MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Liouville's theorem states that the `natural' 2-from is preserved under the Hamiltonian flow. Apparently this leads to an invariant measure $\mu$ as follows

\begin{equation} d\mu = \frac{d\sigma}{|| \nabla H ||} \end{equation} where $H$ is the Hamiltonian and $d\sigma$ an infinitesimal standard volume element. I have never quite understood how one arrives at this and why you cannot consider the 2-form of the Hamiltonian as a measure. The books I have seen up to now seem to label this as trivial but I do not see the connection. Could anybody explain this?

Thanks in advance.

share|cite|improve this question
This was asked without a satisfactory answer also at… – András Bátkai Aug 23 '13 at 10:08
I know that was me. I even put a bounty on it to attract attention but it remained unanswered. :( – Novo Aug 23 '13 at 10:45
In order to define a quantity such as $\|\nabla H\|$, you'd need to have a norm of some kind on (co-)vectors and you'd need to know what the 'standard volume element' is. However, in the general case of a Hamiltonian on a manifold, you don't have those quantities defined, so I don't know what your formula means. The usual invariant volume measure $d\mu$ (as a volume form) is more simply defined directly using the $2$-form $\omega$, by the formula $n!\ d\mu = \omega^n$, which doesn't use any metrics or 'standard volume forms'. – Robert Bryant Aug 23 '13 at 11:40
2-form is not a measure, except in the space of dimension 2. A measure is a volume form, that is an n-form in the space of dimension n. – Alexandre Eremenko Aug 24 '13 at 4:39

You can find a derivation of that formula in Section 7 of Khinchin's book and in Chapter 8 of Pettini's book.

I hope this helps.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.