Page 377 - Electromagnetics

and
˜
c
˜
H = jω˜ ∇× Π e . (5.63)
˜ i
Here J can represent either an impressed electric current source or an impressed polar-
˜ i
˜ i
ization current source J = jωP . The electric Hertzian potential obeys
J ˜ i
2 2 ˜
(∇ + k )Π e =− . (5.64)
jω˜ c
˜ i
˜ i
When J
= 0 and J = 0 we let
m
˜ c ˜
A h = jω ˜µ˜ Π h
and find
˜ ˜
E =− jω ˜µ∇× Π h (5.65)
and
J ˜ i
˜ ˜ 2 ˜ ˜ m
H =∇(∇· Π h ) + k Π h =∇ × (∇× Π h ) − . (5.66)
jω ˜µ
Here J ˜ i can represent either an impressed magnetic current source or an impressed
m
˜ i
˜ i
magnetization current source J = jω ˜µM . The magnetic Hertzian potential obeys
m
J ˜ i
2 2 ˜ m
(∇ + k )Π h =− . (5.67)
jω ˜µ
When both electric and magnetic sources are present we have by superposition
˜ ˜ 2 ˜ ˜
E =∇(∇· Π e ) + k Π e − jω ˜µ∇× Π h
J ˜ i
˜ ˜
=∇ × (∇× Π e ) − − jω ˜µ∇× Π h
jω˜ c
and
˜ c ˜ ˜ 2 ˜
H = jω˜ ∇× Π e +∇(∇· Π h ) + k Π h
J ˜ i
˜
˜
c m
= jω˜ ∇× Π e +∇ × (∇× Π h ) − .
jω ˜µ
5.2.1 Solution for potentials in an unbounded medium: the retarded
potentials
Under the Lorentz condition each of the potential functions obeys the wave equation.
This equation can be solved using the method of Green’s functions to determine the
potentials, and the electromagnetic fields can therefore be determined. We nowexamine
the solution for an unbounded medium. Solutions for bounded regions are considered in
§ 5.2.2.
Consider a linear operator L that operates on a function of r and t. If we wish to solve
the equation

L{ψ(r, t)}= S(r, t), (5.68)
we first solve

L{G(r, t|r , t )}= δ(r − r )δ(t − t )

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