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I am reading the book "C. J. Pethick, H. Smith - Bose-Einstein condensation in dilute gases". It says the Bose-Einstein distribution function is:

$$f(E) = \frac{1}{e^{\frac{E-\mu}{kT}} - 1}$$

It says that at high temperatures, the effects of quantum statistics become negligible, and the Bose-Einstein distribution function is given approximately by the Maxwell-Boltzmann distribution

$$f(E) \approx e^{-\frac{E-\mu}{kT}}$$

Isn't this valid only if $E<\mu$? Do we know for a fact that $E$ is always less than $\mu$?

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The Bose-Einstein distribution reduces to Maxwell-Boltzmann distribution when $E-\mu>>k_BT$. Since in this case, $$\frac{E-\mu}{k_BT}>>1\rightarrow \exp\left(\frac{E-\mu}{k_BT}\right)>>1$$ Hence one can neglect on in the denominator. This limit corresponds to low density ($\mu$ small) and here there are many more states that are thermally accessible to the particles than there are particles; thus double occupancy never occurs and the requirements of exchange symmetry become irrelevant and both fermions and bosons behave like classical particles.

In particular, note that the distribution function for bosons diverges when $\mu=E$. Thus for bosons, the chemical potential must always be below, even if only slightly, the lowest-energy state. If it is not, the lowest-energy state would become occupied with an infinite number of particles which is unphysical.$^\dagger$

$$\mu <E\ \ : \ \text{For bosons}$$


$\dagger$ [1]S. J. Blundell and K. M. Blundell, Concepts in Thermal Physics. OUP Oxford, 2009.

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  • $\begingroup$ Thanks for the explanation. You mentioned that the Bose-Einstein distribution reduces to Maxwell-Boltzmann distribution when $E-\mu>>k_BT$. If $E$ and $\mu$ remain constant, this condition can be violated for sufficiently large $T$ which might suggest that the Maxwell-Boltzmann equation is not applicable at high temperatures (this is contrary to what the book says). Do $E$ or $\mu$ compensate sufficiently to hold the condition $E-\mu>>k_BT$ even when the temperature is large? $\endgroup$
    – Matrix23
    Commented Dec 24, 2021 at 15:17
  • $\begingroup$ If you fix the energy and let temperature increase then it's true that it won't follow the maxwell distribution. $\endgroup$ Commented Dec 25, 2021 at 3:07
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For high temperatures, particles occupie a large range of states (in terms of energy), this means that the average particle number density $n_i$ in state i is very small. Hence, $n_i = \frac{1}{e^{(E_i-\mu)/k_BT}-1}\ll 1$, or equivalently, $e^{(E_i-\mu)/k_BT}-1 \gg 1$, which is the same as $e^{(E_i-\mu)/k_BT} \gg 1$.

Therefore, one can approximate $n_i \approx e^{-(E_i-\mu)/k_BT}$.

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