Page 270 - Electromagnetics

Note that we choose the negative sign in the exponential function and allow the vector
components of k to be either positive or negative as required by the physical nature of
a specific problem. Also note that the magnitude of the wave vector is the wavenumber:
|k|= k.
We may always write the wave vector as a sum of real and imaginary vector components

k = k + jk (4.217)

which must obey

2
2

k · k = k = k − k 2 + 2 jk · k . (4.218)
When the real and imaginary components are collinear, (4.216) describes a uniform plane
wave with
ˆ

k = k(k + jk ).
When k and k have different directions, (4.216) describes a nonuniform plane wave.

We shall find in § 4.13that any frequency-domain electromagnetic field in free space
may be represented as a continuous superposition of elemental plane-wave components of
the type (4.216), but that both uniform and nonuniform terms are required.
The TEM nature of a uniform plane wave. Given the plane-wave solution to
the wave equation for the electric field, it is straightforward to find the magnetic field.
Substitution of (4.216) into Faraday’s law gives
˜
˜
− jk(ω)·r
∇× E 0 (ω)e =− jωB(r,ω).
Computation of the curl is straightforward and easily done in rectangular coordinates.
This and similar derivatives often appear when manipulating plane-wave solutions; see
the tabulation in Appendix B, By (B.78) we have
˜
k × E
˜
H = . (4.219)
ω ˜µ
Taking the cross product of this expression with k, we also have
˜
˜
˜
k × (k × E) k(k · E) − E(k · k)
˜
k × H = = . (4.220)
ω ˜µ ω ˜µ
˜
We can show that k · E = 0 by examining Gauss’ law and employing (B.77):
˜ ρ
˜
˜ − jk·r
∇· E =− jk · Ee = = 0. (4.221)
˜
2
2
c
Using this and k · k = k = ω ˜µ˜ , we obtain from (4.220)
˜
k × H
˜
E =− . (4.222)
ω˜ c
ˆ
Now for a uniform plane wave k = kk, so we can also write (4.219) as
˜
ˆ
k × E ˆ ˜
˜ k × E 0 − jk·r
H = = e (4.223)
η η
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