A Simple Stochastic Dynamic Billiard

Question

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 $(v_x(t), v_y (t)) = \mathbf{v}(t)=\frac{d\mathbf{r}}{dt}(t)$;

(ii) when the point reaches the vertical side of $R$ corresponding to $x=0$ it is reflected diffusely, that is its velocity after the collision has always magnitude $v$, and the oriented angle $\theta$ that the versor $\mathbf{i}=(1,0)$ makes with the velocity $\mathbf{v}$ after the collision is uniformly distributed over $[-\pi/2,\pi/2]$;

(iii) if the point reaches any other side of $R$, then it is reflected elastically, that is in the case of the vertical side corresponding to $x=a$ the $y$-component of $\mathbf{v}$ is preserved, while the $x$-component of $\mathbf{v}$ changes sign, while in the case of either of the two horizontal sides the $x$-component of $\mathbf{v}$ is preserved, while the $y$-component of $\mathbf{v}$ changes sign;

(iv) finally, if the point reaches one of the vertices of $R$, then the reflection must be thought as the combination of two reflections by the the two edges involved; so e.g. if the point hits the vertex $(0,0)$ then it is reflected diffusely, that is its velocity after the collision has always magnitude $v$, and the oriented angle $\theta$ that the versor $\mathbf{i}=(1,0)$ makes with the velocity $\mathbf{v}$ after the collision is uniformly distributed over $[0,\pi/2]$; if instead the point hits e.g. the vertex $(a,0)$ then its velocity $\mathbf{v}$ before the collision is changed in $-\mathbf{v}$.

Consider a time $T> 0$, and let $N(T)$ be the number of time the point touches the vertical side of $R$ corresponding to $x=a$ (you can compute or not in $N(T)$ the times the point touches one of the vertices of this side: it should make no essential difference for what we want to prove, I think). Let $\theta_j$ be the angle that the velocity $\mathbf{v}$ just before the collision makes with $\mathbf{i}$ the $j$-th time the point touches the vertical sides of $R$ corresponding to $x=a$ (or one of its vertices, if you have considered also them in the computation of $N(t)$), and form the random sum \begin{equation} \sum_{j=1}^{N(T)} cos \theta_j. \end{equation} I would like to prove that for any initial conditions $(\mathbf{r}(0), \mathbf{v}(0))$, with $v_{x}(0) \neq 0$, the following equation holds \begin{equation} \lim_{T \rightarrow \infty} \frac{1}{T} E \left[ \sum_{j=1}^{N(T)} cos \theta_j \right] = \frac{v}{4 a}. \end{equation}

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). I formulated a first model of dynamic stochastic billiard in my post Stochastic Dynamic Billiard, but it was considerably more complicated. Then I realized that the simplest model above could be a good formulation of the physical system as well. In the obvious three-dimensional version of this model, in which the point moves inside a parallelepiped, whose face corresponding to $x=0$ reflect diffusely (that is $v$ remains unchanged and $\mathbf{v}/v$ is uniformly distributed over a hemisphere) while all other faces (and all the edges and vertices) reflect elastically, we should get

\begin{equation} \lim_{T \rightarrow \infty} \frac{1}{T} E \left[ \sum_{j=1}^{N(T)} cos \theta_j \right] = \frac{v}{6 a}, \end{equation}

which has a deep thermodynamical meaning (see the reference above and also ter Haar and Wergeland, Elements of Thermodynamics, $\S 5.3.3$).

Answer 1

To avoid trivialities, assume that the initial position of the particle is in the interior of the rectangle. Then the limit is actually $0$. This is trivial in the case when the $x$-coordinate of the initial velocity is $0$.

Otherwise, in a nonrandom finite time, depending only on the initial position and velocity, the particle will hit the side $x=0$. So, without loss of generality let us assume the movement begins right after that hitting time, with the initial velocity of magnitude $v$ at a random angle $\theta_1\in[-\pi/2,\pi/2]$ with the $x$-axis. Note that the horizontal coordinate of the velocity may change in magnitude only right after hitting the side $x=0$; except for this, it may only change in sign right after hitting the side $x=a$. Also, the cosine of the angle between the velocity and the $x$-axis does not vary while the particle is traveling from the side $x=0$ toward the side $x=a$. Therefore, introducing $Y_j:=\cos\theta_j$, $X_i:=\frac{a}{vY_i}$ -- the $i$th traveling time from the side $x=0$ to the side $x=a$, $T_j:=\sum_1^j X_i$, $N_t:=\sum_{j=1}^\infty I\{T_j\le t\}$ (with $I\{\cdot\}$ denoting the indicator function), we see that $N(t)\le N_t$ for real $t>0$; here, we understand $N(t)$ as the number of times the particle hits the vertical side $x=a$ before or at time $t$. So,
\begin{equation} \sum_{j=1}^{N(t)}\cos\theta_j\le R_t:=\sum_{j=1}^{N_t}Y_j. \end{equation} By the Renewal Reward Theorem, we have \begin{equation} \frac1t\,ER_t\underset{t\to\infty}\longrightarrow\frac{EY_1}{EX_1}=\frac{2/\pi}\infty=0. \end{equation} Thus, \begin{equation} 0\le\frac1t\,E\sum_{j=1}^{N(t)}\cos\theta_j\le \frac1t\,ER_t\underset{t\to\infty}\longrightarrow0. \end{equation}