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This question was already asked here:

Degeneracy in Maxwell Boltzmann distribution

But the answer was not very satisfying so I'm asking again.

I can somewhat understand the derivation of Maxwell-Boltzmann statistics from Wikipedia:

https://en.wikipedia.org/wiki/Maxwell–Boltzmann_statistics

The degeneracy is what bothers me. I understand that degeneracy means the number of ways we can have the same energy with different states. Let's say we have two particles with one having speed so that its kinetic energy is $A$ and other having speed such that its energy is $B$. This is a state. Now the situation could be reversed; with the first particle now having energy $B$ and the second particle having energy $A$. This state has a degeneracy of 2, because there are two ways the system of particles can have the same total energy $A + B$. Do I understand the concept correctly?

Now, in the Wikipedia article it is just stated in the end that by "Thomas-Fermi approximation", the degeneracy is the square root of $\epsilon$, the energy of a particular state. I'm having trouble finding much about this step, the Wikipedia article on it seems to talk about Quantum Mechanics. Then in another article on Maxwell-Boltzmann distribution (https://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution), the end result is stated as

$${\displaystyle {\frac {N_{i}}{N}}={\frac {\exp(-E_{i}/kT)}{\sum _{j}\exp(-E_{j}/kT)}}}$$

which seems like the same thing except the degeneracy seems to be taken as $1$. There is also another page where the degeneracy is also taken as $1$: https://universe-review.ca/R15-30-stat.htm.

So my question is: What really is the degeneracy in Maxwell-Boltzmann statistics? If it is one, why would there be only one way the state could have some particular energy? In other distributions I can imagine how QM puts limitations of how many particles can occupy a state, but why are there such limitations in classical physics?

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I'll make some remarks and hope it helps you understand a bit better the degeneracy concept.

First, "This state has a degeneracy of 2" should be replaced by "This energy level has degeneracy 2" (it has actually more than 2 since for example a state particle 1 $\rightarrow \, \frac{1}{2}A$, particle 2 $\rightarrow \, \frac{1}{2}A+B$ will still give the same energy).The state himself has no degeneracy. Instead, if N states have the same energy, we say that the energy level has degenaracy N.

Second, "which seems like the same thing except the degeneracy seems to be taken as 1". No. Here you have the mean number of particles in state i. Now if you want the mean number of particles in the energy level $E=E_i$ you use \begin{equation} \frac{N(E)}{N}=\frac{\sum\limits_{E_k=E}\exp(-E_k/k_bT)}{\sum\limits_{j}\exp(-E_j/k_bT)} \end{equation}

You were simply making a confusion between an energy level (which can have a degeneracy) and a state (which can't have a degeneracy).

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