Why M.Planck used Boltzman distribution to calculate average energy?

[aliinuur]

TL;DR

As far as I understand kT defines scale of Boltzman distribution, i.e. avg. energy and Boltzman distr. are the same thing. And for me, Planck's move to find avg. energy different from 1/2kT from Boltzman distr. is like 1+1=2, 2-1=3.

As far as I understand kT defines scale of Boltzmann distribution, i.e. avg. energy and Boltzmann distr. are the same thing. And for me, Planck's move to find avg. energy different from 1/2kT from Boltzmann distr. is like 1+1=2, 2-1=3.

 

[PeterDonis]

If you think Planck used the Boltzmann distribution, you are quite mistaken. The whole point of Planck's paper was to not use the Boltzmann distribution, because that distribution predicted the ultraviolet catastrophe, which was obviously a very wrong prediction.

I'm not sure what the point is of your post, but per the above you seem to have a basic misunderstanding of this subject.

 

[Demystifier]

Both classical and quantum physics in a canonical ensemble compute the average energy by the formula of the form
$$\langle E\rangle = \frac{\sum_s E_s e^{-E_s/kT}}{\sum_s e^{-E_s/kT}}$$
where the sum is taken over all states $s$, so at a first look it may appear that both are based on Boltzmann distribution. However, classical and quantum physics differ by the set of states $\{ s \}$, i.e. the sum is performed differently. In classical physics the sum is really the integral over phase space, namely, the Boltzmann distribution is a distribution over phase space. By contrast, in quantum physics the sum is performed over a basis of a Hilbert space, so the corresponding distribution is not the Boltzmann distribution, Boltzmann did not know about Hilbert spaces. (Technically, Planck did not know about Hilbert spaces either, but he had the right intuition how to perform the sum in a way different from the classical phase-space integral.) One consequence is that the equipartition theorem, according to which each degree of freedom (appearing quadratically in the Hamiltonian) carries average energy $kT/2$, is valid in classical physics but not in quantum physics.

 

[WernerQH]

As far as I understand kT defines scale of Boltzmann distribution, i.e. avg. energy and Boltzmann distr. are the same thing.

There is a difference between the Maxwell-Boltzmann distribution (applying to particles with mass) and the more general Boltzmann factor $e^{-E/kT}$. The mean energy of a single mode of the radiation field with frequency $\nu$ can be written as a series of Boltzmann factors: $$
{ h\nu \over e^{h\nu / kT} - 1} = h\nu \left( e^{-h\nu / kT} + e^{-2h\nu / kT} + e^{-3h\nu / kT} + \dots \right) \quad,
$$ so the mean energy is $kT$ only in the case $h\nu \ll kT$.

Planck disliked the idea of quanta. For him the radiation field was continuous and could have any energy, in accordance with Maxwell's theory. He hypothesized that the radiating oscillators could only exchange energy in finite amounts $h\nu$ with the electromagnetic field, in order to be able to apply Boltzmann's method of calculating entropy ($S = k \ln W$). As he himself admitted, for him it was an ad hoc idea, introduced in an act of desperation.

 

[aliinuur]

If you think Planck used the Boltzmann distribution, you are quite mistaken. The whole point of Planck's paper was to not use the Boltzmann distribution, because that distribution predicted the ultraviolet catastrophe, which was obviously a very wrong prediction.

I'm not sure what the point is of your post, but per the above you seem to have a basic misunderstanding of this subject.

I mean exp(-E/kT) is Boltzmann distribution

 

[PeterDonis]

I mean exp(-E/kT) is Boltzmann distribution

First, that's not correct, as other posts in this thread have already told you.

Second, that's not the function that Planck used, as other posts in this thread have also already told you.