University of Rhode Island DigitalCommons@URI Classical Dynamics Physics Course Materials 2015 10. Scattering from Central Force Potential Gerhard Müller University of Rhode Island, [email protected] Creative Commons License This work is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 4.0 License. Follow this and additional works at: http://digitalcommons.uri.edu/classical_dynamics Abstract Part ten of course materials for Classical Dynamics (Physics 520), taught by Gerhard Müller at the University of Rhode Island. Entries listed in the table of contents, but not shown in the document, exist only in handwritten form. Documents will be updated periodically as more entries become presentable. Recommended Citation Müller, Gerhard, "10. Scattering from Central Force Potential" (2015). Classical Dynamics. Paper 12. http://digitalcommons.uri.edu/classical_dynamics/12 This Course Material is brought to you for free and open access by the Physics Course Materials at DigitalCommons@URI. It has been accepted for inclusion in Classical Dynamics by an authorized administrator of DigitalCommons@URI. For more information, please contact [email protected]. Contents of this Document [mtc10] 10. Scattering from Central Force Potential • Scattering from stationary central force potential [msl2] • Determination of the scattering angle [mln20] • Total cross section for shower of meteorites [mex 58] • Rutherford scattering formula [mex56] • Scattering from hard spheres [mex55] • Elastic scattering from hard ellipsoids [mex60] • Scattering cross section for inverse square potential [mex59] • Particle experiencing soft Coulomb kick [mex10] • Scattering angle in the laboratory frame [msl3] • Loss of kinetic energy in elastic collision [mex57] • Elastic collision: angle between scattered particles [mex240] • Elastic collision: velocities of scattered particles [mex241] • Mechanical refraction [mex167] • Scattering from a spherical potential well [mex168] • Grazing collision between flat surfaces [mex219] • Absorption cross section of power-law potential [mex242] • Small-angle scattering [mln105] • Small-angle scattering from-power-law potential [mex246] • Classical inverse scattering [mln104] • Classical inverse scattering problem I [mex243] • Classical inverse scattering problem II [mex244] • Classical inverse scattering problem III [mex245] • Decay of particle I [mln102] • Decay of particle II [mln103] • Decay of particle: maximum kinetic energy [mex237] • Decay of particle: directions in lab frame I [mex238] • Decay of particle: directions in lab frame II [mex239] Determination of scattering angle [mln20] r ℓ/mr2 Orbital integral: ϑ = dr . Z∞ 2 − − ℓ2 E V (r) 2 q m 2mr . Periapsis: ϑ(rmin) = ψ ⇒ 2ψ + θ = π. ψ ψ θ ∞ dr/r2 ∞ sdr/r2 ⇒ ψ = = . 2 Zrmin 2m − − 1 Zrmin V (r) s 2 [E V (r)] 2 1 − − q ℓ r q E r2 Substitute u =1/r: umax sdu ⇒ ψ = . Z 0 1 − V (1/u) − s2u2 q E Use energy conservation to determine umax: 2 2 1 2 ℓ ℓ 2 2 E = mr˙ + 2 + V (r)= 2 + V (rmin)= Es umax + V (1/umax). 2 2mr 2mrmin ⇒ 2 2 ⇒ s umax + V (1/umax)/E =1 umax = umax(s, E). umax sdu Scattering angle: θ(s, E)= π − 2 . 2 2 Z0 1 − V (1/u)/E − s u p Total scattering cross section: π s ds smax σ = σ(θ)dΩ=2π sin θ dθ =2π sds. T Z Z Z 0 sin θ dθ 0 Note: In quantum mechanics σT can be finite even if smax is infinite. [mex58] Total cross section for shower of meteorites A uniform beam of small particles with mass m1 and velocity v0 is directed toward a planet of mass m2 and radius R. Calculate the total cross section σT for the particles to be absorbed by the planet. Solution: [mex56] Rutherford scattering formula Derive the scattering cross section κ 2 1 ZZ0e2 σ(θ) = 4 ; κ = 4E sin (θ=2) 4π0 for elastic scattering of particles with electric charge Ze and energy E from stationary atomic nuclei with charge Z0e. Note that σ(θ) does not depend on whether the beam is positively or negatively charged. Solution: [mex55] Scattering from hard spheres (a) Calculate the scattering cross section σ(θ) for elastic scattering from hard spheres of radius a. R 2 (b) Calculate the total scattering cross section σT = d Ω σ(θ). Solution: 1 [mex60] Elastic scattering from a hard ellipsoid Show that the cross section for elastic scattering from a hard ellipsoid described by the equation x2=a2 + (y2 + z2)=b2 = 1 with the incident beam along the x-axis is 1 a2b2 σ(θ) = b2 : 4 [a2 sin2(θ=2) + b2 cos2(θ=2)]2 z ψ b s x a Solution: [mex59] Scattering cross section for inverse square potential Show that the cross section for scattering from the stationary potential V (r) = κ/r2 with κ > 0 is κπ2 π − θ σ(θ) = : E θ2(2π − θ)2 sin θ Solution: [mex10] Particle experiencing soft Coulomb kick A particle with charge Q1 and mass m1 moves at very high velocity v1 along a (nearly) straight line that passes at a distance b from a particle with charge Q2 and mass m2, which is initially at rest. The assumptions are that the two particles interact via a Coulomb central force and that the second particle does not change its position significantly during the encounter. (a) Find the direction in which the second particle will move after the encounter. (b) Find the energy ∆E transferrred from the first to the second particle during the encounter. 