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Strange Hamilton Jacobi equation

  1. Dec 10, 2004 #1
    let be (dS/dt)+(gra(S))^2/2m+(LS)+V(x) where L is the Laplacian Operator and V is the potential...could it be considered as the Hamiltan Jacobi equation for a particle under a potential Vtotal=V(x)+(LS) where S is the action
  2. jcsd
  3. Dec 11, 2004 #2
    I assume you mean to equate to 0, i.e.:

    [tex] \frac{\partial S}{\partial t}\right)+\frac{(\vec\nabla S)^2}{2m}+\vec\nabla S+V(q)=0 [/tex]

    If we compare it to the Hamillton-Jacobi equation for the generating function S (a concept more general than the "action")

    [tex] H\left(q,\frac{\partial S}{\partial q},t\right)+\frac{\partial S}{\partial t}=0 [/tex]

    we find they are compatible provided we let

    [tex] H\left(q,\frac{\partial S}{\partial q},t\right)=\frac{(\vec\nabla S)^2}{2m}+\vec\nabla S+V(q) [/tex]

    Since [tex] p_i=\frac{\partial S}{\partial q_i} [/tex], we can rewrite it as

    [tex] H\left(q_i,p_i,t\right)=\frac{(\sum_i p_i)^2}{2m}+\vec\nabla S+V(q) [/tex]


    [tex] H=T+W [/tex]


    [tex] W=\sum_i p_i+V(q_i) [/tex].

    Here we see that [tex] W=f(p_i,q_i) [/tex], in other words the "potential" W is not conservative and the meaning of W is that of "virtual work". Is that the source of your doubts?
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