One way of calculating the critical temperature (or the critical number of particles) is (as seen in Statistical Mechanics, Pathria, p. 184) this:

Starting with the equation:

$\frac{N}{V} =\frac{1}{V}\frac{z}{1-z} + \frac{1}{\lambda^3}g_{\frac{3}{2}}(z)$

where $N$ is the total number of particles, $V$ the volume, $z$ the fugacity, $\lambda$ the thermal wavelength and $g_\frac{3}{2}$ one of the Bose-Einstein functions. One then associates:

$N_0=\frac{z}{1-z}$ and $N_e=\frac{V}{\lambda^3}g_{\frac{3}{2}}(z)$

with the number of particles in the ground and the excited states, respectively. Then one can replace $g_{\frac{3}{2}}(z)$ for its maximum $g_{\frac{3}{2}}(1)=\zeta(\frac{3}{2}) =2.61$ so that $N_e$ has a maximum too:


This means that if $N>N_{cr}$ there has to be particles in the ground state. My question is: why does this also mean that if $N<N_{cr}$ there are no particles in the ground state at all?

I believe it has something to do with the fact that if z is close to unity (so that $N_0=\frac{z}{1-z}$ diverges) then $N_0$ is comparable to $N$ but for $z<1$ $N_0$ is finite so it gives zero contribution in the thermodynamic limit of $N \rightarrow \infty$.


1 Answer 1


I beleive I found the answer to this problem:

In boltzmann or Fermi Dirac statistics, there's a finite number of particles $N_i$ in any given energy state. Also, the contribution of any particular energy to the number of particles is negligible because we have an energy continuum.

Bose-Einstein condensation, on the other hand, doesn't mean that you have just a lot of particles in the ground state... if you take the thermodynamic limit of $N\rightarrow \infty$ it will still be a negligible contribution to N. The key point of Bose-Einstein condensation is that the number of particles in the ground state $N_0$ diverges. This means that $N_0$ is now of order of magnitude of $N$. So, for the condensation to happen it is necessary that $N_0 = \frac{z}{1-z}$ diverges, so $z$ has to be of order of unity.

For this reason, if you have $N<N_c$ then, obviusly $N_e<N_c$ too. But $N_e$ is equal to $N_c$ only when $z=1$ so if $N_e<N_c$ then $N_0$ doesn't diverge so there is no condensation.

For $N<N_c$ or $T>T_c$ there are still some particles in the ground state, but that's just like in Boltzmann or Fermi-Dirac statistics because $\frac{N_0}{N} \rightarrow 0$.

I hope that someone can check this answer to be sure I solved my doubt. Thanks!

  • $\begingroup$ Is it okey for me to answer my own question if I've found the answer? I hope it's fine. $\endgroup$ Commented Nov 3, 2016 at 0:42

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