The action is not a well defined function

  1. The action of moving from q_a to q_b in time t is not a well defined function

    for example, consider the harmonic oscillator

    if q_a= q_b, then only for t being the multiple of the harmonic oscillator period will the particle return to its initial position

    Therefore, for such kind of q_a and q_b, the action is ill-defined for an arbitrary t

    but people are always taking the derivatives of the action to the time!!!!!!
     
  2. jcsd
  3. vanhees71

    vanhees71 3,594
    Science Advisor
    2014 Award

    Who does this? Of course the action is not a function but a functional, i.e., it maps from an appropriate space of functions into the real numbers. It also depends on whether you work in the Lagrangian formulation (then the action is a functional over the configuration-space trajectories) or the Hamiltonian formulation (then it's a functional over the phase-space trajectories).
     
  4. stevendaryl

    stevendaryl 3,090
    Science Advisor

    That's not true. Suppose the particle is undergoing simple harmonic motion

    [itex]x = A cos(\omega t + \phi)[/itex]

    If [itex]t_1[/itex] and [itex]t_2[/itex] are two different times, then [itex]x(t_1) = x(t_2)[/itex] in the following cases:

    [itex]t_1 = t_2 + \dfrac{2 n \pi}{\omega}[/itex]
    [itex]t_1 = - t_2 + \dfrac{2 (n \pi - \phi)}{\omega}[/itex]

    In the latter case, [itex]t_2 - t_1[/itex] will not be a multiple of the period.
     
  5. you are right.

    however, t2-t1 still can only take discrete values.
     
  6. stevendaryl

    stevendaryl 3,090
    Science Advisor

    No, the constant [itex]\phi[/itex] is arbitrary. By adjusting [itex]\phi[/itex], you can get [itex]t_2 - t_1[/itex] to be anything you like.

    I shouldn't say it's arbitrary; it's determined by the boundary conditions. The harmonic oscillator [itex]m \ddot{x} + m \omega^2 x = 0[/itex] has solutions of the form:

    [itex]x(t) = A cos(\omega t + \phi)[/itex]

    That has two constants, [itex]A[/itex] and [itex]\phi[/itex]. If you pick two times, [itex]t_1[/itex] and [itex]t_2[/itex], and specify that [itex]x(t_1) = x_1[/itex] and [itex]x(t_2) = x_2[/itex], then that uniquely determines [itex]A[/itex] and [itex]\phi[/itex].
     
  7. Examine this expression for the action for a system with 1 generalized coordinate.

    [itex]S = \int_{t_0}^{t_1} \mathcal{L}(q, \dot{q}, t) dt[/itex]

    The action is well enough defined for different functions q(t), even ones that aren't valid solutions to the problem I'm trying to solve! To use your example, the action can have a value even for space endpoints q_a and q_b, even when there is no valid solution to the problem which connects q_a and q_b in time t1 - t0.

    The valid solutions are the ones for which the action is stationary.

    If we find the conditions for the action being stationary, we still haven't specified a particular path q(t) with starting and ending points!

    These would be related the the initial conditions.

    We may get some differential equations by solving the Euler Lagrange equations, which give us the conditions for which the action is stationary. But differential equations don't specify a single exact solution; there can be more than one function which is a solution to a differential equation. But once you fully specify the initial conditions, the values after time t are going to be set.
     
    Last edited: Apr 6, 2013
  8. stevendaryl

    stevendaryl 3,090
    Science Advisor

    Hmm. If [itex]t_1 > t_0[/itex] and the potential is not infinite, isn't there always a solution connecting [itex]q_a[/itex] at time [itex]t_0[/itex] to [itex]q_b[/itex] at time [itex]t_1[/itex]? I think there is, in the one-dimensional case, anyway.
     
  9. stevendaryl

    stevendaryl 3,090
    Science Advisor

    Classically, I think there's always a solution, but if we are doing a relativistic solution, the solution might be physically unrealistic (because it requires a massive particle to go faster than light).
     
  10. I don't really understand your point in your original post, the harmonic oscillator is periodic and hence the particle does return to a position after some time, decided by the boundary conditions, don't see why this makes the action functional ill-defined.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?

0
Draft saved Draft deleted