Conservative mechanical system is Hamiltonian

Question

Given a conservative mechanical system written in (for example) Newtonian form:

$m \ddot x= -\nabla U(x)$ (so the potential has only a "positional" dependence)

then we know that given the Lagrangian: $L(x,\dot x)=T-U \ $ then the admissible motions for the system are given by extremal points of the action of $L \ $.

Now, if we take the Hamiltonian of the Lagrangian (i.e. its Legendre transform) then we have the Hamilton's equations. I have three questions:

1. Is it true that the admissible motions for the system are given by the solutions of Hamilton's equations?
2. If $1$ is true, then is also true that every (Newtonian) conservative system is also volume preserving?
3. This (for now only hypotetic) correspondence between Hamiltonian's solutions and admissible motions, is still valid if the potential is also velocity-dependent or time-dependent (i.e. $U=U(x,\dot x, t)$)?

P.s.I know that probably these are easy questions, but I have them because I've read Liouville preservation volume theorem and also I've read that the Hamiltonian satysfies the hypotesis of Liouville's theorem, and so I am quite impressed that every conservative mech. syst. is volume preserving.

Thank you in advance

Answer 1

1. True.

2. True if you are referring to the canonical volume in the space oh phases. This is the celebrated Liouville theorem.

3. True if the (generalized) potential has a certain form as it happens in particular when dealing with Lorentz force and inertial forces. In particular the potential has to be linear in the velocities.

Obviously the form of the final equations is not $m\ddot{x}= -\nabla V$ but it is more complicated. As a physically important example, consider

$$U(t, x, \dot{x})= e \varphi(t,x) - \frac{e}{c} A(t, x) \cdot \dot{x}$$

inserted in the Lagrangian $L= T-U$ gives rise to the Lorentz force acting on the charge $e$ with position vector $x\in \mathbb R^3$ and velocity $\dot{x}$

$$m \ddot{x} = -\frac{e}{c}\frac{\partial A}{\partial t}(t,x) - e \nabla \varphi(t,x) + \frac{e}{c} \dot{x} \times (\nabla_x \times A(t,x))$$

namely

$$m \ddot{x} = eE(t,x) + \frac{e}{c} B(t,x)\times \dot{x}\:.$$ These coincide with the standard Euler-Lagrange equations

$$\frac{d}{dt} \nabla_{\dot{x}}L(t,x, \dot{x}) - \nabla_x L(t, x, \dot{x})=0$$

In other similar cases, there is a function, called generalized potential, $U=U (t, x, \dot {x}) $ such that $L = T - U $ inserted into Euler-Lagrange gives rise to the correct equation of motion.

There is no obstruction to pass to the Hamiltonian formulation in the usual way, and Liouville theorem holds as well, since it does not depend on the form of the Hamiltonian, just it must exist.