Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 34D (Statistical Physics)

Here $p(n)$ is the probability of microstate $n$, including all occupation data. In the grand canonical ensemble, with energy $E_n$, particle number $N_n$, chemical potential $\mu$ and $T>0$,Substitution in the Gibbs entropy gives $S=k\log \mathcal{Z}+(U-\mu N)/T$, where $U=⟨E⟩$ and $N=⟨N_n⟩$. Differentiating at fixed $\mu ,V$ gives ${\partial }_T\log \mathcal{Z}=(U-\mu N)/(k{T}^2)$. Therefore

For noninteracting fermions, each single-particle level has occupation $n_i\in \{0,1\}$ by the Pauli exclusion principle. A microstate has $E={\sum }_i{ϵ}_i{n}_i$, $N={\sum }_i{n}_i$, so its grand-canonical weight is the product of its independent level weights. The grand potential is $\Omega =-kT\log \mathcal{Z}$; the particle number, entropy and pressure follow from $N=-{\partial }_{\mu }\Omega$, $S=-{\partial }_T\Omega$ and $P=-{\partial }_V\Omega$. This occupation description is the basis for the product, continuum and adiabatic calculations.

Summing separately over the two allowed occupations of every level givesThusThe factorization uses the absence of interactions; each level, including any degeneracy labels, is counted separately.

Replace the sum by the integral against the density of states and set $x=ϵ/(kT)$, $y=\mu /T$:ThereforeFor the given $a>0$, the integral converges at zero and infinity for every finite $y$.

Differentiating the grand potential givesA reversible adiabatic change in an isolated gas keeps $S$ and $N$ fixed. To infer constant $y$, one must check that this ratio is one-to-one, rather than merely assuming it.

Put $q=y/k$, $n(x)=1/(1+e^{x-q})$, $F_j={\int }_0^{\infty }{x}^{j}n(x)dx$ and $M_j={\int }_0^{\infty }{x}^{a+j}n(x)(1-n(x))dx$ for $j=0,1,2$. Integration by parts gives $M_1=b{F}_a$ and $M_2=(b+1)F_{a+1}$, andThe strict inequality is the Cauchy-Schwarz inequality for the positive weight $x^{a}n(x)(1-n(x))$, since $x$ is not constant on its support. This proves strict monotonicity of Fermi gas entropy per particle and hence $\mu /T$ is constant. Then $N=kV{T}^b{f}^{\prime }(y)$ gives $V{T}^b$ constant, and $P=k{T}^{b+1}f(y)$ givesThis adiabatic scaling of a Fermi gas with power-law density of states concerns positive temperature; the zero-temperature limit is a separate limiting case.