4.4 Gravitational Radiation
87
Putting (4.131) and (4.132) together we can solve for k i to find that
k j = u j −
1
E pq E pq
η pij q H
p
k u
q
E
ki .
(4.133)
But
E
pq
E pq = 2|q |
4 , H
p
k E
ki
= |q |
2 (q
p w
i
− q
i w
p ) ,
(4.134)
and so
k j = u j +
η 0jkl q k w l
|q | 2
= u j +
0jβγ q β w γ
|q | 2
.
(4.135)
With 0123 = −1 this gives us
k j =
1, −
q × w
|q | 2
⇔ k
j
= (1, k) with k = ˆ
q × ˆ
w = ˆ
k (say) .
(4.136)
Hence we see that ˆ
q, ˆ
w, ˆ
k form a right-handed orthonormal triad and it follows that
δ αβ = ˆ
q
α
ˆ
q
β
+ ˆ
w
α
ˆ
w
β
+ ˆ
k
α ˆ
k
β .
(4.137)
With (4.130) and the scalar products (4.129) we have
E αβ E βσ = |q |
4 ( ˆ
q α ˆ
q σ + ˆ
w α ˆ
w σ ) = H αβ H βσ ,
(4.138)
and
E αβ H βσ = |q |
4 ( ˆ
q α ˆ
w σ − ˆ
q σ ˆ
w α ) = −H αβ E βσ .
(4.139)
Using the time evolution equations (4.118) and (4.119) we find that
∂
∂t
E αβ E βσ − H αβ H βσ
= |q |
2
E ασ
ˆ
q · ∇ × ˆ
w) + ˆ
w · (∇ × ˆ
q)
+|q |
2
H ασ
ˆ
q · (∇ × ˆ
q) − ˆ
w · (∇ × ˆ
w)
,
(4.140)
and
∂
∂t
E αβ H βσ + H αβ E βσ
= |q |
2
E ασ
ˆ
w · (∇ × ˆ
w) − ˆ
q · (∇ × ˆ
q)
+|q |
2
H ασ
ˆ
q · (∇ × ˆ
w) + ˆ
w · (∇ × ˆ
q)
,
(4.141)