Thursday, 2 May 2019

quantum mechanics - In what limit do we *really* get Maxwell-Boltzmann statistics from Bose-Einstein and Fermi-Dirac?


Fermi-Dirac and Bose-Einstein energy occupation number $n(\epsilon)$ in natural units ($[T]=[\epsilon]$) read $$n(\epsilon) = \frac{D(\epsilon)}{e^{(\epsilon-\mu)/T}\pm 1},$$ where $D(\epsilon)$ is the density of states at the given energy $\epsilon$, $+$ is Fermi-Dirac and $-$ is Bose-Einstein.


The usual answer to the question by which limit to arrive to Maxwell-Boltzmann statistics is: $$\frac{\epsilon-\mu}{T} \gg 1$$ or $$\frac{\epsilon_{min}-\mu}{T} \gg 1$$ This gives you purely formally Maxwell-Boltzmann distribution but is a make-it-work pseudo-argument as when $T \to \infty$, $\frac{\epsilon-\mu}{T} \to 0$ and on the contrary $T \to 0^+$, $\frac{\epsilon-\mu}{T} \to \infty$. Hence this rule would tell us to apply Boltzmann statistics to low temperatures and stick with the quantum statistics at high temperatures. This is obviously a fall into the fiery pit of slurry textbook mumbo jumbo.


To resolve this, I believe we have to presume a growing density of states $D(\epsilon)$ and faithfully take the $T \to \infty$ limit showing that every macroscopic feature (i.e. every moment of the original distribution) is in the limit reproduced by Maxwell-Boltzmann up to $\mathcal O(T^{-2},\sqrt{N},...)$.


The problem in the "textbook" argument as well as with the discussed limit is that $\epsilon-\mu$ actually always passes through values which are both greater and smaller than $T$ in the integration. The growing $T$ just "smears out" the distribution out into larger and larger regions in $\epsilon$, which motivated my guess that $D(\epsilon)$ must be growing so that the "smearing out" makes the high-energetic states dominate.


So how is the limit rigorously done? And are there some extra assumptions which aren't usually mentioned? (My guess is that the junk with $\mu \approx \epsilon$ in Bose-Einstein also needs some handling.)



Answer



I'm gonna do this Q&A style, since I have just realized what the answer is. First of all, I would like to apologize to the people whom I dissed for the $$\frac{\epsilon -\mu}{T} >> 1$$ answer. Without a commentary, it is slightly confusing but actually right, the above relation must hold for all energy states, so most dearly to the lowest energy $\epsilon_{min}$. If we neglect $\epsilon_{min}/T\approx 0$, we have $$T \ll -\mu$$ That is, in certain cases, the Maxwell-Boltzmann distribution indeed can be a low temperature limit (see the last two paragraphs). For a high temperature limit, this hierarchical structure must be always satisfied $$1 \ll T \ll -\mu$$


What is however often mooted out is the fact that we are actually interested in a closed system in equilibrium (canonical ensemble). For that we need to revert the expression $$N = \int_0^\infty \frac{D(\epsilon)}{e^{(\epsilon - \mu(N,V,T,...))/T}\pm 1} d\epsilon $$ To get the function $\mu(N,V,T,...)$. This is however very different for every system, so we cannot just generally say what the result might be. For example, for a flat $D(\epsilon)=1$ we get $$N = \pm T log(\pm e^{\frac{\mu}{T}}+1)$$ Assuming $-\mu/T \gg 1$ and checking self-consistence later, we get $$N \approx \pm T(1 \pm e^{\frac{\mu}{T}}), \to -\frac{\mu}{T} = - log(\frac{N}{T}\pm 1)$$ Now obviously for $T \to \infty$, $-\mu/T \to 0$ for Fermi-Dirac and gets undefined for Bose-Einstein, (self-consistency - no). So it is really hard to invert this equation even for the simple densities and the hierarchical structure may be obtained only if $-\mu/T$ is growing with $T$. I believe that the high temperature limit must be also taken alongside another simultaneous limit such as low number density as mentioned by user tpg2114 and cannot be proven for a generic system.



Now to the cases, where the Maxwell-Boltzmann statistics is actually a low-temperature limit. This is true for example for a multi-component system where "chemical" reactions are taking place (this may include nuclear reactions and particle change as in early cosmology). In these systems, the component chemical potentials are fixed by the energy releases in the reactions. Similarly in an open system, we have a chemical potential fixed by the external reservoir. Then if the difference $(\epsilon_{min}-\mu)$ is positive, we indeed gain the Maxwell-Boltzmann distribution for low temperatures $$T \ll \epsilon_{min}-\mu$$ E.g. for photons where $\mu=0$, we could actually use Maxwell-Boltzmann only for a very small cavity at very low temperature since the lowest mode of the photon $\epsilon_{min} \sim V^{-1/3}$.


To sum it up, in a grandcanonical ensemble with fixed $\mu$ (that is, it doesn't change in the limiting process), Maxwell-Boltzmann statistics can never be used as a high temperature limit.


The low-temperature case is actually done e.g. in particle cosmology. It was when reading this article (eq. 2.20 a) when I stumbled upon the "paradox" that the Maxwell-Boltzmann statistics can be used because the temperature is low and had to ask this question.


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