I will cast my question mostly in words. I usually find Liouville's volume theorem cast in two forms:

  1. Easy way (without differential forms language): Phase space volume remains preserved under Hamiltonian dynamics. Let's call this phase space volume the 'usual volume'.

  2. Sophisticated way (in the differential forms language): Volume 2-form does not change with the Hamiltonian flow.

Now, I have the impression that volume 2-form when acts upon displacement vectors, gives the 'usual volume' and hence volume 2-form is not the same as the 'usual volume'. So aren't the above two ways of stating Liouville's theorem not equivalent? The first talks about the 'usual volume' being unchanged and the second talks about the volume 2-form being unchanged. Please help me see how are these two statements of Liouville's theorem equivalent.

  • $\begingroup$ Have you checked the book by Vladimir Arnol'd? It should address this point. $\endgroup$
    – DanielC
    Jan 24, 2019 at 11:06
  • $\begingroup$ The confusion actually stemmed from reading Arnold's book. $\endgroup$ Jan 24, 2019 at 11:18

1 Answer 1


The symplectic 2-form is $\omega$ and for a $2n$-dimensional phase space, the phase space volume is $\Omega=\omega^n/n!$.
For $\omega= \sum dp\wedge dq$ this is $\Omega=\pm d^np d^nq$.

Liouville says that in a Hamiltonian flow $V$ where $$ dH= -i_V \omega $$ we have that the Lie derivative ${\mathcal L}_V \omega=0$. This follows from the infinitesimal homotopy formula $$ {\mathcal L}_V\omega = (di_V+i_vd)\omega $$ and the requirement that $d\omega=0$. The usual form of Liouville (volume of phase space conserved) is ${\mathcal L}_V \Omega=0$, but ${\mathcal L}_V$ is a derivation, so $$ {\mathcal L}_V\omega^n = n \omega^{n-1} ({\mathcal L}_V\omega)=0. $$ Thus the two-form staement implies the $2n$-form volume statement.


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