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Liouville's theorem states that phase space volume is conserved over time with respect to the dynamical system generated by the Hamiltonian and Hamilton's equations.

However, any given point in phase space will evolve within a submanifold characterized by certain values of the conserved quantities (energy, momentum,...).

It's not obvious to me that the "phase volume" within this submanifold is also conserved over time, since it is a volume of lower dimension than that of tbe phase space.

Is there a result here that you could point me to?

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    $\begingroup$ en.wikipedia.org/wiki/Non-squeezing_theorem maybe is a bit overkill. But the point is this result stems from symplectic geometry. $\endgroup$ Jul 10, 2020 at 10:52
  • $\begingroup$ @guillaumeTrojani, I've tried reading about symplectic geometry, but I don't yet get how it relates to hamiltonian mechanics. What is the symplectic form here? $\endgroup$
    – user56834
    Jul 10, 2020 at 11:02
  • $\begingroup$ So in the simplest case (or locally if you prefer) the symplectic form is canonical i.e: $\omega = dp^i \wedge dq_i$. But if you prefer, this form is what "generates" the poisson structure. I can try to give you a full fledge answer but I suspect someone will come along before (and probably do a better job at it) $\endgroup$ Jul 10, 2020 at 15:20
  • $\begingroup$ (To be clear, I'm not sure what $dp\land dq$ means, and I've asked the question here: physics.stackexchange.com/q/564834). Is it different from the double integral $\int f dpdq$? $\endgroup$
    – user56834
    Jul 10, 2020 at 15:59
  • $\begingroup$ No worries, it is an advanced subject for sure. I think you received very good answers on the other post actually. But it is a difficult subject to first encounter. And about the non-squeezing theorem, i myself don't understand (the proof of) it because as far as i know it uses a technic from algebraic geometry called witten-gromov invariant, so like i said massive overkill to your question. If i had a better answer, i would have written a post. Sorry! $\endgroup$ Jul 10, 2020 at 16:14

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In order to ask if phase volume on the submanifold is conserved, we first need to define phase volume on the submanifold. It's not obvious how to do this - the symplectic form might vanish on the submanifold, or the submanifold might even by odd-dimensional, so we aren't guaranteed to get a natural volume measure from the symplectic form. A better question is "can we define phase volume on a submanifold such that Liouville's theorem holds?

Defining a volume measure over a submanifold is equivalent to defining integration over that submanifold. For Riemannian manifolds, we usually do this by integrating over an $\epsilon$-thickening of the submanifold, then taking the limit as $\epsilon \rightarrow 0^+$. For a sympletic manifold, an $\epsilon$-thickening doesn't make sense, since there's no notion of distance. However, we can sometimes do something similar using orbits. Fortunately, we don't care about defining volume on an arbitrary submanifold. We care about defining volume on the orbit of of some initial point under the Hamiltonian flow.

Let $p$ be the initial point we care about, and let $M$ be the original manifold. Let $U \subset M$ be a neighborhood of $p$. $\dim U = \dim M$, so we know how to integrate over $U$. We also know how to integrate over the orbit of $U$. To integrate over the orbit of $p$, we can integrate over the orbit of $U$, then divide by $\int 1 $ and take the limit as $U$ shrinks to $p$. This integration gives a well-defined volume measure on the orbit of $p$. With respect to this volume measure, Liouville's theorem is satisfied.

Exercises for the reader:

  • Show that the volume measure really is well-defined (i.e. the limit exists)
  • Show that it satisfies Liouville's theorem
  • On further thought, it's not actually obvious to me that the orbit of $U$ always has a well-defined dimension. Are there Hamiltonian systems with fractal orbits?
  • If we have two different Hamiltonians on $M$ with the same orbits, will the associated volume measures be the same? I don't know the answer to this one either.
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