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Physics of Long-Range Interacting Systems$

A. Campa, T. Dauxois, D. Fanelli, and S. Ruffo

Print publication date: 2014

Print ISBN-13: 9780199581931

Published to Oxford Scholarship Online: October 2014

DOI: 10.1093/acprof:oso/9780199581931.001.0001

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(p.387) Appendix D The Differential Cross-Section of a Binary Collision

(p.387) Appendix D The Differential Cross-Section of a Binary Collision

Source:
Physics of Long-Range Interacting Systems
Publisher:
Oxford University Press

In the derivation of the Boltzmann equation in Chapter 7, we introduced the cross-section related to the binary collisions to transform Eq. (7.46) into Eq. (7.48). As underlined in the derivation, the collisions between two particles are computed neglecting the interaction with the other particles in the system, i.e. as if the two particles were isolated. In this Appendix, we explain in more details the binary collision process and the meaning of the relation in Eq. (7.47).

Let us start from the Hamiltonian of a system of two identical particles

(D.1)
H=p122m+p222m+V|q1q2|,

where pi=|pi|. Making the canonical transformation of coordinates defined in (Eqs. 7.35) and (7.36), here rewritten as

(D.2)
Q=12q1+q2
(D.3)
q=q1q2,

the Hamiltonian becomes

(D.4)
H=P24m+p2m+V|q|,

where P and p are the momenta canonically conjugated to Q and q, respectively. We therefore have the free motion of a particle of mass 2m, the centre of mass, plus the dynamics of a particle of mass m/2 (the reduced mass) subject to the potential V. Let us then concentrate on the motion of a particle of mass μ=m/2 in a central potential. We know that such a motion develops in a plane. Using polar coordinates (r,θ) in this plane, the energy of this mechanical problem is

(D.5)
E=μ2r˙2+r2θ˙2+V(r).

The angular momentum =μrθ˙ is a constant of the motion. Substituting in the expression of the energy, we get (p.388)

(D.6)
E=μ2r˙2+V(r)+22μr2,

showing that the problem is equivalent to a one-dimensional motion in the effective potential V(r)+2/(2μr2). We are interested in the case of open orbits, which occur when E>0. Assuming, without loss of generality, that the particle approaches the origin from r= at the angle θ=0, the general solution, expressing θ‎ as a function of r during the approach, is given by (Goldstein, 1980)

(D.7)
θ(r)=r+dr1r22μEV(r)2r2.

The distance of closest approach, rm, is the largest positive root of the expression under square root in the denominator in Eq. (D.7); if we denote by θm the corresponding value of θ‎, the angle of deflection of the particle, i.e. the angle between the direction of approach and the direction along which the particle goes back to r=, is simply related to θm. Again, without loss of generality, we can assume that ℓ is positive; then χ‎ is given by π2θm, i.e.

(D.8)
χ=π2rmdr1r22μEV(r)2r2.

Both the energy E and the angular momentum ℓ are a simple function of the impact parameter b and of the modulus v of the velocity of the incoming particle at infinity. We have

(D.9)
E=12μv2and=μbv.

Substituting in Eq. (D.8), we get

(D.10)
χ=π2rmdr1r2b12V(r)μv2b2r2.

This expression determines χ‎ as a function of b and v; for most of the potentials V(r) of interest, this function, for given v, is a monotonic decreasing function of b.

Let us now recall the definition of the differential cross-section dσ/dΩ in a scattering problem from a centre of force. If there is a flux I of impinging monoenergetic particles (of energy ε determined by their velocity of incidence v) per unit area and per unit second, they will be scattered to different angles according to their impact parameter. The number N(Ω)dΩ of incident particles scattered per unit second into the solid angle element dΩ about the direction Ω‎ defines the differential cross-section by (p.389)

(D.11)
N(Ω)dΩ=Idσ(Ω,ε)dΩdΩ.

On the other hand, we have just seen that the angle of deflection is uniquely determined by the impact parameter b and by v. Considering the cylindrical symmetry of the problem, we then have

(D.12)
N(Ω)dΩ=Ibdbdϕ,

where ϕ‎ is the azimuthal angle defining, together with r and θ‎, the cylindrical coordinates that describe the scattering. We obtain therefore Eq. (7.47):

(D.13)
dσdΩdΩ=bdbdϕ.