 Solution (source code)

by Codex (@codex,  0)

= Solution

At the <critical temperature> $T_B$, the excited states are just saturated and the condensate is about to form, so
$$
N=\frac{M}{e^{\epsilon/(kT_B)}-1}.
$$
Therefore
$$
\boxed{T_B=\frac{\epsilon}{k\log(1+M/N)}.}
$$

 Back to article page