Topics in Classical Electrodynamics

since ∂R = x . Differentiating once again, we get
∂x R
so that
∂ 2
P
∂x ⃗ = − 1 x2
⃗p + 3 2 R3 R ⃗p + 3 x 2 ∂⃗p
5 c R 4 ∂t − 1 ∂⃗p
cR 2 ∂t + 1 x 2 ∂ 2 ⃗p
c 2 R 3 ∂t , 2
3∑
i=1
∂ 2
∂x 2 i
⃗P = 1 ∂ 2 ⃗p
c 2 R ∂t , 2
which represents the spherically symmetric solution of the wave equation.
Consider the retarded potentials
ϕ (t) = −divP ⃗ (t, R) A ⃗
1 ∂P ⃗ (t, R)
(t) ; = ;
c ∂t
⃗H = rotA ⃗ (t) = rot 1 ∂P ⃗ (t, R)
= 1 ∂
c ∂t c ∂t rot P ⃗ (t, R) ;
⃗E = − 1 c
∂ ⃗ A (t)
∂t
− ⃗ ∇φ = − 1 c 2 ∂ 2 ⃗ P (t, R)
∂t 2 − ⃗ ∇div ⃗ P (t, R)
= − 1 c 2 ∂ 2 ⃗ P (t, R)
∂t 2 + ⃗ ∇ 2 ⃗ P (t, R) + rot rot ⃗ P (t, R) .
On the last line the sum of the first two terms is equal to zero by virtue of
the wave equation. This results in
⃗E = rot rot ⃗ P (t, R) . (78)
Assume that the electric moment changes only its magnitude, but not its
direction i.e.,
⃗p (t) = ⃗p 0 f (t) .
This is not a restriction because moment ⃗p of an arbitrary oscillator can be
decomposed into three mutually orthogonal directions and a field in each
direction can be studied separately. Based on this we have
f ( )
t −
⃗P R c
(t, R) = ⃗p 0
R
rot ⃗ P = f R rot ⃗p 0 +
,
[
⃗∇ f R , ⃗p 0
]
50
= ∂
∂R
( (
f t −
r
c
R
))
− 1 [ ]
⃗R, ⃗p0 .
R