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diff --git a/source/know/concept/electromagnetic-wave-equation/index.md b/source/know/concept/electromagnetic-wave-equation/index.md
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--- a/source/know/concept/electromagnetic-wave-equation/index.md
+++ b/source/know/concept/electromagnetic-wave-equation/index.md
@@ -1,7 +1,7 @@
 ---
 title: "Electromagnetic wave equation"
 sort_title: "Electromagnetic wave equation"
-date: 2021-09-09
+date: 2024-09-08 # Originally 2021-09-09, major rewrite
 categories:
 - Physics
 - Electromagnetism
@@ -9,236 +9,281 @@ categories:
 layout: "concept"
 ---
 
-The electromagnetic wave equation describes
-the propagation of light through various media.
-Since an electromagnetic (light) wave consists of
+Light, i.e. **electromagnetic waves**, consist 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.
+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.
 
-
-## 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:
+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 $$\vb{B} = \mu_0 \mu_r \vb{H}$$:
 
 $$\begin{aligned}
-    \vb{D} = \varepsilon_0 \varepsilon_r \vb{E}
-    \qquad \quad
-    \vb{H} = \frac{1}{\mu_0 \mu_r} \vb{B}
+    \nabla \cross \vb{E}
+    = - \pdv{\vb{B}}{t}
+    = - \mu_0 \mu_r \pdv{\vb{H}}{t}
 \end{aligned}$$
 
-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:
+We move $$\mu_r(\vb{r})$$ to the other side,
+take the curl, and insert Ampère's circuital law:
 
 $$\begin{aligned}
-    \nabla \cross \vb{B}
-    = \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{\vb{E}}{t}
+    \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg)
+    &= - \mu_0 \pdv{}{t} \big( \nabla \cross \vb{H} \big)
+    \\
+    &= - \mu_0 \bigg( \pdv{\vb{J}_\mathrm{free}}{t} + \pdvn{2}{\vb{D}}{t} \bigg)
 \end{aligned}$$
 
-Now we take the curl, rearrange,
-and substitute $$\nabla \cross \vb{E}$$ according to Faraday's law:
+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:
 
 $$\begin{aligned}
-    \nabla \cross (\nabla \cross \vb{B})
-    = \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{}{t}(\nabla \cross \vb{E})
-    = - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{B}}{t}
+    \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg)
+    &= - \mu_0 \pdvn{2}{\vb{D}}{t}
 \end{aligned}$$
 
-Using a vector identity, we rewrite the leftmost expression,
-which can then be reduced thanks to Gauss' law for magnetism $$\nabla \cdot \vb{B} = 0$$:
+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:
 
 $$\begin{aligned}
-    - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{B}}{t}
-    &= \nabla (\nabla \cdot \vb{B}) - \nabla^2 \vb{B}
-    = - \nabla^2 \vb{B}
+    \boxed{
+        \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg)
+        = - \mu_0 \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t}
+    }
 \end{aligned}$$
 
-This describes $$\vb{B}$$.
-Next, we repeat the process for $$\vb{E}$$:
-taking the curl of Faraday's law yields:
+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:
 
 $$\begin{aligned}
-    \nabla \cross (\nabla \cross \vb{E})
-    = - \pdv{}{t}(\nabla \cross \vb{B})
-    = - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t}
+    \nabla \cross \vb{H}
+    = \pdv{\vb{D}}{t}
+    = \varepsilon_0 \varepsilon_r \pdv{\vb{E}}{t}
 \end{aligned}$$
 
-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 $$\nabla \cdot \vb{E} = 0$$:
+Taking the curl and substituting Faraday's law on the right yields:
 
 $$\begin{aligned}
-    - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t}
-    &= \nabla (\nabla \cdot \vb{E}) - \nabla^2 \vb{E}
-    = - \nabla^2 \vb{E}
+    \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg)
+    &= \varepsilon_0 \pdv{}{t} \big( \nabla \cross \vb{E} \big)
+    = - \varepsilon_0 \pdvn{2}{\vb{B}}{t}
 \end{aligned}$$
 
