Thermalization of a particle with dissipative collisions.pdf

Thermalization of a particle with dissipative collisions.pdf

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Thermalization of a particle with dissipative collisions

a r X i v : c o n d - m a t / 9 8 1 0 0 7 2 v 1 [ c o n d - m a t .s t a t - m e c h ] 7 O c t 1 9 9 8 Thermalization of a particle by dissipative collisions Philippe A. Martin Institut de Physique The?orique, Ecole Polytechnique Fe?de?rale de Lausanne, CH-1015, Lausanne, Switzerland Jaros law Piasecki Institute of Theoretical Physics, University of Warsaw, Hoz?a 69, 00 681 Warsaw, Poland (February 1, 2008) One considers the motion of a test particle in an homogeneous fluid in equilibrium at temperature T , undergoing dissipative colli- sions with the fluid particles. It is shown that the corresponding lin- ear Boltzmann equation still posseses a stationary Maxwellian veloc- ity distribution, with an effective temperature smaller than T . This effective temperature is explicitly given in terms of the restitution parameter and the masses. PACS numbers: 51.10.+y, 44.90.+c The search for stationary states of granular matter has recently been a subject of interest for both experimental and theoretical reasons [1,2]. Granular matter can be modelized by spherical particles that partially dissipate their kinetic energy at collisions. If (u,v) and (u?, v?) denote the velocities of two spheres of mass m and M before and after the collision, they are related by mu? +M v? = mu+Mv (1) σ? · (v? ? u?) = ?ασ? · (v ? u), 0 α ≤ 1 (2) σ? ⊥ · (v? ? u?) = σ?⊥ · (v ? u) where σ? is a unit vector normal to the surface of the spheres at the point of impact, and σ?⊥ points in the tangent direction: σ?⊥ · σ? = 0. The first relation is the conservation of the center of mass momentum, whereas the second one says that the normal component of the relative velocity reverses its direction with a magnitude reduced by the factor α, the so called restitution parameter (the tangent component remains unchanged). Solving (1), (2) for u? and v? gives u? = u+ (1? μ)(1 + α)(σ? · (v ? u))σ?, v? = v ? μ(1 + α)(σ? · (v ? u))σ? (3) with μ = m/(m+M). The inverse relation is obtained by exchanging the

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