1 files changed, 23 insertions, 23 deletions
diff --git a/source/know/concept/grad-shafranov-equation/index.md b/source/know/concept/grad-shafranov-equation/index.md
index 35d23f4..b86c032 100644
--- a/source/know/concept/grad-shafranov-equation/index.md
+++ b/source/know/concept/grad-shafranov-equation/index.md
@@ -20,7 +20,7 @@ We would like to find the equilibrium state of the plasma
 in the general case of a reactor with toroidal symmetry.
 Using ideal [magnetohydrodynamics](/know/concept/magnetohydrodynamics/) (MHD),
 we start by assuming that the fluid is stationary,
-and that the confining field $\vb{B}$ is fixed:
+and that the confining field $$\vb{B}$$ is fixed:
 
 $$\begin{aligned}
     \vb{u}
@@ -36,7 +36,7 @@ $$\begin{aligned}
     = 0
 \end{aligned}$$
 
-Notice that $\vb{E} = 0$ is a result of the ideal generalized Ohm's law.
+Notice that $$\vb{E} = 0$$ is a result of the ideal generalized Ohm's law.
 Under these assumptions, the relevant MHD equations to be solved are
 Gauss' law for magnetism, Ampère's law, and the MHD momentum equation, respectively:
 
@@ -54,11 +54,11 @@ $$\begin{aligned}
 The goal is to analyze them in this order,
 exploiting toroidal symmetry along the way,
 to arrive at a general equilibrium condition.
-[Cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/) $(r, \theta, z)$
-are a natural choice, with the $z$-axis running through the middle of the torus.
+[Cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/) $$(r, \theta, z)$$
+are a natural choice, with the $$z$$-axis running through the middle of the torus.
 
-As preparation, it is a good idea to write $\vb{B}$
-as the curl of a magnetic vector potential $\vb{A}$,
+As preparation, it is a good idea to write $$\vb{B}$$
+as the curl of a magnetic vector potential $$\vb{A}$$,
 which looks like this in cylindrical polar coordinates:
 
 $$\begin{aligned}
@@ -76,7 +76,7 @@ $$\begin{aligned}
     \end{bmatrix}
 \end{aligned}$$
 
-Here, it is convenient to define the so-called **stream function** $\psi$ as follows:
+Here, it is convenient to define the so-called **stream function** $$\psi$$ as follows:
 
 $$\begin{aligned}
     \boxed{
@@ -85,8 +85,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Such that $\vb{B}$ can be written as below,
-where we will regard $B_\theta$ as a given quantity:
+Such that $$\vb{B}$$ can be written as below,
+where we will regard $$B_\theta$$ as a given quantity:
 
 $$\begin{aligned}
     \vb{B}
@@ -103,7 +103,7 @@ $$\begin{aligned}
 
 Inserting this into Gauss' law,
 we see that it is trivially satisfied,
-thanks to circular symmetry guaranteeing that $\ipdv{B_\theta}{\theta} = 0$:
+thanks to circular symmetry guaranteeing that $$\ipdv{B_\theta}{\theta} = 0$$:
 
 $$\begin{aligned}
     0
@@ -116,8 +116,8 @@ $$\begin{aligned}
     = 0
 \end{aligned}$$
 
-What matters is that we have expressions for the components of $\vb{B}$.
-Moving on, to find the current density $\vb{J}$,
+What matters is that we have expressions for the components of $$\vb{B}$$.
+Moving on, to find the current density $$\vb{J}$$,
 we use Ampère's law and symmetry to get:
 
 
@@ -138,9 +138,9 @@ $$\begin{aligned}
     \end{bmatrix}
 \end{aligned}$$
 
-Where we have assumed that $B_\theta$ depends only on $r$, not $z$ or $\theta$.
+Where we have assumed that $$B_\theta$$ depends only on $$r$$, not $$z$$ or $$\theta$$.
 Substituting this into the MHD momentum equation
-gives the following pressure gradient $\nabla p$:
+gives the following pressure gradient $$\nabla p$$:
 
 $$\begin{aligned}
     \nabla p
@@ -157,8 +157,8 @@ $$\begin{aligned}
     \end{bmatrix}
 \end{aligned}$$
 
-Now, the idea is to focus on this $r$-component to get an equation for $\psi$,
-whose solution can then be used to calculate the $\theta$ and $z$-components of $\nabla p$.
+Now, the idea is to focus on this $$r$$-component to get an equation for $$\psi$$,
+whose solution can then be used to calculate the $$\theta$$ and $$z$$-components of $$\nabla p$$.
 Therefore, we evaluate:
 
 $$\begin{aligned}
@@ -176,8 +176,8 @@ $$\begin{aligned}
     - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta
 \end{aligned}$$
 
-By using the chain rule to rewrite $\ipdv{}{r}= (\ipdv{\psi}{r}) \; \ipdv{}{\psi}$,
-we get $\ipdv{\psi}{r}$ in each term:
+By using the chain rule to rewrite $$\ipdv{}{r}= (\ipdv{\psi}{r}) \; \ipdv{}{\psi}$$,
+we get $$\ipdv{\psi}{r}$$ in each term:
 
 $$\begin{aligned}
     \pdv{\psi}{r} \pdv{p}{\psi}
@@ -185,7 +185,7 @@ $$\begin{aligned}
     - \frac{1}{\mu_0 r} \pdv{\psi}{r} \pdv{(r B_\theta)}{\psi} B_\theta
 \end{aligned}$$
 
-Dividing out $\ipdv{\psi}{r}$ and multiplying by $\mu_0 r^2$
+Dividing out $$\ipdv{\psi}{r}$$ and multiplying by $$\mu_0 r^2$$
 leads us to the **Grad-Shafranov equation**,
 which gives the equilibrium condition of a plasma in a toroidal reactor:
 
@@ -196,13 +196,13 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Weirdly, $\psi$ appears both as an unknown and as a differentiation variable,
+Weirdly, $$\psi$$ appears both as an unknown and as a differentiation variable,
 but this equation can still be solved analytically by
-assuming a certain $\psi$-dependence of $p$ and $r B_\theta$.
+assuming a certain $$\psi$$-dependence of $$p$$ and $$r B_\theta$$.
 
-Suppose that $B_\theta$ is induced by a poloidal electrical current $I_\mathrm{pol}$,
+Suppose that $$B_\theta$$ is induced by a poloidal electrical current $$I_\mathrm{pol}$$,
 i.e. a current around the "tube" of the torus,
-then, assuming $I_\mathrm{pol}$ only depends on $r$, we have:
+then, assuming $$I_\mathrm{pol}$$ only depends on $$r$$, we have:
 
 $$\begin{aligned}
     B_\theta