Fermat's principle and Snell's law

1. Jun 23, 2008

Nick89

Hi,

While reading about Fermat's principle, and how to use it to derive Snell's law, I came across something I can't find any information about...

I had heard about Fermat's principle before; that it said that the path taken by a lightbeam is the path that takes the least time.

Now however, I read that it is not just the least time, but sometimes also the longest possible time. The book in question doesn't elaborate further, it just states this fact...

When does a lightbeam takes the longest possible path, and why?

2. Jun 23, 2008

Andy Resnick

That's interesting- I suppose it's possible because Fermat's principle requires that light rays travel *extremum* paths, which implies that maximum paths are acceptable. I guess they are not considered because they would require infinite time, and thus not really useful. I wonder if one can interfere with a maximum path and observe an effect?

3. Jun 23, 2008

Mapes

Andy is correct; the modern form of Fermat's principle is:

A light ray must traverse an optical path length that is stationary with respect to variations of that path. [Hecht's Optics]

It's quite easy to sketch out an arrangement involving mirrors in which a light ray takes a longer distance than necessary.

There's a fascinating video (I believe it's http://video.google.com/videosearch?q=feynman+new+zealand&hl=en&sitesearch=#" [Broken]) of Richard Feynman explaining Fermat's Principle by describing how the probability amplitudes of photons all add together constructively when the optical path is maximized or minimized (or at a saddle point).

Last edited by a moderator: May 3, 2017
4. Jun 24, 2008

Nick89

Sounds interesting, although I can't watch the video (it says it may not be available..?)

I guess by distance here you mean 'optical path length', right? The actual distance is not what matters but the time it takes, right?

Could you show me an example of a setup where light takes a maximum path length?

5. Jun 24, 2008

dx

Consider a light source in front of a mirror A, and another point on the same side, B. The principle of least time says that the path taken by light is stationary. Clearly there are two such paths, one going straight from A to B, and one going to the mirror and reflecting to B obeying the law of reflection. Both these are stationary paths, and both are taken. Does light take both of them at the same time? No! That's the important thing to realize about the principle of least time. It is not teleological. The photons determine their motion only locally. If a photon is emitted in the direction of the mirror, it will go to the mirror in a straight line (stationary time), and reflect off the mirror with equal angles (again can be shown to be stationary), and then travels in a straight line again. If the photon is emitted in the direction of B, then it will go in a straight line to B.

6. Jun 24, 2008

vanesch

Staff Emeritus
In fact, quantum-mechanically, yes! This is why you can obtain interference between the two paths.

7. Jun 24, 2008

dx

I was assuming the photons were little classical balls :)

8. Jun 24, 2008

dx

Also, the interference of paths in quantum mechanics is not for stationary paths. It's for all paths, and the result of that interference is the stationary paths. So, the final stationary paths don't interfere.

9. Jun 24, 2008

maverick_starstrider

In fact all of physics can basically be phrased in terms of the 'principal of stationary action'.

10. Jun 25, 2008

vanesch

Staff Emeritus
Ok, so how about this: you have a spherical point source at point P, a plane mirror, and a detector at point Q, on the same side of the mirror as point P. As before, we have, according to Fermat, two paths of "minimum length" from P to Q: one hitting the mirror, the other directly. Now, assume that they are 180 degrees out of phase (depends on the positions of P, Q, the mirror, and the wavelength of our monochromatic source). That means, NO light arrives in Q. Well, some will, because the amplitudes are not equal as they diminish in 1/r, so the path via the mirror will have a smaller amplitude than the direct path, but if P and Q are not too far from the mirror, and not too close together, that can be a small difference.

Remove the mirror: you eliminated the "mirror path", and now light arrives at Q. Put a black paper between P and Q: you eliminated the "direct path" and now light arrives at Q via the mirror. Leave the two paths in place, and none (or very little) arrives in Q.

11. Jun 25, 2008

Andy Resnick

I'm not sure that's the correct way to apply extremum paths- after all, the line from source to mirror to detector is a minimum also.

