A Diffusion Limit for a Test Particle in a Random Distribution of Scatterers

G. Basile; A. Nota; M. Pulvirenti

doi:10.1007/s10955-014-0940-z

Journal of Statistical Physics

June 2014, Volume 155, Issue 6, pp 1087–1111 | Cite as

A Diffusion Limit for a Test Particle in a Random Distribution of Scatterers

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Authors and affiliations

G. Basile
A. NotaEmail author
M. Pulvirenti

Article

First Online: 19 February 2014

Received: 12 July 2013

Accepted: 28 January 2014

192 Downloads
6 Citations

Abstract

We consider a point particle moving in a random distribution of obstacles described by a potential barrier. We show that, in a weak-coupling regime, under a diffusion limit suggested by the potential itself, the probability distribution of the particle converges to the solution of the heat equation. The diffusion coefficient is given by the Green–Kubo formula associated to the generator of the diffusion process dictated by the linear Landau equation.

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Notes

Acknowledgments

We are indebted to S. Simonella and H. Spohn for illuminating discussions.

Appendix 1: On the Scattering Problem Associated to a Circular Potential Barrier

The potential energy for a finite potential barrier is given by

$$\begin{aligned} \phi ( r) = \left\{ \begin{array}{l@{\quad }l} u_0 &{} \text {if } r \le 1 \\ 0 &{} \text {if } r >1 \end{array} \right. \end{aligned}$$

(7.1)

The light particle, of unitary mass, moves in a straight line with energy $E=\frac{1}{2}v^2>u_0$, where $u_0>0$. Let $\rho $ be the impact parameter. For small impact parameters the particle will pass through the barrier, for large ones the particle will be reflected. Inside the barrier the velocity is a constant $v=\bar{v}$ ($\bar{v}<v$). The complete trajectory of the light particle which passes through the barrier consists of three straight lines and is symmetrical about a radial line perpendicular to the interior path. For a general reference to the scattering problem, see [12], Chap. 4.

Let $\alpha $ be the angle of incidence (the inside angle between the trajectory and a radial line to the point of contact with the barrier at $r=1$) and $\beta $ the angle of refraction (the corresponding external angle). We assume that the radius of the circle is $r=1$. According to the geometry of the problem $\alpha $ and $\beta $ are such that

$$\begin{aligned} \sin \beta =\frac{v}{\bar{v}}\sin \alpha , \end{aligned}$$

where $\sin \alpha =\rho $.

The angle of deflection is $\theta =2(\beta -\alpha ).$ Thanks to the energy and angular momentum conservation the expression for the refractive index becomes

$$\begin{aligned} n=\frac{\sin \alpha }{\sin \beta }=\frac{\bar{v}}{v}=\sqrt{1-\frac{2u_0}{v^2}} \end{aligned}$$

(7.2)

and so we have a scattering angle defined in the following way:

$$\begin{aligned} \theta (\rho )= \left\{ \begin{array}{l@{\quad }l} 2\left( \arcsin \left( \frac{\rho }{n}\right) -\arcsin (\rho )\right) &{} \text {if } \rho \le n \\ 2\arccos (\rho ) &{} \text {if } \rho >n. \end{array} \right. \end{aligned}$$

(7.3)

In the first case the particle passes through the barrier (for $\rho \le n$), and in the second one the particle is reflected (for $\rho > n$). The maximum scattering angle $\theta _{\max }=2\arccos (n)$ is the angle at which the particle scatters tangentially to the barrier. The differential scattering cross section

$$\begin{aligned} \Psi (\theta )=\left| \frac{\partial \rho }{\partial \theta }\right| \end{aligned}$$

is then:

$$\begin{aligned} \Psi (\theta )= \left\{ \begin{array}{l@{\quad }l} \frac{n[\cos (\theta /2)-n][1-n\cos (\theta /2)]}{(1+n^2-2n\cos (\theta /2))^{3/2}} &{} \text {if }\theta \le 2\arccos (n)\\ {[1-\cos ^2(\theta /2)]}^{1/2} &{} \text {otherwise}\\ \end{array} \right. \end{aligned}$$

(7.4)

Scaling now the potential as $\phi ( r) \rightarrow \varepsilon ^{\alpha }\phi ( r)$, the previous formulas still hold. Thus, according to this scaling, the refractive index becomes

$$\begin{aligned} n_{\varepsilon }=\sqrt{1-\frac{2\varepsilon ^{\alpha }u_0}{v^2}} \end{aligned}$$

(7.5)

to replace into (7.4). The scattering angle (7.3) reads now

$$\begin{aligned} \theta _{\varepsilon }(\rho )= {\left\{ \begin{array}{ll} 2\left( \arcsin \left( \frac{\rho }{n_{\varepsilon }}\right) -\arcsin (\rho )\right) &{}\quad \text {if } \rho \le n_{\varepsilon } \\ 2\arccos (\rho ) &{}\quad \text {if } \rho >n_{\varepsilon }. \end{array}\right. } \end{aligned}$$

(7.6)

Appendix 2: On the Diffusion Coefficient

In this section we show that the diffusion coefficient is divergent for the circular potential barrier (7.1). At this level we assume that $u_0=1$ to simplify the following expressions.

