Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 33C (Statistical Physics)

With $\beta =1/(k_{B}T)$, the canonical partition function and equilibrium probabilities areDifferentiating $T\log Z$, with fixed energy levels, gives $S=k_B\log Z+⟨E⟩/T$. Since $\log p(n)=-\beta E_n-\log Z$, normalization yieldsThe sums include degeneracy by labeling states rather than distinct energy values, and a zero probability contributes zero by continuity.

For occupations $W_n=Wp(n)$ of $W$ distinguishable copies, the number of arrangements is the multinomial coefficient $\mathcal{N}=W!/{\prod }_n{W}_n!$. The ensemble entropy is $S_W=k_B\log \mathcal{N}$. Stirling's formula and $\sum W_n=W$ giveDividing by the number of independent copies and taking the large-ensemble limit recovers $S=-k_B{\sum }_{n}p(n)\log p(n)$. Terms with zero occupation are interpreted by the usual $0\log 0=0$ convention.

Let $x=\mathcal{E}/(k_{B}T)>0$ and $r=e^{-x}$. The one-particle internal partition function is $z=(1-r^q)/(1-r)$, and independence gives $Z=z^N$. The mean level index isThus the full internal entropy isThis describes the specified internal degrees of freedom, with no translational or mixing contribution added. For a positive level spacing, $S\to 0$ as $T\to 0$, since the unique lowest internal level becomes certain, while $S\to N{k}_B\log q$ as $T\to \infty$, since all $q^N$ internal configurations become equally likely. The formula includes $q=1$, where the entropy is identically zero.