# Are time-$t$ maps of a Hamiltonian system with 1 degree of freedom typically twist?

If we take a typical Hamiltonian system $H(q,p)$ with one degree of freedom, and look at its time-$1$ map $(q(0),p(0)) \mapsto (q(t),p(t))$, will it generically satisfy the twist property, e.g. $\frac{\partial q(t)}{\partial q(0)} \neq 0$. I mean generic in a sense that twist property is satisfied except for some isolated phase space points (or certain isolated values of $t$)?

I am thinking this because twist property means that the new position $q(t)$ depends on the initial momentum $p(0)$, which makes sense for a classical mechanical system. An exception that comes to mind is of course when we have a periodic orbit and we take $t$ such that $q(t)=q(0)$.

Are there any general counter examples of Hamiltonians (not necessarily that represent a realistic physical system) that fail the twist property?

• Crossposted from math.stackexchange.com/q/1784128/11127 – Qmechanic May 13 '16 at 21:48
• The question here concerns more physics side of things, rather than mathematics, that's why I posted it here separately @Qmechanic – Alex May 13 '16 at 21:52