Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 34D / b / iii / Solution

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.