Does Fermat's principle ever give the longest time?

[hideelo]

I know that the way we calculate in Fermat's principle in optics is to take the path length (as an integral) and demand that it be stationary to first order. Now this approach is mathematically the same as calculating a geodesic, or finding stationary action, namely we use the calculus of variations. In the last two cases there are times when we use this to find the longest possible path. The example that I am thinking of is in relativity where we find the geodesic by making the "proper time" be as long as possible i.e. the real path (the geodesic) through spacetime will be the one for which the time elapsed for the particle taking this path is longest.

The mathematics of a geodesic through spacetime and Fermat's pinciple however, are the same. I know that in the handfull of experiments that I have done, the light takes the shortest path. However given the math I can't think of a reason why the light can't take the longest path and still satisfy Fermat's principle. Are there cases where this actually happens? If not, why not?

 

[Orodruin]

The mathematics of a geodesic through spacetime and Fermat's pinciple however, are the same.

Not exactly, spacetime is based upon a pseudo-Riemannian metric while Fermat's principle builds on a Riemannian metric. This makes a lot of difference.

 

[hideelo]

Not exactly, spacetime is based upon a pseudo-Riemannian metric while Fermat's principle builds on a Riemannian metric. This makes a lot of difference.

True, but my point still stands, stationary doesn't mean youre at a minima, you could also be at a maxima. It doesn't even have to be global, it can be just local

 

[Orodruin]

The point is that if the space is Riemannian, then you can always make a small bump and end up with a longer curve. It therefore cannot be a local maxima. You might find local saddle points.

 

[stedwards]

The point is that if the space is Riemannian, then you can always make a small bump and end up with a longer curve. It therefore cannot be a local maxima. You might find local saddle points.

Now that you bring it up, I can's see anything wrong with your statement, though it clashes with what I've taken verbatim; that the modern upgrade to Fermat's Principle states that paths are either stationary or extremal (minimum or maximum).As I vaguely recall, the path of light inside an ellipsoid from focus to focus is a stationary path.
For a convex lens, there are multiple paths from object plane to image plane. Are these stationary?

I can't think of a maximal solution without introducing an arbitrary negative sign to the integrand.

 

[hideelo]

The point is that if the space is Riemannian, then you can always make a small bump and end up with a longer curve. It therefore cannot be a local maxima. You might find local saddle points.

With a constant index of refraction, I see your point. But can you tweak the index of refraction in different places so that this happens?

 

[hideelo]

Now that you bring it up, I can's see anything wrong with your statement, though it clashes with what I've taken verbatim; that the modern upgrade to Fermat's Principle states that paths are either stationary or extremal (minimum or maximum).As I vaguely recall, the path of light inside an ellipsoid from focus to focus is a stationary path.
For a convex lens, there are multiple paths from object plane to image plane. Are these stationary?

I can't think of a maximal solution without introducing an arbitrary negative sign to the integrand.

Extremal is stationary.

 

[Orodruin]

With a constant index of refraction, I see your point. But can you tweak the index of refraction in different places so that this happens?

No. The statement is based on the metric being Riemannian, not being constant.

 

[stedwards]

Extremal is stationary.

OK. What do we call it when there a collection of nearby paths that are all equal?

 

[hideelo]

No. The statement is based on the metric being Riemannian, not being constant.

What I meant was that perhaps given some index of refraction as a function of position, making a little "bump" in the path might make the time shorter, so there can be a unique longest path

 

[Orodruin]

What I meant was that perhaps given some index of refraction as a function of position, making a little "bump" in the path might make the time shorter, so there can be a unique longest path

Geodesics in Riemannian geometry give locally minimised paths. In order to get a shorter path, you would need a non-local variation.

 

[stedwards]

This author, https://www.amazon.com/dp/8120341899/?tag=pfamazon01-20, claims a maximal. section 1.6

 

[Orodruin]

Referring to a book which we would have to buy to answer your question is not very effective...

 

[stedwards]

Referring to a book which we would have to buy to answer your question is not very effective...

No you don't have to buy it. I don't have a copy.

 

[hideelo]

Geodesics in Riemannian geometry give locally minimised paths. In order to get a shorter path, you would need a non-local variation.

OK, pushing that forward, do we have physical cases where light travels on geodesics which are locally minimizing but not globally (with respect to time)?

 

[hideelo]

OK. What do we call it when there a collection of nearby paths that are all equal?

An inflection, or saddle point, I think. Extremals are stationary, the converse need not be true

 

[stevendaryl]

OK, pushing that forward, do we have physical cases where light travels on geodesics which are locally minimizing but not globally (with respect to time)?

Light bouncing off a mirror is a case that is not a global minimum.

 

[hideelo]

Light bouncing off a mirror is a case that is not a global minimum.

Sweet. Thanks

 

[stedwards]

If you are looking for an arrangement with a local minimum, in the sense that all nearby paths are longer, the optical path length between two points within a spherical mirrored surface is minimal.

 

[Andy Resnick]

<snip>

The mathematics of a geodesic through spacetime and Fermat's pinciple however, are the same. I know that in the handfull of experiments that I have done, the light takes the shortest path. However given the math I can't think of a reason why the light can't take the longest path and still satisfy Fermat's principle. Are there cases where this actually happens? If not, why not?

Sure- it's a ray that goes in the wrong direction- away from the optical system. It would take an infinite amount of time for the ray to reach the image plane. It's a non-physical result.

 

[hideelo]

Sure- it's a ray that goes in the wrong direction- away from the optical system. It would take an infinite amount of time for the ray to reach the image plane. It's a non-physical result.

That is an epic, perfect answer. It also sums up the fact that for a maximal time, if it's unbounded, the longest time is infinity i.e. It will never get there in any finite time . Thanks