Testing Physics GRE Q: 3D Harmonic Oscillator & Thermal Equilibrium

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The discussion revolves around a Physics GRE question regarding the average total energy of a three-dimensional harmonic oscillator in thermal equilibrium at temperature T. The correct answer is identified as D, which is 3kT. Participants reference the equipartition theorem, which states that each degree of freedom contributes (1/2)kT to the energy. The confusion arises regarding the total degrees of freedom for the system. While initially considering only three degrees of freedom, participants realize that the total energy for a three-dimensional harmonic oscillator actually accounts for both potential and kinetic energy contributions, leading to a total of six degrees of freedom (three for position and three for momentum). The discussion emphasizes the need to understand the Hamiltonian for n-dimensional systems and the application of the equipartition theorem to derive the correct energy expression.
daveyman
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I'm studying for the Physics GRE and I came across this question. The correct answer is D, but I'm not sure quite how to do it. Any help would be greatly appreciated!

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

(A) (1/2)kT
(B) kT
(C) (3/2)kT
(D) 3kT
(E) 6kT
 
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You need to show work in order to get help here. What have you tried so far? What concepts, etc. apply? What are your thoughts?
 
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I'm really not sure where to start. I guess I'll just post this in a different part of the forum then.
 
I'm studying for the Physics GRE and I came across this question. The correct answer is D, but I'm not sure quite how to do it. Any help would be greatly appreciated!

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

(A) (1/2)kT
(B) kT
(C) (3/2)kT
(D) 3kT
(E) 6kT
 
Well the equipartition theorem says that each Degree Of Freedom = 1/2KT. Can you think of why there would be 6 DOF?
 
That's exactly what I was thinking. It seems like the problem basically states that there are 3 degrees of freedom, which would yield an answer of (3/2)kT. However, the answer guide clearly says that the answer is 6kT (this is from an official GRE practice test).

I can't think of why there would be 6 degrees of freedom...
 
Ok let's go back to the 1 dimensional harmonic oscillator... How would you write its total energy? How many DOF is that?
 
Total energy for a 1-dimensional harmonic oscillator is (1/2)kx^2, right? And wouldn't that just be one degree of freedom?
 
Total energy of a 1D SHO is (1/2)kx^2 + (1/2)mv^2...
 
  • #11
daveyman said:
I'm studying for the Physics GRE and I came across this question. The correct answer is D, but I'm not sure quite how to do it. Any help would be greatly appreciated!

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

(A) (1/2)kT
(B) kT
(C) (3/2)kT
(D) 3kT
(E) 6kT

Acording to the Equipartition Theorem, there is a kt/2 contribution to the energy from each degree of quadratic freedom in the Hamiltonian. In equation form, the average total energy is <E> = skT, where s is the degrees of freedom.

What is the expression for the Hamiltonian for any n-dimensional 1-particle system?
 

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