Topics in Classical Electrodynamics

On the other hand,
Thus,
ϕ =
⃗p · ⃗R
R 3
− ⃗ R
R 2 ∂⃗p
∂R .
ϕ = −div ⃗p ( x ′ , t − R c
.
R
Here divergence is taken over coordinates of the point P (x, y, z) where the
observer is located. Using expression (65), the vector potential becomes
⃗A = 1 c
∫
=
∫ ⃗ j ( x ′ , t − R c
)
R
d 3 x ′ =
d 3 x ′ [⃗j ( x ′ , t − R 0
c
)
R 0
− ⃗ R 0 · ⃗R ′
⃗R 0
)
∂ ⃗j ( x ′ , t − R 0 ]
· · · .
∂R 0 R 0
First integral can also be expressed via electric moment, which can be achieved
by using the continuity equation
(
∂
∂t ρ x ′ , t − R ) (
0
= −div ′ ⃗j x ′ , t − R )
0
.
c
c
Multiplying both sides of this equation by time independent R ⃗ ′ , integrating
over entire space and using the definition (76), we can then state that
(
∂
∂t ⃗p x ′ , t − R ) ∫
(
0
= − d 3 x ′ R ⃗ ′ div ′ ⃗j x ′ , t − R )
0
.
c
c
To proceed, let us sidetrack and consider an arbitrary unit vector ⃗a, i.e.
|⃗a| = 1. Then
(
⃗a ⃗ R
′ ) div⃗j = div
(
⃗ j ( ⃗a ⃗ R ′)) − ⃗j · ⃗∇ ′( ⃗a ⃗ R ′)
c
)
= div
(
⃗ j ( ⃗a ⃗ R ′)) − ⃗j · ⃗a ,
where the last step follows from ⃗a being a constant and ∇ ′ R ⃗ ′ = 1. Based on
that we can write
⃗a · ∂ (
∂t ⃗p x ′ , t − R ) ∫ (
0
= − d 3 x ′ div ′ ⃗ j ( ⃗a R
c
⃗ ′)) ∫ (
+ ⃗a · d 3 x ′ ⃗j x ′ , t − R )
0
.
c
48