Page 264 - Electromagnetics

for the dual problem. Substitution of (4.187)–(4.190) into (4.191) and (4.192) gives

˜ ¯ µ

˜ 1 ˜
D 2 = · E 2 , (4.195)
η 2
0
2
˜
˜

B 2 = η ˜ ¯ 1 · H 2 . (4.196)
0
Comparing (4.195) with (4.193) and (4.196) with (4.194), we conclude that
2 2
˜ ¯ µ = η ˜ ¯ 1 , ˜ ¯ 2 = ˜ ¯µ /η . (4.197)
2 0 1 0
For a linear, isotropic medium specified by ˜ and ˜µ, the dual problem is obtained by
replacing ˜ r with ˜µ r and ˜µ r with ˜ r . The solution to the dual problem is then
˜
˜
˜
˜
E 2 = η 0 H 1 , η 0 H 2 =−E 1 ,
as before. The medium in the dual problem must have electric properties numerically
equal to the magnetic properties of the medium in the original problem, and magnetic
properties numerically equal to the electric properties of the medium in the original
problem. Alternatively we may divide Ampere’s law by η = ( ˜µ/˜ ) 1/2 instead of η 0 . Then
˜
˜
˜
˜
the dual problem has J replaced by J m /η, and J m replaced by −ηJ, and the solution is
˜
˜
˜
˜
E 2 = ηH 1 , ηH 2 =−E 1 . (4.198)
There is no need to swap ˜ r and ˜µ r since information about these parameters is incor-
porated into the replacement sources.
We may also apply duality to a problem where we have separated the impressed and
˜
secondary sources. In a homogeneous, isotropic, conducting medium we may let J =
˜
˜ i
J + ˜σE. With this the curl equations become
˜ i
˜
c ˜
∇× ηH = ηJ + jωη˜ E,
˜
˜
˜
∇× E =−J m − jω ˜µH.
c 1/2
The solution to the dual problem is again given by (4.198), except that now η = ( ˜µ/˜ ) .
As we did near the end of § 2.9.2, we can consider duality in a source-free region. We
let S enclose a source-free region of space and, for simplicity, assume that the medium
within S is linear, isotropic, and homogeneous. The fields within S are described by
˜
˜
∇× E 1 =− jω ˜µH 1 ,
˜
˜
∇× ηH 1 = jω˜ ηE 1 ,
˜
∇· ˜ E 1 = 0,
˜
∇· ˜µH 1 = 0.
˜
The symmetry of the equations is such that the mathematical form of the solution for E
˜
is the same as that for ηH. Since the fields
˜
˜
˜
˜
E 2 = ηH 1 , H 2 =−E 1 /η,
˜
˜
also satisfy Maxwell’s equations, the dual problem merely involves replacing E by ηH
˜
˜
and H by −E/η.
© 2001 by CRC Press LLC