Past exam of the thermal and statistical physics course of the University of Cambridge / 2016 / Paper 1 / 1

Use the number-density virial expansion, following the density versus molar second virial convention, with $n=N/V$:Thus $B_2$ is the coefficient of the first density correction and has units of volume. With molar density instead, the molar coefficient is $N_A{B}_2$. For a classical pair potential $u(r)$,Attractive separations contribute negatively and short-range repulsion contributes positively. For inert-gas interactions the attractive contribution dominates at sufficiently low temperature, $B_2$ crosses zero at the Boyle temperature, and repulsion dominates above it.

The soft-core high-temperature second virial coefficient illustrates why the far-high-temperature tail depends on the model. A hard-core estimate $B_2\simeq b-a/(k_{B}T)$ tends to positive excluded volume $b$. A soft repulsive potential, such as the Lennard-Jones potential in the original diagram, allows a positive maximum followed by decay toward zero. The sign change is the main attraction-versus-repulsion feature; neither sketch should be interpreted as measured data for a specified gas. This classical, dilute-gas sketch omits low-temperature quantum-exchange corrections.^[ref]

Take $\mu >0$ to be the magnitude of either spin projection's magnetic moment, so the independent magnetic levels are $E_{\pm }=\mp \mu B$. Their partition function and magnetic energy arewhere $\beta =1/(k_{B}T)$ and $x=\beta \mu B$. The canonical entropy is consequentlyThis independent-spin magnetic entropy counts the two orientation states of each independent moment; translational entropy and interactions are excluded.

Along the magnetic-entropy demagnetization adiabat, constant magnetic entropy fixes $x$, hence $B/T$ is constant. Lowering the field therefore gives the ideal adiabatic demagnetization relationA useful cycle first magnetizes isothermally while releasing heat to a reservoir, then thermally isolates the sample and lowers the field. Cooling the entire sample requires its other heat capacities and spin-lattice equilibration to be accounted for. Magnetic interactions, residual fields and heat leaks prevent extrapolating the independent-moment formula to an attainable zero temperature.^[ref]

For a complete single-particle state, the Fermi-Dirac distribution isHere ${\mu }_{\mathrm{ch}}$ denotes the chemical potential, distinct from the magnetic-moment symbol in (b). At zero temperature it approaches the Fermi energy $E_F$ and the occupation has a sharp edge. At $T\ll T_F=E_F/k_B$, thermal changes occur within an energy interval of order $k_{B}T$ about that edge.

The Fermi-edge heat-capacity counting argument uses the fact that only order $g(E_F)k_{B}T$ electrons can be rearranged because states well below the edge are blocked by the Pauli exclusion principle. Each active excitation carries energy of order $k_{B}T$, so $U(T)-U(0)\sim g(E_F)(k_{B}T{)}^2$ and $C_{V,e}\propto T$. For a smooth density of states, the Sommerfeld expansion givesHere $g$ is the total single-particle density of states, including spin. In a three-dimensional free-electron gas, $g(E_F)=3N/(2E_F)$, giving $C_{V,e}=({\pi }^2/2)N{k}_B(T/T_F)$. This is the low-temperature normal-state electronic contribution, not the lattice heat capacity or a description across a superconducting transition.^[ref]