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Is there a closed smooth planar curve and a direction satisfying the following property?

The closed smooth curve separates the plane into two disconnected domains, one interior the other exterior. Cut a finite connected segment, called opening, out of the curve. "Shoot" a "photon" through the opening into the interior in the aforementioned direction. The photon bounces off of the curve with the same angle as the incident line with respect to the normal vector of the curve. The photon will remain trapped inside the interior after traversing an infinitely long distance.

An extension the question is here.

Hans
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  • With a circle, this will occur with most incoming angles (as soon as the subtending angle of the first ray is an irrational multiple of a full turn). The set of angles for which the ray eventually comes out has null measure. –  Jul 05 '19 at 07:24
  • In blackbody radiation example physicists use a sphere with a hole, isn't it, a planar version? – Nosrati Jul 05 '19 at 07:25
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    @Nosrati: the black body model doesn't assume specular reflection, but absorption or diffusion. –  Jul 05 '19 at 07:27
  • "moves always with a constant speed": why do you specify that ? –  Jul 05 '19 at 07:29
  • If you allow polygons, what is the reflection rule when you hit a vertex ? –  Jul 05 '19 at 07:35
  • Use regular n-sided polygon, I think works. better is for even $n$. – Nosrati Jul 05 '19 at 07:36
  • My intuition tells me that there is no such curve, but I can't explain why. –  Jul 05 '19 at 07:40
  • Oooops, my first remark only holds if the opening is a single point. For a finite aperture, a circle never works. –  Jul 05 '19 at 08:10
  • @YvesDaoust: The constant speed of a photon is to ensure the word "forever" makes sense as a photon with exponentially decaying speed can traverse a finite length with an infinitely long time. An alternative description would be to replace "forever" with "infinite distance". That may be better as it fuse two concept speed and time into one: space. You ask a good question with regard to the reflection rule. I surmise it does not matter that much as the probability measure of hitting the corner of a finite polygon is zero. But it can be made more precise if a solution is constructed. – Hans Jul 05 '19 at 08:32
  • @YvesDaoust: I also lean towards the nonexistence of such a curve. But what is the proof? – Hans Jul 05 '19 at 08:33
  • If you accept the "zero probability measure" idea, then a ray entering a circle through a single points comes out with zero probability measure. But you didn't mention probability in the OP. –  Jul 05 '19 at 08:35
  • Is the curve required to trap photons from every arriving direction, or from some arriving direction? I'm pretty sure the latter's not too tough, but the former may well be impossible (and requires you to say what happens to reflections when a photon hits a point of $C^1$-discontinuity, like a vertex of a polygon, about which you've said nothing so far). – John Hughes Jul 05 '19 at 10:38
  • @JohnHughes: I said certain, synonymous to some, direction in the question. I will though make it even clearer. – Hans Jul 05 '19 at 18:23
  • @exosphere: In that case, how is the photo going to remain trapped in the interior? – Hans Jul 05 '19 at 18:25
  • @YvesDaoust: I have edited the question to exclude the corners. – Hans Jul 05 '19 at 18:28
  • @exosphere: I do not understand what you are trying to assert. Do you intend your procedure to be a solution? I have stated in the question that the segment cut out is supposed to be connected and of finite length. – Hans Jul 06 '19 at 02:35

2 Answers2

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It is possible to make such a curve for which a particular ray of light will keep bouncing back and forth within a confined region even if the curve is not closed.

One specific case is a hyperbola, but I'm sure there are other examples.

The hyperbola has a reflective property, like the other conic sections. If a ray of light is aimed at one focus, the reflection of that ray will be heading towards the other focus. This will then be reflected back and forth indefinitely.

enter image description here

All you need to make a concrete example are the two segments of the hyperbola around the x-axis, and then fill in the rest with anything that does not block the rays.

And, yes, I stole the illustration from the reference I gave as I'm no good at making them. If my answer is not clear enough, I can try making a better drawing.

Einar Rødland
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  • Cool. I knew the reflection property of an ellipse and a parabola, but not of a hyperbola. – Hans Jul 06 '19 at 18:36
  • How is the ray "reflected back and forth indefinitely" ? IMO it will just exit after three reflections. –  Jul 06 '19 at 19:14
  • @YvesDaoust: Why? Let's take the incoming ray marked I in the illustration directed towards focus point B. After being reflected in the B-side mirror, the ray (marked R) is heading towards focus point A. Same thing happens to this ray: it is reflected by the A-side mirror, with the reflection heading towards B. And so on. Ultimately, you will get a ray that is passing back along (closer and closer to) the x-axis. – Einar Rødland Jul 06 '19 at 21:21
  • @EinarRødland: I see. The figure was misleading. –  Jul 07 '19 at 10:28
  • The ellipse works as well. See my answer. – Hans Jul 07 '19 at 22:05
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Inspired by Einar Røland's answer, I found the ellipse works as well. The ellipse has the same reflection property as the hyperbola, namely, the light ray emanating from one focal point will be reflected by the boundary back towards the other focal point. A light ray shot in from a small aperture on the boundary sufficiently away from its intersection with the major axis towards one focal point will converge to the major axis and trapped in the vicinity.

Hans
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