84
4 Bateman Waves
demonstrating the E, B, k form a right handed triad as in the general case (4.15).
If w = r − t and A(w) = da(w)/dw, C(w) = dc(w)/dw then ∂a/∂t =
−A, ∂c/∂t = −C and ∂a/∂r = A, ∂c/∂r = C. Hence we see that we can write
E = −
∂σ
∂t
and B = ∇ × σ ,
(4.106)
with σ having components
σ
1
=
1
r(r − z) 2
c x y − a{r(r − z) − x
2
}
,
(4.107)
σ
2
=
1
r(r − z) 2
a x y − c{r(r − z) − y
2
}
,
(4.108)
σ
3
= −
1
r(r − z)
(a x + c y) .
(4.109)
This 3-potential satisfies
∇ · σ = 0 .
(4.110)
Defining the 3-vector ω = (ω 1 , ω 2 , ω 3 ) by
ω
1
=
1
r(r − z) 2
−a x y − c{r(r − z) − x
2
}
,
(4.111)
ω
2
=
1
r(r − z) 2
c x y + a{r(r − z) − y
2
}
,
(4.112)
ω
3
= −
1
r(r − z)
(−c x + a y) ,
(4.113)
we find that
(σ
α
+ i ω
α ) dx
α
= −(a + i c) d
x − i y
r − z
.
(4.114)
Thus the 3-potential σ is given neatly by the 1-form
σ
α dx
α
= Re
−(a + i c) d
x − i y
r − z
.
(4.115)
Taking the exterior derivative of (4.114) we have
d
(σ
α
+ i ω
α ) dx
α
= −(A + i C) (dr − dt) ∧ d
x − i y
r − z
= −F ,
(4.116)