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While deriving the Planck's radiation formula, why do we use MB statistics when we calculate the average energy of oscillators? Shouldn't we use BE?

Is this because temperatures concerned are very high and quantum stat goes to classical at high temperatures?

But again, in a similar calculation, while calculating the lattice specific heat of the solid using Einstein's model or the Debye model, I have seen books calculate the average energy of oscillators using MB distribution only, but we use this formula to predict specific heats at very low temperatures also. So I don't understand, what is the reason?

Qmechanic
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Shikhar Chamoli
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    Bose-Einstein distribution naturally appears in the modern derivation of the Black body radiation - see derivation in this answer: https://physics.stackexchange.com/a/751324/247642 Of course, Planck didn't know about it at the time, and many texts reproduce his analysis - not for the sake of clarity, but to motivate the need for quantum mechanics. – Roger V. May 15 '23 at 08:23

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Planck's Law is the first place BE statistics shows up, and so it is necessary to have some way to derive BE statistics.

It is not just that we are treating each QHO separately, but that we are also treating each energy level of any one QHO separately, and those follow MB statistics. This is equivalent to treating each QHO with BE statistics; it is how we derive BE statistics in the first place.

Another way to put it is this: I know you want to get $$f_i=\frac1{e^{E_i/k_BT}-1}$$ But with some manipulations, you can see that $$f_i=\frac{e^{-\beta E_i}}{1-e^{-\beta E_i}}=\frac1{1-e^{-\beta E_i}}-1\\ 1-e^{-\beta E_i}=\frac1{1+f_i}\\ e^{-\beta E_i}=1-\frac1{1+f_i}=\frac{f_i}{1+f_i}$$ i.e. Essentially we are temporarily working with this last expression, so that we can apply MB statistics, and thereby derive BE statistics. With a little changes here and there, we can get the same with FD statistics.

This is a very standard derivation, just that it is often not stated this explicitly.

naturallyInconsistent
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  • Ok sir, what i understood is that you wrote the MB distribution in terms of BE distribution. But I don't understand how by temporarily working with this expression $\frac{f_i}{1+f_i}$ we can derive BE? Please explain a bit more. The answer is not clear to me. – Shikhar Chamoli May 15 '23 at 07:27
  • Work backwards. Once you assumed that $\frac{f_i}{1\pm f_i}$ is a thing that follows MB distribution, then you can derive that $f_i$ follows BE or FD distributions. – naturallyInconsistent May 15 '23 at 07:30
  • So this means you can always derive BE or FD using MB? And since this is the case, we can always use MB instead of BE.. is this the point? – Shikhar Chamoli May 15 '23 at 10:33
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    No. I am saying that, before you have derived BE or FD statistics, you can use these contrived stuff and MB, to derive BE and FD. It is VERY awkward to work with these contrived stuff. Once you have derived them, you can let go of the contrived stuff and just reason much more naturally with BE and FD statistics on the things that we are studying, and get the correct physics. Given how lazy we all are, it would not make sense to tell you about BE and FD statistics if we could make do with MB all the time. BE and FD have real physics understanding in them, that MB and these contrived stuff don't – naturallyInconsistent May 15 '23 at 10:43