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I have a one degree of freedom analytic Hamiltonian $H(q,p)$ defined on a semi-infinite cylinder, i.e. $(q,p) \in \mathbb{T} \times \mathbb{R}^{+}$, such that all level sets $H(q,p)=c$ are closed curves. I also impose the condition that there are no equilibria.

I think that then the time-1 map of the flow will be twist (with integrability following from the integrability of $H(q,p)$), but how do I rigorously prove the twist property?

Moreover, could one relax the assumptions on the Hamiltonian (i.e. defined on a cylinder, and all orbits are closed, no equilibria) and still retain the twist property?

The related scientific literature I've seen either deals with 1.5 degree of freedom Hamiltonian, or Poincare maps of 2 d.o.f. systems, not with the type of Hamiltonians I wrote above...

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  • $\begingroup$ Crossposted from math.stackexchange.com/q/1574336/11127 $\endgroup$
    – Qmechanic ♦
    Commented Dec 14, 2015 at 1:01
  • $\begingroup$ Yes, I figure this (now edited by me) question is suited to both forums @Qmechanic $\endgroup$
    – Alex
    Commented Dec 14, 2015 at 1:03
  • $\begingroup$ Note that SE sites are Question & Answer sites, not forums/fora. $\endgroup$
    – Kyle Kanos
    Commented Dec 14, 2015 at 1:37
  • $\begingroup$ Sure, that's what I meant by "forum" -I just couldn't find the right expression... But anyway, this is a question, so I don't see your point? @KyleKanos $\endgroup$
    – Alex
    Commented Dec 14, 2015 at 1:47
  • $\begingroup$ Yes, calling things something they are hardly related to is a good idea, right? Everyone will know full well what you mean when you say, "I'll shoot you a message on my walkie talkie," right? My point was simply that: be specific and don't call something what it isn't. $\endgroup$
    – Kyle Kanos
    Commented Dec 14, 2015 at 1:57

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