1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fermat's principle spec case

  1. Apr 17, 2013 #1
    Fermat's principle is the "path of least time principle" or we can say that "he path a light ray takes is a local minimum". Quantum physics also says that the light take (test) all possible paths and only the minimal remains.
    Is it working in special cases like this too?
    Let's examine the fig. 1-3 below.

    fermat_spec.png

    Fig1. is the normal case, the light goes from A to B and finds the "least time" path through the object.

    Question 1:
    What if we cut the object (Fig. 2), resulting a possible better path with less time? Will the light go unchanged, and arrives to the B as the dotted line shows, or will it find the less-time-path, even if this means that it must go through the object in a different angle?

    Question 2:
    What happens if we cut the object even shorter (Fig. 3), crossing the path of the ray goes normally through the object? Let's assume that in this case the "least time" path is if the light does not enter the object, but goes on the surface. Is this possible, and the path will be the orange one? Or in this case the light will do something else, like the reflecion inside, and will reach the B' ?

    My question in general:
    Is the "least time between two points" principle is an always working principle (rule 1), or the light is "not so smart" and can decide only locally on the entering points and can't "think further" (rule 2).
    If the "rule 2" is the real one, why QP says that the light take all the possible paths?

    Thank you!
     
  2. jcsd
  3. Apr 17, 2013 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    welcome to pf!

    hi zrek! welcome to pf! :smile:
    no, it will always be quicker for the light to go through the glass, very near the corner …

    check it and see! :wink:
     
  4. Apr 17, 2013 #3
    Ok, tiny-tim, I see that on he corner it will enter into the glass, but on a small part of the path it will follow the surface, right? Of course only if this is the case of the principle is working "in general", which was my main question.
    May I understand your answer that you are sure that the principle working "in general", so the light is "smart" and "knows" that the glass will end soon and will choose a different, "unusual" angle (fig 2)?

    This behavior surely different to the explanations of the "general physics", that is about the substance property and the refractive index, this is why I think that this topic is about the Quantum Physics. This behavior explained by the QF and the calculation result of the all possible paths, am I right?

    Unfortunately I have not found answers for my special cases on the Internet, I have found only explanations and examples of the cases where the "least time" principle resulted the same as the "refractive index" calculations. It would be nice if someone shows me a link where I can find an explanation on the net that shows an example where these two are different.

    Please strenghten my tought that in case of the specially cut object (fig2 and fig3) the light will choose a different way and is able "to see before its path" and find always the least time path.

    Thank you!
     
  5. Apr 17, 2013 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    nooo …

    then it would be going through the air round two sides of a triangle, wouldn't it? :wink:

    i think once you've convinced yourself that fermat works for this example, you'll be happy applying it generally :smile:
     
  6. Apr 17, 2013 #5
    I'm sorry but I don't get what do you mean on "round two sides of a triangle" :-(
    I see that it is strange to follow the surface and then suddenly enter the object, but I don't know why, it seems to me that this is the least time path. The other possible explanation is that in this case the light will be reflected, or something else, but it can't reach the "B" (there is no possible path at all).

    I'm not yet convinced myself, I still need some help.

    ...and what about the fig 2? Is the orange path possible? Is it possible that the light goes through in a different angle than in the fig 1?
     
  7. Apr 17, 2013 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    i mean if the light starts at A, goes through the air straight to B, then through the air along the edge of the glass from B to C, then to D E F etc,

    then the light could go quicker if it went ACDEF etc, leaving out the detour via B :smile:
     
  8. Apr 17, 2013 #7
    I really appreciate your help, tiny-tim, but I'm afraid of I don't understand you clearly :confused:
    I'm not native english, maybe I don't get the information behind the sarcasm and I'm not always understand if someone is kidding me. I think I understand only the straight answers...

    So once again I'd like to try to explain what is my problem and where I need some help:

    There are 2 kinds of interpretation of the path of the light:
    1) Based on the substance property, the refractive index: if the light enters to a material with different refractive index, this definies clearly how it will behave, which angle will be the direction of the light.
    2) Based on the QF and the Fermat's principle: from A to B the path will be always the shortest in time.

    I hardly believe that this two explanation results the same in every case. I do believe however the second one, the "least time" is the correct one. Now I'd like to find somewhere an example which shows the difference between this two approach of the problem.

    Let's see this figure:
    fermat_spec2.png

    Between A and B there are several objects with different refraction indexes. As you can see that this is some kind of a maze.
    Let's shot a light ray from the point 'A' to a random direction and examine where it will go...
    Let's try to calculate the path by the simple way: when we reach an object, calculate the new angle by the refraction index, and step forward. The orange line shows this path.

