1000 Solved Problems in Modern Physics

334 6 Special Theory of Relativity

6.4θ∗= 90 ◦

γ=γcγ∗= 1. 667 × 1. 038 = 1. 73

β=(1. 732 −1)^1 /^2 / 1. 73 = 0. 816

tanθ=sinθ∗/(γcβc/β∗)= 1 /(1. 667 × 0. 8 / 0 .268)= 0. 2

θ= 11. 3 ◦

6.5θ= 90 ◦

γ∗=γcγ(1−βcβcosθ)=γcγ

γ=γ∗/γc

γc= 1 /[1−(0.2)^2 ]^1 /^2 = 1. 02

γ∗= 1 /[1−(0.268)^2 ]^1 /^2 = 1. 038

γ= 1. 038 / 1. 02 = 1. 0176

β=(γ^2 −1)^1 /^2 /γ=(1. 01762 −1)^1 /^2 / 1. 0176 = 0. 185

tanθ∗=sinθ/γc(cosθ−βc/β∗)

=−β∗/βcγc(Becauseθ= 90 ◦)

βc=βπ= 0. 2

tanθ∗=− 0. 268 / 1. 02 × 0. 2 = 1. 3137

θ∗= 127 ◦

6.6 The scalar wave equation for the propagation of electromagnetic waves deriv-

able from Maxwell’s equations is:

(∂^2 /∂x^2 +∂^2 /∂y^2 +∂^2 /∂z^2 −(1/c^2 )∂^2 /∂t^2 )φ(x,y,z,t)=0(1)

for theSsystem. We are required to show that inS′system, the equation has

the form:

∂^2

∂x′^2

+

∂^2

∂y′^2

+

∂^2

∂z′^2

−

(

1

c^2

)(

∂^2

∂t′^2

)

φ(x′,y′,z′,t′)= 0

The Lorentz transformations are:

x′=γ(x−βct)(2)

y′=y (3)

z′=z (4)

t′=γ(t−βx′c)(5)

Assume that we have propagation alongx-axis so that the wave func-

tion will depend only onx andt. Now the functionφ(x′,y′,z′,t′) = 0

is obtained fromφ(x,y,z,t) =0 by a substitution of variables. We have

φ(x,t)=φ(x′,t′). Then,

dφ=(∂φ/∂x)dx+(∂φ/∂t)dt=(∂φ/∂x′)dx′+(∂φ/∂t′)dt′ (6)

Differentiating (2) and (5),

dx′=γ(dx−βcdt)(7)

dt′=γ(dt−βcdt)(8)

Substituting (7) and (8) in (6) and equating the coefficients of dxand dt:

(∂φ/∂x)dx+(∂φ/∂t)dt=(γ∂φ/∂x′−γβc∂φ/∂t′)dx+

(γ∂φ/∂t′−γβc∂φ/∂x′)dt