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Poisson Brackets

  1. Nov 13, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]

    f(p(t),q(t)) = f_o + \frac{t^1}{1!}\{H,f_o\}+\frac{t^2}{2!}\{H,\{H,f_o\}}+...

    [/tex]
    Prove the above equality. p & q are just coords and momenta

    How do we do this if we don't know what H is?
    Where do we start?
    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 13, 2009 #2
    Looks like you may not need to know what the Hamiltonian is, just how the Hamiltonian works in the Poisson bracket (assume [itex]f[/itex] is not a constant of motion):

    [tex]
    \frac{df(q,p)}{dt}&=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial q}\frac{dq}{dt}+\frac{\partial f}{\partial p}\frac{dp}{dt}=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial q}\frac{\partial H}{\partial p}-\frac{\partial f}{\partial p}\frac{\partial H}{\partial q}=\frac{\partial f}{\partial t}+\{f,H\}
    [/tex]

    So, using this, it looks like you would expand [itex]f(q,p)[/itex] as a Taylor series. Note that if [itex]f(q,p)[/itex] is a constant of motion, [itex]f_{,t}=0[/itex] so that you have

    [tex]
    \frac{df(q,p)}{dt}=\{f,H\}
    [/tex]
     
  4. Nov 13, 2009 #3
    What's the theorem that says mixed partials are commutative? Not Claurait's....
     
  5. Nov 13, 2009 #4
    Also

    I know

    [tex]
    \frac{\partial }{\partial t}\{H,f\} = \{\frac{\partial H}{\partial t},f\} + \{H,\frac{\partial f}{\partial t}\}
    [/tex]

    What about for ordinary derivatives?
    [tex]
    \frac{d}{d t}\{H,f\} =?
    [/tex]
     
  6. Nov 14, 2009 #5

    gabbagabbahey

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    No, if [itex]f(q,p)[/itex] is a constant of motion, then [tex]\frac{df}{dt}=0[/itex]. The fact that [itex]f[/itex] has no explicit time dependence (it is given as a function of [itex]q[/itex] and [itex]p[/itex] only), tells you that [tex]\frac{\partial f}{\partial t}=0[/itex]
     
  7. Nov 14, 2009 #6

    gabbagabbahey

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    You tell us...expand the Poisson bracket and calculate the derivatives...what do you get?
     
  8. Nov 14, 2009 #7
    [tex]
    \frac{d}{dt}\frac{\partial f}{\partial q}\frac{\partial H}{\partial p}-\frac{d}{dt}\frac{\partial f}{\partial p}\frac{\partial H}{\partial q}
    [/tex]
     
  9. Nov 16, 2009 #8

    gabbagabbahey

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    Okay, now use the product rule...

    [tex]\frac{d}{dt}\left(\frac{\partial f}{\partial q}\frac{\partial H}{\partial p}\right)=[/itex]

    ???
     
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