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}) %= \nabla \cross \Big( \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{\vb{E}}{t} \Big) = \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{t} (\nabla \cross \vb{E}) %= \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{t} \Big( \!-\! \pdv{\vb{B}}{t} \Big) = - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv[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 \pdv[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}) %= - \nabla \cross \pdv{\vb{B}}{t} = - \pdv{t} (\nabla \cross \vb{B}) %= - \pdv{t} \Big( \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{\vb{E}}{t} \Big) = - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv[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 \pdv[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{ \pdv[2]{\vb{E}}{t} - \frac{1}{v^2} \nabla^2 \vb{E} = 0 } \qquad \quad \boxed{ \pdv[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 \pdv[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 \pdv[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 $$\vb{E}_0$$.

## References

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

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.