We need to compute

$$\begin{aligned} \tilde{B}:=\lim _{\varepsilon \rightarrow 0}{\frac{\mu \varepsilon ^{-2\alpha }}{2}|v|\int _{-1}^{1}{\theta _{\varepsilon }^2(\rho )\,d\rho }}. \end{aligned}$$

(8.1)

Thanks to the symmetry for the scattering problem

$$\begin{aligned} \varepsilon ^{-2\alpha }\int _{-1}^{1}{\theta _{\varepsilon }^2(\rho )\,d\rho }=2\varepsilon ^{-2\alpha }\int _{0}^{1}{\theta _{\varepsilon }^2(\rho )\,d\rho }. \end{aligned}$$

(8.2)

According to (7.3):

$$\begin{aligned} 2\varepsilon ^{-2\alpha }\int _{0}^{1}{\theta _{\varepsilon }^2(\rho )\,d\rho }=2\varepsilon ^{-2\alpha }\left( \int _{0}^{n_{\varepsilon }}{\theta _{\varepsilon }^2(\rho )\,d\rho }+\int _{n_{\varepsilon }}^{1}{\theta _{\varepsilon }^2(\rho )\,d\rho }\right) \end{aligned}$$

(8.3)

Our aim is to perform a Taylor expansion of the first branch of $\theta _{\varepsilon }(\rho )$ for $\rho \ge 0$,

$\rho /n_{\varepsilon }<(1-\delta )$, with $\delta >0$. We have

$$\begin{aligned} \arcsin \left( \rho /n_{\varepsilon }\right) =\arcsin (\rho )+\frac{1}{\sqrt{1-\rho ^2}}\left( \frac{\rho }{n_{\varepsilon }}-\rho \right) +R_1\left( \rho /n_{\varepsilon }\right) , \end{aligned}$$

where

$$\begin{aligned} R_1\left( \rho /n_{\varepsilon }\right) =\frac{\bar{\rho }}{2(1-\bar{\rho }^2)^{\frac{3}{2}}}\left( \frac{\rho }{n_{\varepsilon }}-\rho \right) ^2\quad \rho <\bar{\rho }<\frac{\rho }{n_{\varepsilon }}. \end{aligned}$$

(8.4)

Then, looking at the first integral in the r.h.s of (8.3), we have to split it as

$$\begin{aligned} \varepsilon ^{-2\alpha }\int _{0}^{n_{\varepsilon }}{\theta _{\varepsilon }^2(\rho )\,d\rho }=\underbrace{\varepsilon ^{-2\alpha }\int _{0}^{n_{\varepsilon }(1-\delta )}{\theta _{\varepsilon }^2(\rho )\,d\rho }}_{A}+\underbrace{\varepsilon ^{-2\alpha }\int _{n_{\varepsilon }(1-\delta )}^{n_{\varepsilon }}{\theta _{\varepsilon }^2(\rho )\,d\rho }}_{B}. \end{aligned}$$

Thus

$$\begin{aligned} A&= \varepsilon ^{-2\alpha }\int _{0}^{n_{\varepsilon }(1-\delta )}{[2(\arcsin \left( \rho /n_{\varepsilon }\right) -\arcsin (\rho ))]^2\,d\rho }\nonumber \\&\le 4\varepsilon ^{-2\alpha }\left[ \int _{0}^{n_{\varepsilon }(1-\delta )}{\frac{(1-n_{\varepsilon })^2}{n_{\varepsilon }^2}\frac{\rho ^2}{(1-\rho ^2)}\,d\rho }+\int _{0}^{n_{\varepsilon }(1-\delta )}{R_1\left( \rho /n_{\varepsilon }\right) ^2\,d\rho }\right] \nonumber \\&+\,\,4\varepsilon ^{-2\alpha }\left[ \left( \int _{0}^{n_{\varepsilon }(1-\delta )}{R_1\left( \rho /n_{\varepsilon }\right) ^2}\right) ^{\frac{1}{2}}\left( \int _{0}^{n_{\varepsilon }(1-\delta )}{\frac{(1-n_{\varepsilon })^2}{n_{\varepsilon }^2}\frac{\rho ^2}{(1-\rho ^2)}\,d\rho }\right) ^{\frac{1}{2}}\right] . \end{aligned}$$

(8.5)

