Temperature, Kinetic Energy, Boltzmann Factors, and SHO

[gyroscopeq]

I am confused about the following; where am I going wrong here?

1. (1/2)kT is defined as the average kinetic energy of the molecules of a substance at temperature T, right?

2. You can derive the Boltzmann distribution/Boltzmann factors using (1/2)kT as the kinetic energy, making an argument about kinetic energy being converted into potential energy, and solving a differential equation. Feynman does this.

3. Then, we arrive at the ultraviolet catastrophe. We get a power spectrum that goes as the square of the frequency, times kT. If we then say that kT is the classical average energy of the SHO, and replace it with the correct average energy (a sum of hw*(Boltzmann Factors)), we get the right result. However, we found those Boltzmann factors using the wrong (classical) harmonic oscillator energy. What gives?

Thanks for any help! This question has been driving me crazy for the past couple days.

 

[MisterX]

1. (1/2)kT is defined as the average kinetic energy of the molecules of a substance at temperature T, right?

Perhaps you mean $\frac{3}{2}kT$ ? More generally we have the internal energy $U = \frac{f}{2}kT$. In the case of a monotomic gas confined in three dimensions $f=3$. This is only compatible with our other formula if the potential energy stored in the substance is taken to be zero and so the average internal energy and the average kinetic energy are the same thing. In some cases the averages of $U$ and the kinetic energy can be related by the Virial theorem.
Maybe some else can give a fuller response.

 

[gyroscopeq]

Perhaps you mean $\frac{3}{2}kT$ ? More generally we have the internal energy $U = \frac{f}{2}kT$.

Sure, I was just using (1/2)kT per degree of freedom since (I think) that's the minimum you need to assume for my next two questions.

This is only compatible with our other formula if the potential energy stored in the substance is taken to be zero and so the average internal energy and the average kinetic energy are the same thing.

This is something I am still a little confused about. Temperature is always introduced with kinetic theory. Then, they add more degrees of freedom, but these are mostly still just kinetic energy written a different way (vibrational, rotational, etc.). However, they also start adding in the potential energy of the vibrations (which, on average, is the same as the kinetic energy, by the Virial theorem). So is temperature proportional to the total energy? (I don't think this is the case). Or is it proportional to the kinetic energy? (I would have also thought that (3/2)kT would be, by definition, the average kinetic energy per molecule in a 3D gas)

In any case, I am still super confused on part 3 of my question. It is about the most simple case possible for deriving Boltzmann Factors and using them, but I just don't get it. It seems that the method of getting the Boltzmann Factors is incompatible with turning around and using a different expression for the energy of the SHO.

Thanks!

EDIT: Is this correct? I think that temperature could be defined as (3/2)kT = Kinetic Energy, for a monatomic gas. Then, for more complicated things, we get (3/2)kT per atom as the Kinetic Energy. Finally, since we have the Virial Theorem, we can say that, if we only have kinetic energy and vibrational potential energy, we can just tack on a (1/2)kT per bond (oscillator) which will get us up to (7/2) for a diatomic gas, for example.

 

[ZealScience]

I don't think it is conventional to define kinetic energy and derive Boltzmann factor like that. Kinetic energy is by definition always the point particle kinetic energy and the mean is calculated from the distribution. The distribution arises entirely from statistical argument by considering canonical ensemble. Can you give a reliable reference on where you learned your method.

 

[gyroscopeq]

I don't think it is conventional to define kinetic energy and derive Boltzmann factor like that. Kinetic energy is by definition always the point particle kinetic energy and the mean is calculated from the distribution. The distribution arises entirely from statistical argument by considering canonical ensemble. Can you give a reliable reference on where you learned your method.

I don't know if it is the conventional way to do it, but: http://www.feynmanlectures.caltech.edu/I_40.html#Ch40-S2

 

[ZealScience]

I don't know if it is the conventional way to do it, but: http://www.feynmanlectures.caltech.edu/I_40.html#Ch40-S2

I could not find where kinetic energy is defined in this manner. Well, Boltzmann k is already defined, and I am sure that temperature is also defined in another way. I believe that the average kinetic energy has to be the actual average of the molecular kinetic energy. In that sense, you could not define the average kinetic energy using well defined quantities; you can only show that they satisfy this relation, and in fact the way to do this is to use Maxwell-Boltzmann distribution.

 

[gyroscopeq]

I could not find where kinetic energy is defined in this manner. Well, Boltzmann k is already defined, and I am sure that temperature is also defined in another way. I believe that the average kinetic energy has to be the actual average of the molecular kinetic energy. In that sense, you could not define the average kinetic energy using well defined quantities; you can only show that they satisfy this relation, and in fact the way to do this is to use Maxwell-Boltzmann distribution.

In an earlier chapter he defines Temperature by (3/2)kT = Average Kinetic Energy of a molecule. The link I gave is where he uses that to then get the Boltzmann distribution. Is there some other derivation of the Boltzmann distribution? I don't think I've ever seen it derived; only taken as a given.