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

  • I
  • Thread starter Thread starter aliinuur
  • Start date Start date
AI Thread Summary
The discussion centers on the distinction between Boltzmann distribution and Planck's approach to average energy in quantum physics. It emphasizes that while both classical and quantum physics calculate average energy using similar formulas, they differ fundamentally in the sets of states considered. Planck intentionally avoided using the Boltzmann distribution due to its failure to address the ultraviolet catastrophe. The conversation also highlights that Planck's hypothesis of quantized energy exchange was a significant departure from classical views, which treated energy as continuous. Overall, the thread clarifies misconceptions about the relationship between Boltzmann distribution and Planck's work.
aliinuur
Messages
31
Reaction score
0
TL;DR Summary
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 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.
 
Science news on Phys.org
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.
 
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.
 
Last edited:
  • Like
Likes dextercioby, Bandersnatch and div_grad
aliinuur said:
As far as I understand kT defines scale of Boltzman distribution, i.e. avg. energy and Boltzman 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.
 
  • Like
Likes dextercioby and PeterDonis
PeterDonis said:
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 Boltzman distribution
 
aliinuur said:
I mean exp(-E/kT) is Boltzman 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.
 
I need to calculate the amount of water condensed from a DX cooling coil per hour given the size of the expansion coil (the total condensing surface area), the incoming air temperature, the amount of air flow from the fan, the BTU capacity of the compressor and the incoming air humidity. There are lots of condenser calculators around but they all need the air flow and incoming and outgoing humidity and then give a total volume of condensed water but I need more than that. The size of the...
Thread 'Why work is PdV and not (P+dP)dV in an isothermal process?'
Let's say we have a cylinder of volume V1 with a frictionless movable piston and some gas trapped inside with pressure P1 and temperature T1. On top of the piston lay some small pebbles that add weight and essentially create the pressure P1. Also the system is inside a reservoir of water that keeps its temperature constant at T1. The system is in equilibrium at V1, P1, T1. Now let's say i put another very small pebble on top of the piston (0,00001kg) and after some seconds the system...
Back
Top