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diff --git a/source/know/concept/canonical-ensemble/index.md b/source/know/concept/canonical-ensemble/index.md
index dd7fe90..8a96e91 100644
--- a/source/know/concept/canonical-ensemble/index.md
+++ b/source/know/concept/canonical-ensemble/index.md
@@ -12,25 +12,25 @@ layout: "concept"
 The **canonical ensemble** or **NVT ensemble** builds on
 the [microcanonical ensemble](/know/concept/microcanonical-ensemble/),
 by allowing the system to exchange energy with a very large heat bath,
-such that its temperature $T$ remains constant,
-but internal energy $U$ does not.
+such that its temperature $$T$$ remains constant,
+but internal energy $$U$$ does not.
 The conserved state functions are
-the temperature $T$, the volume $V$, and the particle count $N$.
+the temperature $$T$$, the volume $$V$$, and the particle count $$N$$.
 
-We refer to the system of interest as $A$, and the heat bath as $B$.
-The combination $A\!+\!B$ forms a microcanonical ensemble,
-i.e. it has a fixed total energy $U$,
+We refer to the system of interest as $$A$$, and the heat bath as $$B$$.
+The combination $$A\!+\!B$$ forms a microcanonical ensemble,
+i.e. it has a fixed total energy $$U$$,
 and eventually reaches an equilibrium
-with a uniform temperature $T$ in both $A$ and $B$.
+with a uniform temperature $$T$$ in both $$A$$ and $$B$$.
 
 Assuming that this equilibrium has been reached,
-we want to know which microstates $A$ prefers in that case.
-Specifically, if $A$ has energy $U_A$, and $B$ has $U_B$,
-which $U_A$ does $A$ prefer?
+we want to know which microstates $$A$$ prefers in that case.
+Specifically, if $$A$$ has energy $$U_A$$, and $$B$$ has $$U_B$$,
+which $$U_A$$ does $$A$$ prefer?
 
-Let $c_B(U_B)$ be the number of $B$-microstates with energy $U_B$.
-Then the probability that $A$ is in a specific microstate $s_A$ is as follows,
-where $U_A(s_A)$ is the resulting energy:
+Let $$c_B(U_B)$$ be the number of $$B$$-microstates with energy $$U_B$$.
+Then the probability that $$A$$ is in a specific microstate $$s_A$$ is as follows,
+where $$U_A(s_A)$$ is the resulting energy:
 
 $$\begin{aligned}
     p(s_A)
@@ -39,12 +39,12 @@ $$\begin{aligned}
     D \equiv \sum_{s_A} c_B(U - U_A(s_A))
 \end{aligned}$$
 
-In other words, we choose an $s_A$,
-and count the number $c_B$ of compatible $B$-microstates.
+In other words, we choose an $$s_A$$,
+and count the number $$c_B$$ of compatible $$B$$-microstates.
 
-Since the heat bath is large, let us assume that $U_B \gg U_A$.
-We thus approximate $\ln{p(s_A)}$ by
-Taylor-expanding $\ln{c_B(U_B)}$ around $U_B = U$:
+Since the heat bath is large, let us assume that $$U_B \gg U_A$$.
+We thus approximate $$\ln{p(s_A)}$$ by
+Taylor-expanding $$\ln{c_B(U_B)}$$ around $$U_B = U$$:
 
 $$\begin{aligned}
     \ln{p(s_A)}
@@ -53,8 +53,8 @@ $$\begin{aligned}
     &\approx - \ln{D} + \ln{c_B(U)} - \bigg( \dv{(\ln{c_B})}{U_B} \bigg) \: U_A(s_A)
 \end{aligned}$$
 
-Here, we use the definition of entropy $S_B \equiv k \ln{c_B}$,
-and that its $U_B$-derivative is $1/T$:
+Here, we use the definition of entropy $$S_B \equiv k \ln{c_B}$$,
+and that its $$U_B$$-derivative is $$1/T$$:
 
 $$\begin{aligned}
     \ln{p(s_A)}
@@ -63,7 +63,7 @@ $$\begin{aligned}
     &\approx - \ln{D} + \ln{c_B(U)} - \frac{U_A(s_A)}{k T}
 \end{aligned}$$
 
-We now define the **partition function** or **Zustandssumme** $Z$ as follows,
+We now define the **partition function** or **Zustandssumme** $$Z$$ as follows,
 which will act as a normalization factor for the probability:
 
 $$\begin{aligned}
@@ -74,8 +74,8 @@ $$\begin{aligned}
     = \frac{D}{c_B(U)}
 \end{aligned}$$
 
-Where $\beta \equiv 1/ (k T)$.
-The probability of finding $A$ in a microstate $s_A$ is thus given by:
+Where $$\beta \equiv 1/ (k T)$$.
+The probability of finding $$A$$ in a microstate $$s_A$$ is thus given by:
 
 $$\begin{aligned}
     \boxed{
@@ -84,33 +84,33 @@ $$\begin{aligned}
 \end{aligned}$$
 
 This is the **Boltzmann distribution**,
-which, as it turns out, maximizes the entropy $S_A$
-for a fixed value of the average energy $\Expval{U_A}$,
-i.e. a fixed $T$ and set of microstates $s_A$.
+which, as it turns out, maximizes the entropy $$S_A$$
+for a fixed value of the average energy $$\Expval{U_A}$$,
+i.e. a fixed $$T$$ and set of microstates $$s_A$$.
 
