1 files changed, 30 insertions, 30 deletions
diff --git a/source/know/concept/maxwell-boltzmann-distribution/index.md b/source/know/concept/maxwell-boltzmann-distribution/index.md
index 65e983c..ebb2460 100644
--- a/source/know/concept/maxwell-boltzmann-distribution/index.md
+++ b/source/know/concept/maxwell-boltzmann-distribution/index.md
@@ -17,23 +17,23 @@ probability distributions with applications in classical statistical physics.
 
 In the [canonical ensemble](/know/concept/canonical-ensemble/)
 (where a fixed-size system can exchange energy with its environment),
-the probability of a microstate with energy $E$ is given by the Boltzmann distribution:
+the probability of a microstate with energy $$E$$ is given by the Boltzmann distribution:
 
 $$\begin{aligned}
     f(E)
     \:\propto\: \exp\!\big(\!-\! \beta E\big)
 \end{aligned}$$
 
-Where $\beta = 1 / k_B T$.
-We split $E = K + U$,
-with $K$ and $U$ the total kinetic and potential energy contributions.
-If there are $N$ particles in the system,
-with positions $\tilde{r} = (\vec{r}_1, ..., \vec{r}_N)$
-and momenta $\tilde{p} = (\vec{p}_1, ..., \vec{p}_N)$,
-then $K$ only depends on $\tilde{p}$,
-and $U$ only depends on $\tilde{r}$,
+Where $$\beta = 1 / k_B T$$.
+We split $$E = K + U$$,
+with $$K$$ and $$U$$ the total kinetic and potential energy contributions.
+If there are $$N$$ particles in the system,
+with positions $$\tilde{r} = (\vec{r}_1, ..., \vec{r}_N)$$
+and momenta $$\tilde{p} = (\vec{p}_1, ..., \vec{p}_N)$$,
+then $$K$$ only depends on $$\tilde{p}$$,
+and $$U$$ only depends on $$\tilde{r}$$,
 so the probability of a specific microstate
-$(\tilde{r}, \tilde{p})$ is as follows:
+$$(\tilde{r}, \tilde{p})$$ is as follows:
 
 $$\begin{aligned}
     f(\tilde{r}, \tilde{p})
@@ -62,10 +62,10 @@ $$\begin{aligned}
     \:\propto\: \exp\!\big(\!-\! \beta U(\tilde{r}) \big)
 \end{aligned}$$
 
-We cannot evaluate $f_U(\tilde{r})$ further without knowing $U(\tilde{r})$ for a system.
-We thus turn to $f_K(\tilde{p})$, and see that the total kinetic
-energy $K(\tilde{p})$ is simply the sum of the particles' individual
-kinetic energies $K_n(\vec{p}_n)$, which are well-known:
+We cannot evaluate $$f_U(\tilde{r})$$ further without knowing $$U(\tilde{r})$$ for a system.
+We thus turn to $$f_K(\tilde{p})$$, and see that the total kinetic
+energy $$K(\tilde{p})$$ is simply the sum of the particles' individual
+kinetic energies $$K_n(\vec{p}_n)$$, which are well-known:
 
 $$\begin{aligned}
     K(\tilde{p})
@@ -75,7 +75,7 @@ $$\begin{aligned}
     = \frac{|\vec{p}_n|^2}{2 m}
 \end{aligned}$$
 
-Consequently, the probability distribution $f(p_x, p_y, p_z)$ for the
+Consequently, the probability distribution $$f(p_x, p_y, p_z)$$ for the
 momentum vector of a single particle is as follows,
 after normalization:
 
@@ -84,7 +84,7 @@ $$\begin{aligned}
     = \Big( \frac{1}{2 \pi m k_B T} \Big)^{3/2} \exp\!\Big( \!-\!\frac{(p_x^2 + p_y^2 + p_z^2)}{2 m k_B T} \Big)
 \end{aligned}$$
 
-We now rewrite this using the velocities $v_x = p_x / m$,
+We now rewrite this using the velocities $$v_x = p_x / m$$,
 and update the normalization, giving:
 
 $$\begin{aligned}
@@ -114,10 +114,10 @@ $$\begin{aligned}
 ## Speed distribution
 
 We know the distribution of the velocities along each axis,
-but what about the speed $v = |\vec{v}|$?
-Because we do not care about the direction of $\vec{v}$, only its magnitude,
-the [density of states](/know/concept/density-of-states/) $g(v)$ is not constant:
-it is the rate-of-change of the volume of a sphere of radius $v$:
+but what about the speed $$v = |\vec{v}|$$?
+Because we do not care about the direction of $$\vec{v}$$, only its magnitude,
+the [density of states](/know/concept/density-of-states/) $$g(v)$$ is not constant:
+it is the rate-of-change of the volume of a sphere of radius $$v$$:
 
 $$\begin{aligned}
     g(v)
@@ -125,8 +125,8 @@ $$\begin{aligned}
     = 4 \pi v^2
 \end{aligned}$$
 
-Multiplying the velocity vector distribution by $g(v)$
-and substituting $v^2 = v_x^2 + v_y^2 + v_z^2$
+Multiplying the velocity vector distribution by $$g(v)$$
+and substituting $$v^2 = v_x^2 + v_y^2 + v_z^2$$
 then gives us the **Maxwell-Boltzmann speed distribution**:
 
 $$\begin{aligned}
@@ -137,9 +137,9 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Some notable points on this distribution are
-the most probable speed $v_{\mathrm{mode}}$,
-the mean average speed $v_{\mathrm{mean}}$,
-and the root-mean-square speed $v_{\mathrm{rms}}$:
+the most probable speed $$v_{\mathrm{mode}}$$,
+the mean average speed $$v_{\mathrm{mean}}$$,
+and the root-mean-square speed $$v_{\mathrm{rms}}$$:
 
 $$\begin{aligned}
     f'(v_\mathrm{mode})
@@ -176,7 +176,7 @@ $$\begin{aligned}
 
 Using the speed distribution,
 we can work out the kinetic energy distribution.
-Because $K$ is not proportional to $v$,
+Because $$K$$ is not proportional to $$v$$,
 we must do this by demanding that:
 
 $$\begin{aligned}
@@ -187,9 +187,9 @@ $$\begin{aligned}
     = f(v) \dv{v}{K}
 \end{aligned}$$
 
-We know that $K = m v^2 / 2$,
-meaning $\dd{K} = m v \dd{v}$
-so the energy distribution $f(K)$ is:
+We know that $$K = m v^2 / 2$$,
+meaning $$\dd{K} = m v \dd{v}$$
+so the energy distribution $$f(K)$$ is:
 
 $$\begin{aligned}
     f(K)
@@ -197,7 +197,7 @@ $$\begin{aligned}
     = \sqrt{\frac{2 m}{\pi}} \: \bigg( \frac{1}{k_B T} \bigg)^{3/2} v \exp\!\Big( \!-\!\frac{m v^2}{2 k_B T} \Big)
 \end{aligned}$$
 
-Substituting $v = \sqrt{2 K/m}$ leads to
+Substituting $$v = \sqrt{2 K/m}$$ leads to
 the **Maxwell-Boltzmann kinetic energy distribution**:
 
 $$\begin{aligned}