11.13 Microscopic Meaning of the Variables

The new variables introduced in the previous section assume the temperature to be defined, hence there must be thermodynamic equilibrium in some meaningful sense. That is important for identifying their microscopic descriptions, since the canonical expression $P_q$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e^{-{\vphantom' E}^{\rm S}_q/kt}$$\raisebox{.5pt}{$/$}$$Z$ can be used for the probabilities of the energy eigenfunctions.

Consider first the Helmholtz free energy:

\begin{displaymath}
F = E - T S
= \sum_q P_q {\vphantom' E}^{\rm S}_q + T k_...
... \ln\left(e^{-{\vphantom' E}^{\rm S}_q/{k_{\rm B}}T}/Z\right)
\end{displaymath}

This can be simplified by taking apart the logarithm, and noting that the probabilities must sum to one, $\sum_qP_q$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, to give
\begin{displaymath}
\fbox{$\displaystyle
F = - k_{\rm B}T \ln Z
$} %
\end{displaymath} (11.40)

That makes strike three for the partition function $Z$, since it already was able to produce the internal energy $E$, (11.7), and the pressure $P$, (11.9). Knowing $Z$ as a function of volume $V$, temperature $T$, and number of particles $I$ is all that is needed to figure out the other variables. Indeed, knowing $F$ is just as good as knowing the entropy $S$, since $F$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E-TS$. It illustrates why the partition function is much more valuable than you might expect from a mere normalization factor of the probabilities.

For the Gibbs free energy, add $PV$ from (11.9):

\begin{displaymath}
\fbox{$\displaystyle
G
=
- k_{\rm B}T
\left[
\ln...
...t(\frac{\partial \ln Z}{\partial V}\right)_T
\right]
$} %
\end{displaymath} (11.41)

Dividing by the number of moles gives the molar specific Gibbs energy $\bar{g}$, equal to the chemical potential $\bar\mu$.

How about showing that this chemical potential is the same one as in the Maxwell-Boltzmann, Fermi-Dirac, and Bose-Einstein distribution functions for weakly interacting particles? It is surprisingly difficult to show it; in fact, it cannot be done for distinguishable particles for which the entropy does not exist. It further appears that the best way to get the result for bosons and fermions is to elaborately re-derive the two distributions from scratch, each separately, using a new approach. Note that they were already derived twice earlier, once for given system energy, and once for the canonical probability distribution. So the dual derivations in {D.62} make three. Please note that whatever this book tells you thrice is absolutely true.