4.2 Bateman Electromagnetic Waves
77
4.2
Bateman Electromagnetic Waves
An illustration of the use of the shear-free, null geodesic congruence tangent to the
vector field k i in (4.12) is to demonstrate how Maxwell’s equations can be fully
integrated for the case of electromagnetic radiation propagating in a vacuum. We
begin with the null geodesic equation (4.30) which we write as
∂
∂t
(k
1
− i k
2 ) + (k · ∇)(k
i
− i k
2 ) = 0 ,
(4.50)
and
∂k 3
∂t
+ (k · ∇)k
3
= 0 .
(4.51)
Now define the complex-valued function Y(t, x, y, z), with complex conjugate ¯
Y,
by
Y =
k 1 − i k 2
1 − k 3 ⇒ k
3
=
Y ¯
Y − 1
Y ¯
Y + 1
,
(4.52)
with the implication here following from k · k = 1. Introducing new coordinates
ζ, ¯
ζ , u, v via
ζ = x − i y , ¯
ζ = x + i y , u = −t − z , v = t − z ,
(4.53)
we have
k · ∇ = (1 − k
3 )
Y
∂
∂ζ
+ ¯
Y
∂
∂ ¯
ζ
− k
3
∂
∂u
+
∂
∂v
.
(4.54)
Now (4.50) reads
Y v − Y u + (1 − k
3 ) (Y Y ζ + ¯
Y Y ¯
ζ ) − k
3 (Y u + Y v ) = 0 ,
(4.55)
with the subscripts on Y denoting partial differentiation, and (4.51) becomes
∂
∂v
−
∂
∂u
k
3
+ (1 − k
3 )
Y
∂
∂ζ
+ ¯
Y
∂
∂ ¯
ζ
k
3
− k
3
∂
∂u
+
∂
∂v
k
3
= 0 .
(4.56)
Substituting from (4.52) for k 3 these equations simplify to
Y v + Y Y ζ + ¯
Y (Y ¯
ζ − Y Y u ) = 0 ,
(4.57)
77
4.2
Bateman Electromagnetic Waves
An illustration of the use of the shear-free, null geodesic congruence tangent to the
vector field k i in (4.12) is to demonstrate how Maxwell’s equations can be fully
integrated for the case of electromagnetic radiation propagating in a vacuum. We
begin with the null geodesic equation (4.30) which we write as
∂
∂t
(k
1
− i k
2 ) + (k · ∇)(k
i
− i k
2 ) = 0 ,
(4.50)
and
∂k 3
∂t
+ (k · ∇)k
3
= 0 .
(4.51)
Now define the complex-valued function Y(t, x, y, z), with complex conjugate ¯
Y,
by
Y =
k 1 − i k 2
1 − k 3 ⇒ k
3
=
Y ¯
Y − 1
Y ¯
Y + 1
,
(4.52)
with the implication here following from k · k = 1. Introducing new coordinates
ζ, ¯
ζ , u, v via
ζ = x − i y , ¯
ζ = x + i y , u = −t − z , v = t − z ,
(4.53)
we have
k · ∇ = (1 − k
3 )
Y
∂
∂ζ
+ ¯
Y
∂
∂ ¯
ζ
− k
3
∂
∂u
+
∂
∂v
.
(4.54)
Now (4.50) reads
Y v − Y u + (1 − k
3 ) (Y Y ζ + ¯
Y Y ¯
ζ ) − k
3 (Y u + Y v ) = 0 ,
(4.55)
with the subscripts on Y denoting partial differentiation, and (4.51) becomes
∂
∂v
−
∂
∂u
k
3
+ (1 − k
3 )
Y
∂
∂ζ
+ ¯
Y
∂
∂ ¯
ζ
k
3
− k
3
∂
∂u
+
∂
∂v
k
3
= 0 .
(4.56)
Substituting from (4.52) for k 3 these equations simplify to
Y v + Y Y ζ + ¯
Y (Y ¯
ζ − Y Y u ) = 0 ,
(4.57)
