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

Question

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.)

This post imported from StackExchange Physics at 2015-03-23 09:16 (UTC), posted by SE-user Void

Answer 1

I will give you the argument that is presented in the book I learned from, Physical Gas Dynamics. For reference, it's page 104, Chapter 4 section 6. I present it instead because it does not rely on the Boltzmann distribution for the derivation so perhaps there is some additional insight for you. They established in section 5 that:

$$ \frac{N_j^*}{C_j} = \frac{1}{e^{\alpha + \beta \epsilon_j} \mp 1} $$

for the BE and FD statistics respectively. The other notation: $N_j^*$ is the number of particles in the energy state $j$, $C_j$ is the number of possible levels and the coefficients $\alpha$ and $\beta$ have yet to be determined.

At "sufficiently high" temperatures, $C_j \gg N_j^*$ and the only way for that to be true is if the denominator is large. And when the denominator is large, the exponential is much bigger than 1 so that term may be neglected. So both FD and BE statistics lead to the same expression:

$$ \frac{N_j^*}{C_j} = e^{-\alpha - \beta \epsilon_j} $$

Then they apply this to the number of microstates (after Sterling's approximation):

$$ \ln W = \Sigma_j \left[ \pm C_j \ln \left(1\pm\frac{N_j}{C_j}\right)+N_j\ln\left(\frac{C_j}{N_j}\pm 1\right)\right]$$

where the $\pm$ is for BE and FD respectively. So they plug in the expression for $N_j^*/C_j$ and further give the approximation: $\ln\left(1 \pm x\right) \approxeq \pm x$ for $x \ll 1$ to arrive at the Boltzmann limit:

$$ \ln W = \Sigma_j N_j \left( \ln \frac{C_j}{N_j} + 1 \right)$$

So the entire derivation really just hinges on the notion that at "sufficiently large" temperatures, there are considerably more energy states than particles and so the vast majority of them are empty. Furthermore, both FD and BE statistics have the same result because so many states are empty that the likelihood of two particles attempting to occupy the same level is very low.

This entire derivation relies only on information about quantum statistics and does not require the hand-waving needed to reach the same conclusion from classical mechanics like the derivation from the Boltzmann distribution does (so say the authors of the book immediately following the derivation).

It goes on further to say that $\epsilon_j$ has a wide range of values, including zero, and so the only way for the exponential to be large for all states is to have $\alpha \gg 1$. This is equivalent to (and it's left as an exercise):

$$ \frac{V}{N} \times \frac{\left(2 \pi m k T\right)^{3/2}}{h^3} \gg 1$$

which is violated for very small $m$ with a very large number density $N/V$.

Then it's fairly straight forward to eliminate $\alpha$ and come up with an expression with only $\beta$, which is not possible to solve practicably. So then they use the expression for the maximum number of microstates and perturb the solution about $N_j^*$ and neglect terms second or and higher of $\Delta N_j$.

Based on all of that, it is found that $\beta = \frac{1}{k T}$. But this hinges on the assumption that all possible microstates of a system are a priori equally probable.

Obviously I skipped a bunch of steps and equations in the derivation, but I highly recommend this book (or any other that shows the derivation from the quantum mechanics side rather than from the Maxwell-Boltzmann distribution side).

This post imported from StackExchange Physics at 2015-03-23 09:16 (UTC), posted by SE-user tpg2114

Comments

My question is basically asking by which argument we conclude that at high temperatures $C_j\gg N_j^*$ (in your notation). Also $\alpha= -\mu/T \gg 1$ means that the chemical potential must be always negative and it's absolute value much greater than $T$ ($k_B T$ in your units). This is actually needed most painfully for B-E but this is no surprise. I think the mentioned condition for $\alpha \gg 1$ applies only for an ideal gas in a box, though. Still, the $$C_j\gg N_j^*$$ is essential and falls out of nowhere.

This post imported from StackExchange Physics at 2015-03-23 09:16 (UTC), posted by SE-user Void

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Hey, I upvoted you since reading your post made me realize where the problem actually was, but posted my own answer since there is a whole sector where the Maxwell-Boltzmann statistics is actually a low-energy limit.

This post imported from StackExchange Physics at 2015-03-23 09:16 (UTC), posted by SE-user Void

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@Void It's not out of nowhere when you consider it from the QM side of things. Unfortunately I don't have my book with me on campus, but they give an example calculation earlier in the book about how large $C_j$ really is. I want to say it's 6 or 7 orders of magnitude larger than $N_j$.

This post imported from StackExchange Physics at 2015-03-23 09:16 (UTC), posted by SE-user tpg2114

Answer 2

For BE or FD statistics you have the additional constrained that you are dealing with a fixed number of particles (or fixed mean number). That gives you a normalization condition $$ \int d\epsilon \ n(\epsilon) = N $$ Now, for large T $\beta \epsilon$ is small, i.e. a large fraction of the possible energy states is actually accessible, i.e. could possibly contribute to the integral. To ensure that the normalization condition is nevertheless fulfilled you need to have $\mu$ such that $$ e^{\beta (\epsilon - \mu)} \gg 1 $$

So in short, the restriction on the number of particles requires a temperature dependence of the chemical potential which in turn leads to MB statistics at high temperatures.

This post imported from StackExchange Physics at 2015-03-23 09:16 (UTC), posted by SE-user taupunkt

Comments

@void, sorry, I didn't get your answer before typing mine.

This post imported from StackExchange Physics at 2015-03-23 09:16 (UTC), posted by SE-user taupunkt

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Sure, but I gave you an upvote anyways, since this is really the core of my problem.

This post imported from StackExchange Physics at 2015-03-23 09:16 (UTC), posted by SE-user Void

Answer 3

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.

This post imported from StackExchange Physics at 2015-03-23 09:16 (UTC), posted by SE-user Void

Comments

The dilute gas limit, where the MB distribution is recovered corresponds to $T\ll |\mu|$, with $\mu<0$. Note that the temperature can still be high compare to the typical energy ($\epsilon$). Starting from a quantum gas, one can show in a virial expansion that the classical results (using MB) are recovered in this limit.

This post imported from StackExchange Physics at 2015-03-23 09:16 (UTC), posted by SE-user Adam

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Yes, this is already more or less stated in the answer. What I am also saying is that M-B generally holds only for specific systems with specific $D(\epsilon)$ also relying on other parameters of the system, and is not a general limit from "quantum" to "classical" thermodynamics or "low-temperature" to "high-temperature" thermodynamics (the last paragraph shows a counterexample). Therefore, M-B is always just an $X$ limit for system $Y$ (if at all any limiting case) that isn't in any way general.

This post imported from StackExchange Physics at 2015-03-23 09:16 (UTC), posted by SE-user Void