3
$\begingroup$

I'm trying to get some intuition into how the Poincare return map is area-preserving (when there are two momenta and two positions).

Suppose $H=H(q_1,q_2,p_1,p_2)$, and let's suppose the system is integrable - say a Toda Hamiltonian with two positions and two momenta. A collection $X_0$ of initial points on the surface $q_2=0$ with the same energy $E$ will, in general, not lie on the same torus since the second conserved quantity - call it $F_2$ - will in general be different for those points.

Suppose the collection of points defines a curve $C$ on the $(q_1,p_1, q_2=0)$ plane (of course $p_2$'s will be different for different points). We can compute $$ \oint_C (p_1dq_1+p_2dq_2)=\oint_C p_1dq_1 $$ by taking the contour on the surface of the section at $q_2=0$.

The Poincare return map is constructed from points "puncturing" the $q_2=0$ plane at different times so computing the "equal time" invariant $dp_1\wedge dq_1+dp_2\wedge dq_2$ doesn't seem to be helpful. Indeed, Kibble and Berkshire (Classical Mechanics, 5th Ed., Imperial College Press, sec. 14.2) state that

It should be noted that the Poincare return map for a Hamiltonian system can be shown to be area-preserving. This property is related to Liouville’s theorem, treated in §12.5, although it is not derivable from it.

Thus, in what sense is the Poincare return map area-preserving?

Note: The best I can find is from Michael Tabor, Chaos and integrability in nonlinear dynamics, appendix 4.1. It seems one should consider the collection of "first re-intersection points", which will not occur at the same times, but for which we can still obtain a contour $C'$ on the surface of section and show that $$ \oint_C p_1dq_1=\oint_{C'} p^\prime_1dq^\prime_1 $$ with the collection of "first return points" $\{(q^\prime_1,p^\prime_1)\}$ defining $C'$ taken at different reintersection times. This then raises the question: is the Poincare return map only area preserving for the collection of "equal return points" $\{(q^k_1,p^k_1)\}$?

$\endgroup$
1
$\begingroup$

in what sense is the Poincare return map area-preserving?

In the usual sense: its dynamics preserves phase-space volumes.

I'm trying to get some intuition into how the Poincare return map is area-preserving

I'm not very sure I have one myself. But this is a consequence of not only the Hamiltonian flow, but also its Poincaré map (Ref. 2) being symplectic. And, as eloquently explained in Cvitanović's Chaos Book:

for Hamiltonian flows the sum of $D$ oriented areas $\mathcal{V}_i$ bounded by $D$ loops $\Omega\mathcal{V}_i$, one per each $(q_i,p_i)$ plane, is conserved [...]. One can show that also the $4, 6, ..., 2D$ phase-space volumes are preserved.
[...]
The phase space is $2D$-dimensional, but as there are $D$ coordinate combinations conserved by the flow, morally a Hamiltonian flow is $D$-dimensional. Hence for Hamiltonian flows the key notion of dimensionality is $D$, the number of the degrees of freedom (dof), rather than the phase-space dimensionality $d=2D$.

So, all these conserved quantities allow for the dynamics on the Poincaré section to also be conservative.

One can probably gain a feeling for how exactly this happens by going through some proof of it. I didn't find any in a quick search, but Jason Frank's very readable lecture notes on Numerical Modelling of Dynamical Systems does that in one page (Ch. 16, p. 101) for the flow map.

is the Poincare return map only area preserving for the collection of "equal return points" $\{(q^k_1,p^k_1)\}$? [at different reintersection times]

Yes. Because that's the only time that exists for the map: After all, if we're talking about the Poincaré map being conservative or not, we are considering it as a dynamical system of its own - and then the only meaningful time is the discrete iteration number $k$.