It is sufficient to compute the first two integrals. Let $A_1$ and $A_2$ be the first and the second integrals respectively. We have

$$\begin{aligned} A_1&= \varepsilon ^{-2\alpha }\frac{(1-n_{\varepsilon })^2}{n_{\varepsilon }^2}\int _{0}^{n_{\varepsilon }(1-\delta )}{\frac{\rho ^2}{{1-\rho ^2}}\,d\rho }\nonumber \\&= -\frac{\varepsilon ^{-2\alpha }}{2}\frac{(1-n_{\varepsilon })^2}{n_{\varepsilon }^2}[2n_{\varepsilon }(1-\delta )+\log (1-n_{\varepsilon }(1-\delta ))-\log (1+n_{\varepsilon }(1-\delta ))].\qquad \quad \end{aligned}$$

(8.6)

Using that $n_{\varepsilon }=1-\frac{\varepsilon ^{\alpha }}{|v|^2}+o(\varepsilon ^{2\alpha })$, from (8.6) it is clear that

$$\begin{aligned} A_1\simeq -\frac{\varepsilon ^{-2\alpha }}{2}(1-n_{\varepsilon })^2(\log (1-n_{\varepsilon }(1-\delta )))=-\frac{\varepsilon ^{-2\alpha }}{2}\left( \frac{\varepsilon ^{2\alpha }}{|v|^4}\right) \log (\varepsilon ^{\alpha }(1-\delta )+\delta ). \end{aligned}$$

A straight forward computation shows that the right hand side of the previous expression is

$$\begin{aligned} -\frac{\varepsilon ^{-2\alpha }}{2}\frac{\varepsilon ^{2\alpha }}{|v|^4}\Big (\log (\varepsilon ^{\alpha })+\log \Big (1-\delta +\frac{\delta }{\varepsilon ^{\alpha }}\Big )\Big )&= -\frac{1}{2|v|^4}\Big (\log (\varepsilon ^{\alpha })+\delta \Big (1-\frac{1}{\varepsilon ^{\alpha }}\Big )\Big )\nonumber \\&= \frac{\alpha }{2|v|^4}|\log (\varepsilon )|\left( 1+\frac{\delta \left( 1-\frac{1}{\varepsilon ^{\alpha }}\right) }{|\log (\varepsilon ^{\alpha })|}\right) .\qquad \quad \end{aligned}$$

(8.7)

Choosing $\delta =\frac{\varepsilon ^{\alpha }}{|\log {\varepsilon }|^{\gamma }}$ with $\gamma \in (0,\alpha /2)$, it follows $\delta /\varepsilon ^{\alpha }\underset{\varepsilon \rightarrow 0}{\longrightarrow } 0$.

In order to compute $A_2$, we need the following estimate for the remainder term

$$\begin{aligned} |R_1\left( \rho /n_{\varepsilon }\right) |\le \frac{1}{2} \frac{\rho }{n_{\varepsilon }}\frac{1}{\Big (1-\frac{\rho ^2}{n_{\varepsilon }^2}\Big )^{\frac{3}{2}}}\left( \frac{\rho }{n_{\varepsilon }}-\rho \right) ^2. \end{aligned}$$

(8.8)

Then

$$\begin{aligned} A_2&\le \frac{\varepsilon ^{-2\alpha }}{4}\int _{0}^{n_{\varepsilon }(1-\delta )}{ \frac{\rho ^2}{n_{\varepsilon }^2}\frac{1}{\Big (1-\frac{\rho ^2}{n_{\varepsilon }^2}\Big )^3}\left( \frac{\rho }{n_{\varepsilon }}-\rho \right) ^4\,d\rho }\nonumber \\&{\underset{u=\frac{\rho }{n_{\varepsilon }}}{=}}\varepsilon ^{-2\alpha }n_{\varepsilon }\int _{0}^{1-\delta }{\frac{u^2}{2(1-u^2)^3}u^4(1-n_{\varepsilon })^4\,du}\nonumber \\&{\underset{v=1-u}{=}}\frac{\varepsilon ^{-2\alpha }n_{\varepsilon }}{2}\int _{\delta }^{1}{\frac{(1-v)^6}{v^3}(1-n_{\varepsilon })^4\,dv}\simeq \frac{\varepsilon ^{-2\alpha }n_{\varepsilon }}{2}\frac{(1-n_{\varepsilon })^4}{\delta ^2}. \end{aligned}$$

(8.9)

