# A Stochastic Dynamical Billiard

Consider the following stochastic dynamical system.

Fix $a > 0$, $b > 0$ and $v > 0$, and let $\mathbf{r}(t)=(x(t),y(t))$ be the position at time $t$ of a point which moves in the rectangle $R=\{ (x,y) \in \mathbb{R}^2: 0 \leq x \leq a, 0 \leq y \leq b \}$ with velocity of constant magnitude $v$ a according to the following rules:

(i) in the interior of $R$ the point is subject to no force, so that is moves with constant velocity $\mathbf{v}(t)=\frac{d\mathbf{r}}{dt}(t)$;

(ii) when the point reaches one of the vertical sides of $R$ then it is reflected elastically, that is the $y$-component of $\mathbf{v}$ is preserved, while the $x$-component of $\mathbf{v}$ changes sign;

(iii) when the point reaches one of the horizontal sides of $R$ then it is reflected diffusely, that is its velocity after the collision has always magnitude $v$, and the convex angle $\theta$ that $\mathbf{v}$ makes with the versor $\mathbf{i}=(1,0)$ takes any value in $[0,\pi]$ with equal probability (that is $\theta$ is uniformly distributed on $[0,\pi]$);

(iv) finally, if the point reaches one of the vertices of $R$, then its velocity after the collision has magnitude $v$ and the convex angle that $\mathbf{v}$ makes with the versor $\mathbf{i}$, when the point reaches one of vertices $(0,0)$ and $(0,b)$, or rispectively $-\mathbf{i}$, when the point reaches one of the vertices $(a,0)$ and $(a,b)$, assumes any value in $[0,\pi/2]$ with equal probability.

Consider a time $T> 0$, and let $N(T)$ be the number of time the point touches one of the vertical sides of $R$ (you can compute or not in $N(T)$ the times the point touches one of the vertices of $R$: it should make no essential difference for what we want to prove, I think). Let $\theta_j$ be the angle that $\mathbf{v}$ makes with $\mathbf{i}$ the $j$-th time the point touches one of the vertical sides of $R$ (or one of the vertices, if you have considered also them in the computation of $N(t)$), and form the random sum $$\sum_{j=1}^{N(T)} |cos \theta_j|.$$ I would like to prove that for any initial conditions $(\mathbf{r}(0), \mathbf{v}(0))$, the following equation holds $$\lim_{T \rightarrow \infty} \frac{1}{T} E \left[ \sum_{j=1}^{N(T)} |cos \theta_j| \right] = \frac{v}{2 a}.$$

Any help is welcome. For now, I have no idea about a possible proof.

NOTE. This problem has been suggested to me by an interesting proof of a physical law called Wien's displacement law given by Richtmyer, Kennard and Cooper in their book "Introduction to Modern Physics", Sixth Edition, Appendix of Chapter 5 (see in particular p. 145). The equation I would like to prove should be true for compelling physical reasons of thermodynamical character. Actually it should be true more generally also if all the sides of $R$ reflected elastically and only a small segment on one or both of the horizontal sides should reflect in a diffuse way, but the proof of this last statement seems absolutely prohibitive to me. Even in the simplified case I described in my post above a proof seems not trivial at all.

• By a dimensional analysis argument, the limit of interest $L(a,b,v)$ (if it exists) is of the form $cv/(a^αb^β)$ where $α+β=1$. Think about applying the usual billiards trick by replacing the right mirror with a pane of glass looking into a mirrored version of the room. Counting only the hits on the left wall of the original $a$-by-$b$ room gives $L(a,b,v)/2$, which equals counting only the mirror hits in the extended room, i.e. $L(2a,b,v)$. So $L(a,b,v)$ is independent of $b$. The problem might be more manageable with $b$ very large. – MTyson Sep 27 '17 at 18:28

EDIT: As Benoît Kloeckner pointed out, the following extended comment is actually a comment to a different question.

Are you sure about the constant $3$ in the denominator? I get $2$ instead. However, this is a heuristic argument, not a formal proof (so, in particular, this is not an actual answer).

