8

It is well known that $\delta S = 0$ lays the foundation for variational mechanics. But I am confused as to whether or not this S is a minimum, a maximum, or a saddle point. Some books address this issue by using the language of "Stationary Action" instead of the more well-known "Least Action". But that doesn't really solve the problem of identifying different types of possible extrema.

So my question is: under what occasions are $S$ minimum or saddle? Is it possible for $S$ to be a maximum? (I have not encountered a single scenario in which $S$ is a maximum) In other words, can we refine the constraints on the possible extrema of $S$?

DanielSank
  • 24,439
Zhengyan Shi
  • 2,957
  • 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 Sep 27 '15 at 08:30

2 Answers2

4

The action is sometimes a saddle, but it is a minimum over small enough regions, but the reason it is a minimum instead of a maximum is due to a convention.

The action depends on the endpoints, the path and the Lagrangian. So the stationary action(s) and the physical path(s) depend on the endpoints and the Lagrangian.

How do you know which Lagrangian to use? Firstly, it should generate the correct Euler-Lagrange equations. This is a hood ohtsical reason to use a Lagrangian compared to another because this criteria focuses on testable experimental predictions. Secondly, you can use some pure conventions. Because if you multiply a Lagrangian by a nonzero scalar it will generate the same solutions to the Euler-Lagrange equations, and make the same predictions about observations and experiments. So which of the continuum many Lagrangians that are physically the same you choose to use is determined by convention.

Thus you could multiply all the Lagrangians you've seen by a negative one and you'd get the same physics and now all the minimum actions would be maximum actions. So we get minimum actions instead of maximum actions because we make an arbitrary choice unrelated to the observation that is literally chosen to give the least action (this is because the whole game was inspired by the idea of a ray taking the least time, so if you select the action that has the term for the kinetic energy be positive then the free particle action gives a minimum).

But even if you decided to break convention and multiply your Lagrangian by $-1$ you still couldn't say greatest action instead of least action because the action can still be a saddle. So the only terminology that is always correct is stationary action.

When can the action be a saddle? Remember how the action depended on the Lagrangian and the endpoints? Whether you are a saddle or a minimum depends on the endpoints.

Specifically, the action can be a saddle when the endpoints are far away. See for example When action is not least by C.G. Gray and Edwin F. Taylor in the American Journal of Physics Volume 75 Issue 5 pages 434-458 (2007); http://dx.doi.org/10.1119/1.2710480

A free source is, I think available on one of the author's websites www.eftaylor.com/leastaction.html and the other author has a peer-reviewed article on least action at Scholarpedia at http://www.scholarpedia.org/article/Principle_of_least_action

Keep in mind that if you write a Lagrangian as a sum of a bunch of terms, it is only the overall nonzero scalar that doesn't matter, the relative sign and magnitude of each term are totally physical.

Now, let's say you have a Lagrangian and some endpoints and you wonder how to tell whether the stationary action is a minimum or a saddle. In one-dimensional calculus you can take a second derivative and if it is strictly positive you have a minimum. There is a version of the second derivative test in variational calculus and so you can use that. It involves taking a second variation and checking to see if it is strictly positive.

It won't change the physics. The physics won't change in the sense that you get the same Euler-Lagrange equations regardless of whether the action was a minimum or a saddle. Now it is possible to have multiple paths between the same endpoints. You already knew that. You could send a billiard ball from the center of one side to the center of the other side in two different ways that take the same amount of time (shoot left or shoot right) and the same thing happens in physics, you can have two different velocities that take you on two different paths that lead to you starting in the same place and ending up in the same place and taking the same amount of time.

That's life. It happens when that time interval is long enough.

