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Question about the Hamilton Jaccobi Equation

  1. Jul 23, 2014 #1
    Hi, I was wondering about the interpretation of the Hamilton Jaccobi equation.

    Naively we have H + [itex]\partial[/itex]S/[itex]\partial[/itex]t = 0 where H is the Hamiltonian and S is the action. But the action is the time integral of the Lagrangian so you would expect [itex]\partial[/itex]S/[itex]\partial[/itex]t = L

    Thus H + L = KE + PE + KE - PE = 2KE = 0

    (PE = potential energy, KE = Kinetic energy)

    Which simply says that then system is not moving which seems odd for something billed as an equation of motion. Something is clearly very wrong here but what?
     
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  3. Jul 24, 2014 #2

    vanhees71

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    The idea is to find new canonical coordinates [itex](Q,P)[/itex] of phase space such that all trajectories are described by by [itex](Q,P)=\text{const}[/itex]. Thus you look for the corresponding canonical transformation from the original canonical coordinates [itex](q,p)[/itex]. You can easily show that up to "gauge invariance" this is achieved by the demand that
    [tex]H'(t,Q,P)=0.[/tex]
    To that end we introducd the generating function [itex]g(t,q,P)[/itex] for this canonical transformation. The relation between the old and the new quantities is
    [tex]p=\partial_q g, \quad Q=\partial_P g, \quad H'(t,Q,P)=H(t,q,p)+\partial_t g =H \left(t,q,\partial_q g \right)+\partial_t g\stackrel{!}{=}0.[/tex]
    This is the Hamilton-Jacobi equation.

    For the case of a Hamilton function that is not explicitly dependent on time, the Hamiltonian is conserved along the possible trajectories of the system (energy conservation). Then from the Hamilton-Jacobi equation you get
    [tex]H(t,q,p)=E=-\partial_t g=\text{const} \; g=-E t + S(q,E,P_2,\ldots,P_f),[/tex]
    where we use [itex]E[/itex] as one of the new canonical momenta, and [itex]S[/itex] is not explicitly time dependent.

    Now we evaluate the action functional in terms of [itex]g[/itex]:
    [tex]A=\int_{t_1}^{t_2} \mathrm{d} t (\dot{q} \dot{p}-H)=\int_{t_1}^{t_2} \mathrm{d} t (\dot{q} \partial_q g+\partial_t g)=g_2-g_1,[/tex]
    where
    [tex]g_j=g(t=t_j,q(t_j),P).[/tex]
    Note that [itex]P=\text{const}[/itex] according to the construction of the new canonical coordinates. As you see, indeed [itex]g[/itex] is closely related to the action.
     
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