# Does a constant of motion always imply a Hamiltonian formulation?

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?

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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 math.stackexchange.com/q/324348/11127 –  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.

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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