I read Feynman's description of how he developed his concept of "sum over histories". Free space is constructed as a limit- start with a opaque plane, begin adding regularly spaced slits, and let the slit spacing go to zero and the number of slits increase without bound. Then consider the effect os successive planes. In this way, actual paths are interferences of 'virtucal paths'. And the actual path has stationary action.

12. Jun 25, 2008

dx

But that's not interference in the sum over histories sense. That's just ordinary interference of two separate light beams.

13. Jun 25, 2008

vanesch

Staff Emeritus
It's the same ! The "ordinary" interference of two separate light beams is a good analogy (more than that in fact) to the single-particle sum over histories interference.

What you USUALLY find with Fermat's principle is that there is ONE "extremal" path. In that case, it is the same as the classical path. However, in certain cases, you may have "multiple solutions" to the extremal problem, such as in the case of the direct and reflected paths. In that case, it seems that there are now "two classical paths" that "interfere". Or three. Or four. Or... and in the limit of infinity, you find back the sum of histories, if we have all turned them into "small local extrema" (say, by putting up screens with holes in it). That's what Andy Resnick referred to.

14. Jun 25, 2008

dx

Ok, but what I said in my original post was that although there are two possible stationary paths in the situation, any given photon only takes one of them. Different photons can take different paths to the second point, and they can interfere. The same photon will not take both the paths and interfere with itself because all those have already been eliminated by the sum over histories. The result of the sum over histories is that the photon either takes stationary path one or stationary path two. If you point a laser at the mirror, all the photons take only the first stationary path. Why don't they take the other path? It is stationary too, isn't it? The reason is that the principle of least time is not teleological. They don't "know" what possible paths are available to them. If you shoot them in a certain direction, they go in that direction, adjusting their motion only locally to keep every local segment stationary.

15. Jun 25, 2008

Nick89

I feel this discussion is going a bit beyond my knowledge lol, but please carry on ^^.

Although I still don't know if it's actually possible for a lightray to take the longest possible path... You say that if you point a laser in a certain direction, the photons will travel in that direction only and do not know (or 'try') any other directions. They only 'check their position' locally (for example in a lens where the lightray bends to still take the fastest path).

Is there any example of a light ray being pointed at a lens for example, and taking the longest possible path instead of the shortest (time-wise) ?

16. Jun 25, 2008

Mapes

The example used in Hecht's book is a mirror shaped like part of an ellipse, with a source at one focus and the detector at the other focus. Draw a light path from source to detector that reflects off this mirror. Now consider two additional mirrors that are tangential to the original mirror at the point of reflection. The first additional mirror has less curvature than the elliptical mirror, or perhaps it is flat (no curvature at all). The light path you drew takes the shortest possible path that reflects off this mirror. The second additional mirror has a higher curvature than the elliptical mirror (it curves in closer to the source and detector). The light path you drew takes the longest possible path that reflects off this mirror.

17. Jun 26, 2008

vanesch

Staff Emeritus
If that were true, then the number of photons arriving at the point of arrival would be the sum of those that took path 1 and those that took path 2 right ? But that would mean that there are MORE when both paths are open than when only one is open, in direct contradiction with destructive interference. Photons interfere with themselves, not with "other photons" (in most usual optical conditions). This is demonstrated by the appearance of interference phenomena at low photon count rate. The only way to do that in the path formalism is to consider that photons took BOTH paths.
As I said, most of the time, this is not obvious, because most of the time, there is one single "extremal" path, which corresponds then to the classical path, and you can think of the photon "sniffing locally around" to "keep on the extremal track". But in the case of several solutions to the extremal path, as is the case in any interference experiment, the "local sniffing around" can only determine the different alternative paths, and in the end you still have to add together the phases of all the paths to know what is the amplitude of arrival. And those amplitudes are valid for each individual photon by itself.

18. Jun 26, 2008

dx

Ahh.. I understand now. Thanks for the education.