-We thus arrive at the following two (implicitly coupled)
-wave equations for $$\vb{E}$$ and $$\vb{B}$$,
-where we have defined the phase velocity $$v \equiv 1 / \sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}$$:
+And then we insert $$\vb{B} = \mu_0 \mu_r \vb{H}$$ to get the analogous
+electromagnetic wave equation for $$\vb{H}$$:
 
 $$\begin{aligned}
     \boxed{
-        \pdvn{2}{\vb{E}}{t} - \frac{1}{v^2} \nabla^2 \vb{E}
-        = 0
-    }
-    \qquad \quad
-    \boxed{
-        \pdvn{2}{\vb{B}}{t} - \frac{1}{v^2} \nabla^2 \vb{B}
-        = 0
+        \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg)
+        = - \mu_0 \varepsilon_0 \mu_r \pdvn{2}{\vb{H}}{t}
     }
 \end{aligned}$$
 
-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.
+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.
 
-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
+## Homogeneous linear media
 
-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:
+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:
 
 $$\begin{aligned}
-    \vb{D}
-    = \varepsilon_0 \varepsilon_r(\vb{r}) \vb{E}
-    \qquad \quad
-    \vb{B}
-    = \mu_0 \mu_r(\vb{r}) \vb{H}
+    \nabla \cross \big( \nabla \cross \vb{E} \big)
+    &= - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t}
+    \\
+    \nabla \cross \big( \nabla \cross \vb{H} \big)
+    &= - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{H}}{t}
 \end{aligned}$$
 
-Inserting these expressions into Faraday's and Ampère's laws
-respectively yields:
+This can be rewritten using the vector identity
+$$\nabla \cross (\nabla \cross \vb{V}) = \nabla (\nabla \cdot \vb{V}) - \nabla^2 \vb{V}$$:
 
 $$\begin{aligned}
-    \nabla \cross \vb{E}
-    = - \mu_0 \mu_r(\vb{r}) \pdv{\vb{H}}{t}
-    \qquad \quad
-    \nabla \cross \vb{H}
-    = \varepsilon_0 \varepsilon_r(\vb{r}) \pdv{\vb{E}}{t}
+    \nabla (\nabla \cdot \vb{E}) - \nabla^2 \vb{E}
+    &= - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t}
+    \\
+    \nabla (\nabla \cdot \vb{H}) - \nabla^2 \vb{H}
+    &= - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{H}}{t}
 \end{aligned}$$
 
-We then divide Ampère's law by $$\varepsilon_r(\vb{r})$$,
-take the curl, and substitute Faraday's law, giving:
+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:
 
 $$\begin{aligned}
-    \nabla \cross \Big( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \Big)
-    = \varepsilon_0 \pdv{}{t}(\nabla \cross \vb{E})
-    = - \mu_0 \mu_r \varepsilon_0 \pdvn{2}{\vb{H}}{t}
+    \boxed{
+        \nabla^2 \vb{E} - \frac{n^2}{c^2} \pdvn{2}{\vb{E}}{t}
+        = 0
+    }
 \end{aligned}$$
 
-Next, we exploit linearity by decomposing $$\vb{H}$$ and $$\vb{E}$$
-into Fourier series, with terms given by:
-
 $$\begin{aligned}
-    \vb{H}(\vb{r}, t)
-    = \vb{H}(\vb{r}) \exp(- i \omega t)
-    \qquad \quad
-    \vb{E}(\vb{r}, t)
-    = \vb{E}(\vb{r}) \exp(- i \omega t)
+    \boxed{
+        \nabla^2 \vb{H} - \frac{n^2}{c^2} \pdvn{2}{\vb{H}}{t}
+        = 0
+    }
 \end{aligned}$$
 
-By inserting this ansatz into the equation,
-we can remove the explicit time dependence:
+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:
 
 $$\begin{aligned}
-    \nabla \cross \Big( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \Big) \exp(- i \omega t)
-    = \mu_0 \varepsilon_0 \omega^2 \mu_r \vb{H} \exp(- i \omega t)
+    \boxed{
+        \square
+        \equiv \nabla^2 - \frac{1}{c^2} \pdvn{2}{}{t}
+    }
 \end{aligned}$$
 
