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Simple classic action integral

  1. Dec 11, 2012 #1
    I'm trying to solve this simple problem (it's the first problem of Quantum Mechanics and Path Integrals by Feynman, I feel like an idiot not being able to do it....) It's just solving for the action, S, of a free particle (no potential, only kinetic energy..)

    So it should just be [tex]S = \int_{t_a}^{t_b}{\frac{m}{2} (\frac{dx}{dt})^2 dt}[/tex]
    which according to the book is simply [tex]S = \frac{m}{2} \frac{(x_b - x_a)^2}{t_b - t_a}[/tex]

    I've tried a couple of different ways to reason myself into this solution but I can't seem to figure it out.
     
  2. jcsd
  3. Dec 11, 2012 #2

    Mute

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    What have you tried so far? What did you plug in for ##dx/dt##?
     
  4. Dec 11, 2012 #3
    Incredibly wrong stuff, heh..

    Yeah I'm an idiot. I was supposed to just plug in [itex]v = \left ( \frac{x_{b} - x_{a}}{t_{b} - t_{a}} \right )[/itex] because 'v' is constant from the Euler-Lagrange equation..

    Thanks for helping me see what should have been obvious >_< I was hell bent on doing things symbolically and didn't seem to care about the appearance of the end point 'x' values.. These should have been very suggestive.
     
    Last edited: Dec 11, 2012
  5. Dec 11, 2012 #4

    Mute

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    Great! You figured it out! Yeah, with a problem like this it helps to remember that the action is a functional of ##x(t)## and ##\dot{x}(t)##, so you get different answers depending on which function x(t) you use. Of course, varying the action with respect to x(t) (giving the Euler-Lagrange equations) yields the equation of motion for the classical path. The problem wanted the action of a classical path with boundary values ##x(t_a) = x_a## and ##x(t_b) = x_b##.

    It can take some practice seeing these sorts of problems a few times before it clicks. =)
     
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