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### (GR0177 #5) Thermal Equilibrium Question on GR0177

Posted: **Sun Mar 27, 2005 9:10 pm**

by **Crumbles**

ETS Practice Test: GR0177

Question 5

I know that from equipartition theorem, the average energy in each direction is given by kT/2

Now, since a 3D harmonic oscillator has 3 degrees of freedom, that makes it's total average energy = 3kT/2 which is option C. But the answer is D: 3kT

Can someone please explain?

Posted: **Tue Mar 29, 2005 6:52 pm**

by **us.anjan**

The equipartition theorem assigns 1/2 kT to each degree of freedom. For a 3D harmonic oscillator, that's 3 modes for P.E and 3 modes for K.E and so it's a total of 6 modes - which gives you 3 kT.

Does that sound alright?

Posted: **Thu Mar 31, 2005 5:04 pm**

by **jdexter**

Thinking of it as potential could be confusing, since remember degrees of freedom are quadratic forms of energy. In the case of the 3D oscillator, it not only has 3 translational degrees of freedom (x,y,z) it also has 3 vibrational degrees of freedom (spring energy goes like x^2, so it's quadratic). So in this case it's true that spring potential energy contributes the other 3 degrees of freedom, but that's only because it's a quadratic energy.

Posted: **Fri Apr 01, 2005 6:20 pm**

by **Crumbles**

us.anjan wrote:The equipartition theorem assigns 1/2 kT to each degree of freedom. For a 3D harmonic oscillator, that's 3 modes for P.E and 3 modes for K.E and so it's a total of 6 modes - which gives you 3 kT.

Does that sound alright?

Yes, thanks, that kinda makes sense: 3 degrees of freedom for the molecules and another 3 for the oscillator. I wouldn't really call it PE either though, because they are both technically kinetic. I mean 3/2kT is actually the average kinetic energy associated with the translational motion in a direction. Would you agree?

Posted: **Mon Aug 07, 2006 1:03 pm**

by **Sota**

Edit: Wrong, wrong, see my edit below. This describes a flying, spinning, vibrating dumbbell-shaped object. (Removed reference to oscillator so not to confuse others.)

It has 6 degrees of freedom:

1. Translation, x direction

2. Translation, y direction

3. Translation, z direction

4. Rotational, about x axis

5. Rotational, about y axis

6. Vibrational, z axis

We've oriented it along the Z axis so that we can discount rotational degree about same axis, and so that we only have to consider the one vibrational degree.

Edit: I've confused this with another. I came back to correct it and saw someone else had done so for me.

Again, to really simplify it, consider the oscillator's "energy" in one dimension goes like: E = 1/2mv^2 + 1/2kx, where x is displacement. Two degrees here, and three dimensions, for six degrees. See other answers for a more complete explanation.

### 3D Harmonic Oscillator

Posted: **Tue Aug 08, 2006 6:09 pm**

by **quantumc**

To answer Crumbles question to #5 of GR0177, jdexter's answer is right and Sota is incorrect I believe. Sota: you must be confusing this case with the diatomic rotator or something similar, but that is not the case for the 3D harmonic oscillator. The 3D harmonic oscillator does not have rotational degrees of freedom.

The 3D harmonic oscillator has 3 translational degrees of freedom and 3 vibrational degrees of freedom (in the 3 spatial directions).

Each one of these degrees of freedom contributes in the total Hamiltonian of the system with a term that has either a quadratic form in the momentum coordiante component (translational degrees of freedom) or a quadratic form in the position coordinate component (vibrational degrees of freedom). Each one of these terms contributes 1/2(kT) to the total average thermal energy. Since there are six of these terms, the total average thermal energy is 6/2(kT) = 3(kT), hence the answer is D.

Citing the book "Thermal Physics" by Kittel and Kroemer, pgs 77 and 78: "Whenever the hamiltonian of the system is homogenous of degree 2 in a canonical momentum component, the classical limit fo the thermal average kinetic energy associated with that momentum will be 1/2(kT). Further, if the hamiltonian is homogenous of degree 2 in a position coordiante component, the thermal average potential energy associated with that coordinate will also be 1/2(kT). The result thus applies to the harmonic oscillator in the classical limit."

Quantumc

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### Re: (GR0177 #5) Thermal Equilibrium Question on GR0177

Posted: **Wed Sep 23, 2015 11:20 am**

by **pinkfishegg**

I looked a the solution on GREPhysics.NET and don't understand why you would use the Hamiltonian for this:

http://grephysics.net/ans/0177/5
The Hamiltonian is total energy and they multiply total energy by average energy to get total average energy. What is total average energy?

### Re: (GR0177 #5) Thermal Equilibrium Question on GR0177

Posted: **Wed Sep 23, 2015 12:00 pm**

by **slowdweller**

pinkfishegg wrote:I looked a the solution on GREPhysics.NET and don't understand why you would use the Hamiltonian for this:

http://grephysics.net/ans/0177/5
The Hamiltonian is total energy and they multiply total energy by average energy to get total average energy. What is total average energy?

Wow, strong bump. Anyway, this is a simple application of the equipartition theorem: the average internal energy of a system with f degrees of freedom is f/2 kT. A 3d oscillator has 6 degrees of freedom (3 translational, 3 vibrational).