-Because $A\!+\!B$ is a microcanonical ensemble,
+Because $$A\!+\!B$$ is a microcanonical ensemble,
 we know that its [thermodynamic potential](/know/concept/thermodynamic-potential/)
-is the entropy $S$.
-But what about the canonical ensemble, just $A$?
+is the entropy $$S$$.
+But what about the canonical ensemble, just $$A$$?
 
 The solution is a bit backwards.
-Note that the partition function $Z$ is not a constant;
-it depends on $T$ (via $\beta$), $V$ and $N$ (via $s_A$).
+Note that the partition function $$Z$$ is not a constant;
+it depends on $$T$$ (via $$\beta$$), $$V$$ and $$N$$ (via $$s_A$$).
 Using the same logic as for the microcanonical ensemble,
-we define "equilibrium" as the set of microstates $s_A$
-that $A$ is most likely to occupy,
-which must be the set (as a function of $T,V,N$) that maximizes $Z$.
+we define "equilibrium" as the set of microstates $$s_A$$
+that $$A$$ is most likely to occupy,
+which must be the set (as a function of $$T,V,N$$) that maximizes $$Z$$.
 
-However, $T$, $V$ and $N$ are fixed,
-so how can we maximize $Z$?
+However, $$T$$, $$V$$ and $$N$$ are fixed,
+so how can we maximize $$Z$$?
 Well, as it turns out,
 the Boltzmann distribution has already done it for us!
 We will return to this point later.
 
-Still, $Z$ does not have a clear physical interpretation.
+Still, $$Z$$ does not have a clear physical interpretation.
 To find one, we start by showing that the ensemble averages
-of the energy $U_A$, pressure $P_A$ and chemical potential $\mu_A$
-can be calculated by differentiating $Z$.
+of the energy $$U_A$$, pressure $$P_A$$ and chemical potential $$\mu_A$$
+can be calculated by differentiating $$Z$$.
 As preparation, note that:
 
 $$\begin{aligned}
@@ -118,7 +118,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 With this, we can find the ensemble averages
-$\Expval{U_A}$, $\Expval{P_A}$ and $\Expval{\mu_A}$ of the system:
+$$\Expval{U_A}$$, $$\Expval{P_A}$$ and $$\Expval{\mu_A}$$ of the system:
 
 $$\begin{aligned}
     \Expval{U_A}
@@ -141,7 +141,7 @@ $$\begin{aligned}
     = - \frac{1}{Z \beta} \pdv{Z}{N}
 \end{aligned}$$
 
-It will turn out more convenient to use derivatives of $\ln{Z}$ instead,
+It will turn out more convenient to use derivatives of $$\ln{Z}$$ instead,
 in which case:
 
 $$\begin{aligned}
@@ -155,15 +155,15 @@ $$\begin{aligned}
     = - \frac{1}{\beta} \pdv{\ln{Z}}{N}
 \end{aligned}$$
 
-Now, to find a physical interpretation for $Z$.
-Consider the quantity $F$, in units of energy,
-whose minimum corresponds to a maximum of $Z$:
+Now, to find a physical interpretation for $$Z$$.
+Consider the quantity $$F$$, in units of energy,
+whose minimum corresponds to a maximum of $$Z$$:
 
 $$\begin{aligned}
     F \equiv - k T \ln{Z}
 \end{aligned}$$
 
-We rearrange the equation to $\beta F = - \ln{Z}$ and take its differential element:
+We rearrange the equation to $$\beta F = - \ln{Z}$$ and take its differential element:
 
 $$\begin{aligned}
     \dd{(\beta F)}
@@ -198,7 +198,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 As was already suggested by our notation,
-$F$ turns out to be the **Helmholtz free energy**:
+$$F$$ turns out to be the **Helmholtz free energy**:
 
 $$\begin{aligned}
     \boxed{
@@ -209,7 +209,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 We can therefore reinterpret
-the partition function $Z$ and the Boltzmann distribution $p(s_A)$
+the partition function $$Z$$ and the Boltzmann distribution $$p(s_A)$$
 in the following "more physical" way:
 
 $$\begin{aligned}
@@ -220,17 +220,17 @@ $$\begin{aligned}
     = \exp\!\Big( \beta \big( F \!-\! U_A(s_A) \big) \Big)
 \end{aligned}$$
 
-Finally, by rearranging the expressions for $F$,
-we find the entropy $S_A$ to be:
+Finally, by rearranging the expressions for $$F$$,
+we find the entropy $$S_A$$ to be:
 
 $$\begin{aligned}
     S_A
     = k \ln{Z} + \frac{\Expval{U_A}}{T}
 \end{aligned}$$
 
-This is why $Z$ is already maximized:
-the Boltzmann distribution maximizes $S_A$ for fixed values of $T$ and $\Expval{U_A}$,
-leaving $Z$ as the only "variable".
+This is why $$Z$$ is already maximized:
+the Boltzmann distribution maximizes $$S_A$$ for fixed values of $$T$$ and $$\Expval{U_A}$$,
+leaving $$Z$$ as the only "variable".