Timaeus
  • 25,523
  • Interesting POV. Within our convention though, it still seems that maximums don't occur... I guess I will look into the book Mozibur recommended and see if what Fomin says. – Zhengyan Shi Sep 27 '15 at 06:01
  • @ZhengyanShi There are online resources. For instance http://www.scholarpedia.org/article/Principle_of_least_action is peer reviewed and the current curator of that article Chris G. Gray also wrote a fairly accessible article (aimed at nonspecialists in action principles, but still aimed at physics instructors) for the American Journal of Physics called When Action Is Not Least for which a freely available copy seems accessible from the coauthor Edwin F. Taylor's website at http://eftaylor.com/leastaction.html – Timaeus Sep 27 '15 at 06:21
  • @ZhengyanShi And the other answer basically is saying that you look at the equivalent of the second derivative, and check that it is strictly positive, to test that it is a minimum. – Timaeus Sep 27 '15 at 06:26
  • @timaeus: your comment is cleaner than your answer; perhaps you should edit your answer to include it; and perhaps link to where you learnt it... – Mozibur Ullah Sep 27 '15 at 06:59
  • As well as perhaps saying exactly how 'many Lagrangians there are in the universe'; where I can find them; and how I can count them - is the set of them of countable or uncountable cardinality? – Mozibur Ullah Sep 27 '15 at 07:02
  • @MoziburUllah You get a term in the Lagrangian for every interaction, and if the whole Lagrangian is scaled by a nonzero scalar it changes nothing. – Timaeus Sep 27 '15 at 07:18
  • @timaeus: which doesn't answer my question nor is relevant to the OPs question as per your 'critique'... – Mozibur Ullah Sep 27 '15 at 07:29
  • @MoziburUllah The OP asked if the action can be a maximum, it is not a maximum, but that is a convention, there are continuum many total Lagrangian's that give the same EL equations because a nonzero scalar multiple of a Lagrangian gives the same EL equations. Negative one is a nonzero scalar so maximums are possible if you select one of those Lagrangians, but it is a convention that you don't. Whereas saddles happen often as long the endpoints are far. As yes, for far away endpoints you can get multiple solutions too. – Timaeus Sep 27 '15 at 14:04
  • @MoziburUllah I've edited the answer. I don't really know what is unclear to you, but if you think your answer answered the question and mine doesn't then maybe we just read the question differently. I read it as asking for physically what makes the action be a minimum a saddle or a maximum and the saddles physically happen when the time interval is large and when the saddles don't happen we get minimums instead of maximums because of a convention amongst physicists and some historical reasons. – Timaeus Sep 27 '15 at 15:10
3

In analogy to the ordinary calculus you need to look at the second or quadratic variation, Gelfand and Fomins Calculus of Variations do a good job of explaining it with the minimum of fuss.

Mozibur Ullah
  • 12,994
  • Thanks Mozibur! Does the book contain a classification of all the extrema in physical scenarios? Cuz the name sounds like a book of pure math... – Zhengyan Shi Sep 27 '15 at 05:09
  • You're welcome; it's more a book on applied math - his examples are all physical; I think, if I recall correctly the material you are looking for is in the second chapter. – Mozibur Ullah Sep 27 '15 at 05:11
  • Great I will look into that! – Zhengyan Shi Sep 27 '15 at 05:12
  • Glad to be of help; Remember to vote up/and accept if you find the answer useful... – Mozibur Ullah Sep 27 '15 at 05:14
  • @Timaeus: I don't think its a homework question; he's been pointed to a relevant reference; it's also his choice to vote up or accept. – Mozibur Ullah Sep 27 '15 at 06:35
  • I said by analogy; and by the quadratic variation; brief I know, but its enough - when a student has a good understanding of standard calculus; which he should do, given he was asking about the calculus of variations. – Mozibur Ullah Sep 27 '15 at 06:50
  • BTW, I couldn't make head or tail of your answer; perhaps you owe to yourself and the community to make it understandable? – Mozibur Ullah Sep 27 '15 at 06:51
  • And more properly to the community because I'm guessing you understand your own conceptualisations? – Mozibur Ullah Sep 27 '15 at 06:52
  • Besides don't you think students should look at books and learn to understand and interpret them correctly? – Mozibur Ullah Sep 27 '15 at 06:55
  • And then people say philosophy makes no sense; perhaps if people learnt to focus as much attention on philosophy as they do on physics they might find answers forthcoming; and if not, can we blame philosophy or the lack of attention... – Mozibur Ullah Sep 27 '15 at 07:06