90
4 Bateman Waves
From (4.130),
E
αβ
+ i H
αβ
= |q |
2 ( ˆ
q
α
+ i ˆ
w
α )( ˆ
q
β
+ i ˆ
w
β ) ,
(4.160)
and using this in (4.158) gives
+ 0α = −|q |
2 ( ˆ
q
α
+ i w
α )
( ˆ
q
1
+ i ˆ
w
1 ) dx ∧ dt + ( ˆ
q
2
+ i ˆ
w
2 ) dy ∧ dt
+( ˆ
q
3
+ i ˆ
w
3 ) dz ∧ dt − ( ˆ
q
β
+ i ˆ
w
β ) ˆ
k
γ dx
β
∧ dx
γ
.
(4.161)
Since ˆ
q = ˆ
w × ˆ
k and ˆ
w = ˆ
k × ˆ
q it follows that
( ˆ
q
β
+ i ˆ
w
β ) ˆ
k
γ dx
β
∧ dx
γ
= i ( ˆ
q
3
+ i ˆ
w
3 ) dx ∧ dy − i ( ˆ
q
2
+ i ˆ
w
2 ) dx ∧ dz
+i ( ˆ
q
1
+ i ˆ
w
1 ) dy ∧ dz .
(4.162)
As a result (4.161) becomes
+ 0α = −|q |
2 ( ˆ
q
α
+ i ˆ
w
α )
( ˆ
q
1
+ i ˆ
w
1 )(dx ∧ dt − i dy ∧ dz)
+( ˆ
q
2
+ i ˆ
w
2 )(dy ∧ dt + i dx ∧ dz)
+( ˆ
q
3
+ i ˆ
w
3 )(dz ∧ dt − i dx ∧ dy)
.
(4.163)
With (4.160) in (4.159) we have
+ αβ = |q |
2
{ ˆ
k
α ( ˆ
q
β
+ i ˆ
w
β ) − ˆ
k
β ( ˆ
q
α
+ i ˆ
w
α )}( ˆ
q
σ
+ i ˆ
w
σ ) dx
σ
∧ (dt − ˆ
k
γ dx
γ ) .
(4.164)
Making use of (4.162) again and the simplifications
− i ( ˆ
q
3
+ i ˆ
w
3 ) = ˆ
k
1
ˆ
q
2
− ˆ
k
2
ˆ
q
1
+ i ( ˆ
k
1
ˆ
w
2
− ˆ
k
2
ˆ
w
1 ) ,
(4.165)
i ( ˆ
q
2
+ i ˆ
w
2 ) = ˆ
k
1
ˆ
q
3
− ˆ
k
3
ˆ
q
1
+ i ( ˆ
k
1
ˆ
w
3
− ˆ
k
3
ˆ
w
1 ) ,
(4.166)
−i ( ˆ
q
1
+ i ˆ
w
1 ) = ˆ
k
2
ˆ
q
3
− ˆ
k
3
ˆ
q
2
+ i ( ˆ
k
2
ˆ
w
3
− ˆ
k
3
ˆ
w
2 ) ,
(4.167)
(4.164) is given by
+ 12 = −i|q |
2 ( ˆ
q
3
+ i ˆ
w
3 )
( ˆ
q
1
+ i ˆ
w
2 )(dx ∧ dt − i dy ∧ dz)
+( ˆ
q
2
+ i ˆ
w
2 )(dy ∧ dt + i dx ∧ dz)