The key initial electrodynamic equations are:
E = - Ñ · V - ¶A / ¶ t
B = Ñ X A
1/ m 0 Ñ X Ñ X A = - c Ñ V' - cA" + J
And for the Lorentz gauge: Ñ · A = - 1/ c 2 ( ¶ V / ¶ t )
From which we obtain the equations:
i) Ñ2 A = - 1/ c 2 ( ¶2 A / ¶ t2 ) = - m 0 J
2) Ñ2 V - 1/ c 2 ( ¶2 V / ¶ t2 ) = - r / εo
For the static case:
i) Ñ2 A = - m 0 J
ii) Ñ2 V = - r / εo
From which we obtain solutions:
i) A = m 0 / 4 p ò J (r’) dt / R
ii) V ( r) = 1/ 4 p εo ò r(r’) dt / R
R = ‖r - r'‖
We need to reference an earlier time for t in the potentials so make use of a time delay,
t r = t - R /c
Then in terms of t r :
A ( r, t ) = m 0 / 4 p ò J (r’, t r) dt / R
V ( r, t ) = 1/ 4 p εo ò r(r’, t r) dt / R
Since a time delay is involved, both A,V are regarded as retarded potentials.
From the diagram shown we see V1 is a small volume about P.
The total volume is: V = V1 + V 2
The total potential is the superposition of potentials from the sources in
V1 and V 2: e.g.
V = V1 + V2
Where:
V1 = 1/ 4 p εo ò r(r’, t r) dt / R
Similar to static case and corresponds to the solution for:
Ñ2 V1 = - r / εo
Which is exactly true in the limit: V1 -> 0
For calculation purposes we may change the center of coordinates to a point P using the fact that: R = ‖r - r' ‖
Whence we can write:
ÑR (r / R ) = 1/ R 2 ¶ / ¶ R {R ¶ / ¶ R {r / R) }
= 1/ R 2 ¶ / ¶ R {R2/R ) ¶r / ¶ R - R2 r/ R}
= 1/ R 2 { ¶ r/ ¶ R + R ¶r2 /¶ R2 - ¶r / ¶ R} = 1/R ¶r2 /¶ R2
Then :
Ñ2 V2 ( r, t ) = 1/ 4 p εo ò (1/ R ¶r2 /¶ R2 ) dt
Then the total potential: V = V1 + V 2
Ñ2 V = Ñ2 (V1 + V2) = Ñ2 V1 + Ñ2 V2
= - r / εo + 1/ c 2 ( ¶2 V / ¶ t2 )
=> Ñ2 V = - 1/ c 2 ( ¶2 V / ¶ t2 ) = - r / εo
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