Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Assume we have a Hamiltonian system on $(\mathbb{R}^{2n},\omega)$ with Hamiltonian $H = H(q,p)$. In a paper I read, it says, without clarification, that the natural Liouville measure $\mu$ obtained by the volume form $\Omega = \omega \wedge \cdots \wedge \omega $ restricts on the energy level surfaces $N = \left\{(q,p) \in \mathbb{R}^{2n}: H(q,p) = c\right\}$ to

$$\iota(F(q,p))\Omega$$

where F(q,p) is a function that satisfies $\omega(X_{H},F(q,p)) = dH(F(q,p)) = 1$ and $\iota(\cdot)$ is the interior product. A possible choice would be $F(q,p) = \frac{grad H}{\|grad H\|}$.

Why is this so?

Thanks in advance!

share|improve this question
    
It's often helpful to look at the original source; could you give a reference to the paper? –  user53153 Feb 11 '13 at 1:21
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.