How to show that the Hamiltonian $H$ is invariant under flow generated by $F$?

Question

I know usually if I have a transformation of phase space $Q(p,q), P(p,q)$ it is defined to be canonical if and only if its Jacobi matrix is part of the Symplectic Group or equivalently $\{Q^{i}, P_j \}= \delta^{i}_j, \{P_{i}, P_j \}=0, \{Q^{i}, Q^{j} \}= 0 $ (please correct me if some of this is wrong).

Now I have to show that if $F$ is a conserved quantity then the Hamiltonian is invariant under the flow:

$$ \psi^{\lambda}(q^{\alpha},p_{\alpha})=((q^{\lambda})^{\alpha}(p,q), (p^{\lambda})_{\alpha}(p,q)) $$ Characterised by:

$$ \frac{d (q^{\lambda})^{\alpha}}{d \lambda}=\frac{\partial F}{\partial p_{\alpha}}, \ \frac{d (p^{\lambda})^{\alpha}}{d \lambda}=-\frac{\partial F}{\partial q_{\alpha}} $$

If $F$ Is conserved we have $\{F,H \}=0$, I could probably somehow use this fact and the definition of canonical via the poisson bracket I stated above, but I don’t know how to deal with this flow thing and how to interpret $\frac{d}{d \lambda} $ in this context, so I‘d be happy if someone you’ll help me out :)

Answer 1

It follows directly from the poisson bracket relation $\{{F,H}\}=-\{{H,F}\}=0$

This says that $F$ is invariant under the flow of $H$ and $H$ is invariant under the flow of $F$. If you want an explicit calculation then:

$$\dfrac{dH}{d\lambda}=\dfrac{\partial H}{\partial q}\dfrac{dq}{d\lambda}+\dfrac{\partial H}{\partial p}\dfrac{dp}{d\lambda}=\dfrac{\partial H}{\partial q}\dfrac{\partial F}{\partial p}-\dfrac{\partial H}{\partial p}\dfrac{\partial F}{\partial q}= \{{H,F}\}=0$$

Note that $\dfrac{dH}{d\lambda}$ is exactly the rate of change of $H$ along the flow of $F$.