the Galileo transformation B − A = B − A. Conclude that the Newtonian relative vector physical quantities (position, velocity, acceleration, momentum, etc.) are invariant (in the sense that they do not change) under the action of the Galileo transformation. (b) Based on (a) deduce that Newton’s law of the gravitational field between two masses m 1 , m 2 gives F1 2 =
Gm 1 m 2 (r2 − r1 ), r2 − r1 3
and similarly the Coulomb’s law for charges q1 , q2 gives F1 2 = −
K q1 q2 (r2 − r1 ), r2 − r1 3
that is, F1 2 is covariant wrt the action of the Galileo transformation. Generalize to conclude that every law of Newtonian Physics, which is expressed in terms of the relative position, relative velocity, etc., is covariant wrt the Galileo transformation, hence compatible with the Galileo’s Principle of Relativity. 9
10
3 The Position FourVector
3.1.2. A strong laser beam which is rotating in the Earth about a perpendicular axis with angular speed ω = 1 rotations/s reaches the Moon. If the Earth–Moon distance equals 380×103 km, calculate the speed at which the light spot moves on the surface of the Moon. Explain the result.
3.2 The Lorentz Transformation 3.2.1. A particle is moving around a circle on the x–y plane of a RIO Σ with constant angular speed. Describe the world line of the particle in spacetime. 3.2.2. The clocks 1, 2 which are synchronized at the event A (that is, t A = t A = 0) start to move from the origin along the xaxis of the RIO Σ and are met again at the event B. Clock 1 moves with constant speed υ0 so that x = υ0 t, υ0 = constant > 0 and clock 2 moves with constant acceleration a0 so that x=
1 2 a0 t , a0 = constant > 0. 2
Calculate the proper time measured by each clock during the transportation from event A to event B. Compare the results and comment on the result. 3.2.3. It is given that the frames Σ ( , x, y, z) and Σ ( , x , y , z ) are related by the following linear transformation: ⎧ = a + bx + cy + dz ⎪ ⎪ ⎨ x = b + ax , ⎪ y = y ⎪ ⎩ z =z where the coefficients are constants. Determine a, b, c, d so that the above transformation is the (general) Lorentz transformation relating Σ and Σ . Give a kinematic interpretation of the transformation. 3.2.4. In the LCF Σ consider the coordinate transformation Σ (l, x, y, z) → (u, υ, y, z) which is defined by the relations 1 u = √ (l + x), 2
1 υ = √ (l − x). 2
3.2
The Lorentz Transformation
11
(a) Compute the metric ds 2 = −dl 2 + d x 2 + dy 2 + dz 2 in the new coordinates (u, υ, y, z). ∂2 ∂2 (b) Write the box operator ≡ 2 − 2 2 in the new coordinates. ∂x c ∂t (c) Show that this coordinate transformation is not a Lorentz transformation. Comment on the result. 3.2.5. Consider two RIO Σ, Σ in Minkowski space whose space axes are parallel and their relative velocity is u. (a) Determine the Lorentz transformation between Σ, Σ assuming that – Along the direction of the relative velocity Σ, Σ are related by a boost and – The spatial directions perpendicular to the relative velocity u remain the same. (b) Discuss the limit of the transformation equations when β → 0 (Newtonian limit). (c) Determine the change of the temporal and the spatial differences under the action of the Lorentz transformation for this general motion. Discuss the result. 3.2.6. Consider the Euclidean Cartesian frame Σ1 and a direction defined by the unit vector n in Σ1 . Define a rotation R of the coordinates so that the xaxis becomes parallel to the direction n. Use the result to derive the general Lorentz transportation. 3.2.7. The general Lorentz transformation between two RIO Σ, Σ with relative velocity u for the position fourvector is given by the relations l = γ l −
r·u c
,
r = r − γ ut + (γ − 1)
(u · r)u . u2
Show that the second relation can be written as r = γ (r − ut) + (γ − 1)
u × (u × r) . u2
Therefore, r consists of one part in the plane (r, u) and another part normal to the plane (u, r). 3.2.8. Consider a transformation which in a Cartesian frame is represented by the block matrix: 10 L= , 0A where A is a 3 × 3 matrix.
12
3 The Position FourVector
(a) Prove that L is a Lorentz transformation if and only if A is a Euclidean transformation. (b) Show that these Lorentz transformations form a closed subgroup of the Lorentz group. We call as these Lorentz transformations 3Euclidean rotations. Consider a secondorder tensor Ti j and show that under the action of these Lorentz transformations its parts transform as follows: T0 0 = T00 ,
Tμ 0 = ATμ0 ,
Tμ ν = ATμν At .
Deduce that under the action of threeEuclidean rotations a tensor transforms as three independent quantities: An invariant (the T00 ), a Euclidean vector (the Tμ0 ), and a Euclidean tensor of second order (the tensor Tμν ). 3.2.9. It can be shown that the proper Lorentz transformation L ∈ L+↑ leaves the direction of a null fourvector invariant. That is, if α is a null fourvector ( α α = 0) then L +↑ ( α ) = A α , A > 0. A null tetrad consists of four linearly independent four vectors ( α , m α , x α , y α ), such that the only nonvanishing inner products among the various vectors are x α xα = y α yα = m α α = 1. It follows that a null tetrad consists of two null vectors (the α , m α ) and two spacelike unit fourvectors (the x α , y α ). Compute the action of the (proper) Lorentz transformation on the null tetrad ( α , m α , x α , y α ) assuming that the fourvector α is an eigenvector of the (proper) Lorentz transformation. 3.2.10. The proper Lorentz transformation with beta factor β = (β μ ) is given by the matrix expression L +↑ (β) =
γ −γβμ μ −γβμ δνμ + γβ−1 2 β βν
,
or equivalently by the relations
A0 = γ (1 − β · A), A = A +
γ −1 (β β2
· A)β − γ A0 β.
Define the dyadic φ(β) associated with the factor β by the relation φ(β)μν = δνμ +
γ −1 μ β βν . β2
Obviously the quantity φ(β) is a Euclidean tensor (not a Lorentz tensor) of type (1, 1) which in an arbitrary RIO Σ is represented by a 3 × 3 matrix. In terms of φ(β) Lorentz transformation is written as follows:
3.2
The Lorentz Transformation
13
L +↑ (β) =
γ −γβμ −γβμ φ(β)μν
,
or equivalently
A0 = γ (1 − β · A) A = φ(β)A − γ A0 β. Consider two successive Lorentz transformations L +↑ (β 1 ), L +↑ (β 2 ) and show that their composition L +↑ (β 2 )L +↑ (β 1 ) can be written in the form L +↑ (β 2 )L +↑ (β 1 ) =
−γ12 β12μ γ12 μ −γ12 β21μ Γ12ν
,
where γ12 = γ1 γ2 (1 + β 1 · β 2 ) = γ21 , β21μ =
1 1 + β1 · β2
β12μ =
1 1 + β1 · β2
μ
μ
β1μ γ2 − 1 + 1 + 2 (β 1 · β 2 ) β2μ , γ2 β2 γ2
β2μ γ1 − 1 + 1 + 2 (β 1 · β 2 ) β1μ , γ1 β1 γ1
ρ
μ
Γ12ν = φ2ρ φ1ν + γ1 γ2 β2 β1ν . Show that 2 = β 12 · β 12 = β12
(β 1 −β 2 )2 − (β 1 ×β 2 )2 2 = β21 . (1+β 1 ·β 2 )2
Also prove that γ12 (= γ21 ) is the gamma factor for the βfactor β12 (= β21 ), that 1 is, γ12 = 1−β 2 . 12 From relation β12 = β21 deduce that β 12 , β 21 differ only in direction and show that the angle ψ they form (in Euclidean threespace) is given by the relation cos ψ = 1 −
2 sin2 φ , 1 + 2A cos φ + A2
1/2 1 +1)(γ2 +1) 2 where A = (γ (1 ≤ A ≤ ∞) and cos φ = ββ11·β is the angle between (γ1 −1)(γ2 −1) β2 β 1 , β 2 . Considering the angle ψ as a rotation of β 12 → β 21 determine the direction eˆ of the axis of rotation.
14
3 The Position FourVector
Using the fact that the composite Lorentz transformation can be decomposed in the product of a proper Lorentz transformation and a Euclidean rotation R shows μ that the Euclidean tensor Γ12ν of the composite transformation can be written as follows: Γ = φ(β 21 )R, that is, r = φ(β 21 )ρ − γ A0 β 21 ,
A0 = γ12 (1 − ρ·β 21 ), where ρ = RA and φ(β 21 ) = 1 + γ12β 2−1 β 21 ⊗ β 21 . 12 Prove that the rotation R is given by the following formula: R(ˆe, ψ) = 1 + ω1 (β 1 ⊗ β 2 + β 2 ⊗ β 1 ) + ω2 (β 1 ⊗ β 1 + β 2 ⊗ β 2 ) + ω3 (β 1 ⊗ β 2 − β 2 ⊗ β 1 ), where ω1 = ω2 A + ω3 and ω2 =
2 cos φ , 1 + 2A cos φ + A2
ω3 =
−2 sin ψ . = 1 + 2A cos φ + A2 cos ψ
(3.1)
3.3 The Boost 3.3.1. In the RIO Σ(x) the fourvector Ai has components Ai = (0, 0, sin(5x 1 − 3x 0 ), sin(5x 1 − 3x 0 )). Calculate the components of Ai in the RIO Σ (x ) which is moving wrt Σ in the standard configuration with speed u. 3.3.2. A point P leaves the origin of a RIO Σ at the time t = 0 of Σ. After proper time Δτ of P a light signal is sent toward the origin of Σ and reaches the origin at the moment t of Σ. Calculate the speed of P in Σ. 3.3.3. Consider two RIO Σ, Σ which are related in the standard configuration with speed βc. Show that for every moment in Σ there exists only one plane in spacetime on which the clocks ofΣ coincide with those of Σ . Show that this plane moves in 1/2 2 2 . Show that this result stays valid even if Σ with speed u = cυ 1 − 1 − υc2 the origins of Σ, Σ change arbitrarily. 3.3.4. In order to study the boost it is enough to study its action on the basis 1 0 vectors eˆ 0 = , eˆ 1 = (because it is a linear transformation). Consider 0 1 γ βγ L= and show that βγ γ
3.4
Length Contraction
15
1 γ = 0 βγ 0 βγ =L = . 1 γ
eˆ 0,L = L eˆ 1,L
Show that the Lorentz length of e0,L , e1,L equals ±1 whereas their Euclidean length is ≥ 1. Find the locus of the vectors e0,L , e1,L considered as Euclidean vectors on the Euclidean plane! 3.3.5. The RIO Σ, Σ are moving in the standard configuration with speed 3c/5. Draw on the Euclidean plane (x, ) of Σ the axes x , of Σ . 3.3.6. (a) Show that if two events A, B are simultaneous and occur at different threespace points in some RIO Σ, then there does not exist a limit in the time difference of the events in another RIO Σ , whereas their spatial distance varies from infinity to a minimum which is determined from their spatial distance in Σ. (b) If two events A, B occur at the same point in the threespace of a RIO Σ show that their temporal order (that is, the sequence at which they occur or, equivalently, their causal relation) is the same for all RIO.
3.4 Length Contraction 3.4.1. A particle is produced in the laboratory whose lifetime is 10−6 s with constant velocity and speed 2.7 × 108 m/s. (a) Compute the lifetime of the particle in the laboratory. (b) Compute the distance traveled by the particle in the laboratory before it disintegrates. (c) Compare the results with the corresponding Newtonian results. 3.4.2. Two particles A, B are moving along the xaxis of the LCF Σ, each with velocity factor β when they hit a standing wall (in Σ) with a time difference Δt. Calculate the spatial distance of the particles in their proper frames. 3.4.3. The points A, B are at a distance d along the xaxis of the LCF Σ. At some moment in Σ a light signal is emitted simultaneously from each point toward the other. Show that in the LCF Σ which is moving in the standard way wrt Σ with speed u, the emittance of the light signals is not simultaneous and more specifically and when the second the first emittance is done earlier at a time interval γ (u)ud c2 c−u emittance occurs the distance of the points in Σ is d c+u .
16
3 The Position FourVector
3.4.4. In the LCF Σ two light signals are moving along the xaxis so that their distance d remains constant. Calculate the distance d of the signals in an LCF Σ which is moving wrt Σ in the standard configuration with speed u. 3.4.5. A mass m of a gas of molecular weight k rests in a cylindrical container of base radius R and height L. Calculate the average number of molecules per cm3 (this is called the number density) which is counted by an observer Σ who is moving with speed v wrt the cylinder and in direction (a) parallel to the axis of the cylinder and (b) normal to the axis of the cylinder.
3.5 Time Dilation 3.5.1. A photon is emitted from the origin and in the x y plane of the LCF Σ and subsequently it is absorbed at the point (r, θ ) where r, θ are polar coordinates in the x y plane of Σ. Another LCF Σ is moving in the standard configuration wrt Σ with speed u. Calculate the time difference of the emittance and the absorption of the photon in Σ as well as the spatial polar coordinates in Σ of the point where absorption occurs. 3.5.2. A π meson is moving in the outer atmosphere of the Earth when it disintegrates into a μmeson of speed β = 0.985 and a neutrino. If before its disintegration the μmeson moves a distance d = 4 km wrt the observer in the Earth, calculate the lifetime of the μmeson. 3.5.3. The π meson has a lifetime τ0 = 2.6 × 10−8 s. Consider a beam of N0 π mesons of speed β in the LCF Σ and determine the distance l traveled by these particles in Σ at which half of the particles have disintegrated. Plot l(β). 3.5.4. A beam of negative pions passes through a target of liquid hydrogen. With a system of measuring devices, a current of charged particles along the initial direction of the beam of pions is found. One of the pions interacts with a proton of the target at the point whose coordinates in the laboratory are x = −0.500 m, y = 0.500 m, z = 0.200 m, t = 0.000 s. At this event, which we call event 1, a K meson is produced. The K meson is neutral and propagates with constant velocity until it disintegrates into two charged pions (K → π + π − ). This disintegration, which we call event 2, in to the laboratory has coordinates x = 0.880 m, y = 0.480 m, z = 0.250 m, t = 5.31 × 10−9 . (i) Calculate the lifetime of the K meson. (ii) Calculate the speed of K meson in the laboratory. 3.5.5. An ideal clock is realized by fixing two perfect mirrors along a rod and let a photon be reflected elastically between the mirrors. The unit of time (the “second”)
3.5
Time Dilation
17
is measured by a successive reflection of the photon on the same mirror. If the rod slides along the xaxis of an LCF Σ with velocity factor β, calculate the unit of time in Σ. 3.5.6. The Feynmann clock (see Fig. 3.1) consists of an electric bulb Λ which emits a beam of photons which is elastically reflected on a plane mirror and is received by the photocell Φ. The photocell activates with a special electronic device with zero activation time again the bulb Λ when a new beam of photons is emitted and so on.
α y
d
V
F
x
Fig. 3.1 The Feynmann clock
Let Σ be the “proper” observer of the clock and let d1 = distance of the reflecting mirror from Φ, Λ in Σ.
d cos α
be the spatial
(i) Consider an LCF Σ which is moving in the standard configuration along the xaxis with speed u and show that in Σ the clock is slow (time dilation). (ii) Repeat the calculations if Σ moves in the standard configuration along the yaxis. 3.5.7. A beam of unstable particles is propagating with speed u = 35 c along the xaxis of the LCF Σ. Along the path of the beam there have been placed two counters A, B at a distance of 9 m. The first counter measures a flow of 1000 particles per second and the second counter 250 particles per second. Assuming that the reduction in the number of particles is due to decay only, calculate the halflife1 time of the particles and their half lifetime in an LCF Σ , which moves in the standard configuration normal to the direction of the beam with speed v = 35 c.
1 The halflife time is the time required in the rest frame of the radioactive source for half the number of particles to decay.
18
3 The Position FourVector
3.6 Problems with Rods 3.6.1. In the LCF Σ two light signals are moving along the xaxis so that their distance d remains constant. Calculate the distance d of the signals in an LCF Σ which is moving wrt Σ in the standard configuration with speed u. 3.6.2. A particle is produced in the laboratory whose lifetime is 10−6 s with constant velocity and speed 2.7 × 108 m/s. (a) Compute the lifetime of the particle in the laboratory. (b) Compute the distance traveled by the particle in the laboratory before it disintegrates. (c) Compare the results with the corresponding Newtonian results. 3.6.3. A rod is moving along the xaxis of an LCF Σ with constant speed u. It is measured that in Σ the rod requires a time interval T to pass in front of a fixed point P of the xaxis, whereas the corresponding time interval in the proper frame Σ− of the rod is T . Calculate the proper length L 0 of the rod. This experiment shows that one can compute relativistic spatial distances using Newtonian experimental methods. 3.6.4. A rod of (proper) length l is moving with velocity factor β along the positive xaxis of an LCF Σ. A particle moves in the opposite direction along the xaxis with velocity factor β. Compute the time needed (in Σ) in order for the particle to travel across the rod. 3.6.5. A spaceship 2 of length 2 moves antiparallel to another spaceship 1. The observer at the end of spaceship 1 observes that it takes time T1 in order the spaceship 2 to pass in front of him. (a) Determine the relative speed u of the spaceships. (b) If the observer at the end of spaceship 2 measures that it takes time T2 for spaceship 1 to pass in front of him, calculate the length 1 of spaceship 1. 3.6.6. Two rigid rods 1,2 of length L 1 = L 2 = L move along the xaxis of the LCF Σ toward each other with common speed u. Determine the length of each rod in the frame of the other. 3.6.7. From the end points A, B of a train of proper length l which is traveling with constant speed β are emitted two light signals toward its midpoint M. (a) For the observer Σ in the ground the light signals arrive simultaneously at M. Compute the time difference of the emittance of the signals for the observer Σ in the train (i) algebraically and (ii) geometrically.
3.6
Problems with Rods
19
(b) If the light signals are emitted simultaneously for the observer in the train what is the time difference at which they reach the middle M of the train for the observer in the ground? 3.6.8. In the x y plane of the LCF Σ a rod of proper length l slides with speed u so that one of its end points is always on the xaxis while the rod is making a constant angle θ with the xaxis in the proper frame of the rod. Find the length of the rod in Σ. 3.6.9. A rod AB is moving in the x y plane of the LCF Σ with velocity v = (vx , v y , 0). Another LCF Σ is moving wrt Σ in the standard way with velocity u = ui. If at the moment t of Σ the projection of the end points of the rod in Σ and Σ are l x , l y and l x , l y , respectively, show that l x = γ (u) l x 1 − vcx2u , v u l y = l y − γ (u) cy2 l x where −1/2 u2 γ (u) = 1 − 2 . c 3.6.10. A rod is resting along the x axis of the LCF Σ when it starts sliding along the x axis with constant velocity u = uj . Show that in the LCF Σ in which Σ is moving in the standard configuration with speed v, the rod appears to make an angle with the xaxis. θ = tan−1 γvcuv 2 3.6.11. A right circular cone of angle 2θ and contiguous surface area S0 is resting with its axis along the xaxis of the LCF Σ (see Fig. 3.2). Another LCF Σ moves wrt Σ in the standard way with speed β. Calculate the angle 2θ and the contiguous surface S0 of the cone in Σ . Discuss the Newtonian and the ultrarelativistic limit of the results. yy
y
v
y’
x,x’ x x
Fig. 3.2 The right circular cone
20
3 The Position FourVector
3.6.12. A rod AB slides along the xaxis of the LCF Σ with speed β ( c = 1). At the same moment in the proper frame of the rod two light signals are emitted toward the origin of Σ. If the proper length of the rod is L 0 calculate the length of the rod L˜ which sees (not measures!) the observer at the origin of Σ. How this length varies with the speed of the rod in Σ? Compare L˜ with the measured length of the rod in Σ.
Relativistic Kinematics: Theory for Chapters 4 and 5
Consider a relativistic mass point (ReMaP) P with position vector x i and proper time τ . The fourvelocity and the fouracceleration of P are defined by the relations: ui =
dxi , dτ
ai =
du i . dτ
The fourvectors u i , a i satisfy the conditions: u i u i = −c2 , u i ai = 0, a i ai = (a+ )2 , where a+ is the proper acceleration of P.
ct , threevelocity Consider a LCF Σ in which P has position vector x = r Σ v, and threeacceleration a. Then the fourvelocity and the fouracceleration of P have components i
ui =
γc γv
Σ
,
ai =
ca0 a0 v + γ 2 a
Σ
,
(1)
where a0 = c12 γ 4 v · a = c12 γ 4 v · a = γ γ˙ . We call the special onedimensional motion with constant proper acceleration a+ as hyperbolic motion. This motion is covariant, that is, if P executes hyperbolic motion in one LCF Σ then it does so in all LCF. Consider two LCF Σ, Σ with parallel axes and relative velocity u and suppose that in Σ, Σ the ReMaP P has threevelocities v, v and threeaccelerations a, a , respectively. Then under the proper Lorentz transformation which relates Σ and Σ the fourvector of the fourvelocity u i transforms as follows: u · v γv = γu γv c 1 − 2 , c γu v · u 1 v = v+ − 1 γu u , γu Q γu + 1 c 2 where Q = 1 −
(2) (3)
u·v . c2
21
22
Relativistic Kinematics: Theory for Chapters 4 and 5
The angle θ between the threevelocities v, u is tan θ =
v sin θ γu (v cos θ + u)
(4)
and the speed is v =
1
2
1+
uv cos θ 2 c2
! u 2v2 2 (v ) + u + 2v u cos θ − 2 sin θ . c 2
2
(5)
Concerning the fouracceleration of P we have the relation u · v u·a u·a 1 a = −1 u 3 1 − 2 a + 2 v + 2 c c u γu γu2 1 − u·v c2
1
(6)
and for the magnitude " # + 2 = γ 6 a6 − (a × β)2 . a
(7)
In case Σ, Σ are moving in the standard configuration with speed u = βc the boost relating Σ, Σ gives for the fourvelocity: uυ γυ = γu γυ 1 − 2x , c − u υ x υx = , uυ 1 − c2x υy , υ y = uυ γu 1 − c2x υz . υz = uυ γu 1 − c2x
(8)
The last three relations have been named Relativistic Law of composition of threevelocities. Concerning the fouracceleration the boost gives 1 1− 1 a y = γu2 1 − 1 az = γu2 1 − ax =
γu3
ax , uvx 3 2 c
uv y uvx a 1 − ax , + y uvx 2 c2 c2 c2 uvx uvz a 1 − ax . + z uvx 2 c2 c2 2 c
(9)
Chapter 4
FourVelocity
4.1. The LCF Σ moves with velocity u wrt the LCF Σ. A particle P moves in Σ with velocity v , which makes an angle θ with the direction of u (in Σ ). Calculate (a) The velocity v of P in Σ. (b) The angle θ between v and u in Σ. (c) The speed v2 of P in Σ. 4.2. (a) Consider the vectors A and u and show the identity/decomposition A=
A·u 1 u + 2 u × (A × u). 2 u u
Define A = A·u u and A⊥ = u12 u × (A × u) and write A = A + A⊥ . u2 (b) A relativistic mass point has velocity v1 , v2 in the LCF Σ1 , Σ2 , respectively. Let u the velocity of Σ2 wrt the LCF Σ1 . Prove the relations # 1 u · v1 1 " − 1 u= (v1 ) − u , 2 Q u Q 1 u · v1 1 v1 − u = (v2 )⊥ = (v1 )⊥ , γu Q u2 γu Q (v2 ) =
where the and ⊥ refer to the direction of the velocity u and Q = 1 −
(4.1) (4.2) u·v . c2
4.3. The LCF Σ, Σ have parallel axes and Σ is moving wrt Σ with velocity u. A relativistic particle P has threevelocities v, v in Σ and Σ , respectively. (a) Show that v + [ v=
γ u · v + 1] γ u γ + 1 c2 . u · v γ (1 + 2 ) c 23
24
4 FourVelocity
(b) If u is parallel to the xaxis, that is, we have a boost, and the velocity v has components in the plane x − y only, show that tan θ =
v sin θ , cos θ + u)
γ (v
where u, v are the speeds of u, v and θ, θ are the angles made by v, v with the axes x, x , respectively. Show that for photons this relation becomes tan θ =
sin θ . γ (β + cos θ )
Finally, show that the speeds are related as follows: v2 =
1 γ 2 (1 +
1 [v 2 + u 2 + 2uv cos θ − β 2 v 2 sin2 θ ]. cos θ )2
uv c2
4.4. (a) Show that the rapidity α of a particle in the LCF Σ is related with the β factor of the particle in Σ with the relation 1 1+β . φ = ln(γ + γβ) = ln 2 1−β (b) Let (l, x)Σ , (l , x )Σ be the components of the position vector of a particle in the LCF Σ, Σ which are related with a boost L(β). Show that x + l = eφ (x + l). (c) Show that the rapidity of successive boosts in the same direction equals the sum of the rapidities of each boost. 4.5. In the LCF Σ the fourvector Ai has components Ai =
A0 A
. Σ
(a) Calculate the βfactor of Ai in Σ if Ai is timelike. (b) What happens when Ai is spacelike?
(c) A particle is moving in the LCF Σ so that its fourvelocity is u = i
Calculate the βfactor of the particle in Σ.
2 . i+j
4.6. The spaceships A, B are moving in the same direction along the xaxis of the LCF Σ with speeds u, υ, respectively. At some moment a light signal is emitted from A, reflected at B, and then returns to A after a time period T. If the spatial distance of the two spaceships at the moment of emittance of the light signal according to A is l, determine the speed of A in terms of υ, l, T .
4 FourVelocity
25
4.7. An LCF Σ is moving with velocity u wrt the LCF Σ . We consider a particle with velocities v, v in Σ, Σ , respectively, and direction angles θ and θ with the direction of the relative velocity u. Show that the rapidities ξ , ζ , ζ which correspond to the velocities u, v, u satisfy the relations tanh ζ sin θ , cosh ξ (tanh ζ cos θ − tanh ξ ) cosh ζ = cosh ξ cosh ζ − sinh ξ sinh ζ cosh θ. tan θ =
4.8. The observer Γ is moving away from the observer A with speed 0.6 c along a common direction x while observer B approaches A along the same direction with speed 0.8 c (see Fig. 4.1). Γ is emitting continuously a beam of green light (λ = 0, 54 μm) at an angle 170◦ with the direction of its motion. The beam is reflected (elastically) at B and is deflected to A. Calculate (a) The relative speed of B, Γ. (b) The angle with the xdirection and the wavelength of the radiation received by A.
°
Fig. 4.1 The path of the light beam
4.9. In the laboratory two beams of particles with velocities v1 , v2 are produced. (a) Calculate the relative speed of the beams. (b) In the ring of colliding beams (SR) at CERN a researcher wants to produce beams which have the maximum relative speed. Assuming that the beams have the same speed in the ring, calculate the relative orientation of the beams which the researcher must employ. 4.10. In the laboratory frame two electrons have kinetic energy 1MeV each and are moving in opposite directions along the xaxis. Calculate (a) The speed of each electron in the lab system. (b) Their relative velocity. (Mass of electron m = 0.51 MeV/c2 ). 4.11. The relativistic observers 1, 2 are moving along the positive semiaxis O x, O y of the LCF Σ with velocities v and kv, respectively. If 1 sees 2 to move in a direction
26
4 FourVelocity
which makes an angle φ with the positive semiaxis O x, calculate v as a function of k. Discuss the result for various values of the parameter k. 4.12. (a) A particle P of energy E and mass m is moving along the positive semiaxis y of the LCF Σ. An observer O who moves along the negative semiaxis x of Σ sees P to slide in a direction which makes an angle φ with the positive semiaxis x. (a) Calculate the speed of the observer in terms of the angle φ. Also calculate the energy of P measured by the observer. (b) Solve the problem in case m = 0, that is, P is a photon. 4.13. The LCF Σ2 moves in the standard configuration wrt the LCF Σ1 with velocity v21 = v1 i. Consider a third LCF Σ3 which moves in the standard way wrt Σ2 along the common yaxis with velocity v32 = v2 j2 . Suppose that the velocity u13 of Σ3 wrt Σ1 makes an angle θ1 with the axis x1 of Σ1 and θ3 with the x3 axis of Σ3 . (a) Show that tan θ1 =
v2 , v1 γv1
tan θ3 =
v2 γv2 . v1
(b) Compute the deviation δθ = θ3 − θ1 in terms of the velocity factors β1 , β2 . Show that when β1 , β2 1 (Newtonian limit), δθ = 12 β1 β2 . Application In the LCF Σ1 a particle moves in a circular orbit with constant speed β 1. Consider two adjacent positions of the particle along its orbit, the 2, 3 say, and let Σ2 , Σ3 be the instantaneous LCF at the points 2, 3, respectively. Choose Σ2 , Σ3 so that their space axes are parallel (in the Euclidean sense!) and show that after a complete rotation of the particle in Σ1 the proper frame of the particle has deviation πβ 2 in a direction opposite to the threevelocity of the particle. (The phenomenon of the braking of Euclidean parallelism after a complete rotation is called Thomas effect.) 4.14. Consider the linear threedimensional space R3 whose elements are the threevelocities of the relativistic particles, i.e., the position vector of each point of the space is determined/represents a threevelocity. Let us write the fourvelocity v i of a relativistic particle in the LCF Σ, Σ as follows: γc γc vi = , . γ v Σ γv Σ If the relative velocity of Σ, Σ is u show that v and v are related as follows (this is the effect of the Lorentz transformation in the threevelocity space): v =
v+u v·u . 1+ 2 c
(4.3)
4 FourVelocity
27
Introduce in the threevelocity space E3 spherical coordinates (v, χ , ψ), where v is the length of v (the speed) and the angles χ , ψ are defined by the relations (Fig. 4.2). vx = v cos χ , v y = v sin χ cos ψ, vz = v sin χ sin ψ,
(4.4)
so that v = v cos χ i + v sin χ cos ψj + v sin χ sin ψk .
Fig. 4.2 Spherical coordinates in the velocity 3space
(a) Introduce a new coordinate system (v , χ , ψ ) using the transformed velocity v and relations (4.4) and write (4.3) as a coordinate transformation (v, χ , ψ) → (v , χ , ψ ) assuming that Σ , Σ are moving in the standard configuration with velocity u = ui. (b) Compute the Newtonian result by considering u/c → 0. Discuss the result in the threevelocity space E3 considering the v, v , u as position vectors. (c) Show that if v c then v c (assuming always that u c) and when v = c then v = c for all u < c. (d) Show that for each value of the angle χ the angle χ diminishes as the relative speed of Σ, Σ increases. Discuss the case of the photons (v = v = c) and show that in this case the following relation holds: tan χ =
1 sin χ . γ β + cos χ
Show that this relation can be written equivalently: χ tan = 2
$
χ 1−β tan . 1+β 2
Deduce that for each value of χ the angle χ diminishes as the relative speed u (β → 1) of Σ, Σ increases.
28
4 FourVelocity
ct1 4.15. Two particles 1, 2 in the LCF Σ have position vectors x1i = , xi = r1 Σ 2 ct2 cγ1 cγ2 i i and fourvelocities v1 = , v2 = , respectively. Let r2 Σ γ1 v1 Σ γ2 v2 Σ Σ2 the proper frame of particle 2 and that the position fourvector and the suppose ct12 i 12 , v = dr . threevelocity of 1 in Σ2 are x1 = dt12 r12 Σ 12 2
(a) Calculate v12 in terms of v1 , v2 . (b) Show that v12 = −v21 , while v12  = v21 . Calculate the angle between the threevectors v12 , v21 . 4.16. Show that the γ factor of an observer with fourvelocity u i wrt another observer with fourvelocity v i is given by γ (u, v) = − c12 u i vi . Consider a third observer with fourvelocity wi and compute the γ factor γ (u, w) of u i wrt wi in terms of the γ factors γ (u, v) and γ (v, w). 4.17. (a) Show that two successive boosts in the same direction with speeds u 1 , u 2 are equivalent to a single boost in the same direction and with speed u=
u1 + u2 u1u2 . 1+ 2 c
(b) Consider the boost along the xdirection with speed u 1 and the boost in the ydirection with speed u 2 . Show that the result of composition of the two boosts depends on the order they are composed (the composition is not commutative). Also show that the composition of these boosts is not a boost in either case. 4.18. A relativistic particle in the LCF Σ has fourvelocity u = i
the rapidity a(∈ R) of the particle in Σ with the relation
γc γv
. Define Σ
tanh a = β.
(a) Show that γ = cosh a,
βγ = sinh a
i i and conclude that the fourvelocity u in terms of the rapidity is written as u = c cosh a , where eˆ is the unit in the direction of the threevelocity. Let θ, φ c sinh a eˆ Σ be the angles made by eˆ with the xaxis and the zaxis of Σ. Then show that
4 FourVelocity
29
eˆ = sin θ cos φi + sin θ sin φj + cos θ k and use this to write u i in the form ⎛
⎞ c cosh a ⎜ c sinh a sin θ cos φ ⎟ ⎟ ui = ⎜ ⎝ c sinh a sin θ cos φ ⎠ . c sinh a cos θ Σ
(b) Consider two LCF Σ1 , Σ2 with parallel axes which are moving with relative velocity u and a relativistic particle which in Σ1 , Σ2 has fourvelocity ui =
c cosh a1 c sinh a1 eˆ 1
Σ1
,
ui =
c cosh a2 c sinh a2 eˆ 2
Σ2
.
Calculate the relation between a1 , a2 . Examine the special case when Σ1 , Σ2 are moving in the standard configuration. % cos, (c) Define the trigonometric functions cos, % sin, % c% ot with the relations cosi % x = cosh x % x = sinh x sini
cosi % x = tanh x c% oti x = coth x .
Show that these functions satisfy the same relations as the real trigonometric functions sin, cos, tan, cot. Show that if in the complex plane we consider a triangle with sides ia0 , ia1 , ia2 and corresponding angles x0 , x1 , x2 as in Fig. 4.3, then the relations which have been found in the (b) part express the sine and the cosine law for this triangle. Such triangles and angles are the subject of spherical Euclidean geometry. The space of this geometry is the Euclidean sphere (= space of two dimensions with constant curvature equal to 1) whereas the classical Euclidean geometry is based on the Euclidean flat space (space of constant curvature equal to 0).1 1 The space whose points are the threevelocities (the threevelocity space) is a space of constant curvature equal to −1. Such a space is called hyperbolic. The difference between spherical, flat, and hyperbolic geometry is measured with the defect ε, which is defined by the relation ε = π − (x0 + x1 + x2 ). In Euclidean geometry ε = 0 while in spherical geometry ε is negative and its absolute value equals the area of the triangle (012). In hyperbolic geometry ε is positive and can be shown that it is given by the relation 1 (cosh a2 − 1)(cosh a0 − 1) 2 ε sin = sin x2 2 2(1 + cosh a1 )
(and cyclically for the three tips of the triangle). The nonvanishing of ε is due to the curvature of the space and it is a measure of deviation from the Euclidean geometry. Note that ε = 0 implies that a spherical or hyperbolic triangle does not “close” in flat Euclidean geometry.
30
4 FourVelocity
Fig. 4.3 The triangle in spherical Euclidean geometry
4.19. Consider in spacetime the timelike straight line2 z 1 i (τ ) = a1 i + u 1 i τ , where a1 i , u 1 i are constant fourvectors, u 1 i u 1i = −1, and τ is an affine parameter along the line. Let P be a spacetime point outside the straight line. As it is well known at every point of spacetime there exists a unique light cone with vertex at that point. Consider the light cone with vertex at P and let R and A be the intersection points of this light cone with the straight line z 1 i (τ A > τ R ). (a) Calculate the values τ R , τ A . (b) Consider the point Q of z 1 i defined for the parameter value τaver. = 12 (τ R + τ A ) and show that Q P i is perpendicular to z 1 i ; therefore, its length defines the minimum distance of P from the straight line z 1 i . (c) Let z 2 i (τ2 ) = a1 i + u 2 i τ2 (u 2 i u 2i = −1) be another timelike straight line which intersects z 1 i (τ ) at the point a1 i . Assume that P is the point of z 2 i defined by the parameter value τ2,P , that is, z P i = a1 i + u 2 i τ2,P . Compute the normal P Q i and its length in terms of the fourvectors u 1 i , u 2 i and the value τ P . Use these results to the case that z 1 i , z 2 i are the world lines of two relativistic inertial observers (RIO) with fourvelocities u 1 i , u 2 i , respectively, to show that the measure of the relative velocity of these RIO is defined covariantly and in an intrinsic geometric manner.3
2
The results are general; therefore, we write gi j instead of ηi j .
3
See A. Aurilia and F. Rohrlich Am. J. Phys. (1975) 43, 261–264.
Chapter 5
FourAcceleration
5.1. A relativistic mass point (ReMaP) P is moving along the xaxis of the LCF Σ. (a) Assuming that P moves in an arbitrary manner compute the fourvelocity and the fouracceleration of P. (b) Consider the initial conditions v(0) = 0, x(0) = x0 and calculate the motion for the case of hyperbolic motion (i.e., a+ = constant). (c) Assume a + = 2ckτ (k positive constant) and compute the motion of P. 5.2. A relativistic mass point P moves so that its position vector x i is related with its fouracceleration a i with the relation a i = k 2 x i (k > 0). (a) Show that during the motion x i is at all times normal to the fourvelocity vector u i (this “motion” in spacetime is the analogue of uniform circular motion). (b) Calculate the equation of motion of P. (c) Compute the motion of P in an LCF Σ if P starts from rest from the point x0 of xaxis. 5.3. The world line of a relativistic mass point P is described by the equations x 2 − c2 t 2 = a 2 , y = z = 0.
(a) Show that a parametric form of the world line is ct = a sinh φ,
x = a cosh φ,
y = z = 0.
What is the kinematic meaning of the parameter φ and how it is related with the proper time of P? (b) Compute the fourvelocity and the fouracceleration of P and find how the later is related to the position fourvector of P. Comment on the result.
31
32
5 FourAcceleration
(c) If the mass of P is m show that the necessary (inertial) fourforce to sustain the motion of P is F i = am2 x i . Does the energy of P varies during its motion (c = 1)? 5.4. The relativistic point P moves in spacetime so that its fourvelocity u i and the i = a 2 u i (aε R). fouracceleration a i are related by da dτ (a) Show that a 2 is the length of the fouracceleration. (b) Consider an arbitrary LCF Σ in which P has threevelocity u and threeacceleration a and show that in Σ the covariant equation of motion is equivalent to the following two equations: (γ γ˙ ) = a 2 ,
(γ 3 a)· = 0.
Solve the equations of motion in case the motion is taking place along the xaxis. 5.5. A spaceship departs from the origin of the LCF Σ and moves along the xaxis with constant proper acceleration a + = 10 m/s2 until it attains the speed v1 = 0.5 c. Calculate the time the√spaceship moved in Σ and in its proper frame. Repeat the calculations for v1 = 0.9999 c and a + = 20 m/s2 . Assume c = 3 × 108 m/s and that the year has 365 days. 5.6. A relativistic mass particle P moves along the xaxis of the LCF Σ with constant proper acceleration a + . (a) Show that the acceleration of P in Σ is not constant and it is given by the relation γu3 ax = a + , where u is the instantaneous speed of the particle in Σ. (b) An elastic string brakes when its proper frame is elongated to a length equal twice its natural length. Assume that the string is resting along the xaxis of the LCF Σ along its natural length when all its points start accelerating with the same proper acceleration a + . Calculate (in Σ) the time required for the string to brake. 5.7. In the LCF Σ a particle moves uniformly along a circular trajectory of radius R with period T. Calculate the period and the distance covered for one complete rotation in the proper frame of the particle. Application In a space platform an astronaut starts a journey along a circular path of radius R with uniform speed v. When the astronaut returns back to the basis will he be older or younger from another astronaut who stayed in the basis during the trip? 5.8. (a) The fouracceleration of a particle P whose threevelocity and three a0 c i acceleration in the LCF Σ are v, a, respectively, is given by a = a0 v + γ 2 a
5 FourAcceleration
33
where a0 = γ 2 + v·a . Let a+ be the proper acceleration of P. Show the relations. c2 a =
1 + a , γ3
a⊥ =
1 + a , γ2 ⊥
where the parallel and the perpendicular parts are wrt the direction of the threevelocity v. (b) A particle P of charge q and mass m is accelerating in the laboratory along the xaxis under the action of the constant electric field E = Ei. Calculate the motion of P in the lab frame assuming the initial conditions v(0) = 0, x(0) = qcmE . (c) The same particle P enters in the laboratory with speed u normal to a constant magnetic field B. Calculate the motion of the P in the lab frame. It is given that the electromagnetic field in two LCF Σ, Σ whose relative velocity is v transforms as follows: E = E 1 E⊥ = γ E + v × B c ⊥
B = B 1 E⊥ = γ B − v × E . c ⊥
5.9. The LCF Σ, Σ have their spatial axes parallel and the velocity of Σ wrt Σ is u. (a) Show that the transformation of the threeacceleration of a particle P in Σ and Σ is given by the relation
a =
1
γu2 1 −
u·v 3 c2
u · v u·a u·a 1− 2 a+ 2 v+ 2 c c u
1 −1 u . γu
Write this relation in case Σ, Σ are related with a boost along the xaxis with speed u. (b) Show that the parallel and the normal projections a and a⊥ of the threeacceleration of P relative to the direction of the velocity u are transformed as follows: 1 3 a , 1 − u·v 2 c 1 1 a a⊥ = + u × (a × v) . 3 ⊥ c2 γu2 1 − u·v 2 c a =
γu2
5.10. The fouracceleration of a particle P in an LCF Σ in which the threevelocity of the particle are v and a, respectively, is and the threeacceleration ca0 dγ i where a0 = γ dt . a = a0 v + γ 2 a Σ
34
5 FourAcceleration
(a) Prove that a0 = c12 γ 4 v·a. Show & & that if&the& speed of P is constant then the proper &. Also show that if the velocity of P in acceleration a+ has length &a+ & = γ 2 & dv dt & & & &2 & . Σ does not change direction and changes only the speed, then &a+ & = γ 6 & dv dt (b) Show that in general + 2 " # a = γ 6 a6 − (a × β)2 . 5.11. Two rockets are accelerating along the xaxis of the LCF Σ in opposite directions with constant acceleration a. The equation of their world lines are x1 (τ1 ) = −
c2 c2 cosh ψ1 , ct1 = sinh ψ1 a a
and x2 (τ2 ) = − where ψi =
aτi c
,
c2 c2 cosh ψ2 , ct2 = sinh ψ2 , a a
i = 1, 2, is the rapidity of each rocket.
(a) Draw a spacetime diagram of the world lines of the two rockets. (b) Define the fourvector Z i (τ ) = x2i (τ ) − x1i (−τ ), where xii (τ ) is the position vector of the i = 1, 2 rocket at its proper time τ. The fourvector Z i defines a synchronization of the proper times of the rockets (recall that a synchronization between two observers is a 1 to 1 correspondence of their proper time). Prove that – The (Lorentz) length Z i (τ )Z i (τ ) is independent of τ. Explain the geometric significance of this result. – Z i (τ ) is normal to the fourvelocity of rocket 1 at the event x1i (−τ ) and to the fourvelocity of rocket 2 at the event x2i (τ ).
Chapter 6
ThreeMomentum–Energy
6.1 Problems on Energy and ThreeMomentum 6.1.1. In the SI system the charge of the electron equals 1.6021 × 10−19 Cb and its mass 9.1 × 10−28 g. Calculate the mass of the electron in natural units. 6.1.2. Calculate the speed of a particle in an LCF Σ in which (a) The kinetic energy of the particle equals its internal energy. What happens when the particle is a photon? (b) The kinetic energy of the particle equals its mass. 6.1.3. A particle P of mass m in the LCF Σ has energy E and threemomentum p. (a) Show that the velocity of the particle both in Special Relativity and in Newtonian Physics is given by the formula u = ddpE . Is this relation valid for photons? (b) Show that dp = vγ dm + md(γ v), d E = mc2 dγ + γ c2 dm, and conclude that the change in energy and in threemomentum may be due to either a change in γ (equivalently the speed of P) or a change in the mass independent of the change in velocity, e.g., heating of the particle. Note: This implies that there are two types of forces in Special Relativity: – The inertial forces which cause change in velocity and – The (socalled) pure forces, which cause change of the internal energy of the particles only. It is emphasized that change in the mass does not mean that the mass is not invariant. It is another matter to be constant and another to be invariant! 35
36
6 ThreeMomentum–Energy
6.1.4. Show that if in a given motion the threemomentum of a particle in an LCF Σ changes only in direction, then the energy is conserved in Σ. Examine the converse. 6.1.5. In the LCF Σ a particle of mass m is displaced by dr under the action of a 3force f. (a) Show that the work done by f in Σ is dW = d(mc2 γ ). (b) Show that if the change in the kinetic energy T1→2 of a particle in a LCF Σ 'γ between the states 1 and 2 is T1→2 = γ12 dW , then T1→2 = mc2 (γ2 − γ1 ). Conclude that it is required infinite energy for a particle to reach speed c. 6.1.6. (a) Show that if the temporal components of two fourvectors are equal in all LCF, then the fourvectors are equal. (We have shown in another exercise that the same holds for the spatial components.) (b) Apply (a) to the fourmomentum vector and show that in Special Relativity the conservation of energy implies the conservation of threemomentum and the conservation of threemomentum implies the conservation of energy. This result means that – The two conservation laws of Newtonian Physics are contained in the law of conservation of fourmomentum of Special Relativity – In Special Relativity the conservation energy is bound by the conservation of threemomentum and conversely. 6.1.7. Prove that (a) If the spatial parts of two fourvectors are equal in all LCF, then the fourvectors are equal. (b) If the threevelocities of two particles of different masses are equal in one LCF, then they are equal in all LCF. (c) If the fourmomentum of two particles of different masses are equal in one LCF, then they are not equal in any other LCF. 6.1.8. The world line of a particle of energy E in the LCF Σ is described by the equations ( x=
( r cos θ cos φdλ,
y=
( r cos θ sin φdλ,
z=
( r sin θ dλ,
t=
r dλ,
where r, θ, φ are (smooth) functions of the parameter λ. Determine the type of the particle and compute its fourmomentum.
6.2
Relative ThreeMomentum – Colliding Beams
37
6.1.9. Assuming the boost along the xaxis show that the quantity that is,
d3p E
is invariant,
d 3p d 3 p . = E E [The quantity
d3p E
is the invariant volume element in the space of threemomenta].
6.2 Relative ThreeMomentum – Colliding Beams 6.2.1. Two particles 1, 2 of masses m 1 and m 2 , respectively, are connected with a spring of constant k and negligible mass. The two particles are resting on a smooth plane when a photon is emitted from particle 1 toward particle 2 where it is absorbed. Due to this the system of the two masses starts to oscillate with simple harmonic motion. Assuming that the emittance of the photon causes a reduction Δm in the mass of 1 and that the time of traveling of the photon from one particle to the other is negligible compared to the period of oscillation, show that the maximal extension 1 √ + m12 . of the spring is Δx = cΔm m1 k Examine Δx for the limiting values of the constant k.
6.2.2. The linear accelerator of electrons in Stanford has length 3 km and accelerates electrons to a final energy of E = 20 GeV. Given that the energy of the electrons in the accelerator is a linear function of the distance they covered, find the total length of the accelerator “seen” by the electrons (m e = 0.511 MeV). 6.2.3. (a) Show that in Special Relativity the threevelocity v of a relativistic particle P with threemomentum p and energy E (all quantities in the same LCF Σ) 2 is given by the relation v = pcE . (b) Two relativistic mass points 1, 2 have fourvelocities u i1 , u i2 and fourmomentum p1i , p1i , respectively. Define the relative energy E 12 and the relative threemomentum p12 of 1 wrt 2 with the components of the fourvector p1i in the proper frame of the fourvector p2i , that is, p1i
=
E 12 /c p12
Σ+ 2
and show that the relative velocity of 1 wrt 2 is given by the relation v12 = Ep1212 in accordance with part (a) of the problem. Calculate the velocity v21 of 2 wrt 1 and show that v12  = v21  and, in general, v21 = Ep2121 = v12 .
38
6 ThreeMomentum–Energy
6.2.4. In DESY (Hamburg) operates the accelerator HERA, which produces two headon colliding beams of electrons and protons of energy 30 GeV/c2 and 800 GeV/c2 , respectively. Calculate (a) The available energy in the CM of the beams. (b) The speed of the CM in the laboratory frame. (c) The threemomentum of the beams in the CM frame. It is given that m p = 0.94 Gev/c2 , m e = 0.5 Mev/c2 . 6.2.5. In the ring of colliding beams I.S.R. at CERN two beams of protons each of threemomentum p = 25 GeV/c, collide at an angle 15◦ as in Fig. 6.1. Calculate the energy of one beam in the rest frame of the other, assuming that the mass of the proton is 1 Gev/c2 .
Fig. 6.1 Colliding beams
6.2.6. A particle of mass m 1 = 1 Gev and kinetic energy T1 = 25 GeV collides with another particle of mass m 2 which rests in the laboratory. (a) Calculate the available energy for the production of particles if the mass m 2 takes the values 1.5 and 100 GeV. (b) If the value of the available energies is smaller than T1 , explain the deficit in the energy balance. (c) Compute the maximal value of the available energy for this configuration. 6.2.7. Consider two photons 1, 2 which in the LCF Σ are propagating along the directions eˆ 1 , eˆ 2 with energies E 1 , E 2 , respectively. (a) Prove that there are infinite LCF Σ in which the two photons have the same energy. (b) Assume that Σ has parallel axes with Σ and calculate the common energy E of the photons in Σ as well as the βfactor of Σ wrt Σ in the following cases: (b1) The photons are propagating – In Σ in directions which coincide with those in Σ. – In Σ in opposite directions.
6.3
Scattering Densities
39
(b2) – The relative direction of propagation of the photons in Σ is along the positive direction of the bisector in the first quarter of the x–y plane of Σ. – The photons propagate along the positive directions along the x, y axes of Σ. 6.2.8. (a) Show that if two photons are moving along parallel orbits in one LCF they are moving in parallel orbits in all LCF. Based on this result conclude that if two photons are moving along parallel orbits in one LCF their fourmomenta are parallel. (b) A beam of similar photons is observed in two LCF Σ, Σ which are moving in the standard configuration along the common xaxis with speed factor β. If the densities of the beam (= number of photons per unit spatial volume) and the frequencies of the photons in Σ, Σ are ρ, ρ and ν, ν , respectively, show that if the beam propagates along the yaxis of Σ the following relation holds: ρ : ρ = ν : ν. Compute the relation between the densities when the beam propagates along the common xaxis.
6.3 Scattering Densities 6.3.1. A mass dm of proper volume d V0 is moving with velocities v and v in the the mass densities and ρ E = dd VE , LCF Σ and Σ , respectively. Let ρ = ddmV , ρ = ddm V d E ρ E = d V the energy densities of the mass dm in Σ and Σ , respectively. Calculate the relation between the quantities (a) d V and d V , (b) ρ and ρ , (c) ρ E and ρ E . 6.3.2. A beam of atoms of radius R1 is moving along a fixed direction in the laboratory with speed u when enters a region in which there is a gas consisting of atoms of radius R2 . (a) If in the gas there are k atoms per unit of volume and the gas is assumed to be at rest in the laboratory, calculate the percentage of atoms of the beam which are scattered in time t in the laboratory frame. (b) If the atoms of the gas are moving in the laboratory in the opposite direction of the beam with speed V, compute again the percentage of atoms of the beam which are scattered in time t in the laboratory frame assuming that the proper density of the gas is k. (Hint: Reduce all quantities in the lab frame.)
40
6 ThreeMomentum–Energy
6.4 Reflection of Light on Mirror 6.4.1. A plane mirror K is moving in the LCF Σ with constant speed β and in a direction which is (a) Parallel to its plane. (b) Normal to its plane. A photon of frequency ν is reflected on the mirror with angle of incidence θ (see Fig. 6.2). Compute the angle of reflection and the frequency of the reflected photon in both cases of motion of the mirror. Examine if there is a speed of the mirror for which the frequency of the reflected photon remains unchanged in Σ. Assume c = 1.
Fig. 6.2 Reflection on a moving mirror
6.4.2. A particle is moving in the x–y plane of an LCF Σ so that its velocity makes an angle θ with the positive semiaxis x. The particle is reflected elastically on a smooth wall which moves perpendicularly to the xaxis with speed u. Calculate the reflection angle of the particle in Σ. 6.4.3. In the LCF Σ(x, y, z) a mirror moves so that its plane remains always parallel to the y–z plane and its velocity is u = u xˆ . A photon of frequency ν0 in Σ moves along the direction √12 (ˆx + yˆ ) when it is reflected on the mirror. Determine the frequency and the direction of motion of the reflected photon in Σ (a) in Newtonian Physics and (b) in Special Relativity.
Dynamics: Theory of Chapters 7–11
Consider a relativistic mass particle (ReMaP) P of mass m which has a fourvelocity u i . The fourmomentum of P is the fourvector pi = mu i . It has length pi pi = γc and the fourmomentum −mc2 . In an LCF Σ the fourvelocity of P is u i = γv Σ E/c , where E = mγ c2 is the total energy (i.e., kinetic plus “potential”) pi = p Σ of P and p = mv is the threemomentum of P, both in the LCF Σ. The length of pi implies the relation E=
p2 c2 + m 2 c4 .
i i i , p(2) , . . . , p(n) be the fourmomenta of n ReMaPs. The center of momenLet p(1) tum (CM) particle is the (virtual) ReMaP whose fourmomentum is
P = i
n )
i p(A)
=
A=1
* n )
E (A) ,
A=1
n )
+ p(A)
A=1
and its mass is M = − c12 P i Pi . The B and Γ factors of the CM particle are given by the relations ,n ,n A=1 p(A) A=1 E (A) , Γ= . B = ,n Mc2 A=1 E (A) The fourforce F i on a ReMaP P with proper time τ and fourmomentum pi is defined by the relation Fi =
dpi = ma i , dτ
where the second equality applies whenthe mass m of P is constant. In an LCF Σ ca0 i in which P has fouracceleration a = , the four  force F i = ma i a0 v + γ 2 a Σ 41
42
Dynamics: Theory of Chapters 7–11
has decomposition Fi = γ
1 c
f·v f
Σ
,
where f is the threeforce applied to P according to Σ. In an LCF Σ the equations of motion of a ReMaP on which a threeforce f (in Σ!) is applied are dE mc2 3 dβ 2 = mcγ 3 β · a = γ , dt 2 dt dp = mγ 3 (β · a)β + mγ ca. dt In the collision or interaction of a set of ReMaP, the total fourmomentum remains the same (is conserved). The number of the particles is not necessarily conserved and when this is the case we say that the collision is elastic. We say that a reaction occurs at the threshold if in the proper frame of the CM particle (the CM frame) the sum of threemomenta of all the interacting particles equals zero. If three particles with masses m 1 , m 2 , m 3 are involved, the condition for threshold is that the triangle function λ2 (m 1 , m 2 , m 3 ) = (m 21 + m 22 − m 23 )2 − 4m 21 m 22 = 0.
Chapter 7
Relativistic Optics
7.1. Two photons with energies E 1 and E 2 = k E 1 , k < 1, propagate along different directions along the xaxis of an LCF Σ. (a) Calculate the βfactor of another LCF Σ which is moving in the standard configuration wrt Σ along the xaxis in which the two photons have the same frequency. Calculate the common energy of the photons in Σ . (b) An observer is moving with constant speed along the xaxis receding from a fixed light source which emits blue light of frequency ν = 7 × 5 × 1013 Hz and approaching another fixed source which emits red light of frequency ν = 5 × 1013 Hz. Calculate the speed for which the observer sees the light from the two sources to have the same color. Calculate the frequency of the common color. As an application of the above, discuss the following methodology for the determination of the color of a galaxy. Suppose we know the color of the galaxy A which recedes (reference galaxy) and we wish to determine the color of a galaxy B which approaches. According to the above we speed up along the line joining the two galaxies until the colors of the two galaxies appear to be the same. Then we measure our speed with a speedometer and determine the color of the approaching galaxy. Is it possible to develop with electronic methods an artificial Doppler effect which will replace the galaxy A? 7.2. Photons 1, 2 of frequency ν1 , ν2 , respectively, are propagating in the LCF Σ in directions which make an angle θ . (a) Calculate the velocity β of the CM of the photons in Σ. (b) Calculate the velocity β of the CM when the photons propagate along the positive and negative directions of the axes x, y, respectively. What happens in this case if ν1 = ν2 ? (c) Repeat the same if the photons move along the xaxis in opposite directions. 43
44
7 Relativistic Optics
7.3. In the LCF Σ the ReMaP P has fourmomentum pi = mc(20, 8, 6, 0)t where m is a constant. (a) Determine the mass of P. (b) Consider the LCF Σ which moves wrt Σ in the standard configuration with speed u = 0.6c. Determine the energy and the direction of motion of P in Σ . 7.4. (a) Show that two null fourvectors are parallel if and only if their space parts are parallel. (b) In the LCF Σ two photons 1, 2 whose frequencies are ν1 , ν2 (in Σ!), respectively, are emitted from the points A, B along the xaxis of Σ parallel to the yaxis. Compute the relative velocity and the fourmomenta of the photons in an LCF Σ which moves wrt Σ in the standard way along the common x, x axis with speed u. 7.5. The observers 1, 2 with fourvelocities u 1 , u 2 are observing a photon with fourmomentum P i . (a) Show that the ratio ν1 :ν2 of the frequencies observed by the observers in their proper frame equals u 1 P:u 2 P. (b) If u is the velocity of 1 wrt 2 use (a) to compute the Doppler formula assuming that in the proper frame of 2 the angle between u and the spatial propagation of the photon is a. 7.6. During the study of the spectrum of the light from a distant galaxy, it is found ˚ corresponds to the line of the spectrum that the spectral line with wavelength 7300 A ˚ If we assume that the of hydrogen, which in the laboratory has wavelength 4870 A. galaxy is receding along the radial direction and furthermore that the observed shift in the spectrum is only due to the relativistic Doppler effect, calculate the speed of the receding galaxy from the Earth. From other methods we know that the distance of the receding galaxy from our galaxy (the Earth) is 5 million light years. Assuming that at the Big Bang the two galaxies were created simultaneously and since then the foreign galaxy recedes with the same speed, calculate the time of the Big Bang in the Earth and in the proper frame of the receding galaxy. 7.7. In order to compute the receding velocity of a galaxy one measures the wavelength of a line in the spectrum of the light emitted by the galaxy and compares it with the wavelength of the same line in the spectrum of the light in the laboratory. ˚ on the Earth It has been found that the spectral line of Na with wavelength 5890 A ˚ in the spectrum of the galaxy. Determine the speed at which has wavelength 6100 A the galaxy is receding from the Earth. 7.8. A light source is placed at some point along the positive semiaxis x of one LCF Σ and emits monochromatic radiation of frequency ν0 toward the origin of Σ.
7 Relativistic Optics
45
This radiation is observed by another observer Σ who passes through the origin of Σ with a velocity in the plane xy at an angle θ with the positive direction of the axis x. (a) Determine the frequency ν and the direction of the radiation measured by Σ . (b) Determine the angle θ so that the frequency measured by Σ is ν0 . Is this angle uniquely determined? 7.9. A rocket moves with speed u along the xaxis of an LCF Σ. In order the observer in the rocket to measure the speed of the rocket in Σ emits monochromatic radiation of frequency ν0 which is reflected at the two ends of the rocket and returns to the observer. If the returned radiation in Σ has frequency ν¯ and arrives at the observer with a time difference ΔT, compute the speed and the length L of the rocket. The rocket is assumed to be a solid, that is, all its points have the same speed in Σ. 7.10. (a) Consider a photon with fourmomentum f i and an observer with fourvelocity u i and show that the frequency ν measured by the observer is given by the relation ν=−
1 i f ui . h
(b) An emitter E and an receiver R of photons are fixed in a centrifugal machine at a distance r from the axis of rotation and at an angle a, both r and a measured when the machine is resting in the laboratory. The machine starts rotating with constant angular velocity ω as measured in the laboratory. Calculate the redshift z of the photon, which is defined by the relation z=
λR − λE , λE
where λ E is the wavelength of the emitted photon and λ R is the wavelength of the absorbed photon. 7.11. Consider a photon with fourmomentum pi and define the frequency fourvector f i of the photon as follows: pi = hc f i ( f i is a fourvector because both h, c are universal constants, hence invariants). Calculate the components of f i in an LCF in which the photon has energy E and it is propagating along the threedirection with unit vector eˆ : (a) In terms of the frequency of the photon. (b) In terms of the wave vector k which is defined with the relation k = ν eˆ , where ν is the frequency of the photon in Σ, when the photon is considered as an electromagnetic wave.
46
7 Relativistic Optics
Consider a mass particle with fourmomentum pi and define with the same formula the fourvector of frequency f i of the particle. Compute f i in an LCF Σ in terms of the energy and the threemomentum of the the particle in Σ. Finally, compute the threevelocity of the particle in terms of the wave vector (de Broglie wave vector) of the particle. Comment on the results. 7.12. The usual geometric representation of the boost is in terms of the hyperbolic angles in the Euclidean plane via the introduction of the rapidity ψε R with the relations β = tanh ψ. In order to define a geometric representation of the boost in terms of the standard trigonometric functions, we define β = sin θ (−1 ≤ β ≤ 1, −π/2 ≤ θ ≤ π ). (a) Write the boost in terms of the angle θ instead of the parameter β. (b) Express the components of the fourvelocity in terms of θ and making use of a trigonometric identity show that u i u i = −c2 . Calculate the components of fourmomentum and making use of the same trigonometric identity show that E 2 = p 2 c2 + m 2 c4 . (c) Consider a particle with energy E and threemomentum p which moves along a straight line with constant velocity in the LCF Σ. Define for the particle a frequency and one wavelength with the de Broglie relations E = hν, p = h/λ. 2 h , ν0 = mch Subsequently show that λ = λ0 cot θ, ν = ν0 / cos θ where λ0 = mc is the wavelength and the Compton frequency of the particle. Show that the phase speed of the particle u ph = λν equals u ph = sinc θ which dν implies u ph > c. Define the group velocity of the particle u gr = d(1/λ) and show that it equals the velocity of the particle in Σ. Also show that u gr u ph = c2 and explain the result. Finally, making use of the trigonometric identity sec2 θ − tan2 θ = 1 shows that ν 2 = c2 (k02 + k 2 ) (relativistic dispersion formula of de Broglie waves), where k = 1/λ is the wave number of the de Broglie waves (see V. Majernik (1986), Am. J. Phys. 54, 536–538). 7.13. (a) Prove that necessary and sufficient condition for the quotient of the components of two tensorial fourquantities to be (Lorentz) invariant is that the fourquantities are fourvectors. (b) Application The wave nature of matter is expressed with the frequency fourvector, whereas the particle nature of matter is expressed with the fourmomentum. The dual perspective of the wave and the particle nature of matter is expressed by the statement: The frequency fourvector and the fourmomentum are parallel. The zeroth component of the fourmomentum is the energy E of the particle while the corresponding component of the frequency fourvector is the fre
7 Relativistic Optics
47
quency ν of the wave nature. These physical quantities are related by Plank relation E = hν, where h is a universal constant (therefore, invariant in all theories of physics, whatever the spacetime symmetry group). Use these data and part (a) to produce the de Broglie relation which relates the threemomentum p with the wave vector k of a particle. Conclude that the dual nature of relativistic matter is inherent in Special Relativity and does not hold in Newtonian Physics (the electromagnetic field is not invariant under the Galileo group) (see R. Newburgh (1956), Lett. Nuovo Cimento. 29, 195–196; P. Dirac (1924) Proc. Cambridge Philos. Soc. 22, 432).
Chapter 8
FourForce
8.1. In the LCF Σ two mass points 1, 2 are in contact exerting forces of action– reaction f12 and f21 , respectively, to one another. Prove the third Newton’s law f12 = −f21 (Law of action–reaction) assuming that the masses of the two particles remain constant. 8.2. Write the equations of motion of a particle of mass m in the LCF Σ in which the threeforce f is applied on the particle. Write these equations in the following special cases: (a) The threevelocity of the particle is collinear with the threeforce (⇔ the direction of the velocity is constant). (b) The threeforce is always normal to the threevelocity (⇔ the speed is constant). (c) The speed is c. 8.3. A particle of mass m is moving along the xaxis of the LCF Σ under the action 2 of the attractive force 2mc . If at t = 0, x = 2 show that the motion of the particle in x2 . Σ is simple harmonic motion with period 4π c 8.4. A particle of mass m is moving along the xaxis of the LCF Σ under the action of the force: f=−
mc3 ω2 x i, (c2 − ω2 a 2 + ω2 x 2 )3/2
where ω, a are positive constants. If at t = 0 the particle is at the origin of Σ and has velocity v = ωai, show that the speed satisfies the relation v 2 = ω2 (a 2 − x 2 ). What this implies for the motion of the particle? 8.5. A particle of mass m is resting at the position x = a (a > 0) in the LCF Σ when the threeforce f = −mω2 xi is applied to the particle. Show that the speed on √ 2 +ω2 a 2 . the particle when passes the origin equals ωac2c24c +ω2 a 2 49
50
8 FourForce
8.6. A particle of mass m moves with speed υ along the xaxis of the LCF Σ when . Show that the time interval required in Σ in order that the it suffers the reaction mv k # " 5 speed of the particle will be reduced from 45 c to 35 c is ln 32 + 12 k. 8.7. Assuming Special Relativity holds for the gravitational field near the Earth (which is not true!), calculate the “corrections” in the flight time, the range, the maximal height, and the angle of firing of a projectile as it is seen by an observer moving with velocity u = ui. 8.8. A ReMaP P moves in the LCF Σ with velocity v under the action of the threeforce f. (a) Show that the analysis of the fourforce on P in Σ are the following: F i = γ (v)
f · v/c f
Σ
.
