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

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

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

From this we can obtain
[tex]T=-\frac{2\pi\hbar^2N}{mAk_B\log (1-z)}[/tex]
Inserting [itex]z \to 1[/itex] gives [itex]T_c=0[/itex]. 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
[tex]\langle n_0 \rangle =\frac{1}{z^{-1}-1}[/tex]If we equate this with [itex]N[/itex], we would obtain a "critical temperature" of
[tex]
T=\frac{2\pi\hbar^2N}{mAk_B\log (N+1)}
[/tex]
For [itex]1mol[/itex] of "helium" gas in an [itex]1m^2[/itex] area has a critical temperature of [itex]8380K[/itex].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 [itex]n_0[/itex] to reach [itex]N[/itex]?

 

[DrClaude]

From this we can obtain
[tex]T=-\frac{2\pi\hbar^2N}{mAk_B\log (1-z)}[/tex]
Inserting [itex]z \to 1[/itex] gives [itex]T_c=0[/itex]. 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
[tex]\langle n_0 \rangle =\frac{1}{z^{-1}-1}[/tex]If we equate this with [itex]N[/itex], we would obtain a "critical temperature" of
[tex]
T=\frac{2\pi\hbar^2N}{mAk_B\log (N+1)}
[/tex]

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