= Solution
\b[Excited-state saturation.] Consider first a uniform three-dimensional noninteracting gas of identical spinless <bosons> in a large box. Set the one-particle ground <energy> to zero. The <Bose-Einstein distribution> is $\bar n(\epsilon)=[z^{-1}e^{\beta\epsilon}-1]^{-1}$, with <quantum gas fugacity> $z=e^{\beta\mu}\leq1$. For dispersion $\epsilon=p^2/(2m)$, the excited <density of states> is
$$
g(\epsilon)=\frac V{4\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2}\sqrt\epsilon.
$$
Using the <thermal de Broglie wavelength> $\lambda_T=\sqrt{2\pi\hbar^2/(mk_BT)}$, integration gives
$$
N_{\rm ex}=\frac V{\lambda_T^3}\operatorname{Li}_{3/2}(z),\qquad
N_0=\frac z{1-z}.
$$
The <polylogarithm> is $\operatorname{Li}_s(z)=\sum_{\ell\geq1}z^\ell/\ell^s$. Its finite value at $z=1$ implies a maximum excited population $V\zeta(3/2)/\lambda_T^3$, with $\zeta(3/2)\simeq2.612$. Once this is less than $N$, the excess particles occupy the <ground state> macroscopically. This is <Bose-Einstein condensation>, rather than occupation of a state by merely one or a few atoms.
\b[Transition and condensate fraction.] Setting the saturation number equal to $N$ gives the <homogeneous ideal Bose gas condensation> <temperature>
$$
\boxed{T_c=\frac{2\pi\hbar^2}{mk_B}
\left[\frac n{\zeta(3/2)}\right]^{2/3}.}
$$
Below this <temperature>, $\mu$ approaches zero from below in the <thermodynamic limit>, and particle conservation gives
$$
\boxed{\frac{N_0}{N}=1-\left(\frac T{T_c}\right)^{3/2}
\quad(0\leq T\leq T_c).}
$$
Above $T_c$ there is no macroscopic condensate and $\mu<0$ adjusts to keep $N$ fixed. At finite $N$, the <chemical potential> remains slightly below the ground <energy> and the transition is rounded.
\Image[/past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2017-q2-bose-condensation.png]
{title=Original uniform ideal-gas condensate fraction and schematic accumulation in the lowest state below the critical temperature; the exponent is for a three-dimensional box}
\b[Energy and the role of geometry.] The excited-state <internal energy> is
$$
U=\frac32\frac{Vk_BT}{\lambda_T^3}\operatorname{Li}_{5/2}(z).
$$
The zero-<energy> condensate contributes no term to this <energy> in the ideal model. Below $T_c$,
$$
U=\frac32Nk_BT\frac{\zeta(5/2)}{\zeta(3/2)}
\left(\frac T{T_c}\right)^{3/2},\qquad
C_V=\frac{15}{4}Nk_B\frac{\zeta(5/2)}{\zeta(3/2)}
\left(\frac T{T_c}\right)^{3/2}.
$$
Condensation depends on low-<energy> state counting: for a uniform quadratic-dispersion gas, a finite-<temperature> bulk condensate requires spatial dimension greater than two. A trap changes the <density of states> and therefore the <temperature> law; a three-dimensional harmonic trap gives a cubic, rather than $3/2$, condensate-fraction exponent. Finally, an ideal-gas condensate should not automatically be called a <superfluid>: its free quadratic spectrum has zero critical speed by the <Landau criterion>. Interactions matter for the robust collective-flow properties of real dilute condensates.
Back to article page
