# When/why does the principle of least action plus boundary conditions not uniquely specify a path?

A few months ago I was telling high school students about Fermat's principle.

You can use it to show that light reflects off a surface at equal angles. To set it up, you put in boundary conditions, like "the light starts at A and ends at B". But these conditions by themselves are insufficient to determine what the path is, because there's an extra irrelevant stationary time path, which is the light going directly from A to B without ever bouncing off the surface. We get rid of this by adding in another boundary condition, i.e. that we only care about paths that actually do bounce. Then the solution is unique.

Of course the second I finished saying this one of the students asked "what if you're inside an elliptical mirror, and A and B are the two foci?" In this case, you can impose the condition "we only care about paths that hit the mirror", but this doesn't nail down the path at all because any path that consists of a straight line from A to the mirror, followed by a straight line to B, will take equal time! So in this case the principle tells us nothing at all.

The fact that we can get no information whatsoever from an action principle feels disturbing. I thought the standard model was based on one of those!

My questions are

• Is this anything more than a mathematical curiosity? Does this come up as a problem/obstacle in higher physics?
• Is there a nicer, mathematically natural way to state the "only count bouncing paths" condition? Also, is there a "nice" condition that specifies a path in the ellipse case?
• What should I have told that student?
• possible duplicate of Several stationary points of the action functional – ACuriousMind Aug 31 '15 at 14:14
• I'm inclined to leave this open as the other question is in the context of QFT, and it would do well to have general-purpose answers that don't rely on QFT (lack of) intuition. – user10851 Sep 1 '15 at 22:04

Multiple classical solutions to Euler-Lagrange equations with pertinent/well-posed boundary conditions (such solutions are sometimes called instantons) are a common phenomenon in physics, cf. e.g. this related Phys.SE post and links therein.

In optics, it is well-known that already e.g. two mirrors can create multiple classical paths.

• Thanks for the answer! I realize now that multiple valid classical paths is just fine in QM, since they'll just all contribute to the path integral. But just working classically, is it a problem? I thought action principles were equivalent to just Newton, so they should always pin down a path. – knzhou Aug 31 '15 at 8:27
• @KevinZhou Newton's 2nd doesn't always pin down a path, consider e.g. $V(x)=kr^{4/3}$ and $x(0)=v(0)=0.$ The force is zero and so Newton's 1st says it stays at rest but Newton's 2nd is OK with solutions where it merely has an instantaneous acceleration of zero such as moving cubically in time. – Timaeus Aug 31 '15 at 15:27

Actually, the extra path is not irrelevant. If you put a light bulb at A and a $4\pi$ detector (this means $4\pi$ steradian coverage, i.e. it detects incoming light in any direction) at B, the detector will see light along both paths: direct, and bounced off the mirror, which is exactly the result you got from Fermat's Principle. If you want to exclude the direct path, you have to block it with an opaque wall.

The same thing goes for the elliptical mirror: the detector will see light coming in from every direction, which is again just what Fermat's Principle tells you: every path has stationary (i.e. zero first-order variation) travel time, and thus every path is a valid one for the light.

In Lagrangian mechanics, on the other hand, the state of the system contains both position and velocity - it's a state in phase space, not real space. You don't have to deal with reflections in phase space, and that usually rules out these cases where you have an infinite number of paths by which a particle could get from state A to state B.

• I don't know if the paths moving on the straight line between them are as valid for the ellipse as the other paths since it has to pass through the $4\pi$ detector at an earlier time. If you wait to place it until after the light has passed it could work. But then you also need to remove the source to avoid blocking its own light. And since there are cases with multiple solution even with second order systems just saying it is unusual is a bit vague. – Timaeus Aug 31 '15 at 15:36

Is this anything more than a mathematical curiosity?

It is not a curiosity. The light travels all those paths.

Does this come up as a problem/obstacle in higher physics?

