8
$\begingroup$

I posted a similar question about geodesics on Math.SE. Many sources (Wikibooks for instance) claim that the light could maximize the optical path length in some cases. But I don't think it's actually true since between two points I can always imagine a path with arbitrary length. Therefore, or the light minimizes the optical length or it's stationary (neither maximum nor minimum). My question is: is the trajectory in red in the figure below a local minimum or a stationary curve? enter image description here

If I'm wrong, please provide some examples which the light propagates in trajectory of maximum length.

|cite|improve this question
$\endgroup$
  • $\begingroup$ Note that in the cases where light follows a maximum length path (there was an example in Hetch), it is a local maximum, not global. The path in your example is not a minimum neither a stationary curve. $\endgroup$ – jinawee Feb 27 '15 at 14:15
  • 4
    $\begingroup$ In the same site you linked there is an example of maximum length: upload.wikimedia.org/wikibooks/en/1/1b/… $\endgroup$ – jinawee Feb 27 '15 at 14:38
  • 1
    $\begingroup$ @jinawee, even locally, the ellipse example doesn't reach a maximum. I could make a curve with deviations around OB, for example, which would have longer length. Could you please provide the title of the Hetch's book? $\endgroup$ – Mr. K Feb 27 '15 at 14:51
  • $\begingroup$ In that case it might be something like a stationary curve. I'll have a look if Hetch's Optics explains things further. Still, you try to calculate the path for which $\delta l=0$, using Euler-Lagrange equations. $\endgroup$ – jinawee Feb 27 '15 at 16:53
  • $\begingroup$ @jinawee You should turn your comment with the link into an answer. It is exactly what the OP was looking for. $\endgroup$ – Chris Mueller Feb 27 '15 at 17:48
1
$\begingroup$

Answer

Ray in red is the shortest path if and only if $\theta_I=\theta_R$ and light only travels in such path to reach point B via mirror.

It is not a local minimum, because the shortest way, of course is a straight line connecting $A$ and $B$.

Red line with $\theta_I = \theta_R$ is a stationary curve of the group of curves that go through A, B and a point (s) on the mirror.

Extra stuff

In an ellipsoidal case, you have three points

$O$-$D$-$O$

What is the shortest path connects $O$,$D$ and back to $O$? of course, it is a straight line going from $O$ to $D$ and then back to $O$. Although the time travels from $O$ to $O$ is maximum, for the particular path, $O$-$D$-$O$, light chooses the path such that time it travels is the minimum.

In the mirror example, the problem that is in question is not the minimum path between two points ($A$ and $B$). The real question is

Where is a point in the mirror $P$, such that if you connext $A$-$P$-$B$, it gives you the minimum distance?

Then you can make an argument as represented in wikibook page to find the point $P$.

Turns out, empirically, we have enough data and examples to conclude that light travels such path (the path such that time the light travels is the least).

Yes, in the mirror example, light source can be a light bulb, and the light from the light bulb can travel completely opposite to B, reflected off the wall of the room, bounced off 1000000 times, and finally reach the point B. Sure this way takes the most time.
However, the path that light have taken is the path such that connects all 1000000 points of reflection and gives light a least time to travel.

Hope it helps.

|cite|improve this answer
$\endgroup$
-1
$\begingroup$

The mirror isn't drawn right. The angles should be equal.

But it is a local minimum. Fermat's principle says minimum time. If the ray stays in one medium, it travels at constant speed. Minimum time is minimum distance. That is, there are many paths from A to mirror to B. The path a ray takes picks the point on the mirror that has the shortest path.

You can see it like this:

enter image description here

|cite|improve this answer
$\endgroup$
  • 4
    $\begingroup$ The OP's point is that Fermat's principle actually only specifies a stationary point, which could strictly speaking be a maximum or a minimum. However, there often aren't any maxima to trick us, because you could always add some additional deviation and get a longer path length. In addition, it is not true that "minimum time is minimum distance" in general--otherwise we wouldn't be able to derive the law of refraction. $\endgroup$ – zeldredge Feb 27 '15 at 14:32
  • 1
    $\begingroup$ Exactly @SirElderberry. I'm almost sure that maximum path lengths are impossible. I asked if the red trajectory above is a local minimum because if one considers a neighborhood containing the points A, B and the point where the light hits the mirror, then the trajectory from A to B (directly) will be shorter than the red trajectory. So A-mirror-B isn't a minimum path length (not even locally). $\endgroup$ – Mr. K Feb 27 '15 at 14:44
  • $\begingroup$ You are right. I missed the point. $\endgroup$ – mmesser314 Feb 28 '15 at 4:39
-1
$\begingroup$

As with Fermat, light won't "choose" a longer path as that would violate Feynman's principle of least action. It just doesn't happen.

|cite|improve this answer
$\endgroup$
  • 4
    $\begingroup$ Aren't most laws of physics of stationary action, not only minimum? $\endgroup$ – jinawee Feb 27 '15 at 17:14

Not the answer you're looking for? Browse other questions tagged or ask your own question.