6.3 Solutions
333
6.2 The transformation matrix is
Λ =
⎡
⎢
⎢
⎣
γ 0 0 iβγ
0 1 0 0
0 0 1 0
−iβγ 0 0 γ
⎤
⎥
⎥
⎦
Set γ = cosh α and β = tanh α, so that γβ = sinh α, the transformation
matrix becomes
Λ =
⎡
⎢
⎢
⎣
cosh α 0 0 i sinh α
0
1 0
0
0
0 1
0
− i sinh α 0 0 cosh α
⎤
⎥
⎥
⎦ .
Since we can write i sinh α = siniα and cosh α = cosiα, the matrix
Λ corresponds to a rotation through an angle iα in four-dimensional space,
Further the transformation equations can be obtained from
⎡
⎢
⎢
⎣
x
1
x
2
x
3
ict
⎤
⎥
⎥
⎦ =
⎡
⎢
⎢
⎣
cosh α 0 0 i sinh α
0
1 0
0
0
0 1
0
−i sinh α 0 0 cosh α
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
x 1
x 2
x 3
ict
⎤
⎥
⎥
⎦ .
and the inverse transformation equations from
⎡
⎢
⎢
⎣
x 1
x 2
x 3
ict
⎤
⎥
⎥
⎦ =
⎡
⎢
⎢
⎣
cosh α 0 0 −i sinh α
0
1 0
0
0
0 1
0
i sinh α 0 0 cosh α
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
x
1
x
2
x
3
ict
⎤
⎥
⎥
⎦ .
6.3 β
∗
= 0.268 and β c = 0.8
γ c = 1/(1 − β c
2 )
1/2
= 1/(1 − 0.8
2 )
1/2
= 1/0.6 = 1.667
γ
∗
= 1/(1 − β
∗2 )
1/2
= 1/(1 − 0.268
2 )
1/2
= 1.038
γ = γ c γ
∗ (1 + β c β
∗ cos θ
∗ ), β = (γ
2
− 1)
1/2
/γ
tan θ = sin θ
∗
/γ c (cos θ
∗
+ β c /β
∗ )
θ
∗
= 0
γ = (1.667 × 1.038)(1 + 0.8 × 0.268) = 1.4248
β = 0.712
tan θ = 0 (Because θ
∗
= 0)
Therefore, θ = 0