Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

1 files changed, 36 insertions, 36 deletions

diff --git a/source/know/concept/electromagnetic-wave-equation/index.md b/source/know/concept/electromagnetic-wave-equation/index.md
index c4cd9eb..a27fe6f 100644
--- a/source/know/concept/electromagnetic-wave-equation/index.md
+++ b/source/know/concept/electromagnetic-wave-equation/index.md
@@ -21,8 +21,8 @@ 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 $\vb{J}_\mathrm{free} = 0$:
+but we start with Ampère's circuital law for the "free" fields $$\vb{H}$$ and $$\vb{D}$$,
+in the absence of a free current $$\vb{J}_\mathrm{free} = 0$$:
 
 $$\begin{aligned}
     \nabla \cross \vb{H}
@@ -39,9 +39,9 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Which, upon insertion into Ampère's law,
-yields an equation relating $\vb{B}$ and $\vb{E}$.
+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:
+but keep in mind that $$\vb{J}_\mathrm{bound} \neq 0$$ here:
 
 $$\begin{aligned}
     \nabla \cross \vb{B}
@@ -49,7 +49,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Now we take the curl, rearrange,
-and substitute $\nabla \cross \vb{E}$ according to Faraday's law:
+and substitute $$\nabla \cross \vb{E}$$ according to Faraday's law:
 
 $$\begin{aligned}
     \nabla \cross (\nabla \cross \vb{B})
@@ -58,7 +58,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Using a vector identity, we rewrite the leftmost expression,
-which can then be reduced thanks to Gauss' law for magnetism $\nabla \cdot \vb{B} = 0$:
+which can then be reduced thanks to Gauss' law for magnetism $$\nabla \cdot \vb{B} = 0$$:
 
 $$\begin{aligned}
     - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{B}}{t}
@@ -66,8 +66,8 @@ $$\begin{aligned}
     = - \nabla^2 \vb{B}
 \end{aligned}$$
 
-This describes $\vb{B}$.
-Next, we repeat the process for $\vb{E}$:
+This describes $$\vb{B}$$.
+Next, we repeat the process for $$\vb{E}$$:
 taking the curl of Faraday's law yields:
 
 $$\begin{aligned}
@@ -77,8 +77,8 @@ $$\begin{aligned}
 \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 $\nabla \cdot \vb{E} = 0$:
+and then reduced by assuming that there is no net charge density $$\rho = 0$$
+in Gauss' law, such that $$\nabla \cdot \vb{E} = 0$$:
 
 $$\begin{aligned}
     - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t}
@@ -87,8 +87,8 @@ $$\begin{aligned}
 \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 $v \equiv 1 / \sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}$:
+wave equations for $$\vb{E}$$ and $$\vb{B}$$,
+where we have defined the phase velocity $$v \equiv 1 / \sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}$$:
 
 $$\begin{aligned}
     \boxed{
@@ -103,7 +103,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Traditionally, it is said that the solutions are as follows,
-where the wavenumber $|\vb{k}| = \omega / v$:
+where the wavenumber $$|\vb{k}| = \omega / v$$:
 
 $$\begin{aligned}
     \vb{E}(\vb{r}, t)
@@ -115,9 +115,9 @@ $$\begin{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.
+*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,
+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.
 
@@ -125,8 +125,8 @@ 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 $\mu_r(\vb{r})$.
+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:
 
 $$\begin{aligned}
@@ -148,7 +148,7 @@ $$\begin{aligned}
     = \varepsilon_0 \varepsilon_r(\vb{r}) \pdv{\vb{E}}{t}
 \end{aligned}$$
 
-We then divide Ampère's law by $\varepsilon_r(\vb{r})$,
+We then divide Ampère's law by $$\varepsilon_r(\vb{r})$$,
 take the curl, and substitute Faraday's law, giving:
 
 $$\begin{aligned}
@@ -157,7 +157,7 @@ $$\begin{aligned}
     = - \mu_0 \mu_r \varepsilon_0 \pdvn{2}{\vb{H}}{t}
 \end{aligned}$$
 
-Next, we exploit linearity by decomposing $\vb{H}$ and $\vb{E}$
+Next, we exploit linearity by decomposing $$\vb{H}$$ and $$\vb{E}$$
 into Fourier series, with terms given by:
 
 $$\begin{aligned}
@@ -176,9 +176,9 @@ $$\begin{aligned}
     = \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 $c = 1 / \sqrt{\mu_0 \varepsilon_0}$:
+Dividing out $$\exp(- i \omega t)$$,
+we arrive at an eigenvalue problem for $$\omega^2$$,
+with $$c = 1 / \sqrt{\mu_0 \varepsilon_0}$$:
 
 $$\begin{aligned}
     \boxed{
@@ -187,12 +187,12 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Compared to a uniform medium, $\omega$ is often not arbitrary here:
-there are discrete eigenvalues $\omega$,
-corresponding to discrete **modes** $\vb{H}(\vb{r})$.
+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 $\mu_r(\vb{r})$,
+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:
 
 $$\begin{aligned}
@@ -201,7 +201,7 @@ $$\begin{aligned}
     = - \mu_0 \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t}
 \end{aligned}$$
 
-Then, by replacing $\vb{E}(\vb{r}, t)$ with our plane-wave ansatz,
+Then, by replacing $$\vb{E}(\vb{r}, t)$$ with our plane-wave ansatz,
 we remove the time dependence:
 
 $$\begin{aligned}
@@ -209,8 +209,8 @@ $$\begin{aligned}
     = - \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 $\vb{E}(r)$:
+Which, after dividing out $$\exp(- i \omega t)$$,
+yields an analogous eigenvalue problem with $$\vb{E}(r)$$:
 
 $$\begin{aligned}
     \boxed{
@@ -220,13 +220,13 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Usually, it is a reasonable approximation
-to say $\mu_r(\vb{r}) = 1$,
-in which case the equation for $\vb{H}(\vb{r})$
+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 $\vb{E}(\vb{r})$.
+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 $\vb{E}(\vb{r})$
+the solutions $$\vb{H}(\vb{r})$$ and/or $$\vb{E}(\vb{r})$$
 must satisfy the two Maxwell's equations that were not explicitly used:
 
 $$\begin{aligned}
@@ -237,8 +237,8 @@ $$\begin{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 $\vb{E}_0$.
+the wavevector $$\vb{k}$$ must be perpendicular to
+the amplitudes $$\vb{H}_0$$ and $$\vb{E}_0$$.
 
 
 ## References