1
$\begingroup$

Notation: I denote phase space as the symplectic manifold $(M,\omega)$, in which $\omega=\sum_i\mathrm dp_i\wedge\mathrm dq_i$ in canonical coordinates.

In definitions of Hamiltonian vector fields I have seen so far$^{[1][2]}$, they seem to state that for all $H\in C^\infty (M)$ the associated Hamiltonian vector field is defined as the $X_H$ for which: $$\mathrm dH(\bullet)=\omega(X_H,\bullet) \tag{1}.$$

We also know that integral curves generated by Hamiltonian vector fields are the solutions to Hamilton's equations, which define time evolution in the Hamiltonian formalism. I am confused why all smooth functions generate Hamiltonian vector fields, and why the very special role of defining the time evolution is not entirely contained in one specific smooth function that we call the Hamiltonian.

It would make more sense to me that if we choose one given physical system we have to single out one smooth function on phase space that we call the Hamiltonian, and this and only this function has an associated vector field that we call "the Hamiltonian vector field" for this system, whose integral curves define the time evolution of the system.

Perhaps this is what is meant in my links and I have just misunderstood. Perhaps they are considering the most general case in which all smooth functions could hypothetically serve as "the Hamiltonian function" for a given system?

References:

$^{[1]}$: Hamiltonian Vector Fields and Symplectic Geometry, Woit

$^{[2]}$: Wikpedia, Hamiltonian Vector Fields

$\endgroup$

1 Answer 1

1
$\begingroup$

You should not be mislead by the name "Hamiltonian vector field". Basically, you take a (smooth) function on your manifold, make a differential form out of it by applying the exterior derivative, and then pull it from the cotangent space to the tangent space of your initial manifold. This last step is performed by an isomorphism from one space to the other we call $\mathcal{I}$.

Thus, a Hamiltonian vector field is just $\mathcal{I}(dH)$, if $H$ is your smooth function on phase space. This definition does not imply that $H$ is what physicists call the Hamiltonian of the system. It is just that for the special case where you choose your smooth function $H$ to be the Hamiltonian, you get the canonical equations of motion in local coordinates.

For example, on the two-sphere equipped with the canonical 2-form $\sin \theta d\theta \wedge d\phi$, $\partial_\phi$ is a Hamiltonian vector field.

$\endgroup$
4
  • $\begingroup$ So "Hamiltonian vector field" is a generic name for the vector field constructed from a smooth function on phase space, and so the statement "the integral curves of Hamiltonian vector fields solve Hamilton's equations" is not generally true unless we are talking about the specific function that we have designated as "the Hamiltonian"? $\endgroup$
    – Charlie
    Commented Dec 10, 2020 at 14:54
  • $\begingroup$ Indeed, you will always get integral curves of a Hamiltonian vector field, but these will not correspond to the p,q trajectories you want from your physical system. $\endgroup$ Commented Dec 10, 2020 at 14:57
  • $\begingroup$ Got it, thank you very much for your answer! $\endgroup$
    – Charlie
    Commented Dec 10, 2020 at 14:57
  • 1
    $\begingroup$ My pleasure! At the end of the day, you don't get much physics from this abstract formalism, but this can be useful for some proofs, or perturbation theory. See for instance V.I. Arnol'd, Mathematical methods of Classical Mechanics. It's a very tough read, but chapter 10 on perturbation theory is a direct application of the few chapters he spends on this symplectic formulation. $\endgroup$ Commented Dec 10, 2020 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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