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

  • $\begingroup$ 1. true, 3. false (for velocity dependence), 2. what you mean with volume preserving? $\endgroup$
    – Fabian
    Commented Sep 15, 2016 at 17:02
  • $\begingroup$ Thank you! If I have a system on N point-like masses under the action of a conservative force, then I can imagine the polygon formed by "connecting" every mass to the others, now I have an area (in E^2, in general a volume in E^n) at the time t=0 which evolves in a certain way. Can I say that this area (volume) is constant with the hypotesis of question 2? $\endgroup$
    – HaroldF
    Commented Sep 15, 2016 at 17:30
  • $\begingroup$ The volume conservation associated with Liouville's theorem is phase space volume. That doesn't sound like the volume you are talking about. Are you sure that you haven't misunderstood what you were reading? $\endgroup$ Commented Sep 15, 2016 at 18:29
  • $\begingroup$ @LewisMiller oh you're right! I was thinking about a way too exciting result! Yes, so the Hamiltonian preserves the volume in the phase space (so the space of generalized momenta and coordinates). Also the theorem couldn't be true in general (think about gravitational charges around a gravitational center). Now, can i know more about the fact that 3 is false? $\endgroup$
    – HaroldF
    Commented Sep 15, 2016 at 19:04

1 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 $\dot{x}$

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


$$m \ddot{x} = eE(t,x) + \frac{e}{c} B(t,x)\times \dot{x}\:.$$ 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.

  • $\begingroup$ I would be curious to know how the Lagrangian $L=T-U$ generates the equation of motion $m \ddot x = - \nabla U$ for a potential that is velocity dependent. (that is what the OP was asking about) $\endgroup$
    – Fabian
    Commented Sep 16, 2016 at 11:30
  • $\begingroup$ Well, obviously the equations have not the form $m\ddot{x} = -\nabla V$, but is more complicated, think of a charge in a given pair of scalar and vector electromagnetic potential. $\endgroup$ Commented Sep 16, 2016 at 14:02

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