11

In this question it is discussed why by Hamilton's principle the action integral must be stationary. Most examples deal with the case that the action integral is minimal: this makes sense - we all follow the path with the least resistance.

However, a stationary point can be a maximal or minimal extremum or even a point of inflexion (rising or falling). Can anybody give examples from theory or practice where the action integral takes on a maximal extremum or a point of inflexion?

Gerard
  • 6,791
  • 8
    The (timelike) geodesics in space-time (GR) are precisely the curves that maximize the proper time. Also, it depends on your definition: If you change the sign of the action (just changing your convention), what is a minimum turns out now to be a maximum – Ronaldo Nov 16 '10 at 13:17
  • 1
    see also: https://physics.stackexchange.com/a/18804/139 – Gerard Feb 05 '19 at 13:41
  • 2
    What do you mean by "the path with the least resistance"? – HelloGoodbye Nov 04 '21 at 22:45

4 Answers4

4

If you have in mind a mechanical system and if you are really referring to the Hamilton's principle (which tells how to find the actual trajectory the system follows between the fixed initial and final times at fixed initial conditions), then I doubt you can find such an example. The reason is that a mechanical system contains the kinetic term in its lagrangian, which is unbounded from above. That is, you can always construct a trajectory $\{x(t), {\dot x}(t)\}$ with as large action as you wish; so there is no maximum, not even a local one.

In his answer above, Mark refers to an optical example that sort of maximizes, not minimizes the action. The example can be reformulated in mechanical form using small balls rolling on a surface instead of light rays, but I don't think it really refers to the Hamilton's principle. In this example you send light rays (or small balls) from a fixed initial point into different directions and search for caustics in balls' trajectories. That is, you vary the initial condition (the direction of initial velocity) and at the same time you assume that each trajectory (for each choice of the velocity) is already known. That's not what the Hamilton's principle is about (there you fix all the initial conditions and solve for the trajectory).

Igor Ivanov
  • 3,907
  • 1
  • 22
  • 18
  • 2
    It's true that the trajectory of a classical particle never locally maximizes the action corresponding to a Lagrangian of the form $K - U$, but it can be a saddle point. This can occur even for the simple harmonic oscillator. – tparker Feb 04 '19 at 03:03
  • 1
    This answer is in conflict with the answers here and here which explain (with references) how mechanical systems do not necessarily minimize the action over reasonably finite durations, even with the specified boundary conditions. I think your answer makes sense though for infinitesimally short trajectories. – jnez71 Feb 02 '21 at 17:54
2

If you look at (anti-)self-duality, you get exactly inflection points.

Also, you can think in terms of the WKB-approximation of Feynman Path Integrals, and classify your extrema accordingly. Behind this, there's a whole story about Morse Theory and manifold classification/decomposition, etc... but, the bottom line is that you can think of these WKBs of the Path Integral as the tool to classify the extrema of your Action.

Daniel
  • 4,197
1

In the Lagrange formulation, we require $\delta S = 0$ for the physical path. That does not necessarily imply that the trajectory is a minimum. This feature was noted as early as 1662 by Marin Cureau de la Chambre (1594 – 29 December 1669). As an objection to the idea that Nature always follows the shortest path, he pointed out that for a curved mirror, the path that obeys the law of reflection is not necessarily the one with the smallest length.

ZaellixA
  • 1,779
1

This isn't an action from mechanics, but in gravitational lensing we look for stationary points of the time travel of light. GR basically tells us that light travels at different speeds depending on the gravitational potential. (This is distant light, not local right here in our lab.) If light from a distant quasar travels through a galaxy before reaching us, it experiences a varying gravitational potential depending on the mass distribution of the galaxy and the path the light takes. Different paths for the light then take different amounts of time. Looking in the sky, we can associate a travel time with each angle in the sky. Then we make a 2D map of travel time versus point in the sky. Here's an example from a quick search (the source is here)

contour diagram for gravitational lensing

The lines connect parts of the sky that have equal travel time. The different panels show qualitatively different types of gravitational lensing systems. If you look at the upper-right panel, you see five dots, which are five distinct images of a distant quasar. The ones at intersections of contour lines are saddle points. The others are either local minima or maxima of travel time.

The difference in the time it takes light to reach Earth from different images can be dramatic - weeks or even months!

Mark Eichenlaub
  • 52,955
  • 15
  • 140
  • 238