0
$\begingroup$

I just want to do a sanity check on my understanding of Hamiltonian mechanics:

My understanding is: For any number $n$, take the phase space $\mathbb R^{2n}$, and take any arbitrary differentiable function $H:\mathbb R^{2n}\to \mathbb R$ to be the Hamiltonian. Then all of the standard results about Hamiltonian mechanics will apply to the system generated by $\dot q=H_p, \dot p=-H_q$ (In particular Liouville's theorem applies, and Noether's theorem applies to the Lagrangian obtained from the Legendre transform). There are no further regularity conditions needed to do Hamiltonian mechanics on this system.

Is this correct?

$\endgroup$
3
$\begingroup$

Yes, any smooth function on the phase space can be Hamiltonian. And to any Hamiltonian corresponds a Hamiltonian vector field $V_H$, such that $$ i_{V_H} \omega = -dH $$ In the simple case of $\mathbb{R}^{2n}$ the symplectic form is $$ \omega = \sum_i d q_i \wedge d p_i $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you. I actually don't understand either of these equations (I just asked a question about symplectic geometry here: physics.stackexchange.com/q/564834). What is $i_{V_H}\omega$, or $dH$? $\endgroup$ – user56834 Jul 10 at 14:10
  • $\begingroup$ @user56834 $d$ is the exterior derivative which for scalar functions like $H$ is just the gradient, and $i_v$ is the contraction operator parameterized by a vector field $v$. If you are familiar with tensors and index notation, $(dH)_a = \partial H / \partial x^a$, and $(i_v \omega)_a = v^b \omega_{ab}$. $\endgroup$ – Prof. Legolasov Jul 10 at 17:09
0
$\begingroup$
  1. We only need differentiability of the Hamiltonian $H(q,p,t)$ as long as we stay in the Hamiltonian formulation. In particular, Noether's theorem works with the Hamiltonian action, cf. this Phys.SE post.

  2. However to guarantee the existence of a regular Lagrangian formulation in $n$ variables $(q^1,\ldots,q^n)$ (via a Legendre transformation) we need to impose that the Hessian $\frac{\partial^2 H}{\partial p_i\partial p_j}$ has maximal rank, i.e. is invertible.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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