path: root/content/know/concept/canonical-ensemble/index.pdc

---
title: "Canonical ensemble"
firstLetter: "C"
publishDate: 2021-07-10
categories:
- Physics
- Thermodynamics
- Thermodynamic ensembles

date: 2021-07-08T11:01:02+02:00
draft: false
markup: pandoc
---

# Canonical ensemble

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:

$$\begin{aligned}
    \expval{U_A}
    &= \sum_{s_A} p(s_A) \: U_A
    = \frac{1}{Z} \sum_{s_A} U_A \exp\!(- \beta U_A)
    = - \frac{1}{Z} \pdv{Z}{\beta}
    \\
    \expval{P_A}
    &= - \sum_{s_A} p(s_A) \pdv{U_A}{V}
    = - \frac{1}{Z} \sum_{s_A} \exp\!(- \beta U_A) \pdv{U_A}{V}
    \\
    &= \frac{1}{Z \beta} \pdv{V} \sum_{s_A} \exp\!(- \beta U_A)
    = \frac{1}{Z \beta} \pdv{Z}{V}
    \\
    \expval{\mu_A}
    &= \sum_{s_A} p(s_A) \pdv{U_A}{N}
    = \frac{1}{Z} \pdv{N} \sum_{s_A} \exp\!(- \beta U_A) \pdv{U_A}{N}
    \\
    &= - \frac{1}{Z \beta} \pdv{N} \sum_{s_A} \exp\!(- \beta U_A)
    = - \frac{1}{Z \beta} \pdv{Z}{N}
\end{aligned}$$

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**:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            F
            &\equiv - k T \ln{Z}
            \\
            &= \expval{U_A} - T S_A
        \end{aligned}
    }
\end{aligned}$$

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.