**Edited 1.** Some suggestions are added at the end concerning **Q2**.

**Edited 2.** An "explanation" of spikes at $17^0$ is added at the very end...

**Q1** I think that the answer to **Q1** is positive provided the boundary of the tube is smooth. I'll consider the case of dimension $2$.

So, by our assumptions the light is propagating in the strip bounded by two smooth curves $L_+$ and $L_-$ that are both equidistant from the central curve $L$ on distance $\frac{1}{2}$. It is important that the whole strip is foliated by unit intervals orthogonal to all three curves, we call this foliation $F$.

Now consider our ray of light $R(t)$ propagating in the strip and introduce a function $\angle(t)$ that equals the angle between $R(t)$ and the orthogonal to $F$ in the direction of the tube. On the entrance of the tube the angle equals $0$.

**Claim 1.** For any moment $t$ we have $\angle(t)<\frac{\pi}{2}$.

Proof. Indeed, suppose that at some time $\angle(t)=\frac{\pi}{2}$. This means that $R(t)$ at this time goes in the direction of the foliation $F$. But since any segment of $F$ is a periodic ray in the strip, $R(t)$ must coincide with the segment, which is an absurd. END.

So, we see now that $R(t)$ will always propagate in the strip in one direction. So the only possibility for the ray to stay forever in the strip is to accumulate at some point to a segment $F_0$ of the foliation $F$. Let me explain why this is impossible. The main idea is that *this is impossible in the case the curve $L$ is a circle of radius* $r>1$. In this case it is easy to check the statement. The statement for general $L$ roughly follows from the fact that $L$ can be approximated well by the circle at any point.

To spell out the above in more details we can reduce the question to a question of billiards. Indeed, on the two-dimensional set of straight directed segments that join $L_+$ with $L_-$ there is a (partially defined) self map, consisting of two consequent reflection of the segment (first with respect $L_-$ then with respect to $L_+$). All the segments of $F$ are fixed points of the map. We need to show that for $F_0$ there is no a point that tends to it under the infinite iterations of the map. This self-map have three properties: 1) it preserves an area form 2) it fixes a segment (parametrizing the segments of $F$) 3) its linearisation is never identical on the fixed segment.

These 3 properties are sufficient to deduce that everything roughly boils down to the following exercise:

**Exercise.**
Consider a sequence $a_n$, such that
$a_{n+1}=a_n(1-a_n)$, with $a_0$ positive and less than one . Then $\sum_i a_i=\infty$.

**PS**. I think, that we can ask the curves $L_+$, $L_-$ and $L$ to be only $C^3$-smooth, but the proof uses the fact that the curvature of $L$ is strictly larger than $1$. It is not obvious if this condition can be relaxed.

**Q2** This is more of a suggestion rather than an answer. But this suggestion might help to get some clues to the answer. I would suggest you to make one more picture, namely, the picture of the *Phase portrait* - standard thing one does when dealing with a billiard. So, one only needs to consider the trajectory and for each reflection of the trajectory from the upper curve plot the point with two coordinates:

(angle of the ray; $x$-coordinate modulo $2$)

If you plot 1500 points, a certain shape will appear. Probably the points will fill a two-dimensional domain, but according to the histogram, the trajectory will avoid a large part of the phase portrait. This just reflects the fact that this billiard is not ergodic. I think, that to understand why there are no rays with angles in $[19^0, 111^0]$ one should analyse the boundary of the shape that will appear. This boundary might correspond to some "quasi-periodic" trajectory(ies) of the billiard.

**Further on Q2.** I want to add a couple of remarks on **Q2**, that are rather superficial. So, from the experiment of Joseph we see that with some probability it turns out that the original trajectory is quasi-periodic. I.e. the segments constituting the trajectory land on a one-dimensional curve in the 2-dimensional space of all segments. This at least explains the appearance of spikes in the first histogram. Indeed, when you project a measure evenly distributed on a curve on a plane to the $x$ axes - the projected measure will have singularities at points where the vertical lines $x=const$ are tangent to the curve.

Now, I guess, that in order to really answer the question one can indeed try to prove that the initial trajectory is quasi-periodic. The billiard is rather simple of course, but I don't know how hard it will be. And before you prove this, you can not be sure that the trajectory is really quasi periodic...

{CapForm["Butt"], Tube[spiral, 1/2]}, wherespiralis a list of points along the spiral. $\endgroup$ – Joseph O'Rourke Jul 15 '11 at 16:132more comments