Studying analytical mechanics I encountered Liouville's theorem which states:

In phase spase, the Hamiltonian flow preserves volumes

The book I'm studying is Analytical Mechanics - A. Fasano, S. Marmi. They give the following proof which I'm having trouble to follow.

First of all they give the following definition for the canonical Hamilton's equations: $$\dot{\mathbf{x}} = \mathcal{J}\nabla_x \mathcal{H} = \mathbf{X}(\mathbf{x},t)$$ where $$\begin{align}&\dot{\mathbf{x}} = \left(\begin{matrix}\mathbf{p}\\\mathbf{q}\end{matrix}\right)&\mathcal J = \left(\begin{matrix}\mathbf{0}&-\mathbb{I}\\\mathbb{I}&\mathbf{0}\end{matrix}\right)\end{align}$$ and they let us notice that

If $\mathcal{H}$ has two continuous second derivatives with respect to $q_k$ and $p_k$, then $$\nabla\cdot\mathbf{X}(\mathbf{x},t) = 0$$ which clearly follows from Schwartz's theorem.

Then they proceed to give the proof for Liouville's theorem: we need to show that, fora any $t>0$, the image $\Omega(t)$ of any domain $\Omega\subset\mathbb{R}^{2l}$ with a regular boundary has the same measure as $\Omega$. Consider the flow associated with any system of the type $$\dot{\mathbf{x}} = \mathbf{X}(\mathbf{x},t)$$ Then we have that the variation of volume in time $dt$ can be expressed as $$d|\Omega(t)|=\int_{\partial\Omega(t)} \mathbf{X}\cdot\hat{n}\,d\sigma\,dt$$ so that $$\frac{d|\Omega(t)|}{dt} = \int_{\partial\Omega(t)}\mathbf{X}\cdot\hat{n}\,d\sigma = \int_{\Omega}\nabla\cdot\mathbf{X}\,d^{2l}\mathbf{x}=0$$ where $|\Omega(t)|$ is the measure of $\Omega(t)$.

My problem with the proof is the bold statement. Why we can express the change in volume as that integral with the phase flow?. Is it because, in an intuitive way (in two dimensions), as the domain moves along the phase lines the only volume we get is the one defined by the lateral surface drawn by the phase lines in that time $dt$ around $\Omega$? I'm a little bit confused.

P.S. I haven't posted this on Physics stack exchange because my problem seems to be more mathematical than physical.


This is the physicist’s way to motivate that step:

Consider a small piece $d\sigma$ of the boundary $\partial \Omega(t)$, which in time $dt$ evolves to another small piece $d\sigma’$ of the boundary $\partial \Omega(t+dt)$. The evolution of $d\sigma$ into $d\sigma’$ is dictated by the value of the field $\mathbf X$ on $d\sigma$, and approximately defines a very small parallelepiped of base $d\sigma$ and height $\mathbf X\cdot \hat n\ dt$ – to see this, draw the surface element $d\sigma$, its normal unit vector $\hat n$, and the vector $\mathbf X$ which points toward $d\sigma’$. Hence, the volume of the small parallelepiped is $\mathbf X \cdot \hat n\ d\sigma\ dt$. To get the complete variation of the volume, you integrate this expression over the whole lot of $\partial \Omega(t)$.

  • 1
    $\begingroup$ This was very useful! Thanks! $\endgroup$ – Davide Morgante Aug 22 '18 at 11:02
  • $\begingroup$ Glad I could be of help! $\endgroup$ – giobrach Aug 22 '18 at 11:16
  • $\begingroup$ Posso anche ringraziarti in italiano già che ci siamo :) $\endgroup$ – Davide Morgante Aug 22 '18 at 13:05

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.