6.3 Solutions 333
6.2 The transformation matrix isΛ=
⎡
⎢
⎢
⎣
γ 00 iβγ
0100
0010
−iβγ 00 γ⎤
⎥
⎥
⎦
Setγ=coshαandβ=tanhα, so thatγβ=sinhα, the transformation
matrix becomesΛ=
⎡
⎢
⎢
⎣
coshα 00 isinhα
0100
0010
−isinhα0 0 coshα⎤
⎥
⎥
⎦.
Since we can writeisinhα =siniαand coshα= cosiα, the matrix
Λcorresponds to a rotation through an angleiαin four-dimensional space,
Further the transformation equations can be obtained from⎡
⎢
⎢
⎣x′ 1
x′ 2
x′ 3
ict′⎤
⎥
⎥
⎦=
⎡
⎢
⎢
⎣
coshα 00 isinhα
0100
0010
−isinhα0 0 coshα⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
x 1
x 2
x 3
ict⎤
⎥
⎥
⎦.
and the inverse transformation equations from
⎡
⎢
⎢
⎣x 1
x 2
x 3
ict⎤
⎥
⎥
⎦=
⎡
⎢
⎢
⎣
coshα 00 −isinhα
010 0
001 0
isinhα0 0 coshα⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
x 1 ′
x 2 ′
x 3 ′
ict′⎤
⎥
⎥
⎦.
6.3β∗= 0 .268 andβc= 0. 8
γc= 1 /(1−βc^2 )^1 /^2 = 1 /(1− 0. 82 )^1 /^2 = 1 / 0. 6 = 1. 667
γ∗= 1 /(1−β∗^2 )^1 /^2 = 1 /(1− 0. 2682 )^1 /^2 = 1. 038
γ=γcγ∗(1+βcβ∗cosθ∗),β=(γ^2 −1)^1 /^2 /γ
tanθ=sinθ∗/γc(cosθ∗+βc/β∗)
θ∗= 0
γ=(1. 667 × 1 .038)(1+ 0. 8 × 0 .268)= 1. 4248
β= 0. 712
tanθ=0 (Becauseθ∗=0)
Therefore,θ= 0