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Prove the basic rule's of Poisson's Brackets?

  1. Oct 17, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that:
    (i){pi,pj}=0
    (ii){qi,qj}=0
    (iii){qi,pj}=δij


    2. Relevant equations

    {x,Y} = [(∂x/∂q * ∂Y/∂p) - (∂Y/∂q * ∂x/∂p)]

    3. The attempt at a solution

    Right I know first of all that qij and pij are generalized coordinates so that we dont need to worry about what units etc the answers are in.

    I know that if the Poisson Bracket is equal to zero then the point you have used it on is a conserved quantity.

    I think (i) and (ii) are ok but stuck on what to do on (iii). I have a feeling it has something to do with the Levi Civita Tensor as that is the last place I came across Kronecker Delta.

    (i/ii){qi,qj} = [(∂qi/∂q)*(∂qj/∂p) - (∂qj/∂q)*(∂qi/∂p)]

    The ∂qi/∂p and the ∂qj/∂p on either side are just '0' I think as it is a partial derivative of a q component with respect to p and it does not have any p component so QED 0 - 0 = 0

    And that would be basically the same solution for parts (i) and (ii). Now...

    (iii) {qi,pj} = [(∂qi/∂q) * (∂pj/∂p) - (∂pj/∂q)*(∂qi/∂p)]

    Last part is equal to zero for same reasons as part (i) and (ii) leaving me with

    (∂qi/∂q)*(∂pj/∂p)

    Now I dont really have a clue what to do like I said before I think it is something to do with Levi.... any help is appreciated.
     
  2. jcsd
  3. Oct 17, 2011 #2
    Is not q[itex]_{i}[/itex] independent of q[itex]_{j}[/itex] for i being not equal to j?

    So the partial derivative of one wrt the other is 1 if i = j and 0 otherwise.
    Similarly for the p.
     
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