The measure $(\pi a b)^{-1} dx \times dy \times d\theta$ is an invariant measure for the system under discussion. It looks like this measure is ergodic. We claim that $$\lim_{T \to \infty} \frac{1}{T} \sum_{j=1}^{N(T)} \lvert\cos \theta_j\rvert = \frac{1}{\pi a} \int_0^\pi v (\cos \theta)^2 d\theta = \frac{v}{2 a} \, .$$ Let $\Phi_\varepsilon(x,y,\theta) = 1$ if $\theta > \pi/2$ and $x \in [0, -\varepsilon v \cos \theta]$ or $\theta < \pi/2$ and $x \in [a - \varepsilon v \cos \theta, a]$, and $0$ otherwise. In other words, $\Phi_\varepsilon(x, y, \theta) = 1$ if and only if a particle at $(x,y)$ moving at an angle $\theta$ will hit one of the vertical walls within the next $\varepsilon$ units of time. The limit in the left-hand side can be written as $$\lim_{T \to \infty} \frac{1}{T} \sum_{j=1}^{N(T)} \lvert\cos \theta_j\rvert = \lim_{T \to \infty} \frac{1}{T} \lim_{\varepsilon \to 0} \frac{1}{\varepsilon} \int_0^T \Phi_\varepsilon(x(t), y(t), \theta(t)) \lvert\cos \theta(t)\rvert dt .$$ If one can change the order of the limits, then, by ergodicity, \begin{aligned}\lim_{T \to \infty} \frac{1}{T} \sum_{j=1}^{N(T)} \lvert\cos \theta_j\rvert & = \lim_{\varepsilon \to 0} \frac{1}{\varepsilon} \lim_{T \to \infty} \frac{1}{T} \int_0^T \Phi_\varepsilon(x(t), y(t), \theta(t)) \lvert\cos \theta(t)\rvert dt \\ & = \lim_{\varepsilon \to 0} \frac{1}{\varepsilon \pi a b} \int_0^a \int_0^b \int_0^\pi \Phi_\varepsilon(x,y,\theta) \lvert\cos \theta\rvert d\theta \, dy \, dx.\end{aligned} The expression in the right-hand side does not depend on (small) $\varepsilon > 0$, and it simplifies to the expression claimed above.

• Dear mates, thank you for having kindly replied to my question. @Mateusz Your answer seems to be a good idea for a proof. Obviously, the difficult part is to show that actually the measure $(\pi a b)^{-1} dx \times dy \times d \theta$ is actually ergodic. As for the factor 3 in the denominator, you're right: it should be replaced by 2. When I wrote the post, I had still in my mind the three-dimensional version of the problem. – Maurizio Barbato Sep 28 '17 at 11:53
• In the three-dimensional case the point moves in $P=\{ (x,y,z) \in \mathbb{R}^3 : 0 \leq x \leq a, 0 \leq y \leq b, 0 \leq z \leq c \}$ with velocity of constant magnitude $v$, it is reflected elastically on the faces of $P$ corresponding to $x=0$ and $x=a$, while on the other faces the point is diffusely reflected, that is its velocity $\mathbf{v}$ after the collision has always magnitude $v$ and direction $\mathbf{v}/v$ uniformly distributed over a hemisphere (with obvious modifications for the case in which the reflection happens on an edge or a vertex, as in the two-dimensional case). – Maurizio Barbato Sep 28 '17 at 17:06
• Then, the heuristic argument given by Mateusz would give now $\lim_{T \rightarrow \infty} \frac{1}{T} E \left[ \sum_{j=1}^{N(T)} |cos \theta_j| \right] = \frac{v}{3 a}$, since we have $\frac{1}{2 \pi} \int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{2}} \cos^2 \theta \sin \theta d \theta d \phi = \frac{1}{3}$. – Maurizio Barbato Sep 28 '17 at 17:09
• I very much doubt that the Liouville measure (normalized $dx dy d\theta$) is invariant with a uniform reflection law. One uses the Knudsen reflection law to that effect (if a gas -large number of particles is distributed according to Liouville's measure, then in a unit of volume near the boundary, the particles that are headed toward the boundary at a flat angle will be spread out on a larger part of the boundary than those that are headed toward the boundary nearly orthogonally). – Benoît Kloeckner Nov 10 '17 at 21:49
• @BenoîtKloeckner: You are right! Apparently the only finite invariant measure in this case is "vertical motion only". Thanks for pointing this out! – Mateusz Kwaśnicki Nov 13 '17 at 13:07

You just need to check the average is $\frac{v}{2a}$ if you assume the transitive flow is uniformly ergodic.

So the main difficult is to estimate the time average with any original point.i guess you can have some idea with this article.

arXiv:math/0608004 [pdf, ps, other] Unique Ergodicity of Translation Flow. https://arxiv.org/abs/math/0608004