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:

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

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

$$\fbox{$\displaystyle F = - k_{\rm B}T \ln Z $} %$$ (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):

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