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!

What is the Hamilton-Jacobi equation

  1. Jul 24, 2014 #1

    The Hamilton-Jacobi equation is an additional reformulation of classical mechanics, one that uses a function S variously called the action, the generating function, and Hamilton's principal function.

    It is very convenient for problems where one can separate the variables, and it is also convenient for expressing constants of the motion.


    The generating function S is a function of canonical coordinates [itex]q_a(t)[/itex] and constants of the motion [itex]\alpha_a[/itex]. The canonical momenta [itex]p_a(t)[/itex] are given in terms of it by
    [itex]p_a = \frac{\partial S}{\partial q_a}[/itex]

    and some additional constants of the motion [itex]\beta_a[/itex] by:
    [itex]\beta_a = \frac{\partial S}{\partial \alpha_a}[/itex]

    Plugging these canonical-momentum values into the Hamiltonian, we find the Hamilton-Jacobi equation:
    [itex]\frac{\partial S}{\partial t} + H = 0[/itex]

    Extended explanation

    To demonstrate what the Hamilton-Jacobi equation can be used for, here is a solution of the central-force problem with potential V(r) using it, complete with separation of variables. We start out with its Lagrangian
    [itex]L = \frac12 m \left[ \left( \frac{dr}{dt} \right)^2 + r^2 \left( \frac{d\theta}{dt} \right)^2 \right] - V[/itex]
    Its canonical momenta are
    [itex]p_r = m \frac{dr}{dt}[/itex]

    [itex]p_\theta = m r^2 \frac{d\theta}{dt}[/itex]

    and thus its Hamiltonian is
    [itex]H = \frac{(p_r)^2}{2m} + \frac{(p_\theta)^2}{2mr^2} + V[/itex]

    The Hamilton-Jacobi equation becomes
    [itex]\frac{\partial S}{\partial t} + \frac{1}{2m}\left( \frac{\partial S}{\partial r} \right)^2 + \frac{1}{2mr^2}\left( \frac{\partial S}{\partial \theta} \right)^2 + V = 0[/itex]

    Aside from the partial derivatives, this equation has explicit dependence only on r. This suggests a decomposition of S using constants of the motion E (energy) and L (angular momentum):
    [itex]S = - Et + L\theta + S'(E,L,r)[/itex]

    We easily find S' from it:
    [itex]S' = \int \left[ 2m(E - V) - \left( \frac{L}{r} \right)^2 \right]^{1/2} dr[/itex]

    and the remaining constants of the motion, which are zero points:
    [itex]-t_0 = - t + \frac{\partial S'}{\partial E}[/itex]
    [itex]\theta_0 = \theta + \frac{\partial S'}{\partial L}[/itex]

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted