From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 20 Oct 2022 18:25:31 +0200
Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

---
 source/know/concept/alfven-waves/index.md | 70 +++++++++++++++----------------
 1 file changed, 35 insertions(+), 35 deletions(-)

(limited to 'source/know/concept/alfven-waves/index.md')

diff --git a/source/know/concept/alfven-waves/index.md b/source/know/concept/alfven-waves/index.md
index c560c67..31576f3 100644
--- a/source/know/concept/alfven-waves/index.md
+++ b/source/know/concept/alfven-waves/index.md
@@ -10,11 +10,11 @@ layout: "concept"
 ---
 
 In the [magnetohydrodynamic](/know/concept/magnetohydrodynamics/) description of a plasma,
-we split the velocity $\vb{u}$, electric current $\vb{J}$,
-[magnetic field](/know/concept/magnetic-field/) $\vb{B}$
-and [electric field](/know/concept/electric-field/) $\vb{E}$ like so,
-into a constant uniform equilibrium (subscript $0$)
-and a small unknown perturbation (subscript $1$):
+we split the velocity $$\vb{u}$$, electric current $$\vb{J}$$,
+[magnetic field](/know/concept/magnetic-field/) $$\vb{B}$$
+and [electric field](/know/concept/electric-field/) $$\vb{E}$$ like so,
+into a constant uniform equilibrium (subscript $$0$$)
+and a small unknown perturbation (subscript $$1$$):
 
 $$\begin{aligned}
     \vb{u}
@@ -41,7 +41,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 We do this for the momentum equation too,
-assuming that $\vb{J}_0 \!=\! 0$ (to be justified later).
+assuming that $$\vb{J}_0 \!=\! 0$$ (to be justified later).
 Note that the temperature is set to zero, such that the pressure vanishes:
 
 $$\begin{aligned}
@@ -49,8 +49,8 @@ $$\begin{aligned}
     = \vb{J}_1 \cross \vb{B}_0
 \end{aligned}$$
 
-Where $\rho$ is the uniform equilibrium density.
-We would like an equation for $\vb{J}_1$,
+Where $$\rho$$ is the uniform equilibrium density.
+We would like an equation for $$\vb{J}_1$$,
 which is provided by the magnetohydrodynamic form of Ampère's law:
 
 $$\begin{aligned}
@@ -62,14 +62,14 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Substituting this into the momentum equation,
-and differentiating with respect to $t$:
+and differentiating with respect to $$t$$:
 
 $$\begin{aligned}
     \rho \pdvn{2}{\vb{u}_1}{t}
     = \frac{1}{\mu_0} \bigg( \Big( \nabla \cross \pdv{}{\vb{B}1}{t} \Big) \cross \vb{B}_0 \bigg)
 \end{aligned}$$
 
-For which we can use Faraday's law to rewrite $\ipdv{\vb{B}_1}{t}$,
+For which we can use Faraday's law to rewrite $$\ipdv{\vb{B}_1}{t}$$,
 incorporating Ohm's law too:
 
 $$\begin{aligned}
@@ -78,7 +78,7 @@ $$\begin{aligned}
     = \nabla \cross (\vb{u}_1 \cross \vb{B}_0)
 \end{aligned}$$
 
-Inserting this into the momentum equation for $\vb{u}_1$
+Inserting this into the momentum equation for $$\vb{u}_1$$
 thus yields its final form:
 
 $$\begin{aligned}
@@ -86,9 +86,9 @@ $$\begin{aligned}
     = \frac{1}{\mu_0} \bigg( \Big( \nabla \cross \big( \nabla \cross (\vb{u}_1 \cross \vb{B}_0) \big) \Big) \cross \vb{B}_0 \bigg)
 \end{aligned}$$
 
-Suppose the magnetic field is pointing in $z$-direction,
-i.e. $\vb{B}_0 = B_0 \vu{e}_z$.
-Then Faraday's law justifies our earlier assumption that $\vb{J}_0 = 0$,
+Suppose the magnetic field is pointing in $$z$$-direction,
+i.e. $$\vb{B}_0 = B_0 \vu{e}_z$$.
+Then Faraday's law justifies our earlier assumption that $$\vb{J}_0 = 0$$,
 and the equation can be written as:
 
 $$\begin{aligned}
@@ -96,7 +96,7 @@ $$\begin{aligned}
     = v_A^2 \bigg( \Big( \nabla \cross \big( \nabla \cross (\vb{u}_1 \cross \vu{e}_z) \big) \Big) \cross \vu{e}_z \bigg)
 \end{aligned}$$
 
-Where we have defined the so-called **Alfvén velocity** $v_A$ to be given by:
+Where we have defined the so-called **Alfvén velocity** $$v_A$$ to be given by:
 
 $$\begin{aligned}
     \boxed{
@@ -105,24 +105,24 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Now, consider the following plane-wave ansatz for $\vb{u}_1$,
-with wavevector $\vb{k}$ and frequency $\omega$:
+Now, consider the following plane-wave ansatz for $$\vb{u}_1$$,
+with wavevector $$\vb{k}$$ and frequency $$\omega$$:
 
 $$\begin{aligned}
     \vb{u}_1(\vb{r}, t)
     &= \vb{u}_1 \exp(i \vb{k} \cdot \vb{r} - i \omega t)
 \end{aligned}$$
 
