---
title: "Canonical ensemble"
sort_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.