I'm not sure what the "this" is. Yes, people will sometimes wrongly think there is a unique solution when there isn't. There can be multiple geodesics between the same events (even if for small enough regions this isn't true), there can be multiple solutions to $F=ma.$ But how important is it to have a unique solution?

Sure, doing a path integral could be annoying when there isn't a unique path, but if those failures are small in a measure theoretic way it doesn't matter. But people aren't exposed to measure theory early on. So they can get confused by anything, like that something that happens with probability 1 can fail to happen or that something that happens with probability 0 can happen. They can happen but it isn't supposed to be a big deal.

That's a general symptom of people trying to be sold oversimplified stories, it's great to build physical intuition but it isn't right to let people think that their intuitions are more correct than they are.

Is there a nicer, mathematically natural way to state the "only count bouncing paths" condition?

Nope, that is wrong.

Also, is there a "nice" condition that specifies a path in the ellipse case?

You are aiming to prove that a unique path is followed when it isn't.

What should I have told that student?

You could clearly separate an expectation from a requirement. The probability 1 examples might be helpful since many of the non unique paths happen with probability 0 if you randomly pick initial conditions.

If you lay down that the purpose of science is to make predictions and that biases or expectations can bite you, then teaching relativity and quantum mechanics and general relativity later on might go better if people can set aside their baggage and admit that they just want to learn a theory to see what the theory predicts and see if the theory agrees with what we see. And that a theory doesn't have to respect your human opinions about what you think it should or shouldn't do.

In this case the prediction is whether a detector sees light from a particular direction at a particular time. If light is leaving in many direction it is reasonable to think it can arrive from many directions. Since there are beam splitters and half silvered mirrors it can arrive from multiple directions even if it heads out in just one.

You could even play laser maze, it's a real game (ages 8+, no affiliation).

As for your title question I'm interested in an answer because I hear people say there are unique solutions all the time, but they even say that when there are not unique solution so I don't know why they say that. In either sense of why, I don't know what theorem they are thinking of or even why they want such a result. Just .... why?

What should I have told that student?

The Fermat principle is interesting, but when in doubt use the basic laws of reflection and refraction.

It is also good to realize the Fermat principle does not give one actual path which light ray will follow, it gives the possible paths. Which ones will be realized depends also on things like orientation of the light ray source. If the light source in one focus shines in all directions, all those rays from one focus to another are going to be realized.

Fermat's principle says the path with minimum optical path or minimum time is chosen by light. It can be direct or indirect (containing reflections or refractions). But as others said, it's not necessarily unique, because there might be paths all with minimum optical paths, that means, all with equal minimum optical paths. That's exactly what happens in an elliptical mirror. It's the definition of ellipse that all paths between the two foci which include a reflection are equal. That's why the image of the object in one focus is formed in the other focus. If all the paths were not equal there wouldn't necessarily be a constructive interference leading to image formation. They are all minimum paths and they are REALLY chosen by light. But there are always some barriers (like your hand holding the light source) that block some of these paths. Then light won't bypass, because this new path is not a minimum path anymore. As for the direct path between the two foci, it is also a minimum, but with respect to paths in its neighborhood, not the paths mentioned above. So it is a choice too.

There's another extremely conceptually important class of situations in which there are multiple stationary-action trajectories with the same boundary conditions: situations in which the Lagrangian has a symmetry that leaves the boundary conditions invariant but changes the trajectory that connects them.

The simplest example is just a point object confined to the surface of a sphere and feeling the influence of gravity, which starts out initially balanced right at the top of the sphere. Clearly if you wait long enough, it will eventually slip off the top and eventually end up at the bottom. But given the boundary conditions "starts off at the top" and "ends up at the bottom a time $T$ later", which line of longitude does it go down to get there? There obviously can't be a single correct answer by the sphere's axial symmetry, but the object does physically end up sliding down one line. This concept is known as "spontaneous symmetry breaking" and is cornerstone of much of physics. The literature on this subject is truly staggeringly large, so you won't have a problem finding more info.

## protected by Qmechanic♦Aug 31 '15 at 22:02

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