-Inserting this into the above differential equation for $\vb{u}_1$ leads to:
+Inserting this into the above differential equation for $$\vb{u}_1$$ leads to:
 
 $$\begin{aligned}
     \omega^2 \vb{u}_1
     = v_A^2 \bigg( \Big( \vb{k} \cross \big( \vb{k} \cross (\vb{u}_1 \cross \vu{e}_z) \big) \Big) \cross \vu{e}_z \bigg)
 \end{aligned}$$
 
-To evaluate this, we rotate our coordinate system around the $z$-axis
-such that $\vb{k} = (0, k_\perp, k_\parallel)$,
-i.e. the wavevector's $x$-component is zero.
+To evaluate this, we rotate our coordinate system around the $$z$$-axis
+such that $$\vb{k} = (0, k_\perp, k_\parallel)$$,
+i.e. the wavevector's $$x$$-component is zero.
 Calculating the cross products:
 
 $$\begin{aligned}
@@ -149,7 +149,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 We rewrite this equation in matrix form,
-using that $k_\perp^2 \!+ k_\parallel^2 = k^2 \equiv |\vb{k}|^2$:
+using that $$k_\perp^2 \!+ k_\parallel^2 = k^2 \equiv |\vb{k}|^2$$:
 
 $$\begin{aligned}
     \begin{bmatrix}
@@ -161,9 +161,9 @@ $$\begin{aligned}
     = 0
 \end{aligned}$$
 
-This has the form of an eigenvalue problem for $\omega^2$,
+This has the form of an eigenvalue problem for $$\omega^2$$,
 meaning we must find non-trivial solutions,
-where we cannot simply choose the components of $\vb{u}_1$ to satisfy the equation.
+where we cannot simply choose the components of $$\vb{u}_1$$ to satisfy the equation.
 To achieve this, we demand that the matrix' determinant is zero:
 
 $$\begin{aligned}
@@ -171,12 +171,12 @@ $$\begin{aligned}
     = 0
 \end{aligned}$$
 
-This equation has three solutions for $\omega^2$,
+This equation has three solutions for $$\omega^2$$,
 one for each of its three factors being zero.
-The simplest case $\omega^2 = 0$ is of no interest to us,
+The simplest case $$\omega^2 = 0$$ is of no interest to us,
 because we are looking for waves.
 
-The first interesting case is $\omega^2 = v_A^2 k_\parallel^2$,
+The first interesting case is $$\omega^2 = v_A^2 k_\parallel^2$$,
 yielding the following dispersion relation:
 
 $$\begin{aligned}
@@ -188,10 +188,10 @@ $$\begin{aligned}
 
 The resulting waves are called **shear Alfvén waves**.
 From the eigenvalue problem, we see that in this case
-$\vb{u}_1 = (u_{1x}, 0, 0)$, meaning $\vb{u}_1 \cdot \vb{k} = 0$:
+$$\vb{u}_1 = (u_{1x}, 0, 0)$$, meaning $$\vb{u}_1 \cdot \vb{k} = 0$$:
 these waves are **transverse**.
-The phase velocity $v_p$ and group velocity $v_g$ are as follows,
-where $\theta$ is the angle between $\vb{k}$ and $\vb{B}_0$:
+The phase velocity $$v_p$$ and group velocity $$v_g$$ are as follows,
+where $$\theta$$ is the angle between $$\vb{k}$$ and $$\vb{B}_0$$:
 
 $$\begin{aligned}
     v_p
@@ -204,7 +204,7 @@ $$\begin{aligned}
     = v_A
 \end{aligned}$$
 
-The other interesting case is $\omega^2 = v_A^2 k^2$,
+The other interesting case is $$\omega^2 = v_A^2 k^2$$,
 which leads to so-called **compressional Alfvén waves**,
 with the simple dispersion relation:
 
@@ -215,10 +215,10 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Looking at the eigenvalue problem reveals that $\vb{u}_1 = (0, u_{1y}, 0)$,
-meaning $\vb{u}_1 \cdot \vb{k} = u_{1y} k_\perp$,
-so these waves are not necessarily transverse, nor longitudinal (since $k_\parallel$ is free).
-The phase velocity $v_p$ and group velocity $v_g$ are given by:
+Looking at the eigenvalue problem reveals that $$\vb{u}_1 = (0, u_{1y}, 0)$$,
+meaning $$\vb{u}_1 \cdot \vb{k} = u_{1y} k_\perp$$,
+so these waves are not necessarily transverse, nor longitudinal (since $$k_\parallel$$ is free).
+The phase velocity $$v_p$$ and group velocity $$v_g$$ are given by:
 
 $$\begin{aligned}
     v_p
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