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:

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

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):

$$\fbox{$\displaystyle G = - k_{\rm B}T \left[ \ln Z - V \left(\frac{\partial \ln Z}{\partial V}\right)_T \right] $} %$$ (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.61} make three. Please note that what­ever this book tells you thrice is ab­solutely true.