Page 108 - Electromagnetics

we can envision a uniform plane wave as being created by a uniform surface source of
doubly-infinite extent, plane waves are also useful as models for spherical waves over
localized regions of the wavefront.
We choose the plane of field invariance to be the xy-plane and later generalize the
resulting solution to any planar surface by a simple rotation of the coordinate axes. Since
the fields vary with z only we choose to write the wave equation (2.325) in rectangular
coordinates, giving for a source-free region of space 4
2
2
2
2
∂ E x (z, t) ∂ E y (z, t) ∂ E z (z, t) ∂E(z, t) ∂ E(z, t)
ˆ x + ˆ y + ˆ z − µσ − µ
= 0. (2.328)
∂z 2 ∂z 2 ∂z 2 ∂t ∂t 2
If we return to Maxwell’s equations, we soon find that not all components of E are
present in the plane-wave solution. Faraday’s law states that
∂E y (z, t) ∂E x (z, t) ∂E(z, t) ∂H(z, t)
∇× E(z, t) =−ˆ x + ˆ y = ˆ z × =−µ . (2.329)
∂z ∂z ∂z ∂t
We see that ∂ H z /∂t = 0, hence H z must be constant with respect to time. Because
a nonzero constant field component would not exhibit wave-like behavior, we can only
have H z = 0 in our wave solution. Similarly, Ampere’s law in a homogeneous conducting
region free from impressed sources states that
∂D(z, t) ∂E(z, t)
∇× H(z, t) = J + = σE(z, t) +
∂t ∂t
or
∂ H y (z, t) ∂ H x (z, t) ∂H(z, t) ∂E(z, t)
− ˆ x + ˆ y = ˆ z × = σE(z, t) +
. (2.330)
∂z ∂z ∂z ∂t
This implies that
∂E z (z, t)
σ E z (z, t) +
= 0,
∂t
which is a differential equation for E z with solution
σ

.
E z (z, t) = E 0 (z) e − t
Since we are interested only in wave-type solutions, we choose E z = 0.
Hence E z = H z = 0, and thus both E and H are perpendicular to the z-direction.
Using (2.329) and (2.330), we also see that
∂ ∂H ∂E
(E · H) = E · + H ·
∂t ∂t ∂t
1 ∂E σ 1 ∂H
=− E · ˆ z × − H · E + H · ˆ z ×
µ ∂z

∂z
or
∂ σ 1 ∂E 1 ∂H

ˆ
+ (E · H) = ˆ z · E × − z · H × .
∂t
µ ∂z
∂z
We seek solutions of the type E(z, t) = ˆ pE(z, t) and H(z, t) = ˆ qH(z, t), where ˆ p and ˆ q are
constant unit vectors. Under this condition we have E × ∂E/∂z = 0 and H × ∂H/∂z = 0,
giving
∂ σ

+ (E · H) = 0.
∂t
4 The term “source free” applied to a conducting region implies that the region is devoid of impressed
sources and, because of the relaxation effect, has no free charge. See the discussion in Jones [97].

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