Page 257 - Electromagnetics

Our approach is similar to that of § 2.2.1. Consider a simply-connected region of
space V bounded by surface S, where both V and S contain only ordinary points. The
ˇ
phasor-domain fields within V are associated with a phasor current distribution J, which
may be internal to V (entirely or in part). We seek conditions under which the phasor
ˇ
ˇ
ˇ
ˇ
electromagnetic fields are uniquely determined. Let the field set (E 1 , D 1 , B 1 , H 1 ) satisfy
ˇ
Maxwell’s equations (4.128) and (4.129) associated with the current J (along with an
ˇ
ˇ
ˇ
ˇ
appropriate set of constitutive relations), and let (E 2 , D 2 , B 2 , H 2 ) be a second solution.
To determine the conditions for uniqueness of the fields, we look for a situation that
ˇ
ˇ
ˇ
ˇ
results in E 1 = E 2 , H 1 = H 2 , and so on. The electromagnetic fields must obey
ˇ
ˇ
ˇ
∇× H 1 = j ˇωD 1 + J,
ˇ
ˇ
∇× E 1 =− j ˇωB 1 ,
ˇ
ˇ
ˇ
∇× H 2 = j ˇωD 2 + J,
ˇ
ˇ
∇× E 2 =− j ˇωB 2 .
ˇ
ˇ
ˇ
ˇ
ˇ
ˇ
Subtracting these and defining the difference fields E 0 = E 1 − E 2 , H 0 = H 1 − H 2 , and so
on, we find that
ˇ
ˇ
∇× H 0 = j ˇωD 0 , (4.163)
ˇ
ˇ
∇× E 0 =− j ˇωB 0 . (4.164)
Establishing the conditions under which the difference fields vanish throughout V ,we
shall determine the conditions for uniqueness.
ˇ
Dotting (4.164) by H and dotting the complex conjugate of (4.163) by E 0 ,wehave
ˇ ∗
0
ˇ
ˇ
ˇ ∗

ˇ ∗
H · ∇× E 0 =− j ˇωB 0 · H ,
0
0
ˇ ∗ ˇ
ˇ

E 0 · ∇× H =− j ˇωD · E 0 .
ˇ ∗
0 0
Subtraction yields
ˇ
ˇ ∗ ˇ
ˇ
ˇ
ˇ ∗
ˇ ∗

ˇ ∗
H · ∇× E 0 − E 0 · ∇× H 0 =− j ˇωB 0 · H + j ˇωD · E 0
0
0
0
which, by (B.44), can be written as
ˇ
ˇ
ˇ

ˇ ∗
ˇ ∗
∇· E 0 × H = j ˇω E 0 · D − B 0 · H .
ˇ ∗
0 0 0
Adding this expression to its complex conjugate, integrating over V , and using the di-
vergence theorem, we obtain
ˇ ω

ˇ ∗ ˇ
ˇ
ˇ ∗ ˇ
ˇ
ˇ

ˇ ∗
ˇ ∗
Re E 0 × H · dS =− j E · D 0 − E 0 · D + H · B 0 − H 0 · B dV.
ˇ ∗
0 0 0 0 0
S 2 V
Breaking S into two arbitrary portions and using (??), we obtain

ˇ
ˇ
ˇ ∗
Re H · (ˆ n × E 0 ) dS − Re E 0 · (ˆ n × H ) dS =
ˇ ∗
0 0
S 1 S 2
ˇ ω
ˇ ∗ ˇ
ˇ
ˇ ∗ ˇ
ˇ
ˇ ∗
− j E · D 0 − E 0 · D 0 + H · B 0 − H 0 · B 0 dV. (4.165)
ˇ ∗
0
0
2 V
Now if ˆ n × E 0 = 0 or ˆ n × H 0 = 0 over all of S, or some combination of these conditions
holds over all of S, then

ˇ ∗ ˇ ˇ ˇ ∗ ˇ ˇ
ˇ ∗
E · D 0 − E 0 · D + H · B 0 − H 0 · B dV = 0. (4.166)
ˇ ∗
0 0 0 0
V
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