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 Ex 20.8 - what happened to the mixed term? 
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Joined: 26 Mar 2010, 04:39
Posts: 109
Post Ex 20.8 - what happened to the mixed term?
From the starting point of an analytic Hamiltonian with a minimum at p_i = q^i = 0, on what grounds does Penrose get rid of the mixed second order term?

For example, consider:
H(p,q) = p^2 + q^2 + pq

This function has a global minimum at (p,q)=(0,0).

I noticed that Gandalf's solution for 20.8 starts from the assumption of a special form of the Lagrangian which does not allow a pq term in the Hamiltonian. In general this is not the case (although perhaps it may be excluded on physical grounds).

If we don't make such an assumption (and the book does not seem to), then I don't see that it is possible to ignore such terms.


18 Jul 2010, 08:46

Joined: 03 Jun 2010, 15:18
Posts: 136
Post Re: Ex 20.8 - what happened to the mixed term?
A (classical) Hamiltionian has always the form H(p,q)=1/2 P^ij p_i p_j + V(q).
A linearised Hamiltonian (i.e. the Taylor expansion in the neighbour of a configuration) has a mixed term, but it vanishes at equilibrium.
I posted a solution to this exercise with the details.


14 Sep 2010, 17:42

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Ex 20.8 - what happened to the mixed term?
If you assume potential energy (V) is a function of position only (the q^a), then the p_a 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 p_a=0 cannot contain any odd powers of p_a.

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


13 May 2012, 09:17

Joined: 11 Jul 2009, 20:45
Posts: 25
Post Re: Ex 20.8 - what happened to the mixed term?
I still think Vasco must explain why he feels reality can be mapped to as many dimension you want ,, all you must do is a full turn!


25 Aug 2012, 04:06
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Ex 20.8 - what happened to the mixed term?
harried wrote:
I still think Vasco must explain why he feels reality can be mapped to as many dimension you want ,, all you must do is a full turn!

I've replied to your question from a week or so ago. Have you read it?
This one seems to be in the wrong place.


25 Aug 2012, 15:01
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