Past exam of the thermal and statistical physics course of the University of Cambridge / 2012 / Paper 1 / 1

Use thermal de Broglie wavelength ${\lambda }_T=h/\sqrt{2\pi m{k}_{B}T}$ and spacing $a=n^{-1/3}$. Degeneracy onset from thermal wavelength occurs atThe numerical factor depends on the convention for a typical momentum, but the scale ${\hslash }^2{n}^{2/3}/(m{k}_B)$ does not. For fermions the Fermi temperature $T_F={\hslash }^2(3{\pi }^{2}n{)}^{2/3}/(2m{k}_B)$ gives a comparable scale for spin degeneracy two.

Representative estimates, with the assumed densities stated explicitly, are:

Here $u$ is the atomic mass constant. Thus the characteristic orders are $10^5\mathrm{K}$ for electrons, sub-microkelvin for a dilute heavy atomic gas, and $10^{12}\mathrm{K}$ for neutrons. There is no density-independent quantum-degeneracy temperature for atomic gases: for example, taking $n=2.5\times 10^{25}{\mathrm{m}}^{-3}$ and mass $4u$ instead gives about $0.065\mathrm{K}$. These are illustrative input estimates, not values supplied by the paper. Denser neutron-star cores require relativistic and interaction corrections; the estimate identifies the degeneracy scale rather than a precise stellar temperature.

Assume three-dimensional isotropic gapless excitations, zero chemical potential and one independent branch. Momentum-state counting givesSubstituting $x=ϵ/(k_{B}T)$ into the Bose energy integral shows $U/V\propto T^{1+3/n}$ and $C/V\propto T^{3/n}$. The observed exponent therefore gives $n=2$.

For the quadratic-boson heat-capacity coefficient, the required Bose integral isThe equality follows by expanding $(e^x-1{)}^{-1}={\sum }_{r\ge 1}{e}^{-rx}$ and integrating term by term. HenceThe supplied units make its $C$ a heat capacity per volume. Equating its coefficient to $220\mathrm{J}{\mathrm{K}}^{-5/2}{\mathrm{m}}^{-3}$ givesEquivalently $ϵ=p^2/(2m_{\ast })$ has $m_{\ast }=1.09\times 10^{-28}\mathrm{kg}$. For $q_b$ identical branches, the measured coefficient fixes $q_b/{\alpha }^{3/2}$, so instead $\alpha =(4.58\times 10^{27}\mathrm{k}{\mathrm{g}}^{-1})q_b^{2/3}$. The one-branch convention is necessary for the quoted numerical value. The supplied integral with $x^4$ is not the moment arising for the exponent determined here.

For either phase, the entropy differential at fixed composition isThe pressure derivative uses the Maxwell relation from $dG=-SdT+VdP$. Differentiating the equality of entropies along coexistence gives the general relationFor a differentiable coexistence curve with pressure varying, equality of Gibbs free energies also implies $-\Delta SdT+\Delta VdP=0$. Since $\Delta S=0$, the two phases then share $V$. Thus, with $\Delta C_P\ne 0$,This is the Ehrenfest entropy-continuity relation. Entropy equality alone would give $T\Delta (V{\alpha }_T)/\Delta C_P$ without the shared-volume condition; a constant-pressure boundary or vanishing discontinuities needs separate limiting analysis.