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6 Special Theory of Relativity
6.4 θ
∗
= 90
◦
γ = γ c γ
∗
= 1.667 × 1.038 = 1.73
β = (1.73
2
− 1)
1/2
/1.73 = 0.816
tan θ = sin θ
∗
/(γ c β c /β
∗ ) = 1/(1.667×0.8/0.268) = 0.2
θ = 11.3
◦
6.5 θ = 90
◦
γ
∗
= γ c γ (1 − β c β cos θ) = γ c γ
γ = γ
∗
/γ c
γ c = 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 = (1.0176
2
− 1)
1/2
/1.0176 = 0.185
tan θ
∗
= sin θ/γ c (cos θ − β c /β
∗ )
= −β
∗
/β c γ c (Because θ = 90
◦ )
β c = β π = 0.2
tan θ
∗
= −0.268/1.02 × 0.2 = 1.3137
θ
∗
= 127
◦
6.6 The scalar wave equation for the propagation of electromagnetic waves derivable 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 the S system. We are required to show that in S
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 along x-axis so that the wave function will depend only on x and t. Now the function ϕ(x
, y
, z
, t
) = 0
is obtained from ϕ(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 − βc dt)
( 8 )
Substituting (7) and (8) in (6) and equating the coefficients of dx and dt:
(∂ϕ/∂ x) dx + (∂ϕ/∂t) dt = (γ ∂ϕ/∂x
− γβc∂ϕ/∂t
) dx +
(γ ∂ϕ/∂t
− γβc∂ϕ/∂x
) dt