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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.

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This was asked without a satisfactory answer also at math.stackexchange.com/questions/433618/… –  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

1 Answer 1

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.

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