# Classical regime for Fermi-Dirac and Bose-Einstein gases

I'm studying statistical mechanics, in particular classical regime for Fermi Dirac and Bose Einstein gases. Time average value for occupation numbers in FDBE statistics: $$\langle n_\epsilon\rangle_{FB} = \frac{1}{e^{(\epsilon-\mu)\beta}\pm1}$$ For Boltzmann Statistics: $$\langle n_\epsilon \rangle_B = e^{(\mu-\epsilon)\beta}$$ How can one work out a nice condition of classical regime in which $\langle n_\epsilon\rangle_{FB} \rightarrow \langle n_\epsilon\rangle_B$ ?

An obvious option is $e^{\frac{(\epsilon-\mu)}{kT}}\gg1$. However, I don't really like it, since it implies convergence at low temperature. Moreover I'm expecting an $\epsilon$-free asymptotic expression in terms of temperature and density.

@Adam : i've read your comment again and things are much more clear now :)! Here's what i've got:

I'll assume $\beta|\mu|>>1$ and $\mu<0$ or $z \rightarrow 0$.

In terms of z:

$$\langle n_\epsilon\rangle_{FB} = \frac{1}{\frac{e^{\epsilon\beta}}{z}\pm1} \, \,\underrightarrow{z\rightarrow0} \, \,\langle n_\epsilon\rangle_{B}$$

Being $z=\lambda^3_t \rho$, i can say FDBE gases behaviour like classical one when the particle's thermal wavelenght is small if compared to the typical particle distance. Almost the "low density, high temperature" condition i was looking for.

At low temperature Boltzmann statistics lose physical mean (for example it's easy to recover the classical Sackur–Tetrode entropy from his thermodynamic) . Approximating in this scenary, although it may look mathematically legitimate, is conceptually wrong. Quantum statistcs have to be handled carefully on their own.

Am i doing it right :)?

Sorry for the poor english. Thanks you so much

• On recover the classical regime in the limit $\mu<0$ and $|\mu|/T\gg 1$ (dilute gas). See this question, and my answer there : physics.stackexchange.com/questions/83173/… – Adam Jan 14 '14 at 15:01
• sadly i still don't get it :(. maybe you have some link/reference about the subject? – deppep Jan 14 '14 at 15:51
• Why do you expect an $\epsilon$-free asymptotic expression? I mean, you can express $\epsilon$ in terms of temperature and density, but to do so you absolutely need the partition function (I think). – Alex A Jan 14 '14 at 16:34
• @deppep: If you don't tell what you don't get, it will be hard to explain what's unclear... Tell me what you don't understand, and I will try to give more details. – Adam Jan 14 '14 at 20:31
• @Adam , just edited the first post. take a look! – deppep Jan 14 '14 at 23:54