Page 381 - Electromagnetics

As long as the observation point r does not lie within the source region we may take the
derivatives inside the integrals. Using

˜ i
˜ i
˜

∇· J (r ,ω)G(r|r ; ω) = J (r ,ω) ·∇G(r|r ; ω) + G(r|r ; ω)∇· J(r ,ω)

˜ i

=∇G(r|r ; ω) · J (r ,ω)
we have
1
i
˜

E(r,ω) =− jω ˜µ(ω) ˜ i ∇ ∇G(r|r ; ω) · J (r ,ω) dV .
J (r ,ω)G(r|r ; ω) +
V k 2
This can be written more compactly as

˜ i
¯
˜

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

¯ ¯ ∇∇
G e (r|r ; ω) = I + 2 G(r|r ; ω) (5.83)
k
is called the electric dyadic Green’s function. Using
˜ i
˜ i
˜ i
∇× [J G] =∇G × J + G∇× J =∇G × J ˜ i
we have for the magnetic field

˜ ˜ i
H(r,ω) = ∇G(r|r ; ω) × J (r ,ω) dV .
V
Now, using the dyadic identity (B.15) we may show that
¯
¯
i
˜ i
˜ i
J ×∇G = (J ×∇G) · I = (∇G × I) · J .
So

˜ ¯ ˜ i
H(r,ω) =− G m (r|r ; ω) · J (r ,ω) dV
V
where
¯

G m (r|r ; ω) =∇G(r|r ; ω) × I ¯ (5.84)
is called the magnetic dyadic Green’s function.
Proceeding similarly for magnetic sources (or using duality) we have

˜
¯
˜ i

H(r) =− jω˜ c G e (r|r ; ω) · J (r ,ω) dV ,
m
V

˜ ¯ ˜ i
E(r) = G m (r|r ; ω) · J (r ,ω) dV .
m
V
When both electric and magnetic sources are present we simply use superposition and
add the fields.
When the observation point lies within the source region, we must be much more
careful about how we formulate the dyadic Green’s functions. In (5.82) we encounter the
integral

˜ i

J (r ,ω)G(r|r ; ω) dV .
V
© 2001 by CRC Press LLC