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Hamilton Jacobi Equation

  1. Jun 11, 2012 #1
    Hello! General Question about the H-J equation.
    What are the steps to be followed if we are in a conservative system?

    And while answering my question, please in the step after we find S, and when you derive S wrt alpha and place it equals to β. When is alpha Energy? When it is not? i.e is it only Energy?

    Thanks Guys.
     
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  3. Jun 12, 2012 #2

    vanhees71

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    The Hamilton-Jacobi partial differential equation (HJE) aims to find the generator [itex]g(q,P,t)[/itex] of a canonical transformation such that the trajectories are described by [itex](Q,P)=\text{const}[/itex]. This means that the new Hamiltonian must obey

    [tex]\partial_Q H'(Q,P,t)=\partial_P H'(Q,P,t)=0.[/tex]

    This means that the new Hamiltonian is a function of time only:

    [tex]H'(Q,P,t)=H(q,p,t)+\partial_t g(q,P,t)=f(t).[/tex]

    However, without any change of generality you can add an arbitrary function of time only to [itex]g[/itex], and thus you can as well demand that

    [tex]H'(Q,P,t)=H(q,p,t)+\partial_t g(q,P,t)=0.[/tex]

    Since now [itex]p=\partial_q g[/itex], you get

    [tex]H \left (q,\partial_q g,t \right ) + \partial_t g=0.[/tex]

    For a system with [itex]f[/itex] configuration degrees of freedom, any solution of this partial differential equation of a function of [itex]f+1[/itex] independent variables [itex]q[/itex] and [itex]t[/itex] must contain [itex]f+1[/itex] integration constants. One of these is trivial since it just adds a constant to [itex]H'[/itex] which is irrelevant. Thus you have [itex]f[/itex] integration constants that you identify with the new canonical momenta [itex]P[/itex]. The trajectories of the system are then given by

    [tex]Q^k=\frac{\partial g}{\partial P_k}=\text{const}.[/tex]

    If [itex]H=H(q,p)[/itex], i.e., if the Hamiltonian is not explicitly time dependent, due to Noether's theorem it's constant. That means

    [tex]H=E=\text{const}\; \Rightarrow\; \partial_t g=-E \; \Rightarrow \; g(t,q,P)=-E t+S(q,E,P_2,\ldots,P_f).[/tex]

    Here, we have chosen the energy value as one of the new canonical momenta, i.e., set [itex]P_1=E[/itex]. The corresponding new configuration variable then is

    [tex]Q^1=\partial_E g=-t+\partial_E S.[/tex]

    In a similar way you can simplify the task if one of the configuration variables is cyclical, i.e., if the Hamiltonian doesn't depend on it. Then the corresponding canonical momentum is conserved, and you can keep it as one of the new momenta.

    Note that, against the claim in Landau, Lifgarbages Vol. 1, in general there are not enough conservation laws to solve the HJE completely. That's the case only for integrable systems. For a more mathematical treatment of these issues see,

    V. Arnold, Classical Mechanics, Springer.
     
  4. Jun 12, 2012 #3
    Hello thanks for replying.

    I understood what you wrote but I kinda depend on another procedure.
    First I find LaGrangian
    Then I find the momenta from it
    After that, I find Hamiltonian
    Then place H(q,[itex]\partial[/itex]S[itex]_{o}[/itex]/[itex]\partial[/itex]q)
    Then substitute this in Hamiltonian
    After that we place the formula S(q,t)=So-Et
    Next we work to find So then S.. And so on..

    Do you see where am going? If yes, I want to ask why when we have to separate variable (to summation)
    And try to find S[itex]_{2}[/itex] (Where S[itex]_{o}[/itex]=S[itex]_{1}[/itex]+S[itex]_{2}[/itex] perhaps) Why do we cross out the Derivative squared of S[itex]_{1}[/itex] by dx[itex]^{2}[/itex]? And set the whole thing equals to alpha? Is this regular?
     
  5. Jun 13, 2012 #4

    vanhees71

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    I don't understand precisely what you mean. The idea is of course to solve the HJE for the given problem as efficiently as possible. If you have a cyclic variable you keep the conserved momentum of it as a new momentum. Perhaps it's most efficient to work out an example. Perhaps you could take the Kepler problem with a fixed center as the most simple example.

    We simplify it further by putting in that the motion must be in a plane perpendicular to angular momentum. We use polar coordinates in that plane. The Lagrangian reads

    [tex]L=T-V=\frac{m}{2} (\dot{r}^2+r^2 \dot{\varphi}^2)+\frac{K}{r}.[/tex]

    The canonical momenta are

    [tex]p_r=\frac{\partial L}{\partial \dot{r}}=m \dot{r}, \quad p_{\varphi}= \frac{\partial L}{\partial \dot{\varphi}}=m r^2 \dot{\varphi}.[/tex]

    Because the Lagrangian is homogeneous of degree 2 in the time derivatives of the position variables, the Hamiltonian is

    [tex]H=T+V=\frac{p_r^2}{2m} + \frac{l^2}{2m^2 r^2} - \frac{K}{r}.[/tex]

    You immediately see that [itex]\varphi[/itex] is cyclical, i.e., the corresponding canonical momentum, [itex]l[/itex] is conserved, und you keep it as one of the new canonical momentum variables. Since the Hamiltonian is not explicitly dependent on time, the energy is conserved, which you take as another new canonical momentum. Thus the system is integrable, because there are two conserved quantities, which we take as the new canonical momenta. Thus your ansatz for the solution of the HJE reads

    [itex]S(r,\varphi,E,l)=-E t + l \varphi +\tilde{S}(r,E,l).[/itex]

    Now you insert this ansatz into the HJE and solve for [itex]\tilde{S}[/itex]. I hope this makes the whole procedure clear.
     
  6. Jun 14, 2012 #5
    Thank you really. Thank you very much.
     
  7. Jun 14, 2012 #6

    vanhees71

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    You are welcome :-).
     
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