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

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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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  1. Bodineau, T., Gallagher, I., Saint-Raymond, L.: The Brownian motion as the limit of a deterministic system of hard-spheres. arXiv:1305.3397

  2. Boldrighini, C., Bunimovich, L.A., Sinai, Y.G.: On the Boltzmann equation for the Lorentz gas. J. Stat. Phys. 32, 477–D0501 (1983)

    Article  MATH  MathSciNet  ADS  Google Scholar 

  3. Dürr, D., Goldstein, S., Lebowitz, J.: Asymptotic motion of a classical particle in a random potential in two dimensions: Landau model. Commun. Math. Phys. 113, 209–230 (1987)

    Article  MATH  ADS  Google Scholar 

  4. Desvillettes, L., Pulvirenti, M.: The linear Boltzmann equation for long-range forces: a derivation from particle systems. Models Methods Appl. Sci. 9, 1123–1145 (1999)

    Article  MATH  MathSciNet  Google Scholar 

  5. Desvillettes, L., Ricci, V.: A rigorous derivation of a linear kinetic equation of Fokker–Planck type in the limit of grazing collisions. J. Stat. Phys. 104, 1173–1189 (2001)

    Article  MATH  MathSciNet  ADS  Google Scholar 

  6. Erdos, L., Salmhofer, M., Yau, H.-T.: Quantum diffusion of the random Schroedinger evolution in the scaling limit. Acta Math. 200, 211–277 (2008)

    Google Scholar 

  7. Esposito, R., Pulvirenti, M.: From Particles to Fuids. Hand-Book of Mathematical Fuid Dynamics, vol. III, pp. 1–82. North-Holland, Amsterdam (2004)

    Google Scholar 

  8. Gallavotti, G.: Rigorous Theory of the Boltzmann Equation in the Lorentz Gas, p. 358. Nota interna Istituto di Fisica, Università di Roma, Roma (1973)

    Google Scholar 

  9. Kesten, H., Papanicolaou, G.: A limit theorem for stochastic acceleration. Commun. Math. Phys. 78, 19–63 (1981)

    Article  MathSciNet  ADS  Google Scholar 

  10. Kirkpatrick, K.: Rigorous derivation of the Landau equation in the weak coupling limit. Commun. Pure Appl. Math. 8, 1895–D01916 (2009)

    MATH  MathSciNet  Google Scholar 

  11. Komorowski, T., Ryzhik, L.: Diffusion in a weakly random Hamiltonian flow. Commun. Math. Phys. 263, 273–323 (2006)

    Article  MathSciNet  ADS  Google Scholar 

  12. Landau, L.D., Lifshitz, E.M.: Mechanics, Course of Theoretical Physics, vol. 1. Pergamon press, Oxford (1960)

    Google Scholar 

  13. Lorentz, H.A.: The motion of electrons in metallic bodies. Proc. Acad. Amst. 7, 438–453 (1905)

    Google Scholar 

  14. Spohn, H.: The Lorentz flight process converges to a random flight process. Commun. Math. Phys. 60, 277–D0290 (1978)

    Article  MATH  MathSciNet  ADS  Google Scholar 

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

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Correspondence to A. Nota.


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

The potential energy for a finite potential barrier is given by

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

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

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

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

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

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:

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

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

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

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

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

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

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

Thanks to the symmetry for the scattering problem

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

According to (7.3):

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

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

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


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

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

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


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

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

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

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

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

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

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


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

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

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


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

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

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

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

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

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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Basile, G., Nota, A. & Pulvirenti, M. A Diffusion Limit for a Test Particle in a Random Distribution of Scatterers. J Stat Phys 155, 1087–1111 (2014).

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