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.

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