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

Thus the radiated power into solid angle dΩ is ;dPdΩ=q216π 2 ǫc [ˆn × ˆn × ⃗˙β] 2If a charged particle is accelerated by the ⃗ E field of an EM wave we have;⃗E I = E 0I e i(⃗ k·⃗r − ωt) ⃗η 0⃗F(force) = m⃗a = mc ⃗˙β = q ⃗ EI⃗˙β = (q/mc)E 0I e i(⃗ k·⃗r−ωt) ˆη 0Ignore phase differences (take the instantaneous time) and let (ˆη be the polarization direction,Figure 4.〈 dPdΩ 〉 = c ǫ ( q 24πmc 2 ) 2 |ˆη · ˆη 0 | 2The incident flux is the time-averaged Poynting vector of the incident wave so the crosssection is;dσdΩ = 〈(dP/dΩ)〉〈S in 〉dσdΩ = ( q 24πǫmc 2 ) 2 |ˆη · ˆη 0 | 2The evaluation of the factor |ˆη · ˆη 0 | 2 is obtained from Figure 4.ˆη 1 = cos(θ)[ˆxcos(φ) + ŷ sin(φ)] − ẑ sin(θ)ˆη 2 = −ˆx sin(φ) + ŷ cos(φ)Choose ˆη 0 = ˆx and sum over the final polarizations to obtain;|ˆη · ˆη 0 | 2 = cos 2 (θ) cos 2 (φ) + sin 2 (φ)This result can be substiuted above to obtain the scattered differential power. Note we haveneglected the effect of the magnetic field on the motion of the accelerated charge, but thiseffect was worked out to first order in an earlier example.10