    Now let's calculate the all of the possible paths and let's select the "least time" one. The dotted line represents this solution.

    I think that if the task is to find the least time path throught this labyrinth, we can't succeed easily, only by following the refraction index rule. We need much more work to find the way with the shortest time.

    Am I right or wrong?
     
  9. Apr 17, 2013 #8

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    The paths you drew through that coloured maze were not calculated, were they? You just made them up, I think.
    But, if you consider a lens, there are many different paths through, from object to (focussed) image. They are all the same (shortest possible) length so there may well be multiple possible paths through that maze - as long as they all take the same time. Nothing needs to be 'violated'.
     
  10. Apr 17, 2013 #9
    Of course the image is only a demo, not based on calculation.
    "Nothing needs to be 'violated'"
    I too think so, but I feel that I misunderstood something.
    If it is working as I explained, I'm sure that you can't find an optimal way through a maze just by analyzing the next step. This will not work.

    But now I think that maybe I know what is this all about. The priciple is not about finding the least time path between the A and B, but finding the least time path which fullfills the rule of the step-by step way given by the refractive index.
    I mean that the following selection is explained by Fermat:
    From the point "A" we shoot the light to every direction (360 degrees) and calculate the (step-by-stem, refractive index based) path for every possible case that finally leads to "B". Finally we will select only the least time path from these: this will be that path the light will go on. By this, we can determine that which angle the "B" will receive the light from.

    Am I right, or just getting more confused ??? :uhh:
     
  11. Apr 17, 2013 #10

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    You are describing the common 'ray tracing' method that is commonly used in optics designs. And yes you can determine which angle(s plural)) the light will arrive from.
     
  12. Apr 18, 2013 #11
    I slept on it, and still I don't feel that I got answers to my questions. Not easy to tell what is my problem, but in general I don't get what the Fermat principle is trying to explain.
    I found lots of pages about explaining Snell's law by it, and seems that simply the "least time" principle is enough to calculate the path, no other criterion is necessary.
    http://cnx.org/content/m12895/latest/
    Sometimes it seems that they explain some behavior about the the wave-particle duality by it, like here:
    http://en.wikipedia.org/wiki/File:Snells_law_wavefronts.gif
    Also clear that it is working even with lasers:
    http://simple.wikipedia.org/wiki/Snell's_law

    Seems also that the principle should work in difficult cases too, where there are many entering points or inhomogene substance, in this case it is calculated by the calculus of variations:
    http://en.wikipedia.org/wiki/Calculus_of_variations#Fermat.27s_principle

    Not easy to explain what I still don't understand, maybe tomorrow I'll try to show you in a longer post.
     
  13. Apr 18, 2013 #12

    jasonRF

    User Avatar
    Science Advisor
    Gold Member

    Note that for the first example given in that link, the path of least time is the one that goes directly from A to B and does not reflect at all. Yet both rays (AB and ACB) are legitimate paths for the light to take, even though AB clearly takes less time that ACB. Feynman explains this in the Feynman lectures (vol I) - I recommend you read it as he is much clearer than I could ever be. The upshot is that the paths that the light will take are those for which small changes in the path produces "second order" (very small - smaller than "first order") changes in the propagation time; in other words, there are many nearby paths with almost the same propagation time.

    Also, whenever you are solving these kinds of ray tracing problems Snell's law must hold. If you are using some technique to calculate paths that violates Snell's law then you know you are doing something wrong ...

    jason
     
  14. Apr 20, 2013 #13
    Yes, in other examples they used to draw a barrier between A and B to prevent the direct path.
    If I understand it well, this is the point when they used to say that the "interference" kills the other, non-optimal paths, right?
    I'd be happy to agree with you, but I think I missed something. There must be an additional criteria that prevents the non-Snell-like behavior. If I use only the "least time" principle for the calculation, sometimes other path would be optimal compared to the Snell's law says.
    For example in the post #1 the fig2. orange path why is not good?
    I also miss the "critical angle" explained by the Fermat's principle.

    (But these are not my main problems, I'll explain it later on)
    I'd like to see this principle clearly, this is why I asking your opinion, thank you!
     
  15. Apr 20, 2013 #14

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    I think you have the wrong idea about what Fermat is actually saying. If you start off with a 'ray' in a certain direction, Fermat will tell you where it will end up. Fermat doesn't imply that there's only one path between A and B. Where there is partial reflection, Fermat works just as well. Can you think of any reason why it should not apply to Total Internal Reflection?
     
