Page 378 - Electromagnetics

and determine the Green’s function G for the operator L. Provided that S resides within
V wehave

∞
∞

S(r , t )G(r, t|r , t ) dt dV S(r , t ) L{G(r, t|r , t )} dt dV

L =
V −∞ V −∞
∞

= S(r , t )δ(r − r )δ(t − t ) dt dV
V −∞
= S(r, t),
hence
∞

ψ(r, t) = S(r , t )G(r, t|r , t ) dt dV (5.69)
V −∞
by comparison with (5.68).
We can also apply this idea in the frequency domain. The solution to
˜
˜
L{ψ(r,ω)}= S(r,ω) (5.70)
is

˜ ˜
ψ(r,ω) = S(r ,ω)G(r|r ; ω) dV
V
where the Green’s function G satisfies

L{G(r|r ; ω)}= δ(r − r ).
Equation (5.69) is the basic superposition integral that allows us to find the potentials
in an infinite, unbounded medium. We note that if the medium is bounded then we must
use Green’s theorem to include the effects of sources that reside external to the bound-
aries. These are manifested in terms of the values of the potentials on the boundaries
in the same manner as with the static potentials in Chapter 3. In order to determine
whether (5.69) is the unique solution to the wave equation, we must also examine the
behavior of the fields on the boundary as the boundary recedes to infinity. In the fre-
quency domain we find that an additional “radiation condition” is required to ensure
uniqueness.
The retarded potentials in the time domain. Consider an unbounded, homoge-
neous, lossy, isotropic medium described by parameters µ, , σ. In the time domain the
vector potential A e satisfies (5.30). The scalar components of A e must obey
2
∂ A e,n (r, t) ∂ A e,n (r, t)
2 i
∇ A e,n (r, t) − µσ − µ =−µJ (r, t), n = x, y, z.
n
∂t ∂t 2
We may write this in the form
2 ∂ 1 ∂

2
2
∇ − − ψ(r, t) =−S(r, t) (5.71)
2
2
v ∂t v ∂t 2
i
2
where ψ = A e,n , v = 1/µ , = σ/2 , and S = µJ . The solution is
n
∞

ψ(r, t) = S(r , t )G(r, t|r , t ) dt dV (5.72)
V −∞
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