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Consider the following stochastic dynamical system.

Fix $a > 0$, $b > 0$, $c>0$ and $v > 0$, and let $\mathbf{r}(t)=(x(t),y(t),z(t))$ be the position at time $t$ of a point which moves in the parallelepiped $R=\{ (x,y,z) \in \mathbb{R}^3: 0 \leq x \leq a, 0 \leq y \leq b, 0 \leq z \leq c \}$ with velocity of constant magnitude $v$ a according to the following rules (we shall consider the usual spherical coordinates $(\rho, \theta, \phi)$ in $\mathbb{R}^3$, where $\theta$ is the polar angel with respect to the $z$-axis and $\phi$ the azimuthal angle):

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

(ii) when the point reaches the face of $R$ corresponding to $z=0$ it is reflected diffusely, that is its velocity after the collision has always magnitude $v$, and the versor $\mathbf{v} / v$ has polar coordinates $(1,\theta, \phi)$ with $(\theta, \phi)$ distributed over $[0,\pi/2] \times [0,2\pi)$ with probability density law (Knudsen law) $f(\theta, \phi)= \frac{1}{\pi} \cos \theta d \Omega= \frac{1}{\pi} \cos \theta \sin \theta d \theta d \phi$;

(iii) if the point reaches any other side of $R$, then it is reflected elastically: e.g. in the case of the face corresponding to $z=c$ the $x$ and $y$ components of $\mathbf{v}$ are preserved, while the $z$-component of $\mathbf{v}$ changes sign;

(iv) finally, if the point reaches an edge (or a vertex) of $R$, then the reflection must be thought as the combination of reflections of the two (respectively three) faces involved; so if the point hits the edge $\{ (x,0,0): 0 < x < a \}$ it is reflected diffusely, that is its velocity after the collision has always magnitude $v$, and the versor $\mathbf{v} / v$ has polar coordinates $(1,\theta, \phi)$ with $(\theta, \phi)$ distributed over $[0,\pi/2] \times [0, pi]$ with probability density law $g(\theta, \phi)= \frac{2}{\pi} \cos \theta d \Omega= \frac{2}{\pi} \cos \theta \sin \theta d \theta d \phi$; if instead e.g. the point hits the edge $\{ (x,0,a): 0 < x < a \}$ then is reflected elastically, that is the $x$ component is preserved, while the $y$ and $z$ components change sign.

For any time $T> 0$, let $N(T)$ be the number of times the point hits the face $z=c$ (included its edges and vertices) before or at time $T$. Let $(1,\theta_j,\phi_j)$ be the polar coordinates of the versor $\mathbf{v} / v$ just before the $j$-th collision 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 $v_{z}(0) \neq 0$, 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}{6 c}. \end{equation}

NOTE1. 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, and see also ter Haar and Wergeland, Elements of Thermodynamics, $\S 5.3.3$). I formulated a first model of dynamic stochastic billiard in my post Stochastic Dynamic Billiard. This model was too complicated to be carefully analyzed, so I simplified it to get a much simpler version of it in a second post A Simple Stochastic Dynamic Billiard. The answer by Iosif Pinelis to this last post shows that my conjecture was trivially wrong. Anyway, as suggested by Benoît Kloeckner in his insightful comments, this was due to the fact I had assumed random uniform reflection on $z=0$ in my two previous posts. Replacing the random uniform reflection with the Knudsen law should finally give the right answer.

NOTE2. Let me add here a final remark about the physical meaning of the assumption of Knudsen law in our model. It simply says the every little region of the face $z=0$ emits light with constant radiance in any direction (such an ideal isotropic emitter is often called a Lambertian emitter: see Lambert's cosine law. Such a behavior is exactly what we need in the thermodynamical situation (blackbody radiation in cavity in thermodynamical equilibrium) considered in the proof of Wien's displacement law (see Planck, The Theory of Heat Radiation, Part II, Section III, especially $\S 78$).

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This answer is a modification of my answer at 2D billiard.

To avoid trivialities, assume that the initial position of the particle is in the interior of the box. Also, at this point, assume that the first time the particle hits the side $z=0$, it occurs not at a vertex of the box. If $x$- and $y$-coordinates of the initial velocity are $0$, then the limit in question is obviously $0$.

Otherwise, in a nonrandom finite time, depending only on the initial position and velocity, the particle will hit the side $z=0$. 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 $\theta_1$ with the $z$-axis. Note that the vertical coordinate of the velocity may change in magnitude only right after hitting the side $z=0$; except for this, it may only change in sign right after hitting the side $z=c$. Also, the cosine of the angle between the velocity and the $z$-axis does not vary while the particle is traveling from the side $z=0$ toward the side $z=c$.

