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Let $\gamma\colon[a,b]\to \mathbb R^3$ be a smooth curve with curvature $< 1$. Consider a tube, formed by one parameter family of unit circles with center at $\gamma(t)$ in the plane orthogonal to $\dot\gamma(t)$.

A light ray which is comming into tube from one end and bouncing with perfect reflection from the interior walls will emerge from the other end with probability 1; see this question. Let us call a tube with this property an optic fiber. (Note that I want an optic fiber to be bidirectional.)

One can construct an optic fiber along the same lines using any simple close smooth plane curve $(x(\theta),y(\theta))$ instead of circle. To do this one has to choose a parallel normal frame $e_1,e_2$ along $\gamma$ (i.e., such that $\dot e_i(t)\parallel\dot\gamma(t)$ for all $t$) and consider the tube $[a,b]\times\mathbb S^1\hookrightarrow\mathbb R^3$ defined as $$(t,\theta)\mapsto \gamma(t)+x(\theta){\cdot}e_1(t)+y(\theta){\cdot}e_2(t)$$ (The condition that the frame is parallel implies that any normal plane to $\gamma$ cuts tube at right angle.) This way we get an optic fiber with congruent ends.

Question 1. Are there any constructions of optic fibers different from the one described above?

In other words, is it always possible so slice an optic fibers by planes which cut the walls at right angle?

In particular,

Question 2. Is there an optic fiber with noncongruent ends?

Comments

  • I feel that the answer is "NO", but have no idea "WHY".
  • From Liouville's theorem, it is clear that the ends must have the same area.
  • I realized that if the walls are only piecewise smooth then one can make an optic fiber with a pair of equidecomposable figures at the ends. (The construction is the same, but one splits tube into few on the way and then rearrange them back together.)
  • In dimension 2, a line passing through focal points cuts from confocal ellipses an optic fiber. (I learned it from Arseniy Akopyan.) I do not know smooth 3D examples of that type. [One might also impose an additional condition on optic fibers that random ray which starts inside leaves it with probability 1. The described confocal-ellipses-example does not have this property.] enter image description here
  • An extract from the answer of Marcos Cossarini: Note that if one can cut an optic fiber in two pieces in such a way that almost all rays pass the cut at most once then the cut has to be flat and orthogonal to the boundary. After such cut, one gets two optical fibers. Applying a bit of differential geometry the problem can be reformulated in an equivalent way: is it true that any optic fiber admits such a cut.
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Other than that they may not be bidirectional with respect to light transmission, why disallow frustrums (truncated cones)? Gerhard "Ask Me About System Design" Paseman, 2011.10.09 –  Gerhard Paseman Oct 10 '11 at 5:35
    
No, I want optic fiber to be bidirectional (I will make it more clear). But you right, if you just want to send all rays from the first end to the second that you can get many more examples. –  Anton Petrunin Oct 11 '11 at 0:39
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The title of this question always makes me think this would make for a fantastic sci-fi movie. –  Cam McLeman Mar 1 '12 at 14:33

3 Answers 3

This is a suggestion rather than an answer. I've decided to think out loud in the hopes that someone else can take the thinking and run with it.

Consider the picture from the ray's point of view, and let's consider a (possibly not more restrictive) definition of a "good optic fiber", which I will not give but only suggest here. A good optic fiber makes sure that not only can light pass through both ways (i.e. is bidirectional) but doesn't slow down much. I originally thought of the 2 dimensional version of this as "For every point p, make sure reflected light off p goes further down the tube in the same direction", but now I am thinking "for every open patch on the tube whose reflection through p does not intersect with itself, make sure that reflection does not change much." So the idea is the light coming from a part of the wall of the tube (P) hits p, and reflects to hit another part of the tube (Q), that P and Q are disjoint and satisfy some conditions if they are sufficiently small, say if P has area at most epsilon then Q is guaranteed to have area not much different from epsilon times the appropriate linear scaling.

With an appropriate definition of "good optic fiber", you may be able to show that light travels quickly through the tube as long as the entrance ray is less than, oh say 89.9 degrees off the perpendicular axis of the tube. The next thing would be to show that any optic fiber can be epsilon-approximated by good optic fibers (or not), and then if our luck held, show that the answer to both questions would be NO for good optic fibers and that the answers would hold in the limit of the approximations.

Stop me if you heard this one before.

Gerhard "Oh Wait, I'm Done Already" Paseman, 2011.10.13

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I don't think I understand your patch idea, but you probably realize that the go-further-down-the-tube idea runs afoul of Anton's example that I tried to illustrate here: mathoverflow.net/questions/70421/… . In his construction, the lightray always does go "further down the tube," but the distance "further down" diminishes and the ray stops progressing in the limit. –  Joseph O'Rourke Oct 13 '11 at 21:11
    
The hope was to come up with an appropriate nice subset of optic fibers to use and then show this was a dense subset etc. The patch idea was an attempt at having the good fibers be uniform so as to avoid Anton's example, and show that good fibers allowed light to travel in time proportional to something like tangent of the incidence angle to traverse the tube. The patch idea was to say that the geometry involved in a reflection was not too bizarre: small dots reflected through p aren't very distorted by the receiving wall. Gerhard "Ask Me About System Design" Paseman, 2011.10.13 –  Gerhard Paseman Oct 13 '11 at 21:53
    