$\endgroup$
  • $\begingroup$ Thanks for the answer and the references. I'll check this out this week. $\endgroup$ – ZeroTheHero Nov 26 '18 at 18:18
0
$\begingroup$

Typically, a Poincare map $P$ of a smooth dynamical system, given by a vector field $X(x)$ on a manifold $M$, is constructed on a codimesnion 1 embedded transverse cross-section $\Sigma$ to a periodic trajectory $\gamma_0$ (orbit) of the system. Then, the Poincare map $P : \Sigma \to \Sigma$, also called first-return map, is a local diffeomorphism on the cross-section. To construct it:

  1. take any point $x$ on the cross-section $\Sigma$, close enough to the special point $x_0 = \gamma_0 \,\cap \, \Sigma$, where the periodic orbit punctures the cross-section;

  2. take the unique trajectory $\gamma_x$ of the vector field $X(x)$ that passes trough $x$ and follow it in the direction of $X$ until $\gamma_x$ intersects $\Sigma$ for the first time, after $x$. This intersection point is denoted $P(x)$ and is the image of $x$ under the Poincare map.

The trajectory $\gamma_x$ is considered as unparametrized curve, so the orbits starting from $\Sigma$ return again there at different times, but these times actually depend smoothly on the choice of $x \in \Sigma$, i.e. there is a smooth function $t(x)$ on $\Sigma$ such that $t(x)$ is the time that the point $x$ needs to go along $\gamma_x$ and return to $\Sigma$ as $P(x)$. For some systems, $t(x)$ may happen to be a constant, but in general it is not (especially for most non-linear systems).

Now, if the vector field $X(x)$ is Hamiltonian with function $H(x)$ on a symplectic manifold $M$ with a symplectic two-form $\omega$, then $\Sigma$ is odd dimensional and cannot be a symplectic submanifold of $M$, so the volume form $\Theta = \omega \wedge \omega \wedge ... \wedge \omega$ is not a volume form when restricted to $\Sigma$. However, the restriction of $\omega$ on $\Sigma$, although not a symplecitc form, defines Poisson structure on $\Sigma$, i.e. $\Sigma$ is a Poisson manifold and the intersetions of the level surfaces of the Hamiltonain $H$ with $\Sigma$ define a codimesion one symplectic foliation on $\Sigma$, which is invariant under the poincare map $P$. Now, on each of these symplectic leaves, $P$ is s symplectomorphism and is hence area preserving. So basically, you end up with a Poincare map

$$P : \Sigma \, \cap \, \{H=const\} \to \Sigma \, \cap \, \{H=const\} $$ and on top of that $\Sigma \, \cap \, \{H=const\}$ together with the symplectic form $\omega$ restricted to $\Sigma \, \cap \, \{H=const\}$ is a symplectic submanifold of codimesnion 2 of $M$. As $P$ is symplectic on $\Sigma \, \cap \, \{H=const\}$, it is area preserving.

There is another interpretation, where the situation is more straight-forward. If you have a periodic time-dependent Hamiltonain function $H(x,t)$ on the symplectic manifold $M$, say $H(x, t+T) = H(x,t)$, you look at the extended vector field $$J\nabla_x H(x,t) + \frac{\partial }{\partial t}$$ defined on $M \times \big( \mathbb{R}\, /\, T\mathbb{Z} \big)$ and then the cross-section is simply $M\times \{0\} \equiv M \times \{T\}$ so the Poincare map is simply $$P = \Phi^T\big|_{M\times \{0\} } \, : M\times \{0\} \, \to \, M\times \{T\} $$ where $\Phi^s(x,t)$ is the phase flow of the vector field $J\nabla_x H(x,t) + \frac{\partial }{\partial t}$. Thus, as the flow $\Phi^s(x,t)$ preserves the symplectic form on $M$, the map $P$ is a symplectomorphism, and is therefore volume preserving.

An example is a forced pendulum like this one: \begin{align} &\frac{d x}{ dt} = p\\ &\frac{dp}{dt} = - \sin(x) + \cos(t) \end{align} with Hamiltonain $H = \frac{1}{2} p^2 - \cos(x) - x\,\cos(t)$. Then the extended vector field is \begin{align} &\frac{d x}{ ds} = p\\ &\frac{dp}{ds} = - \sin(x) + \cos(t)\\ &\frac{dt}{ds} = 1 \end{align}

$\endgroup$

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.