Give the physical significance of the zeroth component. (b) Prove that F i Vi = 0 where V i is the fourvelocity of P. (c) Consider another LCF Σ which moves wrt Σ in the standard configuration with speed u and compute the components of the fourforce in Σ . (d) Show that a central threeforce f = −kr in Σ is not, in general, transformed to a central threeforce in Σ . 8.9. (a) Using the identity A×(B × C) = (A · C)B − (A · B)C, show that A=
1 B×(B × A) (A · B)B − . B2 B2
(8.1)
This relation decomposes a vector A parallel and normal to a vector B. (b) Let Σ, Σ be two LCF with parallel axes andrelative 0 β. A fourvector velocity 0 A A , . Show that Ai in Σ, Σ is decomposed as follows: Ai = A Σ A Σ A = γ (A − A0 β) + (γ − 1)
β×(β × A) . β2
This relation decomposes A parallel and normal to the relative velocity β of the LCF Σ, Σ . (c) Apply the above decomposition to the fourvector of fourforce and show that the following relations hold:
8 FourForce
51
1 (v − u) · f, Q 1 u×(v × u) f⊥ + f − · f e , f = γu Q Quc
f · v =
where Q = 1− u·v . Comment on the physical meaning of the time component of c2 the fourforce. Concerning the spatial part of the fourforce, show that provided that f ⊥ = 0 this can always be written in the form of a Lorentz force and calculate the “charge,” the “electric field,” and the “magnetic field” due to this force. 8.10. In the LCF Σ the mass point P with mass m moves along a circular orbit of radius R with constant speed u. Calculate the force exerted on m. 8.11. A particle P of mass m and charge e is moving in the field of another particle of charge q of opposite sign (so that we have attraction) fixed at the origin of the LCF Σ. (a) Show that the equation of motion of P in Σ is kr d (γ u) = − 3 , dt r d k 2 (γ c ) = − 3 (u − r), dt r where k = − eq > 0. m (b) Prove that motion is planar. (c) Use polar coordinates (r, θ ) on the plane of motion to show that γ r 2 θ = A = const. and γ c2 rk = w =const. (d) Introduce the coordinate u = r1 > 0 and eliminating the time, show that the equation of motion becomes wk u2 d 2u + 1 − 2 2 u = 2 2. dθ 2 A c A c 2 Subsequently define the quantity p = 1 − Ak2 c2 and show that the orbit of P in Σ can be written 2 1 w c kw 2 1 , =u= cos + c) + − p (ρθ r Ap 2 c4 2 Ac2 where C is a constant. Deduce that the condition for a bound trajectory is −c2 < w < c2 and that the bound trajectory can be described with an ellipse whose one focus is the origin of Σ and precession of the perihelion equals
52
8 FourForce
2π 1p − 1 per rotation. Finally, show that for small velocities this quantity can be approximated as follows: π k2 4π 2 α 2 = , A2 c2 τ 2 e2 (1 − a 2 ) where α is the length of the small major semiaxis, τ is the period of the orbit, and e is the eccentricity of the orbit. 8.12 In the LCF Σ the Lagrangian of a free particle of speed u is defined by L 0,Σ = −αc where α is a quantity which is computed from the requirement that the action t2',Σ L 0,Σ dtΣ is invariant. Show that α = − mγ . integral S = t1 ,Σ
Chapter 9
Plane Waves
9.1. The equation of an electromagnetic wave which is propagating along the direction eˆ = cos αi + sin αj of the x–y plane of the LCF Σ is Ωi = Ai exp
2πiν (x cos α + y sin α − ct) , c
where ν is the frequency of the wave in Σ. Compute the frequency and the angle of propagation of the wave with the x axis in the LCF Σ which moves wrt Σ in the standard configuration with speed u. Consider c = 1.
53
Chapter 10
Relativistic Reactions
10.1 Center of Momentum Energy 10.1.1. (a) Calculate the available energy in the CM frame in the following experiments of proton to proton collisions. In the first experiment a beam of protons is accelerated to energy E = 30 GeV and hits a resting target (e.g., liquid hydrogen). In the second experiment two beams of protons are accelerated to energy E = 15 GeV and then they come onto headon collision. (b) At which energy in the first experiment the bullet beam must be accelerated in order the available energy to equal that of the second experiment? The mass m p = 0.94 GeV/c2 . 10.1.2. In Newtonian Physics the kinetic energy TΣ of a system of particles in the frame Σ is related to the kinetic energy T ∗ of the system in the CM frame as follows: T = T∗ +
1 MV2 , 2
where M is the sum of the masses of the particles and V is the velocity of the CM frame in Σ. This result is the Theorem of Kinetic Energy in Newtonian Physics. Calculate the corresponding relation in Special Relativity and show that for small speeds the relativistic formula reduces to the Newtonian one. 10.1.3. A particle of mass m and kinetic energy E collides with a similar particle which rests in the laboratory. √ Show that if E m, then the maximum available energy in the CM frame is 2Em.
10.2 Elastic Collisions 10.2.1. A smooth elastic sphere 1 which moves with velocity u in the laboratory collides centrally with another similar sphere 2, which rests in the laboratory. After the collision the two spheres 1, 2 move in directions which make angles θ, φ with the line of collision, respectively. Calculate the product tan θ tan φ in terms of the 55
56
10 Relativistic Reactions
velocity of the CM of the system of spheres in the laboratory. Find the Newtonian limit of the result. 10.2.2. A beam of π mesons is scattered elastically on a target of protons which are at rest in the laboratory. Calculate the energy of a π meson in order the corresponding scattered π meson to have momentum 0.8 GeV /c in a direction normal to the direction of the falling π meson. Calculate the maximum and the minimum energy of the scattered π meson for scattering angle 90◦ in terms of the energy of the falling beam in the laboratory. The masses: m π = 0.14 GeV/c2 , m p = 1 GeV/c2 . 10.2.3. A photon of energy E γ collides head on with an electron with momentum pe . As a result of the collision the electron is scattered at an angle θ to its direction of motion prior to collision while the photon attains energy E γ (all quantities refer in the laboratory frame L). Calculate E γ in terms of the other quantities. Derive the expression of E γ when the electron is at rest in the laboratory. 10.2.4. A photon of frequency ν1 is scattered on a free particle of mass m which is at rest in the laboratory (Compton scattering). If the scattering angle of the photon in the laboratory is θ, show that the new frequency ν2 of the photon in the laboratory is given by the relation 1 θ 1 2h − = sin2 . 2 ν2 ν1 mc 2 Calculate the kinetic energy of the particle after the scattering of the photon. If the photon is scattered backwards (θ = 180◦ ), calculate the speed of the particle in terms of its mass. 10.2.5. A photon is absorbed by an atom of mass m which is at rest in the laboratory. Due to the absorption, the atom is excited with one of its electrons changing to a higher energy level by an energy difference E (excitation energy). Calculate the frequency of the photon in terms of the excitation energy E and the mass m. 10.2.6. Compute the energy of the scattered electrons in the elastic electron – proton scattering in the proper frame of the proton, in terms of the energy of the initial electron and the scattering angle. Assume that the initial electrons have very high energies and the masses of the electron and the proton are m, M and m M.
10.3 Reactions of the Form 1 → 2 + 3 (Decays) 10.3.1. An exited nucleus of mass m which rests in the laboratory frame (L) emits a photon of energy hν. Due to the radiative transition, the mass of the excited nucleus is reduced by Δm. Calculate the energy of the emitted photon taking into consideration the rebound of the nucleus.
10.3
Reactions of the Form 1 → 2 + 3 (Decays)
57
10.3.2. An unstable particle of mass M moves along the xaxis of the LCF Σ which disintegrates into two particles 2, 3 of mass m 2 , m 3 , respectively, so that one of the particles moves in a direction perpendicular to the direction of the mother particle. Calculate the speed β1 of the mother particle in Σ in terms of the quantities mM2 , mM3 and the angle φ between the directions of motion of particles 1, 2 in Σ. Apply the result to the special case m 2 = m 3 = m, φ = 90◦ . Consider c = 1. 10.3.3. The particle 1 of mass m 1 and energy E in the laboratory frame L disintegrates giving the particles 2, 3 whose masses are m 2 , m 3 , respectively. Assuming that the direction of propagation of particle 2 with that of particle 1 in L is θ , calculate the energy and the threemomentum of the daughter particles in L in terms of the quantities E, m 1 , m 2 , cos θ . Calculate the energy and the threemomentum of particle 2 when it is emitted at right angle to the direction of motion of particle 1 (c = 1). 10.3.4. A particle 1 of mass m 1 decays in two other particles 2,3 of mass m 2 , m 3 , respectively. Determine how the decay energy ΔE = (m 1 − m 2 − m 3 )c2 is distributed (as kinetic energy) among the daughter particles 2, 3 in the proper frame of particle 1. 10.3.5. Among the products of a highenergy reaction is an unstable neutral particle which disintegrates into κ + and π − at a distance 0.9 mm from the point of creation. From relevant measurements it is found that κ + and π − have momentum 10.122 GeV/c and 1.047 GeV/c, respectively. Furthermore the two particles have velocities, which are in the same plane with the direction of motion of the neutral particle at angles 2.79◦ and 28.13◦ , respectively, above and below that direction. Calculate (a) The mass of the neutral particle. (b) The angle θ ∗ of κ + with the direction of motion of the neutral particle in the CM. (c) The lifetime of the neutral particle. It is given m κ + = 490 MeV/c2 ,
m π − = 140 MeV/c2 .
10.3.6. One method (obsolete by now) to study the reactions of elementary particles is to study their orbits in the bubble chamber in which the particles leave a visible trace of bubbles which is taken on a photographic plate. The neutral particles do not create visible orbits. (a) Justify the following statement: The bigger the energy (equivalently the threemomentum) of a charged particle in the laboratory, the bigger its orbit on the photographic plate. (b) The particles K + and μ+ are at rest in the laboratory when they disintegrate according to the reactions: K + −→ μ+ + ν
,
π + −→ μ+ + ν.
58
10 Relativistic Reactions
The images of the orbits of the particles obtained on the photographic plate for these reactions are the (a) and (b) as shown in Fig. 10.1.
Fig. 10.1 Disintegration in the bubble chamber
Determine which orbit corresponds to which reaction. The masses m K = 0.494 GeV/c2 , m π = 0.140 GeV/c2 , m μ = 0.1057 GeV/c2 are assumed to be known. (c) Do you trust your results and would you consider the method as a reliable criterion for the distinction of similar reactions? 10.3.7. A particle of mass M moves along the xaxis of the laboratory frame with speed factor β when disintegrates in two particles 1, 2 of masses m 1 , m 2 , respectively. (a) Decompose the threemomenta of the daughter particles 1, 2 perpendicular (⊥) p −p and parallel () to the xaxis and define the quantity α = p1 1 + p2 2 . Subsequently show that this quantity satisfies the relation
α−B A
2
+
p⊥ p∗
2 = 1,
where p⊥ is the common length of the perpendicular threemomenta in the laboratory and p ∗ is the common length of the threemomentum in the CM. Calculate the quantities A, B and show their geometric significance by a diagram. (b) Calculate the maximum value of p ∗ and comment on the result. Assume c = 1. 10.3.8. A π 0 moves in the laboratory with kinetic energy T when disintegrates in two photons. (a) If the photons are emitted in opposite directions in the laboratory, compute the energies of the photons in the laboratory. (b) If the photons are emitted at equal angles to the direction of motion of the pion (all three trajectories are on the same plane), calculate the angle between the directions of propagation of the photons (c = 1). Also compute the direction of motion of the photons in the CM. Application: Assume m π = 135 MeV/c2 , T = 1 GeV/c2 .
10.3
Reactions of the Form 1 → 2 + 3 (Decays)
59
10.3.9. Consider the fission A → B + C where the particles A, B, C have masses m A , m B , m C , respectively. (a) In the proper frame of A show that B has energy: EB =
m 2A + m 2B − m C2 2 c . 2m A
(b) An atom of mass M which rests in the laboratory disintegrates emitting a photon and goes to a new state with mass M − δ. Show that the energy hν of the photon is hν < δc2 . Why in the M¨osbauer effect hν = δc2 ? (c) If A moves during its disintegration, find a relation between the angle of emission of B and the energies of the particles A, B. Application The K 0 is moving in the lab frame with kinetic energy T while disintegrates into K 0 → π + + π − . Calculate the maximum energy of the K 0 in order to have production of π + at 90◦ . Assume m K 0 = 0.5 GeV, m π = 0.140 GeV. 10.3.10. Three identical particles of mass m are placed on the vertices of a canonical triangle which rest on the plane x − y of the LCF Σ. At some moment (in Σ!) each of the particles decays in two other particles of mass am and bm, respectively, where a, b are properly defined constants in the interval (0, 1) so that the particle with mass am propagates along the height of the triangle from the corresponding vertex. (a) Show that the daughter particles arrive simultaneously at the mass center of the triangle. (b) Assume that the three daughter particles interact producing a new particle of mass M. Compute M assuming that the interaction occurs at the threshold. 10.3.11. A particle 1 of mass m 1 decays in two other particles 2,3 of mass m 2 = am 1 and m 3 = bm 1 , respectively, where a, b are properly defined constants in the interval (0, 1). (a) Show that the energy and the length of the threemomentum of particle 2 in the proper frame of particle 1 are X m 1 c2 , 2 YZ 2 m 1 c, 1p = 2
2 1E
=
60
10 Relativistic Reactions
where X = 1 + a 2 − b2 , Y 2 = (1 − a)2 − b2 , Z 2 = (1 + a)2 − b2 . (b) Show that the β and the γ factors of particle 2 in the proper frame of particle 1 are 2 1β
=
YZ 2 X , γ = . X 1 2a
(c) After its creation particle 2 decays into two particles 3, 4 of mass m 3 = am 2 and m 4 = bm 2 , respectively. Subsequently the particle 3 decays in particles 5, 6 and so on. Assuming that all decays occur along the same line and that all particles 2n for n = 1, 2, . . . move in the same direction compute the energy of the particle 2n in the proper frame of particle 1. 10.3.12. The most accurate method for the determination of the mass of neutrino (which we consider to be zero) is the study of the reaction π 0 → μ + ν. Consider this reaction and show that the mass of neutrino is given by the formula: m 2ν = (m π − m μ )2 −
1 2m π μπ T, c2
where μπ T is the kinetic energy of the muon in the proper frame of the pion. Recent methods of spectroscopy of Xrays (see R. Shafer (1967) Phys. Rev. 163, 1451) have improved the accuracy of the value of mass of the pion to m π = 139.577 ± 0.013 MeV/c2 . Assuming m μ = 105.659 ± 0.002 MeV/c2 and μπ T = 4.122 ± 0.016 MeV/c2 , calculate the mass of the neutrino and determine the accuracy of the result. 10.3.13. (a) A π meson disintegrates according to the reaction π → μ + ν. Show that the kinetic energy of the emitted μ meson in the proper frame of the π meson (= CM frame) is T =
(m π − m μ )2 2 c . 2m π
Numerical application: m π = 140 MeV/c2 , m μ = 105 MeV/c2 , m ν = 0 MeV/c2 . (b) Consider threeparticle (that is timelike or null) fourvectors Ai , B i , C i such that Ai = B i + C i . Prove the identity: B i Ci =
1 [−A2 + B 2 + C 2 ], 2
where −A2 = Ai Ai , −B 2 = B i Bi , and −C 2 = C i Ci (A, B, C ≥ 0). Assuming B i to be timelike show that in the proper frame Σ B of B i the zeroth component
10.3
Reactions of the Form 1 → 2 + 3 (Decays)
61
of the fourvector C i is given from the relation BC
0
=
1 [A2 − B 2 − C 2 ]. 2A
Is there a relation between the parts (b) and (a) of the problem?
10.3.14. The excited nucleus N returns to its ground state by emitting a photon (radiative transition) as follows: N →N +γ . 1 2 3 (a) In some LCF Σ (say the lab frame) define the quantities (i = 1, 2) ui k φi = γi 1 − , c where γi = 1 −
u i2 c2
−1/2
ψi = γi
ui × k , c
(i = 1, 2 ) and show that the following relations hold: M1 φ1 = M2 φ2 = α, M1 ψ1 = M2 ψ2 = δ,
for all threemomenta and energies of the nucleus. (b) Show the validity of the following identity (i = 1, 2): γi =
1 (1 + φi2 + ψi2 ), 2φi
and from this show that the energy of the produced photon in Σ equals hν =
c2 (M 2 − M22 ). 2α 1
(c) Let ν0 be the frequency of the photon produced from the radiative transition of nucleus and E 0 the average energy of the states of the atom. Show that ν=
E0 ν0 . ac2
(d) Show that if ν1 (respectively ν2 ) is the frequency of the photon in the proper frame of the nucleus before (respectively after) the radiative transition, then ν ν1 = 2, φ1 φ2 that is, the quantity
νi φi
is an invariant.
62
10 Relativistic Reactions
10.3.15. A nucleus X in excited state is resting in the laboratory when, by means of a radiative transition, it emits a photon of energy E γ . If M is the mass of the nucleus at the ground state, calculate the mass of the excited nucleus. Is it possible to use the produced photons to excite other nuclei similar to X which are resting in the laboratory? Justify your answer with calculations.
10.3.16. For every nonnull fourvector Ai we introduce the symmetric tensor h i j (A) = ηi j − Ak1Ak Ai A j which projects normal to Ai , that is, h i j (A)A j = 0. (a) Let pi the fourmomentum of a particle and p1i the fourmomentum of another particle. Show the identity/relation j
p1 p j i j p + h ij ( p) p1 . p k pk
p1i =
This identity defines the 1+3 decomposition of the fourvector p1i wrt the fourvector pi . In this decomposition the first part corresponds to the parallel part i i and the second the perpendicular part p1⊥ . p1 j
Show that the inner product p1 p j = 12 {( pi + p1i )2 − pi pi − p1i p1i }, that is, it is expressed in terms of the lengths of the fourvectors. In order to give physical meaning to this relation, we introduce the masses of the particles by the relations pi pi = −m 2 c2 , p1i p1i = −m 21 c2 and then the inner product equals −m E 1∗ , where E 1∗ is the energy of the particle with fourmomentum p1i in the proper frame Σ∗ of the particle with fourmomentum pi . Set p2i = pi − p1i and show that E 1∗ =
m 1 + m 22 − m 21 2 c , 2m
i where p2i p2i = −m 22 c2 . Also show that the length of the normal part p1⊥ p1⊥i is given by the relation
[m 2 − (m 1 + m 2 )2 ][m 2 − (m 1 − m 2 )2 ] 2m 2 c λ(m, m 1 , m 2 ). c2 = 2m
i p1⊥ p1⊥i ≡ p1 2 c2 = E 1∗2 − m 21 c4 =
Finally, collect the results as the components of the fourvector p1i in Σ∗ as follows: ⎞ ⎛ m 1 +m 22 −m 21 ∗ c 2m E 1 /c ⎟ ⎜ = , p1i = p1 Σ∗ ⎝ [m 2 −(m +m )2 ][m 2 −(m −m )2 ] ∗ ⎠ 1 2 1 2 cˆ e 2M Σ∗ where eˆ ∗ is the unit along the threemomentum in the proper frame of pi .
10.5
Reactions of the Form 1 + 2 → 3 + 4
63
Deduce that reactions of the form 1 + 2 → 3 and 1 → 2 + 3 are determined completely in terms of five parameters. One set of such parameters is the three masses of the particles and the angles of propagation of one particle in the proper frame of the other particle. Needless to say that there are many other sets of appropriate five parameters.
10.4 Reactions of the Form 1 + 2 → 3 10.4.1. Particles 1, 2 with masses m 1 , m 2 have velocities v1 , v2 , respectively, in the LCF Σ when they collide creating a new particle of mass m 3 . Calculate the mass m 3 and show that m 3 ≥ m 1 + m 2 . Comment on the result (c = 1). 10.4.2. The particle Y (a bound state of b − b¯ quarks) has mass 9.46 GeV/c2 and it is produced during the head on collision of e+ , e− beams each of energy 4.73 GeV. If instead of head on colliding beams one uses e− as a standing target, calculate the energy of the beam of e+ in the rest frame of e− for the Y particle to be produced (m e± = 0.5 MeV/c2 , c = 1).
10.5 Reactions of the Form 1 + 2 → 3 + 4 10.5.1. In an experiment for the production of kaons and Λ hyperons according to the reaction π − + p + → k 0 + Λ0 , it has been found that the π − in the laboratory have threemomentum 2.50 GeV/c and the Λ0 have threemomentum 0.60 GeV/c and are produced in a direction making an angle of 45◦ with the direction of the π − . Calculate the threemomentum of k 0 in the laboratory and in the CM frame. It is given that m π = 140 MeV/c2 , m p = 938 MeV/c2 , m k = 498 MeV/c2 , m Λ = 1116 MeV/c2 . 10.5.2. In the study of the reaction 1+2 → 3+4 in the laboratory it has been found that particle 1 has kinetic energy T1 , particle 2 is at rest, while particle 3 is produced at 90◦ to the direction of motion of the bullet particle 1. Assuming that the masses of the particles are known, calculate the energy E 3 of particle 3 in the laboratory. What is the expression of E 3 when m 1 = m 3 = m k , m 2 = m 4 = m p ? Show that in this case when T1 increases without a limit, the energy E 3 remains finite and compute its maximum value. It is given that m k < m p . 10.5.3. An antiproton with kinetic energy 23 GeV in the laboratory falls on a proton at rest. As a result of the collision two photons are produced. If the photons move in the laboratory along the direction of the antiproton, calculate the energy of each photon and the direction of its motion:
64
10 Relativistic Reactions
(a) In the laboratory frame L (b) In the proper frame of the antiproton. The mass of the proton is m p = 1 GeV/c2 , c = 1. 10.5.4. A particle X 1 of mass m which rests in the laboratory frame L absorbs a photon of energy E. Subsequently the particle disintegrates into a particle X 2 and a photon of energy E and in a direction normal to the direction of the reacting photon. Compute the mass of the daughter particle X 2 and its threemomentum in the laboratory frame. Comment on the result. 10.5.5. Consider the reaction a + b → c + X. By definition the active mass of particle X is given from the relation m 2X = ( p1i + p2i )2 , where pai , pbi are the fourmomenta of particles 1, 2. If√we denote the total energy s, show that the maximum of the reacting particles a, b in the CM frame Σ∗ with √ value for the mass of the particle X is m X,max = c12 s − m c . 10.5.6. In an experiment concerning the reaction of protons at rest in the laboratory with antiprotons, it has been found that p + p → π + C. The study of the spectrum of the threemomenta of the charged π mesons showed the production of π mesons of a single energy and threemomentum p = 190 MeV/c in the laboratory along the direction of the motion of the antiprotons. This result suggests the existence of a bound state C. If the energy of the antiprotons is E 1 = 10 GeV, calculate the mass of the bound state C. It is given that m p = m p = 0.938 GeV/c2 , m π = 0.139 GeV/c2 . 10.5.7. In a bubble chamber a K − meson reacts with a proton at rest,1 producing a π + meson and an unknown particle X . The photograph of the reaction gives the orbits of the particles as shown in Fig. 10.2. From the observation of the figure we infer the following: – The orbits of the particles are in a plane perpendicular to the magnetic field of the chamber. – The orbits of the daughter particles are on the same curve, C say, at the point of reaction O. 1 Interact at rest means that the total threemomentum in the CM of the system of particles vanishes and not that the particles are at rest in the CM. Therefore, in the current problem both the proton and the K − meson have nonvanishing threemomenta in CM which, in this case, coincides with the laboratory.
10.5
Reactions of the Form 1 + 2 → 3 + 4
65
Fig. 10.2 Orbits of particles in the bubble chamber
– The orbits of p and K − are along the principal normal of of orbit. – The radius of curvature of the common orbit C at the point O is R = 0.340 m. Given that the magnetic field in the bubble chamber is normal to the plane of the orbit and has intensity B = 1.70 ± 0.07 T . (a) Calculate the threemomentum of K − at the point O given that the threemomentum of the proton is 10 GeV/c . (b) Calculate the mass and the charge of the unknown particle X. It is given: m π = 139.6 MeV/c2 , m p = 938.3 MeV/c2 , m K = 493.8 MeV/c2 , q = 1.6 × 10−19 Cb, 1 Joule = 6.24 × 1013 MeV. 10.5.8. In the reaction e+ +e− → γ +γ , the target particle e− rests in the laboratory frame (L) and the photons in the CM frame are emitted normal to the direction of motion of the bullet particle e+ . (a) If the energy of the e+ in L is E, calculate the fourmomenta of the photons in L . (b) Calculate the energy of the e+ in the laboratory for which the photons are emitted at angle π4 with the direction of motion of the e+ in the laboratory. The mass m of the electron is known. 10.5.9. (a) Show that a photon is impossible to disintegrate in empty space to the pair e− , e+ . Also show that this is possible if a nucleus is present. If m N is the mass of the nucleus, calculate in the proper frame of the nucleus the minimum energy of the photon in order to achieve the production of the pair e− , e+ . Discuss the cases m N m e and m N = m e . (b) Show that a free electron cannot absorb or emit a photon. 10.5.10. A beam of particles of mass m is scattered elastically on a target of particles of mass M, the latter being at rest in the laboratory. If the energy of the beam in the laboratory is E 1 , calculate the scattering angle in the laboratory as a function of the scattering angle in the CM frame.
66
10 Relativistic Reactions
10.5.11. A particle of mass m is scattered elastically on another particle of mass M which rests in the laboratory frame L . Calculate the angle of scattering of the particles after the collision in terms of the scattering angle θ ∗ in the CM. Application Consider the scattered particle to be a neutrino (m = 0) and the target particle to be an electron (M = m e ). More specifically consider the case E 1 = m e c2 and show that θ ∗ cannot equal 0. 10.5.12. A particle 1 of mass m and energy E collides elastically with a similar particle 2 which rests in the laboratory (lab). Show that in the twodimensional Euclidean plane the tip of threemomentum of each of the particles moves on an ellipse whose axes depend on the energy and the mass of the particles.
10.6 Other Reactions 10.6.1. An electron of mass m and threemomentum pe interacts electromagnetically with a nucleus of mass M which is at rest in the laboratory. As a result of this interaction the electron loses energy by emitting a photon [this radiation is called bremsstrahlung radiation (brake radiation)]. Calculate kinematically the range of energy of the emitted photon assuming that the nucleus is very heavy and the electron very fast so that it does not loose energy during the collision. 10.6.2. A particle of mass M is possible to disintegrate either to two particles of mass m or to three particles of mass μ. (a) Calculate the maximum speed β2 which is possible for one of the particles of mass μ. μ m ,y = M so (b) Derive the condition which must be satisfied by the ratios x = M that for the disintegration in two masses, the velocity factor β of the produced particles of mass m is equal to the maximum velocity factor β2 determined in part (a). 10.6.3. A kmeson which rests in the laboratory disintegrates to three π mesons. (a) Show that the trajectories of the three π mesons are coplanar. (b) If the three π mesons are moving in directions which make an angle 120◦ one with another, find their speed in the laboratory. If the lifetime of the pion is 2.6 × 10−8 s how far the pions travel in the laboratory before they disintegrate? It is given that the mass of the kmeson is 494 MeV/c2 and the mass of the π meson is 140 MeV/c2 .
10.7
Threshold
67
10.6.4. Consider the reaction k¯ + p → k¯ + p +nπ where in the laboratory frame the threemomentum of the beam of kmeson is 10 GeV/c. Find the maximum number n of π mesons which can be produced. It is given that m k = 0.494 GeV, m π = 0.135 GeV, m p = 0.938 GeV. 10.6.5. A typical fission reaction is caused by slow neutrons according to the reaction n 1 + 235 92 U → x + y + 2n 2 . Calculate the velocity factor β of n 2 in the laboratory assuming that the reacting particles n 1 , 235 U, and the products x, y are at rest in the laboratory. If the reaction takes place with a certain quantity of 235 U, the products slow down due to collisions with other nuclei until all their kinetic energy is transformed to heat. Assuming that 5 kg 235 U are involved in the process, calculate the thermal energy produced when 141 x = 93 38 Sr55 , y = 54 Xe87 . The masses of the various nuclei involved are: m x = 92.342327 amu, m y = 140.969100 amu, m U = 235.117496 amu, m N = 1.00896 amu. Note: 1 amu = 1.66 × 10−27 kg, the Avogadro number N A = 6.022 × 1023 atoms per mole, and the atomic weight of 235 U = 238.03.
10.7 Threshold 10.7.1. (a) A particle of mass m (bullet particle) reacts with another particle of mass m (target particle) producing a number of various particles of total mass M. If m + m < M, the reaction occurs only when the kinetic energy of the bullet particle in the proper frame of the target particle is higher than a certain critical value (which is called the threshold energy). Calculate that energy. (b) Calculate the threshold energy of the reaction pp → ppp p¯ . It is given that m p = m p¯ = 1 GeV/c2 . 10.7.2. From the reaction of a π meson with a proton which rests in the laboratory, a kmeson and a Λhyperon are produced. Calculate (a) The threshold energy of the reaction. (b) The minimum kinetic energy of the π meson in order the kmeson to be produced at right angles to the direction of the π meson in the laboratory. (c) The kinetic energy of the kmeson produced in (b) when the kinetic energy of the π meson in the laboratory is 1009 MeV. The masses of the particles are m π = 140 MeV/c2 , m p = 940 MeV /c2 , m k = 494 MeV/c2 , m Λ = 1115 MeV/c2 . Take c = 1.
68
10 Relativistic Reactions
10.7.3. A target 1 of mass N is hit by a bullet beam of particles 2 of mass M and the particles 1, 2 and the particle 3 of mass m are produced. (a) Show that the γ factor for the threshold energy of particle 2 is γ =1+
m2 m(N + M) + . NM 2N M
(b) Define the efficiency of the reaction k by the ratio k=
Energy of produced particle 3 in CM . Required kinetic energy of bullet particle 2
Calculate the efficiency k in the above case and show that for large masses N of the target the efficiency of the reaction tends to 1. Application Show that a photon cannot disintegrate spontaneously into a pair of positron– electron. However, this is possible in the presence of a nucleus which acts as a catalyst. If the mass of the nucleus is N and the mass of the electron m, calculate the threshold frequency of the photon. Show that for large N the efficiency of the reaction is 100%; hence, the nucleus tends to act as a pure catalyst. 10.7.4. In the laboratory frame L a photon collides with a nucleon N of kinetic energy 1600 MeV according to the reaction γ + N → N +π . Calculate the threshold energy of the photon in the laboratory. masses m N = 1000 MeV/c2 and m π = 150 MeV/c2 .
10.8 General Problems on Reactions 10.8.1. Consider a linear space V n and let {x i } a coordinate system. The volume element in the coordinates {x i } is defined by d V{x i } = d x 1 d x 2 . . . d x n . Let {x i } be a second coordinate system in V n . Then the element of volume d V{x i } in the i i coordinates {x i } is given from the relation d V{x i } = J xx i d V{x i } , where J xx i i is the Jacobian of the transformation {x i } → {x i } given by the relation J xx i = i ∂x . ∂ xi (a) In the Euclidean space E 3 consider the Euclidean coordinate system {x, y, z} and the spherical coordinate system {r, φ, θ } defined by the coordinate transformation. x = r cos φ sin θ
y = r sin φ sin θ
z = r cos θ.
10.8
General Problems on Reactions
69
Calculate the volume element in spherical coordinates. (b) Let Σ be an LCF in which we consider spherical coordinates and compute the volume element as in (a). From the volume element separate the part which contains only the angles and define the solid angle dΩ in the direction θ with the relation dΩ = sin θ dθ dφ. Let Σ be a second LCF in which the solid angle is dΩ = sin θ dθ dφ . The quantities dΩ , dΩ are related with the Lorentz transformation defined by Σ, Σ . The exact relation depends on the space we consider as the solid angle. For example, the solid angle transforms differently in spacetime and in the momentum space. The same argument applies to the element of volume. In the LCF Σ a beam of photons propagates within the solid angle dΩ. Let Σ be a second LCF which moves in the standard way wrt Σ with speed u = ux (and has parallel axes). Compute the transformation formula dΩ → dΩ . Application Assuming that the stars in the night sky are distributed isotropically and uniformly for the observers on the Earth, calculate their distribution for observers who move with large velocities wrt the Earth.
Electromagnetic Field: Theory
In an LCF Σ the electromagnetic field is expressed by the electric field E and the magnetic field H and satisfies Maxwell equations. In an isotropic and homogeneous medium the electric field E = D where D is the field of electric inductance and the dielectric constant of the medium. Similarly in this medium the magnetic field is related to the magnetic inductance B with the relation B = μH where μ is the magnetic permeability of the medium. For empty space 0 μ0 = c12 . In empty space the fields E, B in terms of the scalar (φ) and the vector (A) potential are given by the expressions E = −∇
∂A φ − , c ∂t
B = ∇ × A.
The potentials φ and A define the fourpotential Ωi = (−φ/c, A). In terms of the fourpotential, Maxwell equations are written as follows: Ωi,i = 0,
ij ,j
Ω
= −μ0 J i ,
where J i is the fourvector of current density and the comma indicates derivation wrt the index that follows. If a charge of density ρ0 has fourvelocity u i , the fourcurrent J i = ρ0 u i + j i where u i ji = 0. ρ0 u i is the conduction current and j i is the convection current. The electromagnetic field tensor is defined in terms of the fourpotential as follows: Fi j = Ω j,i − Ωi, j . If in an LCF Σ the electric field and the magnetic field have components E = (E x , E y , E z ), B = (Bx , B y , Bz ), the electromagnetic field tensor in Σ (in the SI system of units) has components 71
72
Electromagnectic Field: Theory
⎛ ⎜ ⎜ Fi j = ⎜ ⎜ ⎝
0 − Ecx − Ex c Ey c Ez c
Ey c
− Ecz
⎞
⎟ Bz −B y ⎟ ⎟. ⎟ −Bz 0 Bx ⎠ B y −Bx 0 0
In terms of the electromagnetic field tensor Maxwell equations are written as Fi j,k + F jk,i + Fki, j = 0,
F
ij ,j
= μ0 J i .
The fourforce due to the action of the electromagnetic field Fi j on the charge q whose fourvelocity is u i is given by the relation F i = q F μj u j . The covariant form of the equation of motion of a charge q and mass m under the action of the electromagnetic field Fi j is q F ij u j =
d (mu i ) . dτ
In an LCF Σ in which the charge q has threevelocity v, the fourforce gives the threeforce (Lorentz force) F = q (E + v × B) . In an LCF Σ the equations of motion of a charge can be written in various forms. The zeroth component of the fourforce gives dE = q(β · E), dt where E is the work done by the force F on the charge. From this relation we infer that the magnetic field does not produce work (i.e. under the action of magnetic field only the speed and consequently the β, γ factors are constants of motion). The spatial components of the fourforce give d (mγ vx ) = dt d mγ 2 a y = (mγ v y ) = dt d 2 mγ az = (mγ vz ) = dt
mγ 3 ax =
dpx , dt dp y , dt dpz , dt
Electromagnectic Field: Theory
73
or in vector form q(E + 1c v × B) = mγ 3 a , q(E + 1c v × B)⊥ = mγ 2 a⊥ . Consider the LCF Σ which moves wrt the LCF Σ in the standard configuration along the xaxis with velocity factor β. Let an electromagnetic field which in Σ and Σ is described by the fields E = (E x , E y , E z ), B = (Bx , B y , Bz ) and E = (E x , E y , E z ), B = (Bx , B y , Bz ), respectively. The boost relating Σ, Σ relates the components of the fields as follows: ⎧ ⎧ ⎨ Bx = Bx ⎨ Ex = Ex E y /c = γ (E y /c − β Bz ) B y = γ (B y + β E z /c) ⎩ ⎩ E z /c = γ (E z /c + β B y ) Bz = γ (Bz − β E y /c). The vector form of these relations is E = E
E⊥ = γ (E⊥ + βc × B), B = B 1 B⊥ = γ (B⊥ − β × E) c where the parallel and the normal projections are understood wrt the relative velocity u of Σ, Σ . The electromagnetic field has two invariants under the Lorentz transformation defined by 1 X ≡ − F i j Fi j 2
1 Y ≡ − ηi jkl F i j F kl . 8
These invariants in an LCF Σ are expressed in terms of the electric and the magnetic field in Σ as follows: X≡
1 2 E − B2 c2
,
Y ≡
1 E · B. c
The equations of motion under the action of the electromagnetic field Fi j can be obtained from the Lagrangian .
L S R (r,r) = −mc
2
1−
u2 − q(φ − A · u) c2
or the Hamiltonian H = c (p−qA)2 + mc2 + qφ.
Chapter 11
Electromagnetic Field
11.1. (a) Show that the speed of a charge which moves under the action of a magnetic field only is a constant of motion. (b) In the LCF Σ a charge moves under the action of the constant magnetic field B = Bi. Assuming the initial conditions r(0) = y0 j + z 0 k, v(0) = υ0y j + υ0z k, show that the orbit of the charge is the periphery of a circle. Compute the radius (Larmor radius) of the circle and discuss the differences with the corresponding classical orbit. 11.2. In the LCF Σ the observer the time moment t = 1 measures at the position (x = 2, y = 3, z = 4) an electric field E = (0, E, 0) while the magnetic field vanishes. In the LCF Σ , which moves in the standard configuration wrt Σ, the event of measurement of the electromagnetic field has coordinates t = −1/4, x = 7/4, y = 3, z = 4. (a) Calculate the speed of Σ wrt Σ. (b) Calculate the electric and the magnetic field measured by the observer in Σ at the same event. 11.3. In the LCF Σ there is an electromagnetic field whose electric and magnetic fields are B = (0, B y , Bz ) and E = (0, E y , E z ). Given that the invariant Y = 1c B · E = 0, find the velocities of all LCF Σ for which the vector fields E , B are parallel. 11.4. Consider the fourvector Ωi which in the LCF Σ has components Ωi = where Φ, A are scalar and vector fields, respectively.
Φ c
A
(a) Define the vector fields E, B with the relations E = −∇Φ +
∂ A, ∂t
B = ∇ × A, 75
76
11 Electromagnetic Field
and show that the components of the antisymmetric tensor1 Fi j = Ω j,i − Ωi, j in Σ are the elements of the matrix: ⎛ ⎞ E 0 − Ecx − cy − Ecz ⎜ Ex ⎟ ⎜ c 0 Bz −B y ⎟ ⎟, Fi j = ⎜ ⎜ Ey ⎟ ⎝ c −Bz 0 Bx ⎠ Ez B y −Bx 0 c where we follow the convention that the first index counts columns and the second index counts rows. (b) Let Σ be another LCF which moves wrt Σ in the standard way along the xaxis with speed factor β. Compute the components Fi j of the tensor F in Σ . 11.5. A charge q is moving along the zaxis of the LCF Σ with constant speed u. Given that the moment t = 0 of Σ the particle passes through the origin of Σ (a) Calculate the electromagnetic field (E, B) due to the particle in Σ. Find the Newtonian limit of the fields E, B. (b) Show that the electric field is isotropic (i.e., spherically symmetric) in Σ the moment t = 0 only, while for t > 0 the field is anisotropic. Consider spherical coordinates in Σ and write the fields in these coordinates. Show that in Σ the field lines of the electric field are denser normal to the direction of the velocity u of the particle and less dense parallel to the direction of u. Compute explicitly the field lines of the magnetic field. 11.6. Along the xaxis of the LCF Σ the charges +q and −q are at rest at the points x = ±l , respectively (electric dipole). Calculate the electromagnetic field in the LCF Σ in which Σ moves in the standard way along the xaxis with speed u. Calculate the asymptotic values of the field for distances l. 11.7. A conductor is resting along the xaxis of the LCF Σ when at its ends is applied a constant potential difference and the subsequent development of constant current I along the conductor. (a) Show that the conductor remains electrically neutral in Σ after the application of the potential difference. (b) Let Σ be a second LCF which moves wrt Σ in the standard configuration along the xaxis with speed β. Show that in Σ the conductor appears to be charged and compute the charge density across the conductor. Explain the result in geometric terms.
1
The “,” between two indices indicates derivation wrt the index that follows.
11 Electromagnetic Field
77
11.8. Along the xaxis of the LCF Σ a charge is distributed with a uniform constant density ρ0 , while a second point charge q of opposite sign rests at a distance r0 from the xaxis. (a) Calclulate in Σ the force exerted on the charge q. (b) Calculate the same force on another LCF Σ which moves in the standard way along the xaxis with speed u. 11.9. In the LCF Σ rests a straight conductor of infinite length which is charged with a static charge of linear density λ0 > 0. (a) Calculate the electric field in Σ created by the conductor at a point P in space which is at a distance r0 from the conductor. (b) In order to compute the magnetic field created by a current of intensity I, consider the conductor to move along its length in Σ with constant speed u and show that the magnetic field which is created in Σ at a distance r from the conductor is given by the expression B=
μ0 (I × r0 ). 2πr02
(c) A charge q is moving in Σ parallel to the conductor with the same speed u. Calculate the force on the charge. 11.10. A particle of mass m and charge q is moving in the LCF Σ in which there is an electromagnetic field (E, B). (a) Write the equations of motion of the particle in Σ in component form and in vector form parallel and normal to the velocity. (b) Prove that the equations of motion of the particle in Σ can be written in the form d (mγ vx ) = dt d mγ 2 a y = (mγ v y ) = dt d mγ 2 az = (mγ vz ) = dt mγ 3 ax =
dpx dt dp y , dt dpz dt
where a = dv and v = dr is the threeacceleration and the threevelocity of the dt dt particle in Σ. (c) Show that the rate of change of the energy E of the particle in Σ is given by the expression
78
11 Electromagnetic Field
dE = q(β · E) dt and deduce that the magnetic field does no work. (d) Write the equations of motion of the particle in Σ in terms of the proper time of the particle. 11.11. The purpose of this problem is to show that using the Lorentz force one can derive the generalization of Newton’s second law in Special Relativity and the fundamental formula E 2 = c2 p2 + m 2 c4 . (a) Show that if the spatial parts of two fourvectors are equal in all LCF, then the fourvectors are equal. (b) A particle of mass m and charge q is moving in the LCF Σ under the action of the electromagnetic field which in Σ is described by the pair of fields (E, B). We assume that the equation of motion of the particle in Σ is given by the equation d (mγ v), dt
q (E + v × B) =
where t is time in Σ and v is the velocity of the particle in Σ. The LHS of this equation is the Lorentz force. (b1) Show that the RHS of this equation can be written in covariant form as follows: 1 d (mu μ ) γ dτ
μ = 1, 2, 3,
where τ is the proper time of the particle and u i is the fourvelocity of the particle. (b2) The electromagnetic field tensor Fi j in Σ has components ⎛
0 − 1c E x − 1c E y − 1c E z
⎜ 1E ⎜ x Fi j = ⎜ 1c ⎝ c Ey
−Bz
1 E c z
By
0
⎞
Bz −B y ⎟ ⎟ ⎟ . 0 Bx ⎠
−Bx
0
Show that the LHS of the equation of motion can be written in covariant form as follows: q Fμj u j γ
μ = 1, 2, 3.
From (b1) and (b2) conclude that the covariant form of the equation of motion of the charged particle under the action of the electromagnetic field Fi j has the following form:
11 Electromagnetic Field
79 μ
qF ju j =
d (mu μ ) . dτ
This equation contains the spatial components of the fourvectors qc F ji u j and d (mu j ) and it is valid independently of the particular LCF Σ we have employed dτ to derive it. Making use of (a) deduce that the covariant equation of motion of a charge q under the action of the electromagnetic field Fi j is q F ij u j =
dpi , dτ
where pi = mu i is the fourmomentum of the charge. (c) Define the fourvector of the conduction current J i = qu i and the fourvector F i = F ji J j and show that the covariant equation of motion is written as Fi =
dpi , dτ
which is similar to the Second Law of Newton. This result suggests that we i generalize the force law in Special Relativity with the relation F i = dp where dτ F i is the fourforce (specified uniquely for every threeforce F). In addition in the case of a charged particle, we consider the fourvector F i = F i j J j to be the fourforce on the conduction fourcurrent J j due to the electromagnetic field Fi j . (d) Use the zeroth component of the equation of motion to prove the relation E 2 = p2 + m 2 c4 which expresses the total energy of the charged particle in terms of its threemomentum and its mass. 11.12. In the LCF Σ propagates a cylindrical (parallel) beam of electrons of radius R whose charge density in the proper frame of the beam is axially symmetric about the axis of the beam. If the total charge of the beam is I calculate in Σ (a) The force and the velocity of an electron at a distance r from the symmetry axis of the beam. (b) If the length of the beam is L calculate the widening ΔA of the initial diameter of the beam assuming ΔA L. 11.13. Two electrons A, B are projected from a vertical plane with common velocity v = vi from the points A, B, respectively, which are at a distance d apart and are moving toward a vertical screen which is placed at a distance L from the plane of emission (see Fig. 11.1). Due to mutual repulsion the electrons divert during their motion from their initial direction at distances Δy A , Δy B , respectively. This phenomenon is known as space charge defocusing. To compute the degree of defocusing of the two beams,
80
11 Electromagnetic Field
Fig. 11.1 Space charge defocusing
we assume (the same assumption is made in the Newtonian solution to the problem) that the electrons are emitted simultaneously in the laboratory frame and subsequently they are moving always on the same vertical plane so that the xcomponent of their velocities is equal and constant. As a result the two electrons hit the screen simultaneously in the laboratory frame. (a) Calculate the deviation Δy A in terms of the distance L traveled by the electrons and the velocity of the electrons when they hit the screen. What you expect the Newtonian result to be? Discuss the importance of the result for the beams of charged particles. (b) In the laboratory frame calculate the electromagnetic field and the acceleration of the charge A. Compare the result with (a). (c) Transform the fouracceleration from the proper frame of the charges and discuss if the result agrees with those of (a) and (b). 11.14. The electromagnetic field in the LCF Σ is given by the fields E, B. Define in the complex space C3 the complex vector K = E + icB and show that the Lorentz transformation of the fields corresponds to a rotation of the vector K in C3 for the imaginary angle iφ = cos−1 γ . Subsequently show that the only invariants of the electromagnetic field are 1 2 E − B2 and 1c E · B. c2 11.15. In the LCF Σ there is the uniform electromagnetic field (E, B). Show that (a) If E·B = 0, then there are infinitely many LCF Σ in which the electromagnetic field has – Only magnetic field (i.e., E = 0) when E2 < B2 c2 . – Only electric field (i.e., B = 0) when E2 > B2 c2 . (b) If E · B = 0, then there are infinitely many LCF Σ in which the fields E , B are parallel. Note: If the electromagnetic field is not uniform, then the above holds at every point of Σ separately.
11 Electromagnetic Field
81
11.16. In the LCF Σ a particle of mass m and charge q is moving under the action of the constant electric field E = E xˆ . If at the moment t = 0, the particle starts from the origin of Σ with velocity v(0) = v0 yˆ show that the motion takes place on the plane 'x y and compute the equation of the orbit. x = sinh−1 ax , a = 0.] [Hint: √ad2 +x 2 11.17. A particle of mass m and charge q moves freely in the magnetic field B = (0, 0, B). (a) Show that the total energy and the length of the threemomentum are constants of motion. (b) Assume the initial conditions. x(1) = y(1) = 0, z(1) = k, x˙ (1) = λ (λ > 0), y˙ (1) = 0, z˙ (1) = k, where k, λ are real constants and show that the orbit of the particle in space is spiral: x(t) =
λt cos (a ln t − θ ) , 1 + a2
y(t) = −
λt λ sin (a ln t − θ ) − sin θ, 1 + a2 1 + a2
z(t) = kt, where a =
qB and tan θ = a. mγ0
Finally, prove that the particle moves on the surface
λ2 x + y + sin θ 1 + a2 2
2
=
λ2 z2. k 2 (1 + a 2 )
11.18. In the region 0 ≤ x ≤ a of the LCF Σ there is a constant magnetic field B = (0, 0, B). A particle of mass m and charge q is moving with constant speed u along the xaxis of Σ when enters the region 0 ≤ x ≤ a from the side x < 0. (a) Show that the energy of the particle remains constant during its motion in the field B. (b) Show that the orbit of the particle is the circle: x 2 + y 2 + 2ky = 0,
z = 0,
82
11 Electromagnetic Field
where k = muγ . Finally, show that if k < a the charge will be reflected from qB the magnetic field. 11.19. In the LCF Σ there is a uniform electromagnetic field with E = acj and B = 35 ak (a = constant). A particle of mass m and charge q which rests at the point x0 i of the xaxis of Σ is left to move freely. Calculate the time interval required in Σ in order the particle to return to the xaxis. 11.20. In the LCF Σ there is the homogeneous magnetic field B = Bz. A particle of mass m and charge q moves on the plane x − y of Σ. (a) Show that the speed v of the particle is a constant of motion and that with proper choice of initial conditions the motion is a uniform rotational motion B . with angular speed ω = γ q(v)m (b) Consider a second LCF Σ which moves wrt Σ in the standard configuration along the xaxis with speed u. Calculate u and B so that in Σ the electromagnetic field is E = (0, −E 0 , 0) ,
B = (0, 0, B0 ) .
Describe the motion of a charged particle of (a) in Σ and show that the average speed is EB00 and is parallel to the axis x . 11.21. In the LCF Σ the uniform electromagnetic field is decomposed in the magnetic field B = B zˆ and the electric field with intensity E = 4Bc lying in the plane x − z and making an angle θ = 0 with xaxis. (a) Calculate the invariant Y = 1c E · B and draw your conclusions. (b) Calculate the velocities of all LCF Σ for which the fields E , B of the given electromagnetic field are parallel. 11.22. A plane electromagnetic wave of frequency ν which propagates in the x − y plane of the LCF Σ at a direction which makes an angle θ with the the xaxis consists of the fields E = (−AX sin θ, AX cos θ, 0), B = (0, 0, AX/c), sin θ ). Calculate the frequency and the direction of where X = sin 2π ν(t − x cos θ+y c propagation of the electromagnetic wave in another LCF Σ which moves wrt Σ in the standard configuration with velocity u = c cos θ i.
11.23. Let (E, B) and (E , B ) be the electromagnetic field in two LCF Σ and Σ whose axes are parallel and their relative velocity factor β. Prove the transformation equations
11 Electromagnetic Field
83
E = E , E⊥ = γ (E⊥ + βc × B), B = B , 1 B⊥ = γ (B⊥ − β × E). c Application An observer Σ finds that in his proper frame there exists only electric field E. Show that the electromagnetic field (E , B ) in another LCF Σ satisfies the relation B + 1c β × E = 0. Similarly show that if in Σ the magnetic field B = 0, then in Σ the following condition is satisfied E − 1c β × B = 0. Note: From the above we conclude that if one of the fields E, B vanishes in a LCF, then in all other LCF Σ the corresponding fields (of the same electromagnetic field!) are perpendicular. The inverse is also true, that is, if in one LCF the inner product E · B = 0, then there exists an LCF in which there is only electric or only magnetic field. 11.24. (a) Describe the propagation of an electromagnetic wave in a conducting medium with electric permeability , magnetic permeability μ, and conductivity σ . (b) Show that in a conducting medium of electric permeability , magnetic permeability μ in which there are no free charges and conduction currents, an electromagnetic wave propagates with speed √1μ and the fields H and E are given by the relations H(t, r) = H0 ei(ωt−k1 ·r) , E(t, r) = E0 ei(ωt−k1 ·r) , √ where the wave vector k has length k = ω μ. 11.25. Maxwell equations for empty space in terms of the electromagnetic potential Ωi are Ωi,i = 0,
i, j j
Ω
= −μ0 J i ,
where J i is the fourcurrent of charge density. (a) Show that if two fourvectors Ai , ki are such that Ai k i = 0, ki k i = 0, that is, k i is null and Ai is normal to k i , then Ai is not in general a null fourvector. (b) Show that in a region of space in which there are no free charges and electric r currents (that is, J i = 0) one solution of Maxwell equations is Ωi = Ai eikr x where the fourquantities Ai , ki are constant fourvectors normal to each other,
84
11 Electromagnetic Field
i.e., Ai k i = 0 and k i is a null fourvector (i.e., ki k i = 0). We call this solution of Maxwell equations as plane electromagnetic wave. 11.26. Calculate the Lagrangian and the Hamiltonian of a charged particle of mass m and charge e which moves in the electromagnetic field φ, A. Compare the results with the corresponding Newtonian.
Solutions
Chapter 1
Mathematical Background: Problems
(1) The quantity Ai,i is invariant; therefore, we compute in Σ and then we find it in Σ by using the Lorentz transformation relating Σ, Σ . We have x Ai,i = A0,0 + A1,1 + A2,2 + A3,3 = A,x = e x + xe x = (1 + x)e x .
Σ, Σ are related with a boost along the xaxis with speed kc, hence γ = (k < 1). It follows that x = γ (x + βl ) = √
1 1 − k2
√ 1 1−k 2
(x + kl ),
which gives that the divergence of Ai in Σ is given by the expression √ 1 (x +kl ) 1 Ai,i = 1 + √ (x + kl ) e 1−k 2 . 1 − k2 [Note: Compute the same result by transforming first Ai in Σ and then compute the divergence in Σ .] (2) The components of the vector Bν = Tμνμ are Bν = T1ν1 + T2ν2 + T3ν3 . We compute B1 = T111 + T212 + T313 ⇒ B1 = 8x12 + 2(x22 + x32 ) B2 = T121 + T222 + T323 ⇒ B2 = 8x22 + 2(x32 + x12 ) B3 = T131 + T232 + T333 ⇒ B3 = 8x32 + 2(x12 + x22 ). The div divBν =
∂Bν ∂B1 ∂B2 ∂B3 = + + = 16(x1 + x2 + x3 ) ∂ xν ∂ x1 ∂ x2 ∂ x3 87
88
1 Mathematical Background: Problems
and the curl
∂B3 ∂B2 ∂B1 ∂B3 ∂B2 ∂B1 − , − , − ∂ x2 ∂ x3 ∂ x3 ∂ x1 ∂ x1 ∂ x2 = 4(x2 − x3 , x3 − x1 , x1 − x2 ).
curlBν =
(3) (i) Aμν,ν = ∂ (x 2 ∂ x1 μ
=
∂ Aμν ∂ xν
=
∂ Aμ1 ∂ x1
+ x12 ) +
+
∂ Aμ2 ∂ x2
∂ (x 2 ∂ x2 μ
+
∂ Aμ3 ∂ x3
+ x22 ) +
=
∂ (x 2 ∂ x3 μ
+ x32 ) =
= 2x1 (1 + δμ1 ) + 2x2 (1 + δμ2 ) + 2x3 (1 + δμ3 ) ⇒ Aμν,ν = 2
,3
ν=1
xν (1 + δνν ) .
(ii) Aμν,μν = 4
,3
μ=1 δμμ
= 12 .
(4) (a) The matrix of the transformation is ⎡
cosh φ ⎢ − sinh φ L=⎢ ⎣ 0 0
− sinh φ cosh φ 0 0
0 0 1 0
⎤ 0 0 ⎥ ⎥. 0 ⎦ 1
It is easy to show that η = L t ηL where η = diag(−1, 1, 1, 1). It follows that L is a Lorentz transformation (b) The transformation of the contravariant vector V i is given by the relation
∂xi i V = V = L ii V i , ∂xi i
where L = [L ii ]. We compute
V 0 = L i0 V i = L 00 V 0 + L 01 V 1 + L 02 V 2 + L 03 V 3 = − sinh φ, V 1 = L i1 V i = L 10 V 0 + L 11 V 1 + L 12 V 2 + L 13 V 3 = cosh φ, V 2 = L i2 V i = L 20 V 0 + L 21 V 1 + L 22 V 2 + L 23 V 3 = 0, V 3 = L i3 V i = L 30 V 0 + L 31 V 1 + L 32 V 2 + L 33 V 3 = 1. Hence [V i ] = (− sinh φ, cosh φ, 0, 1)t .
1 Mathematical Background: Problems
89
For the tensor Ti j we have
Ti j =
∂xi ∂x j j Ti j = L ii L j Ti j = L i1 L 1j T1 1 + L i3 L 3j T3 3 , i j ∂x ∂x
= L i1 L 1j + L i3 L 3j
where we have used that the only nonzero components of Ti j in Σ (x , y , z ) are the T0 0 = T3 3 = 1. From this relation we compute the components of the tensor Ti j in Σ(x, y, z) . For example, for the component T00 we have
T00 = L 10 L 10 + L 30 L 30 = sinh2 φ. The result of the calculation is given in the following symmetric matrix ⎡
sinh2 φ ⎢ − sinh φ cosh φ [Ti j ] = ⎢ ⎣ 0 0
− sinh φ cosh φ cosh2 φ 0 0
0 0 0 0
⎤ 0 0 ⎥ ⎥. 0 ⎦ 1
Concerning the vector ai = Ti j V j , we have in Σ(x, y, z): a0 = T00 V 0 + T01 V 1 + T02 V 2 + T03 V 3 = − cosh φ sinh φ a1 = T10 V 0 + T11 V 1 + T12 V 2 + T13 V 3 = cosh2 φ a2 = T20 V 0 + T21 V 1 + T22 V 2 + T23 V 3 = 0 a3 = T30 V 0 + T31 V 1 + T32 V 2 + T33 V 3 = 1. Finally, the invariant ai V i = Ti j V i V j is ai V i = a0 V 0 + a1 V 1 + a2 V 2 + a3 V 3 = cosh2 φ + 1. (c) It is left as an exercise for the reader. (5) To find the geodesics we compute the Lagrangian of the metric and then the geodesic equations are the Lagrange equations for each coordinate. The Lagrangian of the metric ds 2 is L=
1 1 gi j x i x j = 2 (−t˙2 + x˙ 2 ). 2 2t
The tcoordinate is ∂L t˙ =− 2, ∂ t˙ t
∂L 1 = − 3 (−t˙2 + x˙ 2 ). ∂t t
90
1 Mathematical Background: Problems
The tgeodesic equation is d ds
∂L ∂ t˙
−
∂L d =0⇒ ∂t ds
1 t˙ − 2 + 3 (−t˙2 + x˙ 2 ) = 0 ⇒ t t
t¨ 2t˙2 t˙2 x˙ 2 t¨ t˙2 x˙ 2 − + − = 0 ⇒ − − =0⇒ t2 t3 t3 t3 t2 t3 t3 t t¨ − t˙2 − x˙ 2 = 0. The x coordinate is x˙ ∂L = 2, ∂ x˙ t
∂L =0 ∂x
Hence1 :
x˙ t2
= 0 ⇒ x˙ = at 2 ⇒ a = const.
Next we integrate the geodesic equations. Dividing the tgeodesic with t 2 we find t¨ t˙2 x˙ 2 − 2 − 2 =0⇒ t t t
. t˙ t
− a x˙ = 0 ⇒
t˙ − ax t
.
= 0.
Hence, t˙ − ax = β, t The x˙ =
dx ds
=
d x dt dt ds
= x t˙, where ≡
d . dt
β = const. Therefore,
at 2 = x t˙ ⇒ t˙ =
at 2 . x
1 From the equation of the geodesics, it is possible to compute the connection coefficients. It turns out that the nonvanishing connection coefficients Γijk are the following:
Γ000 = where t ←→ 0, x ←→ 1.
1 , t3
Γ110 = Γ101 = −Γ011 = −
1 , t3
1 Mathematical Background: Problems
91
Replacing we find at 1 1 − ax = β ⇒ (ax + β)x = at ⇒ ax 2 + βx + γ = at 2 x 2 2
x2 +
(γ = const.).
2 2γ 2β β 2 β 2γ x+ = t2 ⇒ x + . = t2 + − a a a a a
We set β = −x0 , a
2 β 2γ =A − a a
and obtain the equation (x − x0 )2 = t 2 + A, where x0 , a are constants. We conclude that the geodesics are hyperbolae with asymptotes the null lines on the light cone and origin the point x0 . Indeed the last equation gives the solution √ τ x − x0 = A cosh √ A √ τ t = A sinh √ . A These equations represent the world line of a point accelerating along the xaxis with constant proper acceleration √1A . τ is the proper time of the accelerating point. Note: From the result follows that the coordinates (t, x) which define the metric ds 2 = t12 (−dt 2 + d x 2 ) are “inertial” wrt the accelerating point, in the sense that the later moves along geodesics of the metric. This result justifies the Equivalence Principle which is fundamental in the Theory of General Relativity and in a simplified form can be stated as follows: Free fall in a gravitational field is along the geodesics of a nonflat metric with Lorentzian character in a space free of gravity. (6) Let L be a homogeneous2 linear transformation in Minkowski space: L:
x0 xμ
Σ
x0 → xμ
,
(1.1)
Σ
Homogeneous means that we correspond Σ and Σ so that when the origins coincide the Lorentz transformation is the identity. This is usually stated as when t = t = 0, x μ = x μ = 0.
2
92
1 Mathematical Background: Problems
where 0 the small case Greek indices is assumed to take the values 1, 2, 3 and x is a fourvector. In terms of components the transformation is written xμ Σ as follows:
x 0 = L 00 x 0 + L 0μ x μ x
μ
=
μ L0 x0
+
(1.2)
L μμ x μ .
(1.3)
The Lorentz transformation L is a particular linear transformation which is defined by means of certain physical assumptions. The first assumption is the socalled reciprocity principle which is the demand that the inverse transformation is the inverse of the matrix (L aa ) in Σ. This is expressed by the relation
x 0 = L 00 x 0 + L 0μ x μ μ
μ
x μ = L 0 x 0 + L μ x μ . Applying twice the transformation we find
x0 xμ
*
=
* = * =
+
+
L 00 L 0μ μ L 0 L μμ L 00 L 0μ μ L 0 L μμ
x0 xμ
L 00 L 0ν μ μ L 0 L ν
μ
x0 xν
μ
L 00 L 00 + L 0μ L 0 L 00 L 0ν + L 0μ L ν L μ0 L 00 + L μμ L μ0 L μ0 L 0ν + L μμ L μν
+
x0 , xν
from which the following conditions on the components of L follow:
L 00 L 0ν + L 0μ L μν = 0 μ
+
μ L 0μ L 0
+
μ L μμ L ν
μ
L 0 L 00 + L μμ L 0 = 0 L 00
L 00
μ L 0 L 0ν
=1 =
μ δν .
(1.4) (1.5) (1.6) (1.7)
order to state the second requirement we assume that the fourvector In x0 is the position fourvector of a point P. Then the velocity of P in Σ is xμ Σ
uμ =
dx μ . dx 0
1 Mathematical Background: Problems
93
Replacing dx μ , dx 0 from the transformation we find μ
u
μ
L 0 dx 0 + L μμ dx μ L 0 + L μμ u μ dx μ = = = , dx 0 L 00 dx 0 + L 0μ dx μ L 00 + L 0μ u μ
μ
(1.8)
μ
where u μ = dx the velocity of the point in Σ. dx 0 μ The speed of the origin of Σ wrt Σ is u μ 0 = L 0 /L 00 because the velocity μ u of Σ wrt itself vanishes. Similarly the velocity of the origin O of Σ wrt Σ is u μ 0 =
μ
L 0 . L 00
Now we set the second requirement which is the “relativity of motion,” that is, the relative velocity of the origins of Σ, Σ are opposite. This is expressed mathematically by the condition:
Lμ μ u 0 = −u 0 ⇔ 00 = −u μ ⇒ L 0 = −L 00 u μ . L0 μ
μ
(1.9)
We need one more requirement. This is the “synchronization” of the clocks, which is expressed by the following condition:
L 0 = ±L 0 0 = ∓γ ⇒ L 00 = −L 0 0 = ±γ , L 0μ = L μ 0 ,
(1.10) (1.11)
where3 γ is a parameter which has to be determined in terms of the relative velocity of the frames Σ, Σ . Requirement (1.10) gives for the requirement (1.9): L μ0 = ∓γ u μ .
(1.12)
Requirement (1.11) simplifies requirement (4.17) as follows:
μ L μμ L ν
=
μ δν
+
μ L 0 L 0ν
=
μ δν
L μ L ν0 μ + 00 0 L 00 L 00 = δν + u μ u ν γ 2 . L0 L0
(1.13)
From the above requirements we determine the Lorentz transformation L aa relating Σ, Σ .
