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Homework Help: Canonical Transformation Problem

  1. Oct 14, 2008 #1
    1. The problem statement, all variables and given/known data

    Show that the time reversal transformation given by Q = q, P = − p and T = − t, is canonical, in the sense that the form of the Hamiltonian equations of motion is preserved. However, it does not satisfy the invariance of the fundamental Poisson Bracket relations. This is an example when the two criteria are not equal.

    2. Relevant equations
    3. The attempt at a solution

    This I have done...I just ask you to check if the procedure is correct.( ' denotes d/dt )

    Q'=(dQ/dT)=(dQ/dt)(dt/dT)= -(dQ/dt)= -[(∂Q/∂q)q' + (∂Q/∂p)p' + (∂Q/∂t)]= -[q']= -(∂H/∂p)

    Also, (∂K/∂P)=(∂H/∂p)(∂p/∂P)= -(∂H/∂p)...[we write (∂K/∂P)=(∂H/∂p) as Kamiltonian K is a

    function of Q,P,T and Hamiltonian H is a function of q,p,t]

    Thus, Q'= (∂K/∂P)...1st of Hamilton's canonical equations is proved.

    Similarly, P'= (dP/dT)= -(dP/dt)= -[(∂P/∂q)q' + (∂P/∂p)p' + (∂P/∂t)]= p'= -(∂H/∂q)

    Then, (∂K/∂Q)=(∂H/∂q)(∂q/∂Q)=(∂H/∂q)

    Thus, P'= -(∂K/∂Q)

    This shows that the given transformation leads (q,p) to canonically conjugate variables(Q,P)

    Evaluating the Poisson brackets it is easy to show that they do not satisfy fundamental Poisson bracket.

    Can anyone suggest why there is a mismatch between the two aspects?
  2. jcsd
  3. Oct 15, 2008 #2
    Are people not interested for I have not used TeX?

    My attempt:

    [tex]\dot {Q}= (dQ/dT) = (dQ/dt)(dt/dT) = -(dQ/dt) = -[(\partial Q / \partial q) \dot {q} + (\partial Q / \partial p) \dot {p} + (\partial Q / \partial t)] = -\dot {q} = -(\partial H / \partial p)[/tex]

    [tex](\partial K / \partial P)=(\partial H / \partial p)(\partial\ p / \partial P)= - (\partial H / \partial p)[/tex]

    Thus, [tex]\dot {Q}= (dQ/dT) = (\partial K / \partial P) [/tex]


    [tex]\dot {P}= (dP/dT) = (dP/dt)(dt/dT) = -(dP/dt) = -[(\partial P / \partial q) \dot {q} + (\partial P / \partial p) \dot {p} + (\partial P / \partial t)] = \dot {p} = -(\partial H / \partial q)[/tex]

    [tex](\partial K / \partial Q)=(\partial H / \partial q)(\partial\ q / \partial Q)= (\partial H / \partial q)[/tex]

    Thus, [tex]\dot {P}= (dP/dT) = -(\partial K / \partial Q) [/tex]

    The most sensitive part of this observation is ofcourse the identification that [tex]\ K = \ H [/tex] that I have

    assumed.In general,however,there will be a partial time derivative of a generating function F so that the equation lokks like

    [tex]\ K = \ H + (\partial F / \partial t)[/tex]

    I have assumed that F has no explicit time dependence.So the later time derivative is zero and we have Hamilton's equations


    My question is if my assumption is justified for this problem...Note that, the problem already ensures that the transformation does

    not satisfy the fundamental Poisson brackets.My calculation gave the value of Poissson Bracket -1 instead of 1 (the correct

    one). In that sense,it is not a canonical transformation...

    I am not sure, but it looks to me, my assumption that F does not have explicit time dependence may be somehow related to the fact of fundamnental Poisson Bracket not

    being satisfied...

    Can anyone please take interest?
    Last edited: Oct 15, 2008
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