Page 420 - Electromagnetics

Chapter 6

Integral solutions of Maxwell’s equations

6.1 Vector Kirchoff solution: method of Stratton and Chu

One of the most powerful tools for the analysis of electromagnetics problems is the
integral solution to Maxwell’s equations formulated by Stratton and Chu [187, 188].
˜
˜
These authors used the vector Green’s theorem to solve for E and H in much the same
way as is done in static fields with the scalar Green’s theorem. An alternative approach is
to use the Lorentz reciprocity theorem of § 4.10.2, as done by Fradin [74]. The reciprocity
approach allows the identification of terms arising from surface discontinuities, which
must be added to the result obtained from the other approach [187].

6.1.1 The Stratton–Chu formula
Consider an isotropic, homogeneous medium occupying a bounded region V in space.
The medium is described by permeability ˜µ(ω), permittivity ˜ (ω), and conductivity ˜σ(ω).
The region V is bounded by a surface S, which can be multiply-connected so that S is
the union of several surfaces S 1 ,... , S N as shown in Figure 6.1; these are used to exclude
unknown sources and to formulate the vector Huygens principle. Impressed electric and
magnetic sources may thus reside both inside and outside V .
We wish to solve for the electric and magnetic fields at a point r within V . To do this we
employ the Lorentz reciprocity theorem (4.173), written here using the frequency-domain
fields as an integral over primed coordinates:

˜ ˜ ˜ ˜

− E a (r ,ω) × H b (r ,ω) − E b (r ,ω) × H a (r ,ω) · ˆ n dS =
S

˜ ˜ ˜ ˜

E b (r ,ω) · J a (r ,ω) − E a (r ,ω) · J b (r ,ω)− (6.1)
V
˜
˜
˜
˜

H b (r ,ω) · J ma (r ,ω) + H a (r ,ω) · J mb (r ,ω) dV . (6.2)
Note that the negative sign on the left arises from the definition of ˆ n as the inward normal
to V as shown in Figure 6.1. We place an electric Hertzian dipole at the point r = r p
˜
˜
˜
˜
where we wish to compute the field, and set E b = E p and H b = H p in the reciprocity
˜
˜
theorem, where E p and H p are the fields produced by the dipole (5.88)–(5.89):
˜
H p (r,ω) = jω∇× [˜ pG(r|r p ; ω)], (6.3)
1
˜
E p (r,ω) = ∇× ∇× [˜ pG(r|r p ; ω)] . (6.4)
˜ c
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