Introduce $Y_j:=\cos\theta_j$, $X_i:=\frac{c}{vY_i}$ -- the $i$th traveling time from the side $z=0$ to the side $z=c$ (which is equal to the $i$th traveling time from the side $z=c$ back to the side $z=0$), $T_j:=\sum_1^j X_i$, $N_t:=\sum_{j=1}^\infty I\{T_j\le t\}$ (with $I\{\cdot\}$ denoting the indicator function). Do not confuse $N_t$ with $N(t)=\sum_{j=1}^\infty I\{2T_{j-1}+X_j\le t\}$ -- the number of times the particle hits the side $z=c$. Let then \begin{gather*} R_t:=\sum_{j=1}^{N_t}Y_j, \tag{1} \\ S_t:=\sum_{j=1}^{N(t)} \cos \theta_j=\sum_{j=1}^{N(t)}Y_j \tag{2} =\sum_{j=1}^\infty Y_j\,I\{2T_{j-1}+X_j\le t\}, \\ S_t^-:=\sum_{j=1}^\infty Y_j\,I\{2T_j\le t\}=R_{t/2}, \quad S_t^+:=\sum_{j=1}^\infty Y_j\,I\{2T_{j-1}\le t\}. \end{gather*} Then
\begin{equation*} S_t^-\le S_t\le S_t^+; \tag{3} \end{equation*} everywhere here $t$ is a positive real number. By the Renewal Reward Theorem, we have \begin{equation*} \frac1t\,ER_t\underset{t\to\infty}\longrightarrow\frac{EY_1}{EX_1}=\frac{2/3}{2c/v}=\frac{v}{3c}. \end{equation*} So, \begin{equation*} \frac1t\,ES_t^-= \frac1t\,ER_{t/2}\underset{t\to\infty}\longrightarrow\frac{v}{6c}. \tag{4} \end{equation*} Next, by the independence of $Y_j$ and $T_{j-1}$ and, again, by the Renewal Reward Theorem (with $1$ in place of $Y_j$), \begin{multline*} ES_t^+=\sum_{j=1}^\infty EY_j\,EI\{2T_{j-1}\le t\} =EY_1\,E\sum_{j=1}^\infty I\{2T_{j-1}\le t\} =EY_1\,(1+EN_{t/2}) \\ = EY_1\,\Big(1+E\sum_{j=1}^{N_{t/2}}1\Big) = EY_1\,\Big(1+\frac t{2+o(1)}\, \frac{E1}{EX_1}\Big) \sim EY_1\,\frac t2\, \frac1{EX_1} =\frac{tv}{6c} \tag{5} \end{multline*} as $t\to\infty$. Thus, by (2)--(5), \begin{equation*} \frac1t\,E\sum_{j=1}^{N(t)} \cos \theta_j=\frac1t\,ES_t\underset{t\to\infty}\longrightarrow\frac{v}{6c}. \end{equation*}

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Unfortunately, long and cumbursome formulas completely obfuscate very simple nature of what's going on here. First of all, let me rescale the variables, so that $c=1$ and $v=1$. Now, Knudsen's law says that each time the particle hits the bottom, the vertical component $w=\cos\theta$ of the velocity is independently sampled from the distribution with the density $f(\xi)=2\xi$ on the interval $[0,1]$. The time until the particle hits the bottom again is $\tau=2/w$ (whatever happens on edges can be discarded as it has probability 0). Therefore, by the law of large numbers (which can be called many different names), the limit you are interested in is $$ \frac{\mathbf E w}{\mathbf E \tau} = \frac{2/3}{4} = \frac16 \;. $$

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  • $\begingroup$ It is somewhat unclear to me what version of the law of large numbers you are using here, and how. Of course, proofs of the basic facts of the renewal theory use, in particular, the strong law of large numbers. To apply those facts of the renewal theory, small adjustments are needed in the present case. To describe these adjustments in detail, some writing is indeed needed. $\endgroup$ – Iosif Pinelis Nov 12 '17 at 16:58
  • $\begingroup$ Let $T_i$ be the moments when the particle hits the bottom (assume for simplicity that $T_0=0$), then at the moment $T_n$ the corresponding sum of cosines is $\sum_{i=1}^n w_i$ and $T_n=\sum_{i=1}^n 2/w_i$, where $w_i$ are iid sampled from the distribution with the density $f(\xi)=2\xi$. Passing from the moments $T_n$ to arbitrary times is an undergraduate level exercise on limits. One does not have to call these trivialities "basic facts of the renewal theory" and to give them fancy names $\endgroup$ – R W Nov 12 '17 at 17:52
  • $\begingroup$ It is still somewhat unclear to me what version of the law of large numbers you meant to use here, and how. I never said that applying facts of the renewal theory to the case in question was hard. Yet, to describe those applications in detail does take some writing, doesn't it? Also, are you saying that the renewal theory is just a fancy name (for the law of large numbers?) and the work on this theory by such giants as Doob and Cox (referred to at en.wikipedia.org/wiki/Renewal_theory) was just "trivialities"? $\endgroup$ – Iosif Pinelis Nov 12 '17 at 18:28
  • $\begingroup$ What I meant is that the "Renewal Reward Theorem" you are referring to is a totally trivial exercise on the law of large numbers. Your opinion may be (and, as our discussion shows, actually is) different. $\endgroup$ – R W Nov 13 '17 at 6:45
  • $\begingroup$ The term "Renewal Reward Theorem" is not mine; it appears to be standard and generally accepted. The proof of that theorem is of comparable length and level of difficulty as that of the strong law of large numbers (SLLN) itself. Do you find the proof of the SLLN "a totally trivial exercise" as well? More importantly, the only details that you provide in your answer seem to concern the calculation of $Ew$ and $E\tau$ ($EY_1$ and $EX_1$ in my answer) -- but this calculation seems to me the most trivial part of my answer. $\endgroup$ – Iosif Pinelis Nov 13 '17 at 19:56

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