3
We raise and lower the indices with η = diag(−1, 1, 1, 1).
94
1 Mathematical Background: Problems
Determination of the element (00) From (1.10) we have L 00 = ±γ . Therefore, all we have to do is to determine the factor γ in terms of the relative speed of Σ, Σ . To do that we note that condition (10.10) gives
L 00 L 00
L 0μ L μ −1/2 + 0 00 = 1 ⇒ γ 2 + γ 2 u μ (−u μ ) = 1 ⇒ γ 2 = 1 − u 2 . (1.14) L 0 L 0
Determination of the elements (0μ), (μ0) μ This has been done because from (1.12) we have L 0 = ∓γ u μ and condition t μ (1.11) gives4 L 0μ = L 0 . Determination of the element L μμ First we note that condition (4.15) gives L μμ
μ
μ
L L 0 μ μ + L 0 = 0 ⇒ L μμ u μ = −L 0 = − 00 L 00 = ∓γ u μ . L 00 L0
(1.15)
To determine the spatial part L μμ we solve the system of equations (1.13), (4.20). We look for a solution of the form
L μμ
=
aδμμ
uμ uμ +b 2 , u
(1.16)
where u 2 = u ρ u ρ . Substituting in (1.13) we get * aδμμ
uμ uμ +b 2 u
+
μ
aδν + b
uμuν u2
μ
= a 2 δν + 2ab
= δνμ + γ 2 u μ u ν .
μ uμ uν 2 u uν + b = u2 u2
(1.17)
From the last equation we conclude that a 2 = 1, 2ab + b2 = γ 2 u 2 .
(1.18)
Adding these equations and using (1.14), we find (a + b)2 = 1 + γ 2 u 2 = γ 2 ,
4
Where t stands for transpose.
(1.19)
1 Mathematical Background: Problems
95
from which follows b = ±γ ∓ 1, a = ±1.
(1.20)
We examine now (4.20). Substitution of (1.16) in (4.20) gives *
aδμμ
uμ uμ +b 2 u
+
u μ = (a + b) u μ = ±γ u μ ,
(1.21)
from which we find that a + b = ±γ .
(1.22)
This condition gives us nothing new and we have the final solution: L μμ
=
±δμμ
! uμ uμ uμ uμ μ + ±(γ − 1) 2 = ± δμ + (γ − 1) 2 . u u
(1.23)
We conclude that the Lorentz transformation relating Σ, Σ is the following: * L Σ,Σ =
∓γu μ uμ u ∓γ u μ ± δμμ + (γ − 1) u 2 μ
±γ
+ .
(1.24)
We note that we have determined four Lorentz transformations which satisfy the assumed requirements. To distinguish between them we compute the determinant of the transformation. It is an easy exercise to show that detL Σ,Σ = 1 + γ (ε − 1),
(1.25)
where ε = ±1. We call the Lorentz transformation with determinant equal to +1 as the proper Lorentz transformation. Two of the four Lorentz transformations are proper Lorentz transformations. The rest two are improper Lorentz transformations and their determinant is = 1. Another selective rule is defined by means of the sign of the term L 00 . We call the proper Lorentz transformations defined by the value +γ as the orthochronous Lorentz transformations. These transformations preserve the direction of time between Σ, Σ (that is, the dt dt > 0). The orthochronous Lorentz transformations form a subgroup of the totality of Lorentz transformations (because they contain the identity transformation).
96
1 Mathematical Background: Problems
The various Lorentz transformations have different names as follows: The orthochronous Lorentz transformation: * L Σ,Σ =
γ
−γ u μ
−γ u μ δμμ + (γ − 1)
+
uμ uμ u2
.
(1.26)
.
(1.27)
The time inversion: * L Σ,Σ =
μ −γ γ u uμ u γ u μ δμμ − (γ + 1) u 2 μ
+
The space inversion: * L Σ,Σ =
μ −γ u uμ u −γ u μ δμμ − (γ + 1) u 2 μ
γ
+ .
(1.28)
.
(1.29)
The space and time inversion: * L Σ,Σ =
μ −γ γ u uμ u γ u μ −δμμ + (γ + 1) u 2 μ
+
In the literature by proper Lorentz transformation it is understood the orthochronous Lorentz transformation. We shall follow this abuse of terminology in this book.
Chapter 2
Classic Experiments
2.1. (a) The time required for a collinear passing of the light through the medium, according to Fresnel hypothesis, is t1 =
c n
, + κV
where κ is the drag coefficient. Similarly for the opposite course we have t2 =
c n
. − kV
The time difference between the two courses equals: 2 κ V 2 n 2 κ V ΔT = t2 − t1 = 2 ≈ . c 2V 2 c2 − κ n Using the relation T = λ/c relating the wavelength with the period of the wave we have N=
2 kn 2 V Δλ ΔT 2 n 2 κ V N λc = = ≈ ⇒κ= . λ T T c2 λc 2 n 2 V
The experimental measurements agree with the above result thus justifying the Fresnel hypothesis. (b) In order to show that the Fresnel hypothesis results directly from the Theory of Special Relativity without any further (i.e., the dragging) assumption, we consider two RIO Σ, Σ which are related with a boost and are such that
r r
The common x, x is along the direction of the propagation of light. Their relative speed is V , that is, Σ is comoving with the medium.
Then in Σ the refraction index is given by n = vc where v is the speed of light in the comoving (isotropic and homogeneous) medium. The speed of the beam in Σ is given by v = nc and it is computed in terms of v , V by means of 97
98
2 Classic Experiments
the relativistic threevelocity composition rule
v=
v + V , 1 + vc2V
where = 1 when the speed of the beam is collinear with the speed of Σ in Σ , and = −1 differently. Replacing we find 2 ! V εV + εV 1− +O v= = n nc cn 1 + εV nc 2 ! V V εV 1 V2 c c +O . = − 2 + εV − = +ε 1− 2 V +O n n cn cn n n cn c n
+ εV
c
We see that the speed of light in the moving medium to the first approximation equals the speed ( = c/n) in the resting medium augmented by the term (1 − 1/n 2 )V = κ V , where κ = (1 − 1/n 2 ) is the drag coefficient introduced by Fresnel.
2.2. During the rotation of the circular table the speed of each glass tube is v = ωr where r is the distance of the center of the tube from the rotation center. Obviously
r=
1 2 4R − 2 . 2
Hence v = ωr =
ω 2 4R − 2 . 2
The speed of light along the direction of rotation is nc + κv, where κ is the dragging coefficient and in the opposite direction nc − κv. Therefore, the time of traveling time is, respectively,
t1 =
c n
2 + κωr
,
t2 =
c n
2 . − κωr
2 Classic Experiments
99
The phase difference of the light signals in the telescope equals t2 − t1 = 2
c n
1 − − κωr 2κωr
= 2 2 c n
c n
1 + κωr
− κ 2 ω2 r 2
1 4 κωr n 2 κ 2 ω2 r 2 n 2 c2 1 − c2 √ 2 4 κωr n κωn 2 4R 2 − 2 ≈ = . c2 c2 =
This time difference corresponds to a difference in the optical path equal to √ κωn 2 4R 2 − 2 . c(t2 − t1 ) = c 1 (b) For n = 1.33, we have for the drag coefficient κ = 1 − n12 ⇒ κ1 − 1.33 2 = 0.25. From the general formula, we have.
αλ =
2 1 −
1 n2
√ n 2 ω 4R 2 − 2 c
Replacing we calculate the result.
⇒ω=
2
n2
αλc . √ − 1 4R 2 − 2
Chapter 3
The Position FourVector
3.1 The Speed of Light 3.1.2. The speed of the light spot on the surface of the Moon is v = ω · R = 1 s−1 × 380 × 103 km = 380 × 103 kms−1 . We note that v > c! This does not contradict the assumption that the maximal speed is the speed of light because the light spot on the surface of the Moon is not the light beam itself but the image of the light source on the surface of the Moon. Furthermore, the light spot itself does not transport information (the beam does!). A similar case we have in the Galaxy Grab Nebula which is rotating and the ray which emits reaches our telescopes 30 times per second (hence ω = 30 rotations/s) whereas its distance from our Galaxy is many light years. If we consider a spherical surface centered at our Galaxy and radius equal the distance of the Grab Nebula from our Galaxy, then we see that the image of the light it emits moves on the surface with huge speed compared to the speed of light c.
3.2 The Lorentz Transformation 3.2.1. Assume that the center of the circle is at the origin of the coordinates. Then the motion of the particle in Σ is described by the equations x = R cos ωt, y = R sin ωt, z = 0. We infer that the world line of the particle is a helix on a cylinder of radius R and axis along the time axis of Σ.
101
102
3 The Position FourVector
3.2.2. The proper time of clock 1 at the event B is 1 (τ B − τ A )1 = c
(B
1 dτ1 = c
A
(B (t B υ 2 1 0 2 2 2 tB . 1− c dt1 − d x1 = dt = c γ (υ0 ) A
0
Similarly the proper time of clock 2 equals (τ B − τ A )2 =
=
=
a0 c
tB 2
1 c
't B c 2 a0
0
'B
dτ2 =
A
−
=
t 2 dt
'B
1 c
c2 dt22 − d x22 =
't B
A
a0 2c
1−
0
 2 c a0
t
−
t2
+
2 c a0
a0 t 2 c
dt !t B
sin−1 tac0 0
1−
t B a0 2 c
+
c a0
sin−1
t B a0 c
.
In order to compute the time t B , we note that at the event B the two clocks meet; therefore, x B = υ0 t B , x B = 12 a0 t B2 . It follows: υ0 t B =
1 2 2υ0 a0 t B ⇒ t B = . 2 a0
Replacing the value of t B in the previous formulae we compute (τ B − τ A )1 = (τ B − τ A )2 =
υ 2 2υ 0 0 c a0 υ 2 c 0 −1 2υ0 . 1−4 + sin c a0 c
1− 
υ0 a0
The argument of the function sin−1 must be ≤ 1. This means that υ0 must satisfy the constraint 2υc 0 ≤ 1 or υ0 ≤ 2c . The kinematic interpretation of this condition is that if υ0 > 2c , then the clocks 1, 2 do not meet. In order to compare the periods (τ B −τ A )1 and (τ B −τ A )2 , we consider the limiting case υ0 = 2c . Then 1 −1/2 2 γ (υ0 ) = 1 − =√ 4 3 from which follows √ 3 c (τ B − τ A )1 = 2 a0
and
(τ B − τ A )2 =
π c . 2 a0
We note that for this speed (τ B −τ A )2 > (τ B −τ A )1 which was to be expected because clock 1 moves inertially along a straight line (geodesic) in spacetime and this is the
3.2
The Lorentz Transformation
103
(timelike) line with the minimal length between the events A, B. Obviously this conclusion holds for all υ0 ≤ 2c . Note: It is interesting to study the condition υ0 ≤ 2c along physical grounds. This is done if we look at (a) spacetime intervals and (b) relative velocities. (a) Spacetime intervals It must hold dτ12 < 0 , dτ22 < 0 because the world lines of the clocks are timelike. Therefore, we have the conditions: Clock 1: d x12 − c2 dt12 < 0 ⇒ 0 < υ0 < c. Clock 2: d x22 − c2 dt22 < 0 ⇒ 0 < a0 t2 < c. During the course of motion, t2 increases and its maximal value is tB =
2υ0 . a0
Replacing in the last relation we find υ0
0 We calculate for the speed of 2 relative to 1: υ21 =
a0 t − υ0 . 1 − υ0cta2 0
It must hold υ21 < c. Hence we have the condition & & &a t −υ & 0& & 0 & & < c. & 1 − υ0cta2 0 & At the event B we have & & & & & & &a t − υ & & 2υ − υ 0& 0 0 && & 0B & < c ⇒ 2υ02 + cυ0 − c2 < 0. & & < c ⇒& 2 & & 1 − υ0ct B2 a0 & & 1 − 2υ20 & c
The roots of this equation are c υ21 = −c, . 2 The first root −c is rejected and we end up again with the condition 0 < υ0 < 2c . 3.2.3. The Lorentz transformation is defined by two (main) requirements: (a) It is linear. (b) Preserves the Lorentz inner product. The given transformation is linear; therefore, it remains to examine only condition (b). We consider the invariance of the fundamental form x 2 + y 2 + z 2 − 2 = x + y + z − . 2
2
2
2
Replacing the primed coordinates by means of the transformation, we find − 2 + x 2 + y 2 + z 2 = −(a + bx + cy + dz)2 + (b + ax)2 + y 2 + z 2 = (−a 2 + b2 ) 2 + (−b2 + a 2 )x 2 + (−c2 + 1)y 2 + (−d 2 + 1)z 2 − −2ac y − 2ad z + 2(ba − ab)x − 2bcx y − 2bd x z − 2cdyz = − 2 + x 2 + y 2 + z 2
3.2
The Lorentz Transformation
⇒
105
√ −b2 + a 2 = 1 ⇒ a = ± b2 + 1 . c=d=0
Therefore, the Lorentz transformation relating Σ ( , x, y, z) and Σ ( , x , y , z ) is given by the expression ⎧ ⎪ ⎪ ⎨ x ⎪ y ⎪ ⎩ z
√ ⎫ = ε1 b2 + ⎪ √1 + bx ⎪ ⎬ = b + ε1 b2 + 1x , where ε1 = ±1. ⎪ =y ⎪ ⎭ =z
We define two new parameters β ∈ (−1, 1) , γ ∈ (0, +∞) with the relations 1 + b2 =
1 = γ 2 ⇒ b = ε2 γβ , where ε2 = ±1. 1 − β2
Then the transformation is written as follows: ⎧ ⎪ ⎪ ⎨ x ⎪ y ⎪ ⎩ z
⎫ ε1 ε2 γ + ε2 γβx ⎪ ⎪ ⎬ ε2 βγ + ε1 ε2 γ x . y ⎪ ⎪ ⎭ z
= = = =
These relations contain four families of Lorentz transformations depending on the value of the constants ε1 , ε2 . We have ε1 = 1, ε2 = 1 ⎧ ⎪ ⎪ ⎨ x ⎪ y ⎪ ⎩ z
⎫ = γ ( + βx) ⎪ ⎪ ⎬ = γ (x + β ) . =y ⎪ ⎪ ⎭ =z
This type of Lorentz transformation is called time reversal. ε1 = 1, ε2 = −1 ⎧ ⎪ ⎪ ⎨ x y ⎪ ⎪ ⎩ z
= = = =
⎫ γ (− − βx) ⎪ ⎪ ⎬ γ (−x − β ) . y ⎪ ⎪ ⎭ z
This type of Lorentz transformation is called spacetime reversal.
106
3 The Position FourVector
ε1 = −1, ε2 = 1 ⎧ ⎪ ⎪ ⎨ x y ⎪ ⎪ ⎩ z
= = = =
⎫ γ (− + βx) ⎪ ⎪ ⎬ γ (−x + β ) . y ⎪ ⎪ ⎭ z
This type of Lorentz transformation is called space reversal. ε1 = −1, ε2 = −1 ⎧ ⎪ ⎪ ⎨ x y ⎪ ⎪ ⎩ z
⎫ = γ ( − βx) ⎪ ⎪ ⎬ = γ (x − β ) . =y ⎪ ⎪ ⎭ =z
The final transformation is called the proper Lorentz transformation and it is the one we apply, as a rule. The reason for this preference is (among others) that; (a) preserves the causal relation of the events and (b) contains the identity; therefore, the set of all proper Lorentz transformations forms a group. Note: An additional important element of the Lorentz transformations has to do with the orientation (lefthanded and righthanded) of the coordinate frames. The Lorentz transformation preserves the orientation of the frame only if its determinant equals +1, otherwise the orientation changes (a fact we do not want in general). The proper Lorentz transformation does preserve the orientation of coordinate frames in spacetime as well as the orientation of the corresponding frames in the threespace. This means that preserves the causal sequence (i.e., the order of occurrence) of events. 3.2.4. (a) From the equations of transformation, we find 1 x = √ (u − υ), 2
1 l = √ (u + υ), 2
from which follows: 1 dl = √ (du + dυ), 2
1 d x = √ (du − dυ). 2
Replacing in the Lorentz metric, we find: ds 2 = −2dudυ + dy 2 + dz 2 .
3.2
The Lorentz Transformation
107
(b) We have ∂u ∂ ∂υ ∂ 1 ∂ ∂ ∂ = + =√ − , ∂x ∂ x ∂u ∂ x ∂υ ∂υ 2 ∂u ∂2 ∂ ∂u ∂ 1 ∂ ∂υ ∂ 1 ∂ ∂ ∂ ∂ = − + − = √ √ ∂x2 ∂x ∂x ∂ x ∂u ∂υ ∂ x ∂υ ∂υ 2 ∂u 2 ∂u 2 2 2 2 2 2 ∂ ∂ ∂ 1 ∂ 1 ∂ ∂2 1 ∂ − + = = − − − . 2 ∂u 2 ∂u∂υ 2 ∂υ∂u ∂υ 2 2 ∂u 2 ∂υ 2 ∂u∂υ Similarly, we compute 1 ∂2 = 2 ∂l 2
∂ ∂2 + ∂u 2 ∂υ 2
+
∂2 . ∂u∂υ
Therefore, =−
∂2 ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 + + + = −2 + . + ∂l 2 ∂x2 ∂ y2 ∂z 2 ∂u∂υ ∂ y2 ∂z 2
(c) We have ⎛ ⎞ ⎛ ⎞ l u ⎜x ⎟ ⎜v⎟ ⎜ ⎟ = B ⎜ ⎟, ⎝y⎠ ⎝y⎠ z z where ⎛
1 −1 1 ⎜ 1 1 B=√ ⎜ ⎝ 2 0 0 0 0
0 0 1 0
⎞ 0 0⎟ ⎟. 0⎠ 1
0 0 1 0
⎞ 0 0⎟ ⎟, 0⎠ 1
We compute ⎛
0 −2 1⎜ 2 0 t B B= ⎜ 2 ⎝0 0 0 0
hence, B is not a Lorentz transformation. This means that the above transformation is not a Lorentz isometry. This is not a problem because the transformation involves change of coordinates of the same RIO; hence the Principle of Relativity (which involves two observers!) does not apply.
108
3 The Position FourVector
3.2.5. (a) Let P be an event in spacetime with coordinates (l, r) in Σ and (l , r ) in Σ . We decompose r parallel (r ) and normally (r⊥ ) to u using the identity r = r + r⊥ =
r · u r · u u + (r − u). u2 u2
It follows r · u u u = 2 u, r = r · u u u
(3.1)
r · u u. u2
(3.2)
r⊥ = r − r = r −
We transform (l , r ) in the RIO Σ transforming independently the parts l, r , and r⊥ . Because r⊥ is normal to the relative velocity, we have r⊥ = r⊥ . l , r transform with a boost, i.e., r ·u l l r = γ r + u = γ u+ u , c u2 c r ·u r ·u = γ l + . l = γ l + c c We conclude that r = r + r⊥ and l are given by the relations u r · u r = r + (γ − 1) 2 u + γ l , u c u · r l=γ l + . c
(3.3) (3.4)
(Show that the expressions for r , l in terms of r, l are found if we substitute u with −u.) 2 relations If in these expressions we replace u 2 from the identity β 2 = γ γ−1 2 (3.3), (3.4) are written as follows: γ γ r · u + l u, r = r + γ +1 c c u · r . l = γ l + c
(3.5) (3.6)
3.2
The Lorentz Transformation
109
(b) We find the Newtonian limit of the transformation equations (3.5) and (3.6), if we replace l = ct, l = ct , and let c → ∞. For this approximation, we find for γ γ =
1 3 1 = 1 + β 2 + β 4 + . . . = 1 + O(β 2 ). 2 1/2 (1 − β ) 2 8
(3.7)
Replacing in (3.3), we find r = r +
r · u u O(β 2 ) + [1 + O(β 2 )] ut = r + ut u2
and similarly (3.4) gives t = t . Therefore, in the limit β 2 ≈ 0, the (proper) Lorentz transformation becomes r = r + ut , t = t ,
(3.8)
that is, coincides with the Galileo transformation. We infer that for small speeds the relativistic quantities, for which there exists Newtonian analogue, are identical numerically with the corresponding Newtonian quantities. (c) In order to compute the effects of the Lorentz transformation on the special and temporal differences, we differentiate the transformation equations (3.5), (3.6) keeping u constant. We compute γ γ dr · u + dl u, dr = dr + γ +1 c c u · dr . dl = γ dl + c
(3.9) (3.10)
We consider a spatial distance dr which is resting in the RIO Σ , whereas it is measured in the RIO Σ. As it is known from the theory, the measurement of the spatial distance of two events in a RIO Σ equals the difference of the spatial coordinates of the events provided they are simultaneous (that is, they have equal time coordinates) in Σ. Therefore, the measurement of dr in Σ requires dl = 0. The transformation equation (3.10) gives dl + β · dr = 0 that is, −dl = β · dr .
(3.11)
110
3 The Position FourVector
Replacing in (3.9) we find
γ γ dr = dr + (−dl ) + dl γ β = dr + βdl ⇒ γ +1 γ +1 γ dr = dr − (β · dr )β. γ +1
(3.12)
Equation (3.12) is the general expression for the spatial contraction when the two RIO are moving with arbitrary relative velocity u. We note that normal to u we do dr not have contraction, that is, dr⊥ = dr⊥ , whereas parallel to u we get dr = γ (show this by projecting (3.12) parallel to u). We work similarly with the time differences and produce the time dilation formula for a general motion. That is, time measurement in Σ requires dr = 0 (events occurring at the same point in 3space) and the transformation equations (3.9), (3.10) give, respectively, dr = γ β dl , dl = γ dl . The first equation is the expected result dr = udt and it is an identity in Σ, and the second relation is the expression for the time dilation. Note that in both the spatial contraction and the time dilation we identify the measured quantities with coordinate values. Our measurements involve only coordinates and nothing more! 3.2.6. Consider first a rotation about the zaxis with angle φ so that the new x axis is on the plane defined by the zaxis and the direction n ⎛
⎞ cos φ sin φ 0 R1 = ⎝ − sin φ cos φ 0 ⎠ . 0 0 1 Consider a second rotation around the new y axis for an angle π/2 − θ so that the new x axis coincides with the direction n: ⎛
⎞ sin θ 0 cos θ R2 = ⎝ 0 1 0 ⎠ . −cos θ 0 sin θ The composition of the two rotations is ⎛
⎞ sin θ cos φ sin θ sin φ cos θ cos φ 0 ⎠. R = R2 · R1 = ⎝ − sin φ − cos θ cos φ − cos θ sin φ sin θ
(3.13)
3.2
The Lorentz Transformation
111
If n = (n x , n y , n z ), then the angles φ, θ are given by ny , nx
tan φ =
cos θ = n z .
Application: Derivation of the general Lorentz transformation. Consider two coordinate systems Σ1 , Σ2 in spacetime with parallel axes and relative velocity u. We rotate the space axes of Σ1 , Σ2 so that the xaxis is along the 10 direction of u. This rotation defines the 4 × 4 matrix R(u) = , where the 0R 3 × 3 Euclidean matrix R is given in (3.13). Show that – The matrix R(u) is a Lorentz transformation (that is, satisfies the condition R t (u)η R(u) = η). – The transformation L(u) relating Σ1 , Σ2 is given by the relation: L(u) = R −1 (u)L 0 (u)R(u) = R t (u)L 0 (u)R(u), where L 0 (u) is a boost along the direction of u with speed u. – Calculate the transformation L(u) and show that it can be written as follows: ⎛ L(u) = ⎝
⎞
−γ u μ
γ
−γ u μ δνμ +
γ −1 μ u uν u2
⎠,
t μ where u μ = (u x , u y , u z ) , u μ = u x , u y , u z , and δνμ + γu−1 2 u u ν is the symmetric 3 × 3 matrix: ⎛ ⎜ ⎜ ⎜ ⎜ ⎝
1+
γ −1 2 ux u2
γ −1 ux uz u2
γ −1 ux u y u2
1+
γ −1 2 uy u2
γ −1 u y uz u2
1+
⎞ ⎟ ⎟ ⎟. ⎟ ⎠
γ −1 2 uz u2
– Consider a fourvector Ai which in the LCF Σ1 , Σ2 has decomposition
A0 1 A1
Σ1
,
A0 2 A2
Σ2
and prove the relations A1 · u , =γ − c u u·A1 A2 = A1 − γ A0 + (γ − 1) 2 u. c u
A02
A01
112
3 The Position FourVector
– Consider the matrix B = L(u)A, where A is a threeEuclidean transformation defined by the Euclidean 3 × 3 matrix A3 . Show that B is a Lorentz transformation. [Proof: B t ηB = At L t ηL A = At η A = η.] The matrix B describes the most general Lorentz transformation for which the space axes of the LCF Σ1 , Σ2 are not parallel and they become so with the action of the Euclidean threerotation A3 . 3.2.7. Using the identity of the vector calculus A × (B × C) = (A · C)B − (A · B)C on the double cross product, we have u × (u × r) = (u · r)u − u 2 r ⇒
(u · r)u 1 = 2 u × (u × r) + r. 2 u u
Hence r = r − γ ut + (γ − 1)r + (γ − 1) u×(u×r) = γ (r − ut) + (γ − 1) u×(u×r) . u2 u2 3.2.8. (a) We have L t ηL =
10 0 At
−1 0 0 I
10 0A
=
−1 0 0 At A
.
This matrix equals the Lorentz metric ηi j = diag(−1, I3 ) if and only if At A = I3 , that is, A is a Euclidean transformation. (b) For the tensor Ti j we have under the action of the transformation Ti j = L ri Tr s (L sj )t = =
T00 T0μ ATμ0 ATμν
10 0A
T00 T0μ Tμ0 Tμν
10 0 At
=
10 T00 T0μ At = . ATμ0 ATμν At 0 At
3.2.9. The Lorentz transformation is linear; therefore, we can write ⎫ L( α ) = A α ⎪ ⎪ ⎬ L(x α ) = α1 x α + β1 y α + γ1 α + δ1 m α , L(y α ) = α2 x α + β2 y α + γ2 α + δ2 m α ⎪ ⎪ ⎭ L(m α ) = α3 x α + β3 y α + γ3 α + δ3 m α
(3.14)
where we have used that α is an eigenvector of the (proper) Lorentz transformation, the various coefficients being constants. To determine the constants we write the
3.2
The Lorentz Transformation
113
conditions on the basis vectors in terms of the coefficients. We start with the fourvector x α , which satisfies the conditions x α xα = 1 , x α α = x α yα = x α m α = 0. The Lorentz transformation preserves the inner products, that is, these relations are invariant under the action of the transformation. This implies the conditions: L(x α )L(xα ) = 1 , L(x α )L( α ) = L(x α )L(yα ) = L(x α )L(m α ) = 0. Replacing in (3.14) we find the equations δ1 = 0, α1 α2 + β1 β2 + γ1 δ2 = 0,
(3.15) (3.16)
α1 2 + β1 2 = 1,
(3.17)
α1 α3 + β1 β3 + γ1 δ3 = 0.
(3.18)
Similarly for the fourvector y α we have L(y α )L(yα ) = 1, L(y α )L( α ) = L(y α )L(xα ) = L(y α )L(m α ) = 0. Replacing in (3.14) we find δ2 = 0,
(3.19)
α2 + β2 = 1, α2 α3 + β2 β3 + γ2 δ3 = 0.
(3.20) (3.21)
2
2
Finally, the fourvector m α gives the two conditions Aδ3 = 1, 2 α3 2 + β3 2 + γ3 = 0. A
(3.22) (3.23)
Equations (3.15), (3.16), (3.17), (3.18), (3.19), (3.20), (3.21), (3.22), and (3.23) are a system of simultaneous equations with unknowns, the constant coefficients. The solution of the system is
114
3 The Position FourVector
α1 = β2 = α β1 = −α2 = β α3 = r2 γ
(3.24) (3.25) (3.26)
β3 = r2 δ γ1 = −r2 A(αγ + βδ)
(3.27) (3.28)
γ2 = r2 A(αδ − βγ ) 1 γ3 = − A(γ 2 + δ 2 ). 2
(3.29) (3.30)
Therefore, the action of the Lorentz transformation on the vectors of the null tetrad is L( α ) L(x α ) L(y α ) L(m α )
⎫ = A α ⎪ ⎪ ⎬ = αx α + βy α − r2 A(αγ + βδ) α , α α α = −βx + αy + r2 A(αδ − βγ ) ⎪ ⎪ ⎭ = r2 γ x α + r2 δy α − 12 A(γ 2 + r 2 ) α + A1 m α
where α, β, γ , δ, r2 ∈ R and α 2 + β 2 = 1. 3.2.10. Let L +↑ (β) =
γA −γ A β Aμ
−γ A β Aμ φ(β A )μν
(A = 1, 2) be two Lorentz transformations. Then L +↑ (β 2 )L +↑ (β 1 ) = =
γ2 −γ2 β 2μ
γ1 γ2 (1 + β 1 · β 2 ) μ ρ −γ1 γ2 β2 − γ1 φ(β 2 )μρ β1
−γ2 β2μ φ(β 2 )μν
γ1 −γ1 β 1μ
−γ1 β1μ φ(β 1 )μν
−γ1 γ2 β1μ − γ2 φ(β 1 )ρμ β2ρ μ φ(β 1 )μρ φ(β 2 )ρν + γ1 γ2 β1 β2ρ
.
The term (1, μ) can be written as β2μ γ1 − 1 −γ1 γ2 β1μ + + (β 1 ·β 2 )β1μ = −γ1 γ2 (1 + β 1 · β 2 )v21 , γ1 γ1 β22 where v21 =
1 1+β 1 ·β 2
β1 γ2 − 1 . + 1+ (β ·β )β 1 2 2 γ2 γ2 β22
=
3.2
The Lorentz Transformation
115
Similarly the (μ, 1) term is written as −γ1 γ2 (1 + β 1 · β 2 )v12 , where v12 =
1 1+β 1 ·β 2
β2 γ1 − 1 . + 1+ (β ·β )β 1 2 1 γ1 γ1 β12
Finally, the (μ, ν) term gives μ
μ
ρ
μ
Γ12ν = φ2ρ φ1ν + γ1 γ2 β2 β1ν . Based on the above, the composite Lorentz transformation is written as follows L +↑ (β 2 β 1 ) = L +↑ (β 2 )L +↑ (β 1 ) =
γ12 −γ12 β21μ μ μ −γ12 β21 Γ12ν
,
where γ12 = γ1 γ2 (1 + β 1 · β 2 ). The length 2 = β 12 · β 12 = ... = β12 " # 1 = (1 + β 1 · β 2 )2 − (1 − β12 )(1 − β22 ) . 2 (1 + β 1 · β 2 )
Using the identity of vector calculus a × b2 = a2 b2 − (a · b)2 , we compute eventually 2 β12 = β 12 · β 12 =
(β 1 −β 2 )2 − (β 1 ×β 2 )2 . (1+β 1 ·β 2 )2
2 2 2 We note that β12 is symmetric in the β1 , β2 , hence β12 = β21 . It is a simple calculation to show that
γ12 =
1 , 2 1 − β12
that is, the quantity γ12 is the γ factor for the relative velocities v12 , v21 .
116
3 The Position FourVector
Let φ be the angle between the velocities v1 , v2 . Then
γ12 =
⎧ ⎪ ⎨
β 1 +β 2 1+β 1 ·β 2
when v1 v2 (φ = 0)
⎪ ⎩ β 2 + β 2 − β 2 β 2 when v ⊥ v (φ = 1 2 1 2 1 2
π ) 2
.
The angle ψ between the velocities β 12 , β 21 is given by cos ψ =
β 12 ·β 21 2 sin2 φ = 1 − , 2 1 + 2A cos φ + A2 β12
1/2 1 +1)(γ2 +1) (1 ≤ A ≤ ∞). where A = (γ (γ1 −1)(γ2 −1) Considering this angle as the measure of rotation the unit vector e along the axis of rotation is determined from the relation eˆ =
β 2 ×β 1 β 21 ×β 12 = . β 21 ×β 12  β 2 ×β 1 
The angle ψ is called Wigner’s angle.
3.3 The Boost 3.3.1. From the boost relating Σ, Σ we have A0 (x ) = γ (u) A0 (x) − β A1 (x) = 0, A1 (x ) = γ (u) A1 (x) − β A0 (x) = 0, A2 (x 2 (x) = sin 5x 1 − 3x 0 = " # = sin 5 γ (u) x 1 + βx 0 − 3 γ (u) x 0 + βx 1 = 1 " #2 = sin γ (u) (5β − 3)x 0 + (5 − 3β)x 1 , A3 (x ) = A3 (x) = A2 (x) = A2 (x ). Note that in the RHS the expressions are in terms of the components of Σ and not of Σ! 3.3.2. Suppose that P emits the signal when it is at the position x1 along the xaxis of Σ. The time traveled by P in Σ is γ Δτ . If u is the speed of P in Σ, then x1 = uγ Δτ.
3.3
The Boost
117
On the other hand, the traveling time of the signal in Σ is t − γ Δτ and its speed is c. Hence x1 = c(t − γ Δτ ). Equating the two expressions, we find c(t − γ Δτ ) = uγ Δτ ⇒ β =
t 2 − Δτ 2 . t 2 + Δτ 2
3.3.3. The boost relating Σ, Σ gives β t =γ t− x . c
The points of Σ on which the clocks coincide with those of Σ are defined by the condition t = t. This gives t =γ
1 β c 1 β −1 =− x ⇒ x = 1− t. t− x ⇒t c γ c β γ
This is the required plane. Its propagation speed in Σ is dx c u= = dt β
1/2 ! 2 1 c υ2 1− ⇒u= . 1− 1− 2 γ υ c
If we consider t1 = t + a and t1 = t + δ to be two different origins of time, then the boost relating Σ, Σ gives β β t + δ − x = γ t + γ δ − γ x ⇒, c c 1 a c 1− t+ δ− x= β γ γ
t + a = γ
from which follows the same velocity of propagation. 3.3.4. We have  e0,L 2L = ( γ βγ )  e0,L 2E = ( γ βγ )
−1 0 0 1 10 01
Similarly we work with e1,L .
γ βγ
γ βγ
= −γ 2 + β 2 γ 2 = −1,
= γ 2 + β 2 γ 2 = 2γ 2 − 1 > 0.
118
3 The Position FourVector
To find the locus of the vectors e0,L , e1,L for the various values of the parameter x β, we write e1 = and require y Σ t −1 0 x x = ±1 ⇒ x 2 − y 2 = ∓1. 0 1 y y We conclude that the tip of the vectors e0,L , e1,L on the Euclidean plane trace a hyperbola with asymptotes the lines x = ±y, as shown in Fig. 3.1.
1
1
Fig. 3.1 The unit Lorentz vectors
3.3.5. We consider the point A( = 1, x = 0) in Σ . This point defines the four 1 vector which in Σ becomes 0 Σ A = γ (1 + β · 0) = γ , x A = γ (0 + β · 1) = γβ. Because Σ, Σ have common origin the point A defines the axis on the plane (x, ) of Σ; the axis of Σ is represented by a straight line which is defined from the point A and the point (γ , βγ ). In order to draw the x axis we “compute” the image of the point B ( = 0, x = 1) of Σ which in Σ is B = γ (0 + β · 1) = βγ , x B = γ (1 + β · 0) = γ . Therefore, the axes of Σ in the plane (x, ) of Σ are represented with two lines as shown in Fig. 3.2
3.3
The Boost
119
Fig. 3.2 Hyperbolic rotation on the Euclidean plane
The axes , x are (obviously) symmetric about the dichotomous of the axes , x. For the speed given in the problem, we compute β = 0.6 and γ = 1.25; hence from the figure we have tan φ =
γβ = β = 0.6 ⇒ φ = 30◦ . γ
3.3.6. We solve the problem algebraically and geometrically. Algebraic Solution (a) Without restricting the generality we assume that the components of the four vector AB i in Σ and Σ are AB i = (0, x, 0, 0)Σ = (ct , x , 0, 0)Σ (why?). The boost relating Σ, Σ gives x = γ x, t = −
βγ x . c
We consider now x , t as functions of β, with 0 < β < 1. We have for x βx dx = . dβ (1 − β 2 )3/2 The function x (β) has a minimum when the first derivative vanishes (and the second derivative is negative). Condition ddβx = 0 gives β = 0 (that is, Σ coincides with Σ) and the minimum value is xmin = x (0) = x. Concerning the time we have x d 2 dt x =− γ − 1 = − γ 3, dβ c dβ c where we have used that dγ = βγ 3 . dβ This does not vanish in the interval [0, 1]. Hence Δt = t (β) is not bounded for the acceptable values of the parameter β.
120
3 The Position FourVector
(b) In this case we have AB i = (ct, 0, 0, 0)Σ = (ct , x , 0, 0)Σ (why?). The boost relating Σ, Σ gives −1/2 t = γ t = t 1 − β2 . Because (1 − β 2 ) > 0 when β ∈ [0, 1], t and t have the same sign; therefore, the time sequence of the events A, B remains the same for all β, hence for all observers Σ . Note: A second solution to (b) can be given by means of the proper observer. Indeed because A, B occur at the same spatial point of Σ, Σ coincides with the proper frame Σ+AB of the timelike fourvector AB i . The representation of AB i in Σ+AB is (dl + , 0)Σ+AB . The dl + is an invariant because (AB)2 = −(dl + )2 ; therefore, its value is the same in all RIO. This value is the minimum because in any other RIO related to Σ with a boost with factor γ , the spatial distance of the events A, B is γ l + > l + . Geometric Solution (a) Consider Fig. 3.3. Explanation of the figure Events: A, B: Two simultaneous events in the RIO Σ. Σ: World line of the RIO Σ. Σ : World line of the RIO Σ with factor β wrt Σ. Data: AB: Spatial distance of the events A, B in the RIO Σ. Required: AΓ: Temporal difference of the events A, B in Σ . From the triangle AΓB, we have (AΓ) = (AB) sinh φ = (AB)βγ = (AB)(γ 2 − 1)1/2 . But γ ∈ (0, +∞), therefore (AΓ) ∈ (0, +∞).
Fig. 3.3 Spacetime diagram for time measurement
3.4
Length Contraction
121
(b) In this case we have Fig. 3.4 Explanation of the figure Events: Σ, Σ+AB : World lines of the two RIO. Σ+AB : Proper observer of the timelike fourvector AB. A, B: Events occurring at the same spatial point and not simultaneous in Σ+AB . Data: AB : Time difference of the events A, B in Σ+AB . cosh φ = γ . Required: AΓ: Time difference of the events A, B in Σ. From the triangle AΓB we have (AΓ) = (AB) cosh φ = γ (AB). Because γ > 1, the time difference of the events A, Γ is the same with that of A, B .
3.4 Length Contraction 3.4.1. We compute β =
2.7×108 3×108
= 0.9 ⇒ γ = 2.29.
(a) The lifetime t of the particle in the laboratory is t = γ τ = 2.29 × 10−6 s. (b) The distance covered by the particle in the laboratory before disintegrates is = υt = 2.7 × 108 (m/s) × 2.29 × 10−6 s = 6.18 × 102 m.
Fig. 3.4 Spacetime diagram for the measurement of spatial distance
122
3 The Position FourVector
(c) According to Newtonian Physics the particle covers in the laboratory distance N = υτ = 2.7 × 108 (m/s) × 10−6 s = 2.7 × 102 m. This value is 44% smaller than the relativistic and it is possible to be measured in the laboratory. Such measurements validate each day the validity of Special Relativity. 3.4.2. We consider the events: Event A: Particle A (say) hits the target. Event B: Particle B hits the target. Let Σ be the proper frame of A. We have the following table of coordinates: Σ Σ A : (ct A , x A )Σ (ct A , x A )Σ B : (ct A + cΔt, x A )Σ (ct A + cΔt , x A − d )Σ AB i : (cΔt, 0)Σ (cΔt , −d )Σ The boost relating Σ, Σ gives for the fourvector B Ai −d = γ (υ)(0 − βcΔt) ⇒ d = γ (υ)υΔt. But υΔt = d is the distance of the two particles in Σ. Therefore, d = γ (υ)d, the expected result.
3.4.3. Consider the following events: Event A: Emission of light signal from the point A (say). Event B: Emission of light signal from the point B. The table of coordinates for the events A, B is Σ Σ A : (ct A , x A )Σ (ct A , x A )Σ B : (ct A , x A + d)Σ (ct A + cΔt , x A + d )Σ AB i : (0, d)Σ (cΔt , d )Σ The boost relating Σ, Σ gives for the time coordinate cΔt = γ (u)(0 − βd) ⇒ Δt = −
ud γ (u) c2
3.4
Length Contraction
123
and for the spatial d = γ (u)d. The second signal is emitted in Σ after the time period Δt . Therefore, the distance of the two signals in Σ, the moment of emittance of the second signal (in Σ ) – taking into consideration that the speed of the light signal when emitted is c – is $ 1−β ud u d = d − cΔt = γ d − c =d . γ = dγ 1 − c2 c 1+β 3.4.4. Let A, B the events which refer to the measurement of the distance of the light signals. For the measurement of the spatial distance of the events A, B in Σ , we have the following table of coordinates: Σ Σ A : (ct A , x A )Σ (ct A , x A )Σ B : (ct A , x A + d)Σ (ct A + cΔt , x A + d )Σ AB i : (0, d)Σ (cΔt , d )Σ The boost relating Σ, Σ gives $ d = γ (d − βd) = γ d(1 − β) ⇒ d = d
1−β . 1+β
3.4.5. In the rest frame of the cylinder there are N=
m N0 k
molecules (this is an invariant!), where N0 = 6.023 × 1023 molecules/mole is the Lochschmidt constant and m is the mass of the gas. The volume of the gas is V = π R 2 L(cm 3 ); therefore, the average number of molecules per cm3 is n0 =
m N0 1 N = . V k π R2 L
The average number of molecules per cm3 for Σ is n0 =
N , V
where V is the volume of the cylinder for Σ for each case (a) or (b) Therefore, n=
m N0 1 , k S
124
3 The Position FourVector
L
(a)
(b)
Fig. 3.5 Moving cylinder with gas
where S is the surface of the cross section of the cylinder and its height in each case. (a) Figure 3.5 (a). In this case =
L , γ
S 2 = π R 2 , hence n=
(b) Figure 3.5 (b). In this case = L , S =
π R2 γ
m N0 γ = γ n0. k π R2 L
(the proof is given below); therefore, n = γ n0.
Calculation of the cross section area S in case (b). Considering the axes x, y in the plane of the base of the cylinder and the xaxis (yaxis) normal (parallel) to the velocity of the observer, we have x = x, y =
1 y. γ
Therefore, x 2 + γ 2 y 2 = R 2 , that is, the cross section of the cylinder is an ellipsis. The area S of the ellipsis is √
R
S = where z =
'γ
y =0 γ R
y
dy
R 2 −γ 2 y 2
'
x =0
R
dx =
'γ y =0
R 2 − γ 2 y 2 dy =
R2 γ
'1 √ 1 − z 2 dz, 0
(0 ≤ z ≤ 1). We set z = sin θ ,θ ∈ [0, π ] and compute (1 1 − z 2 dz = π ⇒ 0
S =
π R2 . γ
3.5
Time Dilation
125
3.5 Time Dilation 3.5.1. Let l be the distance traveled by the photon in Σ and t the time taken by the photon between its emission (event A) and its absorption (event B) in Σ. Then l = ct and we have the following coordinates for the events A, B : Σ Σ A : (0, 0, 0)Σ (0, 0, 0)Σ B : (l, l cos θ, l sin θ )Σ (ctc , l cos θ , l sin θ )Σ . The boost relating Σ, Σ gives tc = γ (l − βl cos θ ) = γ l(1 − β cos θ ), l cos θ = γ (l cos θ − βl), l sin θ = l sin θ. Dividing the last two relations, we get cot θ =
γ (cos θ − β) γ l(cos θ − β) = . l sin θ sin θ
Squaring and adding the same relations, we find l = l 2 sin2 θ + γ 2l 2 (cos θ − β)2 = l 2 [1 − cos2 θ + γ 2 (cos θ − β)2 ]. 2
3.5.2. We consider the following events A : Creation of the μmeson. B: Decay of the μmeson. If Σ is the Earth frame and Σ the proper frame of the μmeson, we have the following table of coordinates of the events A, B: Σ Σ A : (ct A , x A , 0, 0)Σ (ct A , x A , 0, 0)Σ+ B : (ct A + cΔt, x A + Δx, 0, 0)Σ (ct A + cτμ , x A , 0, 0)Σ+ AB i : (cΔt, Δx, 0, 0)Σ (cτμ , 0, 0, 0)Σ+ The boost relating Σ, Σ+ gives cΔt = γ τμ ,
d = βcΔt.
For β = 0.985 we compute γ = 5.795 and then for the lifetime of the μmeson τμ = 2.35 × 10−6 s.
126
3 The Position FourVector
3.5.3. The decay law in the Σ is t , N = N0 exp − τ where N (t = 0) = N0 and τ is the lifetime of the π meson in Σ. In time t a meson covers a distance l = βct; hence l N l N0 ⇒ ⇒, = exp − = ln N0 βγ cτ0 βγ cτ0 N
l = βγ cτ0 ln
N0 . N
From the data of the problem, we compute ln
N0 = ln 2 = 0.693 N
and cτ0 = 3 × 1010
cm × 2.6 × 10−8 s = 780 cm. s
Replacing we find l = 540.6 β γ
(cm).
The graph l = l(β) in Fig. 3.6. 1600 l 1200
800
400
0.2
Fig. 3.6 Distance travelled in terms of speed
0.4
0.6
0.8
1
3.5
Time Dilation
127
3.5.4. First Solution (i) We compute the components of the fourvector 12 = x i in the laboratory: x i = (2) − (1) = (3 × 108 × 5.31 × 10−9 , 0.880 + 0.500, 0.480 − 0.500, 0.250 −0.200) = (1.593, 1.380, −0.020, 0.050). The length x 2 of the fourvector x i is, " # x 2 = 1.3802 + (−0.020)2 + 0.0502 − 1.5932 = −0.630 m2 . Let τ be the lifetime of the K meson. Then x 2 = −c2 τ 2 from which follows τ = 2.65 × 10−9 s. (ii) The spatial distance Δd of the events 1, 2, in the laboratory, is #1/2 " = (1.382 + 0.022 + 0.052 )1/2 Δd = (x2 − x1 )2 + (y2 − y1 )2 + (z 2 − z 1 )2 = 1.381 m. Therefore, the speed of the K meson in the laboratory is u=
Δd 1.381 m/s = 2.60 × 108 m/s. = Δt 5.31 × 10−9
Second Solution (using time dilation) We calculate the speed of the K meson v = 2.60 × 108 m/s as in (ii). β = 0.867, hence γ (v) = 2.01. The lifetime of the K meson in the laboratory is Δt = t2 − t1 = 5.31 × 10−9 s. The lifetime τ of the K meson in its proper frame is τ=
1 Δt = × 5.31 × 10−9 = 2.64 × 10−9 s. γ 2.01
Third Solution (using the algebraic method) Let Σ be the LCF of the laboratory and Σ the proper frame of the K meson. We have the following table of coordinates for the events 1, 2: Σ Σ 1 : (0.000, −0.500, 0.500, 0.200)Σ (0, 0)Σ 2 : (1.593, 0.880, 0.480, 0.250)Σ (cτ, 0)Σ 21 (1.593, 1.380, −0.020, 0.050)Σ (cτ, 0)Σ
128
3 The Position FourVector
The boost relating Σ, Σ gives 1.593 = γ cτ, 1.380 = βγ cτ from which the previous results follow. 3.5.5. Algebraic Solution Consider the following events: A : Emission of the photon from the mirror 1 (say). B : Reflection of the photon from mirror 1. We have the following table of coordinates in the proper frame of the rod, the Σ , say, and Σ: Σ Σ A : (ct A , x A )Σ (ct A , 0)Σ B : (ct B , x B )Σ (ct A + cT, 0)Σ i AB : (cΔt AB , Δx AB )Σ (cT, 0)Σ The boost relating Σ, Σ gives Δt AB = γ T. Therefore, the unit of time in Σ is γ times the unit of time in Σ . [The other equation of the boost gives Δx AB = βcΔt AB . What is the meaning of this equation?] Geometric Solution We have the following spacetime diagram (Fig. 3.7) Description of diagram Σ: World line of the LCF Σ K 1 , K 2 : World lines of mirrors A, B : Events of emission and reflection of light signal on mirror K 1 φ : Rapidity between the LCF Σ and K 1 . Data: (AB) = cT Required: (AΓ) = cΔt AB . From the triangle ABΓ, we obtain (AΓ) = (AB) cosh φ ⇒ Δt AB = γ T.
3.5.6. Algebraic solution (i) Consider the following events A: nth lighting of the bulb. B: (n + 1)th lighting of the bulb.
3.5
Time Dilation
129
Fig. 3.7 The Feynmann clock in motion
The boost relating Σ, Σ gives for the time coordinate c Δt B = γ (2d1 /c) > (2d1 /c). But 2d1 /c is the period of the clock for Σ and Δt B the period of the clock in Σ . We conclude that the clock is “slow” for Σ . (ii) Working similarly we find Σ Σ A : (0, 0)Σ (ct A , 0, y A , 0)Σ B : (2d1 /c, 0)Σ (ct B , 0, y B , 0)Σ AB i (2d1 /c, 0)Σ (c Δt AB , 0, d , 0)Σ from which follows c Δt = γ (2d1 /c) > 2d1 /c, which shows that the clock is slow for Σ in this case too. Geometric Solution We have the following spacetime diagram (Fig. 3.8) Description of the diagram Σ: World line of bulb (and the observer Σ) Σ : World line of the observer Σ A, B: Events of successive emission of light signals for the bulb Λ Data: (AB): Period of the clock in Σ. Required: AΓ: Period of the clock in Σ .
130
3 The Position FourVector
Fig. 3.8 The Feynmann clock in motion
From the triangle AΓB, we have (AΓ) = (AB) cosh φ = γ (AB) > (AB), which shows that the clock is slow for Σ . Similarly we work for part (ii) of the problem. 3.5.7. We recall that the rate of disintegration ddtN of a beam of radioactive particles in an LCF Σ at every moment is given by the formula 1 dN = − N (t) ⇒ N (t) = N0 e−t/T , dt T where T is the average lifetime of the particles in Σ and N0 is the initial (that is for t = 0) number of particles. [The quantity ddtN is called radioactivity of the beam, it is not an invariant and its value depends on the LCF where the beam is considered.] For the counters A, B, we have N (t A ) = N0 e−t A /T ,
N (t B ) = N0 e−t B /T ⇒
N (t A ) = e(t B −t A )/T , N (t B )
where t A , t B are the time moments (in Σ) at which the beam is passing through the , hence counters A, B, respectively. The t B − t A = (AB) u N (t A ) = e(AB)/uT . N (t B )
3.6
Problems with Rods
131
From the definition of the radioactivity of the beam, we have
dN 1 = − N A (t A ), dt A T dN 1 = − N B (t B ). dt B T Thus dN N A (t A ) 1000 = ddtN A = = 4. N A (t B ) 250 dt B Replacing we find for the average lifetime of particles in Σ: T =
7.5 s. c ln 2
The halflife of the particles in Σ is t 1 and is given by the relation 2
t 1 = T ln2 = 2
7.5 s = 2.5 × 10−8 s. c
In the proper frame of the particles the half lifetime τ is: τ= where γ = 1/ 1 −
u2 c2
1 t1 , γ 2
= 54 . Replacing we find τ = 2 × 10−8 s.
In the LCF Σ which moves normally to the beam with speed v = 35 c ⇒ γ (v) = 54 , the halflife time is t 1 = γ (v)τ = 2.5 × 10−8 s. 2
3.6 Problems with Rods 3.6.1. Let A, B be the events which refer to the measurement of the distance of the light signals. For the measurement of the spatial distance of the events A, B in Σ , we have the following table of coordinates:
132
3 The Position FourVector
Σ Σ A : (ct A , x A )Σ (ct A , x A )Σ B : (ct A , x A + d)Σ (ct A + cΔt , x A + d )Σ AB i : (0, d)Σ (cΔt , d )Σ The boost relating Σ, Σ gives $ d = γ (d − βd) = γ d(1 − β) ⇒ d = d 3.6.2. We compute β =
2.7×108 3×108
1−β . 1+β
= 0.9 ⇒ γ = 2.29.
(a) The lifetime t of the particle in the laboratory is: t = γ τ = 2.29 × 10−6 s. (b) The distance covered by the particle in the laboratory before disintegrates is = υt = 2.7 × 108 (m/s) × 2.29 × 10−6 s = 6.18 × 102 m. (c) According to Newtonian Physics the particle covers in the laboratory distance: N = υτ = 2.7 × 108 (m/s) × 10−6 s = 2.7 × 102 m. This value is 44% smaller than the relativistic and it is possible to be measured in the laboratory. Such measurements validate each day the validity of Special Relativity. 3.6.3. The speed of the rod in Σ equals υ = TL , where L is the length of the rod in Σ and T is the required time to pass in front of the point P. Let A, B be the events of passing of the ends of the rod in front of P. We have the following table of coordinates: Σ− Σ A : (ct A , x A )Σ (ct A , x A )Σ− B : (ct A + cT, x A )Σ (ct A + cT , x A + L 0 )Σ− AB i : (cT, 0)Σ (cT , L 0 )Σ− The boost relating Σ, Σ− gives υ L 0 = γ 0 + cT ⇒ L 0 = γβcT, c υ cT = γ cT + 0 ⇒ T = γ T. c
3.6
Problems with Rods
133
Using the identity β 2 γ 2 = γ 2 − 1, we compute βγ = γ 2 − 1 =

T 2 − 1. T2
Finally, $
L 0 = cT 1 −
T T
2 .
3.6.4. First Solution The length of the rod in Σ is γl . The speed of the particle wrt the rod in Σ is 2βc. Therefore, the required time t is t=
l . 2γβc
Second Solution The speed of the particle wrt the rod is v =
−v − v 1−
= −γ 2 2v.
v2 c2
Therefore, the time required for the particle to trace the rod, in the proper frame of the rod, is 2vγl 2 . Transforming this time in Σ we find t =γ
l l . = 2 2vγ 2βcγ
3.6.5. Algebraic Solution Let AB be the spaceship 1 and ΓΔ be the spaceship 2. The measurement of the length of spaceship 2 from the observer in spaceship 1 requires that the ends Γ, Δ will be simultaneous in spaceship 1. Therefore, we have the following table of the coordinates of the events Γ, Δ: 1 2 Γ : (ctΓ , xΓ )1 (ctΓ , xΓ )2 Δ : (ctΓ + cT1 , xΓA )1 (ctΓ + cT2 , xΓ + 2 )2 ΓΔi : (cT1 , 0)1 (cT2 , 2 )2 The boost relating the two spaceships gives 2 = γ uT1 ⇒ u =
3 2 4 4 5 T1
1+
1
2 cT1
2 .
134
3 The Position FourVector
[Question: How u changes in terms of the parameters 2 and T1 and the quotient 2 ?] T1 (b) The observation of spaceship 1 from spaceship 2 is similar to (a); Therefore, without any further calculations we write 1 = γ uT2 , where 1 is the proper length of spaceship 1. From the above relations it follows that 2 T2 = 1 T1 , from which one computes the proper length 1 . We note that the product (proper length) × (observation time) is independent of the relative speed of the spaceships. Geometric Solution (see Fig. 3.9) Explanation of the figure A, B : World lines of ends of spaceship 1. Γ, Δ: World lines of ends of spaceship 2. E, Z : Events of observation of the length of spaceship 2 from spaceship 1 (E Z = cT1 ). E, K : Events of observation of the length of spaceship 1 from spaceship 2 (E K = cT2 ). (a) Draw Z Z normal to E K . Then Z Z = 2 and from the triangle E Z Z we have 3 2 Z Z 2 2 4 4 = sinh φ = ⇒ βγ = ⇒u= 5 ZE cT1 cT1 T1
1+
1
2 cT1
2 .
(b) Draw E E normal to B H. Then E E = 1 and from the triangle E E K we have sinh φ =
1 . cT2
Hence 1 2 = ⇒ 1 T1 = 2 T2 . cT2 cT1
3.6.6. We calculate first the relative speed u 21 of the rods and then apply the Lorentz contraction formula L/γ21 . One is possible to work in two ways.
3.6
Problems with Rods
135
H
Z
K Z′ cT2
φ
E′
E
l2
l1 A
B
Γ
Fig. 3.9 Spacetime diagram for the measurement of the length of a rod
First method We consider the rod 1 and write for the fourvelocities u i2 , u i1 : u i2 =
γ21 c γ21 u21
Σ1
, u i1 =
c . 0 Σ1
These velocities in Σ are u i2 =
γ2 c γ2 u2
Σ
, u i1 =
γ1 c γ1 u1
Σ
.
The invariant u i2 u 1i gives u i2 u 1i = −γ21 c2 = −γ1 γ2 c2 + γ1 γ2 (u1 ·u2 ) ⇒ u1 ·u2 , γ21 = γ1 γ2 1 − 2 c from which follows γ21 = γ 2 (1 + β 2 ).
136
3 The Position FourVector
Therefore, L 21 =
γ 2 (1
L . + β 2)
Second method We apply the relation which gives the relative speed in terms of the speeds of the mobiles in Σ (see the relevant theory): 2 β21 =1−
(1 − β12 )(1 − β22 ) . (1 − β 1 · β 2 )2
Third method We consider the Lorentz transformation which relates Σ, Σ1 and transfer the velocity of 2 to 1. This method calculates both γ21 and u 21 . We have u i21 =
γ βγ βγ γ
γc −γ u
Σ
=
γ 2c + β 2γ c −βγ 2 c − γ 2 βc
=
γ 2 (1 + β 2 )c −2βγ 2 c
,
from which follows γ21 = γ 2 (1 + β 2 ) and u 21 = −
2βc 2βγ 2 c =− . 2 +β ) 1 + β2
γ 2 (1
3.6.7. (a) Algebraic solution Consider the following events: A, B : Emittance of light signals from the end points of the train. M: Arrival of light signals at the middle point M of the train. Taking into consideration that (i) the light signals for Σ arrive at the middle of the train simultaneously and (ii) the speed of the signals is c, we infer that the signals where emitted simultaneously for Σ. This leads to the following table of coordinates: Σ Σ A : (ct A , x A )Σ (ct A , x A )Σ B : (ct A , x B )Σ (ct A + cΔt , x A + l)Σ AB i : (0, Δx AB )Σ (cΔt AB , l)Σ The boost relating Σ, Σ gives = −γβΔx AB cΔt AB
3.6
Problems with Rods
137
and the inverse transformation ). Δx AB = γ (l + βcΔt AB
Eliminating Δx AB we find cΔt AB = −βγ γ (l + βcΔt AB ) ⇒ cΔt AB = −βl ⇒ cΔt B A = βl.
Geometric Solution We have the following spacetime diagram (Fig. 3.10): Description of diagram (A), (B) : World lines of the end points A, B of the train. (M) : World line of the middle M of the train. A, B : Events of emission of light signals. M : Simultaneous arrival of light signals at M for observer Σ. Data: A A + B B = l, where l is the proper length of the train. Required: Time difference Δt A B = A B . From the triangles A ΔA and B ΔΓ, we have A Δ = A A tanh φ, ΔB = B B tanh φ ⇒ Δt A B = A Δ+ΔB = (A A + B B ) tanh φ = l tanh φ = βl.
Fig. 3.10 Spacetime diagram for the measurement of time difference for observer in the train
138
3 The Position FourVector
Fig. 3.11 Spacetime diagram for the measurement of time difference for observer in the ground
(b) Geometric Solution (Fig. 3.11): Description of spacetime diagram (A), (B), (M), A, B, M : As in the previous diagram Γ: Moment of simultaneous emittance of light signals for Σ from end points of the train. Required: (A1 B1 ). From the triangles B1 ΔB and A1 AΔ, we have ΔB1 = ΔB sinh φ, ΔA1 = ΔA sinh φ ⇒ (B1 A1 ) = (AB) sinh φ = l sinh φ = βγ l ⇒ cΔt AB = βγ l.
3.6.8. See Fig. 3.12 Let Σ− be the proper frame of the rod. The boost relating Σ, Σ− parallel and normal to the xaxis is: (AΔ)Σ = γ (AΔ)Σ− = γ (AB)Σ− cos θ = γ l cos θ. (BΔ)Σ = (BΔ)Σ− = l sin θ.
3.6
Problems with Rods
139
Fig. 3.12
Therefore, (AB)Σ =
(AΔ)2Σ + (BΔ)2Σ = l γ 2 cos2 θ + sin2 θ ⇒ (AB)Σ = l 1 + (γ 2 − 1) cos2 θ .
We note that θ = 0 ⇒ (AB)Σ = γ (AB)Σ− = γ l and θ = (AB)Σ− = l as it is expected.
π 2
⇒ (AB)Σ =
3.6.9. We consider the events A, B to be defined by the end points of the rod at each moment. Then assuming measurement of the length of the rod in Σ we have the following table of coordinates: Σ Σ A : (ct A , x A , y A , 0)Σ (ct A , x A , y A , 0)Σ B : (ct B , x B , y B , 0)Σ (ct A , x B , y B , 0)Σ i B A : (cΔt, x B − x A , y B − y A , 0)Σ (0, x B − x A , y B − y A , 0)Σ where Δt = t B − t A and l x = x B − x A , l y = y B − y A . We assume that the end point A starts its motion from the common origin of Σ, Σ the moment t = t = 0 and we have x A = vx t A y A = vy t A x B = vx t B + l x y B = v y t B + l y . Hence x B − x A = vx Δt + l x y B − y A = v y Δt + l y . Then in Σ the fourvector B Ai is written as follows: (cΔt, vx Δt + l x , v y Δt + l y , 0)Σ .
140
3 The Position FourVector
The boost relating Σ, Σ gives l x = γ (vx Δt + l x − uΔt) l y = y B − y A = v y Δt + l y . Also u u Δt = γ 0 + 2 l x = γ 2 l x . c c Therefore l x = γ l x + (vx − u) γ
u l c2 x
#
⇒ 1 + γ2
u2 c2
− γ2
vx u c2
l x = γ l x ⇒
uv x l x = γ 1 − 2 l x . c Working similarly we compute l y = l y + γ
vy u v yu l ⇒ l y = l y − γ 2 l x . c2 x c
3.6.10. The image of the rod in Σ (this is what we mean by appears) is observed when the ends A, B of the rod are simultaneous in Σ. This gives the following coordinate table: Σ Σ A : (ct, x A , y A , 0)Σ (0, 0, y A , 0)Σ B : (ct, x B , y B , 0)Σ (ct , l , y A + vt , 0)Σ B Ai : (0, Δx, Δy, 0)Σ (ct , l , vt , 0)Σ where l is the proper length of the rod. The boost relating Σ, Σ gives l = γu Δx, ct = −γu βu Δx, Δy = vt . Eliminating t we find Δy = −γu βu βv Δx. The angle θ made by the rod and the xaxis is tan θ = [What is the length of the rod?]
Δy = −γu βu βv . Δx
3.6
Problems with Rods
141
3.6.11. We have in an obvious notation: (AΔ)Σ = γ (AΔ)Σ , (BΔ)Σ = (BΔ)Σ ⇒
tan θ =
(BΔ)Σ (BΔ)Σ = γ (υ) ⇒ tan θ = γ (v) tan θ. (AΔ)Σ (AΔ)Σ
Concerning the limiting cases we have (i) Newtonian limit: β → 0 or β = 0 + O(β 2 ). Then tan θ = tan θ and the cone behaves as a solid, as it should. (ii) Ultrarelativistic limit: β → 1 or β = 1 + O(β 2 ). Then tan θ → ∞, hence θ → π2 , that is, the cone tends to degenerate to a surface (a circular disk) in the plane y , z . We continue with the contiguous surface of the cone. In Σ we have S0 = π (BΔ)Σ (AB)Σ = π (BΔ)Σ
(BΔ)2Σ (BΔ)Σ =π . sin θ sin θ
In Σ we have S0 = 2π
sin θ ( BΔ)2Σ ( BΔ)2Σ = 2π = S0 . sin θ sin θ sin θ
But tan θ γ tan θ = . sin θ = √ 2 1 + tan θ 1 + γ 2 tan2 θ Therefore sin θ cos θ 1 + γ 2 tan2 θ = S0 1 + γ 2 tan2 θ ⇒ γ tan θ γ $ cos2 θ 2 S0 = S0 + sin θ = S cos2 θ − β 2 cos2 θ + sin2 θ ⇒ 0 γ2 S0 = S0 1 − β 2 cos2 θ . S0 = S0
Note: We encourage the reader to solve the problem without the “obvious” assumptions we did above. That is, the reader must take into consideration that in Special Relativity there do not exist rigid bodies (one can define the Born rigidity and this is as far as one can go).
