Page 424 - Electromagnetics

Here we have used d = 4π for the total solid angle subtending the sphere S δ . Finally,

assuming that the volume sources are continuous, the volume integral over V δ vanishes
and we have

i
˜ ˜ i ˜ ρ ˜ i
E(r,ω) = −J ×∇ G + c ∇ G − jω ˜µJ G dV +
m
V ˜
N

˜
˜ ˜
+ (ˆ n × E) ×∇ G + (ˆ n · E)∇ G − jω ˜µ(ˆ n × H)G dS −
n=1 S n
N
1
˜

− (dl · H)∇ G. (6.7)
jω˜ c
n=1
na + nb
˜
A similar formula for H can be derived by placing a magnetic dipole of moment ˜ p m at
r = r p and proceeding as above. This leads to

˜ ρ i
˜ ˜ i m c ˜ i
H(r,ω) = J ×∇ G + ∇ G − jω˜ J G dV +
m
V ˜ µ
N

c
˜ ˜ ˜
+ (ˆ n × H) ×∇ G + (ˆ n · H)∇ G + jω˜ (ˆ n × E)G dS +
n=1 S n
N
1 ˜

+ (dl · E)∇ G. (6.8)
jω ˜µ
n=1
na + nb
We can also obtain this expression by substituting (6.7) into Faraday’s law.
6.1.2 The Sommerfeld radiation condition
In § 5.2.2 we found that if the potentials are not to be influenced by effects that are
infinitely removed, then they must obey a radiation condition. We can make the same
argument about the fields from (6.7) and (6.8). Let us allow one of the excluding surfaces,
say S N , to recede to infinity (enclosing all of the sources as it expands). As S N →∞ any
contributions from the fields on this surface to the fields at r should vanish.

Letting S N be a sphere centered at the origin, we note that ˆ n =−ˆ r and that as

r →∞
e − jk|r−r | e − jkr

G(r|r ; ω) = ≈ ,

4π|r − r | 4πr

1 + jkR
1 + jkr e − jkr
ˆ

∇ G(r|r ; ω) = R e − jkR ≈−ˆ r .
4π R 2 r 4πr
Substituting these expressions into (6.7) we find that

˜
˜
˜

lim (ˆ n × E) ×∇ G + (ˆ n · E)∇ G − jω ˜µ(ˆ n × H)G dS
S N →S ∞
S N

2π π
1 + jkr
˜ ˜
≈ lim (ˆ r × E) × ˆ r + (ˆ r · E)ˆ r

r →∞ r
0 0
e − jkr
2
˜

+ jω ˜µ(ˆ r × H) r sin θ dθ dφ

4πr
2π π e − jkr
˜
˜
˜

≈ lim r jkE + jω ˜µ(ˆ r × H) + E sin θ dθ dφ .

r →∞ 4π
0 0
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