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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?

Qmechanic
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Dave
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  • There are no physically realizable dynamical systems that actually obey the idealization of Lagrangian/Hamiltonian mechanics perfectly. ALL physical formulas are merely approximations. That's the main recipe for the incredible success of physics: we know how to work with "good enough". – CuriousOne Jun 25 '15 at 20:48
  • Possible duplicates: https://physics.stackexchange.com/q/907/2451 , https://physics.stackexchange.com/q/69077/2451 , https://physics.stackexchange.com/q/122486/2451 , https://physics.stackexchange.com/q/144356/2451 and links therein. – Qmechanic Jun 25 '15 at 21:06

2 Answers2

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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').

tok3rat0r
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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.

Zo the Relativist
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