---
title: "Electromagnetic wave equation"
sort_title: "Electromagnetic wave equation"
date: 2024-09-08 # Originally 2021-09-09, major rewrite
categories:
- Physics
- Electromagnetism
- Optics
layout: "concept"
---

Light, i.e. **electromagnetic waves**, consist of
an [electric field](/know/concept/electric-field/)
and a [magnetic field](/know/concept/magnetic-field/),
one inducing the other and vice versa.
The existence and classical behavior of such waves
can be derived using only [Maxwell's equations](/know/concept/maxwells-equations/),
as we will demonstrate here.

We start from Faraday's law of induction,
where we assume that the system consists of materials
with well-known (linear) relative magnetic permeabilities $$\mu_r(\vb{r})$$,
such that $$\vb{B} = \mu_0 \mu_r \vb{H}$$:

$$\begin{aligned}
    \nabla \cross \vb{E}
    = - \pdv{\vb{B}}{t}
    = - \mu_0 \mu_r \pdv{\vb{H}}{t}
\end{aligned}$$

We move $$\mu_r(\vb{r})$$ to the other side,
take the curl, and insert Ampère's circuital law:

$$\begin{aligned}
    \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg)
    &= - \mu_0 \pdv{}{t} \big( \nabla \cross \vb{H} \big)
    \\
    &= - \mu_0 \bigg( \pdv{\vb{J}_\mathrm{free}}{t} + \pdvn{2}{\vb{D}}{t} \bigg)
\end{aligned}$$

For simplicity, we only consider insulating materials,
since light propagation in conductors is a complex beast.
We thus assume that there are no free currents $$\vb{J}_\mathrm{free} = 0$$, leaving:

$$\begin{aligned}
    \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg)
    &= - \mu_0 \pdvn{2}{\vb{D}}{t}
\end{aligned}$$

Having $$\vb{E}$$ and $$\vb{D}$$ in the same equation is not ideal,
so we should make a choice:
do we restrict ourselves to linear media
(so $$\vb{D} = \varepsilon_0 \varepsilon_r \vb{E}$$),
or do we allow materials with more complicated responses
(so $$\vb{D} = \varepsilon_0 \vb{E} + \vb{P}$$, with $$\vb{P}$$ unspecified)?
The former is usually sufficient:

$$\begin{aligned}
    \boxed{
        \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg)
        = - \mu_0 \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t}
    }
\end{aligned}$$

This is the general linear form of the **electromagnetic wave equation**,
where $$\mu_r$$ and $$\varepsilon_r$$
both depend on $$\vb{r}$$ in order to describe the structure of the system.
We can obtain a similar equation for $$\vb{H}$$,
by starting from Ampère's law under the same assumptions:

$$\begin{aligned}
    \nabla \cross \vb{H}
    = \pdv{\vb{D}}{t}
    = \varepsilon_0 \varepsilon_r \pdv{\vb{E}}{t}
\end{aligned}$$

Taking the curl and substituting Faraday's law on the right yields:

$$\begin{aligned}
    \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg)
    &= \varepsilon_0 \pdv{}{t} \big( \nabla \cross \vb{E} \big)
    = - \varepsilon_0 \pdvn{2}{\vb{B}}{t}
\end{aligned}$$

And then we insert $$\vb{B} = \mu_0 \mu_r \vb{H}$$ to get the analogous
electromagnetic wave equation for $$\vb{H}$$:

$$\begin{aligned}
    \boxed{
        \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg)
        = - \mu_0 \varepsilon_0 \mu_r \pdvn{2}{\vb{H}}{t}
    }
\end{aligned}$$

This is equivalent to the problem for $$\vb{E}$$,
since they are coupled by Maxwell's equations.
By solving either, subject to Gauss's laws
$$\nabla \cdot (\varepsilon_r \vb{E}) = 0$$ and $$\nabla \cdot (\mu_r \vb{H}) = 0$$,
the behavior of light in a given system can be deduced.
Note that Gauss's laws enforce that the wave's fields are transverse,
i.e. they must be perpendicular to the propagation direction.



## Homogeneous linear media

In the special case where the medium is completely uniform,
$$\mu_r$$ and $$\varepsilon_r$$ no longer depend on $$\vb{r}$$,
so they can be moved to the other side:

$$\begin{aligned}
    \nabla \cross \big( \nabla \cross \vb{E} \big)
    &= - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t}
    \\
    \nabla \cross \big( \nabla \cross \vb{H} \big)
    &= - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{H}}{t}
\end{aligned}$$

This can be rewritten using the vector identity
$$\nabla \cross (\nabla \cross \vb{V}) = \nabla (\nabla \cdot \vb{V}) - \nabla^2 \vb{V}$$:

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

Which can be reduced using Gauss's laws
$$\nabla \cdot \vb{E} = 0$$ and $$\nabla \cdot \vb{H} = 0$$
thanks to the fact that $$\varepsilon_r$$ and $$\mu_r$$ are constants in this case.
We therefore arrive at:

$$\begin{aligned}
    \boxed{
        \nabla^2 \vb{E} - \frac{n^2}{c^2} \pdvn{2}{\vb{E}}{t}
        = 0
    }
\end{aligned}$$

$$\begin{aligned}
    \boxed{
        \nabla^2 \vb{H} - \frac{n^2}{c^2} \pdvn{2}{\vb{H}}{t}
        = 0
    }
\end{aligned}$$

Where $$c = 1 / \sqrt{\mu_0 \varepsilon_0}$$ is the speed of light in a vacuum,
and $$n = \sqrt{\mu_0 \varepsilon_0}$$ is the refractive index of the medium.
Note that most authors write the magnetic equation with $$\vb{B}$$ instead of $$\vb{H}$$;
both are correct thanks to linearity.