1 r v1 2 θ b Solution: Scattering angle in the laboratory frame [msl3] The scattering experiment is performed in the laboratory frame. • observed scattering angle: θ¯, • observed scattering cross section:σ ¯(θ¯), • projectile of mass m1 and target of mass m2. The theoretical analysis is performed in the center-of-mass frame: • problem reduced to one degree of freedom, • total mass M = m1 + m2, • reduced mass m = m1m2/(m1 + m2), • calculated scattering angle: θ, • calculated scattering cross section: σ(θ). Task #1: establish the relation between θ and θ¯. − vcm v1 − − θ − v1 v0 θ m1 m m1v¯0 =(m1 + m2)v¯cm ⇒ v¯cm = v¯0 = v¯0. m1 + m2 m2 v¯1 sin θ¯ = v1 sin θ, v¯1 cos θ¯ = v1 cos θ +¯vcm. 1 Relative velocity after collision: v = v¯2 − v¯1 = v2 − v1 (frame-independent). Linear momentum in center-of-mass frame: m1v1 + m2v2 =0. m2 m1 ⇒ v1 = − v, v2 = v ⇒ m1v1 = mv. m1 + m2 m1 + m2 v sin θ sin θ m v¯ m v¯ ⇒ tan θ¯ = 1 = , ρ = 0 = 1 0 . v1 cos θ +¯vcm cos θ + ρ m2 v1 m2 v ⇒ cos θ = −ρ(1 − cos2 θ¯) + cos θ¯ 1 − ρ2(1 − cos2 θ¯). q 1 1 Elastic scattering: T = mv¯2 = mv2 (in center-of-mass frame) 2 0 2 ⇒ v¯0 = v ⇒ ρ = m1/m2. Task #2: establish the relation between σ andσ ¯. Number of particles scattered into infinitesimal solid angle: 2πIσ(θ) sin θ|dθ| =2πIσ¯(θ¯) sin θ¯|dθ¯|. sin θ dθ d cos θ ⇒ σ¯(θ¯)= σ(θ) = σ(θ) . ¯ ¯ ¯ sin θ dθ d cos θ 2 ¯ ¯ ¯ 1+ ρ cos(2θ) ⇒ σ¯(θ)= σ(θ) "2ρ cos θ + # . 1 − ρ2 sin2 θ¯ p Special case: elastic scattering between particles of equal mass: θ θ m = m ⇒ cos θ = cos(2θ¯) ⇒ θ¯ = , σ¯(θ¯) = 4 cos σ(θ). 1 2 2 2 2 [mex57] Loss of kinetic energy in elastic collision Consider a particle of mass m1 and incident velocity v¯0 undergoing an elastic collision via central force with a target of mass m2 that is initially at rest. The particle emerges with velocity v¯1 from the collision as viewed in the laboratory frame. The figure shows this velocity in relation to the center-of-mass velocity v¯cm and the final velocity v1 of the particle in the center-of-mass frame. Also shown are the scattering angles θ (center-of mass frame) and θ¯ (laboratory frame). Show that the ratio of the final and initial kinetic energies in the laboratory frame is 2 T1 1 + 2ρ cos θ + ρ m1 = 2 ; ρ = : T0 (1 + ρ) m2 . −vcm v1 −v θ 1 −θ Solution: [mex240] Elastic collision: angle between scattered particles A particle of mass m1 and incident velocity v¯0 undergoes an elastic collision via central force with a particle of mass m2 that is initially at rest. Given the scattering angles θ1, θ2 = π − θ1 in the ¯ ¯ center-of-mass frame, find the sum θ1 + θ2 of the scattering angles in the laboratory frame as a ¯ ¯ function of θ1 and m1=m2. Show that if m1 = m2 then we have θ1 + θ2 = π=2 for 0 < θ1 < π. Solution: [mex241] Elastic collision: velocities of scattered particles A particle of mass m1 and incident velocity v¯0 undergoes an elastic collision via central force with a particle of mass m2 that is initially at rest. Show that the velocities of the scattered particles depend on the scattering angles in the laboratory frmae as follows: v¯2 m ¯ : m1m2 = 2 cos θ2; m = ; v¯0 m2 m1 + m2 s 2 ! v¯1 m ¯ m2 2 ¯ = cos θ1 ± 2 − sin θ1 ; v¯0 m2 m1 where two solutions (±) exist for m1 > m2 and one solution (+) for m1 < m2. Solution: [mex167] Mechanical refraction A particle of mass m moving in the xy-plane is subject to a potential energy which assumes the constant value V1 at y ≥ 0 and the constant value V2 at y < 0. Let us assume that V2 < V1. Use conversation laws to show that if the particle approaches the x-axis with speed v1 at an angle θ1 as shown, it will proceed with a different speed v2 at a different angle θ2 after crossing the line where the potential energy changes abruptly. Show in particular that the relation s sin θ1 2 n ≡ = 1 + 2 (V1 − V2); sin θ2 mv1 between the two angles holds, where n plays the role of index of refraction.