Reversibility of Hamiltonian dynamics

Question

I'm trying to understand a very basic property of Hamiltonian dynamics. I don't have a physics background but I do know some mathematics. I want to understand why negating the momentum is equivalent to reversing time in Hamiltonian dynamics.

Suppose I have a Hamiltonian $H(q, p)$ which satisfies $H(q, p)=H(q, -p)$. The Hamiltonian equations of motion are $$ \dot{p} = -\nabla_q H(q, p) ~~~~~~~~ \dot{q} = \nabla_p H(q, p) $$ That if $(q(t), p(t))$ satisfy Hamilton's equations then so does $(q(-t), -p(-t))$ seems to be an oft-quoted fact in many works on classical mechanics. But how do I convince myself that it is true?

Answer 1

If we set $p\rightarrow -p$ then the Hamilton equations look like $$-\dot p=-\nabla_qH(q,p)~~~~~~~~~~~\dot q=\nabla_{-p}H(q,p)=-\nabla_pH(q,p)$$ Now we use the fact that $-\dot q=-\partial q/\partial t=\partial q/\partial (-t) $ to get the modified equations that look like $$\frac{\partial p}{\partial(-t)}= -\nabla_qH(q,p)~~~~~~~~~~~\frac{\partial q}{\partial(-t)}=\nabla_pH(q,p) $$ These are the same equations albeit $t\rightarrow-t$. Thus we see that Hamilton’s equations are symmetric under time reversal and also that this can be brought about by setting $p\rightarrow-p$.