Page 260 - Electromagnetics

ˇ ˇ ¯ ˇ ˇ ˇ ˜ ¯ ˇ ˇ
− j ˇω E a · ξ · H b + ˜ ¯ · E b − E b · ξ · H a + ˜ ¯ · E a = 0,
which holds if
ˇ
ˇ
ˇ
ˇ
H a · ˜ ¯µ · H b − H b · ˜ ¯µ · H a = 0,
ˇ
ˇ
ˇ
˜ ¯
˜ ¯
ˇ
H a · ζ · E b + E b · ξ · H a = 0,
ˇ
ˇ
ˇ
ˇ
˜ ¯
˜ ¯
E a · ξ · H b + H b · ζ · E a = 0,
ˇ
ˇ
ˇ
ˇ
E a · ˜ ¯ · E b − E b · ˜ ¯ · E a = 0.
These in turn hold if
T T
T T ˜ ¯ ˜ ¯ ˜ ¯ ˜ ¯
˜ ¯ = ˜ ¯ , ˜ ¯ µ = ˜ ¯µ , ξ =−ζ , ζ =−ξ . (4.170)
These are the conditions for a reciprocal medium. For example, an anisotropic dielectric
is a reciprocal medium if its permittivity dyadic is symmetric. An isotropic medium
described by scalar quantities µ and is certainly reciprocal. In contrast, lossless Gy-
†
rotropic media are nonreciprocal since the constitutive parameters obey ˜ ¯ = ˜ ¯ or ˜ ¯µ = ˜ ¯µ †
T T
rather than ˜ ¯ = ˜ ¯ or ˜ ¯µ = ˜ ¯µ .
For a reciprocal medium (4.169) reduces to
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ

∇· (E a × H b − E b × H a ) = E b · J a − E a · J b − H b · J ma + H a · J mb . (4.171)
At points where the sources are zero, or are conduction currents described entirely by
ˇ
ˇ
Ohm’s law J = σE,wehave
ˇ
ˇ
ˇ
ˇ
∇· (E a × H b − E b × H a ) = 0, (4.172)
known as Lorentz’s lemma. If we integrate (4.171) over V and use the divergence theorem
we obtain

ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ

E a × H b − E b × H a · dS = E b · J a − E a · J b − H b · J ma + H a · J mb dV.
S V
(4.173)
This is the general form of the Lorentz reciprocity theorem, and is valid when V contains
reciprocal media as defined in (4.170).
Note that by an identical set of steps we find that the frequency-domain fields obey
an identical Lorentz lemma and reciprocity theorem.
The condition for reciprocal systems. The quantity

ˇ ˇ ˇ ˇ ˇ
f a , ˇ g b = E a · J b − H a · J mb dV
V
ˇ
is called the reaction between the source fields ˇ g of set b and the mediating fields f of an
ˇ
ˇ
independent set a. Note that E a · J b is not quite a power density, since the current lacks
a complex conjugate. Using this reaction concept, first introduced by Rumsey [161], we
can write (4.173) as

ˇ ˇ ˇ ˇ ˇ ˇ
f b , ˇ g a − f a , ˇ g b = E a × H b − E b × H a · dS. (4.174)
S
© 2001 by CRC Press LLC