Light rays bouncing in twisted tubes

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...

Q1. No. The projection of the ray to the curve $c(t)$ has to be monotonic. In particular, no ray can come back, but some ray might stay in the tube forever.

Construction. The curve and ray will be flat. The curve has countable number of slight turnes left-right-left-right..., so the ray approach a direction tangent to one of the discs. Imagine that you fly on the ray and construct (flat) curve at the same time; the curve should turn left when ray goes from left to right wall and vice versa. (This way the angle between ray and the disc decrease roughly as integral of absolute value of curvature of the curve.) The total length of such curve can be made finite; after that you can continue by anything you want.

More formally, let $c(t)$ be the curve and $r(t)$ be the point on the intersection of ray and segment orthogonal to $c$ at $c(t)$. We will construct $r(t)$ and $c(t)$ simultaneously. For given $t$, the ray will go either from right wall to the left wall or vice versa. In the first case make curvature of $c$ to be $+\varepsilon$ and in the second case $-\varepsilon$. If $t_1 < t_2 <\dots$ be the parameters at which the ray hits the wall, then you have something like $$t_{n+1}-t_n < \left(1-\frac\varepsilon7\right){\cdot}(t_{n}-t_{n-1})$$ Hence $t_n$ converge to a finite limit.

This way you construct a $C^{1,1}$ curve $c$, but with little more work you can make it $C^2$.

Comments. It is possible to keep the curvatue of $c$ in such example arbitray close to zero. One can make it to be twice differentiable, but it seems impossible to make it $C^{2,\varepsilon}$ for any $\varepsilon>0$. Dmitri's answer gives a reason why.

Here is my interpretation of Anton's idea to capture the ray. I found it almost impossible to illustrate; I could only show three sections of the tube, separated by two rotations, the first counterclockwise the second clockwise. (The actual turn angles I selected are irrelevant.)