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+---
+title: "Canonical ensemble"
+date: 2021-07-10
+categories:
+- Physics
+- Thermodynamics
+- Thermodynamic ensembles
+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.
+The conserved state functions are
+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 $U$,
+and eventually reaches an equilibrium
+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?
+
+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:
+
+$$\begin{aligned}
+    p(s_A)
+    = \frac{c_B(U - U_A(s_A))}{D}
+    \qquad \quad
+    D \equiv \sum_{s_A} c_B(U - U_A(s_A))
+\end{aligned}$$
+
+In other words, we choose an $s_A$,
+and count the number $c_B$ of compatible $B$-microstates.
+
+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 $U_B = U$:
+
+$$\begin{aligned}
+    \ln{p(s_A)}
+    &= -\ln{D} + \ln\!\big(c_B(U - U_A(s_A))\big)
+    \\
+    &\approx - \ln{D} + \ln{c_B(U)} - \bigg( \dv{(\ln{c_B})}{U_B} \bigg) \: U_A(s_A)
+\end{aligned}$$
+
+Here, we use the definition of entropy $S_B \equiv k \ln{c_B}$,
+and that its $U_B$-derivative is $1/T$:
+
+$$\begin{aligned}
+    \ln{p(s_A)}
+    &\approx - \ln{D} + \ln{c_B(U)} - \frac{U_A(s_A)}{k} \Big( \pdv{S_B}{U_B} \Big)
+    \\
+    &\approx - \ln{D} + \ln{c_B(U)} - \frac{U_A(s_A)}{k T}
+\end{aligned}$$
+
+We now define the **partition function** or **Zustandssumme** $Z$ as follows,
+which will act as a normalization factor for the probability:
+
+$$\begin{aligned}
+    \boxed{
+        Z
+        \equiv \sum_{s_A}^{} \exp(- \beta U_A(s_A))
+    }
+    = \frac{D}{c_B(U)}
+\end{aligned}$$
+
+Where $\beta \equiv 1/ (k T)$.
+The probability of finding $A$ in a microstate $s_A$ is thus given by:
+
+$$\begin{aligned}
+    \boxed{
+        p(s_A) = \frac{1}{Z} \exp(- \beta U_A(s_A))
+    }
+\end{aligned}$$
+
+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 $s_A$.
+
+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 $A$?
+
+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 $s_A$).
+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 $Z$.
+
+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.
+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 $Z$.
+As preparation, note that:
+
+$$\begin{aligned}
+    \pdv{Z}{\beta} = - \sum_{s_A} U_A \exp(- \beta U_A)
+\end{aligned}$$
+
+With this, we can find the ensemble averages
+$\Expval{U_A}$, $\Expval{P_A}$ and $\Expval{\mu_A}$ of the system:
+
+$$\begin{aligned}
+    \Expval{U_A}
+    &= \sum_{s_A} p(s_A) \: U_A
+    = \frac{1}{Z} \sum_{s_A} U_A \exp(- \beta U_A)
+    = - \frac{1}{Z} \pdv{Z}{\beta}
+    \\
+    \Expval{P_A}
+    &= - \sum_{s_A} p(s_A) \pdv{U_A}{V}
+    = - \frac{1}{Z} \sum_{s_A} \exp(- \beta U_A) \pdv{U_A}{V}
+    \\
+    &= \frac{1}{Z \beta} \pdv{}{V}\sum_{s_A} \exp(- \beta U_A)
+    = \frac{1}{Z \beta} \pdv{Z}{V}
+    \\
+    \Expval{\mu_A}
+    &= \sum_{s_A} p(s_A) \pdv{U_A}{N}
+    = \frac{1}{Z} \sum_{s_A} \exp(- \beta U_A) \pdv{U_A}{N}
+    \\
+    &= - \frac{1}{Z \beta} \pdv{}{N}\sum_{s_A} \exp(- \beta U_A)
+    = - \frac{1}{Z \beta} \pdv{Z}{N}
+\end{aligned}$$
+
+It will turn out more convenient to use derivatives of $\ln{Z}$ instead,
+in which case:
+
+$$\begin{aligned}
+    \Expval{U_A}
+    = - \pdv{\ln{Z}}{\beta}
+    \qquad \quad
+    \Expval{P_A}
+    = \frac{1}{\beta} \pdv{\ln{Z}}{V}
+    \qquad \quad
+    \Expval{\mu_A}
+    = - \frac{1}{\beta} \pdv{\ln{Z}}{N}
+\end{aligned}$$
+
+Now, to find a physical interpretation for $Z$.
+Consider the quantity $F$, in units of energy,
+whose minimum corresponds to a maximum of $Z$:
+
+$$\begin{aligned}
+    F \equiv - k T \ln{Z}
+\end{aligned}$$
+
+We rearrange the equation to $\beta F = - \ln{Z}$ and take its differential element:
+
+$$\begin{aligned}
+    \dd{(\beta F)}
+    = - \dd{(\ln{Z})}
+    &= - \pdv{\ln{Z}}{\beta} \dd{\beta} - \pdv{\ln{Z}}{V} \dd{V} - \pdv{\ln{Z}}{N} \dd{N}
+    \\
+    &= \Expval{U_A} \dd{\beta} - \beta \Expval{P_A} \dd{V} + \beta \Expval{\mu_A} \dd{N}
+    \\
+    &= \Expval{U_A} \dd{\beta} + \beta \dd{\Expval{U_A}} - \beta \dd{\Expval{U_A}} - \beta \Expval{P_A} \dd{V} + \beta \Expval{\mu_A} \dd{N}
+    \\
+    &= \dd{(\beta \Expval{U_A})} - \beta \: \big( \dd{\Expval{U_A}} + \Expval{P_A} \dd{V} - \Expval{\mu_A} \dd{N} \big)
+\end{aligned}$$
+
+Rearranging and substituting
+the [fundamental thermodynamic relation](/know/concept/fundamental-thermodynamic-relation/)
+then gives:
+
+$$\begin{aligned}
+    \dd{(\beta F - \beta \Expval{U_A})}
+    &= - \beta \: \big( \dd{\Expval{U_A}} + \Expval{P_A} \dd{V} - \Expval{\mu_A} \dd{N} \big)
+    = - \beta T \dd{S_A}
+\end{aligned}$$
+
+We integrate this and ignore the integration constant,
+leading us to the desired result:
+
+$$\begin{aligned}
+    - \beta T S_A
+    &= \beta F - \beta \Expval{U_A}
+    \quad \implies \quad
+    F = \Expval{U_A} -  T S_A
+\end{aligned}$$
+
+As was already suggested by our notation,
+$F$ turns out to be the **Helmholtz free energy**:
+
+$$\begin{aligned}
+    \boxed{
+        F
+        \equiv - k T \ln{Z}
+        = \Expval{U_A} - T S_A
+    }
+\end{aligned}$$
+
+We can therefore reinterpret
+the partition function $Z$ and the Boltzmann distribution $p(s_A)$
+in the following "more physical" way:
+
+$$\begin{aligned}
+    Z
+    = \exp(- \beta F)
+    \qquad \quad
+    p(s_A)
+    = \exp\!\Big( \beta \big( F \!-\! U_A(s_A) \big) \Big)
+\end{aligned}$$
+
+Finally, by rearranging the expressions for $F$,
+we find the entropy $S_A$ to be:
+
+$$\begin{aligned}
+    S_A
+    = k \ln{Z} + \frac{\Expval{U_A}}{T}
+\end{aligned}$$
+
+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".
+
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.