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Lagrangian mechanics proof

  1. Feb 2, 2005 #1
    Hi, I would like some help in proving the following:

    Consider the action for a particle in a potential U. Show that an extremum path is never that of a local maximum for the action.

    I think what I have to do is look at the second derivative of the action integral. Then I should somehow argue that this value is always greater or equal zero, so that the extremum is never a local maximum. My problem is how to take the second derivative.

  2. jcsd
  3. Feb 4, 2005 #2

    Andrew Mason

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    I don't think you have to find the second derivative of the action. (The action is the integral of the difference between the particle's kinetic energy and its potential along the particle's path, over time). I think you have to show that for any deviation from that path the change in the action is > 0. An extremum path is one in which there is no infinitesimal first order change in the action for an infinitesimal change in the path. If the change in action is second order (ie. proportional to dx^2) it would have to be > 0.

    This is a very difficult area of physics, conceptually. Feynman's lecture in Vol II, Ch. 19 in his Lectures on Physics is very good as he explains the concepts.


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