---
title: "Electromagnetic wave equation"
sort_title: "Electromagnetic wave equation"
date: 2021-09-09
categories:
- Physics
- Electromagnetism
- Optics
layout: "concept"
---

The electromagnetic wave equation describes
the propagation of light through various media.
Since an electromagnetic (light) wave consists of
an [electric field](/know/concept/electric-field/)
and a [magnetic field](/know/concept/magnetic-field/),
we need [Maxwell's equations](/know/concept/maxwells-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 $$\vb{E}_0$$.


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