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The Wikipedia page for Maxwell-Boltzmann statistics states:

In statistical mechanics, Maxwell–Boltzmann statistics describes the average distribution of non-interacting material particles over various energy states in thermal equilibrium and is applicable when the temperature is high enough or the particle density is low enough to render quantum effects negligible.

Here, it mentions that it is applicable when high temperatures are involved. This seems confusing to me since as far as I know the derivation for Boltzmann statistics only requires the following two assumptions:

  1. The particles are distinguishable

  2. The particles are weakly-interacting

The first assumption plays its role while we count the microstates and the second assumption shows up in the fact that the total energy is considered as sum of individual internal energies without any cross-terms. This is enough for deriving the Boltzmann exponential dependency and the constraints give the Lagrange multiplier values.

Now then why is it said on Wikipedia that high temperatures are required?

As a specific example, for a two energy level system of N particles the equation for specific heat is as follows: $$C=\frac {Nk_b(\theta/T)^2exp(-\theta/T)}{[1+exp(-\theta/T)]^2}$$ now is this expression wrong for $T$ tending to zero?

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The quantum effects being referred to in the Wikipedia passage you quote are about distinguishability. Real quantum mechanical particles are indistinguishable, and obey Bose or Fermi statistics. Usually, any results you get by properly accounting for Bose/Fermi statistics will reproduce what you had for Maxwell-Boltzmann statistics in the limit of high temperature. The two-level system isn't a great example of this phenomenon -- you can calculate the heat capacity of a two-level system of bosons or fermions, but it turns out the bosons will always form a Bose-Einstein condensate in the ground state (ie, you'll only ever have O(1) particles in the excited state), while for fermions you can have no more than two particles in your two level system. A better example would be a free gas, where you can calculate the partition function and observables exactly and see that Bose/Fermi systems give you back exactly the Maxwell-Boltmann result at high temperatures.

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  • $\begingroup$ Yes, I understand that indistinguishability makes all the difference since the 2nd assumption is not followed, etc. My contention is, as I have mentioned in the larger first part of my question, is that there is no accountability or constraint on the temperature while deriving Boltzmann statistics so how come temperature plays a role in defining Boltzmann statistics as high temperature limit of quantum statistics? $\endgroup$
    – Lost
    Apr 15, 2021 at 18:55
  • $\begingroup$ My first guess was that the distinguishability assumption in the derivation of Boltzmann statistics implicitly puts constraints on the temperature but that's not clear to me. $\endgroup$
    – Lost
    Apr 15, 2021 at 18:57
  • $\begingroup$ You don't need to assume anything about the temperature to derive Boltzmann statistics. You could assume just fine that your particles are distinguishable, and then the derivation of Boltzmann statistics goes through just fine at any temperature. All Wikipedia is saying is that it's not appropriate to use Boltzmann statistics at low temperatures because the particles in question aren't distinguishable. In other words, Boltmann statistics aren't defined by the high temperature limit of quantum statistics, but the high temperature limit of quantum statistics agrees with Boltzmann statistics. $\endgroup$
    – Zack
    Apr 15, 2021 at 19:05
  • $\begingroup$ Okay, I understand that. How come then it happens that the indistinguishability stops playing a role at high temperatures or in other words how does the inappropriateness of using Boltzmann statistics (arisen since practically no particle is fundamentally distinguishable) only creeps in at low temperatures and not high temps? $\endgroup$
    – Lost
    Apr 15, 2021 at 19:13
  • $\begingroup$ When temperature is very high, there are many different accessible energy levels. As a result, the average occupation number for any given energy level is very small, and it doesn't matter what statistics the particles have. On the other hand, for low temperatures there is low likelihood of occupying high energy states, and you have to decide how to fit the particles into a smaller number of levels. Then you have to worry about entropic differences between Bose/Boltzmann particles, or the hard-core constraint of Fermi particles. $\endgroup$
    – Zack
    Apr 15, 2021 at 19:46

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