---
title: "Electromagnetic wave equation"
firstLetter: "E"
publishDate: 2021-09-09
categories:
- Physics
- Electromagnetism
- Optics

date: 2021-09-09T21:20:31+02:00
draft: false
markup: pandoc
---

# 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](/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:

$$\begin{aligned}
    \nabla \cross (\nabla \cross \vb{B})
    %= \nabla \cross \Big( \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{\vb{E}}{t} \Big)
    = \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{t} (\nabla \cross \vb{E})
    %= \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{t} \Big( \!-\! \pdv{\vb{B}}{t} \Big)
    = - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv[2]{\vb{B}}{t}
\end{aligned}$$

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 $|\vb{k}| = \omega / v$:

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

In fact, thanks to linearity, these solutions 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.


## 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.