Categories: Physics, Thermodynamic ensembles, Thermodynamics.

# Canonical ensemble

The canonical ensemble or NVT ensemble builds on the 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 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 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.