Re: Ex 20.8 - what happened to the mixed term?

If you assume potential energy (V) is a function of position only (the

), then the

only contribute to the kinetic energy term. In this case there's a very clear

physical reason why the mixed term doesn't appear:

Kinetic energy is unchanged if the direction of motion (or momentum) is reversed.So the Taylor expansion around

cannot contain any odd powers of

.

This isn't true, however, if you allow the potential V to depend on velocity, and on the direction of the velocity in particular (not just its magnitude). In that case, you

can get a non-vanishing mixed term in the Hamiltonian's Taylor expansion.

The first two sentences on p.481 strongly suggest that Penrose is assuming V depends only on position - but he doesn't adequately explain or justify this assumption (in my opinion), nor does he clarify under what circumstances (if any) it might fail to hold.

(Note that the single, unmixed terms linear in

and

must always vanish, simply as a consequence of equilibrium, irrespective of the form of the Hamiltonian).