Physics 321 Hour 38 Scattering Cross Sections
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Physics 321 Hour 38 Scattering Cross Sections The Scattering Problem • In the cm frame, two particles approach each other and scatter. Given the interaction, what is the distribution of scattered particles? π π0 Θ Θ π0 π The Inverse Scattering Problem • Given the distribution of scattered particles, what is the interaction? • This is a very hard but very interesting problem. π π0 Θ Θ π0 π The Scattering Problem • The impact need not be hard-sphere scattering. – The distance between the trajectories is the impact parameter, b. Θ b Θ Relating θ to b • The first thing we need to do is find the relationship between θ and b for a given type of interaction. Θ b Θ Hard Sphere Elastic Scattering – b(θ) π π π π cos = sin − = 2 2 2 π π−π π = π sin 2 Coulomb Scattering – b(θ) First we note that the total angular momentum is β = ππΌπ + ππ½π = ππ π2 β2 πΈ=π ∞ = = 2π 2ππ 2 Coulomb Scattering – b(θ) Next we use a result from the central force problem. πΎ2π 2 πΈ = 2 π −1 2β πΎ = ππ π1 π2 Coulomb Scattering – b(θ) φ 1 cos ππππ₯ = − π π π ππππ₯ = + 2 2 For hyperbolic orbits πΆ π π = 1 + π cos π π π π cos + = −sin 2 2 2 1 =− π Coulomb Scattering – b(θ) φ 1 cos ππππ₯ = − π π π ππππ₯ = + 2 2 For hyperbolic orbits πΆ π π = 1 + π cos π π π π cos + = −sin 2 2 2 1 =− π Coulomb Scattering – b(θ) Using π 1 sin = 2 π πΎ2π 2 πΈ = 2 π −1 2β πΎ 2π π 2 = 2 cot 2β 2 π2 − 1 φ π 2 1 Coulomb Scattering – b(θ) πΎ2π π 2 πΈ = 2 cot 2β 2 β2 = 2ππ 2 πΈ 2 2 2 π 2 2ππ πΈ πΈ 4π πΈ 2 cot = = 2 2 πΎ π πΎ2 π 2ππ0 cot = 2 πΎ Differential Cross Section - Definition • Consider a small beam striking a large target. The detector has solid angle ΔΩ and detects all particles scattered into it. detector θ beam target ππ = πΩ # ππ ππππ‘πππππ πππ‘πππ‘ππ/π ππ # π‘πππππ‘ ππππ‘πππππ # ππ ππππ ππππ‘πππππ / sec × × βΩ ππππ Differential Cross Section - Definition • Consider a large beam striking a small target. A detector has solid angle ΔΩ. detector beam θ target ππ # ππ ππππ‘πππππ πππ‘πππ‘ππ/π ππ = πΩ # ππ ππππ ππππ‘πππππ × # π‘πππππ‘ ππππ‘πππππ × βΩ ππππ β π ππ Differential Cross Section - Theoretical • We usually take 1 target particle… detector beam θ target ππ # ππ ππππ‘πππππ πππ‘πππ‘ππ/π ππ = πΩ # ππ ππππ ππππ‘πππππ × 1 × βΩ ππππ β π ππ Differential Cross Section - Theoretical ππ # ππ ππππ‘πππππ πππ‘πππ‘ππ/π ππ = πΩ # ππ ππππ ππππ‘πππππ × 1 × βΩ ππππ β π ππ 1) Find the total number of beam particles per second between π → π + βπ. This is the number in a ring of radius π. # ππ ππππ ππππ‘πππππ π≡ ππππ β π ππ This is: π 2ππ ππ Differential Cross Section - Theoretical ππ # ππ ππππ‘πππππ πππ‘πππ‘ππ/π ππ = πΩ π βΩ 2) Find the total number of scattered particles between ϑ → π + βπ. π=π π ππ = π ′ π ππ This is: π 2ππ ππ = π 2ππ π ′ π ππ Differential Cross Section - Theoretical ππ # ππ ππππ‘πππππ πππ‘πππ‘ππ/π ππ = πΩ πβΩ 3) The number of particles per second scattering into ππ is: π 2ππ π ′ π ππ The number of those hitting the detector is βπ ′ π 2ππ π π ππ × 2π Differential Cross Section - Theoretical ππ π 2ππ = πΩ βπ π ππ × 2π π βΩ π′ 4) Think of the detector as subtending angles βπ and βπ. The solid angle is defined by: βΩ = sin π βπ βπ or more generally βΩ = sin π ππ ππ (If the area of the detector is small, this can be expressed in terms of the detector’s area: π΄ ≈ π 2 βΩ.) Differential Cross Section - Theoretical βπ π ππ × 2π ππ π 2ππ = πΩ π sin π βπ βπ π′ ππ π π′ π = πΩ sin π Or since we want a positive value, we usually write ππ 1 ππ = π πΩ sin π ππ Total Cross Section The total cross section is the differential cross section integrated over all angles. π= ππ πΩ = πΩ ππ sin πππππ πΩ Hard Sphere Scattering (cm) π−π π = π1 + π2 sin 2 ππ 1 π−π = − π1 + π2 cos ππ 2 2 ππ 1 1 π − π π − π = π1 + π2 2 sin cos πΩ sin π 2 2 2 ππ 1 1 π π 2 = π1 + π2 cos sin πΩ 2cos π sin π 2 2 2 2 2 ππ 1 = π1 + π2 2 πΩ 4 Hard Sphere Scattering (cm) ππ 1 = π1 + π2 2 πΩ 4 ππ π= πΩ = π π1 + π2 πΩ 2 Rutherford Scattering (cm) πΎ π π= cot 2π0 2 ππ πΎ 1 π 2 =− csc ππ 2π0 2 2 2 ππ 1 πΎ π1 π = cot csc 2 πΩ 2cos π sin π 2π0 22 2 2 2 2 ππ 1 πΎ πΎ2 = = π πΩ sin4 π 4π0 2 4 16π0 sin 2 2 CM to Lab Conversion • The only thing that changes in lab vs cm is βΩ. ππ = πΩ # ππ ππππ‘πππππ πππ‘πππ‘ππ/π ππ # π‘πππππ‘ ππππ‘πππππ # ππ ππππ ππππ‘πππππ / sec × × βΩ ππππ βΩπππ = −β cos ππππ βππππ βΩππ = −β cos πππ βπππ βππππ = βπππ ππ ππ π cos πππ → = × πΩπππ πΩππ π cos ππππ
