1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Liouville integrability

  1. Jun 27, 2014 #1

    Let a classical particle with unit mass subjected to a radial potential V and moving in a plane.

    The Hamiltonian is written using polar coordinates [itex](r,\phi)[/itex]

    [itex]H(r,\phi) = \frac{1}{2}(\dot{r}^2+r^2\dot{\phi}^2) - V(r)[/itex]

    I consider the angular momentum modulus [itex]C=r^2\dot{\phi}[/itex],
    and I want to show that the system is Liouville integrable (the problem is planar so I have to find a first integral (that is ,C) which commute to the Hamiltonian).

    My question is : when I want to compute the Poisson bracket [itex]{H,C}[/itex], the only variable is r and [itex]\phi[/itex] ? Because the conjuguate variables [itex]\dot{r}[/itex] and [itex]\dot{\phi}[/itex] appeared in H...
    So I should write [itex]H(r,\phi,\dot{r},\dot{\phi})[/itex] and compute the partial derivative with respect to these four variables ?

    Last edited: Jun 27, 2014
  2. jcsd
  3. Jun 28, 2014 #2
    somes explanations ?
  4. Jun 30, 2014 #3
    up up up
  5. Jun 30, 2014 #4


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    In Hamiltonian dynamics you must substitute the generalized velocities with the canonical momenta. Otherwise the formalism doesn't work properly. There's an alternative formalism, where you only perform the Legendre transformation from the Lagrange to the Hamilton formalisms for some generalized coordinates, which goes under the name of Routh, but I've never seen an application of it in practice (if you are interested in that, see A. Sommerfeld, Lectures on Theoretical Physics, Vol. 1).

    In your case the Lagrangian should read
    [tex]L=\frac{m}{2} (\dot{r}^2+r^2 \dot{\phi}^2) - V(r).[/tex]
    The canonically conjugated momenta thus are
    [tex]p_r=\frac{\partial L}{\partial \dot{r}}=m \dot{r}, \quad p_{\phi}=\frac{\partial L}{\partial \dot{\phi}}=m r^2 \dot{\phi}.[/tex]
    The Hamiltonian now is
    [tex]H(q,p)=\dot{q} \cdot p-L=\frac{m}{2} (\dot{r}^2+r^2 \dot{\phi}^2) +V(r) = \frac{1}{2m} \left (p_r^2 +\frac{p_{\phi}^2}{r^2} \right ) + V(r).[/tex]
    Now you have two first integrals:

    (a) the Hamiltonian doesn't depend explicitly on time. Thus it is conserved:
    (b) the variable [itex]\phi[/itex] is cyclic, i.e., the Hamiltonian doesn't depend on it. Thus the canonical momentum conjugate to this variable is conserved too:
    Thus you have two integral of motion for a system with two degrees of freedom. It is thus integrable.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook