Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If a continuous dynamical system has a constant of motion that is a function of all its variables, and is not already evidently Hamiltonian, is it always possible to use a change of variables and obtain a Hamiltonian system?

share|cite|improve this question
Assuming you mean a continuous system? A discrete dyamical system is a trivial example of a non-Hamiltonian system which can have conserved quantities. – Michael Brown Mar 8 '13 at 6:59
Cross-posted from – Qmechanic Mar 8 '13 at 18:44
Dear @user1544418. In general, it is frown upon to cross-post simultaneously, because it may waste potential answerer's time. As a minimum OP should mention the cross-post (on both sites!). The preferred procedure is to not cross-post, and if the post hasn't received an acceptable answer after, say, a couple of days, then OP could flag for migration. – Qmechanic Mar 8 '13 at 18:45
Thanks! I was wondering about that! – Ethan Mar 8 '13 at 20:21

Let us reformulate OP's question as

Does a constant of motion always imply that a system has a Hamiltonian formulation (by possibly introducing additional variables)?

Answer: No. Take a system $M$ that has a constant of motion and another system $N$ that doesn't have a Hamiltonian formulation. Then the combined system $M\times N$ (where the two parts don't talk to each other) will have a constant of motion, but the full system will not have a Hamiltonian formulation.

In general, it can be hard to tell if a given set of equations of motion (eom) are part of a (possibly larger) set of eom that can be put on Hamiltonian (or on Lagrangian) form. See e.g. this and this Phys.SE post.

share|cite|improve this answer
shoot I posted this in math also and forgot to make the important edit. I want the constant of motion to be a function of all the variables in the system. I will add the edit above. – Ethan Mar 8 '13 at 17:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.