-Dividing out $$\exp(- i \omega t)$$,
-we arrive at an eigenvalue problem for $$\omega^2$$,
-with $$c = 1 / \sqrt{\mu_0 \varepsilon_0}$$:
+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:
 
 $$\begin{aligned}
-    \boxed{
-        \nabla \cross \Big( \frac{1}{\varepsilon_r(\vb{r})} \nabla \cross \vb{H}(\vb{r}) \Big)
-        = \Big( \frac{\omega}{c} \Big)^2 \mu_r(\vb{r}) \vb{H}(\vb{r})
-    }
+    \vb{E}(\vb{r}, t)
+    &= \vb{E}_0 e^{i \vb{k} \cdot \vb{r} - i \omega t}
+    \\
+    \vb{B}(\vb{r}, t)
+    &= \vb{B}_0 e^{i \vb{k} \cdot \vb{r} - i \omega t}
 \end{aligned}$$
 
-Compared to a uniform medium, $$\omega$$ is often not arbitrary here:
-there are discrete eigenvalues $$\omega$$,
-corresponding to discrete **modes** $$\vb{H}(\vb{r})$$.
+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.
 
-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:
+
+
+## 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:
 
 $$\begin{aligned}
-    \nabla \cross \Big( \frac{1}{\mu_r} \nabla \cross \vb{E} \Big)
-    = - \mu_0 \pdv{}{t}(\nabla \cross \vb{H})
-    = - \mu_0 \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t}
+    \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg)
+    &= - \frac{\varepsilon_r}{c^2} \pdvn{2}{\vb{E}}{t}
+    \\
+    \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg)
+    &= - \frac{\mu_r}{c^2} \pdvn{2}{\vb{H}}{t}
 \end{aligned}$$
 
-Then, by replacing $$\vb{E}(\vb{r}, t)$$ with our plane-wave ansatz,
-we remove the time dependence:
+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/):
 
 $$\begin{aligned}
-    \nabla \cross \Big( \frac{1}{\mu_r} \nabla \cross \vb{E} \Big) \exp(- i \omega t)
-    = - \mu_0 \varepsilon_0 \omega^2 \varepsilon_r \vb{E} \exp(- i \omega t)
+    \vb{E}(\vb{r}, t)
+    &= \vb{E}(\vb{r}) e^{- i \omega t}
+    \\
+    \vb{H}(\vb{r}, t)
+    &= \vb{H}(\vb{r}) e^{- i \omega t}
 \end{aligned}$$
 
-Which, after dividing out $$\exp(- i \omega t)$$,
-yields an analogous eigenvalue problem with $$\vb{E}(r)$$:
+Inserting this ansatz and dividing out $$e^{-i \omega t}$$
+yields the time-independent forms:
 
 $$\begin{aligned}
     \boxed{
-        \nabla \cross \Big( \frac{1}{\mu_r(\vb{r})} \nabla \cross \vb{E}(\vb{r}) \Big)
-        = \Big( \frac{\omega}{c} \Big)^2 \varepsilon_r(\vb{r}) \vb{E}(\vb{r})
+        \nabla \cross \bigg( \frac{1}{\mu_r} \nabla \cross \vb{E} \bigg)
+        = \Big( \frac{\omega}{c} \Big)^2 \varepsilon_r \vb{E}
     }
 \end{aligned}$$
 
-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 $$\vb{E}(\vb{r})$$.
+$$\begin{aligned}
+    \boxed{
+        \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg)
+        = \Big( \frac{\omega}{c} \Big)^2 \mu_r \vb{H}
+    }
+\end{aligned}$$
 
-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:
+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:
 
 $$\begin{aligned}
-    \nabla \cdot (\varepsilon_r \vb{E}) = 0
-    \qquad \quad
-    \nabla \cdot (\mu_r \vb{H}) = 0
+    \nabla \cross \big( \nabla \cross \vb{E} \big)
+    &= \Big( \frac{\omega}{c} \Big)^2 \varepsilon_r \vb{E}
+    \\
+    \nabla \cross \bigg( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \bigg)
+    &= \Big( \frac{\omega}{c} \Big)^2 \vb{H}
 \end{aligned}$$
 
-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 $$\vb{E}_0$$.
+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