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+---
+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.