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Canonical Transformations

  1. Jul 7, 2013 #1

    For which of the invertible transformations [itex](\mathbf{q}, \mathbf{p}) \leftrightarrow(\mathbf{Q}, \mathbf{P})[/itex]
    [itex]\mathbf{Q}(\mathbf {q}, \mathbf {p}, t)[/itex]
    [itex]\mathbf{P}(\mathbf{q}, \mathbf {p}, t)[/itex]
    is it so that for every Hamiltonian [itex]\mathcal{H}(\mathbf {q}, \mathbf {p}, t)[/itex] there is a [itex]\mathcal{K}[/itex] such that
    [tex]\dot{Q}_i = \frac{\partial\mathcal{K}}{\partial P_i} \;\;\;\;\;\;\;\; \dot{P}_i = -\frac{\partial\mathcal{K}}{\partial Q_i}\; ?[/tex]
    Stationary action should correspond, and that condition is met if

    [itex]\sum p_i\dot{q}_i - \mathcal{H} = \sum P_i\dot{Q}_i - \mathcal{K} + \frac{dF}{dt}[/itex],

    since integrating [itex]\frac{dF}{dt}[/itex] results in something only dependent of the endpoints.

    Consider this part of Goldstein's Classical Mechanics.


    rearranging 9.13 to make this clear:

    [tex]\mathcal{K} = \mathcal{H} + \frac{\partial F_1}{\partial t} + \sum_i \dot{Q}_i\left(P_i - \frac{\partial F_1}{\partial Q_i} \right) + \sum_i \dot{q}_i\left(\frac{\partial F_1}{\partial q_i} - p_i\right)[/tex]
    I guess I might like this explained a more. Why aren't we able to to have the coefficients of [itex]\dot{q}_i[/itex] or [itex]\dot{Q}_i[/itex] be non-zero, and have the difference absorbed into [itex]\mathcal{K}[/itex] ?
  2. jcsd
  3. Jul 7, 2013 #2
    it's because K should not depend on the derivates of Q and q
  4. Jul 8, 2013 #3


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    I would argue as follows. The Hamiltonian version of the action is
    [tex]S[x,p]=\int_{t_1}^{t_2} \mathrm{d} t [\dot{q}^k p_k - H(t,q,p)].[/tex]
    The trajectory in phase space is determined as the stationary point of this functional with the boundary values [itex]q(t_1)[/itex] and [itex]q(t_2)[/itex] fixed.

    Now if you want to determine new phase-space coordinates [itex](Q,P)[/itex] that describe the same system in the new coordinates by the variational principle, i.e., such that the phase-space trajectories are described by the Hamilton canonical equations, you must have
    [tex]\mathrm{d} q^k p_k - \mathrm{d} Q^k P_k - \mathrm{d} t (H-K)=\mathrm{d} f.[/tex]
    From this it is clear that the "natural" independent variables for [itex]f[/itex] are [itex]q[/itex], [itex]Q[/itex], and [itex]t[/itex]. Then comparing the differential on each side leads to
    [tex]p_k=\frac{\partial f}{\partial q^k}, \quad P_k=-\frac{\partial f}{\partial Q^k}, \quad K=H+\frac{\partial f}{\partial t}.[/tex]
    Then you can go over to other pairs of old and new independent phase-space variables in the "generator" using appropriate Legendre transformations.
  5. Sep 23, 2014 #4
    Jul 7, 2013 #1

    For which of the invertible transformations (q,p)↔(Q,P)
    is it so that for every Hamiltonian H(q,p,t) there is a K such that

    Is It show that K is dependent upon (P, Q) not in Derivative of (P, Q)? plz.

    also show that
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