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Homework Help: More symplectic notation

  1. Oct 2, 2012 #1
    I have some trouble understanding the attached section of my book. Basically I can't see why the marked equations are equivalent - that is the first two are contained in the last one. I can follow the derivation but when I do an example for myself where I just have two variables (q,p) being transformed to (Q,P). I get the condition from the symplectic condition that some determinant (actually the Jacobian defined in my last thread) equal 1 which does not seem equivalent to the equations 9.48. If the reader does not remember the matrix J I have attached that too on my own example. I am right in assuming that eqs. 9.48 should follow from the symplectic condition right? It certainly seems from the book that this condition is just a nicer way of writing eqs. 9.48 (although they look pretty nice to me already? - is there a sum or something?) so I would think so.

    Attached Files:

  2. jcsd
  3. Oct 2, 2012 #2
    9.48 and 9.55 are equivalent because they are derived in exactly the same way, just in slightly different notation. As to why the form is different from your worked example, this is because you did not really convert between (q, p) and (Q, P) when you transposed the Jacobian matrix, so you did not have derivatives with regard to different sets of variables as in 9.48.
  4. Oct 3, 2012 #3
    But I transposed correctly right? If I am to make the expressions equivalent should I just switch variables in one of the matrices M and M(transpose)?
  5. Oct 3, 2012 #4
    You can prove that equivalence. But that will involve the kind of tedious partial derivatives gymnastics you disliked in a recent thread :)
  6. Oct 3, 2012 #5
    didn't dislike it, just wasn't very good at it - practise is good.
  7. Oct 3, 2012 #6
    Well, here is your chance to practice that!
  8. Oct 3, 2012 #7
    I'm just rather confused of the sentence: "The algebraic manipulations that lead to eqs. 9.48 can be written in an elegant manner with use of the symplectic notation."
    It seems that the books says that the symplectic condition is another way of writing eqs. 9.48.
    Maybe I don't understand why it is exactly the same approach that is used for the two derivations. In the first you only look at the dQ/dt and dH/dP. In the other you look at the derivatives of the q's.
    Last edited: Oct 3, 2012
  9. Oct 3, 2012 #8
    And that is true. Both methods ultimately deal with the partial derivatives of q, p, Q. P with respect to one another. But the symplectic notation invokes the Jacobian matrix from the beginning, thus avoiding having to deal with all that massaging one derivative into another to get the required condition.
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