Are there physically realizable dynamical systems where the true trajectory is not a minumum action trajectory?

Formally, Lagrangian mechanics only requires that the trajectory be an extremum (or saddle point?), but all of the cases that I'm aware of, it is, in fact, a minimum. Are the other possibilities relevant for modelling any physical systems?


marked as duplicate by Dave, yuggib, Martin, Qmechanic Jun 26 '15 at 9:29

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As demonstrated in this paper, the trajectory can never maximise the action but can in fact lie on a saddle point in cases where the potential has the appropriate spatial variation (at least partially repulsive) and where the final state is taken sufficiently far 'downstream' (beyond what these authors call the 'kinetic focus').

  • $\begingroup$ Why are they no teaching that in class???? I kind of feel cheated now. $\endgroup$ – CuriousOne Jun 25 '15 at 21:37

Not to mention, there are cases where the local extremum of the action isn't a physically realiziable path. Consider the plane with all the points satisfying $y > |x|$ removed. now, consider a start point of $(-2,1)$, and an end point of $(2,1)$ on this manifold, with the Lagrangian $\frac{1}{2}m\left({\dot x}^{2} + {\dot y}^{2}\right)$. Clearly, the minimum of the Lagrangian will take the particle on a path that leaves the manifold.

  • $\begingroup$ Sounds like you are using a Lagrangian that does not take into account that the potential at the edge is infinite? $\endgroup$ – CuriousOne Jun 25 '15 at 21:30
  • $\begingroup$ @CuriousOne: it's not infinite, it just has zero support outside of the domain. What this is really telling you is that there is no physical path between those points. $\endgroup$ – Jerry Schirmer Jun 25 '15 at 21:40
  • $\begingroup$ Got to think about that some more. $\endgroup$ – CuriousOne Jun 25 '15 at 21:42
  • $\begingroup$ and I'll admit that this is a snarky response that technically answers the OP's question, but is kind of unsatisfying. $\endgroup$ – Jerry Schirmer Jun 25 '15 at 22:11
  • $\begingroup$ Not as snarky as mine. :-) $\endgroup$ – CuriousOne Jun 25 '15 at 22:32

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