Past exam of the thermal and statistical physics course of the University of Cambridge / 2013 / Paper 1 / 1

In the classical equilibrium circuit, the capacitor stores $C{V}^2/2$. Equipartition assigns this quadratic term mean energy $k_{B}T/2$, soThe mean voltage is zero and its variance is $k_{B}T/C$. Resistance determines the equilibration time and noise spectrum, not this full-band equilibrium variance. A finite-bandwidth or appreciably loading readout can measure less than the ideal result.

Let $\Delta =\hslash \Omega$ and choose the nondegenerate energy levels as 0 and $\Delta$. For N independent identical units, $Z=(1+e^{-\beta \Delta }{)}^N$; set N=1 for a single two-level system. Differentiating its logarithm givesAt low temperature, $U\sim N\Delta e^{-x}$ and $C_V\sim N{k}_B{x}^2{e}^{-x}$: the upper level is frozen out. At high temperature, the two levels approach equal populations, giving $U\to N\Delta /2$ while $C_V\sim N{k}_B{x}^2/4\to 0$. The energy therefore rises monotonically to a plateau, while the heat capacity has a Schottky peak:Unequal degeneracies would change the populations and peak; the usual equal-degeneracy model is stated here. Shifting both energy levels changes the energy baseline but not heat capacity.

Use a three-dimensional ideal gas at the same number density n for both isotopes. An unpolarized helium-3 gas has spin degeneracy two, henceHelium-4 has a single spin state, and the ideal Bose-gas saturation condition givesEliminating n, using $m_3/m_4\simeq 3/4$ and $\zeta (3/2)=2.612$, yieldsThe fermionic spin degeneracy must be included. This is an ideal-gas estimate; strong interactions in liquid helium prevent identifying it with an exact experimental transition temperature.