Categories: Electromagnetism, Optics, Physics.

# Electromagnetic wave equation

The electromagnetic wave equation describes the propagation of light through various media. Since an electromagnetic (light) wave consists of an electric field and a magnetic field, we need Maxwell’s equations in order to derive the wave equation.

## Uniform medium

We will use all of Maxwell’s equations, but we start with Ampère’s circuital law for the “free” fields $\vb{H}$ and $\vb{D}$, in the absence of a free current $\vb{J}_\mathrm{free} = 0$:

\begin{aligned} \nabla \cross \vb{H} = \pdv{\vb{D}}{t} \end{aligned}

We assume that the medium is isotropic, linear, and uniform in all of space, such that:

\begin{aligned} \vb{D} = \varepsilon_0 \varepsilon_r \vb{E} \qquad \quad \vb{H} = \frac{1}{\mu_0 \mu_r} \vb{B} \end{aligned}

Which, upon insertion into Ampère’s law, yields an equation relating $\vb{B}$ and $\vb{E}$. This may seem to contradict Ampère’s “total” law, but keep in mind that $\vb{J}_\mathrm{bound} \neq 0$ here:

\begin{aligned} \nabla \cross \vb{B} = \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{\vb{E}}{t} \end{aligned}

Now we take the curl, rearrange, and substitute $\nabla \cross \vb{E}$ according to Faraday’s law:

\begin{aligned} \nabla \cross (\nabla \cross \vb{B}) = \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{}{t}(\nabla \cross \vb{E}) = - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{B}}{t} \end{aligned}

Using a vector identity, we rewrite the leftmost expression, which can then be reduced thanks to Gauss’ law for magnetism $\nabla \cdot \vb{B} = 0$:

\begin{aligned} - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{B}}{t} &= \nabla (\nabla \cdot \vb{B}) - \nabla^2 \vb{B} = - \nabla^2 \vb{B} \end{aligned}

This describes $\vb{B}$. Next, we repeat the process for $\vb{E}$: taking the curl of Faraday’s law yields:

\begin{aligned} \nabla \cross (\nabla \cross \vb{E}) = - \pdv{}{t}(\nabla \cross \vb{B}) = - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} \end{aligned}

Which can be rewritten using same vector identity as before, and then reduced by assuming that there is no net charge density $\rho = 0$ in Gauss’ law, such that $\nabla \cdot \vb{E} = 0$:

\begin{aligned} - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} &= \nabla (\nabla \cdot \vb{E}) - \nabla^2 \vb{E} = - \nabla^2 \vb{E} \end{aligned}

We thus arrive at the following two (implicitly coupled) wave equations for $\vb{E}$ and $\vb{B}$, where we have defined the phase velocity $v \equiv 1 / \sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}$:

\begin{aligned} \boxed{ \pdvn{2}{\vb{E}}{t} - \frac{1}{v^2} \nabla^2 \vb{E} = 0 } \qquad \quad \boxed{ \pdvn{2}{\vb{B}}{t} - \frac{1}{v^2} \nabla^2 \vb{B} = 0 } \end{aligned}

Traditionally, it is said that the solutions are as follows, where the wavenumber $|\vb{k}| = \omega / v$:

\begin{aligned} \vb{E}(\vb{r}, t) &= \vb{E}_0 \exp(i \vb{k} \cdot \vb{r} - i \omega t) \\ \vb{B}(\vb{r}, t) &= \vb{B}_0 \exp(i \vb{k} \cdot \vb{r} - i \omega t) \end{aligned}

In fact, thanks to linearity, these plane waves can be treated as terms in a Fourier series, meaning that virtually any function $f(\vb{k} \cdot \vb{r} - \omega t)$ is a valid solution.

Keep in mind that in reality $\vb{E}$ and $\vb{B}$ are real, so although it is mathematically convenient to use plane waves, in the end you will need to take the real part.

