Question

The equipartition theorem states the following:

Let H
be the Hamiltonian describing a system and xi,xj
be canonical variables. Then, for a canonical ensemble with temperature T
, it follows that:

⟨xi∂H∂xj⟩=δijkT.

Now consider a system with two degrees of freedom x1,x2
and canonical momenta p1
, p2
, with the Hamiltonian being
H(x1,x2,p1,p2)=p21+p222m+κ(x1−x2)2.

One can also introduce center of mass coordinates, R=(x1+x2)/2,r=x1−x2
, with the transformed Hamiltonian being:

H(R,r,pR,pr)=p2R2(m+m)+p2r2(m/2)+κr2.

The equipartition theorem tells us in the first case:

2H=p1∂H∂p1+p2∂H∂p2+x1∂H∂x1+x2∂H∂x2⇒⟨H⟩=2kT.

For the second case (center of mass coordinates):

2H=pR∂H∂pR+pr∂H∂pr+r∂H∂r⇒⟨H⟩=(3/2)kT.

Where is my mistake?

Answer 1

The mistake lies in assuming a linear transformation when using the equipartition theorem in the transformed Hamiltonian.

Step-by-step explanation:

The mistake in the calculations lies in the assumption made when transforming the Hamiltonian to center of mass coordinates.

The equipartition theorem assumes that the transformation from one set of canonical variables to another is a simple linear transformation, which is not the case here because the transformation involves a non-linear term (r).

So, the equipartition theorem cannot be directly applied to the transformed Hamiltonian.