# Does spin affect the expression for a Bose-Einstein Condensate's critical temperature?

In an assignment, we were asked to find the critical temperature of a collection of Rubidium-87 atoms. The answer used an expression derived for spin-zero bosons in Schroeder's Thermal Physics (which I have also found elsewhere online):

$$k_B T_c = 0.527 \left( \frac{h^2}{2\pi m} \right) \left(\frac{N}{V}\right)^{2/3}$$ Schroeder uses the expression $g(\epsilon)$ to denote the density of states, and in derivation of the above expression, explicitly assumes the spin of the bosons to be zero. However, the spin of a Rubidium-87 nucleus appears to be $3/2$, which combined with its valence electron, can either form a boson with spin $1$ or $2$. For spin $S$, this would introduce a factor of $(2S+1)$ into $g(\epsilon)$, as the number of available states would increase by this multiple. Since $N\propto g(\epsilon)$, this would decrease the value for $T_c$ by a factor of $(2S+1)^{2/3}$.

Is the above expression valid for all spin values, or should this factor be included? Is there some reason why we can treat all bosons as spin-zero?

$$k_B T_c = [(2s+1) g_{3/2}(1)]^{-2/3}\frac{h^2}{2 \pi m} \left(\frac N V\right)^{2/3}$$
where $g_{3/2}(1)=\zeta(3/2)\simeq2.612$. If $s=0$ your expression is retrieved.
• It is actually not immediate to compute the equivalent spin of an atom (see here). Unfortunately, I couldn't find anything about the equivalent spin of Rb-$87$ atoms; it could also be that that answer is the correct one, and that the equivalent spin is $0$ in this case. But in general, you have to include the $2s+1$ term. – valerio Oct 11 '16 at 8:21