142
3 The Position FourVector
3.6.12. We consider the events: A, B: Emission of light signals from the ends of the rod at the proper moment t − in the proper frame Σ− of the rod. Due to the finite speed of light the events A, B are not simultaneous in Σ; therefore, the length of the rod seen by the observer at the origin of Σ (the apparent length of the rod) does not equal the measured length of the rod in Σ. We define the apparent length of the rod in Σ as the coordinate length, that is, L˜ = x B − x A . We have the following table of the coordinates of the events A, B in Σ and Σ : Σ Σ − − A : (ct + x A , x A )Σ (ct , 0)Σ B : (ct − + x B , x B )Σ (ct − , L 0 )Σ i AB : (x B − x A , x B − x A )Σ (0, L 0 )Σ −
The boost relating Σ− , Σ gives $ L 0 = γ L˜ (1 − β) =
1−β ˜ L. 1+β
Hence $ L˜ =
1+β L 0. 1−β
We see that the coordinate length L˜ differs from the proper length L 0 by the 1−β . Doppler factor k = 1+β
Fig. 3.13 Apparent length and measured length of a rod
3.6
Problems with Rods
143
If we consider β changing in the interval (−1, 1) where β > 0 (β < 0) when the rod approaches (goes away) from the origin, we have the graph of Fig. 3.13 from which we note that when the rod goes away (β > 0) L˜ > L 0 (length “expansion”), whereas when the rod approaches the origin (β < 0) L˜ < L 0 (length “contraction”). The measured length of the rod in Σ is L = L 0 /γ and it is always smaller than L 0 (length contraction). From this example it becomes clear that in Special Relativity one should distinguish carefully between the apparent images of the “bodies” and their measured dimensions.
Chapter 4
Four Velocity
4.1. (a) Let v i be the fourvelocity of the particle. Then we have in the two LCFs the decomposition ⎞ ⎛ ⎞ γc γc v i = ⎝ γ v ⎠ = ⎝ γ v ⎠ , γ v⊥ Σ γ v⊥ Σ ⎛
where ⊥ are taken with reference to the direction of u. From the Lorentz transformation relating Σ, Σ we obtain u · v , γ c = γu γ c 1 + 2 c γ v = γu γ v + u ,
(4.1)
γ v⊥ = γ v⊥ .
(4.3)
(4.2)
Then γ γu γ v + u + v⊥ γ γ γ " # γ γu v + v − v + γu u = v + (γu − 1)v + γu u . = γ γ
v = v + v⊥ =
Replacing v =
v ·u u u2
v= Also the term
γu −1 γu2 u 2
=
(4.4)
we find γ γu − 1 v + γ v · u + 1 γ u . u u γ γu2 u 2 γu −1 (γu2 −1)c2
v=
=
1 ; (γu +1)c2
therefore,
γ γu v · u v + + 1 γ u . u γ (γu + 1)c2 145
146
4 Four Velocity
But we know that γ 1 u · v = 1+ 2 . γ γu c Replacing we find the final result: v=
1 u · v γu v · u 1+ 2 v + + 1 γ u . u γu c (γu + 1)c2
(b) The acute angle θ between u and v in Σ is given by the relation v⊥  (γ /γ ) v⊥  & &⇒ tan θ = & & = &v & (γu /γ ) &γ v + γ u& tan θ =
v sin θ . γu (v cos θ + u)
(4.5)
(c) The speed v2 = v2⊥ + v2 = 2 2 2 " 2 # γ γ γu 2 2 (γ v + γ u) + (v⊥ ) = = γu (v + u)2 + (v⊥ )2 γ γ γ 2 1 2 2 " #2 γ = (v ) sin θ + γu2 (v )2 cos2 θ + u 2 + 2v u cos θ γ 2 " 2 2 # γ = γu (v ) + (v )2 (1 − γu2 ) sin2 θ + γu2 u 2 + 2γu2 v u cos θ γ * + 1 u 2v2 2 2 2 = (4.6) v + u + 2v u cos θ − 2 sin θ . θ 2 c 1 + uv cos 2 c
4.2. (a) Use the identity of vector calculus u × (A × u) = (u · u)A − (u · A)u = u2 A − (u · A)u. (b) From the definition of the parallel projection and (4.6) we have # (u · v2 ) 1 " v1 · u + (u · v1 )(γu − 1) − γu u2 u = u= 2 2 u u γu Q # " # 1 " 1 u · v1 1 γu (u · v1 ) − u 2 u = − 1 u= = 2 (v1 ) − u . 2 u γu Q Q u Q
(v2 ) =
4 Four Velocity
147
Similarly (v2 )⊥ =
1 1 1 u × (v2 × u) = u × (v1 × u) = (v1 )⊥ . u2 γu Qu 2 γu Q
[Note: Prove the second part considering (v2 )⊥ = v2 − (v2 ) and making use of (a).] 4.3. (i) We have v = dr/dt = c dr/dl. Replacing dr, dl in terms of dr , dl using the Lorentz transformation one finds the required expression. A second solution is to consider the fourvelocity in the two LCF and use again the Lorentz transformation. (ii) The threevelocities of the problem have the components u = ui, v = v cos θ i+ v sin θ j and v = v cos θ i + v sin θ j. From (i) we have for the xcomponent: γ u v cos θ + 1+γ v cos θ + γ u v cos θ + u c2 v cos θ = . = uv γ 1 + c2 cos θ 1 + uv cos θ c2 2
2
Similarly we compute v sin θ =
v sin θ . γ
Dividing by parts we obtain the required answer. Concerning the speeds we have * v = (v cos θ ) + (v sin θ ) = 2
2
2
1 cos θ )2 1 1 = 2 uv γ (1 + c2 cos θ )2
=
1 γ 2 (1 +
uv c2
"
v cos θ + u 1 + uv cos θ c2
+2
1 + 2 γ
*
v sin θ 1 + uv cos θ c2
+2
(1 − γ 2 )v 2 sin2 θ + u 2 γ 2 + v 2 γ 2 + 2uv γ 2 cos2 θ
#
v 2 + u 2 + 2uv cos θ − β 2 v 2 sin2 θ .
We note that when v = c, v = c and conversely, as it should be expected. 4.4. (a) From the relations γβ = sinh φ = $ φ
e = γ + γβ =
eφ −e−φ , 2
γ = cosh φ =
1+β ⇒ φ = ln 1−β
$
1+β . 1−β
eφ +e−φ 2
we find
148
4 Four Velocity
Note: We can also work with the relation: tanh φ = β ⇒
1+β 1 1 − e−2φ = β ⇒ φ = ln . 1 + e−2φ 2 1−β
(b) The boost relating Σ, Σ gives x + l = γ (x + βl) + γ (l + βx) = eφ (x + l), where in the last step we used (a) (c) Let φ1 , φ2 be the rapidities of successive boosts L 1 , L 2 along the direction x. Making use of (b) we find x + l = eφ1 (x + l ) = eφ1 eφ2 (x + l) = e−(φ1 +φ2 ) (x + l). But for the composite boost L 2 L 1 we have x + l α = e−φ12 (x + l). It follows φ12 = φ1 + φ2 . + i 4.5. (a) Suppose that proper frame Σ A the fourvector A has the canoni in0+its A . The Lorentz transformation relating Σ, Σ+A has a cal form Ai = 0 Σ+ A
velocity equal to the velocity of Ai in Σ. The transformation equations are ⎫ A = γ β A0+ ⎬ A0 = γ A0+
⎭
⇒β=
A . A0
frame (b) When Ai is spacelike it does not have a proper frame but a characteristic 0 − Σ A in which its reduced (or canonical) form is Ai = . The Lorentz A− Σ− transformation relating Σ and Σ−A gives
A
& & & & A0 & & & & &, A0 = γ β · A− = γβ &A− = β A ⇒ β = & A & which shows that the factor β can be specified only in its measure but not in its direction. (c) From part (a) we have that the βfactor of the particle in Σ equals β=
i+j . 2
4 Four Velocity
149
4.6. The time required (for A) in order that the emitted signal reaches B is T2 and the distance covered is l + x. The speed of the signal is c; therefore, we have the equation x T l + = . c c 2 From the relativistic rule of composition of threevelocities the speed of B wrt A u−υ T is υ B A = 1− ; hence, we have uυ . The spaceship B covers the distance x in time 2 the equation
c2
x=
u−υ 1 − uυ c2
T . 2
Eliminating x from the two equations we find u=
2l − T (c + υ) 2 c . 2lυ − T c(c + υ)
4.7. As it has been shown in problem 4.3 the following relations hold: u · v (a) γ = γu γ 1 − 2 , c
(b) tan θ =
v sin θ . γu (v cos θ − u)
By its definition the rapidity satisfies the relations: tanh ξ =
u u , cosh ξ = γu , sinh ξ = γu . c c
Similar relations hold for the velocities v and v . Replacing in (a) we compute cosh ζ = cosh ξ cosh ζ (1 − tanh ξ tanh ζ cos θ ) = cosh ξ cosh ζ − sinh ξ sinh ζ cosh θ. Similarly replacing in (b) we obtain the second relation. 4.8. (a) The speed of B wrt Γ is found from the relativistic law of composition of threevelocities: β BΓ =
β B A + β AΓ . 1 + β B A β AΓ
From the data of the problem we have β BΓ =
−β B A − βΓA −0.8 − 0.6 ⇒ β BΓ = −0.946. = 1 + β B A βΓA 1 + 0.8 · 0.6
2 −1/2 Using this value of β BΓ we compute γ BΓ = (1 − β BΓ ) = 3.083.
150
4 Four Velocity
(b) For the second part of the exercise we work as follows. We consider a photon emitted by Γ and compute its fourmomentum assuming – without restricting the generality – that the photon propagates in the plane x − y. We transform the fourmomentum to the observer B using the boost along the xdirection with speed β BΓ relating B, Γ and compute the beam received by B. We compute the reflected beam by changing the ycomponent of the fourmomentum. We transform the reflected fourmomentum at A by using the boost relating A, B. From the result, A is able to compute the energy (color) and the direction of the received beam. We realize this scenario below. The fourmomentum of the photon emitted from Γ is ⎛
⎛ ⎞ ⎞ 1 1 ⎜ ⎟ ⎟ hνΓ ⎜ ⎜ sin 170 ⎟ = h ⎜ 0, 174 ⎟ . PΓi = ⎝ ⎝ ⎠ ⎠ cos 170 −0.985 c λΓ 0 0 The fourmomentum received by B is ⎛ ⎞ ⎞ γΓB λhΓ (1 − βΓB 0.985) 0.210 ⎜ ⎜ γΓB h (−0.985 + βΓB 1) ⎟ ⎟ λΓ ⎟ = h ⎜ −0.120 ⎟ . PBi = ⎜ h ⎝ ⎠ ⎝ 0.174 ⎠ 0.174 λΓ λΓ 0.000 0.000 ⎛
The reflected photon at B has fourmomentum ⎛
⎞ 0.210 ⎟ h ⎜ ⎜ −0.120 ⎟ . PBi = ⎝ ⎠ −0.174 λΓ 0.000 The photon received by A has fourmomentum ⎛ ⎞ ⎞ γ B A λhΓ (0.210 + β B A 0.120) 0.510 ⎜ ⎜ γ B A h (−0.120 − β B A 0.210) ⎟ ⎟ λΓ ⎟ = h ⎜ −0.480 ⎟ . PAi = ⎜ h ⎝ ⎠ ⎝ − λΓ 0.174 λΓ −0.174 ⎠ 0.000 0.000 ⎛
The angle at which A receives the radiation is tan θ =
−0.174 = 0.3625 ⇒ θ ∼ = 19◦ 55 32 . −0.480
4 Four Velocity
151
The wavelength of the photons received by A is h h = 0.510 ⇒ λ A = 1.06 μm. λA λΓ 4.9. (a) In the lab frame the fourvelocities of the beams are v1i = γ (v1 )(c, v1 ),
v2i = γ (v2 )(c, v2 ).
In the proper frame of one beam (the 1 say) the fourvelocities are v1i = (c, 0),
v2i = γ (w)(c, w),
where w is the relative velocity of the beams. The quantity v1i · v2i is invariant; hence, it has the same value in all LCF. In the lab frame we have v1i · v2i = γ (v1 )γ (v2 )(v1 · v2 − c2 ) and in the rest frame of 1 v1i · v2i = −γ (w)c2 . Equating the two expressions we find v1 · v2 . γ (w) = γ (v1 )γ (v2 ) 1 − c2 Reversing and squaring we get β= wc :
2 β12 =1−
(1 − β12 )(1 − β22 ) . 1 − β1 · β2
(b) We consider β1 = β2 = β and let φ be the angle between the velocities of the two beams in the lab frame. Then 2 β12 =1−
(1 − β 2 )2 , 1 − β 2 cos φ
because 0 < 1 − β 2 < 1 β12 takes its maximum value when the denominator becomes maximum. This occurs when cos φ = −1, hence φ = 180◦ . We conclude that the researcher must orientate the beams antiparallel so that a headon collision will be achieved.
152
4 Four Velocity
4.10. (a) If T is the kinetic energy of the electron and E its total energy in the lab system, then E = T + mc2 . But E = γ mc2 ; hence, γ =
T T + mc2 =1+ =⇒ γ = 2.96 , β = 0.94. 2 mc mc2
(b) The relative velocity of the electrons is the velocity of one in the proper frame of the other. From the relativistic law for threevelocities we find vrel  =
v+u , 1 + vu/c2
where u is the speed of each electron in the lab system. Replacing we find vrel  =
2u 2 × 0.94 = = 0.998c. 2 2 1 + u /c 1 + 0.942
4.11. The fourvelocity of observers in Σ is ⎛
⎛ ⎞ ⎞ γ1 c γ2 c ⎜ γ1 v ⎟ ⎜ 0 ⎟ i ⎜ ⎟ ⎟ u i1 = ⎜ ⎝ 0 ⎠ , u 2 = ⎝ γ2 kv ⎠ . 0 0 Σ Σ The relative velocity of 2 wrt 1 is ⎛
⎞ (γ2 − γ1 )c ⎜ −γ1 v ⎟ ⎟ u i2 − u i1 = ⎜ ⎝ γ2 kv ⎠ . 0 Σ It is given that 1 sees 2 to move in a direction which makes an angle φ with the positive semiaxis x. This means tan φ =
(u i2 − u i1 )x γ2 k γ2 kv =− = ⇒ γ2 k = −γ1 tan φ. i i −γ1 v γ1 (u 2 − u 1 ) y
From the results we have the condition k tan φ < 0. Squaring the last relation we find k2 tan2 φ k 2 − tan2 φ 2 = ⇒ v = , tan φ = ±1. 1 − k 2v2 1 − v2 k 2 (1 − tan2 φ)
4 Four Velocity
153
The solution must be real. The condition for this is (k 2 −tan2 φ)(1−tan2 φ) > 0. This inequality has the following solutions (together with the condition k tan φ < 0): −k > tan φ , φ
0, 0 < φ < π2 , hence P is seen from O to move away in a direction in the first quarter of the plane x − y of Σ.
154
4 Four Velocity
In order to compute the energy of P wrt O, we note the general result that the energy of a particle of fourmomentum pi wrt an observer with fourvelocity u i is given by the relation E u = − pi u i .1 The quantity pi u i is invariant; hence, we can compute it in any LCF we consider most convenient. Because we know both fourvectors in Σ we choose to work in Σ and have E = −mu i vi = −m (γu c, 0, γu u, 0)Σ · (γv c, −γv v, 0, 0)Σ = mγu γv c2 ⇒ $
E = mγu c
2
1+
E 2 − m 2 c4 . E 2 tan2 φ
Second Solution The fourvelocity of P in Σ is u i = (γu c, 0, γu u, 0)Σ . We compute it in the proper frame of O using the boost relating Σ and O. If u i = U0 , Ux , U y , Uz O are the components of u i in O we have Ux = γv [γv v + (γu − γv )cβv ] = γu γv v,
U y = γu u,
and the solution continues as before. The energy of P is possible to be computed from the transformation of the fourmomentum. Indeed the fourmomentum of P in Σ is pi = mu i = m (γu c, γu u, 0)Σ . The energy of P for O is the zeroth component of the transformed fourmomentum. From the first boost equation we find $ E = γv mγu c2 = mγv γu c2 = mγu c2 1 +
E 2 − m 2 c4 . E 2 tan2 φ
(b) In case P is a photon, we can no longer work with fourvelocities and we have to use the fourmomentum. The fourmomentum of P in Σ is given by pi = Ec (1, 0, 1, 0)Σ . Suppose that in O the pi = Ec (1, cos φ, sin φ, 0) O . The boost relating Σ, O gives tan φ =
1 , γv v
E = Eγv v =
E . tan φ
E u /c c , ui = . Then pi · u i = −E u where E u is the energy on the 0 Σ p Σu u proper frame of the observer with fourvelocity u i .
1
Indeed let pi =
4 Four Velocity
155
We note that the angle φ is independent of the energy while the energy of the photon for O depends on the angle φ. Finally, we note again that tan φ > 0, hence 0 < φ < π2 , that is, the photon appears to travel away from O in a direction within the first quarter of the plane x − y of Σ. Also when φ → π2 then γv → 1 ⇒ v = 0 (Newtonian observers) while when φ → 0 then γv → ∞ ⇒ v → c (relativistic speeds). The effect of the dependence of the direction of propagation of a photon from the relative velocity of the observers is called aberration of light. 4.13. We calculate the velocity v13 of the instantaneous LCF Σ3 wrt Σ1 . The threevelocity of Σ3 wrt Σ2 is v32 = (0, v2 , 0). From the relativistic composition rule for the threevelocities we find taking into account that v21 = (v1 , 0, 0): v2 v31 = v1 , ,0 . γv1 We note that v31 is on the plane x1 , y1 . The angle θ1 which v31 makes with the axis x1 is tan θ1 =
(v31 ) y v2 = . (v31 )x γv1 v1
We calculate now the velocity v13 of Σ1 wrt Σ3 . The velocity v12 of Σ1 wrt Σ2 is v12 = (0, −v2 , 0). Taking the same composition rule as before and considering v21 = (v1 , 0, 0) we find v13
v1 = − , −v2 , 0 . γv2
We note that v13 lies in the plane x3 , y3 of Σ3 . The angle it makes with the axis x3 is tan θ3 =
(v13 ) y γv v2 = 2 . (v13 )x v1
Obviously v13 is not parallel to v31 . The same is true for the axes of Σ1 , Σ3 (Thomas effect). This result does not bother us because the Euclidean parallelism is not a relativistic covariant concept. The difference δθ = θ3 − θ1 is tan δθ = tan(θ3 − θ1 ) =
tan θ3 − tan θ1 . 1 + tan θ3 tan θ1
Replacing tan θ1 , tan θ3 one finds δθ = tan
−1
−1 + γv1 γv2 β1 β2 . β12 γv1 + β22 γv2
156
4 Four Velocity
The term in brackets is written if one uses the identity β 2 γ 2 + 1 = γ 2 : −1 + γv1 γv2 = β12 γv1 + β22 γv2
1 − γv1 γv2 1−γv21 γv 1
+
1−γv22 γv 2
γv1 γv2 (1 − γv1 γv2 )γv1 γv2 = . (γv1 + γv2 )(1 − γv1 γv2 ) γv1 + γv2
=
Hence
δθ = tan−1
γv1 γv2 β1 β2 γv1 + γv2
3 2 4 2 4 −1 5 γv1 β1 (γv2 β2 ) = tan 2 γv1 + γv2
3 4 4 1 − γv21 1 − γv22 5 = . 2 γv1 + γv2 We note that the RHS is a symmetric function of γv1 , γv2 . In terms of the speeds we have β1 β2 δθ = tan−1 . 1 − β12 + 1 − β22 For small speeds the following approximations apply: ∂ 1 = 1 − β2 = 1 + ( 1 − β 2 )0 β + O(β 3 ) γ ∂β 1 =1+ 2
−2β 1 − β2
! β + O(β 2 ) = 1 + O(β 2 ), 0
from which follows that in the Newtonian limit
−1
δθ = tan
δθ =
γv1 γv2 β1 β2 = tan−1 γv1 + γv2
β1 β2 + O(β12 β22 ). 2
β1 β2 1 + γ1v γv 1
2
! ≈ tan−1
β1 β2 ⇒ 2
4 Four Velocity
157
Application We compute the velocity of Σ3 wrt Σ2 . We have (Fig. 4.1) v31 = v(cos Δφ, sin Δφ, 0), v12 = v(−1, 0, 0).
Fig. 4.1 The Thomas effect
The relativistic composition rule for the threevelocities gives v cos Δφ − v , 1 − β 2 cos Δφ v sin Δφ , (v32 ) y = γv (1 − β 2 cos Δφ) (v32 )z = 0.
(v32 )x =
For small Δφ the cos Δφ ≈ 1, sin Δφ ≈ Δφ and v32 = (0, γv vΔφ, 0). This implies that the threevelocity of Σ3 wrt Σ2 is v2 = vΔφ. Furthermore, we have v1 = vΔφ. Thus, Δθ ≈
1 2 β Δφ. 2
For a complete rotation Δφ = 2π , therefore (Δθ )rotation =
1 2 β 2π = πβ 2 . 2
4.14. (a) We have v = v cos χ i + v sin χ cos ψ j + v sin χ sin ψ k .
158
4 Four Velocity
Replacing v, v in terms of i, j, k and equating the coefficients of i, j, k we find
v=
2 1/2 v 2 + u 2 + 2uv cos χ − uv sin χ c 1+
tan χ =
uv c2
cos χ
,
1 v sin x . γu u + v cos x ψ = ψ.
(4.7)
(4.8) (4.9)
(b) The Newtonian result follows if we set u/c = 0. We find 1/2 2 v = v + u 2 + 2uv cos χ , v sin χ , u + v cos χ ψ = ψ.
tan χ =
These relations give the triangle of the Fig. 4.2 in the threevelocity space:
Fig. 4.2 The velocity triangle in the threevelocity space
From the figure we conclude the linearity of the space of Newtonian threevelocities as well as the Newtonian composition rule for threevelocities. This does not hold in Special Relativity where the space of the threevelocities is not the Euclidean space E 3 , but a linear space with a hyperbolic metric of constant curvature. (c) Relation (4.7) is written as follows: 2 γu 1 − vc2 v2 1− 2 = . c 1 + uv cos χ c2 Because v, v ∈ R the v, v < c. When v = c then the LHS vanishes and follows that v = c.
4 Four Velocity
159
(d) From (4.8) we note that when u = v , χ = π , hence χ = 0, π , which shows that there are velocities which are along u (“forward”) in Σ while having opposite direction (“backward”) in Σ. For χ = π2 we have tan χ = +∞, while tan χ = γ1 vu ∈ (0, +∞); thus χ < π2 . Because χ is a continuous function of its arguments we infer that for every angle χ the corresponding angle χ diminishes as u increases. This means that as a beam of particles propagates it focuses more and more toward the direction of the velocity of its center of momentum. Especially for the photons we have v = v = c, hence, tan χ =
1 sin χ . γ β + cos χ
This relation is the general expression of the aberration of light and it is applied in the calculation of the distribution of photons and neutrinos in the decay of many particles (e.g., π 0 → γ + γ , π +→ μ+ + νμ , Σ0 → π + γ ). Taking into account the last part concerning the definition of the spherical coordinates we have cos χ =
1 u + c cos χ β + cos χ vx = = . uc c c 1 + c2 cos χ 1 + β cos χ
from which follows sin χ =
sin χ . γ (1 + β cos χ )
Using the trigonometric identity tan χ2 = required result follows.
sin χ 1+cos χ
and replacing sin χ , the
4.15. (a) The position fourvector of 1 in Σ and Σ2 is x1i =
ct1 r1
Σ
=
ct12 r12
Σ2
.
These expressions are related by the Lorentz transformation defined by Σ and Σ2 . Hence r12 = r1 − v2 t1 + (γ2 − 1) t12
v2 · r1 − v22 t1
v2 , r1 · v2 v1 t1 · v2 = γ2 Qt1 , = γ2 t1 − = γ2 t1 − c2 c2 v22
160
4 Four Velocity
where Q = 1 −
v1 ·v2 . c2
The relative velocity
v12 =
dr12 dt1 dr12 1 dr12 = = . dt12 dt12 dt1 γ2 Q dt1
But dr12 v2 = v1 − v2 + (γ2 − 1) 2 v2 · v1 − v22 dt1 v2 v1 · v2 = v1 − v2 − (γ2 − 1) 1 − v2 . v2 2 Therefore, v12 =
1 v1 · v2 v2 . v1 − v2 − (γ2 − 1) 1 − γ2 Q v2 2
(b) v21 is calculated from v12 if one interchanges v1 ←→ v2 . Therefore, we have v21
1 v1 · v2 v1 . v2 − v1 − (γ1 − 1) 1 − = γ1 Q v1 2
From these results we see that v12 = −v21 . The vectors v12 , v21 are in the spacelike plane which is normal to the plane defined by the timelike fourvectors v1i , v2i . Their angle is cos ψ =
v12 · v21 . v12  v21 
We calculate the length v12 . We set α2 =
1 γ2 Q
and have
2 v2 2 v12 = α2 2 (v1 − v2 )2 + (γ2 − 1)2 24 v1 · v2 − v22 v2 1 2 +2(γ2 − 1) 2 (v2 · (v1 − v2 )) v1 · v2 − v2 v2 2 ! 2 2 v1 · v2 − v22 2 2 v1 · v2 − v2 2 + 2 (γ2 − 1) = α2 (v1 − v2 ) + (γ2 − 1) v22 v22 ! 2 v1 · v2 − v22 = α2 2 (v1 − v2 )2 + (γ2 − 1) (γ2 + 1) v22
4 Four Velocity
= α2
161
2
(v1 − v2 ) + 2
γ22 β2 2
2 ! v1 · v2 − v22 β2 2 c2
γ2 = α2 2 (v1 − v2 )2 + 22 [v2 · (v1 − v2 )]2 . c Using the identity a × b2 = a2 b2 − (a · b)2 we find # γ2 2 " 2 2 2 2 v12  = α2 (v1 − v2 ) + 2 − v2 × (v1 − v2 ) + v2 (v1 − v2 ) c γ2 2 = α2 2 1 + β2 2 γ2 2 (v1 − v2 )2 − 2 v1 × v2 2 = c 2 γ 2 = α2 2 γ22 (v1 − v2 )2 − 2 v1 × v2 2 = c 1 = α2 2 γ2 2 (v1 − v2 )2 − 2 v1 × v2 2 c 2
2
because 1 + β2 2 γ2 2 = γ2 2 . Therefore, 2 v12 =
(v1 − v2 )2 − c12 v1 × v2 2 . 2 1 − v1c·v2 2
2 is symmetric in v1 , v2 , hence v21  = v12 . We note that v12 We continue with the calculation of the angle ψ between v12 and v21 . Then
2 2 v1 · v2 − v22 v1 · v2 − v22 1 2 + (γ1 − 1) cos ψ = − (v1 − v2 ) + (γ2 − 1) γ1 γ2 Q 2 v22 v12 v1 · v2 + (γ1 − 1) (γ2 − 1) v1 · v2 − v22 v1 · v2 − v12 . v1  v2  [Note: Write v1 · v2 = v1 v2 cos φ and show that cos ψ is simplified as follows: cos ψ = 1 − where ρ 2 =
2 sin2 φ , 1 + 2ρ cos φ + ρ 2
(γ1 +1)(γ2 +1) . (γ1 −1)(γ2 −1)
Another equivalent form is tan
ψ sin φ = . 2 ρ + cos φ
162
4 Four Velocity
Observe that # " (i) The angle ψ is in the interval π2 , π (prove this). (ii) It is symmetric wrt v1 , v2 as it should be (why?).] 4.16. Let Σv be the proper frame of v i and suppose that in Σv ui =
γ (u, v)c γ (u, v)uv
Σv
,
vi =
c , 0 Σv
where uv is the threevelocity of u i in Σv . Then u i vi = −γ (u, v)c2 ⇒ γ (u, v) = −
1 i u vi . c2
Let Σw the proper frame of wi and suppose that in that frame we have the decompositions: u = i
γ (u, w)c γ (u, w)uw
Σw
,
v = i
γ (v, w)c γ (v, w)vw
Σw
.
Then γ (u, v) = −
1 i 1 u vi = − 2 γ (u, w)γ (v, w)(−c2 + uw · vw ) ⇒ 2 c c γ (u, v) . γ (v, w)(1 − uwc·v2 w )
γ (u, w) =
Second Solution Consider the LCF Σw , Σv which have relative velocity vw and assume that the observer in these frames has velocities u w , u v , respectively. Suppose that the fourvelocity of the observer u i in Σw , Σv has decomposition ui =
γ (u, w)c γ (u, w)uw
Σw
,
ui =
γ (u, v)c γ (u, v)uv
Σv
.
The two expressions of u i are related with the Lorentz transformation relating the LCF Σw , Σv . The zeroth component of the fourvector gives uw · vw vw γ (u, w)uw = γ (v, w)γ (u, w) 1 − , γ (u, v)c = γ (v, w) γ (u, w)c − c c2
4 Four Velocity
163
from which follows γ (u, w) =
γ (u, v) . γ (v, w) 1 − uwc·v2 w
4.17. (a) First Solution Consider the geometric representation of the Lorentz transformation in spacetime:
Fig. 4.3 Spacetime diagram for the combination of Lorentz transformations
The angle φ1 corresponds to the first transformation from (x, ) → (x , ) with speed u 1 , hence tanh φ1 = β1 =
u1 . c
Similarly the angle φ2 corresponds to the second transformation (x , ) → (x , ) with speed u 2 . The transformation (x, ) → (x , ) is a boost because the axes x, x and , form equal angles φ1 + φ2 . The speed of this boost, u 3 say, is given by
u 3 = c tanh(φ1 + φ2 ) = c
tanh φ1 + tanh φ2 u1 + u2 ⇒ u3 = . 1 + tanh φ1 tanh φ2 1 + uc1 u2 2
164
4 Four Velocity
Second Solution The matrix representing the boost (x, ) → (x , ) is ⎛
γ1 ⎜ β1 γ1 L(β1 ) = ⎜ ⎝ 0 0
β1 γ1 γ1 0 0
0 0 1 0
⎞ 0 0 ⎟ ⎟. 0 ⎠ 1
β2 γ2 γ2 0 0
0 0 1 0
⎞ 0 0 ⎟ ⎟. 0 ⎠ 1
and that representing (x , ) → (x , ): ⎛
γ2 ⎜ β2 γ2 ⎜ L(β2 ) = ⎝ 0 0
The composite transformation (x, ) → (x , ), the L(β3 ) say, is L(β3 ) = L(β2 ) · L(β1 ). Replacing we find ⎛
γ1 γ2 (1 + β1 β2 ) ⎜ −γ1 γ2 (β1 + β2 ) L(β3 ) = ⎜ ⎝ 0 0
−γ1 γ2 (β1 + β2 ) γ1 γ2 (1 + β1 β2 ) 1 0
0 0 0 0
⎞ 0 0 ⎟ ⎟. 0 ⎠ 1
From the form of the matrix L(β3 ) we conclude that the composite transformation is a boost with velocity factors γ3 = γ1 γ2 (1 + β1 β2 ) ≥ 1 and β32 = 1 − 1/γ32 = (β1 +β2 )2 1 1 − γ 2 γ 2 (1+β 2 = (1+β β )2 . 1 2 1 β2 ) 1 2 (b) The boost along the xaxis is ⎛
γ1 ⎜ −β1 γ1 L(x, β1 ) = ⎜ ⎝ 0 0
0 0 1 0
⎞ 0 0 ⎟ ⎟ 0 ⎠ 1
−β2 γ2 0 γ2 0
⎞ 0 0 ⎟ ⎟. 0 ⎠ 1
−β1 γ1 γ1 0 0
and that along the yaxis ⎛
γ2 ⎜ 0 L(y, β2 ) = ⎜ ⎝ −β2 γ2 0
0 1 0 0
We compute ⎛
γ1 γ2 ⎜ −β1 γ1 γ2 ⎜ L(x, β1 )L(y, β2 ) = ⎝ −β2 γ2 0
−γ1 β1 γ1 γ2 0
−β2 γ1 γ2 γ1 γ2 β1 β2 0 0
⎞ 0 0 ⎟ ⎟. 0 ⎠ 1
4 Four Velocity
165
⎛
γ1 γ2 ⎜ −β1 γ1 L(y, β2 )L(x, β1 ) = ⎜ ⎝ −β2 γ1 γ2 0
−γ1 γ2 β1 γ1 γ1 γ2 β1 β2 0
−β2 γ1 0 γ2 0
⎞ 0 0 ⎟ ⎟= L(x, β1 )L(y, β2 ). 0 ⎠ 1
Because the matrices L(x, β1 )L(y, β2 ) and L(y, β2 )L(x, β1 ) are not symmetric we conclude that there is no direction in which either composite transformation is a boost. 4.18. (a) The solution is left to the reader. (b) Let u be the relative velocity of Σ1 , Σ2 . From the general Lorentz transformation we have 1 u · (c sinh a1 eˆ 1 ) , c cosh a2 = γu c cosh a1 − c 1 γu ˆ u · (c sinh a ) − e c sinh a2 eˆ 2 = c sinh a1 eˆ 1 + (γu − 1) c cosh a 1 1 1 u⇒ u2 c sinh a2 eˆ 2 = sinh a1 eˆ 1 + [(γu − 1) sinh a1 (uˆ · eˆ 1 ) − γu β cosh a1 ] u. Let cosh a0 = γu , sinh a0 = uc γu . Then cosh a2 = cosh a0 cosh a1 − sinh a0 sinh a1 (ˆe1 · u), sinh a2 eˆ 2 = sinh a1 u + [(cosh a0 − 1) sinh a1 (ˆe1 · u) − sinh a0 cosh a1 ]u. Consider the angles x1 = ∠(ˆe1 , u) , x2 = ∠(ˆe2 , u) , x0 = ∠(ˆe1 , eˆ 2 ) and write the above relations in terms of the angles. The first relation is written as cosh a2 = cosh a0 cosh a1 − sinh a0 sinh a1 cos x1 .
x2 Σ1
Fig. 4.4 The velocity triangle in the three velocity space
x1 Σ2
(4.10)
166
4 Four Velocity
Concerning the second relation we project it parallel and normal to u. We have Normal to u: sinh a2 eˆ 2 × u = sinh a1 eˆ 1 × u ⇒ sinh a2 sin x2 = sinh a1 sin x1 .
(4.11)
Parallel to u: sinh a2 cos x2 = sinh a1 cos x1 + [(cosh a0 − 1) sinh a1 cos x1 − sinh a0 cosh a1 ] ⇒ − sinh a2 cos x2 = cosh a0 sinh a1 cos x1 − sinh a0 cosh a1 .
(4.12)
In case Σ1 , Σ2 are moving in the standard configuration eˆ 1 u; hence, x0 = 0, x1 = π, x2 = 0 and relations (4.10), (4.11),(4.12) read cosh a2 = cosh a0 cosh a1 + sinh a0 sinh a1 , sinh a2 = sinh a1 + (cosh a0 − 1) sinh a1 + sinh a0 cosh a1⇒ sinh a2 = cosh a0 sinh a1 + sinh a0 cosh a1 .
(4.13)
The first relation gives cosh a2 = cosh(a0 + a1 ) and the second sinh a2 = sinh(a0 + a1 ). From these we find the simple relation a2 = a0 + a1 . In order to write this relation in terms of the speeds we consider tanh a2 and write tanh a2 = tanh(a0 + a1 ) =
tanh a0 + tanh a1 1 + tanh a0 tanh a1
or v2 =
v1 + u , 1 + vu1 c2
which is the wellknown relativistic formula for the composition of threevelocities. 4.19. The points of intersection of the light cone with the straight line z 1 i are defined by the condition (z 1 i −z 2 i )2 = 0. Replacing z 1 i (τ ) = a1 i +u 1 i τ we find (see Fig. 4.5)
4 Four Velocity
167
Fig. 4.5 Spacetime diagram for the retarded and advanced point
" # (a1 i + u 1 i τ − z P i )2 = 0 ⇒ τ 2 − 2 u 1 j (a1 j − z P j ) τ − (a1 i − z P i )2 = 0 . The roots of this quadratic equation are τ R = u 1 i (a1 − z P )i − τ A = u 1 i (a1 − z P )i +
√ √
Δ, Δ,
(4.14)
where #2 " Δ = (a1 − z P )i (a1 − z P )i + u 1 i (a1 − z P )i = gi j (a1 − z P )i (a1 − z P ) j + u 1i u 1 j (a1 − z P )i (a1 − z P ) j = (gi j + u 1i u 1 j )(a1 − z P )i (a1 − z P ) j = h i j (u 1 )(a1 − z P )i (a1 − z P ) j
(4.15)
and h i j (u 1 ) = gi j + u 1i u 1 j . is the symmetric tensor h i j (u 1 ) which projects normal to u 1 i for all of its indices, that is, h i j (u 1 )u 1 j = h i j (u 1 )u 1 i = 0.2 2 We recall the definition of the projective tensor associated with a nonnull fourvector. For any nonnull fourvector v i the symmetric projection tensor is defined as follows: h i j (v k ) = gi j − (v) v v , where (v) is the sign of v k and v 2 the length of v k . When h i j (v k ) is contracted with any v2 i j tensor, it annihilates the components of that tensor along the direction of v k . The projection tensor has the following properties:
168
4 Four Velocity
The quantity Δ > 0, because the fourvector h i j (u 1 )(a1 − z P ) j is spacelike. Therefore, the two roots τ R , τ A are real and well defined. Moreover, the values τ R , τ A are expressed in terms of covariant quantities, hence, they are covariant, i.e., their value is the same in all LCF. The geometric significance of the values τ R , τ A is the following: " #1 τ R,A = u 1 i (a1 − z P )i ∓ h i j (u 1 )(a1 − z P )i (a1 − z P ) j 2 = (projection of (a1 − z p )i along the straight line z 1 i )∓ ∓ (projection of (a1 − z p )i normally to the straight line z 1 i ) . We consider the quantity 1 (τ A + τ R ) 2 = u 1i (a1 − z P )i .
τav =
(4.16)
The τav defines a new value of τ and a new point Q(τav ) on the line z 1 i . The vector z Q i − z P i is normal to the line z 1 i . Indeed " # Q P i = z Q i − z P i = a1 i + u 1 i u 1 j (a1 − z P ) j − z P i = g i j (a1 j − z P j ) + u 1 i u 2 j (a1 − z P ) j = h i j (u 1 )(a1 − z P ) j . We note that Q P i is the normal projection of the fourvector a1i − z iP on z 1 i . The minimum distance of the point P from the straight line z 1 i is given by the length (P Q). We compute (P Q)2 = h i j (u 1 )(a1 − z P ) j h ir (u 1 )(a1 − z P )r = h rj (u 1 )(a1 − z P ) j (a1 − z P )r = h i j (u 1 )(a1 − z P )i (a1 − z P ) j = Δ > 0 .
(4.17)
We conclude that the fourvector P Q i is spacelike. The point Q is the midpoint of the distance τ R , τ A . Indeed we have 1 τ R = τ Q − (τ A − τ R ), 2 1 τ A = τ Q + (τ A − τ R ) . 2
j
h ii (v a ) = 3, h i j (v a )h k (v a ) = h ik (v a ), h i j (v a )h jk (v a ) = δik . In the special case that v k is a unit timelike fourvector, then (v) = −1, v 2 = 1, and h i j (v k ) =gi j + vi v j .
4 Four Velocity
169
Fig. 4.6 Spacetime diagram for the shortest distance of a point from a line
But 12 (τ A − τ R ) =
√ Δ = (P Q), hence τ R = τ Q − (P Q) , τ A = τ Q + (P Q) .
(4.18)
In Figure 4.6 the distances of the events P, Q, A, R are shown. We calculate next the fourvectors (P R)i , (P A)i , and their lengths. For (P R)i we have (obviously) (P R)i = (P Q)i −
√
Δu 1 i
and for (P A)i (P A)i = (P Q)i +
√ Δu 1 i .
The lengths (P R)2 = (P A)2 = (P Q)2 − Δ = 0 as it is to be expected because the spacetime points R, A are the intersection of the light cone defined by the point P. (c) As we have shown3 (P Q)i = h 1 ij (a1 j − z P j ) = h 1 ij (a1 j − a1 j − u 2 j τ2 p ) = −h 1 ij u 2 j τ2 p = −u 2 i τ P − (u 1 · u 2 )u 1 i τ2 p .
3
h 1ab = ηab + u 1a u 1b .
170
4 Four Velocity
The length #2 " (P Q)2 = u 2 i + (u 1 · u 2 )u 1 i τ22p # " = −1 + (u 1 u 2 )2 + 2(u 1 · u 2 )2 τ2P 2 ⇒ (P Q)2 = − (1 − u 1 · u 2 )2 τ2P 2 .
r r
(4.19)
From the above we conclude the following: As the point P moves along z 2 i , the point Q moves along z 1 i so that (P Q)i is always parallel to itself (prove it). The point Q is defined uniquely for each event P (for given u 1 , u 2 ); therefore, it defines a synchronization between the RIO u i1 , u i1 (a synchronization is a 1 to 1 correspondence between the affine parameters of the straight lines z 1 i , z 2 i ). In order to compute the synchronization we note that4 τ Q = τav = u 1 i (a1 − z P )i = u 1 i (u 2 τ2P )i = (u 1 · u 2 )τ2P ,
therefore, the correspondence we are looking for is τ2P ↔ τ Q = (u 1 · u 2 )τ2P .
(4.20)
(Can you see that this relation is essentially the time dilation formula?) The distance P Q in terms of u 1 , u 2 is (P Q)2 = h i j (u 1 )(a1 i − z P i )(a1 j − z P j ) = h i j (u 1 )(u 2i τ2P )(u 2 j τ2P ) = h i j (u 1 )u 1 i u 2 j τ2P 2 # " = −1 + (u 1 · u 2 )2 τ2P .
(4.21)
Another form occurs if we replace u 1 · u 2 in terms of the proper times τ Q , τ2P . We find ! τQ 2 2 (P Q) = −1 + τ2P 2 τ2P = τ2P 2 − τ Q 2 .
(4.22)
In order P Q i to be spacelike the P Q > 0 which gives the condition τ2,P > τ Q . This is indeed the case because as we have seen τ Q = (u 1 · u 2 )τ2P .
4
To see this relation note that the point Q is the projection of a i − z iP on u i1 .
4 Four Velocity
171
We work now with the relative fourvectors. In the proper frame of u 1 i we have u1i =
u 10 + 0
,
Σ+ 1
u2i =
u 21 β21
Σ+ 1
from which follows u 1 i u 2i = −u 21 u 10 + ,
(4.23)
2 2 Δ = (P Q)2 = −1 + (u 21 u 1 + 0 ) τ2P .
(4.24)
hence
In the proper frame of Σ2 we have similarly u1 = i
u 12 β12
Σ+ 2
,
u2 = i
u 20 + 0
Σ+ 2
,
therefore (u 1 · u 2 )Σ+2 = −u 12 u 2 + 0 . The u 1 · u 2 is invariant, hence u 12 u 20 + = u 21 u + 10 .
(4.25)
This relation is important because it relates the lengths of the threevectors β12 , β21 . Indeed we have −1 = −u 212 + β12 2 = −u 21 2 + β21 2 . and making use of (4.24) β12 = 2
u1+ 0 u2+ 0
2
! − 1 u 20 2 + β21 2
Application We consider u 1 i , u 2 i to be fourvelocities. Then (c = 1) + u1+ 0 = u 2 0 = 1,
β12 2 = γ12 2 v12 2 = 1 − γ12 2 , β21 2 = 1 − γ21 2 .
(4.26)
172
4 Four Velocity
Equation (4.26) gives γ12 2 = γ21 2 ⇔ v12 2 = v21 2 which implies that the relative threevelocities have the same speed but different direction. In order to compute the relative speed we use (4.24) and find (c = 1): (P Q) = (−1 + γ21 )τ2P ⇒ γ12 = 1 + 2
2
2
2
(P Q) τ2P
2 ! .
Obviously, this relation depends on the point P and the straight line z 1 i , that is, the concept of the relative velocity is a covariant concept which can be defined geometrically.5
5
The Lorentz invariant (P Q)2 can be used to define a metric in the threespace of the threevelocities. Then one can study the relative motion in a geometric manner. This space (the threevelocity space) is nonflat and nonEuclidean (it can be proved that it is a space of constant curvature).
Chapter 5
FourAcceleration
5.1. (a) Suppose the position fourvector x i of P in Σ has components (we ignore the y, z components because they vanish) x (τ ) = i
f 0 (τ ) f 1 (τ )
Σ
,
where f 0 , f 1 are smooth functions of the proper time of P. Then the fourvelocity of P is u i (τ ) =
˙f 0 (τ ) ˙f 1 (τ )
Σ
and the fouracceleration a (τ ) = i
¨f 0 (τ ) ¨f 1 (τ )
Σ
,
where a dot over a symbol indicates derivation wrt τ . Because u i u i = −c2 the functions f 0 , f 1 must satisfy the condition −( ˙f 0 )2 + ( ˙f 1 )2 = −c2 . This condition implies ˙f 0 (τ ) = c cosh θ (τ ), ˙f 1 (τ ) = c sinh θ (τ ), where θ is a smooth function of τ (Note: The function θ (τ ) is an invariant!). Replacing we find u (τ ) = c i
cosh θ (τ ) sinh θ (τ )
Σ
,
a (τ ) = cθ˙ (τ ) i
sinh θ (τ ) cosh θ (τ )
Σ
.
173
174
5 FourAcceleration
These relations are general, i.e., they hold for all motions of P along the xaxis (equivalently θ (τ ) is arbitrary). The length a i ai equals (a+ )2 , therefore (a+ )2 = c2 θ˙ 2 (− sinh2 θ (τ ) + cosh2 θ (τ )) ⇒
(a+ )2 = c2 θ˙ 2 .
(b) Let a + = g = constant. Then from the last relation we find θ (τ ) =
g τ + θ (0) . c
From the initial condition v(0) = 0 we compute θ (0) = 0, thus θ (τ ) = gc τ. cosh gc τ i . Concerning the This means that the fourvelocity is u (τ ) = c sinh gc τ Σ position fourvector we have ( x i (τ ) =
u i (τ ) dτ =
c2 g
sinh gc τ + c1 cosh gc τ + c2
Σ
=
c2 i c , a (τ ) + 1 c2 Σ g
where c1 , c2 are constants. Because x i (τ ), a i (τ ) are fourvectors, the same holds for the fourquantity ci = (c1 , c2 )Σ . This means that the constants c1 , c2 are not arbitrary and satisfy the constraint ci = fourvector. 2 A choice is c1 = c2 = 0 which corresponds to the condition x i (0) = cg a i (0). This is the standard choice found in the literature. (c) When a + = 2ckτ the a + = c2 θ˙ 2 . Thus, 2
4c2 k 2 τ 2 = c2 θ˙ 2 (τ ) ⇒ θ˙ (τ ) = 2kτ ⇒ θ (τ ) = kτ 2 + θ (0) . 2 The initial condition v(0) = 0 gives θ (0) = 0 which implies θ (τ ) = kτ . 2 cosh kτ and the fourFrom this we compute the fourvelocity u i = c sinh kτ 2 Σ sinh kτ 2 . Finally, the position fourvector is acceleration a i = 2ckτ cosh kτ 2 Σ
( x = i
c u dτ = k i
' 2 1 ' cosh kτ 2 dτ = 2 2 ai . sinh kτ dτ Σ 4k τ
5.2. (a) From the property a i u i = 0 we have immediately x i u i = 0. (b) The (covariant) equation of motion of P is d2xi − k 2 x i = 0, dτ 2
5 FourAcceleration
175
where τ is the proper time of P. The solution of this equation is x i (τ ) = Ai cosh kτ + B i sinh kτ, where Ai , B i are constant fourvectors which will be determined from the initial conditions x i (0) and u i (0). (c) In Σ we have the initial conditions ⎛ ⎞ ⎛ ⎞ 0 c ⎟ ⎜ ⎜ ⎟ x 0 0⎟ ⎟ , u i (0) = ⎜ x i (0) = ⎜ ⎝0⎠ ⎝0⎠ . 0 Σ 0 Σ Replacing in x i (τ ) we find ⎛
c sinh kτ k ⎜ x0 cosh kτ ⎜
x i (τ ) = ⎝
0 0
⎞ ⎟ ⎟ . ⎠ Σ
5.3. (a) From the equation of the world line and the definition of the parameter φ we compute ds 2 = d x 2 − c2 dt 2 = −a 2 dφ 2 . But ds 2 = −c2 dτ 2 = −a 2 dφ 2 ⇒ τ =
a φ + c (φ = rapidity). c
(b) The position fourvector of P in the LCF Σ is xi = a (sinh φ, cosh φ)Σ , hence the fourvelocity ui =
d x i dφ c d dxi = = (a sinh φ, a cosh φ) = c (cosh φ, sinh φ)Σ dτ dφ dτ a dφ
(check that u i u i = −c2 ). Concerning the fouracceleration we compute ai =
c2 du i dφ c2 = (sinh φ, cosh φ)Σ = 2 x i . dτ dτ a a
Because the relation between a i , and x i is covariant, it is independent of the LCF Σ. In Special Relativity we call this motion as hyperbolic motion which corresponds to the linear uniformly accelerating motion of Newtonian Physics. 2 x i . Concerning the energy of P we recall (c) From (b) we have F i = ma i = mc a2 that the power consumed is given by the invariant F i u i . We compute
176
5 FourAcceleration
F i ui =
mc2 i mc2 x u = a (sinh φ, cosh φ)tΣ (cosh φ, sinh φ)Σ = 0 i a2 a2
from which it follows that the energy is a constant of motion. 5.4. (a) From the orthogonality condition we have a i u i = 0 if we differentiate wrt the proper time of P: da i u i = a i ai = 0. dτ Replacing
da i dτ
from the equation of motion we find a 2 u i u i + a i ai = 0 ⇒ a i ai = a 2 .
Because a i ai = a+2 we infer that the proper acceleration has constant measure. (b) Let Σ be an LCF in which P has threevelocity v and threeacceleration a. In Σ the fourvelocity and the fouracceleration of P have decomposition u = i
γc γv
Σ
,
a = i
cγ γ˙ γ γ˙ v + γ 2 a
.
Σ
We compute da i da i =γ =γ dτ dt
c(γ γ˙ )· · (γ γ˙ ) v + 3γ γ˙ a + γ 2 a˙
Σ
and the equation of motion gives a 2 = (γ γ˙ )· , a 2 v + 3γ γ˙ a + γ 2 a˙ = a 2 v. Multiply the second with γ to find (γ 3 a)· = 0. The last two equations are the equations of motion of P in Σ. In case the motion is taking place along the xaxis only the equation of motion gives (ax = a) (γ 3 a)· = 0 ⇒ γ 3 a = constant. The constant is independent of the LCF considered; therefore, if we go to the proper frame of P we find γ = 1, hence the constant= a(= measure of threeacceleration of P in its proper frame). The equation of motion reads
5 FourAcceleration
177
γ3
dv dv =a⇒ = adt ⇒ 2 dt (1 − vc2 )3/2 v/c v0 /c − = at ⇒ 2 v2 1 − vc2 1 − c02
βγ = β0 γ0 + at ⇒ γ 2 − 1 = A2 , where A = β0 γ0 + at. Hence γ =
$ 1 A =√ ⇒ γ2 1 + A2 β0 γ0 + at v(t) = c. 1 + (β0 γ0 + at)2
A2 + 1,
β=
1−
Similarly, we compute x(t) = x0 +
c 1 + A 2 − γ0 . a
We note that when t → ∞ the speed v(t) → c as expected, whereas in the Newtonian limit (v c, α 1) v(t) → v0 + at and the x(t) = x0 + v0 t + 12 at 2 . Note: The equation of motion can be found from the transformation of the components of the fouracceleration and the invariant γ 3 ax = γv3 ax . 5.5. The equation of motion of the spaceship in Σ is γ 3 ax = constant = a + ⇒ 1 t1 = + a
(
v1 0
dv (1 −
v 2 3/2 ) c2
dv v 2 3/2 ) c2
(1 − v1 = a+ 1 −
v12 c2
= a + dt ⇒ = 0.55 years.
To calculate the same period in the proper frame of the spaceship we use the fact that the differential change in proper time dτ is related to the differential change in Σ as follows: γ dτ = dt. (This relation is derived if one considers the instantaneous locally inertial observer.) From the equation of motion we have γ 3 ax = a + γ dτ. Therefore, the total proper time for the astronaut in the spaceship is
178
5 FourAcceleration
τ1 =
1 a+
(
v1
dv 1−
0
v2 c2
=
& &1 + c & ln 2a + & 1 −
& & &. &
v1 c v1 c
Replacing a + = 10 m/s 2 and v1 = 0.5 c we compute τ1 = 0.52 years. For v1 =
√
0.9999 c and a + = 20 m/s 2 we calculate t1 = 47.56 years,
τ1 = 2.52 years,
that is, the difference in the times increases from 6 to 18%. This shows that a usual claim that one could realize long journeys provided one can attain high accelerations is not reasonable in practice. 5.6. (a) The fouracceleration of P in Σ and in its proper frame Σ+ is a = i
a0 c a0 v + γ 2 a
0 , + a Σ
Σ+
.
The boost relating Σ, Σ+ (to be exact between Σ and the instantaneous inertial observer comoving with Σ+ ) gives a0 c = γβa + a0 v + γ 2 a = γ a + . Eliminating a0 we find γ 2 a = γ a + − γβ 2 a + =
1 + 1 a ⇒ a = 3 a+, γ γ
which is the required result. (b) According to the first part the equation of motion of each point of the string in Σ is a = γ13 a + . The velocity u(t) is given by the relation du (1 −
u 2 3/2 ) c2
= a + dt ⇒
u/c 1−
u2 c2
=
a+ t + constant. c
Because the string starts from rest u(0) = 0, hence a+t u(t) = . + 2 1 + (ac2t)
(5.1)
5 FourAcceleration
179
This speed is common to all points of the string (that is, in Σ the string moves as a rigid body); hence, the position of an arbitrary point P of the string at the moment t of Σ is ( c2 (a + t)2 + C1 = f (t) + C1 , x P (t) = u(t)dt = + 1 + a c2 where f (t) =
c2 a+
1+
(a + t)2 . c2
For t = 0 we have
c2 c2 + C1 ⇒ C1 = x P (0) − + . + a a
x P (0) =
For the two ends A, B say of the string we have x A (t) = f (t) + x A (0), x B (t) = f (t) + x B (0). Hence, the length of the string at each moment of Σ is x B (t) − x A (t) = x B (0) − x A (0) = L where L is the natural length of the string. We conclude that in Σ the length of the string remains fixed and equal to its natural length – emphasizing once more that in Σ the string moves as a rigid body. However, it does not do so in the proper frame of the string where its length is γ (u)L and where u is the instantaneous speed of the points of the string. The condition of the problem is that the string brakes when its length in its proper frame becomes 2L. Therefore, we have 2L = γ (u)L ⇒ γ (u) = 2. In order to compute the time moment, tb say, in Σ where the string brakes we compute the velocity of the points of the string when γ (u) = 2. We have β2 = 1 −
3 1 1 =1− = . 2 γ 4 4
Replacing in (5.1) we find a + tb /c
1+
(a + tb )2 c2
√
√ 3 3c ⇒ tb = + . = 2 a
5.7. The relation between the differential of the proper time dτ of the astronaut and that of the basis dt is
180
5 FourAcceleration
dt = dτ γ0 , where 1 1 γ0 = = . 2 2 ω2 R 2 1 − c2 1 − 4πT 2 cR2 For a complete rotation the proper time ( τ=
T
0
dt 1 4π 2 R 2 = T = T 1− 2 2 . γ0 γ0 T c
Fig. 5.1 World lines of astronauts A and B
The length traveled by the astronaut in his proper frame is obviously zero Application The total time for the astronaut who leaves the platform is τ = γ10 T is smaller from the time T passed for the astronaut who remained in the platform. Therefore, the astronaut who traveled will be “younger” than the one who remained in the base. The asymmetry in the age of the two astronauts is shown in the spacetime diagram of their world lines (Fig. 5.1). 0 where a+ a+ Σ+ is the proper acceleration of P. The Lorentz transformation gives
5.8. (a) The fouracceleration of P in its proper frame is a i =
a0 c = γ β · a+ , a0 v + γ 2 a = γ a+ , γ 2 a⊥ = a+ ⊥. The third equation gives a⊥ =
1 + a . γ2 ⊥
5 FourAcceleration
181
Replacing a0 c from the second into the third we find + 2 γ (β · a+ )β + γ a = γ a .
This can be written as + 2 2 γβ 2 a+ + γ a = γ a ⇒ γ a = γ
(b) The solution is x(t) = grating we find
'
1 + 1 a ⇒ a = 3 a+ . γ2 γ
v(t) dt with the initial condition x(c/a + ) = 0. Inte
$ + 2 a t c2 c2 = + cosh ψ . x(t) = + 1 − a c a We note that the orbit x(t) satisfies the equation x −c t = 2
2 2
c2 a+
2 .
The analogous Newtonian problem has the solution x(t) =
1 c2 + a+t 2 . + a 2
The graph in the plane (x, t) showing the difference of the two results is shown in Fig. 5.2.
Fig. 5.2 Newtonian and relativistic hyperbolic motion
182
5 FourAcceleration
The velocity v(t) is given by dv qE qE 1 qE dv v/c =a= 3 ⇒ dt ⇒ t. 3/2 = 1/2 = 2 2 dt γ m m mc v v 1 − c2 1 − c2 Solving for v and taking into account the initial condition v (0) = 0, we find a+t v(t) = a + t 2 , 1+ c where a + = qmE . + We note that when t → +∞ , v → c. Also when ac t 1, the velocity v(t) = a + t as expected from the Newtonian limit. Finally, if we set tanh ψ = a+ t , then c v(t) = c sinh ψ (from which follow the same results for ψ → 0, +∞). (c) Let E+ , B+ = 0 the electromagnetic field in the proper frame of P. From the transformation equations we find E+ = E = E v E+ ⊥ =γ E− B c ⊥
B+ = B = 0 1 + B⊥ = γ B + v × E = 0, c ⊥
that is, in the proper frame there is only electric field E+ = E. This field causes the proper acceleration a+ =
qE q + q E = E= xˆ = constant . m m m
From the first part we infer that the corresponding acceleration in the laboratory frame is a = a =
1 qE xˆ . γ3 m
The rest of the solution is left as an easy exercise for the reader. 5.9. First Solution From the definition of the threeacceleration and the transformation of the threevelocity (i.e., the relativistic law of composition of threevelocities) we find
5 FourAcceleration
183
dv dt dv = = dt 6dt dt 7 u · v dt d 1 v+ = (γu − 1) − γu u = dt dt γu 1 − u·v u2 c2 u · v u·a 1 dt 1− 2 a+ 2 v = 2 dt γu 1 − u·v2 c c u·a c u · v u · a u · v +(γu − 1) 1 − 2 u + (γ − 1) − γ u =⇒ u u c u2 u2 c2
a =
a =
8 1−
u·v c2
a+
u·a v c2
+ u u·a c2
− 1) 1 − 3 γu2 1 − u·v c2
c2 (γ u2 u
u·v c2
+
u·v (γu u2
− 1) − γu
9 .
The last term is written as c2 c2 u · v u·v c2 (γ − 1) − (γ − 1) + (γ − 1) − γ = (γu − 1) − γu u u u u u2 u 2 c2 u2 u2 2 c c2 c2 1 = − 1 γ − = −1 . u u2 u2 u 2 γu Replacing we find the required expression a =
u · v u·a u·a 1 1 − a + v + − 1 u . 3 c2 c2 u2 γu γu2 1 − u·v 2 c 1
In the special case of the boost along the xaxis we have dv dt d ax = x = dt dt dt
*
vx − u x 1 − uv c2
+ .
The term d dt
*
vx − u x 1 − uv c2
+
ua
x ax 2 = = c 2 (vx − u) + x uvx 1 − uv 1 − c2 c2 ua uvx ax 1 x 1 − = (v − u) + a = x x 2 2 uvx 2 2 c c 1 − c2 γu 1 −
uvx 2 c2
.
184
5 FourAcceleration
Hence ax =
1 γu 1 −
uvx c2
γu2
ax 1−
uvx 2 c2
=
γu3
ax 1−
uvx 3 c2
.
Similarly we prove the remaining two relations. Second solution (a) We compute a using the Lorentz transformation. Suppose the fouracceleration of P in Σ and Σ has the form ca0 ca0 i a = , . a0 v + γv2 a Σ a0 v + γv2 a Σ The two expressions are related with a proper Lorentz transformation. Therefore 8 u · v 9 v+ (γu − 1) − γu u − γv2 a 2 u u·v u·a = a0 v + γv2 a + (γu − 1) 2 a0 + γv2 (γu − 1) 2 − γu a0 u =⇒ u ! u6 u · v a0 a0 (γ − 1) − γ v + − a 0 u u u2 γu 1 − u·v γu 1 − u·v c2 c2 u·v u·a − (γu − 1) 2 a0 − γv2 (γu − 1) 2 + γu a0 u + γv2 a = γv2 a. u u
a0 v =
1 γu 1 −
u·v c2
The coefficient of v becomes if we replace a0 −
u·a γu γ 2 γv2 u · a v u·v 2 = − . 1 − u·v c2 γu 1 − c 2 c c2
The coefficient of u is written as 1 γu 1 −
γu u·v c2
u · v u · a u · v 1 − 2 a0 − γv2 2 − (γ − 1) − γ u u c c u2 u·v u·a − (γu − 1) 2 a0 + γu a0 − γv2 (γu − 1) 2 u u
5 FourAcceleration
u · v γv2 u·a c2 − (γ − 1) − γ (γ − 1) − γ u u u u u2 1 − u·v u2 c2 u·v u·a −(γu − 1) 2 a0 + γu a0 − γv2 (γu − 1) 2 u u * 6 + 7 u2 u · v u · a 2 c γv2 2 −(γu − 1) − (γu − 1) − γu u 1 − u·v u2 c2 u · v u2 u · v u2 γv2 u · a (γu − 1) 1 − 2 + 2 2 (γu − 1) − 2 γu − 1 − u·v u2 c c u c c2 2 2 2 γv u · a u γv u · a u2 − γ (γ − 1 − γ ) = − (1 − ) − 1 u u u 1 − u·v u2 c2 1 − u·v u2 c2 c2 c2 γv2 u · a 1 − −1 . 1 − u·v u2 γu c2
= a0
= = = =
185
u · v
Replacing we find γv2 u · a γv2 u · a 1 = + v+ − 1 u =⇒ 1 − u·v c2 1 − u·v u2 γu c2 c2 2 u · v u·a 1 γv u·a 1 1 − a = a + v + − 1 u γv 1 − u·v c2 c2 u2 γu c2 γv2 a
But
γv γv
=
γv2 a
1 γu 1− u·v 2
hence:
c
a =
1
γu2 1 −
u·v 3 c2
u · v u·a u·a 1− 2 a+ 2 v+ 2 c c u
1 −1 u . γu
Setting u = (u, 0, 0) and projecting in the directions x , y , z , we obtain the required transformation equations of the threeacceleration under boosts. (b) Making use of the first part we find
a =
u · a 1 u= 2 2 u2 u γu 1 −
u·v 3 c2
a·u 1 =⇒ a = 2 3 γu u γu 1 −
u·v 3 c2
a · u.
186
5 FourAcceleration
Also a⊥ = a − a =
u · v u·a u·a 1 −1 u 3 1 − 2 a + 2 v + 2 c c u γu γu2 1 − u·v c2 1 − 3 (a · u)u 2 3 u γu 1 − u·v 2 c u2 1 2 = 3 u a − (u · a)u + 2 ((u · v)a − (u · a)v) . c (u · u)γu2 1 − u·v c2 1
But (u · u)a − (u · a)u = (u × a) × u = u 2 a⊥ and (u · v)a − (u · a)v = u × (a × v) therefore 1 1 a⊥ = 3 a⊥ + 2 u × (a × v) . c γ 2 1 − u·v2 u
c
5.10. (a) The orthogonality relation α ui = 0 ⇒ i
ca0 a0 v + γ 2 a
Σ
·
cγ0 γv
Σ
= 0 ⇒ γ˙ =
1 3 γ (a · v) c2
· a. If the speed of P is constant dγ = γ˙ = 0, and dt 0 . The length of the fourvector a i in the because a0 = 0 we find a i = γ 2a & &2 proper frame of P is &a+ & , therefore which gives a0 =
1 4 γ v c2
& & & dv & (a+ )2 = γ 4 a2 = γ 2 && && . dt When the velocity of P changes only in speed we write v = vˆe where eˆ is the constant unit threevector along the direction of the velocity. The threeacceleration a= where v˙ =
dv . dt
The a0 =
1 4 γ v c2
dv dv = eˆ = v˙ eˆ , dt dt
·a=
1 4 γ v v˙ c2
and
1 4 2 γ v v˙ eˆ + γ 2 v˙ eˆ c2 2 2 1 v2 4 v 4 v + 2 eˆ = v˙ γ + 1 − 2 eˆ = v˙ γ c2 γ c2 c
a0 v + γ 2 a =
= v˙ γ 4 eˆ .
