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Poisson brackets.

  1. Sep 19, 2007 #1
    [SOLVED] Poisson brackets.

    1. The problem statement, all variables and given/known data

    Show that, if Poisson brackets (g,h) = 1, then (g[tex]^{n}[/tex],h) = ng[tex]^{n-1}[/tex]
    where g = g(p,q) and h = h(p,q)

    p and q are canonical coordinates

    3. The attempt at a solution
    I suppose that this is purely mathematical, but I am still searching for a detailed example in literature.
    I also would like to ask - what book/author can you recommend, where alike problem is discussed.

    Thank You!
    P.S. I tried search function, but found nothing similar.

    Solutions:

    (g,h) [tex]\equiv[/tex] [tex]\frac{\delta g}{\delta q}[/tex][tex]\frac{\delta h}{\delta p}[/tex] - [tex]\frac{\delta g}{\delta p}[/tex][tex]\frac{\delta h}{\delta q}[/tex] = 1

    so, from here we have

    (g[tex]^{n}[/tex],h) [tex]\equiv[/tex] [tex]\frac{\delta g^{n}}{\delta q}[/tex][tex]\frac{\delta h}{\delta p}[/tex] - [tex]\frac{\delta g^{n}}{\delta p}[/tex][tex]\frac{\delta h}{\delta q}[/tex]

    => ng[tex]^{n-1}[/tex][tex]\frac{\delta g^{n}}{\delta q}[/tex]*[tex]\frac{\delta h}{\delta p}[/tex] - ng[tex]^{n-1}[/tex][tex]\frac{\delta g}{\delta p}[/tex]*[tex]\frac{\delta h}{\delta q}[/tex] =>

    ng[tex]^{n-1}[/tex]([tex]\frac{\delta g}{\delta q}[/tex][tex]\frac{\delta h}{\delta p}[/tex] - [tex]\frac{\delta g}{\delta p}[/tex][tex]\frac{\delta h}{\delta q}[/tex])
    and the part in brackets is = 1 as we know from given Poisson bracket =>

    (g[tex]^{n}[/tex],h) [tex]\equiv[/tex] ng[tex]^{n-1}[/tex]


     
    Last edited: Sep 19, 2007
  2. jcsd
  3. Sep 19, 2007 #2

    nrqed

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    What have you already seen on Poisson brackets? I could give you the answer but it will make more sense to you if you can get the answer from what you have learned.

    First of all, have you seen the definition of a PB?
     
  4. Sep 19, 2007 #3
    Yes, I have seen the definition of PB, and brief explanation of its properties.
    Should I involve partial integration by time?
    I will try to work out what (g^2,h) will give.
     
  5. Sep 19, 2007 #4

    nrqed

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    There are no integrations involved, just partial derivatives.

    Actually, I take back what I wrote. If you know the definition, just plug in g^n and you will directly get the final answer!!
     
  6. Sep 19, 2007 #5
    The important thing is that the Poisson bracket is like a derivative operator. Use the Leibnez rule.
     
  7. Sep 19, 2007 #6
    Well, here is what I have:

    (g,h) [tex]\equiv[/tex] [tex]\frac{\delta g}{\delta q}[/tex][tex]\frac{\delta h}{\delta p}[/tex] - [tex]\frac{\delta g}{\delta p}[/tex][tex]\frac{\delta h}{\delta q}[/tex] = 1

    so, from here we have

    (g[tex]^{n}[/tex],h) [tex]\equiv[/tex] [tex]\frac{\delta g^{n}}{\delta q}[/tex][tex]\frac{\delta h}{\delta p}[/tex] - [tex]\frac{\delta g^{n}}{\delta p}[/tex][tex]\frac{\delta h}{\delta q}[/tex]

    => ng[tex]^{n-1}[/tex]*[tex]\frac{\delta h}{\delta p}[/tex] - ng[tex]^{n-1}[/tex]*[tex]\frac{\delta h}{\delta q}[/tex] (??)





    I am actually ashamed of being unable to do such an easy task, which is basically depends on elementary calculus.
     
  8. Sep 19, 2007 #7

    nrqed

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    You are applying the chain rule, so you should have written, for example,
    [tex] \frac{\delta g^n}{\delta q} = n g^{n-1} \frac{\delta g}{\delta q} [/tex] (you forgot the delta g/delta q)
    and the same for the derivative with respect to p.

    Then you will see that it works out.

    Patrick
     
  9. Sep 19, 2007 #8
    (g,h) [tex]\equiv[/tex] [tex]\frac{\delta g}{\delta q}[/tex][tex]\frac{\delta h}{\delta p}[/tex] - [tex]\frac{\delta g}{\delta p}[/tex][tex]\frac{\delta h}{\delta q}[/tex] = 1

    so, from here we have

    (g[tex]^{n}[/tex],h) [tex]\equiv[/tex] [tex]\frac{\delta g^{n}}{\delta q}[/tex][tex]\frac{\delta h}{\delta p}[/tex] - [tex]\frac{\delta g^{n}}{\delta p}[/tex][tex]\frac{\delta h}{\delta q}[/tex]

    => ng[tex]^{n-1}[/tex][tex]\frac{\delta g^{n}}{\delta q}[/tex]*[tex]\frac{\delta h}{\delta p}[/tex] - ng[tex]^{n-1}[/tex][tex]\frac{\delta g}{\delta p}[/tex]*[tex]\frac{\delta h}{\delta q}[/tex] =>

    ng[tex]^{n-1}[/tex]([tex]\frac{\delta g}{\delta q}[/tex][tex]\frac{\delta h}{\delta p}[/tex] - [tex]\frac{\delta g}{\delta p}[/tex][tex]\frac{\delta h}{\delta q}[/tex])
    and the part in brackets is = 1 as we know from given Poisson bracket =>

    (g[tex]^{n}[/tex],h) [tex]\equiv[/tex] ng[tex]^{n-1}[/tex]

    ?

    Simon
     
  10. Sep 19, 2007 #9
    Yep! Or, using the fact that [tex]\{.,h\}[/tex] acts like a derivative, via the Leibniz rule:

    [tex]\{ab,h\} = a\{b,h\} + \{a,h\}b[/tex]

    So,

    [tex]\{g^n,h\} = g^{n-1}\{g,h\} + \{g^{n-1},h\}g[/tex]

    Giving you a recursive relation, that should be solvable.
     
  11. Sep 19, 2007 #10
    That is beautiful. Thank You everybody who took part in this!
     
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