Scattering 1 Classical scattering of a charged particle (Rutherford ...

dσdΩ = − sin(θ) sdθdsAssuming a completely elastic collision, energy is conserved during the motion of the charge.Thus in spherical coordinates, the conservation of energy is given by;(1/2)mṙ 2 + (1/2) L2mr 2 + U = constant = T 0The angular momentum, l for a central force (Coulomb) is also conserved. This is;l = mr 2 ωThis allows an exchange of the variable of integration from t to θ.dθ ′ =l drmr 2√ (2/m)[t 0 − U r − l 2 /(2mr 2 )]This is integrated after using the Coulomb potential, U =4πǫ 1 zZe2 r . Here z and Z are thecharge on the particle and scattering center, respectively, and e the electronic charge. Thesolution is;(1/r) = − mqQL 2 (1 + η cos(θ) ′ )This is the equation of a hyperbola or an ellipse depending on the sign of the energy. Theexcentricity, η, is√η = 1 + 8πǫE 0L 2mzZ(e) 2Now Change the angle θ ′ to the scattering angle θ as observed in Figure 2, where θ = π−2θ ′ .This results in the equation;cot(θ/2) = 8πǫT 0szZe 2s = zZe28πǫT 0cot(θ/2)Substitution into the equation for the cross section gives;dσdΩ = [ zZe24πǫT 0] 2 [ 1sin 4 (θ/2) ]In reality the scattering is not elastic as the charge is accelerated and loses energy. Thisloss could be determined by a perturbation series as it is small. However, to first order thetrajectory does not change so the velocity and acceleration of the charge q can be foundusing the acceleration as determined from the equation of motion to get the radiated power.2