74
4 Bateman Waves
A further important equation, in addition to (4.30), satisfied by the null geodesic
vector field k i is available by expanding k α,β + k β,α , where k α = −k α , on
the orthonormal triad k, e, b. Such an expansion of any A αβ follows from the
identity A αβ = δ μα δ νβ A μν after substitution for the Kronecker deltas from (4.19).
Proceeding in this way we obtain
k α,β + k β,α = (k α e β + k β e α ) k μ,λ e
μ k
λ
+ (k α b β + k β b α ) k μ,λ b
μ k
λ
+(e α b β + e β b α )(k μ,λ b
μ e
λ
+ k μ,λ e
μ b
λ ) + 2 e α e β (k μ,λ e
μ e
λ )
+2 b α b β (k μ,λ b
μ b
λ ) .
(4.31)
We use (4.24) and (4.25) to simplify this expression. First we note that
e · (∇ × b) = e
α αβγ b
γ
,β = e
α αβμ δ μγ b
γ
,β
= e
α αβμ (k
μ k
γ
+ e
μ e
γ
+ b
μ b
γ ) b
γ
,β by (4.19)
= −e
α αβμ k
μ k
γ
,β b
γ (since b
γ b
γ
= 1 and k
γ b
γ
= 0)
= −e
α αβμ μρσ e
ρ b
σ k
γ
,β b
γ (since k
μ
= μρσ e
ρ b
σ )
= −b
β b
γ k
γ
,β by (4.17)
= k γ ,β b
γ b
β .
(4.32)
In similar fashion we find that
b · (∇ × e) = −k γ ,β e
γ e
β , b · (∇ × b) = −k γ ,β b
γ e
β
+ k
γ e
β b
β
,γ ,
(4.33)
and
e · (∇ × e) = k γ ,β e
γ b
β
− k
γ b
β e
β
,γ .
(4.34)
Substituting these into (4.24) and (4.25) (and using b β e β = 0 in (4.25)) we have
k γ ,β e
γ e
β
= k γ ,β b
γ b
β and k γ ,β b
γ e
β
+ k γ ,β e
γ b
β
= 0 .
(4.35)
When these are introduced into (4.31) the result is
k α,β + k β,α = k α
(k μ,λ e
μ k
λ ) e β + (k μ,λ b
μ k
λ ) b β
+k β
(k μ,λ e
μ k
λ ) e α + (k μ,λ b
μ k
λ ) b α
+2 k μ,λ e
μ e
λ (e α e β + b α b β ) .
(4.36)
But by the first of (4.35)
2 k μ,λ e
μ e
λ
= k μ,λ (e
μ e
λ
+ b
μ b
λ ) = k μ,λ (δ μλ − k
μ k
λ ) = k μ,μ = −k
μ
,μ ,
(4.37)
4 Bateman Waves
A further important equation, in addition to (4.30), satisfied by the null geodesic
vector field k i is available by expanding k α,β + k β,α , where k α = −k α , on
the orthonormal triad k, e, b. Such an expansion of any A αβ follows from the
identity A αβ = δ μα δ νβ A μν after substitution for the Kronecker deltas from (4.19).
Proceeding in this way we obtain
k α,β + k β,α = (k α e β + k β e α ) k μ,λ e
μ k
λ
+ (k α b β + k β b α ) k μ,λ b
μ k
λ
+(e α b β + e β b α )(k μ,λ b
μ e
λ
+ k μ,λ e
μ b
λ ) + 2 e α e β (k μ,λ e
μ e
λ )
+2 b α b β (k μ,λ b
μ b
λ ) .
(4.31)
We use (4.24) and (4.25) to simplify this expression. First we note that
e · (∇ × b) = e
α αβγ b
γ
,β = e
α αβμ δ μγ b
γ
,β
= e
α αβμ (k
μ k
γ
+ e
μ e
γ
+ b
μ b
γ ) b
γ
,β by (4.19)
= −e
α αβμ k
μ k
γ
,β b
γ (since b
γ b
γ
= 1 and k
γ b
γ
= 0)
= −e
α αβμ μρσ e
ρ b
σ k
γ
,β b
γ (since k
μ
= μρσ e
ρ b
σ )
= −b
β b
γ k
γ
,β by (4.17)
= k γ ,β b
γ b
β .
(4.32)
In similar fashion we find that
b · (∇ × e) = −k γ ,β e
γ e
β , b · (∇ × b) = −k γ ,β b
γ e
β
+ k
γ e
β b
β
,γ ,
(4.33)
and
e · (∇ × e) = k γ ,β e
γ b
β
− k
γ b
β e
β
,γ .
(4.34)
Substituting these into (4.24) and (4.25) (and using b β e β = 0 in (4.25)) we have
k γ ,β e
γ e
β
= k γ ,β b
γ b
β and k γ ,β b
γ e
β
+ k γ ,β e
γ b
β
= 0 .
(4.35)
When these are introduced into (4.31) the result is
k α,β + k β,α = k α
(k μ,λ e
μ k
λ ) e β + (k μ,λ b
μ k
λ ) b β
+k β
(k μ,λ e
μ k
λ ) e α + (k μ,λ b
μ k
λ ) b α
+2 k μ,λ e
μ e
λ (e α e β + b α b β ) .
(4.36)
But by the first of (4.35)
2 k μ,λ e
μ e
λ
= k μ,λ (e
μ e
λ
+ b
μ b
λ ) = k μ,λ (δ μλ − k
μ k
λ ) = k μ,μ = −k
μ
,μ ,
(4.37)
