# bose einstein phase transition

From Carter's book Thermodynamics and Statistical Mechanics, the partition function of a bose-einstein gas in $d$ dimensions is $$\ln(Z) = -\int_{0}^{\infty}\ln(1-e^{\beta(\mu-\epsilon)})c_d\epsilon^{d/2-1}d\epsilon \ ,$$ where $c_d$ is independent of $\mu$ and $\beta$. The heat capacity is $$C_V =k_B\beta^2 \dfrac{\partial^2 \ln(Z)}{\partial \beta^2} = \dfrac{k_Bc_d}{4} \int_{0}^{\infty} \dfrac{\epsilon^{d/2+1}d\epsilon}{\sinh^2(\beta(\mu-\epsilon)/2)} \ .$$ Carter states that the presence of singularities in the heat capacity is a measure of the existence of a phase transition. From the integral above, how does one identify the temperature at which the heat capacity is infinite?

• $\mu$ is implicitly a function of temperature. The phase transition occurs at $\mu = 0$ (for a non-interacting Bose gas). – Mark Mitchison Jul 24 '15 at 7:37
• @MarkMitchison When $\mu=0$, the singularity of the $C_V$'s integrand becomes $\epsilon^{d/2+1}/\epsilon^{2}=\epsilon^{d/2-1}$. For $d=(1,2,3)$, the integrand's singularity equals $(\epsilon^{-1/2},\epsilon^{0},\epsilon^{1/2})$, which all are convergent integral singularities. The only way any of the integrals can diverge is when $\mu>0$. – linuxfreebird Jul 24 '15 at 13:02