Scattering Cross Sections

Consider the elastic collision between two spheres of radius R, one fixed rigidly in place. From the figure we see that b=R sinα, and also that 2α + θ = π. Use the second expression to eliminate sinα from the first. We have sinα = sin(π/2 -θ/2) = cos(θ/2), so that

b = Rcos(θ/2).

Solving for θ we find
θ(b) = 2cos^{-1}(b/R).

If the impact parameter lies in the ring between b and b+db, then it will have a scattering angle between θ and θ+dθ. The initial ring the particle passes through has area dσ = 2πb db, and it scatters into a solid angle dΩ = 2πsinθ d&theta = -2πd(cosθ). (θ lies between 0 and π such that dΩ is always positive.) We call the ratio dσ/dΩ the differential cross section, and in this example it is

dσ/dΩ = (b/sinθ)(db/dθ)

Differentiating the expression for b(θ) we have

dσ/dΩ = Rb sin(θ/2)/2 sinθ = R^{2} cos(θ/2)sin(θ/2)/2 sinθ = R^{2}/4 .

The total cross section is obtained by integrating over the full solid angle
σ = Int{(dσ/dΩ)dΩ} = Int {R^{2}/4 dΩ} = πR^{2}.

This is exactly the answer we expect for this example.
The classic(al) scattering experiment is Rutherford's scattering experiment where α particles (^{4}He nuclei) emitted by radioactive decay are collimated into a narrow beam impinging on a thin gold foil.
The α particles scatter from the gold nuclei and are detected.
In previous courses you have probably worked out the 1/sin^{4}(θ/2) distribution of scattered particles.
Let's review how this is done to make connection with quantities we will use.

The relation between impact parameter and scattering angle for a charge q_{1} with initial kinetic energy T scattering in the electrostatic potential from a fixed charge q_{2} is

b = (q_{1}q_{2}/2T)cot(θ/2).

The differential cross section is
dσ/dΩ = {q_{1}q_{2}/(4Tsin^{2}(θ/2))}^{2}.

The total cross section is
σ = 2π (q_{1}q_{2}/4T)^{2} Int{ (sinθ dθ)/sin^{4}(θ/2)} .

which is infinite.
This shouldn't cause concern since Rutherford's actual experiment used gold In scattering experiments we characterize the beam by its luminosity, L, defined as the number of beam particles per unit area per second passing through an imaginary plane. The number of interactions per second is N = Lσ. The number of interactions scattering into solid angle dΩ per second is dN = L(dσ/dΩ)dΩ. We can also write dσ/dΩ = (1/L)dN/dΩ.

In the general case of particles 1 and 2 colliding and producing particles 3, 4, ... , N

1 2 --> 3 4 ... N

the differential cross section is given the formula
dσ = |Mif|^{2} hbar^{2}S/4Sqrt{(p_{1}·p_{2})^{2} - (m_{1}m_{2}c^{2})^{2}} {Π_{j=1}^{nf} [(c d^{3}**p**_{j})/((2π)^{3} 2E_{j})]} (2π)^{4} δ^{4}(p_{i} - Σ_{j} p_{j})

Consider a two-body to two-body scattering process

1 2 --> 3 4

in the CM frame.
Assume the matrix element of this interaction is M, and determine the differential cross section.
In the CM, **p**_{1} = -**p**_{2}, therefore p_{1μ}p_{2}^{μ} = E_{1}E_{2} + **p**_{1}^{2}.
After some massaging,

Sqrt{(p_{1}·p_{2})^{2} - (m_{1}m_{2}c^{2})^{2}} = (E_{1} + E_{2})|**p**_{1}|/c .

Inserting this into the expression for dσ yields:
dσ = (hbarc/8π)^{2} {S|M|^{2}c/(E_{1} + E_{2})|**p**_{1}|} (d^{3}**p**_{3}/E_{3}) (d^{3}**p**_{4}/E_{4}) δ^{4}(p_{1} + p_{2} - p_{3} - p_{4})

The delta function becomes
δ^{4}(p_{1} + p_{2} - p_{3} - p_{4}) = δ(E_{1} + E_{2} - E_{3} - E_{4}) δ^{3}(-**p**_{3} - **p**_{4})

Consider a two-body to two-body scattering process

A B --> C D

in which a parallel beam of particles A, moving with speed vConsider the target to be of thickness dx, and to have n_{B} B particles per unit volume.
The flux of incoming particles (particles/cm^{2}/sec) is Φ = n_{A}v_{i} where n_{A} is the density of A particles in the beam and v_{i} is their velocity.
For a single nucleus interaction cross section, σ, the fraction of the target area covered by the cross section of B particles is σ n_{B} dx.
The scattering rate is the rate that A particles will hit this area

f s n_{B} dx

The reaction rate per target particle is
W = f s

Copyright © Robert Harr 2003