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A particle starts a brownian walk in the middle of a long tunnel in the plane, at one end of the tunnel is a region Y of given area A.

Does the shape of region Y affect average time for the particle to exit through the other end of the tunnel?

If yes, what shapes (qualitatively) Y would maximize or minimize this time?

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    $\begingroup$ If I understand the question correctly, the circular region $X$ is irrelevant. You could as well just have that end of the tunnel open and ask how $Y$ affects the average time for the particle to reach the open end. $\endgroup$ Aug 29, 2014 at 13:57
  • $\begingroup$ @AndreasBlass Yea, sorry about that complication. $\endgroup$
    – user57600
    Aug 29, 2014 at 13:58
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    $\begingroup$ I have found some relevant articles. See links: Holcman and Schuss 2011, Schuss 2012, and Holcman and Schuss 2014. It appears that the relevant timescale may be $AL/WD$, where L is the length of the tunnel, W its width. This differs from Carlo Beenakker's answer, but I do believe that the case of Brownian motion should differ from that of ballistic motion (between collisions with walls). $\endgroup$ Sep 3, 2014 at 22:17

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The region $Y$ introduces a delay time $\tau$, the average time between entry and exit. Let me first consider the case that the dynamics in the two-dimensional region $Y$ is a Brownian motion (diffusion) with diffusion coefficient $D$. The region $Y$ has area $A$, perimeter $P$, and $W$ is the width of the tunnel. The mean time between collisions with the perimeter of $Y$ can be estimated as $A/D$, and on average about $P/W$ collisions occur between entry and exit, so we have the estimate

$$\tau=C\times\frac{AP}{DW}.\quad\quad[1]$$

The shape dependence enters in the coefficient $C$ of order unity, but different shapes with the same area and perimeter will have the same $\tau$ in order of magnitude.

We can do better and obtain a precise expression for $\tau$ including the numerical coefficient, if we assume that $Y$ is a billiard, in which the particle moves along a straight line (velocity $v$), bouncing from the walls until it finally exits. (See diagram.) This problem has been analyzed by Lewenkopf and Vallejos under the assumption of ``weak ergodicity'', which basically means that the particle explores the whole phase space of the billiard before exiting. (This assumption is weaker than that of chaotic dynamics, it also applies to, say, a circular billiard which has regular dynamics.) Under this assumption one has simply

$$\tau=\frac{\pi A}{vW}\quad\quad [2]$$

without any shape dependence.

The assumption of a constant velocity is actually not needed, if only the energy $E$ of the particle is constant but its velocity varies, then one has

$$\tau=\frac{1}{\tilde{\Omega}}\frac{d\Omega}{dE}\quad\quad [3]$$

where $\Omega$ is the area of phase space in the billiard and $\tilde{\Omega}$ the area of phase space connected to the tunnel. The simpler expression $\tau=\pi A/vW$ for constant velocity (momentum $p=mv$) follows from $\Omega=\pi p^2 A=2\pi mE A$ and $\tilde{\Omega}=2pW$. In this way one also obtains the generalization to a three-dimensional billiard (volume $V$) connected to a tunnel with cross-sectional area $S$: $\Omega=\frac{4}{3}\pi p^3 V=\frac{4}{3}\pi(2mE)^{3/2}V$ and $\tilde{\Omega}=\pi p^2 S$, so $$\tau=\frac{1}{\tilde{\Omega}}\frac{d\Omega}{dE}=\frac{4V}{vS},\quad \quad[4]$$ again without any shape dependence.

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  • $\begingroup$ Since the question is about Brownian motion, do these expressions for $\tau$ apply? If so, with what value of $v$? $\endgroup$ Aug 29, 2014 at 15:29
  • $\begingroup$ I am not familiar with the terms "diffusive", "ballistic". Can you please define? Also, what does "ergodic inside Y" mean? Finally, what regularity conditions do you impose on Y (measurability, openness, etc). I do not see how those are used in the proof. $\endgroup$
    – Boris Bukh
    Aug 29, 2014 at 20:48
  • $\begingroup$ @BorisBukh -- I have added a figure and included a reference where the result for the average dwell time is derived in some detail. $\endgroup$ Aug 29, 2014 at 21:28
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    $\begingroup$ Sorry, I still do not follow. You claim that C "is of order of unity". If I were to follow your proof, I should be able to write down explicit numeric constant c1 and c2 such that c1<C<c2, but I simply do not see how to do that. What steps am I missing? $\endgroup$
    – Boris Bukh
    Aug 29, 2014 at 22:45
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    $\begingroup$ The situation you have described (let's call it the "ballistic" case) has a tau that is independent of the length of the tunnel. That makes sense for ballistic motion, where once the particle has entered the tunnel it will bounce its way to the other side without turning back. However, with Brownian Motion it will make a random number of excursions back into the cavity. The length of the tunnel will affect how many excursions. That may be why the results in the papers I link to in my comment (on the question) seem to have L (length of tunnel) instead of P. $\endgroup$ Sep 3, 2014 at 22:29

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