Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

An electromagnetic wave is called a plane wave when its sources are at infinity, that is, when the region of interest contains no sources at all. In phasor form, with the time factor $e^{j ω t}$ suppressed, Maxwell's equations of Section 5.6 then lose their right-hand sides:

with sources \overset{◯}{1} \nabla \times E + j ω B = 0 \overset{◯}{2} \nabla \times H - j ω D = J \overset{◯}{3} \nabla \cdot D = ρ_{v} \overset{◯}{4} \nabla \cdot B = 0 \to \to \to \to source-free \nabla \times E + j ω B = 0 \nabla \times H - j ω D = 0 \nabla \cdot D = 0 \nabla \cdot B = 0 (6.1.1)

Taking the curl of $\overset{◯}{1}$ , inserting $\overset{◯}{2}$ and using $\nabla \cdot E = 0$ from $\overset{◯}{3}$ , exactly as in Section 5.5, gives the homogeneous Helmholtz equations

\nabla^{2} E + k^{2} E = 0, \nabla^{2} H + k^{2} H = 0, k^{2} = ω^{2} μ ε . (6.1.2)

Every Cartesian component of $E$ and of $H$ therefore obeys the same scalar equation $\nabla^{2} u + k^{2} u = 0$ .

In practice "at infinity" means far away. A small source radiates spherical wavefronts, but an observer who looks at a region much smaller than its distance from the source cannot tell the curved front from a plane (Figure 6.1.1). How far is far enough? Across a width $D$ a sphere of radius $r$ departs from its tangent plane by the sagitta $D^{2} /8 r$ . The front may be treated as plane when this is a small fraction of a wavelength, and the usual criterion, $D^{2} /8 r < λ /16$ , gives

r > \frac{2 D ^{2}}{λ} . (6.1.3)

A receiving dish of $D = 1 m$ at $10 GHz$ ( $λ = 3 cm$ ) sees a plane wave from any source farther than about $67 m$ , a broadcast antenna a few kilometres away is far beyond that, and the sun is so far that sunlight is an excellent plane wave over any instrument we can build.

Far from its source every wave is a plane wave. Over the small dashed window of width D at the distance r the spherical wavefronts are practically flat and parallel, and the wave travels along a single direction \vec a_k. — **Figure 6.1.1.** Far from its source every wave is a plane wave. Over the small dashed window of width $D$ at the distance $r$ the spherical wavefronts are practically flat and parallel, and the wave travels along a single direction $a_{k}$ .

The one-dimensional solution

Let the field depend on one coordinate only, $u = u (z)$ . The Helmholtz equation becomes an ordinary differential equation with constant coefficients,

\frac{d ^{2} u ( z )}{d z ^{2}} + k^{2} u (z) = 0, (6.1.4)

whose general solution is the sum of two exponentials,

u (z) = C_{1} e^{- j k z} + C_{2} e^{j k z} . (6.1.5)

In a lossy medium $k$ is complex (Section 5.6), and we write it as

k = β - j α, (6.1.6)

so that $e^{- j k z} = e^{- j (β - j α) z} = e^{- α z} e^{- j β z}$ and

u (z) = C_{1} e^{- α z} e^{- j β z} + C_{2} e^{α z} e^{j β z} . (6.1.7)

$β [rad/m]$ is the phase constant and $α [Np/m]$ the attenuation constant. Back in the time domain, with real constants $C_{1}$ and $C_{2}$ ,

u (z; t) = Re {u (z) e^{j ω t}} = towards + z C_{1} e^{- α z} cos (ω t - β z) + towards - z C_{2} e^{α z} cos (ω t + β z) . (6.1.8)

The first term is a wave travelling towards $+ z$ and the second one a wave travelling towards $- z$ , as the next paragraph shows. Each decays in its own direction of travel: the factor $e^{α z}$ of the second wave grows with $z$ , which means that it shrinks as the wave advances towards $- z$ (Figure 6.1.2). In a lossless medium $α = 0$ and $β = k$ .

The forward wave C_1\mathrm e^{-\alpha z}\cos(\omega t-\beta z) at two instants, drawn with \alpha\lambda=0.5. The crests move towards +z while the amplitude follows the envelope \mathrm e^{-\alpha z}, which does not move. — **Figure 6.1.2.** The forward wave $C_{1} e^{- α z} cos (ω t - β z)$ at two instants, drawn with $α λ = 0.5$ . The crests move towards $+ z$ while the amplitude follows the envelope $e^{- α z}$ , which does not move.

Phase velocity and wavelength

Follow a point of constant phase of the forward wave, a crest for instance:

ω t - β z = const ⟹ ω d t - β d z = 0. (6.1.9)

The point moves with the phase velocity

v_{p} = \frac{d z}{d t} = \frac{ω}{β} [m/s] . (6.1.10)

At a fixed instant the pattern repeats when $β z$ grows by $2 π$ , so the wavelength and the phase constant are related by

β = \frac{2 π}{λ} [rad/m] . (6.1.11)

For the backward wave $ω t + β z =$ const gives $d z / d t = - ω / β$ : the same speed towards $- z$ . Figure 6.1.3 shows the two snapshots.

Phase velocity and wavelength in a lossless medium

In a lossless medium $α = 0$ and $β = k = ω μ ε$ , so

v_{p} c = \frac{1}{μ ε} = \frac{c}{μ _{r} ε _{r}}, λ = \frac{2 π}{β} = \frac{v _{p}}{f}, = \frac{1}{μ _{0} ε _{0}} = 3 \times 1 0^{8} m/s . (6.1.12)

In plain words

A wave is a pattern that slides along without changing shape. It covers one wavelength in one period, so its speed is $λ f$ . In a material the pattern is slower than in vacuum by the factor $μ_{r} ε_{r}$ , and at the same frequency it is shorter by the same factor.

Phase velocity and wavelength. In one eighth of a period the crest moves on by \lambda/8, so v_p=\lambda/T=\omega/\beta. — **Figure 6.1.3.** Phase velocity and wavelength. In one eighth of a period the crest moves on by $λ /8$ , so $v_{p} = λ / T = ω / β$ .

Two speeds, one wave

$v_{p}$ is the speed of the pattern, not of a particle and, in a dispersive medium, not of the energy either. The speed of energy and of signals, the group velocity, is the subject of Section 6.2. In vacuum the two coincide and both equal $c = 1/ μ_{0} ε_{0} = 3 \times 1 0^{8} m/s$ .

Quick check

A $600 MHz$ wave travels in a lossless dielectric with $ε_{r} = 4$ , $μ_{r} = 1$ . Find $v_{p}$ , $λ$ and $β$ .

$v_{p} = c / 4 = 1.5 \times 1 0^{8} m/s$ , $λ = v_{p} / f = 25 cm$ (against $50 cm$ in air) and $β = 2 π / λ = 8 π = 25.1 rad/m$ .

6.1.1 Plane waves and the Helmholtz equation

The one-dimensional solution

Phase velocity and wavelength