I read these two statement in the notes of my teacher that seem to me opposing. Let $H$ an Hamiltonian and let $\Phi$ an integral of motion of $H$, so that $\Phi$ keeps constant value along the orbit of $H$.
The two statements are the following:
The flow of $\Phi$ sends orbits of $H$ into orbits of ˆ$H$.
The Hamiltonian is invariant under the flow generated by $\Phi$.
My question is: how can the Hamiltonian be costant along the flow of $\Phi$ if the flow of $\Phi$ send the orbits (that is the level curve of $H$) into other orbits? Where am I wrong?
I think you are talking about Hamiltonian in symplectic geometry:
Let $(M,\omega)$ be a symplectic manifold, and $H:M\rightarrow\mathbb{R}$ a differentiable function, the Hamiltonian vector field associated to $H$ is defined by $\omega(X_H,.)=dH$.
Let $\phi_t$ the flow of $X_H$, ${d\over{dt}}H(\phi_t)=dH.X(\phi_t(x))=\omega(X_H(\phi_t(x),X_H(\phi_t(x))=0$ since $\omega$ is alternated, we deduce that $H$ is constant along its orbits.
The orbit of $x$ is $\phi_t(x),t\in\mathbb{R}$ which is invariant by $\phi_t$, $\phi_t$ preserves each orbit.