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Is Action Always "Locally" Least?
In general, I know it's true that the Principle of Least Action is more properly called the Principle of "Stationary" Action. However, there are results which seem to suggest that for sufficiently short trajectories of a physical system, the action is always least. It is only past the "kinetic focus" of the trajectory of a physical system that the action can fail to be minimized. (e.g. When action is not least in Classical Mechanics and When action is not least for orbits in General Relativity). In her recent book "The Lazy Universe: An Introduction to the Principle of Least Action", Jennifer Coopersmith suggests that the Principle of Least Action isn't just a misnomer for these reasons.
In Landau & Lifshitz' books on Mechanics and Classical Field Theory they also say things like this (e.g. "the action must be a minimum for infinitesimal displacements").
I would like to know how general this is. Are there general results which show that the Action is always minimum when considering sufficiently short trajectories (in classical mechanics, classical field theory, general relativity, quantum mechanics, etc.)? What is the best resource which goes in depth about this sort of stuff?
classical-mechanics lagrangian-formalism variational-principle action
Qmechanic♦
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$\begingroup$ At least for bounded motion this seems false, see e.g. this answer. But this assumes "small but finite" displacements. $\endgroup$ – Ruslan Mar 8 '19 at 17:03
Introduction. It is well-known that the stationary solution to a simple harmonic oscillator (SHO) (with Dirichlet boundary conditions) is not a local minimum but a saddle point beyond the first caustics, i.e. if the lapsed time $$ \Delta t~:=~t_f-t_i~>~ \frac{T}{2}\tag{1} $$ is more than half the period $T$ of the SHO, cf. e.g. my Phys.SE answer here.
OP is essentially asking the following interesting question.
Question: Given an action with Dirichlet boundary conditions, is for sufficiently short lapsed time $\Delta t\in]0,\epsilon[$ the stationary solution a local minimum?
Answer: No, not necessarily. A counterexample is a system of infinitely many SHOs, with periods $T_n\to 0$ for $n\to \infty$. For any finite $\Delta t>0$ the highest modes would then be beyond their first caustics, and hence saddle points.
Such a system can e.g. be realized by a vibrating string of length $L$ with Lagrangian density ${\cal L}~=~\frac{1}{2}(\partial_t\phi^2-\partial_x\phi^2)$, which has infinitely many overtones.
Qmechanic♦Qmechanic
You linked to an article by C. G. Gray and Edwin F. Taylor, that is available from Taylors website. Taylors website is my main source of information about the principle of least action.
A crucial aspect of the principle of least action is pointed out in a quote from Richard Feynman, that Taylor gives in one of his articles.
Along the true path, S is a minimum. Let's suppose that we have the true path and that it goes through some point a in space and time, and also through another nearby point b.
Now if the entire integral from t1 to t2 is a minimum, it is also necessary that the integral along the little section from a to b is a minimum. It can't be that the part from a to b is a little bit more. Otherwise you could fiddle with just that piece of path and make the whole integral a little lower. So every subsection of the path must also be a minimum. And this is true no matter how short the subsection. Therefore, the principle that the whole path gives a minimum can be stated also by saying that an infinitisimal section of path also has a curve such that it has minimum action.
Quote can be found in:
Feynman Lectures, book II, chapter 19, the principle of least action
I take it as given there are also cases where the true trajectory corresponds to a maximum of the action.
The same reasoning then applies: if the action is a maximum for the global trajectory it must be a maximum for every subsection of that global trajectory, down to subsections that are infinitisimally short.
[Later edit, 4 hours after initial posting]:
A maximum of the action
First off: my understanding is limited to cases where the potential energy is a function of spatial coordinate only; cases where the potential is also a function of time is out of scope for me.
The following reasoning is based on the understanding that is presented in my answer to the stackexchange question:
What is the physical content of the principle of least action
The range of possible force laws consists of the various powers of the spatial coordinate.
Of course, the two best known forces, gravity and the Coulomb force, are inverse square forces; force proportional to $r^{-2}$
For force proportional to $r^{-2}$ the action corresponding to the true trajectory is a minimum.
For a force that is the same at every spatial coordinate the action corresponding to the true trajectory is a minimum.