Perhaps I mean secant rather than tangent. Anyway perhaps Anton can find a dense subset of fibers for which he can get nice results, and then get closure somehow (pun intended). Gerhard "I Guess I Wasn't Done" Paseman, 2011.10.13 –  Gerhard Paseman Oct 13 '11 at 21:56

Suppose that the normal frame along $\gamma$ twists. Are you sure it still constitutes an optic fiber? I am imagining, for example, smoothly twisting an ellipse of nonzero eccentricity.
           Twisted ellipse

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Тhe frame should be parallel; i.e., $\dot e_i(t)\parallel \dot\gamma(t)$ for all $t$. The curve $\gamma$ and the form of one end completely describes the tube. –  Anton Petrunin Oct 8 '11 at 20:01
    
My apologies for not understanding that frame-parallel notation; now I do. Still, it seems gentle(?) twisting might be a way to achieve a nonstandard optic fiber, a possible answer to your Q1. –  Joseph O'Rourke Oct 8 '11 at 21:26
    
The only way to prove that something is an optic fiber which I know is to slice it by planes, so that the planes always orthogonal to the walls. Any other way would be an answer to Q1. –  Anton Petrunin Oct 9 '11 at 0:11

I don't know why this question appeared yesterday in my main mathoverflow screen, since the last comment appears to be from March 1st. I also think the conjecture is true, but my argument has plenty of holes. Here's what I thought.

Let $\Omega$ be an optic fiber that is the interior region of a compact connected $\mathcal C^1$ surface in $\mathbb R^3$, that is the union of a reflective surface $R$, a starting surface $F_0$ and an ending surface $F_1$. I assume (0) all optic fibers are like this.

Assume (1) that you can foliate $\Omega$ by a uniparametric family of plane surfaces $(F_s)_{s\in[0,1]}$, such that every ray starting at $F_0$ with positive initial $\dot s$ keeps a positive $\dot s$, until it reaches $F_1$. Let's call this a foliated optic fiber. Assume (2) that all the $F_s$ are plane surfaces and call it planely foliated, and let's call the fiber Petruninean if each $F_s$ is orthogonal to the reflective surface $R$ along its border.

Claim (3): Every planely foliated optic fiber is Petruninean.

Let my say why I believe it.

Let $m$ be the last value such that the fiber is Petruninean for $s\in[0,m]$. By finding a foliation of curves that is orthogonal to $F$ we can identify the points of $F_s$ at different values of $s$, and think of $F_s$ as a uniparametric family of regions in $\mathbb R^2$.

Conjecture (4): $F_s$ has constant area.

I don't know why this should be true, but it seems to be clear to Anton (if I understood well his comment concerning the Liouville theorem), and I would like to know why. Is it evident after studying billiard problems? I'll assume it true.

Can $F_s$ be nonconstant? It is constant until $s=m$, and then it starts to change. Since the area is conserved, it must advance at some points of its border and recede at other points. Find a point where it has just started receding. It represents a point $P\in R$. Fire a ray from $P$ into $\Omega$, with initial velocity orthogonal to $R$. Because of how we chose $P$, it will have negative initial $\dot s$, and by adjusting our choice of $P$ I would like (5) to be able to ensure that it reaches $F_m$. After that the optical fiber is Petruninean, and the ray makes its way to $F_0$. By reversing this ray, we obtain a ray that starts at $F_0$, reaches $P$, and bounces back to $F_0$, so $\Omega$ was not an optic fiber, after all.

The holes in the argument suggest more questions:

Question 0: Is every foliated optic fiber planely foliated?

If $\Omega$ is foliated, the phase space is the disjoint union of $T\Omega_+=\{\dot s>0\}$, $T\Omega_-\{\dot s<0\}$, and their zero-measure common border $T\Omega_0=\{\dot s=0\}$. Does every phase in $T\Omega_+$ correspond to a ray that came from $F_0$ and is going to $F_1$? If this is true (and also the analogous statement for $T\Omega_-$), then it is easy to see (by shooting rays between points of the same $F_s$) that every $F_s$ must be locally convex, and hence plane.

Question 1: Is every optic fiber foliated?

We can express the phase space as a union of the set $T\Omega_+$ of phases corresponding to rays that go from $F_0$ to $F_1$, the set $T\Omega_-$ defined analogously (so that $\dot x\in T\Omega_+$ iff $-\dot x\in T\Omega_-$), and the set $T\Omega_0$ of phases corresponding to rays that remain inside $\Omega$ eternally (in both directions of time). Observe how for each $x\in\Omega$, the classification of phases partitions $T_x\Omega$ into two opposite cones $T_x\Omega_+$, $T_x\Omega_-$ and a selfopposite cone $T_x\Omega$.

Does $T\Omega_0$ have measure zero? Are the cones convex? I have no idea.

Can we totally order the phases of $T\Omega_+$, so that it is then possible to assign them a scalar parameter, that makes possible to apply supremum arguments to prove things? At least we can partially order it. Do supremum arguments work in posets?

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"why this question appeared yesterday": Because the MO bot reposts questions that have no upvoted answer. Anton should be honored: his name is now an adjective! Petruninean! –  Joseph O'Rourke Apr 13 '12 at 1:06

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