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Proving Some Poisson Bracket identities - a notational question

  1. Nov 3, 2013 #1
    Proving Some Poisson Bracket identities -- a notational question

    I need some help just understanding notation, and while this might count as elementary it has to do with Hamiltonians and Lagrangians, so I posted this here.

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

    Prove the following properties of Poisson's bracket:

    [A,A] = 0 and [A,B] = [-B,A]

    Now honestly this looks a lot like matrix math to me, and I know it's related. But I a jut trying to make sure I get what it is that [A,A] wants me to do – add the two quantities? Multiply them? And understand what it is I am seeing when I see that notation.

    So I know that Poisson's bracket is:
    [tex]
    \sum_i \left(\frac{\partial f}{\partial q_i} \frac{\partial \mathcal H}{\partial p_i} - \frac{\partial f}{\partial p_i} \frac{\partial \mathcal H}{\partial q_i} \right)
    [/tex]
    and the poisson bracket would be [itex][f, \mathcal H][/itex]. So does that mean the above question would be, prove that

    [tex]
    \sum_i \left(\frac{\partial A}{\partial q_i} \frac{\partial A}{\partial p_i} - \frac{\partial A}{\partial p_i} \frac{\partial A}{\partial q_i} \right) = 0
    [/tex]
    and
    [tex]
    \sum_i \left(\frac{\partial A}{\partial q_i} \frac{\partial B}{\partial p_i} - \frac{\partial B}{\partial p_i} \frac{\partial A}{\partial q_i} \right) = \sum_i -\left(\frac{\partial B}{\partial p_i} \frac{\partial A}{\partial q_i} + \frac{\partial B}{\partial p_i} \frac{\partial A}{\partial q_i} \right)
    [/tex]

    Am I reading this correctly? Maybe this sounds terribly elementary but notation I often find confusing rather than illuminating.


    Anyhow, if someone could let me know if I have read this right that would go a long way.
     
  2. jcsd
  3. Nov 3, 2013 #2

    TSny

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    Yes, that's what you want to prove, except you didn't quite get the signs correct in the last equation.
     
  4. Nov 3, 2013 #3

    TSny

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    Oops, it's more than just the signs that are wrong in the last equation.
     
  5. Nov 3, 2013 #4
    Did I screw up the order too? (I was thinking it should have been [itex]\frac{\partial B}{\partial q_i}[/itex] and [itex]\frac{\partial A}{\partial p_i}[/itex] in the summation but let me know)
     
  6. Nov 3, 2013 #5

    TSny

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    Go back to your original definition of Poisson bracket for [f, H]. Note that the order of f and H is the same in the two terms, but the order of q and p is switched. That should help you see how to set up [-B, A].

    Just curious: Did they ask you to show [A, B] = [-B,A] or did they ask you to show [A,B] = -[B,A]? Not that it really matters.
     
  7. Nov 3, 2013 #6
    it was -[B.A].

    That's why I sort of wondered if there was a way to set it up as a matrix... just using q1 and p1 et cetera in place of x1, x2, x3 (would it be a matrix with q1... q_i as the first row, and I guess p1... p_i as the second? ) I just feel it's almost easier to do that, but if I am speaking of irrelevancies then please do let me know.
     
  8. Nov 3, 2013 #7

    TSny

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    OK, so you want to show [A,B] = -[B,A]. In your first post your expression for [A,B] was not quite correct. Can you see how to correct it?
     
  9. Nov 3, 2013 #8
    SO it should look like this, then:

    [tex]
    \sum_i \left(\frac{\partial A}{\partial q_i} \frac{\partial B}{\partial p_i} - \frac{\partial B}{\partial q_i} \frac{\partial A}{\partial p_i} \right) = \sum_i -\left(\frac{\partial B}{\partial q_i} \frac{\partial A}{\partial p_i} - \frac{\partial B}{\partial p_i} \frac{\partial A}{\partial q_i} \right)
    [/tex]

    I think that should be the right order (and sign)..
     
  10. Nov 3, 2013 #9

    TSny

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    Your expression is correct. But it seems a little odd the way you wrote it. On the left side, your numerators (A and B) switch order in the two terms but the denominators (q and p) do not switch order. That's fine. On the right you decided to not switch the order of the numerators but you switched the order of the denominators. That's fine too. But it just seems a little odd that you would not use the same pattern on both sides of the equation.
     
  11. Nov 3, 2013 #10
    that has more to do with just not wanting to deal with the quirks of LATEX :-)
     
  12. Nov 3, 2013 #11
    One last thing: could this be set up as a matrix? The way I intuited it.
     
  13. Nov 3, 2013 #12

    TSny

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    I don't see how matrices would enter here. I could be wrong. How would you construct a matrix ##M## that would be relevant? What would be the entry in the ##i##th row and ##j##th column: ##M_{ij}##?
     
  14. Nov 4, 2013 #13
    I was looking t it as thought the ps and qs were x and y, and thinking that the identities kind-a-sorta looked like the way matrices operate, like if you have a 2-dimensional expression in cartesian coordinates you can set it up as a matrix. That's all.
     
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