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I know that for a particle under the potential

$$V(x,y,z)=\frac{k}{2}(x^2+y^2+z^2)$$

the equipartition theorem says that it contributes to the mean energy to $\frac{3k_BT}{2} $ (one half for each degree of freedom).

My question is: let's say we have a potential like this one, for instance 2 particles with an harmonic interaction potential, where $r_{eq}$ stands for the equilibrium point:

$$V(x,y,z)=\frac{k}{2}(\sqrt{x^2+y^2+z^2}-r_{eq})^2$$

Note that in this case the coordinates $(x,y,z)$ are coupled in the potential.

How would one compute the contribution of such a potential to the mean energy? Is there any analogy to the equipartition theorem in a case like this?

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Is r here a vector, or is it a constant distance? If it is a vector it's inconsistent, and if it's a constant you'll have a ring of minima rather than a point. –  Nuclear_Wizard Apr 5 at 12:22
    
It is a constant. –  gunbl4d3 Apr 5 at 12:33
    
To correct Nuclear_Wizard, you would have a spherical surface, since there are three dimensions. –  fibonatic Apr 5 at 20:30

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