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 remains constant,
but internal energy does not.
The conserved state functions are
the temperature , the volume , and the particle count .
We refer to the system of interest as , and the heat bath as .
The combination forms a microcanonical ensemble,
i.e. it has a fixed total energy ,
and eventually reaches an equilibrium
with a uniform temperature in both and .
Assuming that this equilibrium has been reached,
we want to know which microstates prefers in that case.
Specifically, if has energy , and has ,
which does prefer?
Let be the number of -microstates with energy .
Then the probability that is in a specific microstate is as follows,
where is the resulting energy:
In other words, we choose an ,
and count the number of compatible -microstates.
Since the heat bath is large, let us assume that .
We thus approximate by
Taylor-expanding around :
Here, we use the definition of entropy ,
and that its -derivative is :
We now define the partition function or Zustandssumme as follows,
which will act as a normalization factor for the probability:
The probability of finding in a microstate is thus given by:
This is the Boltzmann distribution,
which, as it turns out, maximizes the entropy
for a fixed value of the average energy ,
i.e. a fixed and set of microstates .
Because is a microcanonical ensemble,
we know that its thermodynamic potential
is the entropy .
But what about the canonical ensemble, just ?
The solution is a bit backwards.
Note that the partition function is not a constant;
it depends on (via ), and (via ).
Using the same logic as for the microcanonical ensemble,
we define “equilibrium” as the set of microstates
that is most likely to occupy,
which must be the set (as a function of ) that maximizes .
However, , and are fixed,
so how can we maximize ?
Well, as it turns out,
the Boltzmann distribution has already done it for us!
We will return to this point later.
Still, does not have a clear physical interpretation.
To find one, we start by showing that the ensemble averages
of the energy , pressure and chemical potential
can be calculated by differentiating .
As preparation, note that:
With this, we can find the ensemble averages
, and of the system:
It will turn out more convenient to use derivatives of instead,
in which case:
Now, to find a physical interpretation for .
Consider the quantity , in units of energy,
whose minimum corresponds to a maximum of :
We rearrange the equation to and take its differential element:
Rearranging and substituting
the fundamental thermodynamic relation
We integrate this and ignore the integration constant,
leading us to the desired result:
As was already suggested by our notation,
turns out to be the Helmholtz free energy:
We can therefore reinterpret
the partition function and the Boltzmann distribution
in the following “more physical” way:
Finally, by rearranging the expressions for ,
we find the entropy to be:
This is why is already maximized:
the Boltzmann distribution maximizes for fixed values of and ,
leaving as the only “variable”.
- H. Gould, J. Tobochnik,
Statistical and thermal physics, 2nd edition,