Page 380 - Electromagnetics

The retarded potentials in the frequency domain. Consider an unbounded, ho-
˜
c
mogeneous, isotropic medium described by ˜µ(ω) and ˜ (ω).If ψ(r,ω) represents a scalar
potential or any rectangular component of a vector or Hertzian potential then it must
satisfy
2
˜
2
˜
(∇ + k )ψ(r,ω) =−S(r,ω) (5.75)
c 1/2
where k = ω( ˜µ˜ ) . This Helmholtz equation has the form of (5.70) and thus

˜ ˜
ψ(r,ω) = S(r ,ω)G(r|r ; ω) dV
V
where
2
2
(∇ + k )G(r|r ; ω) =−δ(r − r ). (5.76)

This is equation (A.46) and its solution, as given by (A.49), is
e − jkR

G(r|r ; ω) = . (5.77)
4π R
2
Here we use v = 1/ ˜µ˜ and = ˜σ/2 in (A.47):

1 ˜ σ
2
c
˜
˜
k = ω − j2ω = ω µ ˜ − j = ω µ˜ .
v ω
The solution to (5.75) is therefore
e
− jkR
˜ ˜
ψ(r,ω) = S(r ,ω) dV . (5.78)
V 4π R
When the medium is lossless, the potential must also satisfy the radiation condition

∂
˜
lim r + jk ψ(r) = 0 (5.79)
r→∞ ∂r
to guarantee uniqueness of solution. In § 5.2.2 we shall showhowthis requirement arises
from the solution within a bounded region. For a uniqueness proof for the Helmholtz
equation, the reader may consult Chew[33].
We may use (5.78) to find that
˜ µ e − jkR
˜ ˜ i
A e (r,ω) = J (r ,ω) dV . (5.80)
4π V R
Comparison with (5.74) shows that in the frequency domain, time retardation takes the
form of a phase shift. Similarly,
1 e − jkR
˜ i
φ(r,ω) = c ˜ ρ (r ,ω) dV . (5.81)
4π ˜ V R
The electric and magnetic dyadic Green’s functions. The frequency-domain elec-
tromagnetic fields may be found for electric sources from the electric vector potential
using (5.60) and (5.61):
jω ˜µ(ω)

˜ i
˜ i
˜

E(r,ω) =− jω ˜µ(ω) J (r ,ω)G(r|r ; ω) dV − ∇∇ · J (r ,ω)G(r|r ; ω) dV ,

V k 2 V

˜ ˜ i
H =∇ × J (r ,ω)G(r|r ; ω) dV . (5.82)
V
© 2001 by CRC Press LLC