  16. Apr 21, 2013 #15
    I'm pretty sure that I miss something about the Fermat's principle, but I don't know what. :frown: :smile:
    It is clear for me that there may be several proper paths between A and B but they must be finally all optimal, with the same minimal time. (am I right?)
    I assume that it is working even if we are thinking of the light as wave, and Fermat finally determines if it is arrived to B, how it is appear as "ray", particle.
    http://en.wikipedia.org/wiki/File:Snells_law_wavefronts.gif

    I think I understand this part.
    --------
    I still don't get why the orange path between A and B is not allowed (post #1 fig.2)
    As I see the orange path is time-optimal, the normal path (Snell's law) takes more time in that case.
    Am I right?
     
  17. Apr 21, 2013 #16

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    Of course there can be many paths. Just think of how a paraboloid works. The focus us where all path lengths are equal.

    Have you proved that the orange path is shortest time? But the shortest time (that's minimum phase delay) from the point of incidence is given by Smells law. Fermat will operate over each infinitessimal step and give Snell. There can always be shorter ways through that involve extra paths that the light can't know about - but you can't see round corners; you have to use a mirror. ;-)
     
  18. Apr 21, 2013 #17

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    I just thought. Is it possible that you are assuming that Fermat works 'globally'? In fact, it works in small steps and it just has the effect of looking as thought it worked it all out on a large scale. In the example of your figure, as always, you need to satisfy phase coherence as well as Fermat. There is no way that the direction could suddenly take a sharp left turn, any more than it could go into the glass at any other than the Snell predicted direction or be reflected asymetrically.
    I think you are assuming that Fermat is the prime idea here but imo, it is only a principle which helps in making predictions. Remember, it is pretty ancient!
     
  19. Apr 22, 2013 #18

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    ah …

    there is a path that goes via the endpoint, C, of the barrier …

    light is diffracted at C, and some of it does get to B ! :wink:

    (remember, fermat's principle doesn't say that all light from A gets to B)
     
  20. Apr 22, 2013 #19
    Yes, in several examples I saw that they suggested that (the modern version of) Fermat is (1) general and (2) working in every case, and not only in case of 2 homogene substances. If the path is going through difficult material combination, they use the calculus of variations ( http://en.wikipedia.org/wiki/Calculus_of_variations#Fermat.27s_principle ) to determine the path.
    I agree, but my problem is that they explain the Snell with Fermat but I have not found that where the Fermat says that this is impossible. I miss an an additional restriction that prevents this behavior. Is it come from the phase coherence that you mentioned? But how?
    I don't agree with you in this point, since as I read, the Fermat is the deeper rule and the base of the others (this principle drew Feynmann's attention to use it in QP too) Snell's law even more ancient.
    "There can always be shorter ways through that involve extra paths that the light can't know about"
    This is what I assume: the Fermat is all about that the light will always "knows" the shortest time path between any A and B (if there was a light travelled between A and B, then that path was surely the least time available of the all the possible)

    I know that my knowledge is not perfect about this, and maybe soon I'll find what is missing.

    The wiki says:
    "...modern statement of the principle is that rays of light traverse the path of stationary optical length with respect to variations of the path.In other words, a ray of light prefers the path such that there are other paths, arbitrarily nearby on either side, along which the ray would take almost exactly the same time to traverse"

    I think I know what the "variations of the path" means, but what they mean on "stationary optical length" and "arbitrarily nearby" exactly? It would be nice of you if you try to explain it to me (or show me a link for it), thank you.

    (I'll open a new thread here, in the next post, but please try to answer this too, thank you! :smile:)
     
  21. Apr 22, 2013 #20
    Here is an example, shows that the Fermat is better than Snell.
    View attachment 58104
    fermat_spec3.jpg
    (The fig. 21 shows that I assumed that the orange path is not available, since in the point B0 the path is not "least time optimal". But I changed my mind, this would be not relevant, only the final result counts: in the point B we measured the light ray, so it must be optimal to that point and irrelevant if during the path there are "seemingly not optimal" points)

    Let's see the fig. 22.
    To make it simple, the A is far in the infinity.
    If we are not counting with the strangely cut part, the path#1 is the correct one, fits to Snell's law.
    But if you take a look at the path#2, you can see that it is the least time path (and still fits to Snell!).
    Am I right that the ray of light will go in the path#2?
     
    Last edited: Apr 22, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Fermat's principle spec case
Loading...