Does Fermat's principle ever give the longest time?

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SUMMARY

The discussion centers on the application of Fermat's principle in optics and its relationship with geodesics in spacetime. Participants clarify that while both concepts utilize calculus of variations, Fermat's principle operates under a Riemannian metric, whereas geodesics are based on a pseudo-Riemannian metric. The consensus is that light typically follows the shortest path, and while local maxima can exist, they do not represent practical solutions in physical scenarios. The longest path, theoretically, could lead to infinite time, which is deemed non-physical.

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  • Understanding of Fermat's principle in optics
  • Knowledge of geodesics in pseudo-Riemannian geometry
  • Familiarity with calculus of variations
  • Concept of Riemannian versus pseudo-Riemannian metrics
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hideelo
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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?
 
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hideelo said:
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.
 
Orodruin said:
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
 
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.
 
Orodruin said:
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.
 
Orodruin said:
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?
 
stedwards said:
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.
 
hideelo said:
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.
 
hideelo said:
Extremal is stationary.

OK. What do we call it when there a collection of nearby paths that are all equal?
 
  • #10
Orodruin said:
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
 
  • #11
hideelo said:
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.
 
  • #13
Referring to a book which we would have to buy to answer your question is not very effective...
 
  • #14
Orodruin said:
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.
 
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  • #15
Orodruin said:
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)?
 
  • #16
stedwards said:
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
 
  • #17
hideelo said:
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.
 
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  • #18
stevendaryl said:
Light bouncing off a mirror is a case that is not a global minimum.

Sweet. Thanks
 
  • #19
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.
 
  • #20
hideelo said:
<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.
 
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  • #21
Andy Resnick said:
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
 

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