5 FourAcceleration
187
Therefore & &2 4 & & & + &2 &a & = (a0 v + γ 2 a)2 − a 2 c2 = v˙ 2 γ 8 − v γ 8 v˙ 2 = γ 6 v˙ 2 = γ 6 & dv & . 0 & dt & 2 c (b) We replace a0 =
1 4 γ v c2
· a in the general expression of a i and take α = i
1 4 γ v c2 4
·a 1 γ (v · a)v+γ 2 a c2
. Σ
The length γ8 γ8 2γ 6 2 2 2 4 2 + v + γ a + · v) · v) (a (a (a · v)2 c2 c4 c2 γ6 γ 2 v2 4 2 2 2 = γ a + 2 (a · v) −γ + 2 + 2 . c c
α i αi = −
The term in parenthesis is written as −γ 2 + γ 2 β 2 + 2 = 1 − γ 2 + γ 2 β 2 + 1 = −γ 2 β 2 + γ 2 = γ 2 1 − β 2 = 1. Hence α i αi = γ 4 α 2 + γ 6 (a·β)2 . Making use of the identity of vector calculus1 (A × B)2 = A2 B2 − (A · B)2 , we find " # α i αi = γ 4 a2 + γ 6 a2 β 2 − (a × β)2 = γ 4 a2 1 + γ 2 β 2 − γ 2 (a × β)2 " # = γ 6 a2 − (a × β)2 . 2 But α i αi = a+ , therefore
1
a+
2
= γ 6 a2 − (a × β)2 .
To prove this identity use the other wellknown identity (A × B) · (C × D) = (A · C)(B · D) − (A · D)(B · C).
188
5 FourAcceleration
5.11. (a) The world lines of the two rockets are two hyperbolae in the x > 0 and x < 0 parts of the x–ct plane with asymptotes the light cone lines x = ±ct. (b) The position vector x1i (τ1 ) = hence the fourvelocity (ψ1 =
c2 a
sinh ψ1 − cosh ψ1
,
aτ1 ) c
u i1 (τ1 )
=c
cosh ψ1 − sinh ψ1
,
and similarly for rocket 2 x2i (τ2 ) =
c2 a
sinh ψ2 cosh ψ2
, u i2 (τ2 ) = c
cosh ψ2 sinh ψ2
.
We note that x1i (τ1 )x1i (τ1 ) = x2i (τ2 )x2i (τ2 ) = The fourvector Z i (τ ) is (ψ = Z (τ ) = i
x2i (τ )
−
x1i (−τ )
c2 = a
c2 a
2 = constant.
aτ ) c
sinh ψ cosh ψ
c2 − a
− sinh ψ − cosh ψ
c2 =2 a
sinh ψ cosh ψ
= 2x2i (τ ). The length 2 2 c Z i (τ )Z i (τ ) = 4x2i (τ2 )x2i (τ2 ) = 2 = constant > 0. a We infer that Z i (τ ) is a spacelike vector. The inner product Z i (τ )u i1 (−τ ) = 2
c2 a
sinh ψ cosh ψ
cosh ψ c =0 sinh ψ
Similarly we find that Z i (τ )u i2 (τ ) = 0. The fact that the length of the connecting vector Z i (τ ) is constant means that as the rockets “move” in spacetime, the line connecting the events x2i (τ ), x1i (−τ ) on the corresponding world lines rotates about the origin so that its middle point is 2 always at the origin. Its Lorentz length is constant and equal to 2 ca (that is, equal to the distance of the rockets at the time moment t = 0 of Σ).
Chapter 6
ThreeMomentum – Energy
6.1 Problems on Energy and ThreeMomentum 6.1.1. The mass of the electron in natural units equals its energy. Thus m = E = m S I c2 = 9.1 × 10−28 g × (3 × 108 ms−1 )2 = 8.19 × 10−14 J. But 1 eV = 1 V × 1.6021 × 10−19 Cb = 1.6021 × 10−19 J, hence m=
8.19 × 10− 14 eV = 0.51 × 106 eV = 0.51 MeV 1.6021 × 10−19
6.1.2. (a) The total energy of the particle in Σ is E = T + mc2 = mγ c2 where T is the kinetic energy. When T = mc2 = internal energy of the particle, we find √ 3 . mγ c = 2mc ⇒ γ = 2 ⇔ β = 2 2
2
In the case of photons the “internal energy” mc2 equals zero and the above do not apply. This is not a problem because for photons we do not define kinetic energy but only energy with the Planck relation E = hν. (b) We have E Kin = mγ c2 − mc2 ⇒ mγ c2 − mc2 = mc2 ⇒ γ = 2 ⇒ u =
√ 3 c. 2
6.1.3. (a) Newtonian Physics E=
1 2 d dE mu ⇒ = 2 dp d (mu)
d 1 2 du 1 2 1 mu = mu = mu = u. 2 du 2 d (mu) m
189
190
6 ThreeMomentum – Energy
(b) Special Relativity From the relation E =
p2 c2 + m 20 c4 we have
dE 1 2pc2 pc2 mγ uc2 = u. = = = dp 2 E E mγ c2 Concerning the photons we have that the threemomentum p = (E/c)ˆe where eˆ is the unit in the direction of propagation of the photon in Σ. Hence E = cp · eˆ and dE = c eˆ . dp The RHS is exactly the velocity of the photon in Σ. We conclude that the formula does hold for photons. (b) Obvious. 6.1.4. We have E 2 = p2 c2 + m 2 c4 ⇒ Ed E = p · dpc2 = pdpc2 =
1 2 dp . 2
Therefore, p =constant implies E = constant. The converse is obviously true. 6.1.5. (a) From the definition of the work of a threeforce we have dW = f · dr = v · dp = vm · (dγ v + γ dv). But v · dv = vdv = c2
dγ , γ3
hence, 2 v dγ 1 dW = m v 2 dγ + γ c2 3 = mc2 dγ = d(mγ c2 ). + γ c2 γ2 (b) Obvious. 6.1.6. (a) Let Ai , Bi be two fourvectors which in the LCF Σ have representation (A0 , A), (B0 , B). Let Σ be another LCF which is related with Σ with a Lorentz transformation with relative velocity β. Let (A0 , A ) and (B0 , B ) be the representation of the fourvectors in Σ . The Lorentz transformation gives
6.1
Problems on Energy and ThreeMomentum
191
γ (A · β)A0 γ β, A0 = γ (A0 + β · A ), γ +1 γ (B · β)B0 γ β, B0 = γ (B0 + β · B ). B = B + γ +1
A = A +
If A = B and A = B these relations imply A0 = A0 and B0 = B0 which proves the required result. (Similarly if we consider A0 = A0 , B0 = B0 we prove that A = B and A = B ). (b) Consider the fourmomentum of a system which reacts from the state 1 to state 2. If P1i , P2i are the fourmomenta of the system at the states 1, 2, respectively, the change in fourmomentum is ΔP i = P2i − P1i . In an arbitrary LCF ΔE/c , where ΔE = E 2 − E 1 and ΔP = P2 − P1 all quantities Σ ΔP i = ΔP being in Σ. If the energy is conserved then ΔE = 0 in all LCF and part (a) implies ΔP i = 0 hence ΔP = 0, that is, the threemomentum is also conserved in Σ. Because Σ is arbitrary we infer that the threemomentum is conserved in all LCF. Similarly we work with the conservation of the threemomentum. 6.1.7. (a) Consider two LCF which are related with a boost (this is not a restriction because a Lorentz transformation can be decomposed in the product of a boost and a rotation) and let Ai , B i be two fourvectors which have the same spatial part in Σ, Σ . The vector Ai − B i in Σ, Σ has the form (x, 0)Σ , (x , 0)Σ . The two expressions are related with a boost; therefore, x = γ x, 0 = γ x ⇒ x = x = 0 that is, Ai − B i = 0 ⇒ Ai = B i . (b) When the threevelocities are equal the γ factors are equal, hence the fourvelocities are equal. (c) Let p = p be the threemomenta of the two particles in the LCF Σ. The energies E, E of the particles in Σ are different because E=
p 2 c2 + m 2 c4 ,
E =
p 2 c2 + m 2 c4 .
Therefore, the difference of the fourmomenta in Σ does not vanish, and because it is a fourvector and the Lorentz transformation is homogeneous it does not vanish in any other LCF. 6.1.8. We calculate the length ds 2 along the world line of the particle. From the equations of the curve we have d x = r cos θ cos φdλ,
dy = r cos θ sin φdλ,
dz = r sin θ dλ,
dt = r dλ.
192
6 ThreeMomentum – Energy
Hence ds 2 = −dt 2 + d x 2 + dy 2 + dz 2 = (−r 2 + r 2 )dλ2 = 0. We conclude that the particle is a photon. The fourmomentum of the photon is pi =
E c
1 , eˆ Σ
where eˆ is the unit normal along the direction of propagation of the photon in Σ. To find this direction we compute the threevelocity of the photon in Σ. We have (c = 1) υx =
dx = cos θ cos φ, dt
υy =
dy = cos θ sin φ, dt
υz =
dz = sin θ, dt
from which follows eˆ = (cos θ cos φ, cos θ sin φ, sin θ )Σ . 6.1.9. First Solution We have dpx dp y dpz γ (dpx − βd E/c)dp y dpz d 3 p = = E E cγ (E/c − βpx ) dpx dp y dpz 1 − (β/c)(∂ E/∂ px ) . = E 1 − βc( px /E) But E 2 = p2 c2 + m 2 c4 ⇒ Ed E = c2 ( px dpx + p y dp y + pz dpz ) ⇒
px ∂E = c2 . ∂ px E
The term 1 − (β/c)(∂ E/∂ px ) 1 − (β/c)c2 ( px /E) = =1 1 − βc( px /E) 1 − βc( px /E) from which follows d 3 p d 3p . = E E Second solution A more formal proof is the following. The volume d 3 p under the transformation ( px , p y , pz ) → ( px , p y , pz ) transforms as a tensor density of weight. 1. Hence
6.2
Relative Threemomentum – Colliding Beams
193
d 3 p =J d 3 p , where J is the Jacobian of the transformation J
px , p y , pz p x , p y , pz
& ∂ p & x & ∂ px & ∂ p = && ∂ pxy & ∂ pz & ∂p
∂ px ∂ py ∂ py ∂ py ∂ pz ∂ py
x
∂ px ∂ pz ∂ py ∂ pz ∂ pz ∂ pz
& & & & &. & & &
From the boost we have
J = det diag
∂ px , 1, 1 ∂ px
=
∂ px . ∂ px
The partial derivative ∂ px ∂ =γ ∂ px ∂ px But E =
p2 + m 2 ⇒
dE dpx
px − 2
β E c
= 12 2 px EE =
∂ px =γ ∂ px
px E
=γ
β dE 1− . c dpx
, thus
E E β px = ⇒J= . 1− c E E E
Finally, d 3 p =
E 3 d 3 p d 3p d p⇒ . = E E E
6.2 Relative Threemomentum – Colliding Beams 6.2.1. The emittance of the photon causes a change Δm in the mass m 1 , hence a change in energy Δmc2 . If ν is the frequency of the emitted photon, then hν = Δmc2 . The change Δ p of the threemomentum of m 1 equals the momentum of the emitted photon, hence Δp =
hν = m 1 Δu 1 . c
194
6 ThreeMomentum – Energy
The change of the kinetic energy ΔT1 of the mass m 1 equals ΔT1 =
1 Δm 2 m 1 Δu 1 2 = . 2 2m 1 c2
Similarly for the change of the kinetic energy of the mass m 2 we have Δm 2 . 2m 2 c2
ΔT2 =
The total change in the kinetic energy of the system of the two masses is ΔT = ΔT1 + ΔT2 =
Δm 2 c2
1 1 + m1 m2
.
This energy is stored in the string as potential energy 12 kΔx 2 . Therefore, 1 1 ⇒ + m1 m2 Δm 1 1 1/2 Δx = √ + . m2 c k m1
1 Δm 2 kΔx 2 = 2 2 c
When k → 0, that is, the string essentially does not exist, the Δx → ∞, which means that the two particles are moving away permanently. When k → ∞, that is, the string becomes a rigid rod, Δx → 0, and the relative distance of the particles does not change. 6.2.2. In the lab frame the total energy of the electron when it has covered a distance x in the accelerator is E x = E 0 + ax, where E 0 = m e c2 and a is a proportionality constant. On the other hand, E x = γx m e c2 = γx E 0 . These two relations give γx =
E 0 + ax a ⇒ γx = 1 + x. E0 E0
We consider next two adjacent positions x, x + d x along the trajectory of the beam in the accelerator in the lab frame. These points in the proper frame of the electron have a distance d x equal to
6.2
Relative Threemomentum – Colliding Beams
dx =
195
dx . γx
Integrating we find the total length of the accelerator as “seen ” by the electrons
(L
L =
(L
dx = 0
0
dx = γx
(L 0
E0 a dx = ln 1 + L . 1 + Ea0 x a E0
The constant a is fixed by the requirement at x = L, the E x = E. It follows a=
E − E0 , L
and finally we have L = L
E0 ln E − E0
E E0
.
Numerical application. Replacing the given values we calculate L = 81 cm.
We note that L L and more specifically that LL = 0.27, which shows how drastically different can be the Newtonian and the relativistic calculations for high speeds. 6.2.3. (a) We have v=
p pc2 = . mγ E
(b) In the proper frame of 2 we have p1i
=
E 12 /c p12
Σ+ 2
,
p2i
=
m2c 0
The inner product p1i p2i = −E 12 m 2 ⇒ E 12 = −
p1i p2i . m2
Σ+ 2
.
196
6 ThreeMomentum – Energy
We consider the length of p1i and have 2 p1i p1i = −m 21 c2 = −E 12 /c2 + p212 ⇒ i 2 # p1 p2i 1 " 2 /c2 − m 21 c2 = − m 21 c2 = 2 2 ( p1i p2i )2 − m 21 m 22 c4 . p212 = E 12 m2c m2c
The measure of the threevelocity of 1 wrt 2 according to the above is given by v212 =
# " i # 1 " c2 p212 c4 m 2 c4 ( p1 p2i )2 − m 21 m 22 c4 . = i 2 2 2 2 ( p1i p2i )2 − m 21 m 22 c4 = i 2 2 E 12 ( p1 p2i ) m 2 c ( p1 p2i )
This relation is covariant, hence it stays valid in all LCF. In terms of the fourvelocities of 1, 2 it is written as follows: m 2 m 2 c4 m 2 m 2 c4 c4 2 = c 1 − . v212 = v221 = c2 1 − 1i 2 2 = c2 1 − 2 21 2i ( p1 p2i ) m 1 m 2 (u 1 u 2i )2 (u i1 u 2i )2 We note that this expression is symmetric in u i1 , u i2 , hence v12  = v21  . 6.2.4. (a) Let pei , pip be the fourmomenta of the particles of the two beams. Then in the laboratory frame we have p1i =
E 1 /c p1 eˆ
,
p2i =
L
E 2 /c − p2 eˆ
. L
In the CM frame of the two beams the fourmomenta have the following form: p1i =
E 1∗ /c p ∗ eˆ ∗
Σ∗
,
p2i =
E 2∗ /c − p ∗ eˆ ∗
Σ∗
.
The quantity p1i + p2i is the same in both frames. We compute p1i + p2i =
(E 1 + E 2 )/c ( p1 − p2 )ˆe
Σ
=
(E 1∗ + E 2∗ )/c 0
Σ∗
.
The available energy in the CM frame is the quantity E 1∗ + E 2∗ which we denote with M. From the invariance of the length of the fourvector p1i + p2i and the above expressions, we find −M 2 = −(E 1 + E 2 )2 + (p1 − p2 )2 . The lengths p A  = E 2A − m 2A . (A = 1, 2). Because m A E A without significant error we consider p A  = E A and have
6.2
Relative Threemomentum – Colliding Beams
197
M 2 (E 1 + E 2 )2 − (E 1 − E 2 )2 = 2E 1 E 2 . Replacing the data we compute M 310 GeV/c2 . (b) The βCM factor in the laboratory frame is found by the quotient of the total threemomentum in the laboratory frame over the total energy in the same frame, that is, β CM =
p1 eˆ − p2 eˆ E1 − E2 eˆ ⇒ β CM  = 0.928. E1 + E2 E1 + E2
(c) In order to compute the common threemomentum p∗ of the beams in the CM frame, we use the boost relating the CM frame with the laboratory frame along the common direction of propagation of the beams. For the beam of electrons we find ∗ E1 E1 = p1i = E 1 eˆ L p ∗ eˆ ∗ CM and $ p ∗ ex∗
= γCM (E 1 − βCM E 1 ) =
p ∗ e∗y = 0
1 − βCM E1, 1 + βCM
p ∗ ez∗ = 0,
CM from which follows ex∗ = 1 and p ∗ = 1−β E . 1+βCM 1 Introducing the given data we compute p ∗ = 154.60 GeV/c. [Note: If we work with the beam of protons we find p ∗ = 155.24 GeV/c. What is the cause in the difference of the two results?] 6.2.5. In the lab frame (L) the two beams have the same energy (because they have the same threemomentum) which is E = p 2 c2 + m 2 c4 ≈ pc = 25 GeV. Assuming that the beam 1 (say) is propagating along the negative direction of the xaxis we write for the fourmomenta of the two beams in the lab frame
198
6 ThreeMomentum – Energy
⎞ E/c p2i = ⎝ p cos 15◦ ⎠ , p sin 15◦ L
⎛
⎞ E/c p1i = ⎝ − p ⎠ 0 L
⎛
from which follows p1i p2i = −
E2 − p 2 cos 15◦ ≈ − p 2 (1 + cos 15◦ ). c2
In the proper frame of one of the beams the inner product p1i p2i is given by p1i p2i
=
E /c p
1
mc 0
= −E m.
1
Because p1i p2i is invariant its value is the same in all LCF; therefore, E =
p2 (1 + cos 15◦ ) ≈ 1250 GeV. m
6.2.6. (a) Let p1i , p2i be the fourmomenta of the two particles. The mass M ∗ of the CM particle is −M ∗2 c2 = ( p1i + p2i )2 = −(m 21 + m 22 )c2 + 2 p1i p2i . In the lab frame we have p1i p2i
=
E 1 /c p
L
m2c 0
= −E 1 m 2 , L
hence M ∗2 = m 21 + m 22 +
E1 m 2 = (m 1 + m 2 )2 + c2
T1 2 c − m1
m2.
The available energy (in the lab frame) for the production of particles is $ ∗
E Δ /c = M − (m 1 + m 2 ) = 2
(m 1 + m 2 )2 + 2
T1 m 2 − (m 1 + m 2 ). c2
For m 2 = 1.5 GeV, 100 GeV we compute E Δ (1.5) = 6.51 GeV,
E Δ (100) = 22.2 GeV.
6.2
Relative Threemomentum – Colliding Beams
199
(b) We note that in both cases E Δ < T1 and furthermore as the mass m 2 increases the same happens to E Δ . (To prove this formally you can show that the deriva2 EΔ tive ∂∂m = m 1 +m22 +T1 /c T1 1/2 − 1 ≥ 0.) 2 [(m 1 +m 2 ) +2
m 2 c2
]
In order to explain the difference T − E Δ we consider the speed β ∗ of the CM frame in the lab frame , T1 (T1 + 2m 1 c2 ) p T1 ∗ , ≈ . β = = T1 + m 1 c2 + m 2 c2 T1 + m 1 c2 + m 2 c2 E We note that as m 2 → ∞ , β ∗ → 0; therefore, the difference T1 − E Δ is spend for the kinetic energy (speed) of the CM particle in the lab frame. (c) The maximal value for E Δ is E Δ = T1 . This occurs when m 2 → ∞ and β ∗ → 0, that is, the CM particle is produced at the threshold in the lab frame. 6.2.7. Suppose the fourmomenta p1i , p2i of the photons in Σ and Σ are (I = 1, 2) EI E i i c pI = EI , p I = E c , where E is the common energy of the photons eˆ eˆ c I Σ c I Σ in Σ . The fourvector p1i − p2i in Σ and Σ is p1i
−
p2i
=
E 1 −E 2 c E1 eˆ − Ec2 eˆ 2 c 1
= Σ
0 A eˆ c
, Σ
where A eˆ = E (ˆe1 − eˆ 2 ),
(6.1)
and eˆ 1 , eˆ 2 are the unit vectors along the direction of propagation of the photons in Σ . We note that the fourvector p1i − p2i is spacelike; therefore, it is not the fourmomentum of a photon. In order to compute the quantity A we consider the identity ( p1i − p2i )2 = 2 2 p1i + p2i − 2 p1i p2i and compute the LHS in Σ and the RHS in Σ. From the above decompositions of the fourvectors we find easily A = 2E 1 E 2 (1 − eˆ 1 · eˆ 2 ). 2
(6.2)
Concerning eˆ we consider the general Lorentz transformation which relates the components of the fourvector ( p1i − p2i ) in Σ and Σ . The zeroth coordinate gives E1 − E2 E1 − E2 A eˆ . = γβ · ⇒ β · eˆ = c c γ A
(6.3)
200
6 ThreeMomentum – Energy
The space part gives 2 γ A eˆ E1 E2 A eˆ + β· β eˆ 1 − eˆ 2 = c c c γ +1 c A eˆ γ E1 − E2 = + β⇒ c γ +1 c 1 γ (E 1 − E 2 )β . eˆ = E 1 eˆ 1 − E 2 eˆ 2 − A 1+γ
(6.4) (6.5)
We note that β is fixed uniquely in terms of the eˆ and the directions of propagation of the photons in Σ. This implies that the LCF Σ is not fixed uniquely from the condition of equal energies of the photons and one has to give also the direction of propagation of the photons either in Σ or in Σ . The same applies to the common energy E of the photons, which in order to be specified needs the knowledge of the velocity factor β. A different method to tackle the problem is the following. The Lorentz transformation for the fourmomentum of the photons is (see (6.3), (6.5) I = 1, 2) E = γ E I (1 − β · eˆ I ), γ β · eˆ I − γ β . E eˆ I = E I eˆ I + 1+γ
(6.6) (6.7)
From the second relation we infer that if one is given the quantities eˆ I , E I then β is determined in terms of the quantities eˆ I , E I . The first relation specifies E in terms of β. In addition if we eliminate E we take a relation/constraint between the components of β which is the following β · eˆ 1 = 1 −
E2 (1 − β · eˆ 2 ), E1
(6.8)
that is, condition (6.3). Finally, there remain only two scalar equations which determine β in terms of eˆ I or the inverse. Practically this means that the direction of propagation of the photons in Σ as well as the value of the common energy depends on the velocity β of Σ . From the second equation we infer that E (ˆe1 − eˆ 2 ) = (E 1 eˆ 1 − E 2 eˆ 2 ) −
γ (E 1 − E 2 )β, 1+γ
(6.9)
hence β is specified and specifies the “relative direction” of propagation of the photons in Σ , that is, the two relations we have already mentioned above (ˆe is unit, therefore requires only two relations in order to be specified).
6.2
Relative Threemomentum – Colliding Beams
201
Collecting results we have the following system of equations for the current problem A eˆ = E (ˆe1 − eˆ 2 ),
(6.10)
2
A = 2E 1 E 2 (1 − eˆ 1 ·ˆˆe2 ), E2 β · eˆ 1 = 1 − (1 − β · eˆ 2 ) , E1 1 γ (E 1 − E 2 )β]. eˆ = [E 1 eˆ 1 − E 2 eˆ 2 − A 1+γ
(6.11) (6.12) (6.13)
(b) We consider the special cases of propagation of the photons in Σ assuming E1 > E2 . (b1) In this case we have the following data eˆ 2 = eˆ 2 . Propagation of the photons in Σ : eˆ 1 = eˆ 1 , Propagation of the photons in Σ: eˆ 1 = −ˆe2 . Without restricting the generality we consider eˆ 1 = xˆ . Equation (6.11) gives A2 = 2E 1 E 2 (1 + 1) = 4E 1 E 2 and (6.10) A eˆ = E (ˆe1 + eˆ 1 ) ⇒ E =
E1 E2,
eˆ = eˆ 1 .
Condition (6.9) gives γ E (ˆe1 + eˆ 1 ) = (E 1 eˆ 1 + E 2 eˆ 1 ) − (E 1 − E 2 ) β ⇒ 1+γ √ 1 + γ E1 + E2 2 E1 E2 β= − eˆ 1 . γ E1 − E2 E1 − E2 Therefore, Σ moves in the standard way along the xaxis of Σ with speed β. In order to compute β we use condition (6.8) and the fact that β y = βz = 0 E 1 (1 − β) = E 2 (1 + β) ⇒ E1 − E2 . β= E1 + E2 From the equation of transformation of the common energy E we have
E = γ (E 1 − β E 1 ) = γ
E1 − E2 E1 − E1 E1 + E2
E1 + E2 ⇒γ = √ . 2 E1 E2
(The same result is obtained if one calculates γ directly from β).
202
6 ThreeMomentum – Energy
We conclude that for the given propagation of the photons, the LCF Σ moves 2 and wrt Σ in the standard way along the common x, x axis with speed β = EE11 −E +E 2 √ that the common energy of the photons is E = E 1 E 2 . (b2) In this case we have Propagation of the photons in Σ The relative direction of propagation of the photons in Σ is the bisector of the first quarter in the plane x − y of Σ, therefore xˆ + yˆ eˆ = √ . 2 Propagation of the photons in Σ eˆ 1 = xˆ ,
eˆ 2 = yˆ .
From (6.11) we have A2 = 2E 1 E 2 . Equation (6.9) gives if we replace A , eˆ β=
1 γ +1 (E 1 − E 1 E 2 )ˆx − (E 2 + E 1 E 2 )ˆy . γ E1 − E2
This relation does not evaluate β because γ is unknown. However, we do have one more equation. Indeed condition (6.8) gives γ +1 γ +1 1 1 (E 1 − E 1 E 2 ) = E 2 1 + (E 2 + E 1 E 2 ) , E1 1 − γ E1 − E2 γ E1 − E2 from which we compute γ +1 (E 1 − E 2 )2 = 2 √ 2 γ E 1 + E 2 − (E 1 − E 2 ) E 1 E 2 and √ E 12 + E 22 − (E 1 − E 2 ) E 1 E 2 γ = . √ −2E 1 E 2 + (E 1 − E 2 ) E 1 E 2 Introducing the first result in the expression for β we find eventually β=
(E 1 − E 2 ) (E − E E )ˆ x − (E + E E )ˆ y . √ 1 1 2 2 1 2 E 12 + E 22 − (E 1 − E 2 ) E 1 E 2
6.2
Relative Threemomentum – Colliding Beams
203
Finally, we compute the common energy E in Σ from the Lorentz transformation as before. We have E = γ (E 1 − E 1 βx ) and replacing γ , βx we find
E =
√ √ E 1 (E 12 + E 22 − (E 1 − E 2 ) E 1 E 2 (E 1 − E 2 )(E 1 − E 1 E 2 ) 1− 2 , √ √ −2E 1 E 2 + (E 1 − E 2 ) E 1 E 2 E 1 + E 22 − (E 1 − E 2 ) E 1 E 2
which is simplified as follows E =
E 1 E 2 (E 1 + E 2 ) . √ −2E 1 E 2 + (E 1 − E 2 ) E 1 E 2
6.2.8. (a) Suppose the frequency of the photons in two LCF Σ and Σ are ν1 , ν2 and ν1 , ν2 , respectively. Then the fourmomentum of the photon 1 (say) in Σ, Σ is p1i = hν1 (1, eˆ )Σ = hν1 (1, eˆ 1 )Σ , where eˆ , eˆ 1 are the directions of propagation of the photon in Σ, Σ . Similarly the fourmomentum of the second photon in Σ, Σ is p2i = hν2 (1, eˆ )Σ = hν2 (1, eˆ 2 )Σ . Comparing the last two relations (or using the Lorentz transformation) we find eˆ 1 = eˆ 2 that is, the photons are propagating parallel in Σ . Concerning the fourmomenta we have p2i = νν21 p1i , that is, they are parallel. (b) The number of the photons in the LCF Σ, Σ is the same. Hence n = ρΔsΔl = ρ Δs Δl . Δs ⊥u ⇒ Δs = Δs ⇒ ρΔsΔl = ρ Δs γ1 Δl ⇒ ρρ = γ . 1 Δl x ⇒ Δl = γ Δl For the fourmomentum of the photons of the beam when propagating in the yaxis of Σ we have But
pi = hν(1, 0, 1, 0) = hν (1, 0, 1, 0).
204
6 ThreeMomentum – Energy
The boost relating Σ, Σ gives hν = hν ⇒
ρ ν =γ = . ν ρ
If the beam propagates along the common xaxis, then pi = hν(1, 1, 0, 0) = hν (1, 1, 0, 0) ⇒ hν = γ (hν − βhν) ⇒ ν ρ ν = γ ν(1 − β) ⇒ = . ρ ν(1 − β)
6.3 Scattering Densities 6.3.1. (a) The volume d V of the mass dm in the LCF Σ can be written d V = d ds⊥ , where d is the dimension of dm parallel to the relative velocity u of Σ, Σ and ds⊥ is its cross section normal to u. The Lorentz transformation gives d =
d 0 γ
,
ds0⊥ = ds⊥ ,
where d 0 , ds0⊥ are the corresponding quantities in the rest frame of dm. The volume dV =
d 0 d V0 ds0⊥ = . γ γ
Similarly for Σ we write dV =
d V0 γ
because d V0 is invariant. Finally, we find γ d V = γ d V . (b) From the invariance of the mass dm we have dm = ρd V = ρ d V from which follows ρ ρ = . γ γ
6.3
Scattering Densities
205
(c) The energy E of the mass dm in Σ is E = dmγ c2 where γ is the velocity factor c2 of dm in Σ. Writing E = ρ E d V we find ρ E = dmγ = ργ 2 c2 . Similarly for dV Σ we write ρ E = ρ γ 2 . From these relations it follows that ρE ρ E = . γ2 γ 2 6.3.2. (a) The atoms of the beam of radius R1 are scattered from the atoms of the gas of radius R2 in a cross section area S = π (R1 + R2 )2 as in the Fig. 6.1
Fig. 6.1 Scattering of atoms
In time dt the atoms of the beam trace a volume d V = Sdl = π (R1 + R2 )2 udt in which there are kd V atoms of gas, hence the number of the scattered atoms of the beam in time dt is1 d N (t) = −N (t)kd V ⇒
d N (t) = −kπ (R1 + R2 )2 udt. N (t)
Integrating we find ln
N (t) = −kπ (R1 + R2 )2 ut N0
⇒
N0 − N (t) = 1 − exp{−π (R1 + R2 )2 kut} , N0
t > 0.
(b) This case is similar to (a) if we reduce the data in the lab frame. The velocity of the atoms of the gas is normal to the scattering S, hence S remains the same. Therefore, Number of scattering centers=kgas d VLAB = kgas SdlLAB =
k gas SdlLAB , γ (v)
where kgas is the proper density of the gas (kgas = k) and S = π (R1 + R2 )2 as before. The number of scattering centers is invariant. The speed of the beam in the lab frame is
1 Because (reduction of the number of atoms)=(number of atoms of the beam) × (number of scattering centers in time dt).
206
6 ThreeMomentum – Energy
u =
u+v , 1 + uv c2
hence dl L AB = u dt =
u+v dt. 1 + uv c2
The number of scattered atoms of the beam is d N (t) = −N (t)kgas π (R1 + R2 )2
u + v dt . 1 + uv γ (v) c2
Integrating, we find v2 N (t) 2 u+v 1− 2 t ⇒ ln = −kgas π (R1 + R2 ) N0 1 + uv c c2 7 6 2 v N0 − N (t) u + v 1 − 2 t , t ≥ 0. = 1 − exp −π (R1 + R2 )2 kgas N0 1 + uv c c2
6.4 Reflection of Light on Mirror 6.4.1. (a) Suppose that the photon before the reflection in the LCF Σ has fourmomentum2 p1i = (E 1 , p1 )Σ and in the rest frame of the mirror, Σ say, four = (E 1 , p1 )Σ . For the reflected photon assume correspondingly momentum p1i = (E 2 , p2 )Σ . The boost relating Σ, the fourmomenta p2i = (E 2 , p2 )Σ and p2i Σ gives tan θ1 = tan θ2 =
γ ( p1,x + β E ) p1,x = , p1,y p1,y
+ β E 2 ) γ ( p2,x p2,x = . − p2,y − p2,y
The reflection is governed by the reflection law of Snell which is stated as follows In the proper frame of the mirror Σ (and in that frame only!) – The angle of incidence equals the angle of reflection. – The energy of the falling photon equals the energy of the reflected photon.
2 When we lower the zeroth component we change the sign. Here we do not lower the component but we simply write the covariant form of the fourmomentum, which is a 1 × 4 matrix.
6.4
Reflection of Light on Mirror
207
Therefore the following relations hold: E 1 = E 2 , = p2,x , p1,x p1,y = − p2,y . Replacing in the former relations we find tan θ2 =
γ ( p1,x + β E 1 ) p1,y
= tan θ1 ⇒ θ1 = θ2 .
) = E 2 . We infer that, in case the mirror moves parallel Also E 1 = γ (E 1 + βp1,x to its plane in Σ, Snell law applies in Σ. Obviously the frequency of the photon remains unchanged during reflection for all speeds of the mirror.
(b) In this case the fourmomentum of the falling photon in Σ is p1i = (E 1 , 0, E 1 sin θ1 , −E 1 cos θ1 )Σ and suppose that in Σ p1i = (E 1 , p1,x , p1,y , p1,z )Σ .
The boost relating Σ, Σ gives E 1 = γ (E 1 + β E 1 cos θ1 ) = γ E 1 (1 + β cos θ1 ), p1,x =0
p1,y = E 1 sin θ1 ,
= γ (−E 1 cos θ1 − β E 1 ) = −γ E 1 (β + cos θ1 ). p1,z
From Snell’s law we have that the fourmomentum of the reflected photon is p2,i = (E 1 , px,1 , p y,1 , − pz,1 ) = (γ E 1 (1 + β cos θ1 ), 0, E 1 sin θ1 , γ E 1 (β + cos θ1 )) .
Transforming this in Σ with the inverse boost we find E 2 = γ (E 2 + βpz,2 ) = γ (γ E 1 (1 + β cos θ1 ) + βγ E 1 (β + cos θ1 )) ,
= γ 2 E 1 (1 + β 2 + 2β cos θ1 ), px,2 = px,2 =0 p y,2 = p y,2 = E 1 sin θ1 , pz,2 = γ ( pz,2 + β E 2 ) = γ [γ E 1 (β + cos θ1 ) + βγ E 1 (1 + β cos θ1 )]
= γ 2 E 1 [cos θ1 + 2β + β 2 cos θ1 ].
208
6 ThreeMomentum – Energy
The angle of reflection in Σ is cos θ2 =
pz,2 2β + (1 + β 2 ) cos θ1 pz,2 γ 2 E 1 (cos θ1 + 2β + β 2 cos θ1 ) = = = . 2 2 p2  E2 γ E 1 (1 + 2β cos θ1 + β ) 1 + β 2 + 2β cos θ1
Concerning the frequency of the photon in Σ we have ν2 =
E2 = γ 2 1 + β 2 + 2β cos θ1 ] ν1 . h
The condition that the frequency of the photon remains unchanged during reflection is ν2 = ν1 , that is, γ 2 (1 + β 2 + 2β cos θ1 ) = 1 ⇒ cos θ1 = −β ⇒ θ1 >
π , 2
which is impossible. 6.4.2. The collision is elastic only in the rest frame of the wall and not in Σ. Therefore, we have to transform all data in the rest frame of the wall. We decompose the threemomentum of the particle normal and along the velocity of the wall in Σ and have (c = 1) Σ
Σ
Reflection
Σ
⎛
⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ E E E E ⎝ p ⎠ ⎝ p ⎠ ⎝ −p ⎠ ⎝ p ⎠ . p⊥ p⊥ p⊥ p⊥ The p⊥ remains unchanged whereas for the energy and p we have E = γ (E − βp ) E = γ (E − βp )
p = γ ( p − β E) p = γ (− p + β E ).
Replacing E , p we find " # E = γ 2 "(1 + β 2 )E − 2βp # p = γ 2 −(1 + β 2 ) p + 2β E . The angle φ between the reflected direction and the xaxis in Σ is p cot φ = = γ2 p⊥ But Ep = the result
1 υ
E −(1 + β ) cot θ + 2β . p sin θ 2
(c = 1) where υ is the speed of the particle in Σ. Therefore, we have
6.4
Reflection of Light on Mirror
209
cot φ = γ 2
−(1 + β 2 ) cot θ +
2u . υ sin θ
6.4.3. (a) The velocity of the photon in the proper frame of the mirror before the reflection is vM
c c c = √ (ˆx + yˆ ) − u xˆ = √ − u xˆ + √ yˆ . 2 2 2
After the reflection, in the proper frame of the mirror the frequency of the photon remains the same and its velocity becomes c c vM = − √ + u xˆ + √ yˆ . 2 2 The frequency and the velocity of the reflected photon in Σ is found from the Galileo transformation relating the proper frame of the mirror with Σ. The frequency being Euclidean scalar remains the same, i.e., ν0 whereas the velocity c c vΣ = vM − u xˆ = − √ xˆ + √ yˆ . 2 2 (b) The fourmomentum of the photon in Σ is ⎛ pi =
1
⎞
√1 ⎟ hν0 ⎜ ⎜ 12 ⎟ . c ⎝ √2 ⎠ 0 Σ
We compute the fourmomentum of the photon in the proper frame of the mirror using the boost relating Σ with the mirror. We find ⎛
γ −βγ hν0 ⎜ −βγ γ ⎜ p = 1 c ⎝ i
⎞⎛
1
⎞
⎛
⎜ ⎜ ⎟ ⎜ √1 ⎟ ⎟ ⎜ 12 ⎟ = hν0 ⎜ ⎠⎝ √ ⎠ c ⎜ ⎝ 2 1 0 Σ
√ ⎞ γ 1 − β/ 2 √ ⎟ −γ β − 1/ 2 ⎟ ⎟ . ⎟ √ ⎠ 1/ 2 0 M
After the collision the fourmomentum of the photon in the proper frame of the mirror is
210
6 ThreeMomentum – Energy
√ ⎞ γ 1 − β/ 2 ⎜ √ ⎟ hν0 ⎜ i 2 ⎟ γ β − 1/ ⎟ , ⎜ p = ⎟ ⎜ √ c ⎝ ⎠ 1/ 2 0 M ⎛
therefore, the fourmomentum of the reflected photon in Σ is √ ⎞ γ 1 − β/ 2 ⎟ ⎜ ⎟ ⎜ γ β − 1/√2 ⎟ ⎟ ⎟⎜ ⎟ √ ⎠⎜ ⎠ ⎝ 1/ 2 1 0 M ⎞ ⎛ √ 2 2 γ (1 + β ) − β 2 ⎜ ⎟ ⎜ hν0 ⎜ 2 1 2 ⎟ √ ⎟ . 2β − (1 + β ) γ = ⎟ √2 c ⎜ ⎠ ⎝ 1/ 2 0 Σ ⎛
γ βγ ⎜ hν βγ γ 0 ⎜ p i = ⎝ 1 c
⎞
⎛
From this relation we infer that the frequency and the direction of propagation of the reflected photon in Σ are √ ν = ν0 γ 2 (1 + β 2 ) − β 2 , √ β 2 − 2 2β + 1 1 eˆ = − √ √ xˆ + √ √ yˆ . 2 2 1+β −β 2 2 1 + β2 − β 2
Chapter 7
Relativistic Optics
7.1. The fourmomenta of the photons in the LCF Σ are (we ignore the y, z coordinates) P1i =
E1 c
1 , 1 Σ
P2i =
E2 c
1 −1
Σ
.
If in the LCF Σ the common energy of the photons is E , then P1i
E = c
E 1 1 i ,P = . 1 Σ 2 c −1 Σ
The fourvector c(P1i − P2i ) in Σ and Σ is
E1 − E2 E1 + k E1
Σ
,
0 2E
Σ
.
P1i − P2i is a null fourvector, hence 0 = γ [E 1 − k E 1 − β (E 1 + k E 1 )] ⇒ β =
1−k (k < 1). 1+k
The two expressions of P1i − P2i in Σ and Σ are related with a boost with factor β. The spatial component gives 2E = γ [E 1 + k E 1 − β (E 1 − k E 1 )] = γ E 1 ⇒ E =
4k (1 + k)2 − (1 − k)2 = γ E1 1+k 1+k
2k γ E1. 1+k
211
212
7 Relativistic Optics
The factor γ =
1 1−( 1−k 1+k )
2
=
1+k √ 4k
=
1+k √ . 2 k
Replacing we find
√ 2k E = √ E1 = k E1. 2 k (b) We have k= The β =
5 1− 7.5 5 1+ 7.5
=
2.5 12.5
E1 νred 5 . = = E2 νblue 7.5
= 0.2. The common frequency is 
hν =
7.5 5 νblue hνblue ⇒ ν = √ =√ · 1013 Hz. 7.5 1.5 1.5
Second solution The photon with energy E 1 > E 2 must recede the observer (redshift) and the photon with energy E 2 must approach the observer (blueshift). Let ν be the common frequency of the $photons. The Doppler relation gives
1−β ν1 1+β $ 1+β ν2 Blueshift: ν = 1−β Equating we find Redshift: ν =
1−β ν2 1−k = . =k⇒β= 1+β ν1 1+k Replacing in any of the previous expressions follows
ν =
1−
1−k 1+k
1+
1−k 1+k
ν1 =
√ 2k ν1 = kν1 . 2
7.2. The fourvectors of the photons 1, 2 in Σ are, respectively, p1i = hν1
1 p1
Σ
, p2i = hν2
1 p2
Σ
.
7 Relativistic Optics
213
The fourmomentum of the CM particle is pi = p1i + p2i = h
ν1 + ν2 ν1 p1 + ν2 p2
Σ
.
The β factor of the CM frame in Σ is β=
ν1 eˆ 1 + ν2 eˆ 2 . ν1 + ν2
The length of β is ν1 eˆ 1 + ν2 eˆ 2  β= = ν1 + ν2
(ν1 eˆ 1 + ν2 eˆ 2 ν1 + ν2
)2
=
ν12 + ν22 + 2ν1 ν2 cos θ ν1 + ν2
.
Concerning the direction of β relative to the direction of motion of photon 1 (say) in Σ, we compute the inner product β · eˆ 1 . We have β · eˆ 1 =
ν1 + ν2 cos θ ν1 + ν2 cos θ . ⇒ cos φ = ν1 + ν2 β(ν1 + ν2 )
(b) In this case we have θ = 90◦ . Hence β=
ν12 + ν22
ν1 + ν2
,
ν1 cos φ = ν12 + ν22
or β=
ν1 eˆ 1 + ν2 eˆ 2 . ν1 + ν2
When ν1 = ν2 the β = 12 (ˆe1 + eˆ 2 ), that is, the velocity of the CM frame is √ along the bisector of the first quarter of the plane x–y with speed c/ 2. (c) In this case eˆ 2 = −ˆe1 = eˆ , where eˆ is the unit in the direction of propagation of the photons. Hence β=
ν1 eˆ − ν2 eˆ ν1 − ν2 = eˆ . ν1 + ν2 ν1 + ν2
7.3. (a) We have p 2 = −M 2 c2 = m 2 c2 (−400 + 64 + 36) = −300 m 2 c2 ⇒ M =
√
300 m
214
7 Relativistic Optics
(b) From the speed u = 0.6c of Σ we compute β = 0.6, γ = 54 . Therefore, the fourmomentum of P in Σ is ⎛ ⎞ ⎞⎛ ⎞ 19 20 00 ⎜ −5 ⎟ ⎜ ⎟ 00 ⎟ ⎜ ⎟ ⎟⎜ 8 ⎟ ⎝ 0 0 1 0 ⎠ ⎝ 6 ⎠ = mc ⎝ 6 ⎠ . 0 Σ 0 0 0 01 ⎛
pi =
5 −3 4 4 ⎜ −3 5 mc ⎜ 4 4
We conclude that the energy of P in Σ is E = 19 mc2 and the direction of its 6 = −1.2. velocity the angle makes with the x axis is tan φΣ = −5 7.4. (a) Obvious.
1 (b) The photons 1, 2 in Σ have fourvelocities = =c . Therefore, in j Σ i i Σ the fourvector A1 − A2 = 0 and that is true in all LCF. Hence the relative velocity of the photons in Σ is zero. A different approach to the same problem is the following. According to (a) because the two photons are propagating in parallel directions in Σ, their fourvectors are parallel. This means that they also propagate in parallel directions in Σ . Because the direction of propagation is normal to the relative velocity of Σ, Σ their relative distance will not be affected by the boost relating Σ, Σ , hence it will be constant in Σ . This implies that their relative velocity in Σ vanishes. Ai1
Ai2
Note: The frequencies of the photons do change and the same applies to the fourmomenta. Suppose the fourmomenta of the photons in Σ are ⎛ ⎞ 1 ⎜0⎟ i ⎟ p1 = hν1 ⎜ ⎝1⎠ , 0 Σ
⎛ ⎞ 1 ⎜0⎟ i ⎟ p2 = hν2 ⎜ ⎝1⎠ . 0 Σ
We compute the fourmomenta in Σ using the boost relating Σ, Σ . We have for photon 1 hν1 = γ (hν1 − β0) = γ hν1 ⇒ ν1 = γ ν1 , hν1 ex = γ (0 − βhν1 ) ⇒ ex = 1/γ , hν1 ey = hν1 ⇒ ey = 1/γ .
7 Relativistic Optics
215
Therefore, ⎛
⎞ 1 ⎜ −β ⎟ ⎟ p1i = hγ ν1 ⎜ ⎝ 1/γ ⎠ . 0 Σ Due to part (a) the photon 2 will be described by a parallel fourmomentum which will have the same form. Therefore, ⎛ ⎞ 1 ⎜ −β ⎟ ⎟ p2i = hγ ν2 ⎜ ⎝ 1/γ ⎠ . 0 Σ 7.5. (a) In his proper frame observer 1 observes the photon to have frequency ν1 and propagate along the spatial direction defined by the unit vector e1 . Then the fourmomentum of the photon in the proper frame of 1 is hν1 P = c
i
1 eˆ 1
Σ+ 1
.
The fourvelocity of 1 in this frame is u 1i = (0, c)Σ+1 . Therefore, the invariant 1 P i u 1i = −hν1 ⇒ ν1 = − P i u 1i . h Similarly for the observer 2 we have 1 ν2 = − P i u 2i . h It follows ν1 P i u 1i = i . ν2 P u 2i In Σ+ 2 the threevelocity of 1 is u, hence the fourvelocity u i1
c = γ (u) . u Σ+ 2
The invariant P i u 1i =
hν2 c
1 eˆ 2
Σ+ 2
·
γ (u)c γ (u)u
Σ+ 2
u = −hν2 γ (u) 1 − cos a . c
216
7 Relativistic Optics
But P i u 1i = −hν1 , therefore (β = uc ) ν2 =
v1 γ (u)(1 − β cos a)
This is the wellknown Doppler formula which relates the frequencies of the photon observed by the observers 1, 2. 7.6. The Doppler relation gives λ v = = λ v
$
1 − βr , 1 + βr
where βr = u/c and u is the speed of the receding galaxy. Solving for βr we find 2 1−(λ /λ) βr = 1+(λ /λ)2 . The ratio λ /λ =
4870 7300
= 0.67 from which follows βr =
0.55 = 0.38. 1.45
5 The distance 5 × 106 light years with speed 0.38c requires a period Δt = 0.38 × 6 6 10 = 13×10 light years (in the Earth’s frame). In the proper frame of the receding galaxy this time is
Δt = γ
−1
Δt =
1
√ 1 − 0.382
−1
× 13 × 106 ly = 12 × 106 ly.
Obviously these numbers are very rough and should be taken lightly! 7.7. The change in the wavelength of the spectral line of Na is due to the phenomenon Doppler. If λ0 is the wavelength of the spectral line on the Earth and λ the wavelength in the spectrum of the galaxy as measured on the Earth (!) then the Doppler formula gives λ0 = λ
$
1−β ⇒ 1+β
λ0 λ
2 =
1 − (λ0 /λ)2 1−β ⇒β= = 0.035. 1+β 1 + (λ0 /λ)2
We infer that the galaxy recedes from the Earth with speed u = 0.035c. 7.8. First Solution The fourmomentum of the radiated radiation and the fourvelocity of the observer Σ in the LCF Σ are, respectively,
7 Relativistic Optics
217
⎛
⎞ 1 ⎟ hν0 ⎜ ⎜ −1 ⎟ , fa = ⎝ 0 ⎠ c 0 Σ
⎛
⎞ γu ⎜ γ u cos θ ⎟ ⎟ ua = ⎜ ⎝ γ u sin θ ⎠ . 0 Σ
Suppose that in the proper frame of Σ the fourmomentum of the radiation is fa =
hν c
1 , e Σ
where ν is the frequency measured by the observer Σ and e the unit along the direction of the radiation in Σ . The invariant f a u a gives in the two LCF hν hν0 (−γ u − γ u cos θ ) = − c ⇒ ν = ν0 γ (1 + β cos θ ). c c
(7.1)
The condition that the frequency measured by observer Σ coincides with that radiated from the source in Σ is γ (1 + β cos θc ) = 1,
(7.2)
from which follows cos θc =
1−γ . βγ
Obviously, this relation is satisfied by the two values ±θc . Second solution In order to compute the frequency and the direction of the radiation for the observer Σ we use the Lorentz transformation relating Σ and Σ . We note that the velocity of Σ wrt Σ is u = u cos θ xˆ + u sin θ yˆ ; therefore, the Lorentz transformation is not a boost. It is easy to show that the Lorentz transformation for the velocity u is given by the following matrix ⎛
γ −γβ cos θ −γ u sin θ 0 γ −1 2 2 2 ⎜ −γβ cos θ 1 + γ −1 u cos θ u cos θ sin θ 0 2 2 u u Lu = ⎜ γ −1 2 2 2 ⎝ −γ u sin θ − γ −1 u cos θ sin θ 1 + u sin θ 0 u2 u2 0 0 0 1 ⎞ ⎛ γ −γβ cos θ −γ u sin θ 0 ⎜ −γβ cos θ 1 + (γ − 1) cos2 θ (γ − 1) cos θ sin θ 0 ⎟ ⎟ =⎜ ⎝ −γ u sin θ −(γ − 1) cos θ sin θ 1 + (γ − 1) sin2 θ 0 ⎠ . 0 0 0 1
⎞ ⎟ ⎟ ⎠
218
7 Relativistic Optics
Applying this transformation to the fourmomentum fourvector in Σ we find ⎛
γ ⎜ −γβ cos θ ⎜ ⎝ −γ u sin θ 0 ⎛
−γβ cos θ 1 + (γ − 1) cos2 θ −(γ − 1) cos θ sin θ 0
−γ u sin θ (γ − 1) cos θ sin θ 1 + (γ − 1) sin2 θ 0 ⎞
⎞ 0 0 ⎟ ⎟ hν0 0 ⎠ c 1
⎛
⎞ 1 ⎜ −1 ⎟ ⎜ ⎟ ⎝ 0 ⎠ 0 Σ
γ + γβ cos θ hν0 ⎜ γβ cos θ − 1 − (γ − 1) cos2 θ ⎟ ⎜ ⎟, = ⎝ −γ u sin θ + (γ − 1) cos θ sin θ ⎠ c 0 that is, ⎛
⎛ ⎞ ⎞ γ + γβ cos θ γ + γβ cos θ ⎟ hν0 ⎜ −γβ cos θ − 1 − (γ − 1) cos2 θ ⎟ hν ⎜ Ax ⎜ ⎜ ⎟= ⎟. ⎠ Ay c ⎝ c ⎝ −γβ sin θ + (γ − 1) cos θ sin θ ⎠ 0 0 The zeroth coordinate gives again relation (7.2) and the spatial components give the vector ⎛
⎞ ⎛ ⎞ Ax −γβ cos θ − 1 − (γ − 1) cos2 θ 1 ⎝ A y ⎠ = ⎝ −γβ sin θ + (γ − 1) cos θ sin θ ⎠ . γ (1 + β cos θ ) 0 Σ 0
(7.3)
The direction of propagation of the radiation for Σ is given by the unit vector ⎛
⎞ Ax ⎝ A y ⎠ . n= 2 2 Ax + A y 0 Σ 1
Replacing from (7.3) we find ⎛ ⎜ ⎜ ⎜ ⎝
−βγ cos θ−(γ −1) cos2 θ−1
⎞
(βγ cos θ+(γ −1) cos2 θ+1)2 +sin2 θ(βγ −γ cos θ+cos θ)2 ⎟ ⎟ ⎟. ⎠ sin θ(βγ −γ cos θ+cos θ) − 2 2 2 2 (βγ cos θ+(γ −1) cos θ+1) +sin θ(βγ −γ cos θ+cos θ)
Note: Instead of using the Lorentz transformation in matrix form one can use the vector form of Lorentz transformation
7 Relativistic Optics
219
A0 = γ (A0 − β · A), γ −1 A = A+ 2 (β · A)β − γ A0 β, β with the appropriate velocity u and recover the same results. 7.9. We compute first the speed of the rocket using the change in the frequency (Doppler shift). The fourmomentum of the emitted photons and the fourvelocity of the rocket in Σ are hν0 1 γc , ui = . fi = 1 Σ γu Σ c In the proper frame Σu of the rocket these fourvectors are fi =
hν c
1 c , ui = , 1 Σ 0 Σ u
u
where ν is the frequency of the emitted radiation. The invariant inner product f i u i when computed in the two frames gives −
hν0 hν c= γ c(−1 + β) ⇒ ν = γ (1 − β)ν0 . c c
Concerning the reflected radiation we assume that Snell’s law applies, that is, the frequency of the reflected radiation in the proper frame Σu of the rocket remains the same while its sense of propagation is reversed.1 We infer that the fourvectors of the reflected radiation and the fourvelocity v i of Σ in Σ, Σu are as follows: h ν¯ f = c i
hν f = c i
1 −1
1 −1
c ,v = , 0 Σ Σ
i
,v = i
Σu
γc −γ u
Σu
,
where ν¯ is the frequency of the reflected radiation in Σ. The invariant f i vi when computed in Σ and Σu gives the equation hν h ν¯ 1−β (−c) = (−γ c + γ u) ⇒ ν¯ = νγ (1 − β) = γ 2 (1 − β)2 ν0 = ν0 . c c 1+β 1 It is irrelevant at which end of the rocket we consider the reflection, because the rocket is considered to be a solid in Σ, therefore the speed of all its points it is the same (in Σ).
220
7 Relativistic Optics
Solving we find the speed of the rocket in Σ: β=
−¯ν + ν0 . ν¯ + ν0
Second Solution We solve the problem using the Lorentz transformation. We have for the various fourvectors Fourmomentum of emitted radiation in Σ hν0 1 i . f = 1 Σ c Lorentz transformation from Σ to Σu : L=
γ −βγ −βγ γ
.
Fourmomentum of emitted radiation in Σu : hν0 hν0 1 γ −βγ 1 i = . fe = γ (1 − β) 1 −βγ γ 1 Σ c c Σ u Fourmomentum of reflected radiation in Σu : f ai
hν0 1 γ (1 − β) = . −1 Σu c
Lorentz transformation from Σu to Σ, γ βγ L= . βγ γ Fourmomentum of reflected radiation in Σ: hν0 γ βγ 1 1 i 2 2 hν0 γ (1 − β) fa = = γ (1 − β) , βγ γ −1 Σu −1 Σ c c from which follows ν¯ = γ 2 (1 − β)2 ν0 . Having computed the speed of the rocket in Σ we use the time difference of the arrival of the reflected signals at the ends of the rocket as measured by Σ in order to estimate the length of the rocket. Let A and B be the events of reflection for the observer Σ. We have
7 Relativistic Optics
221
Radiation which is reflected at the nearer end of the rocket Suppose that the radiation is emitted at the time moment t = 0 of Σ and it is received by the observer in Σ the time moment 2t A (of Σ). Then the coordinates of the event A for Σ are ct A (1, 1)Σ . Radiation which is reflected at the further end of the rocket Working similarly we find the coordinates for the event B in Σ are ct B (1, 1)Σ . We conclude that in Σ the fourvector AB i = c(t B − t A )(1, 1)Σ . In the proper frame of the rocket the components of the fourvector AB i are computed using the Lorentz transformation relating Σ to Σu . We find AB i = c(t B − t A )
1 1 = c(t B − t A )γ (1 − β) . 1 Σ 1 Σ
γ −βγ −βγ γ
u
The time difference of the reflection of the radiation at the two ends of the rocket is ΔT AB (Σu ) = (t B − t A )γ (1 − β). But we also have ΔT AB =
L . c
Equating the last two expressions we compute $
1−β L = c(t B − t A )γ (1 − β) = c(t B − t A ) = cΔT 1+β

ν¯ . ν0
7.10. (a) Suppose we have a photon which in the proper frame Σu of the observer u i has components hν f = c i
1 . e Σ u
The fourvelocity in Σu is ui =
c . 0 Σu
Therefore, the invariant f i u i = −hν ⇒ ν = −
1 i f ui . h
(7.4)
222
7 Relativistic Optics
(b) Applying this result we find z+1=
νE f i u iE = . νR f i u iR
(7.5)
We have to compute the invariant quantities f i u iE , f i u iR . Because they are invariant they can be computed in any LCF we wish and we choose the laboratory frame, Σ say. In Σ the velocities of the emitter u and the receiver v are normal to the radius and have common speed ωr . This implies u = v = ωr, γu = γv = γ . Next we note that when the emitter emits a photon, that photon reaches the receiver after time period Δt (in Σ). The chord which connects the position of the emitter at the moment of emission and the receiver at the moment of reception (all time moments in Σ!) subtracts the angle a + ωt. The length of the . Therefore, we have chord (in Σ!) is R E = cΔt and also R E = 2r sin a+ωΔt 2 the equation (essentially this is the rigidity condition) sin
cΔt a + ωΔt = . 2 2r
(7.6)
We choose the xaxis to be defined by the position of the emitter the moment (of Σ!) the photon is emitted and the yaxis normal to that axis. Then the velocity of the emitter and the receiver are (see Fig. 7.1)
Fig. 7.1 Redshift of rotating observers
7 Relativistic Optics
223
u E = ωr j, u R = ωr (− cos φ R i + sin φ R j), where φ R = π2 − (a + ωt). This means that the fourvelocity of the emitter and the receiver are ⎛
⎞ γc ⎜ 0 ⎟ γc ⎟ = where e E = j, u iE = ⎜ ⎝ γ ωr ⎠ γ ωr e E 0 Σ ⎞ ⎛ γc ⎜ −ωr cos φ R ⎟ γc ⎟ = where e R = − sin(a + ωΔt)i+ cos(a + ωΔt)j. u iR = ⎜ ⎝ ωr sin φ R ⎠ γ ωr e R 0 Σ Suppose that the fourmomentum of the photon in Σ is as follows fi =
hν0 c
1 , e Σ
where π − a − ωΔt π − a − ωΔt ER = − cos i + sin j ER 2 2 a + ωΔt a + ωΔt = − sin i + cos j. 2 2
e=
Then we have 1 ν0 1 γc f i u iE = − · = γ ν0 (1 − e · e E ), γ eE , Σ h c e Σ 1 ν0 1 γc ν R = − f i u iR = − · = γ ν0 (1 − e · e R ). γ eR Σ h c e Σ
νE = −
Replacing we find z=
1 − e · eE νE −1= − 1. νR 1 − e · eR
(7.7)
224
7 Relativistic Optics
We compute a + ωΔt , 2 a + ωΔt a + ωΔt e · eR = sin sin(a + ωΔt) + cos cos(a + ωΔt) 2 2 a + ωΔt . = cos 2
e · e E = cos
Replacing in (7.7) we find z = 0. that is, the redshift equals zero. 7.11. Suppose that in Σ the fourmomentum of the photon is pi =
E hν (1, eˆ )Σ = (1, eˆ )Σ . c c
Then f i = ν(1, eˆ )Σ . We note that f i is a null vector. In terms of the wave vector k = ν eˆ of the photon we have f i = (k, k)Σ where k = ν is the length of k in Σ. We note that the wave vector k describes completely the photon in Σ. We consider now a mass particle which in Σ has fourmomentum pi = ( Ec , p)Σ and define the frequency fourvector f i of the particle by the same relation pi = h i f . Associating a wave vector with the particle by the relation f i = (ν, k)Σ we c find E h E = hk ,p (de Broglie relations). = (k, k)Σ ⇒ p = hc k c c Σ These relations define for a particle a wave with wave vector k and are the foundation stone of the wave theory of matter. We note that they follow in the most natural way from the relativistic theory. In general it is assumed that a continuous family of “waves” of wave vector k is associated with a particle so that the “frequency” of the particle is ω = ω(k). The motion of the particle is described by the superposition of these “waves.” The threevelocity of the particle in Σ is given by v=
dE dk c dω = = . dp dk 2π dk
This relation is known as the dispersion formula.
7 Relativistic Optics
225
7.12. (a) From the definition of the angle θ we have β = sin θ ⇒ γ = sec θ,
βγ = tan θ.
Hence the boost along the xaxis with velocity β is written as x = sec θ x − tan θ ct, ct = sec ct − tan θ x, y = y, z = z. (b) The components of the fourvelocity of a particle which moves along the xaxis with speed β = sin θ are ui =
γc γu
Σ
=
c sec θ c tan θ
Σ
.
From the trigonometric identity sec2 θ − tan2 θ = 1 we find u i u i = c2 (tan2 θ − sec2 θ ) = −c2 . For the fourmomentum of the particle we have p = mu = i
i
E/c p
Σ
=
mc sec θ mc tan θ
Σ
⇒ E = mc2 sec θ,
p = mc tan θ.
Using again the identity sec2 θ − tan2 θ = 1 we find E 2 − p 2 c2 = m 2 c4 . (c) The frequency associated with the particle in Σ is ν=
mγ c2 mc2 E = = sec θ ⇒ ν = ν0 sec θ, h h h
2
where ν0 = mch is the de Broglie frequency of the particle (i.e., the frequency in the proper frame of the particle). Similarly for the wavelength we compute λ = λ0 cot θ, where λ0 =
h mc2
is the de Broglie wavelength of the particle. The phase velocity vph = λν =
c c, sin θ
that is, the phase velocity of de Broglie waves is higher than c and equals c for the photons only. The group velocity
226
7 Relativistic Optics
vgr =
dν/dθ dν = = c sin θ = cβ = v d(1/λ) d(1/λ)/dθ
equals with the velocity of the particle in Σ. We note that v ph vgr =
c c sin θ = c2 . sin θ
It follows that for photons v ph = vgr = c. Making use of the relations ν = ν0 sec θ, k = k0 tan θ where k = 1/λ, k0 = 1/λ0 and replacing in the trigonometric identity sec2 θ − tan2 θ = 1 we find
ν ν0
2
−
k k0
2 = 1 ⇒ ν 2 = c2 (k02 + k 2 ),
which is the relativistic dispersion formula of de Broglie waves. 7.13. (a) In an arbitrary LCF Σ consider the fourquantities (not necessarily fourvectors) (a0 , a1 , a2 , a3 ) and (b0 , b1 , b2 , b3 ) which are such that ai = Abi , i = 0, 1, 2, 3. Let Σ be another LCF which is related to Σ with the Lorentz transformation L ii . Under the action of the transformation L ii the quantities (a0 , a1 , a2 , a3 ), (b0 , b1 , b2 , b3 ) transform to (a0 , a1 , a2 , a3 ), (b0 , b1 , b2 , b3 ), respectively, according to the relation (because they are tensorial) j
ai = L i a j
j
bi = L i b j .
But the quotient abii i = 0, 1, 2, 3 is invariant; therefore, its value is independent of the LCF considered, hence in Σ holds ai = Abi , i = 0, 1, 2, 3. This implies the relation (a0 , a1 , a2 , a3 ) = A(b0 , b1 , b2 , b3 ). This relation is covariant hence the fourquantities ai , bi are fourvectors which are parallel to one another. The inverse is obvious. (b) Application The particle nature of a particle in the LCF Σ is characterized by the fourmomentum of the particle Σ, i.e., pi = (E/c, p)Σ . Concerning the wave character of a particle we know that a relativistic wave in an LCF Σ is characterized by the following quantities  The amplitude (a)  The frequency (ν)  The phase velocity (w) or the wave vector (k = w/w).
