Page 388 - Electromagnetics

∂G(r|r ; ω) ∂ψ(r ,ω)

ψ(r ,ω) − G(r|r ; ω) dS .
∂n ∂n
S 0 +S ∞
We note that ˆ n points outward from V , and G is the Green’s function (5.77). By
inspection, this Green’s function obeys the reciprocity condition

G(r|r ; ω) = G(r |r; ω)

and satisfies
2
2

∇ G(r|r ; ω) =∇ G(r|r ; ω).
˜
2
2 ˜
2
˜
2

Substituting ∇ ψ =−k ψ − S from (5.75) and ∇ G =−k G − δ(r − r ) from (5.76)
we get

˜ ˜
ψ(r,ω) = S(r ,ω)G(r|r ; ω) dV −
V
˜

∂G(r|r ; ω) ∂ψ(r ,ω)

˜

− ψ(r ,ω) − G(r|r ; ω) dS .
∂n ∂n
S 0 +S ∞
˜
˜
Hence ψ within V may be written in terms of the sources within V and the values of ψ
and its normal derivative over S 0 + S ∞ . The surface contributions account for sources
excluded by S 0 .
Let us examine the integral over S ∞ more closely. If we let S ∞ recede to infinity, we
expect no contribution to the potential at r from the fields on S ∞ . Choosing 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

∂G(r|r ; ω) ∂ e − jkr e − jkr

= ˆ n ·∇ G(r|r ; ω) ≈ =−(1 + jkr ) .
∂n ∂r 4πr 4πr

Substituting these, we find that as r →∞
˜ 2π π ˜ − jkr
∂G ∂ψ 1 + jkr 1 ∂ψ e 2
˜
ψ ˜ − G dS ≈ − ψ − r sin θ dθ dφ

∂n ∂n r 2 r ∂r 4π

S ∞ 0 0
2π π
∂ψ ˜ e − jkr
˜
˜

≈− ψ + r jkψ + sin θ dθ dφ .
0 0 ∂r 4π
Since this gives the contribution to the field in V from the fields on the surface receding
to infinity, we expect that this term should be zero. If the medium has loss, then the
exponential term decays and drives the contribution to zero. For a lossless medium the
contribution is zero if
˜
lim ψ(r,ω) = 0, (5.96)
r→∞
˜
∂ψ(r,ω)
˜
lim r jkψ(r,ω) + = 0. (5.97)
r→∞ ∂r
This is called the radiation condition for the Helmholtz equation. It is also called the
Sommerfeld radiation condition after the German physicist A. Sommerfeld. Note that
© 2001 by CRC Press LLC