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

The derivation of Section 5.5.2 assumed a region without charges and currents. Let us now keep both, in a medium that also conducts ( $σ \neq = 0$ ). The current density then has two parts of very different nature,

J = σ E + J_{i} . (5.5.24)

The first, $σ E$ , is the conduction current that the wave itself drives in the medium by Ohm's law. The second, $J_{i}$ , is the impressed or source current, prescribed from outside, for instance the current fed into a transmitting antenna. (The lecture slides write $J_{v}$ for it. We use $J_{i}$ to keep $J_{v}$ for the convection current of Section 4.2.) The medium is still simple: $ε$ , $μ$ and $σ$ are constants.

For E. Take the curl of Faraday's law, exchange curl and $\partial / \partial t$ as in Eq. (5.5.4), and substitute the generalised Ampère law with Eq. (5.5.24), $\nabla \times H = σ E + J_{i} + ε \partial E / \partial t$ :

\nabla \times \nabla \times E = - μ \frac{\partial}{\partial t} (\nabla \times H) = - μ \frac{\partial}{\partial t} (σ E + J_{i} + ε \frac{\partial E}{\partial t}) . (5.5.25)

The divergence of $E$ is no longer zero. Gauss's law gives $\nabla \cdot E = ρ_{v} / ε$ , so the identity Eq. (5.5.6) turns the left-hand side into $\nabla (ρ_{v} / ε) - \nabla^{2} E$ . Moving all field terms to the left,

\nabla^{2} E - μ ε \frac{\partial ^{2} E}{\partial t ^{2}} - μ σ \frac{\partial E}{\partial t} = \frac{1}{ε} \nabla ρ_{v} + μ \frac{\partial J _{i}}{\partial t} . (5.5.26)

For H. Take the curl of the generalised Ampère law. Each of its three terms gives one term, $\nabla \times (σ E) = σ \nabla \times E$ , $\nabla \times J_{i}$ , and $ε \partial (\nabla \times E) / \partial t$ , and Faraday's law replaces $\nabla \times E$ by $- μ \partial H / \partial t$ :

\nabla \times \nabla \times H = \nabla \times J_{i} - σ μ \frac{\partial H}{\partial t} - ε μ \frac{\partial ^{2} H}{\partial t ^{2}} . (5.5.27)

Here $\nabla \cdot H = 0$ holds as always, so $\nabla \times \nabla \times H = - \nabla^{2} H$ , and

Wave equations with sources and conduction

In a simple medium ( $ε$ , $μ$ , $σ$ constant) with charge density $ρ_{v}$ and impressed current density $J_{i}$ ,

\nabla^{2} E - μ ε \frac{\partial ^{2} E}{\partial t ^{2}} - μ σ \frac{\partial E}{\partial t} = \frac{1}{ε} \nabla ρ_{v} + μ \frac{\partial J _{i}}{\partial t}, (5.5.28)

\nabla^{2} H - μ ε \frac{\partial ^{2} H}{\partial t ^{2}} - μ σ \frac{\partial H}{\partial t} = - \nabla \times J_{i} . (5.5.29)

For $σ = 0$ , $ρ_{v} = 0$ and $J_{i} = 0$ they reduce to the wave equations Eq. (5.5.13).

In plain words

Read the equation term by term. $\nabla^{2}$ : how the field bends in space. $μ ε \partial^{2} / \partial t^{2}$ : the displacement current, which makes it a wave. $μ σ \partial / \partial t$ : the conduction current, which makes it a loss. The right-hand side: the sources, which launch the wave.

In the lossless case $σ = 0$ the speed is again $v = 1/ μ ε$ , the speed of light $c = 1/ μ_{0} ε_{0}$ in free space. Note that the source terms of the two equations are different. The equation for $H$ is driven by the curl of the impressed current alone, the one for $E$ by the time derivative of the current and the gradient of the charge, which are tied together by the continuity equation. Section 5.5.8 shows that the potentials lead to simpler equations, each driven by one source only.

5.5.6 Wave equations with sources and conduction