7 Relativistic Optics
227
These quantities are related with the equation F(x i ) = a(x i )P(x i f i ), k)Σ is the frequency fourvector of the wave, x i is the point where f i = (ν, νc w in spacetime where we study the wave and the function P(x i f i ) is the phase of the wave. The amplitude a(x i ) is a Lorentz tensor field (it is not necessarily an invariant) defined by the nature (i.e., scalar, vector) of the wave. The dual description of the nature of matter is described by the definition of a correspondence between the components of the fourmomentum and the frequency fourvector. The Planck relation E = hν defines a correspondence between the zeroth coordinate of the fourvectors. We extend this correspondence by demanding that the two fourvectors are parallel, that is, we demand pi =
h i f . c
This requirement induces the following relation between the spatial parts of the fourvectors p=
h νc k. c w
Due to (a) one finds the same result by demanding that the threemomentum of the particle in Σ is parallel to the wave vector k, the constant of proportionality being appropriately chosen, and then conclude the Planck relation E = hν. We note that in Special Relativity the de Broglie relations are one relation (the pi = hc f i ) while in Newtonian Physics they are two independent relations (E = hν and p = hc νc k). Finally, we note that these relations concern the phase w and not the amplitude of the wave. This is important because it allows us to describe de Broglie waves in nonhomogeneous and anisotropic media.
Chapter 8
FourForce
8.1. Since the two mass points are touching they have common threevelocity, say u, in Σ, hence the same fourvelocity u i . The fourforce on each mass point is F1i = γu
1
f · c 12 f12
u
Σ
, F2i = γu
1
f · c 21 f21
u
Σ
.
The threeforces of action and reaction do not produce work because their application point does not move. Since the masses of the two particles do not change we infer that the total energy of the system of particles is constant (closed system). This statement is equivalent to the equation (F1i + F2i )u i = 0 (recall that the zeroth component of the fourforce measures the power produced by the threeforce). This equation is covariant, i.e., independent of the particular LCF Σ and furthermore the fourvelocity u i is arbitrary (can be any). We conclude that the zeroth component of the fourvector F1i + F2i vanishes in all LCF. From the theorem of the vanishing of the zeroth component we infer that the fourvector F1i + F2i = 0. This implies in Σ 1 c
(f12 + f21 ) · u (f12 + f21 )
Σ
= 0 ⇒ f12 + f21 = 0.
[Note: The proof is valid only if the particles are in contact. If they are not then one has to consider the fact that the interactions in Special Relativity are transmitted with finite speed (maximum c) whereas in Newtonian Physics the interactions are done with infinite speed (action at a distance). In such cases one has to consider the field of interaction between the two particles and there is no meaning in considering the action–reaction concept.] 229
230
8 FourForce
8.2. In the LCF Σ the threeforce defines the fourforce1 mγ 4 β · a i F = . mγ 4 (β · a)β + γ 2 mc dβ dt Σ From the relativistic second law we have dpi dpi F = =γ =γ dτ dt i
1 dE c dt dp dt
.
Σ
Equating the two expressions we find the equations of motion in Σ mc2 3 dβ 2 dE = mcγ 3 β · a = γ , dt 2 dt dp = mγ 3 (β · a)β + mγ ca. dt
(a) When the threevelocity of the particle is collinear with the threeforce the direction of motion of the particle is constant and only the speed changes. If we write v = vˆe where eˆ is the unit along the direction of motion the threeacceleration a = dv eˆ . Replacing in the equations of motion we find dt mc2 3 dβ 2 dE = γ , dt 2 dt dp dβ dβ dβ dβ = mcγ 3 β 2 + mγ c eˆ = mγ c (γ 2 β 2 + 1)ˆe = mγ 3 c eˆ . dt dt dt dt dt Obviously the normal components p⊥ of the threemomentum remains constant. (b) When the threevelocity is normal to the threeforce the speed is constant, hence 2 2 dv = 0 and β · a = cβ · dβ = c2 dβ = 0. Therefore, in this case the equations dt dt dt of motion become dE = 0 ⇒ E = constant, dt dp dβ = mγ c = mγ a. dt dt
1
Proof F i = ma i and a i =
expression for F i .
ca0 ca0 β + γ 2 a
where a0 = γ γ˙ = 1c γ 4 β · a. Replacing we find the given
8 FourForce
231
(c) When v c we approximate γ ≈ 1 and β · a = cβ · equations of motion become
=
dβ dt
c2 dβ 2 2 dt
≈ 0 and the
dE = 0 ⇒ E = constant, dt dv dp =m = ma. dt dt But E = mγ c2 ≈ mc2 ; therefore, the first equation is trivial (m = constant) and the second reduces to the Newtonian Second Law.
8.3.
d dt
*
+
mv
d 2c2 =− 2 m⇒ x dx
*
+
v
dx 2c2 =− 2 ⇒ dt x
1 − v 2 /c2 1 − v 2 /c2 (v (x dx 2 2 −1/2 2 = −2c vd v 1 − v /c ⇒ x2 0
2
&x & 2 2 −1/2 2 1& v 1 − v /c dv = 2c ⇒ 0 x &2 0 &v 1 1 c2 v2 & 2 1 2 2 − ⇒ + 1 − v /c & = 2c 0 2 −1/2 + 1 x 2 1 − v 2 /c2 (x (t 2 x d x dx x2 2 2 v =c 1− ⇒ =c 1− ⇒ = cdt ⇒ 2 4 dt 4 1− x −1/2 &&v v 1 − v 2 /c2 & −
(v
2
2
4
π ct x &&x x ⇒ 2 arcsin & = ct ⇒ arcsin − = 2 2 2 2 2 c x(t) = 2 cos t. 2
0
We see that the motion is simple harmonic motion with angular frequency ω = 2c , . hence period T = 4π c
232
8 FourForce
8.4. d dt
*
dx d dt d x (v ωa
*
+
mv 1 − v 2 /c2 v
=− +
1 − v 2 /c2
mc3 ω2 x ⇒ (c2 − ω2 a 2 + ω2 x 2 )3/2
=−
(c2
mc3 ω2 x ⇒ − ω2 a 2 + ω2 x 2 )3/2
(x c3 d(c2 − ω2 a 2 + ω2 x 2 ) 2 2 −1/2 =− vd v 1 − v /c ⇒ 2 (c2 − ω2 a 2 + ω2 x 2 )3/2 0
1/2 1/2 = + c2 1 − v 2 /c2 − 1 − ω2 a 2 /c2
v ω a 1/2 − 1/2 2 2 1 − v /c 1 − v 2 /c2 −1/2 2 −1/2 1 c3 c2 − ω2 a 2 + ω2 x 2 ⇒ =− − c − ω2 a 2 2 −3/2 + 1 2 c2 1 − vc2 = c2 − ω2 a 2 + ω2 x 2 ⇒ 2
2 2
v 2 = ω2 a 2 − x 2 . This relation implies that the motion of the particle is bounded because v 2 ≥ 0 ⇒ (a 2 − x 2 ) ≥ 0 ⇒ −a ≤ x ≤ a. 8.5. In an obvious notation we have d dt
ud (u ud 0
u2 1 − u 2 /c2
c2 1 − u 2 /c2
=
+ c2
mu 1 − u 2 /c2 u 1 − u 2 /c2 u 1 − u 2 /c2
! = −mω2 x ⇒ ! = −ω2 xd x ⇒ !
(0 = −ω
xd x ⇒
2 a
ω2 a 2 ⇒ 1 − u 2 /c2 − 1 = 2
2c2 + ω2 a 2 4c4 ⇒ 1 − u 2 /c2 = 2 ⇒ 2 2c2 + ω2 a 2 √ ωac 4c2 + ω2 a 2 . u= 2c2 + ω2 a 2
8 FourForce
233
8.6. The equation of motion of the particle in Σ is ⎞
⎛ d ⎝ mv dt 1−
⎠ = − mv ⇒ 2 k v c2
(v2 v1
⎛ v 1 ⎝ d v 1−
⎞ ⎠ = −1 2 k v c2
(t dt ⇒ 0
&v2 & (v2 & 1 v 1 1 & − v − t= d ⇒ k v 1 − v2 && v v2 1 − v1 c2 v1 c2 &v2 v & (2 & 1 dv 1 & + . − t= & k v2 & v2 1 − c2 v 1 − v1 c2 v1
The integral (
v 1−
√ √ c + c2 − v 2 c + c2 − v 2 1 ln = − ln . =c − √ c v v v c2 − v 2
(
dv
v2 c2
dv
=c
Hence − k1 t
=
− k1 t =
1
1−( 35 ) 5 4
−
5 3
2
−
− ln
1
1−( 45 ) 1+ 45 3 5
2
− ln
+ ln
1+ 35 4 5
2 c+ c2 −( 35 ) c2
2 c+ c2 −( 45 ) c2
+ ln 4 5c "5 3 # ⇒ t = k 12 + ln 2 . 3 5c
⇒
8.7. Let Σ be the LCF of the observer on the Earth and Σ the LCF of the observer who moves with velocity u = ui. The acceleration of the projectile in Σ is a = 2u (0, −g, 0)Σ and the flight time T = a yy . The boost relating Σ, Σ gives 1/2 1 − u 2 /c2 = 3 ax = 0 1 − uvx /c2 uv /c2 a y + 1−uvy x /c2 ax −g a y = 2 = − 2 β 2 1 − uvx /c2 β 2 1 − uvx /c2 uvz /c2 az + 1−uv a 2 x x /c az = 2 = 0. β 2 1 − uvx /c2
ax
Also vx =
vx − u , 1 − uvx /c2
v y =
vy , β 1 − uvx /c2
vz = 0.
234
8 FourForce
Therefore
T =
2v y
=
a y
2β
vy
(1−uvx /c2 ) ay 2
β 2 (1−uvx /c2 )
uvx T. ⇒ T = β 1 − c2
We note that T < T , that is, we have reduction of the flight time in Σ . Concerning the range we have R = vx T . Hence R = vx T =
vx − u β 1 − uvx /c2 T = β(vx − u)T ⇒ 2 1 − uvx /c R = β(R − uT ).
[Note: The above results are possible to be derived differently as follows. At the moment t = t = 0 the LCF Σ, Σ coincide and at the moment T the observer in Σ observes the event (R, 0, 0, T )Σ which the observer in Σ observes at the time moment T and attributes the coordinates (R , 0, 0, T )Σ . The boost relating Σ, Σ gives R = β R − βuT = β(R − uT ),
uvx T = β 1 − uvx /c2 T ]. T = βT − β Rx u/c2 = βT − β c2
(8.1) (8.2)
Concerning the maximal height we have the event (0, H, 0, T2 )Σ which gives H = H , that is, the maximal height is the same. Indeed
H =
v 2y
(v 2y )
=
2(a y )
β2
2
(1−uvx /c2 )2 ay 2
β 2 (1−uvx /c2 )
=
vy = H. 2a y
For the angle of firing we have R 4 cot θ = H
+ 2vx v y /a y vx = 4 = 4 cot θ , = 2 v y /2a y vy
*
hence R T R β(R − uT ) = β( − u ) = H* H H +H 2v y /a y = β 4 cot θ − u 2 = β 4 cot θ − 4u/v y ⇒ v y /2a y 4 cot θ = 4β cot θ − u/v y . 4 cot θ =
8 FourForce
235 i
8.8. (a) The fourforce is defined by the relation F i = dp where pi is the fourdτ momentum and τ is the proper time of P. In Σ we have pi =
mγ (v)c mγ (v)p
Σ
⇒ Fi =
dt dpi f · v/c = γ (v) . f dτ dt Σ
Concerning the physical significance of the zeroth component we have 1 1 dp c d E2 dE f·v= ·p= = , c mγ c dt 2E dt dt where E is the total energy of P in Σ. It follows that f · v/c measures the work done on the ReMaP by the threeforce f in Σ. (b) m d(−c2 ) dV i m d(V i Vi ) Vi = = = 0. dτ 2 dτ 2 dτ f · v /c i (c) Suppose that in Σ the fourforce is F = γ (v ) . The boost relatf Σ ing Σ, Σ gives F i Vi = m
γ (v )f ·v = γ (u)γ (v)(f · v − u f x ), u γ (v ) f x = γ (u)γ (v)( f x − 2 f · v), c γ (v ) f y = γ (u)γ (v) f y , γ (v ) f z = γ (u)γ (v) f z . But we know from the transformation of the zeroth component of the fourvelocity that the γ factors are transformed as follows γ (v ) = γ (u)γ (v)(1 −
u·v ). c2
Replacing in the transformation equations, we find after a standard calculation 1 u ( f x − 2 f · v), (1 − u·v ) c c2 1 fy, f y = γ (u)(1 − u·v ) c2 1 f z = fz . γ (u)(1 − u·v ) c2 f x =
236
8 FourForce
(d) Without restricting generality (why?) we consider that in Σ the central motion takes place in the plane y–z. Then the threeforce in Σ is f = (0, f y , f z ) and vx = 0. From the transformation equations we compute the threeforce in the LCF Σ (note that u · v = 0) u f · v = 0, c2 f y = f y /γ (u)= 0,
f x = −
f z = f z /γ (u)= 0. We see that in Σ the threeforce is not a central force. This result is important due to the crucial role of central motions in physics. 8.9. Questions (a) and (b) are obvious. (c) We consider a ReMaP which in Σ, Σ has velocity v, v and suffers the force f, f , respectively. The decomposition of the fourforce in Σ, Σ is as follows Fi =
1
γ f· c v γv f
v
Σ
,
1
γ f · c v γv f
v
Σ
Lorentz transformation for the time component gives 1 γv f · u 1 γv f · v = γu γv f · v− ⇒ c c c 1 f · v = (v − u) · f Q
(8.3)
This relation means that the work done by f in Σ is proportional to the relative velocity of the ReMaP in Σ. Concerning the spatial part Lorentz transformation gives u · γv f 1 γv f = γv f + (γ u − 1) 2 u − 2 γu [γv f · v]u u c u u · γv f v = γv f − u+γu − 2 ·f u u2 u2 c 8 u 9 v − 2 ·f u , = γv f⊥ +γu u2 c where f⊥ = f − the quantity
u·γv f u u2
is the normal projection of f on the velocity u. We define
R=
u v ·f − u2 c2
8 FourForce
237
and have finally γv γu f = γv
But γv = γu γv Q where Q = 1 −
u·v . c2
1 f = Q
1 f⊥ +Ru . γu
Therefore 1 f⊥ +Ru . γu
(8.4)
This relation decomposes f parallel and normal to the velocity u. This relation can be written differently. We have f =
f ⊥ γu Q
1 u e + Re ⊥ , γu2 f ⊥ γu
(8.5)
where f⊥ = f ⊥ e⊥ . The R gives (use (8.3) for v): β v·u u×(v × u) ·f u+ u R = f − c u2 u2 v · u 1 u×(v × u) = 1 − 2 f − · f. c u c2 Replacing we find the final expression 1 u×(v × u) f⊥ + f − · f e . f = γu Q Quc2
(8.6)
In order to show that the spatial part of all fourforces can be written as a Lorentz force for some charge, electric field, and magnetic field provided f ⊥ = 0 we work as follows. We introduce the vectors h = u × e⊥ , V = 2 u + fc⊥ γRu e⊥ and calculate R c2 R e⊥ × (u × e⊥ ) = c2 −β 2 e⊥ + u . V×h= u+ f ⊥ γu f ⊥ γu But from (8.5) we have that f ⊥ γu R 2 f = (1 − β )e⊥ + u . Q f ⊥ γu
Comparing the last two expressions we conclude that the force f is written as follows
238
8 FourForce
f =
f ⊥ γu 1 e⊥ + 2 V × h . Q c
(8.7)
This expression is similar to the Lorentz force2 with the following characteristics
r r r
The electric field is normal to the relative velocity u of Σ, Σ . Its physical dimensions are L 0 M 0 T 0 , therefore, it is a pure kinematic quantity not related to any dynamic field. The “magnetic field” is normal to both the electric field and the relative velocity u of Σ, Σ and it is also a geometric quantity depending on the velocities u, v not related to a dynamic field. The “charge” depends on the normal projection of the threeforce on the relative velocity and varies with the velocity v of the ReMaP in Σ.
8.10. The threeacceleration of P in Σ is a = a⊥ = −
u2 eˆ r , R
where eˆ r is the unit along the radial direction. Because of P in Σ is 0 ai = . γ 2a Σ
dγ dt
= 0 the fouracceleration
+ In the proper frame Σ+ of P the acceleration is the proper acceleration a and the 0 fouracceleration is a i = . The Lorentz transformation gives a+ Σ+
a+ = γ 2 a = −γ 2
u2 eˆ r ⇒ R
F+ = ma+ = −γ 2
mu 2 eˆ r . R
Second Solution The equation of motion of P in Σ is F=
dp u2 = mγ a = −mγ eˆ r . dt R
Hence the force in the proper frame of P is F+ = γ F = −γ 2
mu 2 eˆ r . R
2 A special case of this general result can be found in D. Bedford and P. Krumm (1985), Am. J. Phys. 53, 889.
8 FourForce
239
8.11. See theory. 8.12. We have S=
(t2 ,Σ (2 L 0,Σ dtΣ = −αcγΣ dτ . t1 ,Σ
1
Because dτ is invariant the quantity −αcγΣ (u) must be invariant, hence α=−
k γΣ (u)
where k is a constant depending only on the particular particle. Then L 0,Σ = −kc 1 − u 2 /c2 . But L 0,Σ must have the dimensions of action, hence the constant k has dimensions (mass)× (velocity). The only quantity which depends only on the particle is its mass m. As velocity we consider c which is independent of the particle. Therefore, we have finally L 0,Σ = − which is the required result.
m 2 c , γ
Chapter 9
Plane Waves
" # 9.1. The exponent (phase) 2πiν (x cos α + y sin α − ct) of the wave is an invariant c quantity (exponentials and logs are always invariant quantities), hence
2πiν 2πiν x cos α + y sin α − ct = (x cos α + y sin α − ct) . c c
The boost relating Σ, Σ gives x = γ (x − ut),
y = y,
t = γ (t − ux).
Replacing we find ux ν γ (x − ut) cos α + y sin α − cγ (t − 2 ) = ν (x cos α + y sin α − ct) ⇒ c " # u ν γ cos α + x +ν sin α y − ν γ (u cos α + c) t = νx cos α +ν sin α −νct. c This relation must hold for all x, y, t, therefore u ν γ cos α + = ν cos α, c ν sin α = ν sin α, ν γ (u cos α + c) = νc. Dividing the first and the third we find cos α =
cos α − u/c . 1 − u cos α/c
Replacing in the first we get ν = γ (1 − β cos α)ν.
241
Chapter 10
Relativistic Reactions
10.1 Center of Momentum Energy 10.1.1. (a) Let E ∗ be the (common) energy of the beams in the CM in both experiments. Then E ∗2 = −( p1i + p2i )2 = 2m 2 c2 − 2 p1i p2i . c2 The difference in the two experiments is in the term 2 p1i p2i . In the first experiment we have in the laboratory frame 2 p1i p2i
=2
E/c p
· L
m pc 0
= −2Em p , L
hence E 1∗2 = 2m p c2 (E + m p c2 ). Replacing the given data we compute E 1∗ ≈ 7.6 GeV. In the second experiment we calculate the product 2 p1i p2i in the CM frame. We have 2 p1i p2i
=2
E /c p
· L
E /c −p
*
L
E 2 = 2 − 2 − p2 c
+ = −2
2E 2 + 2m 2p c2 . c2
From the given data we find that the available energy is E 2∗ = 2E = 30 GeV. 243
244
10 Relativistic Reactions E∗
We note that E1∗ = 0.25, that is, a difference of 25% between the available 2 energies in the two experiments. (b) In order to have E 1∗ = E 2∗ = 30 GeV we compute p1i p2i in the two cases and have m p c2 E = 2E 2 − m 2P c4 ⇒ E =
2E 2 − m p c2 . m p c2
Replacing we find E ≈ 478 GeV.
i /c E (α) , α = 1, 2, . . . , n the fourmomenta of the particles p(α) Σ of the system in the LCF Σ. The fourmomentum of the CM particle is
i 10.1.2. Let p(α) =
P = i
)
i p(α)
α
, i ∗ E (α) /c E /c α , = = . 0 α p(α) CM Σ
The Lorentz transformation relating Σ and the CM gives )
i E (α) = ΓE ∗ ,
α
where Γ is the velocity factor of the CM in Σ. But ) α
i E (α) =
) α
i T(α) +
)
m (α) c2 = TΣ + Mc2 ,
α
where TΣ is the kinetic energy of the particles in Σ. Replacing in the transformation formula we find TΣ = ΓE ∗ − Mc2 . The energy of the system of particles in the CM frame is E ∗ = T ∗ + Mc2 , where T ∗ is the kinetic energy of the system of particles in the CM. We replace in the last relation and obtain the “Theorem of Conservation of Kinetic Energy” in Special Relativity TΣ = ΓT ∗ + (Γ − 1)Mc2 .
10.1
Center of Momentum Energy
245
For small speeds Γ = 1 + 12 B 2 + O(B 4 ) and TΣ = (1 +
1 2 ∗ 1 2 1 B )T + V M = T ∗ + M V 2 + O(B 2 ), 2 2 2
which coincides with the corresponding Newtonian expression. Note: The velocity V of the CM frame in Σ is defined by the quotient V =
, cp ,α i(α) . α E (α)
Hence small speed of the CM means small momentum for large energy, that is, “heavy” particles which move “slowly”. 10.1.3. We consider the collision as the reaction 1 + 2 → 3 + 4. Conservation of fourmomentum gives ( p1i + p2i )2 = ( p3i + p4i )2 . We compute the LHS in the laboratory frame. We have p1i
=
E 1L /c p1L
,
p2i
=
L
mc 0
⇒
( p1i
+
p2i )2
=
(p1L )2
−
L
E 1L + mc c
The RHS we compute in the CM frame ( p3i
+
p4i )2
=−
CM E total c
2 .
Replacing in the conservation equation we find 2 C M 2 E + mc = − total ⇒ c L 2 C M 2 E E (p1L )2 − c1 − (mc)2 − 2E 1L m = − total . c (p1L )2 −
But (p1L )2 −
E 1L c
2
E 1L c
= −m 2 c2 , hence
C M 2 E total − 2mc = − ⇒ c $ √ mc2 CM E total = c 2 E 1L m + m 2 c2 = c 2m E 1L 1 + L . E1
−2E 1L m
2
2 .
246
10 Relativistic Reactions
If mc2 E 1L ⇒
mc2 E 1L
0 then, CM = c 2m E 1L . E total
10.2 Elastic Collisions 10.2.1. Let v = vx be the velocity of the CM of the spheres in the laboratory. The CM is related to the laboratory frame L (say) with a boost with speed v. For sphere 2 we have the following fourvelocities before and after the collision in L and CM, respectively (we work only with fourvelocities because the masses are equal) (Fig. 10.1) Before ⎛ ⎞ c ⎜0⎟ ⎜ ⎟ ⎝0⎠ 0 L ⎛
⎞ γ (u ∗ )c ⎜ −γ (u ∗ )c ⎟ ⎜ ⎟ ⎝ ⎠ 0 0 CM
After γ (u 2 )c ⎜ γ (u )u cos θ 2 2 ⎜ ⎝ γ (u 2 )u 2 sin θ 0 ⎛
⎞ ⎟ ⎟ ⎠ L
⎞ γ (u ∗ )c ⎜ −γ (u ∗ )u ∗ cos θ ∗ ⎟ ⎟ ⎜ ⎝ −γ (u ∗ )u ∗ sin θ ∗ ⎠ . 0 CM ⎛
Fig. 10.1 Collision of spheres in the CM and the laboratory frame
The boost relating L , CM gives γ (u ∗ ) = γ (v) ⇒ v = u ∗ .
10.2
Elastic Collisions
247
After the collision (taking into account the above result) we have " # γ (u ∗ ) = γ (v) γ (u 2 ) − vc γ (u 2 ) cos θ " # −γ (u ∗ )u ∗ cos θ ∗ = γ (v) γ (u 2 ) cos θ − vc γ (u 2 )c
−γ (u ∗ )u ∗ sin θ ∗ = −γ (v)γ (u 2 ) sin θ. The first equation gives
∗
γ (u ) =
γ (v)γ (u 2 )
v cos θ 1− c
and the other two tan θ =
sin θ ∗ 1 . γ (v) cos θ ∗ − 1
Due to symmetry the result holds for sphere 1 if we replace θ ∗ with π − θ ∗ . Therefore, we have tan φ =
sin(π − θ ∗ ) 1 − sin θ ∗ 1 = . γ (v) cos(π − θ ∗ ) − 1 γ (v) 1 + cos θ ∗
Hence tan θ tan φ =
1 − sin2 θ ∗ 1 = 2 . 2 2 ∗ γ (v) cos θ − 1 γ (v)
We note that tan θ tan φ < 1, that is, the recoil directions of the spheres have sum < π2 (why?). In the Newtonian approximation γ (v) = 1 and θ + φ = π2 always, that is, in Newtonian Physics the spheres recoil always at normal directions. In order to calculate the product tan θ tan φ in terms of u we apply the boost to the zeroth coordinate of the fourvelocity of sphere 1 before collision and take into account that the speeds of the two spheres in the CM frame are equal (because the masses are equal), say v. Then v2 v = 2γ 2 (v) − 1, γ (u)c = γ (v) γ (v)c + γ (v)v = γ 2 (v) 1 + c c hence γ 2 (v) =
1 (1 + γ (u)) . 2
248
10 Relativistic Reactions
Replacing we find 2 . 1 + γ (u)
tan θ tan φ = 10.2.2. We consider the reaction
π + p→π + p . 1 2 3 4 Conservation of fourmomentum gives p4i = p1i + p2i − p3i . Squaring −m 2p c2 = −2m 2π c2 − m 2p c2 + 2 p1 p2 − 2 p1 p3 − 2 p2 p3 . The fourmomenta in the laboratory frame are (we consider the x, y components only) ⎛
⎞ ⎛ ⎛ ⎞ ⎞ E 1 /c m pc E 3 /c E 4 /c p1i = ⎝ p1 ⎠ , p2i = ⎝ 0 ⎠ , p3i = ⎝ 0 ⎠ , p4i = . p4 Σ 0 0 p 3 Σ Σ Σ We compute the inner products: p1 p2 = −E 1 m p ,
p1 p3 = −
E1 E3 , c2
p2 p3 = −E 3 m p .
Replacing in the conservation equation it follows easily that E1 =
E 3 m p − m 2π c2 mp −
= 4.213 GeV/c2 .
E3 c2
The energy E3 =
p32 c2 + m 2π c4 =
0.82 + 0.142 GeV/c2 = 0.812 GeV/c2 .
To compute the limiting values of the energy of the scattered π meson (particle 3) we solve the last relation in terms of E 3 . We find E3 =
m 2π E 1 /c2 mp E 1 /c2
mp + 1+
c2 .
10.2
Elastic Collisions
249
For high speeds of the falling beam E 1 /c2 m p , m π ⇒ E 3 m p c2 . For low values of the falling beam E 1 /c2 m π , hence E3
m 2π mπ mp mπ
mp + 1+
c2 = m π c2 ,
that is, the π mesons interchange their energies. 10.2.3. The collision is described in the following figure
Conservation of fourmomentum gives pei = pei + pγi − pγi . Squaring −m 2e = −m 2e + 2 pei pγ i − 2 pei pγ i − 2 pγi pγ i ⇒ pei pγ i − pei pγ i − pγi pγ i = 0. In the laboratory frame we have for the fourmomenta pei =
Ee pe eˆ
, pγi =
Eγ −E γ eˆ
, pγi =
E γ E γ eˆ
where eˆ · eˆ = cos θ . We compute the inner products pei pγ i = −E e E γ − pe E γ , pei pγ i = −E e E γ + pe E γ cos θ, pγi pγ i = −E γ E γ − E γ E γ cos θ. Replacing in the conservation equation we get E γ =
E γ (E e + pe ) , (E γ + E e ) − (E γ − pe ) cos θ
which is the answer because E e =
pe2 + m 2e .
,
250
10 Relativistic Reactions
When the electron is at rest in the laboratory, pe = 0 and E e = m e . Therefore E γ =
Eγ m e = (E γ + m e ) − E γ cos θ 1+
Eγ Eγ (1 me
− cos θ )
,
which is the case of the classical Compton scattering. 10.2.4. In the laboratory the scattering is represented in Fig. 10.2. The fourmomenta of the particles involved are Fourmomentum of bullet photon: p1 = hνc 1 , hνc 1 eˆ L Fourmomentum of particle before scattering: 0) L p2 = (mc, Fourmomentum of scattered photon: p3 = hνc 2 , hνc 2 eˆ L Fourmomentum of particle after scattering: p4 = (E/c, p) L .
Fig. 10.2 Compton scattering
Conservation of fourmomentum gives p4i = p1i + p2i − p3i ⇒ p42 = p12 + p22 + p32 + 2 p1 p2 − 2 p1 p3 − 2 p2 p3 ⇒ h 2 ν1 ν2 h 2 ν1 ν2 cos θ + 2 + 2mhν2 ⇒ 2 c2 c θ h 2h ⇒ m(ν1 − ν2 ) = 2 ν1 ν2 (1 − cos θ ) = 2 ν1 ν2 sin2 c c 2 −m 2 c2 = −m 2 c2 − 2hν1 m − 2
1 θ 1 2h − = sin2 . 2 ν2 ν1 mc 2 Conservation of energy (the first component of the fourmomentum) gives hν1 hν2 E + mc = + ⇒ h(ν1 − ν2 ) = E − mc2 = T. c c c
10.2
Elastic Collisions
251
from which follows: T = h(ν1 − ν2 ). In the case of backward scattering the direction of motion is parallel to the direction of the bullet photon and the conservation of 4momentum gives p=
h (ν1 − ν2 )ˆe. c
The length of the threemomentum is p =
h (ν1 − ν2 ). c
When the angle of scattering is θ = π ⇒ sin2
θ 2
= 1, hence
1 1 2h mν1 c2 − = ⇒ ν = . 2 ν2 ν1 mc2 mc2 + hν1 Replacing we find pc =
2h 2 ν12 . mc2 + hν1
The speed of the electron is u=c
u=
pc pc =c =E c p 2 + m 2 c2
1+
1
mc2 pc
2 c ⇒
1 c= 3 * +2 c. 4 2 hν1 mc2 (mc2 +hν1 ) 4 1+ 2 51 + 1+ 2h 2 ν12 mc
1
2
hν1 mc2
2
10.2.5. Let eˆ be the direction of motion of the photon in the laboratory frame L . We have the following fourmomenta in L: 1 Fourmomentum of the photon: hν c eˆ L mc Fourmomentum of the atom prior to excitation: 0 L ∗ E /c Fourmomentum of the excited atom: p∗ L
252
10 Relativistic Reactions
Conservation of fourmomentum gives hν eˆ = p∗ c hν + mc2 = E ∗ . The excitation energy E δ = (energy of excited atom) − (energy of atom at normal state with the same threemomentum), that is, Eδ = E ∗ −
p ∗2 c2 + m 2 c4 = hν + mc2 − c p ∗2 + m 2 c2 .
Replacing p∗ in the last relation and taking E δ = E follows: $ hν E + mc − = c c
hν c
2 + m 2 c2 .
Squaring and solving for hν we find hν =
1− 1
E 2mc2 − mcE 2
E.
10.2.6. e+ p→e+ p 1
2
3
4
.
Conservation of fourmomentum gives
p4i
2
2 = p1i + p2i − p3i ⇒
−M 2 = −2m 2 − M 2 + 2 p1i p2i − 2 p1i p3i − 2 p2i p3i . In the proper frame of particle 2 we have p1i
=
E 1 /c p1
,
p2i
=
2
Mc 0
,
p3i
=
2
E 3 /c p3
2
from which follows p1i p2i = −E 1 M,
p1i p3i = −
E1 E3 + p1 · p3 , c2
p2i p3i = −E 3 M.
10.3
Reactions of the Form 1 → 2+3 (Decays)
253
We replace in the conservation equation and find m 2 + E1 M =
E1 + M E 3 − p1 · p3 . c2
For very fast electrons we may consider c p1  E 1 , c p3  E 3 (that is, the electrons behave practically as photons). Then p1 · p3 =
E1 E3 cos θ13 . c2
Replacing in the last relation and solving for E 3 we find E3 =
m mM + E1
1+
2E 1 sin2 Mc2
θ13 2
.
But the ratio m/M 1 and can be ignored. Finally, with these approximation we end up with E3 =
E1 1+
2E 1 sin2 Mc2
θ13 2
.
10.3 Reactions of the Form 1 → 2+3 (Decays) 10.3.1. The radiative transition is N∗ → γ + N. Conservation of fourmomentum gives p1i2 = ( pγi + p2i )2 ⇒ −m 2 c2 = −m c2 − 2E γ m , 2
where m is the mass of the nucleus after the radiative transition and the product pγ p2 has been computed in L . Solving for E γ we find Eγ =
m2 − m2 2 c . 2m
Using Δm = m − m it follows Δm 2 c . E γ = Δm 1 − 2m
254
10 Relativistic Reactions
10.3.2. Conservation of fourmomentum in the CM (= the proper frame of the mother particle) gives M = E 2∗ + E 3∗ p ∗ = p2∗ = p3∗ ⇒ E 2∗2 − m 22 = E 3∗2 − m 23 ⇒ m 2 −m 2 E 2∗ − E 3∗ = 2M 3 . From the sum and the difference of E 2∗ , E 3∗ we calculate E 2∗ =
M 2 + m 22 − m 23 , 2M
E 3∗ =
M 2 + m 23 − m 22 . 2M
The common threemomentum is ∗
p = =
E 2∗ − m 22 =
[M 2 − (m 2 + m 3 )2 ]1/2 [M 2 − (m 2 − m 3 )2 ]1/2 2M
1 λ(M, m 2 , m 3 ). 2M
The energies of the daughter particles 1, 2 in Σ are found from the boost equations E 2 = γ1 (E 2∗ − β1 0) = γ1 E 2∗ , E 3 = γ1 E 3∗ , where γ1 is the γ factor of CM frame in Σ. Squaring the conservation equation we find p1i2 = ( p2i + p3i )2 ⇒ −M 2 = −m 21 − m 22 + 2 p2i p3i ⇒ M 2 − m 21 − m 22 = 2(E 2 E 3 − p2 p3 cos φ). The threemomenta p2 = p3 =
E 22 − m 22 =
γ12 E 2∗ − m 22 ,
γ12 E 3∗ − m 23 .
Replacing we find 1 2 M − m 21 − m 22 = γ12 E 2∗ E 3∗ − 2
(γ12 E 2∗ − m 22 )(γ12 E 3∗ − m 23 ) cos φ.
10.3
Reactions of the Form 1 → 2+3 (Decays)
255
The only unknown quantity in this equation is γ1 (the E 2∗ , E 3∗ have been determined in terms of the masses). Solving we find the factor γ1 and from this β1 =
1−
m2 m3 , . M M
1 . γ12
Dividing with M appear the terms In the special case m 2 = m 3 = m, φ = 90◦ ; the general expression we have found simplifies to γ12 =
M 2 − 2m 2 . 2E 2∗ E 3∗
Furthermore, in this case E 2∗ = E 3∗ = M/2, hence γ12 =
M 2 − 2m 2 2
M2
β12 = 1 −
1 γ12
=2
4
m 2 M 2 − 2m 2 = 2 − 4 . M2 M
The β1 factor is =1−
1 2−4(m/M)2
$ 1 β1 = √ 2 where k =
=
1 1−4(m/M)2 2 1−2(m/M)2
⇒
1 − 4k 2 , 1 − 2k 2
m . M
10.3.3. We calculate the required quantities in the proper frame of particle 2 and then transform the results to L by using the boost along the direction of particle 2. The threemomenta of the particles are coplanar because conservation of fourmomentum implies p1 = p2 + p3 . The energy 21 E of 2 in the proper frame of 1 equals 2 1E
=
1 (m 2 + m 22 − m 23 ). 2m 1 1
Therefore, the length of the threemomentum 21 p (this is the triangle function of the three masses m 1 , m 2 , m 3 ) 2 1p
= =
2 1E
− m 22 =
#1/2 " #1/2 1 " (m 1 + m 2 )2 − m 23 (m 1 − m 2 )2 − m 23 4m 1
1 λ(m 1 , m 2 , m 3 ). 2m 1
256
10 Relativistic Reactions
We note that 21 p is independent of the angle θ. Obviously 21 p = 31 p. The fourvector of particle 2 in the proper frame of particle 1 has components ⎛ 2 i 1p
⎞
2 1E
⎜ 2 p cos θ ⎟ 1 ⎟ =⎜ ⎝ 21 p sin θ , ⎠ , 0 Σ∗
where we have taken the plane defined by p1 , p2 , p3 to be the plane x − y and we have identified the direction of motion of the mother particle 1 with the xaxis. Let β, γ be the velocity factors of 1 in L . Then E , γ = m1
β=
E 2 − m 21 E
.
The boost relating particles 1, 2 gives 2 2 2 L E = γ (1 E + β 1 p cos θ ), 2 2 2 L px = γ (1 p cos θ + β 1 E), 2 2 L p y = 1 p sin θ, 2 L pz = 0.
From the first equation follows 2 LE
where 21 E = L is
m 21 +m 22 −m 23 . 2m 1
= γ (21 E + β cos θ
2 2 1E
− m 22 ),
From the rest we calculate that the threemomentum of 2 in
⎞ γβ 21 E + γ cos θ 21 E 2 − m 22 ) ⎜ ⎟ 2 ⎜ ⎟ . Lp = ⎝ ⎠ sin θ 21 E 2 − m 22 0 L ⎛
For particle 3 we have the same results if we replace 2 → 3 and θ → φ where φ is the angle of motion of particle 3 in the laboratory with respect to the direction of motion of the mother particle. The angles θ, φ are not independent but they are related with the perpendicular projection of the threemomentum, that is, 2 L
p⊥ = 3L p⊥ ⇒ 2L p sin θ = 3L p sin φ ⇒
2 2 LE
− m 22 sin θ =
3 2 LE
− m 23 sin φ.
10.3
Reactions of the Form 1 → 2+3 (Decays)
257
When particle 2 is moving perpendicular to the direction of motion of particle 1 the angle θ = 90◦ and from the above follows 2 LE
= γ 21 E =
E m 21 + m 22 − m 23 . m1 2m 1
Note that in this case β = 21 p/E ⇒ 21 p = β E, therefore ⎛
⎞ ⎛2 ⎞ 2 γ β1 E 1 E/m 1 ⎜ ⎟ 2 2 2 2⎝ 2 2 2 1 ⎠ . L p = ⎝ 1 E − m2 ⎠ = 1 E − m2 0 0 L L 10.3.4. The energy of particle 2 in the proper frame of particle 1 is given by 2 1E
=
m 21 + m 22 − m 23 2 c . 2m 1
Therefore, its kinetic energy is 2 1T
1 [m 1 − m 2 ]2 − m 23 c2 2m 1 1 ΔE m2 − m3 2 = (m 1 − m 2 − m 3 )(m 1 − m 2 + m 3 ) = c . 1− 2m 1 2 m1 =21 E − m 2 c2 =
Similarly we show for particle 3 that 3 1T
=
ΔE 2
m2 − m3 2 c . 1+ m1
We infer that if m 2 ≥ m 3 , then 31 T ≥21 T , that is, the particle with the lesser mass gains higher kinetic energy and in case the masses of the daughter particles are equal then both particles have equal kinetic energy. 10.3.5. (a) From the masses we compute the energy of κ + (say particle 2) and π − (particle 3) in the laboratory frame L. We find E2 = E3 =
p22 c2 + m 22 c4 = 10.134 GeV p32 c2 + m 23 c4 = 1.056 GeV
Conservation of fourmomentum gives p1i = p2i + p3i ⇒ −M 2 c2 = −m 22 c2 − m 23 c2 + 2 p2i p3i .
258
10 Relativistic Reactions
In the laboratory we have p2i p3i = (E 2 /c, p2 ) L (E 3 /c, p3 ) L = p2 ·p3 −E 2 E 3 /c2 = p2 p3 cos(θ2 +θ3 )−E 2 E 3 /c2 . Replacing we find p2i p3i =
1 (9.092 − 10.702) (GeV/c)2 ⇒ p2i p3i = −1.610 (GeV/c)2 . c2
From the last relation follows M = 1.865 GeV/c2 (b) We compute first the speed β ∗ of the CM in L (along the direction of motion of the neutral particle). We find β∗ =
(p2 + p3 ) · (u/u) p2x + p3x p2 cos θ2 + p3 cos θ3 c= c= c ⇒ β ∗ = 0.986. E2 + E3 E2 + E3 E2 + E3
The boost in the xdirection gives ∗ p2x = γ ∗ p2x − β ∗ Ec2 , ∗ p2y = p2y . Hence tan θ ∗ ≡
∗ p2y ∗ p2x
=
γ∗
p2y p2x −
β ∗ Ec2
=
γ∗
p2 sin θ2 p2 cos θ2 − β ∗ Ec2
,
where γ ∗ ≡ (1 − β ∗2 )−1/2 = 5.997. Replacing we calculate tan θ ∗ = 0.697 ⇒ θ ∗ = 340 52 14 . (c) The neutral particle moves in L with speed v = β ∗ c, hence in the laboratory frame its lifetime is t=
d , β ∗c
where d = 0.9 mm. In the proper frame of the particle this time period is the lifetime of the particle. We have τ= from which follows
1 d t ⇒τ = ∗ ∗ , ∗ γ γ β c
10.3
Reactions of the Form 1 → 2+3 (Decays)
259
τ = 5.074 × 10−13 s. 10.3.6. (a) Let A,B be the events of creation and disintegration of an unstable particle in the laboratory frame L . If the speed of the particle in L is u and its lifetime τ, the distance it covers in L (which we consider that practically is the same as the length of the trace of the orbit we see on the photographic plate) is l = ut = uγ τ . But the speed √
pc2 = u= E
E 2 − m 2 c4 m 2 c4 c =c 1− , E E2
hence l = cγ τ 1 −
m 2 c4 . E2
We see that as E increases so does l, therefore the claim of the problem is justified. (We have assumed that the orbit of the particle is a straight line which in general is not true. If we do not make this assumption, then we need to know the curvatures of the orbit (Serret–Frenet formulae), and integrate the arc length along the orbit.) (b) We compute the energy of the muon μ+ in the two reactions. We have – K + disintegration Eμ =
m 2K + m 2μ − 0 2m K
= 0.258 GeV
– π + disintegration Eμ =
m 2π + m 2μ − 0 2m K
= 0.110 GeV
According to (a) and the figure (= image of orbits on the photographic plate) we infer that (A) represents the disintegration of K + and (B) the disintegration of π + . (c) We have assumed implicitly that the length of the orbit of the particle equals the length of the curve on the photographic plate. This can be invalid for fast particles or for particles with long lifetime. Furthermore, we have assumed that the orbit is a straight line which is not true for charged particles. Indeed in the bubble chamber magnetic fields are employed which change the direction of motion of charged particles. In conclusion, the results above must be considered as indicative and must be treated very cautiously in practice. However, they do give some indication on the nature of the elementary particle disintegration.
260
10 Relativistic Reactions
10.3.7. (a) The disintegration in the laboratory and in the CM is shown in Fig. 10.3 The boost which relates the laboratory frame and the CM frame gives Particle 1 1 = − p ∗ sin θ ∗ , p⊥ 1 p = γ (− p ∗ cos θ ∗ + β E 1∗ ).
Particle 2 2 = p ∗ sin θ ∗ , p⊥ 2 p = γ ( p ∗ cos θ ∗ + β E 2∗ ),
where γ , β refer to the mother particle M. We compute the quantity α = in terms of the quantities in the CM. We find
p1 − p2 p1 + p2
γ ( p ∗ cos θ ∗ + β E 2∗ ) − γ (− p ∗ cos θ ∗ + β E 1∗ ) γ ( p ∗ cos θ ∗ + β E 2∗ ) + γ (− p ∗ cos θ ∗ + β E 1∗ ) E 2∗ − E 1∗ 2 p∗ ∗ cos θ = + . β(E 1∗ + E 2∗ ) E 2∗ + E 1∗
α=
We define A=
2 p∗ β(E 1∗ + E 2∗ )
B=
E 2∗ − E 1∗ E 2∗ + E 1∗
and it follows that α = A cos θ ∗ + B ∗ . Let p⊥ =  p1⊥  =  p2⊥  be the common length of the normal projection of the threemomentum in the laboratory. From the boost we have p⊥ = p ∗ sin θ ∗ . Eliminating θ ∗ from the last two relations we find
α−B A
2
+
p⊥ p∗
2 = 1.
In the plane ( p⊥ , α) this equation represents an ellipse with one focal point at B, minor semiaxis p ∗ , and major semiaxis A as it is shown in Fig. 10.4. We note that the maximal value of p⊥ is p ∗ . Also B = 0 if m 1 = m 2 . (This diagram is called the ( p⊥ , α) diagram of Armenteros – Podolanski.)
10.3
Reactions of the Form 1 → 2+3 (Decays)
261
Fig. 10.3 Disintegration in the laboratory and in the CM frame
Fig. 10.4 Amanteros – Podolanski diagram
(b) The maximal value p⊥ = p ∗ occurs when θ = 90◦ . In order to evaluate p ∗ we work as follows. From the conservation of fourmomentum we have −M 2 = −m 21 − m 22 + 2 p1i p2i . In the CM frame the fourmomenta are p1i
=
E 1∗ p∗
p2i
=
CM
E 2∗ −p∗
. CM
The invariant p1i p2i = −E 1∗ E 2∗ + p ∗2 . Replacing the product p1i p2i from the first relation we get E 1∗ E 2∗ =
1 2 (m + m 22 − M 2 ) − p ∗2 . 2 1
But E 1∗ =
p ∗2 + m 21
E 2∗ =
p ∗2 + m 22 ,
262
10 Relativistic Reactions
hence ( p ∗2 + m 21 )( p ∗2 + m 22 ) =
1 2 (m + m 22 − M 2 )2 + p ∗4 − (m 21 + m 22 − M 2 ) p ∗2 . 4 1
Solving for p ∗2 we find (M 2 − m 22 − m 21 )2 − 4Mm 22 . 4M 2
p ∗2 =
From this result we deduce that maximum value occurs at the threshold. c Indeed p ∗2 = 2M λ(M, m 1 , m 2 ), where λ(M, m 1 , m 2 ) is the triangle function and we know that threshold occurs when λ(M, m 1 , m 2 ) = 0. 10.3.8. π0 → γ + γ . 1 2 3 In the CM of the reaction (=proper frame of π 0 ) the threemomenta of the photons are p∗2 = −p∗3 which implies E 2∗ = E 3∗ = p∗2  = p∗3 . Conservation of energy gives E 2∗ = E 3∗ =
1 mπ . 2
The velocity factor of the pion is p = β= E
√
E 2 − m2 = T +m
√
(T + 2m)T . T +m
The fourmomenta of the photons in the CM are p2i = E 2∗ (1, 1, 0, 0) = m2π (1, 1, 0, 0), p3i = E 3∗ (1, −1, 0, 0) = m2π (1, −1, 0, 0). Using the boost relating the CM frame and the laboratory we find $ E2 = where k =
1−β 1+β
γ (E 2∗
−
β E 2∗ )
=
E 2∗
mπ 1−β = 1+β 2
$
1 1−β = km π 1+β 2
is the Doppler factor. Similarly we compute E3 =
1 mπ . 2k
10.3
Reactions of the Form 1 → 2+3 (Decays)
263
Application Using the given data we compute √
√
= 101135127 = 0.993, β = (TT+2m)T +m 1−0.993 k = 1+0.993 = 5.93 × 10−2 . 2
Hence E 2 = 12 135 × 5.93 × 10−2 MeV = 4 MeV, 135 E 3 = 12 5.93 × 102 MeV = 1138 MeV. (b) In the CM the energies of the photons are again equal to 12 m π . The fourmomenta of the photons in the laboratory are p2i = E 2 (1, eˆ 2 ) L
p3i = E 3 (1, eˆ 3 ) L .
Conservation of fourmomentum gives p1i = p2i + p3i . Squaring we find −m 2π = 2 p2i p3i = 2E 2 E 3 (−1 + cos φ) ⇒ cos φ = 1 −
m 2π . 2E 2 E 3
Conservation of threemomentum along the yaxis gives E 2 = E 3 . Hence E2 = E3 =
1 mπ γ 2
and cos φ = 1 −
m 2π 2 = 1 − 2. 1 2 2 γ 2 4 mπ γ
Application From the given data we find γ = (1 − β 2 )−1/2 = (1 − 0.993)−1/2 = 11.95, 2 ◦ cos φ = 1 − 11.95 2 = 0.986 ⇒ φ 10 . The fourmomentum of the photons in the laboratory is t φ φ p2i = E 2 1, cos , sin , 0 , 2 2
t φ φ p3i = E 2 1, cos , − sin , 0 2 2
264
10 Relativistic Reactions
where E 2 = 12 m π γ . We transform in the CM frame with the appropriate boost and find (motion is in the x − y plane) E 2∗ = γ E 2 − β E 2 cos φ2 , E 2∗ cos θ ∗ = γ E 2 cos φ2 − β E 2 = γ E 2 cos φ2 − β , E 2∗ sin θ ∗ = E 2 sin φ2 . Dividing we get tan θ ∗ =
sin φ2 . γ cos φ2 − β
Then 1 φ cos φ = 1 − 2 sin2 = 1 − 2 2 ⇒ 2 γ
1 φ sin = 2 γ
and
φ cos = 2
$ 1−
1 = β. γ2
Replacing we find tan θ ∗ = ∞ ⇒ θ ∗ =
π . 2
10.3.9. (a) In the lab frame L we have the components piA =
m Ac 0
,
piB =
L
E B /c pB
,
pCi =
L
E C /c pC
. L
Conservation of fourmomentum gives pC2 = p 2A + p 2B − 2 p A p B ⇒ −m C2 c2 = −m 2A c2 − m 2B c2 + 2m A E B ⇒
EB =
m 2A + m 2B − m C2 2 c . 2m A
(b) We apply (a) with m A = M, m B = 0 (photon), and m C = M − δ. The energy E B = hν is then M 2 − (M − δ)2 2 δ2 2 c = δ − c2 < δc2 . hν = 2M 2M The reason for the inequality hν < δc2 is that some energy is consumed for the reaction of M − δ, which is necessary for the fourmomentum to be preserved. In M¨osbauer effect the feedback is shared among 1023 atoms, hence the reaction energy is negligible.
10.3
Reactions of the Form 1 → 2+3 (Decays)
265
(c) In this case the conservation of fourmomentum gives EA EB −m C2 c2 = −m 2A c2 − m 2B c2 − 2 p A  p A  cos θ + 2 2 ⇒ c 2 1/2 2 1/2 m A + m 2B − m C2 c4 = 2E A E B − 2 E 2A − m 2A c2 E B − m 2B c2 cos θ, which is the required relation. Application It is given θ = 900 ⇒ cos θ = 0 and m B = m C . Replacing in the last relation we find m 2 c4 2E A E B = m 2A + m 2B − m C2 c4 = m 2A c4 ⇒ E A = A . 2E B The maximum value of E A occurs when E B takes its minimum value, which is E B = m B c2 . This implies E A,max =
m 2A c4 m 2A c2 m 2K c2 = = . 2m B c2 2m B 2m π
10.3.10. (a) Suppose that the particles 1, 2, 3 are placed in the vertices of the canonical triangle (see Fig. 10.5). Then we have the decays n → (2n + 2) + (2n + 3) n = 1, 2, 3, where the proper frame of particles 1, 2, 3 coincide. Because the particles 1, 2, 3 are identical and decay in the same manner, we consider only particle 1 which 2
6
30°
A
8
4
30°
1 Fig. 10.5 Triangulardecay
30°
3
266
10 Relativistic Reactions
decays as 1 → 4 + 5, where particle 4 has mass am and particle 5 has mass bm. The energy and the length of the threemomentum of particle 4 in the proper frame of particle 1 is (see a previous exercise) X 2 mc , 2 YZ 2 mc, 1p = 2
2 1E
=
where X = 1 + a 2 − b2 , Y 2 = (1 − a)2 − b2 , Z 2 = (1 + a)2 − b2 , and the βand the γ factors in the same frame are 2 1β
=
YZ 2 X ,1 γ = . X 2a
We see that the βfactor is independent of the mass, hence it is common to all daughter particles. Therefore, all particles of mass am arrive simultaneously in the center of mass of the triangle.1 (b) We consider the reaction: 4 + 6 + 8 → A, where the mass of each of the particles 4, 6, 8 are equal to am and the mass of the formed particle A is M. Conservation of fourmomentum gives p 2A = 3 p42 + 2 p4i · p6i + 2 p4i · p8i + 2 p6i · p8i . We compute the inner products in Σ. We have (see Fig. 10.5)
1 Of course one could reach this conclusion directly from the symmetry of the situation. That is, one assumes that they do not arrive simultaneously and by rotating the triangle by 120◦ about its center reaches contradiction.
10.3
Reactions of the Form 1 → 2+3 (Decays)
⎛
⎞ ⎛2 ⎞ 2X 1 E/c √ ⎟ ⎜ 2 p cos 30 ⎟ ⎜ 0 ⎟ mc ⎜ mc i 1 ⎜ 3Y Z ⎟ ⎜ ⎟ ⎟ p4i = ⎜ ⎝ 21 p sin 30 ⎠ = 4 ⎝ Y Z ⎠ , p6 = ⎝ −21 p ⎠ = 4 0 0 0 Σ Σ Σ ⎛ ⎛ 2 ⎞ ⎞ 2X E/c 1 √ ⎜ −2 p cos30 ⎟ mc ⎜ − 3Y Z ⎟ i 1 ⎜ ⎜ ⎟ ⎟ p8 = ⎝ 2 ⎠ = 4 ⎝ YZ ⎠ . 1 p sin 30 0 0 Σ Σ ⎛
2 1 E/c
⎞
267
⎛
⎞ 2X ⎜ 0 ⎟ ⎜ ⎟ ⎝ −2Y Z ⎠ , 0 Σ
Therefore, m 2 c2 (−4X 2 − 2Y 2 Z 2 ), 16 m 2 c2 (−4X 2 − 2Y 2 Z 2 ), p4i · p8i = 16 m 2 c2 (−4X 2 − 2Y 2 Z 2 ). p6i · p8i = 16 p4i · p6i =
Replacing we find m 2 c2 (−4X 2 − 2Y 2 Z 2 − 4X 2 − 2Y 2 Z 2 − 4X 2 − 2Y 2 Z 2 ) 16 m2 3m 2 (−12X 2 − 6Y 2 Z 2 ) = 3a 2 m 2 + (2X 2 + Y 2 Z 2 ) = 3a 2 m 2 − 8 4 3m 2 2 4a + 2X 2 + Y 2 Z 2 . = 4
M 2 = 3a 2 m 2 − 2
We note that the total threemomentum vanishes in Σ. Indeed we have ⎞ ⎞ ⎛√ ⎛ ⎞ ⎛ √ 0 − 3Y Z mc ⎝ 3Y Z ⎠ mc ⎝ mc ⎝ Y Z ⎠ = 0, −2Y Z ⎠ + p4 +p6 +p8 = + YZ 4 4 4 0 0 0 Σ Σ Σ therefore, the reaction occurs (in Σ!) at the threshold. This means that the particle A is created at rest in Σ. 10.3.11. The energy of particle 2 in the proper frame of particle 1 is given by 2 1E
=
m 21 + m 22 − m 23 2 X c = m 1 c2 , 2m 1 2
(10.1)
where X = 1 + a 2 − b2 . From this we compute the linear momentum 21 pc = 1 2 2− 2 2 2 2 2 2 2 2 2 1 E m 2 c = 2 Y Z m 1 c where Y = (1 − a) − b , Z = (1 + a) − b .
268
10 Relativistic Reactions
(b) We have 2 1β
=
2 1 pc 2 1E
=
YZ , X
2 1γ
=
2 1E
m 2 c2
=
X . 2a
(10.2)
(c) Consider the decay 2 → 4 + 5, in which the mass of the daughter particles are m 4 = am 2 , m 5 = bm 2 . By question (a) above the energy and the momentum of particle 4 in the proper frame of particle 2 are X m 2 c2 =21 Ea, 2 YZ 4 m 2 c =21 pa. 2p = 2
4 2E
=
To compute the energy and the momentum of particle 4 in the proper frame of particle 1 we use the boost relating particle 2 and 1. We have 4 1E
=21 γ
4
2E
−21 βc24 p =21 γ a 21 E −21 βc12 p = am 2 c2 = a 2 m 1 c2 ,
because the term 21 γ 21 E −21 βc12 p is the energy of particle 2 in its proper frame and this energy is the mass m 2 c2 . Concerning the threemomentum we have 4 2 pc
=21 γ
4
2 pc
−21 β24 E = 0,
which means that particle 4 is at rest in the proper frame of particle 1. We conclude that after the 2nth decay the produced particle 4n n = 1, 2, ... will have mass a 2n m 1 c2 and will be at rest in the proper frame of particle 1. 10.3.12. Conservation of fourmomentum gives pνi = pπi − pμi . Squaring we find m 2ν = m 2π + m 2μ − 2
1 i p · pi . c2 π μ
In the proper frame of the pion the 2 pπi · pμi = 2m π πμ E = 2m π [m μ c2 + πμ T ]. Replacing we find the required answer
10.3
Reactions of the Form 1 → 2+3 (Decays)
269
m 2ν = (m π − m μ )2 −
1 2m π μπ T. c2
For the given values we calculate m ν = −0.24206 MeV/c2 . In order to determine the accuracy of the result we assume that the experimental determination of the value of the quantities m π , m μ , μπ T obeys the normal distribution. Then according to the law of Gauss for the propagation of error and assuming that m ν = f (m π , m μ , μπ T ) we have $ δm ν = ±
∂m ν δm π ∂m π
2
+
∂m ν δm μ ∂m μ
2
+
∂m ν μ μ δπ T ∂π T
2
We calculate μ T mπ − mμ ∂m ν = − π 2 = −123.091, ∂m π mν mν c
∂m ν mπ − mμ =− = 140.12, ∂m μ mν ∂m ν mπ =− = −576.612. μ m ν c2 ∂π T Replacing we find δm ν = ±9.361 MeV/c2 . The conclusion is m ν = 0 ± 9 MeV/c2 . 10.3.13. (a) π → μ+ ν . 1 2 3 Conservation of fourmomentum gives p32 = ( p2 − p1 )2 ⇒ −m 2μ − m 2π − 2
1 i p p1i = 0. c2 2
.
270
10 Relativistic Reactions
In the proper frame of 1 the invariant p1i p2i = −2m π E where E is the total energy of μ, which equals E = T + m μ c2 . Replacing we find that the kinetic energy of the μmeson is T =
m 2π + m 2μ
c2 − m μ c2 =
2m π
(m π − m μ )2 2 c . 2m π
Numerical application: Replacing the values of the masses we find T = 4.37 MeV. (b) Squaring Ai = B i + C i one obtains the required relation. In the proper frame Σ B of B i we have for the invariant B i Ci B i Ci =
B 0
ΣB
0
BC
BC
ΣB
= −B B C 0
and B 2 = −(B 0+ )2 . Therefore, BC
0
=
1 2 A − B2 − C 2 . 2B
Part (a) is a special case of part (b) when the fourvectors Ai , B i , C i are the fourmomenta of the particles π, μ, ν with masses m π , m μ , and m ν , respectively. 10.3.14. (a) Conservation of fourmomentum gives p1i − p2i = p3i .
(10.3)
In an arbitrary LCF Σ this equation gives
E 1 /c p1
Σ
p1 − p2 =
− hν k, c
E 2 /c p2
Σ
=
hν c
1 ⇒ k Σ
E 1 − E 2 = hν.
Take Σ so that the xaxis is along the direction of propagation k of the photon and yaxis is normal to that direction. Then the last equations give Parallel to k (xaxis) p1 · k − p2 · k = Normal to k (yaxis)
hν . c
10.3
Reactions of the Form 1 → 2+3 (Decays)
271
p1 × k − p2 × k = 0.
(10.4)
If M1 , M2 are the masses of the nucleus before and after the radiative transition, the conservation of energy gives
M 1 γ1 − M 2 γ2 =
hν c2
(10.5)
and the conservation of threemomentum gives hν , c M1 γ1 u 1 sin θ1 = M2 γ2 u 2 sin θ2 .
M1 γ1 u1 · k − M2 γ2 u2 · k =
(10.6) (10.7)
Equation (10.6) can be written as
M 1 γ1
hν u1 · k u2 · k − M 2 γ2 = 2 c c c
and using (10.5):
u1 · k u2 · k − M 1 γ1 = M 2 γ2 − M 2 γ2 , c c u1 · k u2 · k = M 2 γ2 1 − , M 1 γ1 1 − c c M1 φ1 = M2 φ2 = α,
M 1 γ1
(10.8)
where φi = γi 1 − uci k i = 1, 2. Concerning the second relation we want to prove, we note that ui  sin θi = ui × k subsequently (10.22) takes the following form M1 γ1 u1 × k = M2 γ2 u2 × k ⇒ M1 ψ1 = M2 ψ2 = δ, where ψi = γi ui c×k , i = 1, 2. θi θi (b) Writing φi = γi 1 − ui  cos , ψi = γi ui  sin we calculate c c
(10.9)
272
10 Relativistic Reactions 2 2 ui 2 cos2 θi 2 ui  cos θi 2 ui  sin θi 1 + φi2 + ψi2 = 1 + γi2 + γi2 − 2γ + γ i i c2 c c2 2 2 ui  ui  ui  cos θi + γi2 2 + γi2 − 2γi2 = γi2 1 − 2 c c c ui  cos θi = 2γi φi ⇒ = 2γi2 1 − c 1 γi = (1 + φi2 + ψi2 ). (10.10) 2φi
Noting that Mi2 (1 + φi2 + ψi2 ) = Mi2 + Mi2 φi2 + Mi2 ψi2 = α 2 + δ 2 + Mi2 we get, if we consider (10.10), α 2 + δ 2 + M12 = 2α M1 γ1 , α 2 + δ 2 + M22 = 2α M2 γ2 . Replacing in the conservation of energy equation (10.5) we find hν 1 (M 2 − M22 ) ⇒ = M 1 γ1 − M 2 γ2 = 2 c 2α 1 c2 (M 2 − M22 ). hν = 2α 1 (c) Using the last relation we find hν =
c2 (M1 − M2 )(M1 + M2 ), 2α
(10.11)
but M1 c2 − M2 c2 = hν0 , hence M1 + M2 hν0 ⇒ 2α E0 ν = 2 ν0 . αc
hν =
(10.12) (10.13)
2c where E 0 = M1 c +M . 2 (d) The conservation equation (10.3) gives 2
2
pi · p1i = p12 − p1i · p2i To compute the inner product p1i · p2i we square (10.3) and get
(10.14)
10.3
Reactions of the Form 1 → 2+3 (Decays)
273
p12 + p22 − 2 p1i · p2i = ( p3i )2 = 0 ⇒ −M12 c2 − M22 c2 − 2 p1i · p2i = 0 ⇒ 1 p1i · p2i = − M12 + M22 c2 . 2 Replacing in (10.14) we find pi · p1i = −M12 c2 +
1 2 1 M1 + M22 c2 = −M12 + M22 c2 . 2 2
Let us choose the proper frame Σ1 , say, of the mother nucleus. In this frame the fourmomenta pi , p1i are as follows p1i =
M1 c 0
Σ1
,
pi =
hν1 c
1 k Σ1
from which follows p1i · pi = −hν1 M1 . From the two expressions giving p1i · pi we compute ν1 =
c2 (M 2 − M22 ). 2h M1 1
(10.15)
Now let us consider the proper frame of the nucleus in ground state, say Σ2 . In this frame the fourmomentums p2i , pi would be p2i
=
M2 c 0
hν , p = 2 c Σ2 i
1 , k Σ 2
respectively. Conservation of fourmomentum gives pi = p1i − p2i , thus pi · p2i = p1i · p2i − p2i2 ⇒ 1 = − M12 + M22 c2 + M22 c2 ⇒ 2 1 2 M2 − M12 c2 . = 2 But pi · p2i = −h M2 ν2 , therefore (10.16) gives
(10.16)
274
10 Relativistic Reactions
1 2 M1 − M22 c2 ⇒ 2 c2 2 ν2 = M1 − M22 . 2h M2
h M2 ν2 =
(10.17)
Dividing (10.15), (10.17) we get
ν1 ν2
c2 (M 2 − M22 ) ν1 M2 φ1 2h M1 1 = = ⇒ = 2 ν M1 φ2 c (M12 − M22 ) 2 2h M2
ν1 ν = 2. φ1 φ2 10.3.15. During the radiative transition we have in the laboratory frame (L) 1∗
→1 +
γ,
(E ∗ , 0) L → (E, p) L + (E γ , pγ ) L . & & Obviously p = &pγ & = E γ , hence E ∗ = M ∗ c2 = E + E γ =
E γ2 + M 2 c4 + E γ .
(b) When a nucleus at ground state at rest in L absorbs a photon of energy E γ (in L) has total energy
M 2 c4 + p 2 c2 =
M 2 c4 + E γ2 .
Therefore, the required excitation energy (in L) is ΔE =
M 2 c4 + E γ2 − M.
