# Action max, min, or saddle?

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 $S$?

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.

• 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 5:09
• 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 5:11
• Great I will look into that! – Zhengyan Shi Sep 27 '15 at 5:12
• Glad to be of help; Remember to vote up/and accept if you find the answer useful... – Mozibur Ullah Sep 27 '15 at 5:14
• @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 6:35

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 multiple all the Lagrangians you've seen by 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 arbitrarily choice unrelated to observation that is literally choosen to give a 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 you Lagrangian by $-1$ you still couldn't say greatest action instead of least action because 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 website 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 is 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.

• 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 6:01
• @ZhengyanShi There are online resources. For instance 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 eftaylor.com/leastaction.html – Timaeus Sep 27 '15 at 6:21
• @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 6:26
• @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 6:59
• 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 7:02