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Let $H$ the Hamiltonian of a system and $\gamma $ an integral curve of the Hamiltonian vector field, i.e. if $\gamma (t)=(q(t),p(t))$ and $H(p,q)$ is the Hamiltonian, then $$\begin{cases} \dot p=-H_q\\ \dot q= H_p\end{cases}.$$

In wikipedia the say that $H$ is constant along $\gamma $. Why is this true ? i.e. why $H(\gamma (t))$ is constant ?

user623855
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1 Answers1

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For $H(p,q,t) = H(p(t),q(t))$ think that

$$ \frac{d}{dt}H(p(t),q(t)) = H_p \dot p + H_q \dot q $$

Cesareo
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  • Indeed, it make sense :) But I'm confusion with wikipedia notation, what they mean by $<dH,\dot \gamma >=0$ ? Because $\omega $ is not a scalar product (is not a riemann manifold) – user623855 Jan 11 '19 at 16:03
  • It's the pairing by evaluation between a covector and a vector (or if you wish one-form on the curve and a vector field along the curve). – Max Mar 10 '19 at 21:41