A 3D harmonic oscillator is thermal equilibrium

[lepori]

hay guys,

A three-dimensional harmonic oscillator is in
thermal equilibrium with a temperature reservoir
at temperature T. Finde The average total energy of the
oscillator

I have no idea, how can I solve this problem,
can you hint me please

 

[BruceW]

You can use Maxwell-Boltzmann statistics as long as the thermal deBroglie wavelength of the particles is small enough. (In other words, the classical limit applies). And then I suppose you could use a continuous number of energy states, and use integration of the energy times the distribution to get the total energy.

Or if the particles obey Fermi-Dirac (or Bose-Einstein) statistics, you should use the model of the quantum harmonic oscillator.

Search wikipedia for some of these words. Have you been taught in class how to find it? Because its pretty tricky to do without learning about it first..

 

[Bill_K]

lepori, Never mind the Fermi-Dirac/Bose-Einstein comments - they don't apply. In fact there are no "particles" in this problem at all. All you have is one 3-D harmonic oscillator at temperature T.

I suspect if you have been assigned this problem, your teacher probably worked a similar problem already with a 1-D oscillator. Just follow the same pattern - enumerate all of the possible excited states for a 3-D oscillator, being especially careful to count how many states you have at each energy level. Then write down for temperature T the probability that the oscillator is in each state, and use them to calculate the average energy.

 

[lepori]

Bruce, Bill thank you

 

[Ken G]

There is an easier way-- simply use the fact that the motion of all 3 dimensions is independent of each other. Then the 3D problem is just like the 1D problem in 3 independent dimensions at once. That means the 3D SHO is just like 3 independent 1D SHOs, and so the average energy is 3 times what you get with one 1D SHO.

 

[lepori]

There is an easier way-- simply use the fact that the motion of all 3 dimensions is independent of each other. Then the 3D problem is just like the 1D problem in 3 independent dimensions at once. That means the 3D SHO is just like 3 independent 1D SHOs, and so the average energy is 3 times what you get with one 1D SHO.

I solved this problem like that: Molecules in thermal equilibrium have the same average energy associated with each independent degree of freedom of their motion and that the energy is 1/2*K*T

A 3D harmonic oscillator has 6 degrees of freedom [3 - 3D movement , 2 - rotational, 1 vibrational]
so, 6*(1/2KT) = 3KT

 

[Ken G]

Maybe there's a way to do it like that, it certainly gets the right answer, but that would not normally be considered the way to go. The 3D harmonic oscillator has only 3 degrees of motional freedom, and they are all vibrational. I think you are thinking of diatomic molecules, and their vibrational degree of freedom is often not excited, so they only "count" as 5 degrees of freedom in most applications. But you get the right answer of 3kT-- the reason is that each of the 3 degrees of freedom counts twice, once in momentum space and once in position space. That's because a different momentum in each direction counts as a different state, and a different position in each direction also counts because there is potential energy associated with those displacements-- free particles have a different way of counting positional states, it depends on the volume allowed per particle.

Maybe the best way to see this is the full result. The average energy is a probability weighted sum over all the possible states of the oscillator, where the relative probability of each state is the Boltzmann factor, e^{-(p_x2+p_y2+p_z2+x2+y2+z2)/kT}. (where for simplicity I've scaled p² to 2m and x² to 2/k, rescaling by a constant is no big thing). Counting states in phase space is like integrating over phase-space volume, so that's over dp_xdp_ydp_zdxdydz, and the weighted average of the energy requires that we multiply the integrand in the numerator by the energy, p_x²+p_y²+p_z²+x²+y²+z². The denominator is the normalization of the weighted sum, so is exactly like the integral in the numerator but without that last energy factor. Each of these integrals is completely separable, so just do a change of variables in each one that sends either something that looks like p²/kT or like x²/kT into a dummy variable like z². The energy factor that is only in the numerator then ends up looking like 6kTz², and the action of the integration acts only on the z variables. 5 of the 6 integrals always cancel top and bottom, and the one that doesn't cancel gives 1/2. So you end up with 6kT/2 = 3kT. Thus it is often said that every degree of freedom that shows up as a quadratic in the energy contributes kT/2 to the average energy.

Note that neither k nor m matters, we scaled them out and they never came back. That's why your way might be equivalent, you are in effect doing it in spherical rather than Cartesian coordinates, and scaling out the m and k makes it ambiguous as to which are kinetic energies and which are potential. Your 3 degrees of translational freedom map into my 3 degrees of momentum, and your 2 degrees of rotational energy map into 2 of my vibrational degrees of freedom. I think the correspondence might just be coincidental, however.