(10.18)
The excitation energy ΔE can be obtained if the excited nuclei which emit the photons are moving in L. Indeed in this case due to the Doppler effect the exciting photon in the laboratory has higher energy. Let β be the speed of the excited nucleus in L . If the frequency of the photon in L is ν and in the proper frame of the nucleus is ν0 the Doppler formula gives
10.4
Reactions of the Form 1 + 2 → 3
275
hν0 ⇒ γ (1 − β cos θ ) 1 ΔE = h(ν − ν0 ) = hν0 −1 . γ (1 − β cos θ ) hν =
But θ = 0 ⇒ cos θ = 1, hence √ √ 1−β 2 1−β 2 √ √ ΔE = hν0 − 1 = Eγ −1 ⇒ 2 2 1+β
1+β
√ 1−k β=√ 1+k where k = (1 − ΔE/E γ )2 and ΔE is given by (10.18). Obviously ΔE < E γ which implies 0 < k < 1. 10.3.16. The solution is straightforward and it is left to the reader.
10.4 Reactions of the Form 1 + 2 → 3 10.4.1. The reaction is 1 + 2 → 3. Conservation of fourmomentum gives −m 23 = ( p1i + p2i )2 = −m 21 − m 22 + 2 p1i p2i . The fourmomenta p1i , p2i of 1, 2 in Σ are, respectively, p1i =
m 1 γ1 c m 1 γ1 υ1
Σ
,
p2i =
m 2 γ2 c m 2 γ2 υ2
Σ
.
Hence v1 · v2 p1i · p2i = −m 1 m 2 (c2 − γ1 γ2 v1 · v2 ) = −m 1 m 2 γ1 γ2 c2 1 − . c2 Replacing we find m 23 = m 21 + m 22 + 2m 1 m 2 γ1 γ2 c2 Q, where Q = (1 − v1 · v2 /c2 ).
276
10 Relativistic Reactions
The term Q ≥ 0 because v1 · v2 v 1 v 2  = cos θ12 ≤ 1. 2 c c2 Also the term 2m 1 m 2 γ1 γ2 ≥ 2m 1 m 2 because γ1 , γ2 ≥ 1. Therefore m 23 ≥ m 21 + m 22 + 2m 1 m 2 = (m 1 + m 2 )2 . The inequality m 3 − (m 1 + m 2 ) ≥ 0 means that the kinetic energy of the particles 1, 2 in the proper frame of particle 3 has been transformed into the mass m 3 .The equality holds at the “threshold” of the reaction, that is, the particles 1, 2 do not “move” when they collide. 10.4.2. The reaction is e+ + e− → Y . 1
2
3
Conservation of fourmomentum gives p1 + p2 = p3 ⇒ −m 23 c2 = −m 21 c2 − m 22 c2 + 2 p1i p2i . In the rest frame of the e− we have 2 p1i p2i = −2E 1 m 2 , where E 1 is the energy of the beam of e+ . Replacing we find E1 =
m 23 − m 21 − m 22 2 c . 2m 2
Using the values of the masses we calculate E 1 = 89492 GeV. The huge difference between E 1 and 4.73 GeV for the headon collision makes clear why in the highenergy reactions we use headon colliding beams and not standing targets.
10.5 Reaction of the Form 1 + 2 → 3 + 4 10.5.1. The reaction is (see Fig. 10.6) π − + p + → k 0 + Λ0 . 1 2 3 4
10.5
Reaction of the Form 1 + 2 → 3 + 4
277
CM Fig. 10.6 Disintegration in the laboratory and in the CM frame
The fourmomenta of the particles in the laboratory are ⎞ ⎞ ⎞ ⎛ ⎛ ⎛ ⎞ /c E 4√ E 1 /c m pc E 3 /c ⎜ p1 ⎟ ⎜ 0 ⎟ ⎜ p3 cos φ ⎟ ⎜ p4 / 2 ⎟ i i i ⎟ ⎟ ⎜ ⎜ ⎜ ⎟ √ ⎟ p1i = ⎜ ⎝ 0 ⎠ , p2 = ⎝ 0 ⎠ , p3 = ⎝ − p3 sin φ ⎠ , p4 = ⎝ p4 / 2 ⎠ . 0 0 0 0 L L L L ⎛
Conservation of fourmomentum gives E 3 + E 4 = E 1 + m p c2 , √ p3 cos ϕ = p1 − p4 / 2, √ p3 sin ϕ = p4 / 2. We calculate E1 = E4 =
p1 2 + m 2π c4 = 2.504 GeV, p4 2 + m 2Λ c4 = 1.267 GeV.
Replacing in the conservation equations we find E 3 = 2.175 GeV,
p3 = 2.117 GeV/c,
φ ≈ 11◦ 30 .
In order to compute the physical quantities of k ◦ in the CM frame we compute the velocity factor βCM of the CM in the laboratory and then apply the boost relating CM and the laboratory frame. We have β CM
, p1 + p2 p 2.50 i = 0.726 i. =, = = E E 1 + m p c2 2.504 + 0.938
278
10 Relativistic Reactions
Then γCM = 1.455. The transformation equations read E 3∗ = γCM (E 3 − βCM p3 cos ϕ), E3 p3∗ cos ϕ ∗ = γCM ( p3 cos ϕ − βCM ), c p3∗ sin ϕ ∗ = p3 sin ϕ, from which follows E 3∗ = 2.021 GeV,
p3∗ =
φ ∗ = 12◦ 36 .
E 3∗2 − m 2k c4 = 1.852 GeV/c
10.5.2. (a) The reaction is 1 + 2 → 3 + 4. Conservation of fourmomentum gives p42 = ( p1 + p2 − p3 )2 = −m 21 c2 − m 22 c2 − m 23 c2 + 2 p1i p2i − 2 p12 p3i − 2 p22 p3i = = − m 21 + m 22 + m 23 c2 − 2E 1 m 2 + 2E 1 E 3 + 2m 2 E 3 ,
because p1i p3i = − c12 E 1 E 3 . Replacing E 1 = T1 + m 1 c2 we find m 2 + m 22 + m 23 − m 24 + 2m 2 E3 = 1 2 Tc21 +m 1 + m 2
T1 c2
+m 1
c2 .
(b) When m 1 = m 3 = m k , m 2 = m 4 = m p the energy E 3 reads E3 =
T1
mp
c2
+m k + m 2k
T1 +m k c2
+ mp
c2 = m p + T1
m 2k − m 2p
+m k + m p c2
c2 .
From this expression we find that when T1 → 0, E 3 → E 3,min = m k and when T1 → ∞, E 3 → E 3,max = m p . 10.5.3. (a) The reaction is p + p¯ → γ + γ . 1 2 3 4 In the laboratory frame the fourmomenta of the involved particles are p1i = ( p1 , E 1 )t ,
p2i = (0, m p )t ,
p3i = E 3 (1, eˆ )t ,
p4i = E 4 (1, εˆe)t ,
10.5
Reaction of the Form 1 + 2 → 3 + 4
279
where ε = ±1 if the photons are moving cocurrently or inversely. Conservation of fourmomentum gives p1i + p2i = p3i + p4i . Squaring we find −m 2p − m 2p − 2E 1 m p = +2(−1 + ε)E 3 E 4 ⇒ (−1 + ε)E 3 E 4 = −(E 1 + m p )m p . Because E 3 E 4 > 0 ⇒ ε = −1, that is, the photons are moving in opposite directions in the laboratory. Replacing E 1 = T1 + m p we find for the product of the unknown energies E3 E4 =
1 (T1 + 2m p )m p . 2
From the equation of conservation of energy we have for the sum of these energies E 3 + E 4 = E 1 + E 2 = T1 + 2m p . Since we know the sum and the product of E 3 , E 4 we infer that the energies E 3 , E 4 are the positive roots of the equation (in GeV units) 1 x 2 − (T1 + 2m p )x + m p (T1 + 2m p ) = 0. 2 Replacing T1 =
2 3
GeV, m p = 1 GeV/c2 we find 4 8 x 2 − x + = 0. 3 3
Solving the equation we find 2 x = 2, . 3 Obviously the threemomentum of the photon which moves in the same direction with the antiproton (3 say) is larger than the energy of the other photon (4). Because for a photon the threemomentum equals the energy we infer that E 3 > E 4 . Therefore, E 3 = 2 GeV,
E4 =
2 3
GeV.
(b) In the proper frame of the antiproton the above analysis applies literally, because the masses of the reacting particles proton and antiproton are equal and the same
280
10 Relativistic Reactions
applies to the products of reaction. The only difference is that the photon which moves in the same direction with the p is the number 4 and not the number 3. Hence, E3 =
2 GeV, 3
E 4 = 2 GeV.
10.5.4. The reaction is γ + X1 → X2 + γ . 1 2 3 4 Conservation of fourmomentum gives p3 = p1 + p2 − p4 . Let M be the mass of the daughter particle X 2 . Squaring the conservation equation we find −M 2 c2 = −M 2 c2 + 2 p1i p2i − 2 p1i p4i − 2 p2i p4i . We compute the invariant scalar products in the laboratory frame. The components of the fourvectors in the laboratory frame are p1i
E = c
E 1 1 Mc E 3 /c i i i ,p = ,p = , p4 = , p eˆ L 2 0 L 3 c nˆ L L
where eˆ · nˆ = 0. We compute p1i p2i = −E M, p1i p4i = −
E2 , p2i p4i = −E M. c2
Replacing we find −M 2 c2 = −M 2 c2 + 2 M 2 = M 2 − 2
E2 + 2E M − 2E M ⇒ c2
E2 . c4
The threemomentum of the daughter particle is p3 =
E (ˆe − eˆ 1 ) . c
10.5
Reaction of the Form 1 + 2 → 3 + 4
281
We note that the mass of the daughter particle M < M. This is due to the fact that a certain amount of the mass of the mother particle is transformed into kinetic energy of the daughter particle. 10.5.5. Conservation of fourmomentum gives p 2X = [( pai + pbi ) − pci ]2 ⇒ −m 2X c2 = ( pai + pbi )2 − m c c2 − 2( pai + pbi ) pci . In the CM Σ∗ of the particles a, b the pai + pbi = ( pai + pbi )2 =
√s . Thus
c
0
Σ∗
−s . c2
Suppose that in Σ∗ the fourmomentum of particle c is pc = ( pai + pbi ) pci = −
E/c p
. Then Σ∗
√ √ sE E s =− 2 . c c c
Replacing we find m 2X c2 = m 2X c2 =
s c2
+ m 2c c2 −
√
s c
− mcc
2
√ 2E s , c2
−
√ 2 s [E c
− m c c2 ].
But E − m c c2 = T , where T is the kinetic energy of the particle c in Σ∗ . Therefore, the mass m X has its maximum value in Σ∗ when T = 0, that is, c is produced at rest in Σ∗ . This implies √ m x,max =
s
c2
− mc.
10.5.6. Consider the reaction p+ p→π +C . 1 2 3 4 Conservation of fourmomentum gives p42 = ( p1i + p2i − p3i )2 = p1i2 + p2i2 + p3i2 + 2 p1i · p2i − 2 p1i · p3i − 2 p2i · p3i ⇒ −m C2 c2 = −2m 2p c2 − m 2π c2 + 2 p1i · p2i − 2 p1i · p3i − 2 p2i · p3i .
282
10 Relativistic Reactions
We calculate the inner products in the laboratory frame. We have p1i · p2i = p1i
·
p3i
=
E 1 /c p1
· L
E 3 /c p3
p3i
·
p2i
=
E 1 /c p1
· L
=− L
E 3 /c p3
·
= −E 1 m p , L
E1 E3 E1 E3 + p1 · p3 = − 2 + p1 p3 , c2 c
L
m pc 0
m pc 0
= −E 3 m p . L
Replacing we find m C2 c2 = 2m 2p c2 + m 2π c2 + 2E 1 m p − 2
E1 E3 + 2 p1 p3 − 2E 3 m p ⇒ c2
m C2 c2 = (2m 2p + m 2π )c2 + 2E 1 m p − 2
E 1 + m p c2 2 2 2 2 p3 c + m 2π c4 + E 1 − m 2p c4 p3 . 2 c c
Introducing the given numerical data we compute m C = 4.497 GeV/c2 . 10.5.7. (a) Obviously the threemomentum of K − is 10 GeV/c (why?). (b) The reaction is p + L− → π+ + X 1 2 3 4. Conservation of fourmomentum gives ( p3i + p4i )2 = ( p1i + p2i )2 .
(10.19)
In the laboratory frame (where the bubble chamber rests) the K − , p react at rest, hence ( p1i + p2i )2 = −(m K + m p )2 c2 . Also ( p3i + p4i )2 = −m 2π c2 − m X c2 + 2 p3i · p4i .
10.5
Reaction of the Form 1 + 2 → 3 + 4
283
Replacing the above in (10.19) we find the equation 2 i p · pi = (m K + m p )2 . c2 3 4
m 2X + m 2π −
(10.20)
We compute the invariant p3i · p4i in the laboratory, The threeforce on π − is F = qv B eˆ , where e is the unit normal to the plane defined by the direction of motion of π − and the magnetic field in the chamber and q is the charge of electron. The motion at the event O is radial (i.e., the acceleration is always normal to the velocity) with radius of curvature R. Therefore, F = mγ
p3 v v2 e= e. R R
Equating the two expressions for the threeforce we find p3 = q B R. Because the charge is conserved we deduce that the charge of the particle X is −e. Conservation of fourmomentum gives for the length of the threemomenta p3  = p4 . This implies that in the laboratory the energy of the π − is E3 =
p23 c2 + m 2π c4 =
(q B Rc)2 + m 2π c4
and that of the unknown particle is E4 =
(q B Rc)2 + m 2X c4 .
The above gives that the value of the invariant p3i · p4i in the laboratory is p3i · p4i =
E 3 /c p3
· L
E 4 /c −p3
=− L
E3 E4 E3 E4 − p23 = − 2 − (q B R)2 . c2 c
Replacing in (10.20) we find the equation m 2X +
2 c2
E3 E4 2 = (m K + m p )2 − m 2π ⇒ + (q B R) c2
m 2X
2 + 2 E 3 (q B R/c)2 + m 2X = A, c
284
10 Relativistic Reactions
where A = (m K + m p )2 − m 2π −
2 (q B R)2 . c2
This relation is written as
4E 32 [(q B R/c)2 + m 2X ] = A2 + m 4X − 2Am 2X ⇒ c4 2 4E 32 2E 3 4 2 mX − 2 + A m − (q B R/c)2 = 0, (10.21) X c4 c4 which is a quadratic equation in the unknown m X , all other quantities being known. Second Solution In this solution we assume that the particles K − and p + have zero speed at the reaction, therefore γ K = γ p = 1. Then conservation of fourmomentum in the laboratory gives
mK p
L
mp + −p
L
m X γX = m X γX β X
L
m π γπ + m π γπ βπ
⇒ L
m K + m p = m X γ X + m π γπ , m X γ X β X = m π γπ βπ .
(10.22) (10.23)
Multiplying the first with β X and adding we find βX =
m π γπ βπ . m p + m K − m π γπ
Then #− 1 " γ X = 1 − β X2 2 =
m K + m p − m π γπ
(m p + m K − m π γπ )2 − (m π γπ βπ )2 m K + m p − m π γπ = . (m K + m p − 2m π γπ )(m K + m p ) + m 2π
From (10.22) we have m K + m p − m π γπ = (m K + m p − 2m π γπ )(m K + m p ) + m 2π . γX (10.24) The only extra unknown is γπ . This we compute from the data of the trajectory. The equation of motion of the π − gives as in the first solution mX =
m π γπ
βπ2 c2 q B R2 = q B Rβπ c ⇒ βπ γπ = , R mπ c
10.5
Reaction of the Form 1 + 2 → 3 + 4
285
from which follows γπ2 = 1 + βπ2 γπ2 = 1 +
q B R2 mπ c
2 .
Replacing in (10.24) and solving we find 3⎛ ⎞ $ 4 4 q B R2 2⎠ 4⎝ m X = 5 m K + m p − 2m π 1 + (m K + m p ) + m 2π . mπ c [Verify that the above expression is an acceptable root of the quadratic equation (10.21)]. 10.5.8. e+ + e− → γ + γ . 1 2 3 4 In Fig. 10.7 we show the reaction in the laboratory frame L and in the CM frame Because in the CM frame the photons are emitted normal to the direction of motion of the bullet particle e+ , in the CM they must have equal threemomenta and energies (why?). Let E ∗ be the common energy of the photons in the CM frame. Then p3i + p4i =
2E ∗ /c 0
⇒ ( p3i + p4i )2 = − CM
4E ∗2 . c2
In the laboratory frame L we have ( p1i
+
p2i )2
=
p12
+
p22
+
2 p1i p2i
Fig. 10.7 Two photon final states
= −2m c + 2 2 2
E c
p
mc 0
= −2m 2 c2 − 2Em.
286
10 Relativistic Reactions
Conservation of fourmomentum gives ( p1i + p2i )2 = ( p3i + p4i )2 . We have computed the invariant ( p1i + p2i )2 in L and the invariant ( p3i + p4i )2 in the CM frame. Replacing we find 4E 2∗ = 2m 2 c2 + 2Em ⇒ c2 ∗
E =
1+γ mc2 , 2
where γ is the γ factor of e+ in the laboratory. We infer that the fourmomentum of the photons in the CM is ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 1 1 1 ⎜ ⎟ ⎜ ⎟ ⎟ E∗ ⎜ ⎜ 0 ⎟ = 1 + γ mc ⎜ 0 ⎟ , pi = 1 + γ mc ⎜ 0 ⎟ . p3i = 4 ⎝ ⎝ ⎝ ⎠ ⎠ 1 1 −1 ⎠ c 2 2 0 CM 0 CM 0 CM In order to calculate the fourmomentum of the photons in L we use the boost relating the CM with L . The velocity factor βCM of the CM in the laboratory equals β CM
$ √ , p1 + p2 E 2 − m 2 c4 E − mc2 pc = , = c= i= i 2 E E1 + E2 E + mc E + mc2
and the γ factor2 γCM =
2
E1 + E2 E + mc2 . = √ Mc2 mc2 2(1 + γ )
To calculate the γ factor if the CM we use the formula γ =
Total energy Mass c2
The total energy is E 1 + E 2 and the mass M is −Mc2 = ( p1 + p2 )2 = ( p3 + p4 )2 = −
4E ∗2 ⇒ Mc2 = 2E ∗ . c2
10.5
Reaction of the Form 1 + 2 → 3 + 4
287
⎛
⎛ 4 ⎞ ⎞ E L /c E L3 /c ⎜ px3 ⎟ ⎜ px4 ⎟ i i ⎜ ⎟ ⎟ Let p3 = ⎜ ⎝ p 3y ⎠ , p4 = ⎝ p 4y ⎠ the fourmomenta of the photons in L . 0 0 L L Then E L3 = γCM (E ∗ − βCM 0) = γCM E ∗ = E L4 . Therefore, the energies of the photons in L are equal to E L3 = E L4 = γCM E ∗ =
1 (E + mc2 ). 2
It remains to calculate the angles θ, φ. We have tan θ =
p 3y px3
=
p 3∗ y γCM ( px3∗ −
∗ βCM Ec )
=
E ∗ /c 1 . =− ∗ γCM βCM γCM (0 − βCM Ec )
The RHS is independent of the photon, hence tan θ = − tan ϕ = −
1 . γCM βCM
Instead of the tan of the angles we compute the sin . We find sin θ = √
tan θ 1+
tan2
γCM βCM γCM βCM =− = −βCM = − γCM 2 2 θ 1 + γCM βCM
which implies $ θ = −ϕ = − arcsin
E − mc2 . E + mc2
We see that the photons are emitted symmetrically about the direction of motion of
E−mc π the bullet particles e+ in L at an angle arcsin E+mc 2 < 2 . We note that the bigger + the energy of the e in L, the more the angle between the directions of the emitted photons tends to π. (b) The photons are emitted at an angle π4 in the laboratory when 2
$
E π4 − mc2 E π4 +
mc2
= sin
1 π =√ . 4 2
288
10 Relativistic Reactions
Solving for E we find E π4 = 3mc2 . 10.5.9. (a) Suppose that the following reaction is possible γ → e− + e+ . Then conservation of fourmomentum gives in the CM frame of e− , e+ p∗γ = p∗e− + p∗e+ = 0 which is impossible because the photon has nonvanishing momentum in all frames. We consider a nucleus N and the reaction γ + N → e− + e+ + N . 1 2 3 4 5 Conservation of fourmomentum gives ( p1i + p2i )2 = ( p3i + p4i + p5i )2 . We calculate the LHS in the proper frame of the nucleus and the RHS in the CM. We have ( p1i + p2i )2 = −m 2N c2 − 2Em N , where E is the energy of the photon in the proper frame of the nucleus. For the RHS we have ( p3i + p4i + p5i )2 =
t 2 (E 3∗ + E 4∗ + E 5∗ )/c, p∗3 + p∗4 + p∗5 = −(E 3∗ +E 4∗ +E 5∗ )2 /c2 ,
because in the CM p∗3 + p∗4 + p∗5 = 0. Replacing in the conservation equation and solving for E we find E=
(E 3∗ + E 4∗ + E 5∗ )2 − m 2N c4 . 2m N c2
The minimum value of E occurs at the threshold of the reaction, that is, when the kinetic energy of each of the products in the CM vanishes (they are at rest). Then E 3∗ = E 4∗ = m e c2 ,
E 5∗ = m N c2
10.5
Reaction of the Form 1 + 2 → 3 + 4
289
and E=
(2m e c2 + m N c2 )2 − m 2N c4 2m e (m e + m N ) 2 2m 2e 2 = c = 2m e c2 + c . 2 2m N c mN mN
When m N m e the E min = 2m e c2 and when m N m e the E min = 4m e c2 . The physical meaning of the above limits is the following. In the first case the “heavy” nucleus rebounds with small speed absorbing little energy while in the second case the “light” electron (the catalyst) absorbs double energy because it rebounds with higher speed. (b) We have to show that the following reactions is impossible to occur e+γ →e
e → γ + e.
Because they are similar it suffices to show that one of them, the first say, is impossible. Conservation of fourmomentum gives pei + pγi = pei . Squaring we find −m 2e = −m 2e + 2 pei pγ i ⇒ pei pγ i = 0. In the proper frame of the electron Σ+ e , say, we have the following decomposition of the fourmomenta pγi =
E p
Σ+ e
,
pei =
mc 0
Σ+ e
,
hence pγi pei = 0 ⇒ E = 0, which is impossible because the energy of the photon does not vanish. 10.5.10. We have the following picture for the scattering in the laboratory and in the CM frame The β ∗ factor of the CM in the laboratory is , p1 + p2 pc ∗ β = , = c= E E1 + E2
E 21 − m 2 c4
E 1 + Mc2
eˆ
290
10 Relativistic Reactions
Fig. 10.8 Disintegration in the laboratory and in the CM
and the γ ∗ factor 1 E 1 + Mc2 γ∗ = = . 1 − β ∗2 (m 2 + M 2 )c4 + 2E 1 Mc2 Let p ∗ be the common threemomentum of the particles after the collision in the CM. The fourmomenta of the daughter particles in the laboratory and in the CM have the following components ⎞ ⎛ ⎞ ⎞ ⎛ ⎞ ⎛ E 3 /c E 3∗ /c E 4 /c E 4∗ /c p3i = ⎝ p3 ⎠ , ⎝ p ∗ cos θ ∗ eˆ ⎠ p4i = ⎝ p4 ⎠ , ⎝ − p ∗ cos θ ∗ eˆ ⎠ . p3⊥ L p ∗ sin θ ∗ eˆ ⊥ CM − p ∗ sin θ ∗ eˆ ⊥ CM −p4⊥ L ⎛
The boost relating the two frames gives E 3 = γ ∗ (E 3∗ + β ∗ cp ∗ cos θ ∗ ), p3 = γ ∗ ( p ∗ cos θ ∗ eˆ + β ∗ E 3∗ /c), p3⊥ = p ∗ sin θ ∗ eˆ ⊥ . The angle φ1 is given by the relation tan ϕ1 = But β3∗ =
p∗ c E 3∗
sin θ ∗ p3⊥ p ∗ sin θ ∗ = . = ∗ ∗ ∗ p3 γ ( p cos θ ∗ + β ∗ E 3 /c) γ ∗ (cos θ ∗ + β ∗ E 3∗ / p ∗ c)
where β3∗ is the βfactor of particle 3 in the CM. Hence tan ϕ1 =
sin θ ∗ . γ ∗ (cos θ ∗ + β ∗ /β3∗ )
It remains to compute β3∗ in terms of the energy E 1 and the angle θ ∗ . Squaring the conservation of fourmomentum we obtain
10.5
Reaction of the Form 1 + 2 → 3 + 4
291
p1i p2i = p3i p4i . Replacing components we calculate ⎛
⎞ ⎛ ⎞ E 1 /c Mc p1i · p2i = ⎝ p1 ⎠ ⎝ 0 ⎠ = −E 1 M, 0⊥ 0⊥ L L # 1 1" i E ∗2 ( p3 + p4i )2 − ( p3i )2 − ( p4i )2 = − 2 + m 2 c2 + M 2 c2 2 2 c # 1 " 2 = 2 (m + M 2 )c4 − E ∗2 , 2c
p3i p4i =
where E ∗2 is the total energy of the particles 3, 4 in the CM. The conservation equation gives3 E ∗2 = (m 2 + M 2 )c4 + 2E 1 Mc2 . But 2 p ∗2 c2 + m 2 c4 + p ∗2 c2 + M 2 c4 = 2 p ∗2 c2 + (m 2 + M 2 )c4 + 2 p ∗2 c2 + m 2 c4 p ∗2 c2 + M 2 c4
E ∗2 = (E 3∗ + E 4∗ )2 =
from which follows p ∗2 c2 + m 2 c4 p ∗2 c2 + M 2 c4 = M E 1 c2 − p ∗2 c2 . Squaring and making standard calculations we find the equation # " 2 (m + M 2 )c4 + 2M E 1 c2 p ∗2 c2 = M 2 E 12 c4 − m 2 M 2 c8 ⇒
3
The same result follows if we note that the fourmomentum of the particle in the CM has the form ⎞ ⎛ ∗ ⎞ ⎛ E /c (E 1 + Mc2 )/c ⎠ , ⎝ 0 ⎠ , ⎝ p1 0⊥ 0⊥ L CM and the boost equations give E 1 + Mc2 = γ ∗ E ∗ ⇒ E ∗ =
(m 2 + M 2 )c4 + 2E 1 Mc2 .
292
10 Relativistic Reactions
p ∗2 =
M 2 (E 12 − m 2 c4 ) . (m 2 + M 2 )c2 + 2M E 1
The energy E 3∗ =
p ∗2 c2 + m 2 c4 .
Therefore, the βfactor of particle 3 β3∗ =
p∗ c p∗ c = E 3∗ p ∗2 c2 + m 2 c4
is known in terms of the energy E 1 and the masses of the particles. Finally, the scattering angle φ1 is given by the relation tan ϕ1 =
sin θ ∗ √ √ ∗2 2 2 4 . p c +m c E 1 −m 2 c4 ∗ ∗ γ cos θ + E1 +Mc2 p∗ c
10.5.11. We have the following picture for the reaction in the laboratory and in the CM
The fourmomenta of the particles before and after the collision in the laboratory L and in the CM are Before Lab ⎛ ⎞ E 1 /c 1: ⎝ p1 ⎠ 0 ⎛ ⊥⎞ L Mc 2: ⎝ 0 ⎠ 0⊥ L
After CM Lab CM ⎛ ∗ ⎞ ⎞ ⎛ ∗ ⎞ ⎛ E 1 /c E 3 /c E 3 /c ⎝ p∗1 ⎠ 3: ⎝ p3 ⎠ ⎝ p∗ ⎠ ∗ p ⎛ 0∗⊥ ⎞CM ⎛ 3⊥ ⎞L ⎛ p ∗⊥ ⎞CM E 2 /c E 4 /c E 4 /c ⎝ −p∗1 ⎠ 4: ⎝ p4 ⎠ ⎝ −p∗ ⎠ −p3⊥ L −p∗ ⊥ CM 0⊥ CM
10.5
Reaction of the Form 1 + 2 → 3 + 4
293
We calculate the velocity factors of the CM in L . We have , p1 + p2 pc β∗ = , = c= E E1 + E2
E 21 − m 2 c4
E 1 + Mc2
e ,
1 E 1 + Mc2 = . γ∗ = 1 − β ∗2 (m 2 + M 2 )c4 + 2E 1 Mc2
(10.25)
(10.26)
Next we consider the conservation of fourmomentum. We find p1i · p2i = p3i · p4i . We compute the inner products in L . We have ⎛
⎞ ⎛ ⎞ E 1 /c Mc p1i · p2i = ⎝ p1 ⎠ ⎝ 0 ⎠ = −E 1 M, 0⊥ 0⊥ L L p3i · p4i =
E 3 /c p3
L
E 4 /c p4
=− L
E3 E4 + p3 p4 cos θ, c2
where p3 = p3 , p4 = p4 . Replacing in the conservation equation we find cos θ =
E 3 E 4 − E 1 Mc2 . p3 p4 c 2
(10.27)
We note that for the calculation of cos θ the energies E 3 , E 4 are required. Conservation of energy gives Mc2 + E 1 = E 3 + E 4 , from which follows that only one of the E 3 , E 4 is required. We compute E 3 in terms of the quantities E 3∗ , p3∗ , cos θ ∗ using the boost relating the frames L , CM. The zeroth component gives E 3 = γ ∗ (E 3∗ + β ∗ cp3∗ cos θ ∗ ),
(10.28)
∗ where γ ∗ is given by (10.26). Because θ is assumed to be known it is enough to compute p3∗ because E 3∗ = p3∗2 c2 + m 2 c4 . In the CM we compute (p∗3 = −p∗4 )
p3i
·
p4i
=
E 3∗ /c p∗3
CM
E 4∗ /c −p∗3
=− CM
E 3∗ E 4∗ − p3∗2 . c2
294
10 Relativistic Reactions
Using the conservation equation we find E 3∗ E 4∗ = − p3∗2 c2 + E 1 Mc2 . Replacing the energies in terms of the threemomentum we have ( p3∗2 c2 + m 2 c4 )( p3∗2 c2 + M 2 c4 ) = (− p3∗2 c2 + E 1 Mc2 )2 ⇒
p3∗2 =
E 12 M 2 − m 2 M 2 c4 . (m 2 + M 2 )c2 + 2E 1 M
(10.29)
Replacing in (10.28) we find for the energy E 3 ⎡ E3 =
E 1 + Mc
2
(m 2 + M 2) c4 + 2E 1 Mc2
⎣ p3∗2 c2 + m 2 c4 +
E 12 − m 2 c4
E 1 + Mc2
⎤ cp3∗ cos θ ∗ ⎦ . (10.30)
Finally, replacing in (10.27) we get cos θ =
E 3 (E 1 + Mc2 − E 3 ) − E 1 Mc2 , E 32 − m 2 c4 (E 1 + Mc2 − E 3 )2 − M 2 c4
where the energy E 3 is given from (10.30) and p3∗ from (10.29). Application We replace m = 0, M = m e in (10.29) and have p3∗2 =
E 12 m 2e . m e 2 c2 + 2E 1 m e
Similarly (10.30) gives E 1 + m e c2 E1 ∗ E3 = p3∗ c 1 + cos θ E 1 + m e c2 m e 2 c4 + 2E 1 m e c2 E1 E 1 m e (E 1 + m e c2 ) ∗ 1+ cos θ = m e 2 c2 + 2E 1 m e E 1 + m e c2 E 1 (E 1 + m e c2 ) E1 ∗ = 1 + . cos θ m e c2 + 2E 1 E 1 + m e c2
10.5
Reaction of the Form 1 + 2 → 3 + 4
295
Finally cos θ =
E 3 (E 1 + m e c2 − E 3 ) − E 1 m e c2 . E 3 (E 1 + m e c2 − E 3 )2 − m e 2 c4
(10.31)
When E 1 = m e c2 we find E3 =
E1 2 + cos θ ∗ . 3
Replacing in (10.31) it follows after a standard calculation: cos θ =
−1 − cos2 θ ∗ + 2 cos θ ∗ . (2 + cos θ ∗ ) (4−)2 − 9
In order the quantity under the root to be positive the cos θ ∗ = 1 ⇒ θ ∗ = 0. 10.5.12. Without restricting generality we assume that in the laboratory the particle 1 moves along the xaxis with energy E and threemomentum p 2 c2 = E 2 − m 2 c4 . The fourmomenta of particles in the lab are ⎛
⎛ ⎞ ⎞ E/c mc ⎜ p ⎟ ⎜ 0 ⎟ i ⎜ ⎟ ⎟ p1i = ⎜ ⎝ 0 ⎠ , p2 = ⎝ 0 ⎠ . 0 0 L L In order to exploit the assumption that the scattering is elastic, we transfer the four momenta in the CM frame. The βfactor of this frame in the lab is $ p B = x= E + mc2 ∗
E − mc2 x E + mc2
from which we compute4 Γ∗ = (1 − B∗2 )−1/2 =
E+mc2 . 2mc2
The calculation of Γ∗ can be done directly from the relation Γ∗ = Etotal = E+mc where M is M M the mass of the CM particle. The mass M is computed as the length of the total fourmomentum, that is, ⎛ ⎞2 E/c + mc ⎜ ⎟ p ⎟ = −2mc2 (E/c2 + m) ⇒ M = 2m(E/c2 + m). −M 2 c2 = ( p1i + p2i )2 = ⎜ ⎝ ⎠ 0 0 L
4
2
296
10 Relativistic Reactions
The common energy of particles 1, 2 in the CM frame is ∗
∗ 2
E = mΓ c =
1 2 mc (E + mc2 ) = mc2 2

E + mc2 = Γ∗ mc2 2mc2
and the (common) length of the threemomentum $ 1 mc2 1 ∗2 1 mc2 (E − mc2 ) = p E − m 2 c4 = p = c c 2 2(E + mc2 ) $ 1 2mc2 p = p= . 2 2 E + mc 2Γ∗ ∗
These are preserved because the scattering is assumed to be elastic. Assuming further that the scattering occurs in the x, y plane and the fact that p∗1 = −p∗2 in the CM frame we find for the fourmomenta in the CM frame after the collision ⎛ ⎛ ∗ ⎞ ⎞ E ∗ /c Γ mc ⎜ p ∗ cos φ ⎟ ⎜ p ⎟ ⎟ = ⎜ 2Γp∗ cos φ ⎟ p1i = ⎜ ∗ ⎝ p sin φ ⎠ ⎝ ∗ sin φ ⎠ 2Γ 0 0 CM ⎛ ⎛ ⎞ ⎞ E ∗ /c Γ∗ mc ⎜ − p ∗ cos φ ⎟ ⎜ − p ∗ cos φ ⎟ ⎜ 2Γ ⎟ ⎟ p2i = ⎜ ⎝ − p ∗ sin φ ⎠ = ⎝ − p ∗ sin φ ⎠ . 2Γ 0 0 CM We transfer these momenta to the lab frame. We work first with particle 1. We have ⎞ ⎛ ⎞⎛ ∗ Γ∗ B ∗ 0 0 Γ∗ Γ mc ⎟ ⎜ Γ∗ B ∗ ⎜ p Γ∗ 0 0 ⎟ ⎟ ⎜ 2Γp∗ cos φ ⎟ p1i = ⎜ ⎝ 0 0 1 0 ⎠ ⎝ 2Γ∗ sin φ ⎠ 0 0 0 0 1 CM ⎞ E + mc2 p 2 c2 1 + cos φ ⎜ ⎟ 2c (E + mc2 )2 ⎜ ⎟ p ⎜ ⎟ (1 + cos φ) ⎜ ⎟ ⎜ ⎟ , 2 $ = ... = ⎜ ⎟ ⎜ ⎟ mc2 ⎜ ⎟ p sin φ 2 ⎝ ⎠ 2(E + mc ) 0 L ⎛
p where we have used that B ∗ Γ∗2 = 2m . It follows that in the laboratory the threemomentum of scattered particle 1 has components
10.6
Other Reactions
297
$ p1,x
p = (1 + cos φ), 2
mc2 p sin φ 2(E + mc2 )
= p1,y
from which follows
2 p1,x
p
2 2(E + mc2 ) 2 −1 + p1,y = 1. p 2 mc2
We conclude that the tip of the threemomentum vector of particle 1 in the 2 p 2 p Euclidean plane with axes p1,x , p1,y traces an eclipse with center point (1, 0) and 2mc2 1 semiaxes a = 1, b = E+mc 2 = Γ∗ . Obviously the same result applies to the threemomentum of particle 2. We leave the details to the reader.
10.6 Other Reactions 10.6.1. The reaction is e + N → γ + e∗ + N ∗ . 1 2 3 4 5 Squaring the conservation equation of fourmomentum we find 2 2 2 p5i = ( p1i + p2i )2 + p3i + p4i − 2( p1i + p2i ) p3i − 2( p1i + p2i ) p4i + 2 p3i p4i ⇒
−M 2 c2 − ( p1i + p2i )2 = −m 2 c2 − 2( p1i + p2i )( p3i + p4i ) + 2 p3i p4i . The fourmomenta of the particles in the laboratory are p1i
=
p4i
=
E 1 /c p1 E 4 /c p4
,
p2i
=
L
, L
p5i
=
Mc 0
E 5 /c p5
,
p3i
L
=
E3 c
1 eˆ L
. L
The LHS gives −M 2 c2 + m 2 c2 + M 2 c2 − 2 p1i p2i = m 2 c2 + 2M E 1 . We set pi · p j = pi p j cos θi j (i, j = 1, 2, 3, 4)
(10.32)
298
10 Relativistic Reactions
and the RHS is written as " # − m 2 c2 − 2 p1i p3i + p1i p4i + p2i ( p3i + p4i ) + 2 p3i p4i E3 E3 E4 = −m 2 c2 + 2 − 2 + p4 cos θ34 c c E1 E3 E3 E1 E4 cos θ13 − 2 + p1 p4 cos θ14 − M(E 3 + E 4 ) − 2 − 2 + p1 c c c " # 2 = −m 2 c2 + 2 E 1 + Mc2 − E 4 + cp4 cos θ34 − cp1 cos θ13 E 3 c 2 + 2 E 1 E 4 − c2 p1 p4 cos θ14 + Mc2 E 4 . c Replacing in (10.32) and solving for E 3 we find E3 =
m 2 c4 + M E 1 c2 −(E 1 + Mc2 )E 4 + c2 p1 p4 cos θ14 . E 1 + Mc2 − E 4 + cp4 cos θ34 − cp1 cos θ13
We consider now that the nucleus is very heavy. This means that practically it will stay still after the collision, that is, p5 = p2 = 0. Then the conservation of threemomentum in the laboratory gives p1L = p3L + p4L =
E3 E3 eˆ + p4L ⇒ p4L = p1L − eˆ . c c
Next we assume that the electron is very fast so practically it does not lose energy during the collision, i.e., $ E 1 = E 4 ⇒ p4 =
E 12 − m 2 c 2 = p1 . c2
With the above simplifying assumptions we find the following relations for the angles of the threemomenta. The inner product p4L · p3L = p1L · p3L −
E3 eˆ · p3L c
⇒
p4 p3 cos θ43 = p1 p3 cos θ13 −
p4 cos θ43 = p1 cos θ13 −
E3 . c
Working similarly with the inner product p4L · p1L we find the relation p4 cos θ41 = p1 −
E3 cos θ13 . c
E3 p3 ⇒ c
10.6
Other Reactions
299
Replacing the expression for the energy E 3 we get E3 =
m 2 c4 + E 1 Mc2 + c2 p12 − c2 Ec3 p1 cos θ13 − (E 1 + Mc2 )E 1 E 1 + Mc2 − E 1 + cp1 cos θ13 − E 3 − cp1 cos θ13
=
m 2 c4 + c2 p12 − E 12 − cp1 E 3 cos θ13 cp1 E 3 cos θ13 =− ⇒ Mc2 − E 3 Mc2 − E 3 Mc2 E 3 − E 32 = − p1 cE 3 cos θ13 .
We assume E 3 = 0, therefore E 3 = c(Mc + p1 cos θ13 ). We infer that the minimum value of E 3 is Mc2 − p1 c
for
θ13 = π,
and the maximum value of E 3 is & & 2 & Mc + p1 c&
for θ13 = 0.
10.6.2. (a) The reaction is M→μ+μ+μ . 1 2 3 4 Squaring the conservation equation of fourmomentum we find ( p1i − p2i )2 = ( p3i + p4i )2 ⇒ −M 2 c2 − μ2 c2 − 2 p1i p2i = ( p3i + p4i )2 . In the proper frame Σ M of M p1i p2i = −2M E 2 , where E 2 is the energy of the particle 2 in Σ M . Replacing in the equation of conservation and solving for E 2 we find E2 =
M 2 c2 + μ2 c2 + ( p3 + p4 )2 . 2M
E 2 is maximum when the term ( p3i + p4i )2 is maximum (because ( p3i + p4i )2 < 0). This occurs when the threemomenta of the particles 3, 4 vanish in Σ M . Hence
300
10 Relativistic Reactions
( p3i + p4i )2 = −(μ + μ)2 c2 = −4μ2 c2 and E 2,max =
M 2 − 3μ2 2 c . 2M
But under this condition E 2,max = μγ2 c2 ⇒ γ2 =
M 2 − 3μ2 . 2μM
The velocity factor β2 corresponding to this energy is β22 = 1 − =
1 (2μM)2 (M 2 − 9μ2 )(M 2 − μ2 ) = 1 − = (M 2 − 3μ2 )2 (M 2 − 3μ2 )2 γ22
(1 − 9y 2 )(1 − y 2 ) , (1 − 3y 2 )2
that is, $ β2 = where we have set y = must be satisfied:
μ . M
(1 − 9y 2 )(1 − y 2 ) , (1 − 3y 2 )2
In order for β2 to be real the following condition
(1 − 9y 2 ) ≥ 0 ⇒ 1/3 ≥ y ⇒ M ≥ 3μ, which is true. (b) The reaction is M→m+m . 1 2 3 Conservation of energy in the proper frame of M gives E 1 + E 2 = M. Because the daughter particles have equal masses they will have equal speeds β in the proper frame of M. If E = mγ is their common energy, then 2mγ = M ⇒ γ =
1 M = . 2m 2x
10.6
Other Reactions
301
We want this γ to be equal to γ2 we calculated in part (a), therefore, the condition for this is M 2 − 3μ2 y M = ⇒x= . 2m 2μM 1 − 3y 2 10.6.3. (a) The reaction is k→π +π +π . 1 2 3 4 In the laboratory frame the fourmomenta are p1i =
mk c 0
, p2i = L
E 2 /c p2
, p3i = L
E 3 /c p3
, p4i = L
E 4 /c p4
. L
Conservation of fourmomentum gives p2 + p3 + p4 = 0, E 2 + E 3 + E 4 = m k c2 . The first relation implies that the threemomenta p2 , p3 , p4 are coplanar. Because pi ui (i = 2, 3, 4) the threevelocities and subsequently the orbits of the threepions are on the same plane. (b) Because the three 3momenta make equal angles one with another we infer that p2 = p3 = p4 = p. The energy E i (i = 2, 3, 4) of each of the pions is given by the relation E i = p 2 c2 + m 2π c4 ⇒ E 2 = E 3 = E 4 = E. Replacing in the conservation equation we find 3E = m k c2 ⇒ E =
m k c2 = 164.67 MeV. 3
The length of the threemomentum is p =
(164.67)2 − (140)2 MeV/c = 86.7 MeV/c.
But pc = m π γβ ⇒ β 2 =
p2 p2 c2
+m 2π
⇒ β = 0.53.
The distance l which is covered by a π meson before it disintegrates is l = ut, where t is the lifetime of the pion in the laboratory. If τ is the lifetime of the pion (in its proper frame) we find
302
10 Relativistic Reactions
2.6 × 10−8 t = γτ = √ = 3.07 × 10−8 s. 1 − 0.532 Hence l = (0.53c) × (3.07 × 10−8 s) = 4.9 m. 10.6.4. The reaction is k¯ + p → k¯ + p + nπ . 1 2 3 4 5 Conservation of fourmomentum gives
p1i + p2i = p3i + p4i +
n )
piA .
A=1
Squaring we find * ( p1i
+
p2i )2
=
p3i
+
p4i
+
n )
+2 piA
.
A=1
We compute the LHS in the laboratory frame (which is defined by the requirement p2 = 0) and the RHS in the CM frame (which is defined by the requirement n , piA = 0). We have p3 + p4 + A=1
( p1i
+
p2i )2
=
1 L E 1 + E 2L , p1 c
2 = p21 − L
2 1 L E 1 + E 2L . 2 c
Also * p3i + p4i +
n ) A=1
+2 piA
* * + +2 n n ) ) 1 = E CA M , p3 + p4 + pA E 3C M + E 4C M + c A=1 A=1 CM * +2 n ) 1 = − 2 E 3C M + E 4C M + E CA M . c A=1
10.6
Other Reactions
303
Replacing in the conservation equation we find
* E 3C M
* +2 n ) 1 1 2 E iC M ⇒ p21 − 2 E 1L + E 2L = − 2 E 3C M + E 4C M + c c i=1 +2 n ) 2 + E 4C M + E iC M = E 1L + E 2L − c2 p21 . i=1
The number n of π mesons is maximum when the kinetic energies of the products vanish in the CM (this is the threshold of the reaction). This implies 2 2 m k c2 + m p c2 + nm π c2 = E 1L + m p c2 − c2 p21 ⇒ n=
1 mπ
1/2 1 L 2 2 2 2 E . + m c − c p − m + m p k p 1 1 c2
Numerical application We calculate 2 E 1L = p21 c2 + m κ c2 = 102 + 0.4942 GeV = 10.012 GeV so that n=
9 #1/2 1 8" − (0.494 + 0.938) = 22.4. (10.012 + 0.938)2 − 102 0.135
We conclude that the maximum number of π mesons which can be produced in this reaction is n max = 22. 10.6.5. Conservation of energy in the laboratory gives m N + m U = m x + m y + 2E ⇒ E = 12 (m n + m U − m x − m y ). Hence, the γ factor of the daughter neutrons in the laboratory is γ =
mN + mU − m x − m y E = . mn 2m n
Introducing the values of the masses we calculate γ = 1.395.
304
10 Relativistic Reactions
The kinetic energy produced per atom of uranium is T2n = 2E − 2m n = 2γ m n − 2m n = 2(γ − 1)m n . The Avogadro number NA gives the number of atoms in 1 mole of the substance (1 mole is the quantity of the material equal to the atomic weight). For 5 kg of 235 U we have 5000 × NA = 1.26 × 1025 atoms 235 U. 238.03 Hence the total kinetic energy is T = 1.26 × 1025 × T2n . In order to calculate T in units of energy we change the mass of neutron m n from amu in units of energy. We have mn = = = =
1.00896 amu = 1.00896 × (1.66 × 10−27 ) kg 1.674 × 10−27 kg = 1.674 × 10−27 × 9 × 1016 J 1.507×10−10 8 1.507 × 10−10 J = 1.6021×10 −19 eV = 940.9 × 10 eV 940.9 MeV.
and T2n = 2 × (1.395 − 1) × 940.9 MeV = 743.3 MeV Finally, then T = 1.26 × 1025 × 7.433 × 102 MeV = 9.36 × 1027 MeV = 9.36 × 1027 × 106 × 1.602 × 10−19 J = 1.499 × 1015 J. This is the produced thermal energy. In kWh this is T = 1.499 × 1015
1 kWh = 4.16 × 108 kWh. 3, 6 × 106
10.7 Threshold 10.7.1. (a) The reaction is 1 + 2 → 1 + 2 + . . . .
10.7
Threshold
305
Conservation of fourmomentum gives ( p1i
+
=
p2i )2
* )
+2 piA
.
A
Both sides of this equation are invariant; hence, each can be computed independently of the other. We compute the LHS in the proper frame (L say) of the target particle and the RHS in the CM frame. We have LHS: * ( p1i
+
p2i )2
= −m c − m c +
2 p1i
= −(m + m )c −
2E 1L m,
2 2
2 2
2
2
2
·
p2i
= −m c − m c + 2 2 2
2 2
E 1L c p1
+
mc 0
L
RHS: * )
+2 piA
A
=
E ∗ /c 0
2
= −E ∗2 /c2 .
CM
E ∗ is the total energy of the particles 1 , 2 , . . . in the CM frame. Replacing in the conservation equation we find E 1L =
E ∗2 − (m 2 + m 2 )c4 . 2mc2
But E 1L = T1 + mc2 where T1 is the kinetic energy of the bullet particle in L , hence T1 =
E ∗2 − (m + m)2 c4 . 2mc2
The minimum value of T1 occurs when E ∗ takes its minimum value, that is, Mc2 , where M is the sum of the masses of the particles 1 , 2 , . . . Therefore, T1,thrd =
M 2 − (m + m)2 2 (M − m − m)(M + m + m) 2 c = c . 2m 2m
(b) We number the particles sequentially 1,2,3,4,5,6 p1i + p2i → p3i + p4i + p5i + p6i . The mass of the product particles is ( p3 + p4 + p5 + p6 )2 = −(4mc)2 . The threshold energy is Tthrd = 6mc2 = 6 GeV.
L
306
10 Relativistic Reactions
10.7.2. (a) The reaction is π + p→k+Λ . 1 2 3 4 The threshold occurs when the products of the reaction on the CM are produced at rest, that is, p∗3 = p∗4 = 0. Conservation of fourmomentum gives p1i + p2i = p3i + p4i ⇒ 2 2 p1i + p2i + 2 p1i p2i = − (m κ + m Λ )2 c2 ⇒ −m 2π c2 − m 2p c2 + 2 p1i p2i = − (m κ + m Λ )2 c2 . In the laboratory frame we have p1 = (E thrd /c, p) ,
p2 = m p c, 0 ⇒ p1i p2i = E thrd m p
from which follows E thrd =
(m κ + m Λ )2 − m 2π − m 2p 2m p
c2 .
Introducing the given masses we calculate E thrd = 897 MeV Tthrd = 756.6 MeV. (b) We consider again the conservation equation, solve in terms of p4 , and square in order to eliminate the kinematic data of the Λhyperon. We have 2 −m 2Λ c2 = p1i + p2i − p3i = −m 2π c2 −m 2p c2 −m 2κ c2 +2 p1i p2i −2 p1i p3i −2 p2i p3i . We compute the inner products in the laboratory system (L). We have in that system p1iL = (E/c, p) L , p2iL = m p c, 0 , p3iL = E /c, p L . Hence p1i p2i = p1Li p2iL = −Em p , p1i p3i = p1Li p3iL = p · p − E E , p2i p3i = p2Li p3iL = −E m p . Because we are looking for the minimum of the energy of the kmeson normal to the direction of motion of the pion we consider p · p = 0. Replacing in the conservation equation follows
10.7
Threshold
307
−m 2Λ c2 = − m 2π + m 2p + m 2κ c2 − 2m p E + 2E E + 2E m p ⇒ 2 m Λ − m 2π − m 2p − m 2κ c2 + 2E m p E= . 2 m p − E The minimum value of E occurs when E takes its minimum value. [Note: Although this is obvious (because the numerator becomes minimum when the denominator becomes maximum) one can prove this assertion formally by computing the derivative d E/d E and show that it does not vanish when E > 0 and furthermore that the second derivative d 2 E/d E 2 is positive for E > 0.] The minimum value of E is m k c2 . Therefore, in the laboratory frame E min
2 m Λ − m 2π − m 2p − m 2k + 2m p m k = . 2 m p − mk
Introducing the given masses we find E min = 1149 MeV. Then Tmin = E min − m π c2 1008.7 MeV. (c) The kinetic energy of the kmeson in the laboratory when the π meson in the laboratory has kinetic energy 1008.7MeV is obviously zero (why?). Second Solution We consider the intermediate CM particle whose mass we assume to be M and write π + p → M → k + Λ. The masses of the particles at threshold for the reaction M → k + Λ considered as −k + (−Λ) → (−M) are M, m k , m Λ ; hence, the vanishing of the triangle function λ(m 2 , m 21 , m 22 ) gives λ(m 2 , m 2k , m 2Λ ) =
1 2 2 2 2 m = 0 ⇒ M = mk + mΛ. − m − m − 4m m k Λ k Λ m2
Conservation of fourmomentum for the reaction π + p → M gives 2 −M 2 = pπi + pip = −m 2π − m 2p + 2 pπi · pip . In the proper frame Σ+p of the proton the invariant t pπi · pip = (E thrd , p)tΣ+π m p , 0 Σ+p = −m p E thrd .
308
10 Relativistic Reactions
Replacing we find E thrd =
(m κ + m Λ )2 − m 2π − m 2p 2m p
as before. [Note: In the second solution we see the role played by the CM particle. That is, it is possible to reduce a reaction of many particles to a number of reactions of the form A + B → C. This is important because it is much easier to compute the various quantities involved for this type of simple reaction.] 10.7.3. (a) The reaction is 1 + 2 → 1 + 2 + 3. Conservation of fourmomentum gives ( p1i + p2i )2 = ( p1i + p2i + p3i )2 . We calculate the RHS in the CM frame. In this frame at threshold the threemomenta of the products are zero, hence, ( p1i + p2i + p3i )2 = −(M + N + m)2 c2 . We calculate the ( p1i + p2i )2 in the proper frame of the target (laboratory frame). In this frame we have p1i = (N c, 0) ,
p2i = (Mγ c, p) ,
where γ is the required velocity factor of the bullet particle. Then ( p1i + p2i )2 = −N 2 c2 − M 2 c2 − 2N Mγ c2 . Replacing in the conservation equation and solving for γ we find γ =1+
m2 m(N + M) + . NM 2N M
(b) In order to calculate the efficiency of the reaction we compute the energies of particles 2, 3 in the laboratory at the threshold of the reaction. The kinetic energy of the bullet particle 2 in the laboratory is Mγ c2 − Mc2 = (γ − 1) Mc2 .
10.7
Threshold
309
The energy required for the production of the particle 3 at the CM frame is mc2 . Therefore k=
mc2 m = 2 m(N +M) (γ − 1)Mc + NM
k= 1+
1 2M+m 2N
m2 2N M
⇒ M
.
We note that when N → ∞, k → 1, that is, for target particles with large mass the efficiency of the reaction tends to 100% (all kinetic energy of the bullet particle is transformed into mass of the produced particle). Application Consider the reaction γ → e+ + e. . 1 3 4 Conservation of fourmomentum gives 2 p2i = ( p1i − p3i )2 ⇒ −m 2 c2 = −m 2 c2 + 2m E photon = 0
which is absurd. Therefore, the reaction is impossible. Consider the reaction γ + N → e+ + e + N . 1 2 3 4 5 Conservation of fourmomentum gives p1i + p2i = p3i + p4i + p5i ⇒ ( p1i + p2i )2LAB = ( p3i + p4i + p5i )2CM . At the threshold of the reaction we have −N 2 c2 − 2N E thrd = − (m + m + N )2 c2 ⇒ 2m(m + N ) 2 E thrd = c . N The efficiency k of the reaction at the threshold is k=
Energy of e+ , e− in the CM Kinetic energy at threshold in the laboratory
310
10 Relativistic Reactions
Replacing we find k=
2mc2 = E thrd
2mc2 2m(m+N ) 2 c N
=
1 1+
m N
.
We note that when N m, k → 1, i.e., the efficiency tends to 100%. 10.7.4. The reaction is γ +N →N +π . 1 2 3 4 Conservation of fourmomentum gives ( p1i + p2i )2 = ( p3i + p4i )2 . The LHS in the laboratory gives
( p1i
+
p2i )2
E 1 /c = + · = +2 (E 1 /c)ˆe E1 E2 E 1 p2 2 2 cos θ , = −m N c + 2 − 2 + c c −m 2N c2
2 p1i
p2i
−m 2N c2
L
E 2 /c p2
L
where θ is the collision angle. The RHS in the CM frame (which is different for the laboratory frame) gives ( p3i + p4i )2 =
(E 3∗ + E 4∗ )/c 0
2 =− CM
(E 3∗ + E 4∗ )2 , c2
where E 3∗ , E 4∗ are the energies of the products in the CM. Replacing in the conservation equation we find E1 =
−m 2N c4 + (E 3∗ + E 4∗ )2 . 2(E 2 − p2 c cos θ )
Because E 2 > p2 c the energy E 1 is minimum when the term (E 3∗ + E 4∗ )2 is minimum, that is, the products are at rest in the CM. Then E 3∗ = m 3 c2 , E 4∗ = m 4 c2 , and the required threshold energy is

Fig. 10.9 Parallel and headon collision
10.8
General Problems on Reactions
311
E 1,thrd =
m π (m π + 2m N ) 4 c . 2(E 2 − p2 c cos θ )
We have still to calculate p2 c. We have
p2 2 c2 + m 2N c4 = T2 + m N c2 ⇒ ( p2 c)2 = T2 (T2 + 2m N c2 ) = 2400 MeV.
The angle θ = 0◦ or 180◦ depending on the relative direction of motion of the photon and the nucleon (see Fig. 10.9). For each case we compute E 1,thrd (θ = 0) = 806.25 MeV , E 1,thrd (θ = 180) = 32.25 MeV. We note that the threshold energy for headon collision is of the energy required for parallel collision.
32.25 806.25
= 0.04, that is, 4%
10.8 General Problems on Reactions 10.8.1. (a) The transformation of the volume element is x, y, z d V(r,θ,φ) dV = J r, θ, φ where J
x, y, z r, θ, φ
& ∂x & & ∂r & = & ∂∂ry & ∂z & ∂r
∂x ∂θ ∂y ∂θ ∂z ∂θ
∂x ∂φ ∂y ∂φ ∂z ∂φ
& & & & & = r 2 sin θ dr dθ dφ. & &
We define the solid angle dΩ in the direction θ with the relation dΩ = sin θ dθ dφ. Then the element of volume is written as d V = r 2 dr dΩ. (b) Consider two LCF Σ, Σ moving in the standard configuration with velocity u = u xˆ and let v, v the velocities of a photon in Σ, Σ , respectively. Considering spherical coordinates in both Σ, Σ we have vx = c cos θ vx = c cos θ v y = c sin θ cos φ v y = c sin θ cos φ vz = c sin θ sin φ vz = c sin θ sin φ .
312
10 Relativistic Reactions
The boost relating Σ, Σ gives vx + u , Q 1 vy , vy = γ Q 1 vz , vz = γ Q
vx =
where Q = 1 +
uvx c2
. We note that vz vz = ⇒ tan φ = tan φ ⇒ φ = φ . vy vy
For the angle θ we have cos θ =
vx − u cos θ − β vx cos θ − β = = = c Qc 1 − β cos θ 1 − c coscθ−u
(10.33)
from which follows 1 1 − cos2 θ = 1 − β 2 − (1 − β 2 ) cos2 θ 1 − β cos θ sin θ (1 − β 2 ) sin θ = . = 1 − β cos θ γ (1 − β cos θ )
sin θ =
From (10.33) we find by taking the differential 1 − β2 sin θ dθ ⇒ sin θ dθ = − (1 − β cos θ )2 γ (1 − β cos θ ) 1 dθ. dθ = γ (1 − β cos θ )
− sin θ dθ = −
Therefore dΩ = sin θ dθ dφ = dΩ = Note: Verify that directions.
'
γ 2 (1
1 sin θ dθ dφ ⇒ γ 2 (1 − β cos θ )2
1 dΩ. − β cos θ )2
dΩ =
'
(10.34)
dΩ = 4π where the integration is taken over all
10.8
General Problems on Reactions
313
Application Let dΩ be a solid angle for the observer on the Earth and let d N be the number of stars which are visible (assuming the same fixed luminosity) in this angle. Then the isotropic distribution of the stars implies N0 dN = constant = , dΩ 4π where N0 is the number of stars in the night sky. Let Σ be another LCF which moves with high speed wrt the Earth. Then the distribution of stars in the corresponding direction in the sky is (obviously the total number of stars is independent of the observer) 1 dN d N dΩ N0 = = . 2 dΩ dΩ dΩ 4π γ (1 − β cos θ )2
Chapter 11
Electromagnetic Field
11.1. (a) The equation of motion of the charge in Σ is dp dp = qv × B ⇒ p· = qp · (v × B)= 0 ⇒ p2 = const. ⇒ dt dt E 2 = p2 c2 + m 2 c4 = const. But E = mγ c2 ⇒ γ = const. ⇒ v2 = const. (b) From (a) we have p = mγ v = Ev/c2 . Replacing in the equation of motion we get dv qc2 = (v × B). dt E The xcomponent gives dvx = 0 ⇒ vx = constant = vx (0) = 0. dt Therefore, the motion is taking place in the plane y − z with constant speed 2 v20 = v 2y,0 + vz,0 = v 2y + vz2 . The y, zcomponents are dv y qc2 = vz B, dt E dvz qc2 = (−v y B). dt E Multiplying the second with i and adding we get d q Bc2 (v y + ivz ) = (vz − iv y ) = −iω(v y + ivz ), dt E 315
316
11 Electromagnetic Field
where ω =
q Bc2 . E
The solution of this differential equation is
v y + ivz = α0 e−iωt+α
(α0 , α = constants).
The real and the imaginary parts give v y = α0 cos(ωt + α),
vz = −α0 sin(ωt + α)
from which follows v 2y + vz2 = α02 ⇒ α0 = v0 . Integrating once more we find v0 sin(ωt + α), ω v0 cos(ωt + α). z = z0 + ω
y = y0 +
Eliminating the time we find the equation of the trajectory (y − y0 )2 + (z − z 0 )2 =
v02 , ω2
from which follows that the orbit of the charge is a circle centered at r (0) = v0 E y0 j + z 0 k of radius r = vω0 = qc 2B . 11.2. (a) The event of measurement of the electromagnetic field in Σ and Σ has coordinates (1, 2, 3, 4)Σ and (−1/4, 7/4, 3, 4)Σ , respectively. These coordinates are related with the boost relating Σ, Σ . Therefore, −
1 = γ (1 − 2β), 4 7 = γ (2 − β), 4
from which follows β = 35 , γ = 45 . (b) The electric field in Σ is E = Ej and the velocity of Σ wrt Σ is u = the equations of transformation of the fields E = E , B = B ,
E ⊥ = γ (E⊥ + u × B), 1 B ⊥ = γ (B⊥ − 2 u × E), c
we compute E = γ Ej and B = − 5c3 γ Ek.
3c i. 5
From
11 Electromagnetic Field
317
11.3. We consider the LCF Σ whose velocity wrt Σ is β = βci (i.e., normal to the plane defined by the fields B, E). The equations of transformation of the fields E, B give Ex = Ex Bx = Bx E y /c = γ (E y /c − β Bz ) B y = γ (B y + β E z /c) E z /c = γ (E z /c + β B y ) Bz = γ (Bz − β E y /c). The condition we have is E B ⇔ E × B = 0. We compute E × B = (E y Bz − B y E z )i ⇒ E y Bz − B y E z = 0. Using the transformation equations this condition in Σ becomes γ
Ey Ey Ez Ez − β Bz γ Bz − β − γ By + β γ + β B y = 0. c c c c
Rearranging terms we find β 2 − 2kβ + 1 = 0, where 2k =
1 c2
E 2 +B 2
 1c E×B
> 0. The accepted root of this equation is β=k−
k 2 − 1.
Because β must be real we have the condition k>1⇒
& & & &1 1 2 2 & E × B& . E + B > 2 & & 2 c c
Finally, the velocity of Σ in Σ is u =c k − k 2 − 1 i.
Now we observe that in all LCF which move wrt Σ with velocity v = λ EE the fields 1c E, B are parallel. The velocity v of these LCF wrt Σ is found from the relativistic rule for the composition of threevelocities. We have then v=
u · v 1 , v +u (γ − 1) + γ u u γu Q u2
318
11 Electromagnetic Field
where Q = 1 −
u·v . c2
But u · v = 0 hence Q = 1 and v=
1 λ (λˆe + uγu ) = u+ eˆ , γu γu
where eˆ is the unit along the common direction of the fields E , B in Σ . Second Solution The electromagnetic field tensor in Σ is ⎞ 0 0 − 1c E y − 1c E z ⎜ 0 0 −B y Bz ⎟ ⎟. [Fi j ] = ⎜ ⎝ E y Bz 0 0 ⎠ 0 Bz −B y 0 ⎛
Consider the LCF Σ with velocity β = βci. If L ii is the Lorentz transformation which relates Σ and Σ we have j
j
[Fi j ] = [L ii L j Fi j ] = [L ii ][Fi j ][L j ]t . Matrix multiplication gives [Fi j ] ⎛
⎞ 0 0 −γ 1c E y + βγ Bz −γ 1c E z − βγ B y ⎜ 0 0 −γ Bz + βγ E y γ B y + βγ 1c E z ⎟ ⎟. =⎜ 1 ⎝ γ E y − βγ Bz ⎠ γ Bz − βγ E y 0 0 c 0 0 γ 1c E z + βγ B y −γ Bz − βγ 1c E z
It follows 1 1 1 E = 0, γ E y − βγ Bz , γ E z + βγ B y c c c 1 B = 0, γ B y + βγ E z , γ Bz − βγ E y c and the solution continues as before. Third solution We consider again the LCF Σ . The Lorentz transformation of the fields E, B gives 1 E =γ c
1 E+β × B , c
1 B =γ B−β × E . c
The condition E B ⇔ E × B = 0. We replace and find
11 Electromagnetic Field
319
1 1 E × B γ2 c 1 1 = E+β × B × B − β × E c c 1 1 1 1 = E × B− E × β × E − B × (β × B) − (β × B) β × E = 0. c c c c Using the identity of the vector calculus A × (B × C) = (A · C)B − (A · B)C and taking into account that β·B =β·E = 0 we compute (β×B) × B = −B 2 β, 1 1 E2 E × β × E = − 2 β, c c c 1 1 E × B β 2. (β × B) × (β × E = c c Replacing in the condition of parallelism we obtain (1 + β 2 )
1 1 E × B − β 2 E 2 + B 2 = 0. c c
From this relation follows that β = β eˆ where eˆ = 1 E × B) c
1 c E×B  1c E×B
(= the unit along the
and the speed β is the acceptable solution (under the direction of the vector condition β < 1) of the quadratic equation β2 −
1 c2
E 2 + B2
 1c E × B 
β + 1 = 0.
