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differential equation applications 223V(x,y)yv'bvxFigure 9.8 Left: A classic pinball machine in which multiple scatterings occur from the roundstructures on the top. Right: Scattering from the potential V(x,y)=x 2 y 2 e −(x2 +y 2) . The incidentvelocity is v in the y direction, and the impact parameter (x value) is b. The velocity of thescattered particle is v ′ , and its scattering angle is θ.Here N scatt (θ) is the number of particles per unit time scattered into the detectorat angle θ subtending a solid angle ∆Ω, N in is the number of particle per unit timeincident on the target of cross-sectional area ∆A in , and the limit in (9.58) is forinfinitesimal detector and area sizes.The definition (9.58) for the cross section is the one that experimentalists use toconvert their measurements to a function that can be calculated by theory. We astheorists solve for the trajectories of particles scattered from the potential (9.57) andfrom them deduce the scattering angle θ. Once we have the scattering angle, wepredict the differential cross section from the dependence of the scattering angleupon the classical impact parameter b [M&T 03]:σ(θ)= ∣∣ dθdbb. (9.59)∣ sin θ(b)The surprise you should find in the simulation is that for certain parameters dθ/dbhas zeros and discontinuities, and this leads to a highly variable, large cross section.The dynamical equations to solve are just Newton’s law for the x and y motionswith the potential (9.57):F = ma− ∂V∂x î − ∂V∂y ĵ = xmd2 dt 2 , (9.60)[ ]∓ 2xye −(x2 +y 2 )y(1 − x 2 )î + x(1 − y 2 )ĵ = m d2 xdt 2 î + ymd2 ĵ.dt2 (9.61)−101COPYRIGHT 2008, PRINCET O N UNIVE R S I T Y P R E S SEVALUATION COPY ONLY. NOT FOR USE IN COURSES.ALLpup_06.04 — 2008/2/15 — Page 223