## Non-uniform medium

A useful generalization is to allow spatial change in the relative permittivity $\varepsilon_r(\vb{r})$ and the relative permeability $\mu_r(\vb{r})$. We still assume that the medium is linear and isotropic, so:

\begin{aligned} \vb{D} = \varepsilon_0 \varepsilon_r(\vb{r}) \vb{E} \qquad \quad \vb{B} = \mu_0 \mu_r(\vb{r}) \vb{H} \end{aligned}

Inserting these expressions into Faraday’s and Ampère’s laws respectively yields:

\begin{aligned} \nabla \cross \vb{E} = - \mu_0 \mu_r(\vb{r}) \pdv{\vb{H}}{t} \qquad \quad \nabla \cross \vb{H} = \varepsilon_0 \varepsilon_r(\vb{r}) \pdv{\vb{E}}{t} \end{aligned}

We then divide Ampère’s law by $\varepsilon_r(\vb{r})$, take the curl, and substitute Faraday’s law, giving:

\begin{aligned} \nabla \cross \Big( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \Big) = \varepsilon_0 \pdv{}{t}(\nabla \cross \vb{E}) = - \mu_0 \mu_r \varepsilon_0 \pdvn{2}{\vb{H}}{t} \end{aligned}

Next, we exploit linearity by decomposing $\vb{H}$ and $\vb{E}$ into Fourier series, with terms given by:

\begin{aligned} \vb{H}(\vb{r}, t) = \vb{H}(\vb{r}) \exp(- i \omega t) \qquad \quad \vb{E}(\vb{r}, t) = \vb{E}(\vb{r}) \exp(- i \omega t) \end{aligned}

By inserting this ansatz into the equation, we can remove the explicit time dependence:

\begin{aligned} \nabla \cross \Big( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \Big) \exp(- i \omega t) = \mu_0 \varepsilon_0 \omega^2 \mu_r \vb{H} \exp(- i \omega t) \end{aligned}

Dividing out $\exp(- i \omega t)$, we arrive at an eigenvalue problem for $\omega^2$, with $c = 1 / \sqrt{\mu_0 \varepsilon_0}$:

\begin{aligned} \boxed{ \nabla \cross \Big( \frac{1}{\varepsilon_r(\vb{r})} \nabla \cross \vb{H}(\vb{r}) \Big) = \Big( \frac{\omega}{c} \Big)^2 \mu_r(\vb{r}) \vb{H}(\vb{r}) } \end{aligned}

Compared to a uniform medium, $\omega$ is often not arbitrary here: there are discrete eigenvalues $\omega$, corresponding to discrete modes $\vb{H}(\vb{r})$.

Next, we go through the same process to find an equation for $\vb{E}$. Starting from Faraday’s law, we divide by $\mu_r(\vb{r})$, take the curl, and insert Ampère’s law:

\begin{aligned} \nabla \cross \Big( \frac{1}{\mu_r} \nabla \cross \vb{E} \Big) = - \mu_0 \pdv{}{t}(\nabla \cross \vb{H}) = - \mu_0 \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t} \end{aligned}

Then, by replacing $\vb{E}(\vb{r}, t)$ with our plane-wave ansatz, we remove the time dependence:

\begin{aligned} \nabla \cross \Big( \frac{1}{\mu_r} \nabla \cross \vb{E} \Big) \exp(- i \omega t) = - \mu_0 \varepsilon_0 \omega^2 \varepsilon_r \vb{E} \exp(- i \omega t) \end{aligned}

Which, after dividing out $\exp(- i \omega t)$, yields an analogous eigenvalue problem with $\vb{E}(r)$:

\begin{aligned} \boxed{ \nabla \cross \Big( \frac{1}{\mu_r(\vb{r})} \nabla \cross \vb{E}(\vb{r}) \Big) = \Big( \frac{\omega}{c} \Big)^2 \varepsilon_r(\vb{r}) \vb{E}(\vb{r}) } \end{aligned}

Usually, it is a reasonable approximation to say $\mu_r(\vb{r}) = 1$, in which case the equation for $\vb{H}(\vb{r})$ becomes a Hermitian eigenvalue problem, and is thus easier to solve than for $\vb{E}(\vb{r})$.

Keep in mind, however, that in any case, the solutions $\vb{H}(\vb{r})$ and/or $\vb{E}(\vb{r})$ must satisfy the two Maxwell’s equations that were not explicitly used:

\begin{aligned} \nabla \cdot (\varepsilon_r \vb{E}) = 0 \qquad \quad \nabla \cdot (\mu_r \vb{H}) = 0 \end{aligned}

This is equivalent to demanding that the resulting waves are transverse, or in other words, the wavevector $\vb{k}$ must be perpendicular to the amplitudes $\vb{H}_0$ and $\vb{E}_0$.

1. J.D. Joannopoulos, S.G. Johnson, J.N. Winn, R.D. Meade, Photonic crystals: molding the flow of light, 2nd edition, Princeton.