Is there a "critical temperature" for 2D Boson gas?

[Zhang Bei]

Homework Statement

Use the fact that the density of states is constant in d = 2 dimensions to show that
Bose-Einstein condensation does not occur no matter how low the temperature. (From David Tong's Lectures)

Relevant Equations

$$\langle N \rangle = \int_0^\infty dE\,\frac{g(E)}{z^{-1}\exp(\beta E)-1}$$

The density of states of a 2D gas in a box is
$$g(E)=\frac{Am}{2\pi\hbar^2}\quad.$$

From this we can obtain
$$T=-\frac{2\pi\hbar^2N}{mAk_B\log (1-z)}$$
Inserting $z \to 1$ gives $T_c=0$. We conclude that the 2D boson gas doesn't form BEC.However, on the other hand, according to the Bose-Einstein distribution, the ground state has an expected occupancy number of
$$\langle n_0 \rangle =\frac{1}{z^{-1}-1}$$If we equate this with $N$, we would obtain a "critical temperature" of
$$T=\frac{2\pi\hbar^2N}{mAk_B\log (N+1)}$$
For $1mol$ of "helium" gas in an $1m^2$ area has a critical temperature of $8380K$.What went wrong? Does this mean the BE distribution needs to be modified in this case? Since z can be pushed arbitrarily close to 1, aren't we always able to force $n_0$ to reach $N$?

 

[DrClaude]

From this we can obtain
$$T=-\frac{2\pi\hbar^2N}{mAk_B\log (1-z)}$$
Inserting $z \to 1$ gives $T_c=0$. We conclude that the 2D boson gas doesn't form BEC.However, on the other hand, according to the Bose-Einstein distribution, the ground state has an expected occupancy number of
$$\langle n_0 \rangle =\frac{1}{z^{-1}-1}$$If we equate this with $N$, we would obtain a "critical temperature" of
$$T=\frac{2\pi\hbar^2N}{mAk_B\log (N+1)}$$

How did you get that last equation from the first 2?