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Classical Hamiltonian

  1. Dec 23, 2008 #1

    KFC

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    If the classical Hamiltonian is define as

    [tex]H = f(q, p)[/tex]

    p, q is generalized coordinates and they are time-dependent. But H does not explicitly depend on time. Can I conclude that the energy is conserved (even q, p are time-dependent implicitly)? Namely, if no matter if p, q are time-dependent or not, if H does not contains [tex]t[/tex] explicitly, I find that the Poisson bracket

    [tex]\left\{H, H\right\} \equiv 0[/tex]

    so the energy is conserved, right?

    But what about if H explicitly depend on time? According to the definition of Poisson bracket, [tex]\left\{H, H\right\} \neq 0[/tex] ????
     
    Last edited: Dec 23, 2008
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  3. Dec 23, 2008 #2

    jambaugh

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    Short answer is YES!

    Long answer. Once the hamiltonian is defined the the time evolution of any quantity is given by:

    [tex] \frac{d}{dt}F = [[H,F]] + \frac{\partial}{\partial t}F[/tex]
    where [[ ]] is the Poisson bracket and the partial derivative applies to explicit time dependence. Since:
    [tex] [[H,H]] = 0[/tex]
    and there is no explicit time dependence you get:
    [tex]\frac{d H}{dt} = 0[/tex]

    Expand this in terms of p and q time dependence and you'll get a relationship which becomes the Jacobi identity when you introduce Hamilton's equations.
     
  4. Dec 23, 2008 #3

    KFC

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    Thank you so much. I look up the evolution relation in some textbook, but it reads

    [tex] \frac{d}{dt}F = [[F,H]] + \frac{\partial}{\partial t}F[/tex]

    I wonder if I should exchange F, H in poisson bracket?
     
    Last edited: Dec 23, 2008
  5. Dec 23, 2008 #4

    jambaugh

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    No actually [tex] [[F,H]] = - [[H,F]][/tex]

    But there is a convention choice which is not quite uniform in the sign of the Poisson bracket. Its a matter of whether you define:

    [tex] [[A,B]] = \frac{\partial A}{\partial p}\frac{\partial B}{\partial q}-\frac{\partial B}{\partial p}\frac{\partial A}{\partial q}[/tex]
    vs.
    [tex] [[A,B]] = \frac{\partial A}{\partial q}\frac{\partial B}{\partial p}-\frac{\partial B}{\partial q}\frac{\partial A}{\partial p}[/tex]
    (note p and q are reversed.)

    To conform with your reference reverse my Poisson brackets. (I recommend you stick to convention and do this.)

    I prefer to (buck convention and) reverse the sign/order so that for canonical conjugate variables U and V:

    [tex] \frac{d}{dV} F = [[U,F]][/tex]

    This fits better with the conventions for generators in Lie algebras.
     
  6. Dec 23, 2008 #5

    KFC

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    Got u. Thanks a lot. X'mas
     
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