In a vacuum, where $$n = 1$$, these equations are sometimes written as
$$\square \vb{E} = 0$$ and $$\square \vb{H} = 0$$,
where $$\square$$ is the **d'Alembert operator**, defined as follows:

$$\begin{aligned}
    \boxed{
        \square
        \equiv \nabla^2 - \frac{1}{c^2} \pdvn{2}{}{t}
    }
\end{aligned}$$

Note that some authors define it with the opposite sign.
In any case, the d'Alembert operator is important for special relativity.

The solution to the homogeneous electromagnetic wave equation
are traditionally said to be the so-called **plane waves** given by:

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

Where the wavevector $$\vb{k}$$ is arbitrary,
and the angular frequency $$\omega = c |\vb{k}| / n$$.
We also often talk about the wavelength, which is $$\lambda = 2 \pi / |\vb{k}|$$.
The appearance of $$\vb{k}$$ in the exponent
tells us that these waves are propagating through space,
as you would expect.

In fact, because the wave equations are linear,
any superposition of plane waves,
i.e. any function of the form $$f(\vb{k} \cdot \vb{r} - \omega t)$$,
is in fact a valid solution.
Just remember that $$\vb{E}$$ and $$\vb{H}$$ are real-valued,
so it may be necessary to take the real part at the end of a calculation.



## Inhomogeneous linear media

But suppose the medium is not uniform, i.e. it contains structures
described by $$\varepsilon_r(\vb{r})$$ and $$\mu_r(\vb{r})$$.
If the structures are much larger than the light's wavelength,
the homogeneous equation is still a very good approximation
away from any material boundaries;
anywhere else, however, they will break down.
Recall the general equations from before we assumed homogeneity:

$$\begin{aligned}
    \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg)
    &= - \frac{\varepsilon_r}{c^2} \pdvn{2}{\vb{E}}{t}
    \\
    \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg)
    &= - \frac{\mu_r}{c^2} \pdvn{2}{\vb{H}}{t}
\end{aligned}$$

In theory, this is everything we need,
but in most cases a better approach is possible:
the trick is that we only rarely need to explicitly calculate
the $$t$$-dependence of $$\vb{E}$$ or $$\vb{H}$$.
Instead, we can first solve an easier time-independent version
of this problem, and then approximate the dynamics
with [coupled mode theory](/know/concept/coupled-mode-theory/) later.

To eliminate $$t$$, we make an ansatz for $$\vb{E}$$ and $$\vb{H}$$, shown below.
No generality is lost by doing this;
this is effectively a kind of [Fourier transform](/know/concept/fourier-transform/):

$$\begin{aligned}
    \vb{E}(\vb{r}, t)
    &= \vb{E}(\vb{r}) e^{- i \omega t}
    \\
    \vb{H}(\vb{r}, t)
    &= \vb{H}(\vb{r}) e^{- i \omega t}
\end{aligned}$$

Inserting this ansatz and dividing out $$e^{-i \omega t}$$
yields the time-independent forms:

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

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

These are eigenvalue problems for $$\omega^2$$,
which can be solved subject to Gauss's laws and suitable boundary conditions.
The resulting allowed values of $$\omega$$ may consist of
continuous ranges and/or discrete resonances,
analogous to *scattering* and *bound* quantum states, respectively.
It can be shown that the operators on both sides of each equation
are Hermitian, meaning these are well-behaved problems
yielding real eigenvalues and orthogonal eigenfields.

Both equations are still equivalent:
we only need to solve one. But which one?
In practice, one is usually easier than the other,
due to the common approximation that $$\mu_r \approx 1$$ for many dielectric materials,
in which case the equations reduce to:

$$\begin{aligned}
    \nabla \cross \big( \nabla \cross \vb{E} \big)
    &= \Big( \frac{\omega}{c} \Big)^2 \varepsilon_r \vb{E}
    \\
    \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg)
    &= \Big( \frac{\omega}{c} \Big)^2 \vb{H}
\end{aligned}$$

Now the equation for $$\vb{H}$$ is starting to look simpler,
because it only has an operator on *one* side.
We could "fix" the equation for $$\vb{E}$$ by dividing it by $$\varepsilon_r$$,
but the resulting operator would no longer be Hermitian,
and hence not well-behaved.
To get an idea of how to handle $$\varepsilon_r$$ in the $$\vb{E}$$-equation,
notice its similarity to the weight function $$w$$
in [Sturm-Liouville theory](/know/concept/sturm-liouville-theory/).

Gauss's magnetic law $$\nabla \cdot \vb{H} = 0$$
is also significantly easier for numerical calculations
than its electric counterpart $$\nabla \cdot (\varepsilon_r \vb{E}) = 0$$,
so we usually prefer to solve the equation for $$\vb{H}$$.



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