Degrees of freedom of harmonic oscillator

[McLaren Rulez]

Homework Statement

A three-dimensional harmonic oscillator is in thermal equilibrium with a temperature reservoir at temperature T. The average total energy of oscillator is
A. ½kT
B. kT
C. ³⁄₂kT
D. 3kT
E. 6kT

Homework Equations

Equipartition theorem

The Attempt at a Solution

So I know the answer is D. The justification given is that a harmonic oscillator has two degrees of freedom ($1/2kx^2$ and $1/2mv^2$) in each dimension so 6 degrees of freedom. The equipartition theorem assigns ½kT to each.

I don't understand why the $1/2kx^2$ and $1/2mv^2$ are independent degrees of freedom. Aren't they related by the total energy in that dimension i.e. $1/2kx^2 + 1/2mv^2 = constant$? Thank you.

 

[kuruman]

Homework Statement

A three-dimensional harmonic oscillator is in thermal equilibrium with a temperature reservoir at temperature T. The average total energy of oscillator is
A. ½kT
B. kT
C. ³⁄₂kT
D. 3kT
E. 6kT

Homework Equations

Equipartition theorem

The Attempt at a Solution

So I know the answer is D. The justification given is that a harmonic oscillator has two degrees of freedom ($1/2kx^2$ and $1/2mv^2$) in each dimension so 6 degrees of freedom. The equipartition theorem assigns ½kT to each.

I don't understand why the $1/2kx^2$ and $1/2mv^2$ are independent degrees of freedom. Aren't they related by the total energy in that dimension i.e. $1/2kx^2 + 1/2mv^2 = constant$? Thank you.

Indeed the sum of the kinetic and potential energy is a constant, but look at it another way. Say you put 12 Joules worth of energy into a 3d harmonic oscillator. How is this energy divided among all the possible ways it can be divided into?

You spend 6 J to stretch it in the x, y and z -directions and at that point you have an oscillator at rest. The remaining 6 J go into kinetic energy of the center of mass in the x, y and z-directions.

 

[DrClaude]

Because both position and momentum are needed to uniquely define the state of a harmonic oscillator.

 

[McLaren Rulez]

You spend 6 J to stretch it in the x, y and z -directions and at that point you have an oscillator at rest. The remaining 6 J go into kinetic energy of the center of mass in the x, y and z-directions.

How is this different to allocating all 12J to the oscillator at rest and then releasing it?

Because both position and momentum are needed to uniquely define the state of a harmonic oscillator.

Am I then correct in transforming that to state that the amplitude and the phase are the two degrees of freedom here? I can understand how more energy leads to a larger amplitude but I'm struggling to see why the phase is affected.

 

[kuruman]

If you add 12 J to the oscillator at rest (I suppose you mean the CM is at rest) then when you release it, the oscillator will just oscillate but the CM will not move. However, note that when you add 12 J to an atom or molecule, you have no choice how these 12 J are distributed. If there are six degrees of freedom each one will receive 2 J. This is known as equipartition of energy.

The amplitude and phase of a harmonic oscillator are not related to its degrees of freedom. Degrees of freedom are the possible places where energy can be tucked away when added to a molecule. If you add energy to a molecule that already has some, then the additional energy will be manifested in all the degrees of freedom. In your case, this means that all three amplitudes of oscillation and all three components of the velocity will increase.

 

[McLaren Rulez]

I think I see what you mean. You're saying that the entire oscillator system moves with some energy (and this is translational kinetic energy) and the oscillations themselves carry a different and unrelated energy. This makes sense - earlier, I assumed that the kinetic energy you were talking about was also of the oscillations.

Somehow, when I read the question, I incorrectly assumed a fixed oscillator where the only degree of freedom was the energy of the oscillations (and that the CM was fixed).

Do correct me if I'm wrong but if not, thank you for the help!

 

[kuruman]

You are correct.