Which Thermodynamic variables are averages?

Question

For the various thermodynamic potentials, we get a variety of relations,

\begin{align} dE &= TdS - p dV + \mu dN \\ dF &= -S dT - p dV + \mu dN \\ dH &= T dS + V dp + \mu dN \\ dG &= -S dT - V dp + \mu dN \end{align}

I have noted that they call the ones that are fixed, in the statistical sense, are called natural variables. Are all the other variables statistical averages? For example, is it proper to write the following? \begin{align} \langle T \rangle &= \frac{\partial E(S, V, N)}{\partial S} \\ \langle S \rangle &= -\frac{\partial F(T, V, N)}{\partial T} \\ \langle \mu \rangle &= \frac{\partial F(T, V, N)}{\partial N} \end{align}

If some of the "non-natural" variables are not averages for given potentials/ensembles, then which ones?

Answer 1

Different ensembles hold different quantities to be fixed. Quantities that are not fixed a priori (and by extension, functions of those quantities) typically fluctuate. These quantities are represented as ensemble averages.

Answer 2

Every ensemble is characterized by a probability distribution on a suitable space of states. Such space of states coincides with the dynamical phase space in some cases (microcanonical or canonical ensembles) but it may be an enlarged phase space in others. For instance, in the isobaric ensemble the random variables the distribution depends on are positions and momenta of the particles and the volume the system is confined in. Or, in the grancanonical case, the number of particle has to be added to the phase space point.

Only starting from the distribution function, it is meaningful to speak about averages and variances (fluctuations) of observables depending on the random variables (or part of them) the distribution is function of. For example, in the case of the microcanoncal ensemble, even if energy does not fluctuate, since it is a fixed parameter characterizing the ensemble, it also coincides with the average value of Hamiltonian: $$ \langle H \rangle = \frac{\int H(p,q) \delta(E-H(p.q))d^{3N}pd^{3N}q}{\int \delta(E-H(p.q))d^{3N}pd^{3N}q}. $$ Of course the variance is zero. So, in order to understand which quantity can be seen as an average, the criterion of looking for fluctuation is a sufficient but not necessary condition.

In general, to answer the question if a thermodynamic quantity can be evaluated as an average in a given ensemble, one has to look for a possible rewriting of that expression as an ensemble average of a suitable functions of the random variables characterizing the ensemble.

For example, temperature and excess pressure (pressure minus the corresponding ideal gas pressure) can be obtained as averages of the kinetic energy and of the virial. An important thing to notice is the partition function of an ensemble cannot be obtained as an average on the ensemble. This is a trivial, but sometimes neglected consequence of the fact that the partition function is the normalization constant of the probability distribution.

As a consequence, in the canonical ensemble we cannot get entropy as an average just playing with the thermodynamic definition in term of free energy $$ S=-\left.\frac{\partial{F}}{\partial{T}}\right|_{V,N} $$ because, by using $F=-k_BT \log Z(T,V,N)$, we would get an additive term $k_B |log Z$, which cannot be written as an average. Actually, in the case of entropy, no computable function of the phase space is known whose average is equal to $S$.

A final note on the "natural variables". For each thermodynamic potential, natural variables are the variables which allow to get all the thermodynamic quantities by expressions containing only the natural state variables and the thermodynamic potential with its partial derivatives. It turns out that there is no direct relation between natural variables ad the possibility of expressing thermodynamic quantities as averages.