Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 35D (Statistical Physics)

Measure energy from the single-particle ground state and write $z=e^{\mu /(k_{B}T)}\le 1$. The Bose-Einstein distribution gives the number in excited levels asHere ${\mathrm{Li}}_s(z)={\sum }_{j\ge 1}{z}^j/j^s$, obtained directly from the supplied integral identity. At the Bose-Einstein condensation threshold, $z\uparrow 1$ and the excited states are just able to hold all $N$ particles. ThusThe ground-state population is treated separately from the continuum density of states.

Below $T_c$ the fugacity is $z=1$ in the continuum ideal-gas approximation. The zero-energy condensate contributes no internal energy, so the same Bose integral with one additional factor of energy givesEliminate $B$ using $N=B\Gamma (9/2)\zeta (9/2)(k_B{T}_c{)}^{9/2}$ and $\Gamma (11/2)=(9/2)\Gamma (9/2)$:The excited fraction is $(T/T_c{)}^{9/2}$ and the remaining particles occupy the Bose-Einstein condensate. For fixed $B$, the condensed-state energy at fixed $T$ is independent of how many additional particles are placed in the zero-energy ground state.

In the classical limit the fugacity is small and ${\mathrm{Li}}_s(z)=z+O(z^2)$. The Bose-Einstein distribution approaches the Maxwell-Boltzmann distribution, givingTaking the ratio eliminates both $z$ and $B$, yielding $\boxed{E\simeq \frac{9}{2}N{k}_{B}T}$. The coefficient $9/2$ follows from this density of states, rather than the $3/2$ of a free three-dimensional gas.