11.4. (a) We have in an obvious notation (Greek indices take the values 1, 2, 3) Ω j = (− Φc , Aμ ). The tensor Fi j = Ω j,i − Ωi, j is antisymmetric (Fi j = −F ji ); therefore, it suffices to calculate the components for i > j. We have1
We calculate the components F01 , F02 , F03 as well as the components Fμν simultaneously. This is not necessary. For example, for the component Bx we have & & & i j k & & & B1 = (∇ × A)1 = && ∂ x ∂ y ∂z && = A3,2 − A2,3 = −(A2,3 − A3,2 ) = −F2 3 . & A1 A2 A3 & 1
1
320
11 Electromagnetic Field
1 1 1 ∂ Φ,μ + Aμ,0 = − (∇Φ − A)μ = − E μ c c ∂t c = −Aμ,ν + Aν,μ = −μνρ (∇ × A)ρ = −μνρ B ρ .
F0μ = −Ω0,μ + Ωμ,0 = Fμν = −Ωμ,ν + Ων,μ
The components Fi j of the tensor F can be written in the form of a matrix as follows: Ey Ez ⎞ Ex 0 c c c ⎜ − E x 0 −Bz B y ⎟ c ⎟ [Fi j ] = ⎜ ⎝ − E y Bz 0 −Bx ⎠ . c − Ecz −B y Bx 0
⎛
(b) The boost relating Σ, Σ is written in the form of a matrix ⎛
⎞ γ βγ 0 ⎠. L ii = ⎝ βγ γ 0 I2 The components Fi j of the tensor F in the LCF Σ are found from the relation j
Fi j = L ii L j Fi j . We find for the various values of the indices i , j j
j
j
F0 1 = L i0 L 1 Fi j = L 00 L 1 F0 j + L 10 L 1 F1 j = L 00 L 11 F0 1 + L 10 L 01 F1 0 = Ex Ex Ex Ex + βγβγ = −(γ 2 − β 2 γ 2 ) =− = −γ γ c c c c Ey Ey j j + βγ Bz = −γ − β Bz F0 2 = L i0 L 2 Fi j = L 00 F0 2 + L 10 L 1 F1 2 = −γ c c E Ez z i 3 0 3 1 3 − γβ B y = −γ + β By F0 3 = L 0 L 3 Fi3 = L 0 L 3 F0 3 + L 0 L 3 F1 3 = −γ c c Ey Ey + γ Bz = −γ β − Bz F1 2 = L i1 L 22 Fi2 = L 01 F0 2 + L 11 F1 2 = −γβ c c Ez Ez i 3 0 1 − γ B y = −γ β + By F1 3 = L 1 L 3 Fi3 = L 1 F0 3 + L 1 F1 3 = −γβ c c F2 3 = L 22 L 33 F23 = Bx .
Assuming that the tensor F in Σ is represented with a similar matrix in terms of the vector fields E = (E x , E y , E z ) and B = (Bx , B y , Bz ), we find the following transformation relations among the components of the threevectors E, B and E , B
11 Electromagnetic Field
321
⎧ ⎧ ⎨ Ex = Ex ⎨ Bx = Bx E y /c = γ (E y /c − β Bz ) B y = γ (B y + β E z /c) . ⎩ ⎩ E z /c = γ (E z /c + β B y ) Bz = γ (Bz − β E y /c) Second Solution Especially for tensors of order 2 (covariant or contravariant) we can compute the transformation of the components faster and easier using multiplication of matrices, because these tensors can be represented in terms of matrices. Indeed j the transformation relation Fi j = L ii L j Fi j in terms of matrices is written [Fi j ] = [L ii ][Fi j ][[L j ]−1 ]t , where [L ii ] is the matrix representation of the −1 t Lorentz transformation (not necessarily a boost). For boosts L ii = L ii and we have j
Fi j = ⎛
γ ⎜ βγ ⎜ =⎜ ⎝ 0 0 ⎛ ⎜ ⎜ =⎜ ⎝ ⎛ ⎜ ⎜ =⎜ ⎝
βγ γ 0 0
0 0 1 0
⎞⎛ ⎞⎛ Ey Ez Ex 0 0 γ c c c ⎜ E ⎟⎜ 0⎟ ⎟ ⎜ − cx 0 −Bz B y ⎟ ⎜ βγ ⎟⎜ ⎟⎜ 0 ⎠ ⎝ − Ecy Bz 0 −Bx ⎠ ⎝ 0 1 0 − Ecz −B y Bx 0
βγ γ 0 0
0 0 1 0
⎞ 0 0⎟ ⎟ ⎟ 0⎠ 1
βγ E x /c −γ E x /c −γ E y /c + βγ Bz −γ E z /c − βγ B y γ E x /c −βγ E x /c −βγ E y /c + γ Bz −βγ E z /c − γ B y E y /c −Bz 0 Bx E z /c By −Bx 0
⎞⎛
γ ⎟ ⎜ βγ ⎟⎜ ⎟⎜ ⎠⎝ 0 0
βγ γ 0 0
0 0 1 0
⎞ 0 0⎟ ⎟ ⎟ 0⎠ 1
0 E x /c γ (E y /c − βγ Bz ) γ (E z /c + βγ B y ) 0 −γ (Bz − β E y /c) γ (B y + β E z /c) −E x /c 0 −Bx −γ (E y /c − β Bz ) γ (Bz − β E y /c) −γ (E z /c + β B y ) −γ (B y + β E z /c) Bx 0
⎞ ⎟ ⎟ ⎟. ⎠
Replacing Fi j in terms of E = (E x , E y , E z ) and B = (Bx , B y , Bz ) we obtain the previous transformation equations. 11.5. In its proper frame, say Σ , the particle creates only electric field given by the expression E =
1 q r, 4π ε0 cr 3
where r = (x , y , z ) is the position vector of an arbitrary point P in Σ and r 2 = x 2 + y 2 + z 2 . Obviously E is spherically symmetric (isotropic) in Σ . Σ is related to Σ with a boost along the zaxis with velocity u = uk. From the transformation equations of the electromagnetic field we find the electromagnetic
322
11 Electromagnetic Field
field in Σ. For the electric field we have 1 q k, 4π ε0 cr 3 1 q E⊥ = γ E⊥ = γ 3 (x i + y j), 4π ε0 cr E = E =
which implies that the electric field in Σ is E = E + E⊥ =
1 q q 1 γ ∗3 (xi+yj + (z − ut)k) = γ ∗3 (r − ut) , 4π ε0 cr 4π ε0 cr
where r ∗ = x 2 + y 2 + γ 2 (z − ut)2 and r = (x, y, z) is the position vector of P in Σ. We note that the electric field E in Σ is anisotropic and specifically it is stronger normal to the direction of the velocity u and weaker along the direction of u. Indeed we have E⊥  = γ E  x 2 + y 2 ≥ E . This result is different from the Newtonian one for which the electric field in Σ is isotropic following the isotropy of the threespace (see Fig. 11.1). Concerning the magnetic field we have B = B = 0, 1 1 B⊥ = γ 2 u × E⊥ = 2 u × E⊥ , c c from which follows B=
1 γ 2 qu μ0 γ 2 βq (−yi + xj) = (−yi + xj). 4π ε0 cr ∗3 4π r ∗3
Fig. 11.1 The electric field of a moving charge
11 Electromagnetic Field
323
The magnetic field can be considered as due to the current created by the moving charge q in Σ; therefore, the above expression can be considered as the generalization of Biot–Savart Law in Special Relativity. We compute explicitly the field lines of the magnetic field B in Σ. These lines are defined by the equations dy dz dx = = . Bx By Bz Because Bz = 0 the field lines are on the plane x–y which is normal to the direction of u (i.e., normal to the direction of the current in Σ). The equation of these lines in this plane are dx dy = ⇒ x 2 + y 2 = constant y −x that is, cocentric circles with center on the line of current. Let R be the position vector of a point P with application point at the charge (that is, R = QP where Q is the position of the charge). Then R = xi + yj + (z − ut)k. We consider spherical coordinates (R, φ, θ ) in Σ where θ is the angle between the vectors R and u (that is, the zaxis). Then R · k = cos θ ⇒ z − ut = R cos θ. The length R 2 = x 2 + y 2 + (z − ut)2 = x 2 + y 2 + R 2 cos2 θ ⇒ x 2 + y 2 = R 2 sin2 θ. Also r ∗2 = x 2 + y 2 + γ 2 (z − ut)2 = R 2 sin2 θ + γ 2 R 2 cos2 θ = R 2 γ 2 (1 − β 2 sin2 θ ). Replacing we find q R 1 , 2 2 2 3/2 4π ε0 γ (1 − β sin θ ) R 3 qβ μ0 B= (yi − xj). 4π γ R 3 (1 − β 2 sin2 θ )3/2 E=
From these expressions the anisotropy of the fields E, B in Σ is apparent. Indeed the intensity of the electric field is
324
11 Electromagnetic Field
E =
q 1 . 2 2 4π ε0 R γ (1 − β 2 sin2 θ )3/2
Parallel to u (θ = 0, π ) we have E =
q 1 2 4π ε0 R γ 2
and normal (θ = π/2) to u E⊥ =
q q 1 1 = γ 3 2 2 = γ 3 E . 2 2 2 3/2 4π ε0 R γ (1 − β ) 4π ε0 R γ
The Newtonian limit of the fields E, B is found when β → 0. We compute q R (Coulomb law) 4π ε0 R 3 μ0 q u×E (Biot–Savart law). BN = 4π R 3 EN =
We note that E is independent of the angle φ. This means that in the plane normal to the velocity u (in our case the plane x–y) the electric field E can be drawn for φ = 0 and then be found for all φ with rotation of the plane x–y around the zaxis. A different way to calculate the electromagnetic field (E, B) in Σ is to use the fourpotential. Indeed in Σ we have 1 q , 4π ε0 r Vector potential: A = 0, 1 q Fourpotential: Ωi = 4πε0 cr . 0 Σ Scalar potential:
φ =
The fourpotential in Σ is found from the boost which relates Σ and Σ . We have 1 1 1 1q 1 q φ = γ φ = γ ⇒φ= γ ∗, c c 4π ε0 c r 4π ε0 r φ φ A = γβ k = β k. c c The fields E, B are computed from the relations φ ∂A 1 γq E = −∇ − = ... = (xi+yj + (z − ut)k) , c ∂t 4π ε0 r ∗3 μ0 βγ q B = ∇ × A = ... = (yi − xj). 4π r ∗3
11 Electromagnetic Field
325
11.6. First Solution We compute first the fourpotential in Σ and then the electromagnetic field in Σ by transforming the fourpotential in Σ using the boost relating Σ and Σ (see Fig. 11.2).
Fig. 11.2 The electric dipole
Consider the point P with coordinates P(x , y , z ) in Σ . The scalar and the vector potential at P are −q +q 1 1 + , 4π 0 (x + l )2 + y 2 + z 2 4π 0 (x − l )2 + y 2 + z 2 A = 0.
φ (P) =
Hence the fourpotential in Σ is2 Ωi = (− φc , A ). We compute the fourpotential in Σ using the boost relating Σ, Σ . We have φ φ + βA = γ c c ⎡ −q γ 1 ⎣ = + c 4π 0 γ 2 (x + γl − ut)2 + y 2 + z 2 γ 2 (x − φ 1 + A⊥ = βφ. A = A + A⊥ = γ A + β c c
φ =γ c
⎤ +q l γ
− ut)2 + y 2 + z 2
⎦,
2 We note that on the plane x = 0 the fourpotential vanishes, hence the fourpotential and the Lorentz transformation make no sense. We extend the results on this plane by demanding the continuity of the fourpotential (as a fourvector field – equivalently of the electric field – at the values x = 0+ and x = 0−.
326
11 Electromagnetic Field
The fourpotential Σ has been computed in terms of the coordinates in Σ . Using the boost relating Σ, Σ we express the fourpotential in terms of the coordinates (x, y, z) in Σ and then compute the electric and the magnetic field in Σ from the relations ∂A , ∂t
E = −∇φ −
B = ∇ × A.
For the electric field in Σ we find ⎞ ⎛ l l 2 2 γ γ x + x − − ut − ut γ γ 1 y y z z E=− γq ⎝ − , 3 − 3 , 3 − 3 ⎠, 3 3 4π 0 R+ R− R+ R− R+ R− where we have set R± = γ 2 (x ± for the magnetic field in Σ
l γ
1 γ qβ B= 4π 0 c
− ut)2 + y 2 + z 2 . Working similarly we find
1 1 − 3 3 R+ R−
(0, z, −y).
Second Solution We calculate in Σ the electric field (the magnetic field vanishes) and use the transformation of the fields E, B to compute the electromagnetic field in Σ. We have 1 q 1 q E (P) = E+q (P) + E−q (P) = (x −l −ut, z ) − (x + l −ut, z ) 3 3 4π 0 R+ 4π 0 R− x − l − ut q x + l − ut y y z z , = − , − , − 4π 0 R+3 R−3 R+3 R−3 R+3 R−3 where R± = (x ± l − ut)2 + y 2 + z 2 . Transforming in Σ using the boost relating Σ, Σ we find E = E =
q 4π 0
q = 4π 0
*
x − l − ut x + l − ut − 3 R+ R−3 l − γ 3 R+
γ (x −
−
l − γ 3 R−
γ (x +
ut)
+ ,
y q y y z z y z z = 0, 3 − 3 , 3 − 3 , 3 − 3 , 3 − 3 4π 0 R+ R− R+ R− R+ R− R+ R− 1 γ qβ 1 1 1 B = B⊥ = 2 u × E⊥ = − 3 (0, z, −y). c 4π 0 c R+3 R−
E⊥ = E⊥ =
q 4π 0
ut)
0,
11 Electromagnetic Field
327
Third Solution We compute the electromagnetic field tensor in terms of its components in Σ and use the boost relating Σ, Σ to transfer the results in Σ (via the transformation equa j tions Fi j = L ii Fi j L j ). 11.7. (a) The conductor consists of fixed positive ions (which make up the lattice of the conductor) and free electrons. The positive ions have a charge density ρ+ and the free electrons a charge density ρ− . Before the application of the potential difference the conductor is electrically neutral, hence ρ− = ρ+ . After the application of the potential difference the free electrons start to move in the conductor through the source so that their number within the conductor remains constant. Therefore, in Σ the conductor continues to appear electrically neutral although electric current flows though it. If the velocity of the electrons in the conductor is v, then the conduction current in Σ is j = ρ− v and the charge density ρ = ρ− − ρ+ = 0. This implies that the fourcurrent in Σ is ⎛
0
⎞
⎜ ρ− v ⎟ ⎟ Ji = ⎜ ⎝ 0 ⎠ . 0 Σ (b) In the LCF Σ which moves wrt Σ in the standard configuration along the xaxis with speed β assume that the fourcurrent is ⎛
⎞ ρc ⎜ ρ v ⎟ ⎟ Ji = ⎜ ⎝ 0 ⎠ . 0 Σ The boost relating Σ and Σ gives ρ = γρ−
βv = 0. c
We conclude that in Σ the density of charge does not vanish and the conductor appears charged. Concerning the conduction current the boost gives j = γ j, that is, the current is “dilated” in Σ . In order to explain the above result geometrically we consider the world lines of the positive ions and the free electrons before (Fig. 11.3a) and after (Fig.11.3b) the application of the potential difference at the end points of the conductor.
328
11 Electromagnetic Field
Fig. 11.3 The conductor (a) before and (b) after the application of the potential
Explanation of figure A, B : End points of conductor in Σ. G, B: End points of conductor in Σ . Σ, Σ : World lines of Σ, Σ . +, − : World lines of the positive ions and the free electrons. For the angle φ of the world lines we have cosh φ =
v . c
The conductor appears to be charged in an LCF if the number of world lines of the positive ions which intersect the conductor is different from the corresponding number of world lines of the electrons. In Fig. 11.3(a) we note that the number of world lines of positive ions which intersect the conductor in Σ and Σ (this number is determined from the world lines which intersect the lines AB and G B, respectively) equals the number of world lines of free electrons and the conductor appears to be electrically neutral. This is not the case in Fig. 11.3(b) when the electric current flows through the conductor. 11.8. (a) We consider on the xaxis the element of length d x = l dθ where l = r0 cos θ and θ is the angle with tip at the charge and sides defined by the end points of d x. The element d x has charge ρ0 d x = ρ0l dθ and applies on the charge q the threeforce F(θ ) =
qρ0 cos θ dθ (− sin θ i + cos θ j) . r0
The total force on the charge q due to the charge along the xaxis is
11 Electromagnetic Field
( F=
π −π
329
( F(θ ) dθ =
π
−π
qρ0 2qρ0 (− sin θ cos θ i − cos2 θ j) dθ = − j. r0 r0
(b) In Σ we have from the Lorentz transformation of the fourforce F =
1 u·F 1 F + (γu − 1) 2 u− 2 γu (v · F) , γu Q u c
where v is the velocity of the charge in Σ. In this problem v = 0, hence Q = 1. Furthermore, u · F =0, hence F =
1 2qρ0 j. F=− γ γ r0
Second solution The charge along the xaxis at a distance r0 from the axis creates the electric field E=−
2ρ0 j. r0
From the transformation of the electromagnetic field we have in Σ E = γ (E + β × B) = γ E = γ E⊥ = γ E ⊥ j, 1 1 B = γ B − 2 β × E = − γβ E ⊥ k . c c The force on the charge q in Σ is the Lorentz force F = q(E + u × B ) = q(γ E ⊥ j − γβ 2 E ⊥ i × k) q = qγ (1 − β 2 )j = E ⊥ j γ 2qρ0 =− j . γ r0 11.9. (a) We consider a cylindrical surface of height h and base radius r0 (r0 h) whose axis is along the straight conductor (see Fig. 11.4). Due to the symmetry the electric field created by the conductor is normal to the conductor and has direction from the conductor to the cylindrical surface (i.e., along the outward normal to the surface, note that the conductor has positive charge). Gauss law gives for the electric field on the surface of the cylinder : E dS = S
1 Q. 0
330
11 Electromagnetic Field
Fig. 11.4 The electric field of a line conductor
Due to symmetry only the integral on the cylindrical surface is contributing and we have E · 2πr0 · h =
λh λ ⇒E= j. 0 2π 0r0
(b) Let Σ be the proper frame of the conductor and choose the x axis of Σ to be along the straight conductor. Then in Σ the conductor creates only electric field which is the one we computed in (a). In the following we denote this field with E . In Σ the electromagnetic field is computed from the transformation equations of the electromagnetic field. We have then E = E = 0, E⊥ = γ E = γ
λ j, 2π 0r0
B = B = 0, 1 B⊥ = γ u × E . c2 But λu = λ
dq dr i= i=I dt dt
and
r0 = r0 j.
Replacing we find B=γ Then 0 c2 =
1 , μ0
1 I k=γ (I × r0 ). 2π 0 c2r0 2π 0 c2r02
hence finally B=γ
μ0 (I × r0 ). 2πr02
11 Electromagnetic Field
331
For small speeds γ ≈ 1 and this equation is written as B=
μ0 (I × r0 ). 2πr02
which coincides with the Biot–Savart law of Newtonian Physics. (c) The charge q moves in Σ under the action of the uniform electromagnetic field whose electric and magnetic fields are E = γ 2πλ0 r0 j and B = γ 2π0Ic2 r0 k which we computed in (b). The force on the charge is the Lorentz force F = q(E + u × B), hence F=q
γλ qλ γI γ qλ 1 − β2 j = j+ ui × k = j. 2 2π 0r0 2π 0 c r0 2π 0r0 2π 0 γ r0
11.10. (a) Without restricting the generality we assume the xaxis of Σ along the direction of velocity of the particle. Furthermore, we consider the proper frame Σ+ of the particle with axes parallel to the axes of Σ so that Σ, Σ+ are related with a boost along the xaxis. In the proper frame of the particle the force exerted on the particle is due to the electric field E+ = (E x+ , E y+ , E z+ ). From the transformation equations of the electromagnetic field we have that the fields (E, B) in Σ are related to the electromagnetic field in Σ+ as follows E x+ = E x , E y+ = γ (E y − vcx Bx ), E z+ = γ (E z + vcx Bz ). In Σ+ the equation of motion of the particle is qE+ = ma+ , where a+ is the proper acceleration of the particle. But the proper acceleration a+ is related to the threeacceleration a in Σ with the formula (see relevant theory) ax+ = γ 3 ax , 2 a+ y = γ ay , + az = γ 2 az . It follows that the equations of motion of the particle in Σ in component form are
332
11 Electromagnetic Field
q E x = mγ ax , vx q(E y − Bz ) = mγ a y , c vx q(E z + B y ) = mγ az . c From these equations we find the vector form of the equations parallel and normal to the velocity v of the particle3 q(E + 1c v × B) = mγ 3 a , q(E + 1c v × B)⊥ = mγ 2 a⊥ . This form of the equations of motion is useful in the solution of problems. (b) We consider first the term mγ 3 ax . From the relation γ = √ 1 2 we compute 1−β
1 ˙ 3 = β · βγ ˙ 3 = βγ 3 β. ˙ γ˙ = − [−2(β · β)]γ 2 Therefore, the term ˙ + β 2 γ 2 ) = mc(γ β˙ + γ 3 β ββ) ˙ mγ 3 ax = mγ 3 cβ˙ = mγ cβ(1 x = mc(γ β˙ + γ˙ β) = mc(γβ). = (mγ vx ). = dp . dt Similarly we prove the other two equations. (c) From the relation E 2 = p2 c2 + m 2 c4 we have dp dE = 2p · = 2p · q(E + β × B) = 2qmγ c2 (β · E) = 2qE(β · E) ⇒ dt dt dE = q(β · E). dt γβ · F i (d) The fourforce on the particle in the frame Σ is F = , where F = γF Σ dt dr dr q(E + β × B). The term γ β= dτ = dτ , where τ is the proper time of the dt particle. Therefore, the four force in Σ is written as 2E
Fi =
·F γ q(E + β × B) 1 dr c dτ
Σ
=
· qE q d(ct) q dr E + ×B c dτ c dτ 1 dr c dτ
. Σ
3 We note that the coefficient of the acceleration is different parallel and perpendicular to the threevelocity. For this reason the term mγ 3 has been named longitudinal mass and the term mγ 2 transverse mass. This distinction adds nothing and in fact has been abandoned.
11 Electromagnetic Field
333
The equation of motion of the particle is F i = ma i = m
d2xi , dτ 2
where a is the fouracceleration of the particle and x = i
i
ct r
vector. Replacing in the equation of motion we find *
d 2 (ct) dτ 2 d2r dτ 2
+ = Σ
q c
·E d(ct) dr E + ×B dτ dτ dr dτ
is the position Σ
. Σ
Therefore, the equations of motion in terms of the proper time are d 2 (ct) q dr · E, = dτ 2 c dτ d 2r dr q d(ct) E + × B . = dτ 2 c dτ dτ 11.11. (d) Let E/c = p 0 be the zeroth component of the fourmomentum. Then in Σ E/c = mγ c ⇒ E = mγ c2 . The quantity F0 =
q 0 j q F j u = E · γ v = q (E + v × B) · v = F · v γ γ
where F is the Lorentz force on the charge. This implies that the zeroth component of the equation of motion is dE =F·v. dt equals the change of the total energy of the charge, We conclude that the quantity dE dt hence E is the total energy of the charged particle in Σ. We compute dE = F · vdt = F · dr =
1 1 1 2 dp c2 · dr = dp · v = dp · p = dp = dp2 ⇒ dt mγ mγ 2 2E 2EdE = c2 dp2 .
334
11 Electromagnetic Field
Integrating we find E 2 = c2 p2 + constant. The constant equals the total energy E when p = 0. From the relation E = mγ c2 we find that this constant equals mc2 where m is the mass of the particle. Replacing we find the required formula. 11.12. (a) Let Σ be the proper frame of the beam. We calculate the current which is due to a cylindrical surface of radius r < R, where R is the radius of the beam. Because the distribution of the charge in Σ is symmetric about the axis of the beam the charge density ρ depends only on the distance r from the symmetry axis. Based on that we consider two cylindrical surfaces at distances r and r +dr (r < R) between which we consider the charge density to be constant. The charge dq enclosed by these surfaces equals dq = ρ (r )d V = ρ (r )2πr dr . Therefore, the charge in a cylinder of radius r ≤ R is Q(r ) = 2π
(
r
ρ(r )r dr .
0
Suppose that in Σ the speed of the electrons of the beam parallel to the axis l of symmetry is u. Then the length of the beam in Σ is l /γ = ut ⇒ t = uγ . Subsequently the current which corresponds to the charge Q(r ) of the beam is I (r ) =
Q(r ) = 2π γ u X, t
'r where X = 0 ρ (r )r dr . We calculate now the field E in the proper frame of the beam. Applying Gauss law for a closed cylindrical surface S of radius r and height L r we have :
E dS = 4π Q.
In this relation E is the electric field in the elementary surface dS of a closed surface which contains the charge Q(r ). Due to the axial symmetry (and assuming that L R ) the field E is radial, hence : E ds = E⊥ 2πr = 2π E⊥ r ut = 2πr uE⊥ t,
11 Electromagnetic Field
335
where we have used that E⊥ = γ E⊥ , t = γt , and r = r because r is normal to u. But Q(r ) = I (r )t, thus E⊥ (r ) =
2I eˆ r . ur
The force on an electron in the LCF Σ at a distance r from the axis of the cylindrical distribution is F (r ) = eE⊥ (r ) = e
E⊥ (r ) , γ
where e is the charge of the electron and 0 < r ≤ R (R is the radius of the beam). In Σ this force is F(r ) =
F (r ) 1 2eI = 2 eE⊥ (r ) = eˆ r . γ γ ur γ 2
In order to compute the velocity of the electrons of the beam we decompose the motion in to two parts: one motion parallel to the beam (the axis of symmetry) and a second normal to it. The motion parallel to the axis is caused by the potential difference which accelerates the electrons of the beam. If v is the final speed (due to both motions!) of an electron, then E − mc2 = eV ⇒ mγ c2 − mc2 = eV ⇒ γ =
eV + 1. mc2
From γ we compute the speed u of (the parallel motion of) the electrons of the beam eV 1/2 eV −1 2eV 1/2 1 + u = 1+ . mc2 mc2 2mc2 (b) The motion of the electrons normal to the axis of symmetry is due to the repulsive action from the rest of the electrons of the beam. This force, F⊥ (r ) say, generates the acceleration a⊥ (r ) =
1 1 2eI F⊥ (r ) = . mγ mγ ur γ 2
If the electron is under the action of the force F⊥ (r ) for a period of time t (in Σ) the deviation ΔA of the electron is ΔA =
1 1 L2 a⊥ t 2 = a⊥ 2 . 2 2 u
336
11 Electromagnetic Field
Because the speed of the electrons u along the direction of the symmetry axis is much larger than the speed normal to it, we approximate u = v (recall that v is the speed of the electron due to both motions!) and write ΔA =
1 L2 L Δa 1 ⇒ a⊥ = a⊥ 2 . 2 v2 L 2 u
But we have by assumption ΔA/L 1 ⇒ a⊥
L L 1 ⇒ a⊥ u ⇒ a⊥ t u < c, 2 u u
which justifies our assumption that the velocity for the motion normal to the axis of the beam can be neglected. Finally, we have that the deviation of the electrons at distance r from the axis of symmetry is ΔA =
eI L 2 . mr γ 3 u 3
Hence the “opening” of the beam in Σ is ΔA =
eI L 2 . m Rγ 3 u 3
11.13. (a) We consider the proper frame of the charge B. In this frame the charge produces only electric field, E+ (B) say, which is found from Coulomb’s law. If the charges at some moment t in the laboratory have deviated to the positions + y+ A and y B , respectively, then the electric field acting on the charge A is E+A =
e yˆ . (y + − y B+ )2 A
But y+ A
−
y B+
+
=d +
Δy + A
+
Δy B+
+ Δy + A + Δy B 1+ =d d+ +
+ where Δy + A , Δy B are the ydeviations of the charges A and B, respectively, + from their firing line and d + is their initial distance. We assume Δy + A , Δy B + + e + d which implies E A = d + 2 yˆ , that is, E A is constant. We assume that the repulsive potential which accelerates the electrons apart is small and does not effect the xcomponent of the velocity, i.e., v =constant. This permits us to assume that the proper frame Σ+ of each of the electrons is related with the laboratory frame with a boost with speed v. The force on the charge A is (d = d + why?)
eE+A =
e2 yˆ d2
11 Electromagnetic Field
337
and its acceleration a+A =
e yˆ . md 2
Let T ∗ be the traveling time of A in its proper frame. Then the deviation of the charge A from its original position in Σ+ is Δy + A =
1 +2 +2 e2 2 T+ . aA T = 2 2md 2
1 ∗ t where t ∗ is the time required for the electrons to reach The time T + = γ (w) the screen in the laboratory and γ = γ (w) where w is the speed of A when hits the screen. Finally,
e2 ∗2 1 t . 2γ 2 (w) md 2
Δy + A =
But Δy + A = Δy A because the boost is along the xaxis, which implies that Δy A =
1 e2 t ∗2 . γ (w)2 2md 2
We have assumed that the xcomponent v of the velocity of the electrons remains constant. This means w = v and L = vt ∗ . Solving for t and replacing in the last expression we find Δy A =
e2 L 2 1 . γ (v)2 2md 2 v 2
The distance of the electrons when they hit the screen is ΔyS R = Δy A + Δy B = 2Δy A = 2
1 e2 L 2 . γ (v)2 md 2 v 2
2
e L The Newtonian result is Δy N = md 2 v 2 , that is, the distance between the points at which the particles hit the screen depends only on the distance traveled by the particles and their initial velocity along the xaxis. The relativistic result is related to the Newtonian one as follows
ΔyS R =
1 Δy N γ (v)2
and shows that the defocusing is less by the γ (v)factor. This means that in Special Relativity the orbits of two beams of charged particles with high speed (e.g., the ones produced in high energy accelerators) focus more than expected
338
11 Electromagnetic Field
by Newtonian Physics. This phenomenon is very important and finds many applications in practice. (b) From the equations of transformation of the electromagnetic field we have for the field in Σ at the charge A E A,x = E + A,x = 0, E A,y = γ E + A,y = γ
e e =γ , 2 (y − y A )2 (y B+ − y + ) B A
E A,z = γ E + A,z = 0, B A,x = 0, B A,y = −γβ E + A,z = 0, B A,z = γβ E + A,y = γβ
e . (y A − y B )2
The Lorentz force on the charge A in the laboratory frame Σ is 1 e2 1 F A = e(E A + v × B) = yˆ . c γ (y A − y B )2 Hence the acceleration of the charge A in Σ is a=
1 e2 1 FA = 2 yˆ , mγ γ (y A − y B )2
which coincides with the result we found in (a). 11.14. Let Σ be a second LCF in which the electromagnetic field is represented with the fields E , B and the vector K = E + icB . We consider first the special case that Σ moves wrt Σ in the standard configuration along the xaxis with velocity factor β. From the equations of transformation of the fields E, B we have for the vector K K x = E x + icBx ,
v E z ), c2 v K z = γ (E z + v B y ) + icγ (Bz − 2 E y ). c
K y = γ (E y − v Bz ) + icγ (B y +
These relations can be written in the form of matrices as follows ⎛ ⎞ ⎛ ⎞ 1 0 0 1 0 0 K = ⎝ 0 γ iγ vc ⎠ K = ⎝ 0 cosh φ i sinh φ ⎠ K 0 −iγ vc γ 0 −i sinh φ cosh φ where cosh φ = γ ( φ =rapidity).
11 Electromagnetic Field
339
We introduce the (complex) trigonometric functions cos(iφ) = cosh φ, sin(iφ) = i sinh φ. These functions are well defined and satisfy all the relations of the corresponding real functions in R3 . For example, we have the identity 2
cos2 (iφ) + sin (iφ) = cosh2 φ + (i sinh(iφ))2 = cosh2 φ − sinh2 φ = 1. With the introduction of the angle iφ the transformation of the complex vector K can be written as ⎛ ⎞ 1 0 0 K = ⎝ 0 cos(iφ) sin(iφ) ⎠ K. 0 −sin(iφ) cos(iφ) This transformation corresponds to a rotation by an angle iφ in the space C3 because its determinant equals 1 (so that K 2 = K2 , that is the length of the vector K is preserved).4 Replacing in the identity K2 = Kt K the fields we have computed, we find the relations 1 1 2 E − c2 B2 = 2 E2 − c2 B2 , 2 c c 1 1 E · B = E · B. c c Because K characterizes completely the electromagnetic field, we infer that these are the only invariants of the electromagnetic field. We come now to the case of the general Lorentz transformation. From the equations of transformation of the fields E, B under the action of the Lorentz transformation we have
4
Indeed
⎧⎛ ⎞2 ⎞⎫ T ⎛ ⎞⎛ ⎞ ⎞⎛ K x 1 0 0 Kx 0 Kx ⎬ ⎨ 1 0 ⎝ K y ⎠ = ⎝ 0 γ iγβ ⎠ ⎝ K y ⎠ ⎝ 0 γ iγβ ⎠ ⎝ K y ⎠ ⎭ ⎩ Kz Kz 0 −iγβ γ 0 −iγβ γ K z ⎞T ⎛ ⎞T ⎛ ⎞⎛ ⎞ ⎛ 1 0 0 1 0 0 Kx Kx ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎝ Ky ⎠ 0 γ iγβ 0 γ iγβ = Ky Kz Kz 0 −iγβ γ 0 −iγβ γ ⎛ ⎞ Kx = K x K y K z ⎝ K y ⎠ = K · K = K2 . Kz ⎛
340
11 Electromagnetic Field
K = E + icB = E + E⊥ + icB + icB⊥ 1 = E + γ (E⊥ + β × cB⊥ ) + icB + icγ (B⊥ − β × E⊥ ) c = E + icB + γ [E⊥ + icB⊥ − iβ × (E⊥ + icB⊥ )] = K + γ [K⊥ − iβ × K⊥ ] ⇒ K = K
K⊥ = γ (K⊥ − iβ × K⊥ ) . The length " 2 # 2 2 2 K2 = K2 K⊥ − β 2 K2⊥ = K2 + K2⊥ = K2 . + K⊥ = K + γ The rest of the solution follows as above. 11.15. (a) Consider the LCF Σ with velocity u normal to the plane defined by the vectors E, B which are such that E · B =0. We write u = aE × B,
(11.1)
where the parameter a > 0 has to be determined. Due to the choice of u we have E = E⊥ , B = B⊥ . The electromagnetic field (E , B ) in Σ is found from the equations of transformation of the electromagnetic field E = E = 0 ⇒ E = E⊥ , B = B = 0 ⇒ B = B⊥ . Concerning the normal components we have a E = E⊥ = γ (E⊥ + cβ × B⊥ ) = γ (E + cβ × B) = γ E + (E × B) × B c a a a = γ E − B × (E × B) = γ E − (B · B) E + (B · E)B c c c a = γ (1 − B 2 )E. c 1 aE2 1 B = B⊥ = γ B⊥ − β × E⊥ = γ B − β × E = . . . = γ 1 + 2 B. c c c But E · B = 0 ⇒ E × B = E B, hence from (11.1) follows a = in the above expressions we find
cβ . EB
Replacing
11 Electromagnetic Field
341
E = γ (1 − β
Bc )E, E
B = γ (1 − β
E )B. Bc
Let Bc > E (i.e., the invariant X < 0). Then if we define β = E/Bc < 1, E = 0; thus the LCF ΣE with velocity is u=
1 1 E c (E × B) = 2 (E × B) Bc E B B
and the electromagnetic field has only magnetic field which equals B =
1 B. γ
When E > Bc (i.e., the invariant X > 0) we have that in the LCF ΣB whose velocity in Σ is u=−
c2 (E × B) E2
the electromagnetic field has only electric field which equals E =
1 E. γ
The LCF we have found are not unique. Indeed if in the first case (E > Bc) we consider all LCF ΣE1 (say) with velocity u1 B wrt Σ E (not to Σ!), the transformation of the fields between Σ E and Σ E1 gives E1 = γ u1 × B = 0, B1 = γ1 B = γ1 γ B, that is, the electromagnetic field has only magnetic field. Similarly we show that in all LCF ΣB1 whose velocity in Σ B is u1 E there exists only electric field. (b) In the case E · B = 0, we consider the LCF Σ whose velocity u in Σ is normal to the plane of the fields E, B and write u = aE × B, where a > 0 is to be determined. The equations of transformation of the fields give in Σ E = γ (E + u × B), 1 B = γ (B − 2 u × E). c
342
11 Electromagnetic Field
We demand E B ⇔ E × B = 0 and find (E · u = B · u = 0) (E + u × B) × (B − E × B − B × (u × B) −
1 u × E) = 0 ⇒ c2
1 E × (u × E) c2
1 (u × B) × (u × E) = 0 ⇒ c2 1 1 E × B − B 2 u + (B · u)B − 2 E 2 u + 2 (E · u)E− c c 1 − 2 [(u × B) · E]u = 0 c 1 2 1 2 E × B − (B + 2 E )u − 2 [u · (B × E)]u = 0. c c −
Replacing E × B = a1 u in this equation we find 1 1 u·u 1 u − (B 2 + 2 E 2 )u + 2 ( )u = 0 ⇒ a c c a 1 1 1 + β2 (1 + β 2 ) = B 2 + 2 E 2 ⇒ a = . a c B 2 + c12 E 2 β But β = uc = a 1c E × B ⇒ a =  1 E×B . From the last two relations we infer c that β is computed from the quadratic equation
β2 −
B2 +
1 c2
E2
 1c E × B
β + 1 = 0.
The roots of this equation are β=
1 2 2 Q ± Q −4 , 2
where Q=
B2 +
1 c2
E2
 1c E × B
.
The requirement for real solutions gives Q ≥ 2 ⇒ B2 + The direction of the velocity is
E×B . E×B
1 2 E ≥ 2E × B. c2
11 Electromagnetic Field
343
It remains to compute the fields B , E in Σ in terms of the fields B, E in Σ. From the equations of transformation of the fields we have 1 1 2 1 B = γ B − 2 a(E × B) × E = γ (1 − a 2 E )B+aY E , c c c
where Y = 1c E · B is the invariant of the electromagnetic field. With a similar calculation we find " # E = γ (1 − a B 2 )E+aY cB . The LCF Σ is not the only one but there are infinitely many LCF in which the electric and the magnetic field are parallel. Indeed if we consider all LCF with velocity (wrt Σ not to Σ!) u2 parallel to the common direction of the fields E , B (in Σ ), then u2 ×E = u2 ×B = 0 and the equations of transformation of the electromagnetic field give E2 = γ2 E B2 = γ2 B
⇒ E2 B2 .
11.16. The equation of motion of the charge is dp = q(E + v × B) = qE ⇒ p˙ x = q E, dt
p˙ y = 0,
p˙ z = 0 .
Integrating and using the initial conditions we find px = q Et + px,0 = q Et, p y = p y,0 , pz = 0 . The energy E of the charge is E=
p2 c2 + m 2 c4 =
q 2 E 2 t 2 c2 + p 2y,0 c2 + m 2 c4 = E02 + q 2 E 2 t 2 c2 ,
where E02 = p 2y,0 c2 + m 2 c4 is the energy of the charge at t = 0. From the threemomentum and the energy we compute the velocity of the charge ⎛ ⎞ 2 2 c p q Etc pc2 y,0 = ⎝ v(t) = , , 0⎠ . E 2 2 2 2 2 2 2 2 2 2 E +q E t c E +q E t c 0
The position vector
0
344
11 Electromagnetic Field
( r(t) = =
⎛ v(t) dt = ⎝ 1 qE
(
E02
+
⎞
(
q Etc
d(ct), p y,0 c
q 2 E 2 t 2 c2
E02 + q 2 E 2 t 2 c2 − E0 ,
cp y,0 sinh−1 qE
d(ct) E02
+ q Ect ,0 . E0
, 0⎠
q 2 E 2 t 2 c2
From this result we find the parametric equation of the orbit 1 ( E02 + q 2 E 2 t 2 c2 − E0 ), qE cp y,0 q Ect sinh−1 y(t) = , qE E0 z(t) = 0.
x(t) =
Obviously the motion takes place in the x − y plane. In order to find the equation of the orbit on that plane we eliminate t from the second and replace in the first equation. We have in turn qEy E0 sinh , q Ec cp y,0 ⎞ ⎛$ 2 E q E y 1 ⎝ 2 x= − E0 ⎠ E0 + q 2 E 2 c2 2 02 2 sinh2 qE q E c cp y,0 + *$ E0 qEy E0 2 qEy = cosh −1 . 1 + sinh −1 = qE cp y,0 qE mγ0 v0 c t=
11.17. (a) The force on the particle is the Lorentz force F = q (E + u × B) = qu × B, which is normal to the velocity u, hence F · p = 0. But F = dp ⇒ p · dp = dt dt 2 2 2 2 2 2 4 0 ⇒ p = constant. Then the energy E = p c + m c = E0 =constant. Also the velocity factor γ = E02 /mc2 =constant. (b) The equation of motion of the particle is qu × B = mγ0 a ⇒ mγ0 a = q B u y , −u x , 0 , from which follows u˙ x = au y , u˙ y = −au x , u˙ z = 0. qB where a = mγ . The solution of this system of simultaneous equations under 0 the given initial conditions is
11 Electromagnetic Field
345
u x (t) = λ cos (a ln t) ,
u y (t) = −λ sin (a ln t) , u z (t) = k.
From the velocity components we compute the position x(t), y(t) (the z(t) = kt) of the particle. ' We shall need the integral I = sin a ln t dt with t ≥ 1. We compute I
=
'
' sin a ln t dt = t sin a ln t − t at cos a ln t dt =
= t sin a ln t − ta cos a ln t − a 2 I ⇒ ⇒I =
1 t a 2 +1
sin a ln t −
a t a 2 +1
cos a ln t .
Let tan θ = a . Then I = t cos θ sin a ln t − t sin θ cos a ln t = = t sin a ln t − θ + c. For the coordinate y(t) we have ( ( y = u y dt = −λ sin (a ln t) dt = − λt sin (a ln t − θ ) + c where tan θ = a. For t = 1 the initial condition gives y(1) = 0 hence c = −λ sin θ . Replacing we find y(t) = −λt sin (a ln t − θ ) − λ sin θ. Working similarly we compute x(t) = λt cos (aln t − θ ) . To find the surface generated by the orbit we eliminate the parameter t from the parametric equation of the orbit, i.e., the x(t), y(t). We compute x 2 + (y + λ sin θ)2 = λ2 t 2 = λ2
z2 λ2 2 = z k2 k2
from which follows that the equation of the surface that the orbit generates is x 2 + (y + λ sin θ)2 =
λ2 2 z . k2
11.18. (a) Has been solved in a previous problem. (b) The force on the particle is the Lorentz force F = qv × B. The equation of motion is
346
11 Electromagnetic Field
qB vy , γm qB v˙ y = − vx , γm v˙ z = 0. v˙ x =
where γ = γ0 = constant because according to (a) the speed is constant. From the third equation and the initial conditions we find z(t) = 0; therefore, the motion takes place on the plane x − y. Concerning the remaining two equations we multiply the second with the imaginary unit i and add to the first to get v˙ x + i v˙ y = Ωv y − iΩvx ⇒ (vx + iv y ˙) + iΩ(vx + iv y ) = 0, where the constant Ω = γq mB is called the cyclotron frequency. The solution of the linear homogeneous ordinary differential equation is vx + iv y = Ae−iΩt , A ∈ C. If we write the complex constant A in polar form as A = be−iφ , where b, φ ∈ R the solution is written as vx + iv y = be−i(Ωt+φ) . The real and the imaginary parts give vx = b cos(Ωt + φ), v y = −b sin(Ωt + φ). The real constants b, φ are determined from the initial conditions. We have vx (0) = u v y (0) = 0
⇒
b cos φ = u sin φ = 0,
from which follows b = u , φ = 0 . Finally x˙ = u cos Ωt, y˙ = −u sin Ωt . Integrating these equations we find u sin Ωt, Ω u y = (cos Ωt − 1) , Ω
x=
11 Electromagnetic Field
347
and the parametric equations of the orbit of the particle are u sin Ωt, Ω u y = (cos Ωt − 1), Ω z = 0.
x=
In order to find the equation of the trajectory we square the first two and add to get u 2 x2 + y − = k2, Ω where k=
mu u = Ω qB
−1/2 u2 1− 2 . c
Fig. 11.5 Orbit of a charge entering a magnetic field
The orbit is a circle with center at (0, −k) and radius k. The particle is reflected from the field (i.e., the direction of its motion changes by π ) if and only if k < a (see Fig. 11.5). 11.19. We compute in Σ the invariants of the electromagnetic field. We find Y =
1 E · B = 0, c
X=
1 2 E − B2 < 0 . c2
348
11 Electromagnetic Field
From these values we infer that there are infinite many LCF Σ in which the electromagnetic field is E = 0, B = γ1 B. From the theory we know that the velocity u of one of these, which in the following we shall refer to as Σ , is normal to the plane of E, B, hence u = ui. In order to calculate the speed of u we consider the condition E = 0 and write E + u × B = 0 ⇒ acj + (ui) ×
3c 5a k = 0 ⇒ u = i. 3 5
It follows γ = 54 . In Σ there is only magnetic field B = γ1 B = 45 53 ak = 43 ak ; therefore, we apply the method of solution applied for this case. The equations of motion of the particle in Σ are dvx = ω v y , dt
dv y dt
= −ω vx ,
= 4aqc is the cyclotron frequency in Σ and E is the initial energy where ω = qcB E 3E of the particle in Σ . Because the particle starts to move from rest in Σ we have the initial conditions v(0) = 0, E(0) = mc2 ; hence, E = γ (u)E(0) = 54 mc2 and ω =
4a qc 16 qa . = 3 2 3 4 mc 15 mc
The solution of the equations of motion in Σ is vx (t ) = A cos(ω t + B ), v y (t ) = −A sin(ω t + B ) . In order to compute the constants of integration A , B we must transfer the initial conditions from Σ to Σ . We have in Σ t = 0 , v(0) = 0 , x(0) = x0 i. From the transformation law of threevelocities and the position threevector we compute in Σ 3 v (0) = (−u, 0, 0) = (− c, 0, 0), 5 5 r (0) = γ x0 i = x0 i . 4
11 Electromagnetic Field
349
Replacing in the solution we find 3 vx (0) = − c = A cos B , 5
v y (0) = 0 = −A sin B ⇒ B = 0, A =
3 c. 5
Finally, the solution in Σ is 3 vx (t ) = − c cos ω t, 5 3 v y (t ) = c sin ω t . 5 ' The coordinate z = vz dt = 0, hence the motion is circular in the x –y plane. Because y = y the particle returns to the xaxis when y = 0, that is, after a period. The angular speed in Σ is ω , hence the period of rotation in Σ is T =
15π mc 2π 15mc = . = 2π ω 16qa 8qa
This time period we transfer in Σ using the time dilation formula (i.e., the Lorentz transformation relating Σ, Σ ) is T = γT =
75 π mc 5 15π mc = . 4 8qa 32 qa
Note: In order to check the result we compute y and require that it vanishes. We have y (t ) =
(
v y dt = −
3c cos ω t + k . 5ω
For t = 0 y (0) = 0 = −
3c 3c +k ⇒k = . 5ω 5ω
Hence y (t ) =
3c (1 − cos ω t ) . 5ω
This vanishes when cos ω t = 1 ⇒ ω t = 2kπ ⇒ T =
2π . ω
350
11 Electromagnetic Field
11.20. (a) The force on the particle in Σ is the Lorentz force & & & i j k& & & F = q(v × B) = q && v x v y 0 && = q v y B, −vx B, 0 . & 0 0 B& The force F is normal to v, thus F · v = 0. This implies p2 =constant because dp dp2 = · p = F · (mγ v) = 0. dt dt Then E 2 = p 2 c2 + m 2 c4 = constant and v = p c2 = constant and γ = E γ0 =constant. Taking the above into consideration the equations of motion read in Σ mγ0 v˙ x = q Bv y mγ0 v˙ y = −q Bvx
⇒
v˙ x = ωv y v˙ y = −ωvx ,
qB . where ω = mγ 0 To solve the system of these equations we multiply the second with i and add to get
d vx + iv y = −iω vx + iv y ⇒ w(t) = w(0)e−iωt , dt where w(t) = vx + iv y . The real and the imaginary parts of this equation give vx (t) = vx (0) cos ωt + v y (0) sin ωt, v y (t) = vx (0) sin ωt − v y (0) cos ωt. We consider the initial conditions vx (0) = v, v y (0) = 0 (motion starts from rest) and find vx (t) = v cos ωt,
v y (t) = v sin ωt.
Integrating and taking into account the initial conditions x(0) = 0, y(0) = − ωv we compute for the coordinates x(t) =
v sin ωt, ω
y(t) = −
v cos ωt. ω
It follows that in Σ the motion is a uniform rotational motion on the x–y plane with angular speed ω = γq0Bm . (b) We compute the electric and the magnetic field in Σ by means of the transformation equations for the electric and the magnetic field. We have
11 Electromagnetic Field
351
E 1 = E 1 = 0 E 2 = γ (u) (E 2 − u B3 ) = −γ (u)u B , E 3 = γ (u) (E 3 + u B2 ) = 0 B1 = B1 =0 B2 = γ (u) B2 + B3 = γ (u) B3 −
u c2 u c2
E3 = 0 E 2 = γ (u)B.
We demand −γ (u)u B = −E 0 γ (u)B = B0
⇒u=
E0 . B0
The condition uc < 1 ⇔ 1c E 0 < B0 ⇔ X = c12 E 02 − B02 < 0, where X is one 0 of the invariants of the electromagnetic field. Also the γ factor γ (u) = √B−X , √ thus B = −X . In order to describe the motion of the charged particle in Σ we determine first its motion in Σ and then transfer the results to Σ using the boost relating Σ and Σ . In (a) we have found that the motion in Σ under the specified initial conditions we have chosen is described by the curve x(t) =
v sin ωt, ω
y(t) = −
v cos ωt. ω
Therefore, in Σ we have x (t ) = γ (u)
v
ω
sin ωt − ut ,
y (t ) = − ωv cos ωt, t = γ (u) t −
u v c2 ω
sin ωt .
If we eliminate the time t from these relations we find the orbit in Σ in the parametric form x (t ), y (t ). The average velocity of the particle in Σ v¯ = 0 because in Σ the motion is periodic. The corresponding average velocity in Σ is computed from the relativistic rule of composition of threevelocities. We find v¯ x + u E0 =u= , B0 1 + ucv¯2x v¯ y = 0, v¯ y = 1 + ucv¯2x γ (u) v¯ z v¯ z = = 0, 1 + ucv¯2x γ (u) v¯ x =
352
11 Electromagnetic Field
that is v¯ =
E0 x, B0
which should be expected. 11.21. The components of the fields E, B in Σ are E x = E cos θ = 4Bc cos θ, E y = 0, E z = E sin θ = 4Bc sin θ, Bx = B y = 0, Bz = B. Therefore, the fields are 1 E = 4B(cos θ, 0, sin θ ), c
B = B(0, 0, 1).
We compute Y =
1 E · B = 4B 2 sin θ. c
Because Y = 0 we infer that there are infinitely many LCF Σ in which the fields E , B are parallel.
(b) The velocity of the LCF Σ in which the fields E , B are parallel is of the form 1 E×B v = βcˆe where eˆ =  1c E×B is the unit vector along the direction of the vector 1 E c
× B. We compute
c
1 E × B = 4B 2 cos θ j ⇒ v = βcj. c With this velocity Σ is moving in the standard configuration wrt Σ along the yaxis. From the equations of transformation of the electric and magnetic fields we calculate the fields E , B . We have
11 Electromagnetic Field
353
E x = γ (E x − βcBz ) = γ Bc(4 cos θ − β), E y = E y = 0, E z = γ (E z + βcBx ) = 4γ Bc sin θ, 1 Bx = γ (Bx + β E z ) = 4βγ B sin θ, c B y = B y = 0, 1 Bz = γ Bz − β E x = γ B(1 − 4β cos θ ), c or in vector form 1 E = γ B(4 cos θ − β, 0, 4 sin θ ), c
B = γ B(4β sin θ, 0, 1 − 4β cos θ ).
The condition E B ⇔ E × B = 0. We compute E × B  = γ 2 B 2 (4β 2 cos θ − 17β + 4 cos θ ), so that β is determined from the equation 4β 2 cos θ − 17β + 4 cos θ = 0. The accepted (β < 1) solution of this equation is β=
17 −
√ 289 − 64 cos2 θ . 8 cos θ
We infer that the velocities of all LCF Σ in which the fields E , B are parallel are given by the relation * v=c
17 −
where λ is a real parameter and
+ √ 289 − 64 cos2 θ E j+λ , 8 cos θ E  E E 
is the unit in the common direction of E , B .
11.22. The electromagnetic field in Σ is found from the transformation equations E = E = −AX sin θ i, E⊥ = γ (E⊥ + βc × B) =γ (AX cos θ j − AX cos θ j) = 0, B = B = 0, AX 1 sin2 θ k. B⊥ = γ B⊥ − β × E = γ c c
354
11 Electromagnetic Field
√ The γ factor γ = 1/ 1 − β 2 = 1/ 1 − cos2 θ = 1/ sin θ. Replacing we find AX i, γ AX B = k. γc E = −
(Note that the invariants of the electromagnetic field E · B and B2 − c12 E2 have the same value (i.e., zero) both in Σ and Σ as it should be). The fields E , B we have found are not the answer because they are expressed in terms of the coordinates of Σ. We use the boost relating Σ, Σ and have (recall that β = cos θ, γ = 1/ sin θ ) y cos θ β X = sin 2π ν γ t + x − sin θ − γ (x + βct ) c c c γ y = sin 2π ν γ (1 − β cos θ )t − sin θ − (cos θ − β)x c c y 2π ν 2π ν = sin 2π ν sin θ t − sin θ = sin sin θ (ct − y ) = sin (ct − y ). c c cγ Hence 2π ν A sin (ct − y )i , γ cγ 2π ν A sin (ct − y )k . B = γc cγ
E = −
Concerning the frequency ν of the wave in Σ we note from the equations we have found that ν = γν , hence 2π ν A sin (ct − y )i , γ c 2π ν A sin (ct − y )k . B = γc c
E = −
From these relations we conclude that the electromagnetic wave in Σ is propagating along the direction of the y axis (the direction of propagation is defined by the vector E × B = i × k ) with frequency ν = γν = ν sin θ , amplitude γA = A sin θ , and the electric and magnetic fields in phase. 11.23. We consider the LCF Σ and Σ which are related with a boost along the direction of the xaxis with speed u. Consider an electromagnetic field which in Σ and Σ is represented, respectively, with the fields E = (E x , E y , E z )
11 Electromagnetic Field
355
B = (Bx , B y , Bz ) and E = (E x , E y , E z ), B = (Bx , B y , Bz ). Consider first the electric field. The transformation of the field gives E·u , u E y = γ (E y + u x Bz ), E z = γ (E z + u x B y ), E x = E x =
which implies that the electric field in Σ is E·u i + γ (E y + u x Bz )j + γ (E z + u x B y )k u = E + γ (E⊥ + βc × B),
E = E x i + E y j + E z k =
u and E⊥ = E y j + E z k is the parallel and the normal projection where E = E·u u u of the field along the relative velocity of Σ and Σ . Considering the corresponding projections for E we write E = E , E⊥ = γ (E⊥ + βc × B). These relations contain threevectors, therefore, they are invariant under rotation of the spatial axes. This implies that they are valid for arbitrary relative motion of Σ and Σ and they are not restricted by the assumption that the Σ and Σ are moving in the standard configuration. Working similarly we prove the relations B = B , 1 B⊥ = γ B⊥ − β × E . c The inverse relations from Σ to Σ are obtained if we change the sign of β. That is, we have E = E ,
E⊥ = γ (E⊥ − β × B ), B = B , 1 B⊥ = γ B⊥ + β × E . c Remark We note that for small speeds the terms containing β 2 are ignored (Newtonian limit) and the relations of transformation of the fields become
356
11 Electromagnetic Field
E = E + βc × B, 1 B = B − β × E. c From these relations the asymmetry introduced by the Lorentz transformation along and normal to the direction of the relative velocity as well as the isotropy of the threespace of Newtonian Physics becomes apparent Application From the given data, we have for the electromagnetic field (E , B ) in Σ E = E = 0, because u⊥E,
E⊥ = γ (E⊥ + βc × B) =γ E⊥ =γ E, B = B = 0, 1 1 1 1 B⊥ = γ (B⊥ − β × E) = − γ β × E = − β × E⊥ ⇒ B⊥ + β × E⊥ = 0. c c c c 11.24. (a) In a conducting medium the density of charge is ρ = 0 (there are no free charges) and by Ohm’s law the conduction current j = σ E, where E is the electric field of the wave and σ the conductivity of the medium. With these data Maxwell equations are written as follows ∂H = 0, ∂t ∇ · E = ρ, ∂E ∇ ×H= + σ E, ∂t ∇ · H = 0.
∇ ×E+μ
The LHS of the third equation gives ∇ × (∇ × H) = ∇(∇ · H ) − ∇ 2 H = −∇ 2 H and the RHS ∂E ∂ ∂H ∂ 2H ∂H ∇ × + σ (∇ × E) = (∇ × E) + σ −μ = −μ 2 − σ μ . ∂t ∂t ∂t ∂t ∂t Equating the two expressions we find μ
∂ 2H ∂H = 0, − ΔH+σ μ 2 ∂t ∂t
11 Electromagnetic Field
357
where ΔH = ∇ 2 H. We are looking for solutions of the form H(t, r ) = H0 ei(ωt−k·r) . We compute ∂ 2H = −ω2 H ∂t 2 and ΔH = −(k · k)H. Replacing into the equation we find the following relation for the wave vector k k · k − μω2 + iσ μω = 0. Obviously k is a complex vector. We write k = k1 + ik2 and calculate for the real vectors k1 and k2 the conditions k12 − k22 = μω2 , σ μω , k1 k2 = − 2 cos θ where cos θ = cos(k1 , k2 ) and π2 < θ < π (because k1 k2 > 0). We multiply the first with k22 and use the second to obtain the equation k14 − μω2 k12 +
σ μω 2 = 0. 2 cos θ
The accepted solution of this equation is ! 2 σ 1− . ω cos θ




μ k1 = ω 1+ 2
Similarly we compute μ k2 = ω 1− 2
! 2 σ 1− . ω cos θ
We conclude that the electromagnetic wave is given by the fields H(t, r) = H0 ek2 ·r ei(ωt−k1 ·r) , E(t, r) = E0 ek2 ·r ei(ωt−k1 ·r) .
358
11 Electromagnetic Field
(b) In case there are no free charges and the conduction current vanishes we apply the results of (a) assuming σ = 0. The propagation equation for the magnetic field is ΔH = μ
∂ 2H ∂t 2
and similarly for the electric field ΔE = μ
∂ 2E . ∂t 2
From these two equations it follows that the speed of propagation of the wave √ equals √1μ . Also we compute k2 = 0 and k1 = ω μ = k. A0 1 and k i = K , where eˆ Σ A Σ e is the unit along the direction of propagation of k i in Σ. The inner product Ai ki = − A0 K + A · eˆ =− A0 K + A cos φ = 0 ⇒ A0 = A cosK φ . The condition for Ai to be null is A0 = A which implies K = cos φ ≤ 1. We conclude that if K > 1 A0 > A ⇒Ai is spacelike. If K ≤ 1, then Ai is a null fourvector provided it propagates in a direction eˆ 1 which makes an angle φ with the direction eˆ of propagation of k i in Σ. (b) We consider the fourquantities Ai , ki to be constant and have from the first Maxwell equation
11.25. (a) In an LCF Σ assume that Ai =
r
j
r
r
Ω , j = A j (ikr x r ), j eikr x = i A j kr x r , j eikr x = i Ar k r eikr x = 0 ⇒ Ar k r = 0. Similarly from the second Maxwell equation it follows r
r
Ωi, j = i Ai k j eikr x ⇒ Ωi, j j = −Ai k j k j eikr x = −k j k j Ωi = 0 ⇒ k j k j = 0 . Next we prove that the fourquantities Ai , ki are fourvectors. To do that we consider the Lorentz transformation L ii and write r
Ωi = L ii Ωi = L ii Ai eikr x . But
r
Ωi = Ai eikr x . Comparing these relations we conclude
Ai = L ii Ai , kr x r = kr x r , that is, Ai , ki are fourvectors.
11 Electromagnetic Field
359
11.26. The force on the particle is the Lorentz force F = e(E + u × B). The electric and the magnetic fields in terms of the potentials φ, A are given by the relations E = −∇φ −
∂A , ∂t
B = ∇ × A.
Replacing in the expression of the force we find ∂A F = e −∇φ − + u× (∇ × A) . ∂t
(11.3)
Using the identity ∇(A · B) = B · ∇A + A · ∇B + A× (∇ × B) + B× (∇ × A) of vector calculus and taking into account the relations (why?) ∇ · u = 0, ∇ ×u = 0 we compute ∇(A · u) = u · ∇A + A · ∇u + A× (∇ × u) + u× (∇ × A) = u · ∇A + u× (∇ × A) . Therefore u× (∇ × A) = ∇(u · A) − u(∇ · A). The total derivative of the potential A(t, r) is ∂A dA ∂A dA = + u · (∇ · A) ⇒ u · (∇ · A) = − . dt ∂t dt ∂t Therefore, the term u× (∇ × A) = ∇(u · A)+ ∂A − ∂t the final expression
dA . dt
Replacing in (11.3) we find
dA . F = e −∇ (φ − u · A) − dt
(11.4)
Lagrange equations are d ∂T ∂T = Fq , . − dt ∂ q ∂q
q = x, y, z,
(11.5)
360
11 Electromagnetic Field
where Fq are the components of the generalized force. Because we have employed Cartesian coordinates the components of the generalized force are the x, y, z components of the threeforce. Setting q = x, y, z and replacing Fq from (11.4) we have ∂ d ∂T + e Aq − [T − e (φ − u · A)] = 0. dt ∂ q˙ ∂q The term ∂ ∂ T + e Aq q˙ = (T + eA · u − eφ) ∂ q˙ ∂ q˙ and finally d ∂L ∂L − = Fq dt ∂ q˙ ∂q
(q = x, y, z)
where .
L(r,r) = T − e(φ − A · u). We conclude that the electromagnetic field corresponds to the generalized potential .
U (r,r) = e(φ − A · u). In order to complete the Lagrangian we have to replace the kinetic energy of the free particle. In Newtonian Physics we have T = 12 mu2 , therefore we take .
L Newton (r,r) =
1 2 mu − e(φ − A · u). 2
In Special Relativity the kinetic energy is given by the expression T = −mc
2
1−
u2 . c2
Hence .
L S R (r,r) = −mc
2
1−
u2 − e(φ − A · u). c2
In order to compare the two Lagrangians we Taylor expand the relativistic kinetic energy in terms of β. We have
11 Electromagnetic Field .
361
L S R (r,r) = −mc
2
1 u2 1− 2 − c 8
u2 c2
!
2
+ . . . − e(φ − A · u)
.
= L Newton (r,r) + O(β 2 ). Having computed the Lagrangian we compute the Hamiltonian from the wellknown relation HS R = p · u−L S R , where the generalized momentum is defined by p =
∂ LSR . ∂u
We compute
⎛ ⎞ u −2 1 ∂ L SR 2 = mc2 ⎝− c ⎠ + eA = mγ u+eA, p= ∂u 2 1 − u2 c2
therefore HS R = mγ u +eA · u+mc 2
2
1−
u2 + e(φ − A · u) = mγ c2 + eφ. c2
We replace the term mγ c2 in terms of the conjugate momentum p and get the final expression H = c (p−qA)2 + mc2 + eφ.