Also in this case, the only significant contribution is given by

$$\begin{aligned} \frac{\varepsilon ^{-2\alpha }(1-n_{\varepsilon })^4}{\delta ^2}\simeq \frac{\varepsilon ^{-2\alpha }\varepsilon ^{4\alpha }}{\delta ^2}\underset{\varepsilon \rightarrow 0}{\longrightarrow } 0 \end{aligned}$$

again for $\delta =\frac{\varepsilon ^{\alpha }}{|\log {\varepsilon }|^{\gamma }}$ with $\gamma \in (0,\alpha /2)$. This shows that

$$\begin{aligned} A=A_1(1+\mathcal O(\varepsilon )). \end{aligned}$$

Now we compute $B$ in (8.3), namely

$$\begin{aligned} B&= \varepsilon ^{-2\alpha }\int _{n_{\varepsilon }(1-\delta )}^{n_{\varepsilon }}{\Big [2(\arcsin \left( \rho /n_{\varepsilon }\right) -\arcsin (\rho ))\Big ]^2\,d\rho }\nonumber \\&= \varepsilon ^{-2\alpha }\int _{n_{\varepsilon }(1-\delta )}^{n_{\varepsilon }}\left( \int _{\rho }^{\frac{\rho }{n_{\varepsilon }}}\,dx \frac{1}{\sqrt{1-x^2}}\right) ^2\,d\rho . \end{aligned}$$

(8.10)

Since

$$\begin{aligned} \int _{\rho }^{\frac{\rho }{n_{\varepsilon }}}\,dx\frac{1}{\sqrt{1-x^2}}&= \int _{\rho }^{\frac{\rho }{n_{\varepsilon }}}\,dx\frac{1}{\sqrt{(1-x)}\sqrt{(1+x)}}\\&\le \frac{1}{1+\rho }\int _{\rho }^{\frac{\rho }{n_{\varepsilon }}}\,dx\frac{1}{\sqrt{(1-x)}}\underset{u=1-x}{=}\int _{1-\frac{\rho }{n_{\varepsilon }}}^{1-\rho }\frac{1}{\sqrt{u}}\\&= \frac{1}{\sqrt{(1-\rho )}}\left( \sqrt{(1-\rho )}-\sqrt{(1-\rho /n_{\varepsilon })}\right) \end{aligned}$$

in (8.10) we have

$$\begin{aligned} B&=\varepsilon ^{-2\alpha }\int _{n_{\varepsilon }(1-\delta )}^{n_{\varepsilon }}\frac{1}{(1-\rho )}\left( \sqrt{(1-\rho )}-\sqrt{(1-\rho /n_{\varepsilon })}\right) ^2\,d\rho \nonumber \\&\underset{(1-\frac{\rho }{n_{\varepsilon }}<1-\rho )}{\le }\varepsilon ^{-2\alpha }\int _{n_{\varepsilon }(1-\delta )}^{n_{\varepsilon }}(\rho /n_{\varepsilon }-\rho )\,d\rho \nonumber \\&=\varepsilon ^{-2\alpha }n_{\varepsilon }(1-n_{\varepsilon })[1-(1-\delta )^2]\simeq \varepsilon ^{-2\alpha }\varepsilon ^{2\alpha }\delta . \end{aligned}$$

Again, with the previous choice for $\delta $, this term vanishes in the limit for $\varepsilon \rightarrow 0$.

The second integral in the right hand side of (8.3) reads

$$\begin{aligned}&\varepsilon ^{-2\alpha }\int _{n_{\varepsilon }}^{1}{\theta _{\varepsilon }^2(\rho )\,d\rho }=\varepsilon ^{-2\alpha }\int _{n_{\varepsilon }}^{1}{(\pi -2\arcsin (\rho ))^2\,d\rho }\nonumber \\&\simeq \varepsilon ^{-2\alpha }(1-n_{\varepsilon })^2\simeq \varepsilon ^{-2\alpha }\frac{\varepsilon ^{2\alpha }}{|v|^2}=\frac{1}{|v|^2}. \end{aligned}$$

(8.11)

Therefore the only contribution in the limit is the one given by (8.7) and we obtain

$$\begin{aligned} \tilde{B}:=\lim _{\varepsilon \rightarrow 0}{\frac{\mu \varepsilon ^{-2\alpha }}{2}|v|\int _{-1}^{1}{\theta _{\varepsilon }^2(\rho )\,d\rho }}=\lim _{\varepsilon \rightarrow 0}\mu \left[ \frac{2\alpha }{|v|^3}|\log (\varepsilon )|\right] =+\infty , \end{aligned}$$

(8.12)

and finally

$$\begin{aligned} B:=\lim _{\varepsilon \rightarrow 0}{\frac{\mu \varepsilon ^{-2\alpha }}{2|\log \varepsilon |}|v|\int _{-1}^{1}{\theta ^2(\rho )\,d\rho }}=\frac{2\alpha }{|v|^3}\mu . \end{aligned}$$

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G. Basile
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A. Nota
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M. Pulvirenti
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1.Dipartimento di Matematica “Guido Castelnuovo”Sapienza Università di RomaRomaItaly

A Diffusion Limit for a Test Particle in a Random Distribution of Scatterers

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