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Does Fermat's principle ever give the longest time?

  1. Jul 12, 2015 #1
    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 cant 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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  3. Jul 12, 2015 #2

    Orodruin

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    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.
     
  4. Jul 12, 2015 #3
    True, but my point still stands, stationary doesnt mean youre at a minima, you could also be at a maxima. It doesnt even have to be global, it can be just local
     
  5. Jul 13, 2015 #4

    Orodruin

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    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.
     
  6. Jul 13, 2015 #5
    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.
     
  7. Jul 13, 2015 #6
    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?
     
  8. Jul 13, 2015 #7

    Extremal is stationary.
     
  9. Jul 13, 2015 #8

    Orodruin

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    No. The statement is based on the metric being Riemannian, not being constant.
     
  10. Jul 13, 2015 #9
    OK. What do we call it when there a collection of nearby paths that are all equal?
     
  11. Jul 13, 2015 #10
    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
     
  12. Jul 13, 2015 #11

    Orodruin

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    Geodesics in Riemannian geometry give locally minimised paths. In order to get a shorter path, you would need a non-local variation.
     
  13. Jul 13, 2015 #12
    Last edited by a moderator: May 7, 2017
  14. Jul 13, 2015 #13

    Orodruin

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    Referring to a book which we would have to buy to answer your question is not very effective...
     
  15. Jul 13, 2015 #14
    No you don't have to buy it. I don't have a copy.
     
    Last edited: Jul 13, 2015
  16. Jul 13, 2015 #15
    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)?
     
  17. Jul 13, 2015 #16
    An inflection, or saddle point, I think. Extremals are stationary, the converse need not be true
     
  18. Jul 13, 2015 #17

    stevendaryl

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    Light bouncing off a mirror is a case that is not a global minimum.
     
  19. Jul 13, 2015 #18
    Sweet. Thanks
     
  20. Jul 13, 2015 #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.
     
  21. Jul 13, 2015 #20

    Andy Resnick

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    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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