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 H\vb{H} and D\vb{D}, in the absence of a free current Jfree=0\vb{J}_\mathrm{free} = 0:

×H=Dt\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:

D=ε0εrEH=1μ0μrB\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 B\vb{B} and E\vb{E}. This may seem to contradict Ampère’s “total” law, but keep in mind that Jbound0\vb{J}_\mathrm{bound} \neq 0 here:

×B=μ0μrε0εrEt\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 ×E\nabla \cross \vb{E} according to Faraday’s law:

×(×B)=μ0μrε0εrt(×E)=μ0μrε0εr2Bt2\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 B=0\nabla \cdot \vb{B} = 0:

μ0μrε0εr2Bt2=(B)2B=2B\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 B\vb{B}. Next, we repeat the process for E\vb{E}: taking the curl of Faraday’s law yields:

×(×E)=t(×B)=μ0μrε0εr2Et2\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 ρ=0\rho = 0 in Gauss’ law, such that E=0\nabla \cdot \vb{E} = 0:

μ0μrε0εr2Et2=(E)2E=2E\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 E\vb{E} and B\vb{B}, where we have defined the phase velocity v1/μ0μrε0εrv \equiv 1 / \sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}:

2Et21v22E=02Bt21v22B=0\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 k=ω/v|\vb{k}| = \omega / v:

E(r,t)=E0exp(ikriωt)B(r,t)=B0exp(ikriωt)\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(krωt)f(\vb{k} \cdot \vb{r} - \omega t) is a valid solution.

Keep in mind that in reality E\vb{E} and B\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 εr(r)\varepsilon_r(\vb{r}) and the relative permeability μr(r)\mu_r(\vb{r}). We still assume that the medium is linear and isotropic, so:

D=ε0εr(r)EB=μ0μr(r)H\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:

×E=μ0μr(r)Ht×H=ε0εr(r)Et\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 εr(r)\varepsilon_r(\vb{r}), take the curl, and substitute Faraday’s law, giving:

×(1εr×H)=ε0t(×E)=μ0μrε02Ht2\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 H\vb{H} and E\vb{E} into Fourier series, with terms given by:

H(r,t)=H(r)exp(iωt)E(r,t)=E(r)exp(iωt)\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:

×(1εr×H)exp(iωt)=μ0ε0ω2μrHexp(iωt)\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(iωt)\exp(- i \omega t), we arrive at an eigenvalue problem for ω2\omega^2, with c=1/μ0ε0c = 1 / \sqrt{\mu_0 \varepsilon_0}:

×(1εr(r)×H(r))=(ωc)2μr(r)H(r)\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, ω\omega is often not arbitrary here: there are discrete eigenvalues ω\omega, corresponding to discrete modes H(r)\vb{H}(\vb{r}).

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

×(1μr×E)=μ0t(×H)=μ0ε0εr2Et2\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 E(r,t)\vb{E}(\vb{r}, t) with our plane-wave ansatz, we remove the time dependence:

×(1μr×E)exp(iωt)=μ0ε0ω2εrEexp(iωt)\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(iωt)\exp(- i \omega t), yields an analogous eigenvalue problem with E(r)\vb{E}(r):

×(1μr(r)×E(r))=(ωc)2εr(r)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 μr(r)=1\mu_r(\vb{r}) = 1, in which case the equation for H(r)\vb{H}(\vb{r}) becomes a Hermitian eigenvalue problem, and is thus easier to solve than for E(r)\vb{E}(\vb{r}).

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

(εrE)=0(μrH)=0\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 k\vb{k} must be perpendicular to the amplitudes H0\vb{H}_0 and E0\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.