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4.11.3 Plane waves in a homogeneous, isotropic, lossy material
The plane-wave field. In later sections we will solve the frequency-domain wave
equation with an arbitrary source distribution. At this point we are more interested in
the general behavior of EM waves in the frequency domain, so we seek simple solutions
to the homogeneous equation
2
2 ˜
(∇ + k )E(r,ω) = 0 (4.214)
that governs the fields in source-free regions of space. Here
c
2
2
[k(ω)] = ω ˜µ(ω)˜ (ω).
Many properties of plane waves are best understood by considering the behavior of a
monochromatic field oscillating at a single frequency ˇω. In these cases we merely make
the replacements
ˇ
˜
ω → ˇω, E(r,ω) → E(r),
and apply the rules developed in § 4.7 for the manipulation of phasor fields.
For our first solutions we choose those that demonstrate rectangular symmetry. Plane
waves have planar spatial phase loci. That is, the spatial surfaces over which the phase
of the complex frequency-domain field is constant are planes. Solutions of this type may
be obtained using separation of variables in rectangular coordinates. Writing
˜
˜
˜
˜
E(r,ω) = ˆ xE x (r,ω) + ˆ yE y (r,ω) + ˆ zE z (r,ω)
we find that (4.214) reduces to three scalar equations of the form
2 2
˜
(∇ + k )ψ(r,ω) = 0
˜
˜
˜
˜
where ψ is representative of E x , E y , and E z . This is called the homogeneous scalar
Helmholtz equation. Product solutions to this equation are considered in § A.4. In
rectangular coordinates
˜
ψ(r,ω) = X(x,ω)Y(y,ω)Z(z,ω)
where X, Y, and Z are chosen from the list (A.102). Since the exponentials describe
propagating wave functions, we choose
˜
ψ(r,ω) = A(ω)e ± jk x (ω)x ± jk y (ω)y ± jk z (ω)z
e
e
2
2
2
2
where A is the amplitude spectrum of the plane wave and k + k + k = k . Using this
z
x
y
˜
solution to represent each component of E, we have a propagating-wave solution to the
homogeneous vector Helmholtz equation:
˜
˜
E(r,ω) = E 0 (ω)e ± jk x (ω)x ± jk y (ω)y ± jk z (ω)z , (4.215)
e
e
where E 0 (ω) is the vector amplitude spectrum. If we define the wave vector
k(ω) = ˆ xk x (ω) + ˆ yk y (ω) + ˆ zk z (ω),
then we can write (4.215) as
˜
˜
E(r,ω) = E 0 (ω)e − jk(ω)·r . (4.216)
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