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I Two stationary values for an action

  1. Feb 26, 2016 #1
    Is it possible for an action (the integral of a Lagrangian) to have more than one stationary value? Why or why not?
     
  2. jcsd
  3. Feb 26, 2016 #2
    If there are more than one real path through which the physical system can move in phase space (q,p) then it can have multiple stationary values -but i have yet to see an example of the same
     
  4. Mar 3, 2016 #3
    If there are multiple stationary values, then a point in phase space (q,p) may have multiple trajectories. Does that violate Liouville's theorem: the density of system points in the vicinity of a given system point travelling through phase space is constant with time?

    And Lioville's theorem, I think, also states that a "volume" of system points in phase space must maintain its topology under time evolution. So a simply connected volume must remain simply connected. Assume all except one point P in a simply connected "volume" has exactly one trajectory, while point P has two (say, path A and path B). Only one path (say, A) allows the volume's density and topology to remain unchanged, while the other path (B) either increases the density of the volume (if the path ends within the volume) or makes a hole in the volume (if the path ends outside the volume). Using the same argument, we conclude that if a point has exactly one trajectory, then all its neighbouring points must have exactly one trajectory.

    EDIT: Liouville's theorem shows that if a system A in an initial state X has exactly one trajectory determined by a set S of equations, then the same system A but in any other initial state Y must also have exactly one trajectory determined by the same set S of equations. But it does not show that there could only be one stationary value of the action. It does not show that the Hamilton's principle solves for a unique set S of equations.
     
    Last edited: Mar 3, 2016
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