A force that increases in linear proportion to the spatial coordinate is the critical case:
As we know, an attractive force that is proportional to $r$ is referred to as Hooke's law. As we know, when the force is Hooke's law the resulting motion is harmonic oscillation.
When evaluating the action: the kinetic energy varies in quadratic proportion to the variation of the trajectory (since the kinetic energy is proportional to the square of the velocity.) When the force is proportional to $r$ then the potential energy (integral of force over distance) is proportional to $r^2$ That is: in the case of Hooke's law both the kinetic energy and the potential energy vary in quadratic proportion to the variation of the trajectory. That makes it the critical case.
Next up is when the force increases in proportion to the square of the spatial coordinate: then the potential energy is proportional to the cube of the spatial coordinate. (And of course you can keep going up in power.)
The progression is from sub-critical, to critical, to super-critical. For force proportional to $r^{-2}$ $r^{-1}$ and $r^0$ change of potential energy in response to variation of the trajectory is smaller than the change of kinetic energy, hence the true trajectory corresponds to minimum of the action. I haven't done an actual mathematical verification, but on the above grounds it seems certain to me that in the case of a force that is proportional to the square of the spatial coordinate the action corresponding to the true trajectory will be a maximum.
We can see that the vast majority of the cases that we readily encounter in Nature are sub-critical cases, for which the true trajectory corresponds to a minimum of the action.
Still, the laws of physics do not prohibit a restoring force that increases in larger proportion than Hooke's law. Hence my claim that cases where the true trajectory corresponds to a maximum of the action are physically possible.
[Second additional edit, 17 hours after initial posting.]
In their article When action is not least Gray and Taylor write something very, very odd.
The following intuitive proof by contradiction was given briefly by Jacobi [...] Consider an actual worldline for which it is claimed that S [...] is a true maximum. Now modify this worldline by adding wiggles somewhere in the middle. These wiggles are to be of very high frequency and very small amplitude so that they increase the kinetic energy K compared to that along the original worldline with only a small change in the corresponding potential energy U. The Lagrangian L=K−U for the region of wiggles is larger for the new curve and so is the overall time integral S. The new worldline has greater action than the original worldline, which we claimed to have maximum action. Therefore S cannot be a true maximum for any actual worldline.
This attempt at a proof by Gray and Taylor makes no sense.
Let me make a comparison: suppose that you are computing both the line integral of some function and the area integral (area enclosed between graph line and x-axis). As we know: in the case of computing a line integral adding high frequency small amplitude wiggles to the function graph will increase the line integral value, with minor impact on the area integral value.
It seems as if Gray and Taylor are treating the kinetic energy as being evaluated as a line integral, and treating the potential energy as being evaluated as an area integral.
But we all know that the Lagrangian L=K-U is evaluated as a single area integral. The wiggles in the graph that Gray and Taylor suggest will shift the value of the action away from the point where the action is stationary, just like any other type of variation of the trajectory.
So: the attempted proof that 'S cannot be a true maximum for any actual worldline' is refuted.
CleonisCleonis
$\begingroup$ I can see why the conditional claim "if the whole path is minimum, then each subsection is also minimum" is true. But this does not establish the general claim that action is always minimized in sufficiently short distances, since it is know that action in general is not always minimized (or maximized). $\endgroup$ – Taro Mar 8 '19 at 20:52
$\begingroup$ @DavidB. Indeed: a general claim that in sufficiently short distances action is always minimized is refuted. We have that there are cases where the true trajectory corresponds to a maximum of the action. This implies that for such a trajectory every subsection, down to infinitisimally short subsections, the action is a maximum. That is what is pointed out in my answer: there are physically realistic trajectories where on each infinitisimal subsection the action is a maximum. $\endgroup$ – Cleonis Mar 8 '19 at 21:05
$\begingroup$ Ah ok. In the two published articles I linked to, they both claim that the action is never maximized. "Misconceptions concerning the stationary nature of the action abound in the literature. Even Lagrange wrote that the value of the action can be maximum, a common error of which the authors of this paper have been guilty." Which examples are you thinking of? $\endgroup$ – Taro Mar 8 '19 at 21:24
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