Recall from last lecture:

Lifetimes and Cross Sections

Specifically, for a decay of initial state i (an unstable particle) to final state f containing nf particles, we can write
f = |Mif|2 (S/2hbar mi) {Πj=1nf [(c d3pj)/((2π)3 2Ej)]} (2π)4 δ4(pi - Σj pj)
where S is a statistical factor to account for cases with identical particles in the final state, and to account for different spin states.

Example: π0 --> γγ

Example: spin averaged two-body decay

Calculating Mif

Here is our set of rules for estimating matrix elements for Feynman diagrams:

  1. Momenta: Label incoming and outgoing 4-momenta p1, p2, ... Label the internal 4-momenta q1, q2, ... Put an arrow on each line (internal and external) to keep track of the "positive" direction (the choice is arbitrary for internal lines).
  2. Vertex factors: Each vertex gets a factor -ig where g is the total coupling constant at that vertex. For QED vertices, g is the electric charge coupling to the photon (q sqrt(αEM). For QCD vertices, g is the strong charge coupling constant (sqrt(αS). For charged weak vertices, g is the weak charge times the appropriate CKM matrix element. For neutral weak vertices, g is the weak charge.
  3. Propagator: Each internal line gets a factor i/(q2j - m2j). Recall that these virtual particles can be "off the mass shell" so that q2j is not required to be equal to m2j.
  4. Energy and momentum conservation: For each vertex, write a delta function of the form
    (2π)4 δ4(k1+k2+k3)
    where the k's are the three 4-momenta coming into the vertex (if the arrow on a line points away from the vertex, then k is the negative of the 4-momentum assigned to that line. This factor imposes energy momentum conservation at each vertex.
  5. Integration over internal momenta: For each internal line, write down a factor
    (2π)-4 d4qj
    and integrate over all internal momenta.
  6. Cancel the delta functions: The result will include a delta function for overall energy momentum conservation:
    (2π)4 δ4(pi - Σj pj)
    Remove this factor and what remains is -iMif.

Scattering Cross Sections

What is a scattering cross section?

Consider a situation where you throw darts at a dart board, while blindfolded! Not knowing the precise location of the dart board, you throw the darts in a uniformly random pattern across the wall the board is mounted on. While chance will cause fluctuations in the outcome, on average the fraction of darts that hit the board is proportional to the size of the board; the bigger the board, the more darts that will strike it. The relevant parameter is the area of the board, or in particle physics lingo, the cross section.

This is roughly the situation that occurs in particle physics. It is impossible to accurately locate a nucleus, let alone a proton or an electron. Therefore experimenters aim a beam or particles (typically of order 109 particles in a few nanosecond burst) at a target. The target could be a collection of atoms (solid, liquid, or gas), or another beam of particles. For simplicity, let's assume the target to be a thin foil of material, for instance, gold.

The cross section for the dart board example is easily visualized. After all, a dart either hits the board or misses it -- the board is a "hard" target. When scattering electrons off protons, say, can we define the cross section in a similar way? The proton is more of a "soft" target. The electron need only get close enough for electromagnetic interaction to occur, causing it to be deflected. We can define an "effective" cross section, σ, for a "soft" target, one that is proportional to the probability for an interaction to occur.

For elementary particles, σ depends on the target and the projectile. The cross section for electrons scattering off protons is orders of magnitude larger than the cross section for neutrinos scattering off protons and orders of magnitude smaller than the cross section for pions scattering off protons. So whenever you quote a cross section, you must specify both target and projectile.

Also, in partile physics, the final state particles can be different from the initial state particles. To help keep track of all the possibilities, several types of cross section are commonly used:

The total (inclusive) cross section is the sum of the elastic and inelastic cross sections, and the sum of all possible exclusive cross sections:
σtot = σel. + σinel. = Σi=1n σi

Imagine the situation of scattering from a fixed source (i.e. a heavy charged nucleus). We define the scattering angle θ as the angle between the initial and final momenta. The scattering angle depends on the impact parameter, b, the minimum distance between the projectile and target if no scattering occurred. The relation between impact parameter and scattering angle, θ(b), depends on the type of interaction involved. (We normally deal with potentials that are symmetric in φ. This is the situation that I will assume throughout.)

Hard-sphere scattering

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 = R2 cos(θ/2)sin(θ/2)/2 sinθ = R2/4 .
The total cross section is obtained by integrating over the full solid angle
σ = Int{(dσ/dΩ)dΩ} = Int {R2/4 dΩ} = πR2.
This is exactly the answer we expect for this example.

Rutherford scattering

The classic(al) scattering experiment is Rutherford's scattering experiment where α particles (4He 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/sin4(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 q1 with initial kinetic energy T scattering in the electrostatic potential from a fixed charge q2 is

b = (q1q2/2T)cot(θ/2).
The differential cross section is
dσ/dΩ = {q1q2/(4Tsin2(θ/2))}2.
The total cross section is
σ = 2π (q1q2/4T)2 Int{ (sinθ dθ)/sin4(θ/2)} .
which is infinite. This shouldn't cause concern since Rutherford's actual experiment used gold atoms, not gold nuclei, and gold atoms have orbiting electrons which cancel the field of the nucleus beyond some radius. If we integrate out to a finite impact parameter (translated to scattering angle θ) then the cross section is finite.

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.

Fermi's Golden Rule for Scattering

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 hbar2S/4Sqrt{(p1·p2)2 - (m1m2c2)2} {Πj=1nf [(c d3pj)/((2π)3 2Ej)]} (2π)4 δ4(pi - Σj pj)

Example: Two-body scattering in the CM frame

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, p1 = -p2, therefore pp2μ = E1E2 + p12. After some massaging,

Sqrt{(p1·p2)2 - (m1m2c2)2} = (E1 + E2)|p1|/c .
Inserting this into the expression for dσ yields:
dσ = (hbarc/8π)2 {S|M|2c/(E1 + E2)|p1|} (d3p3/E3) (d3p4/E4) δ4(p1 + p2 - p3 - p4)
The delta function becomes
δ4(p1 + p2 - p3 - p4) = δ(E1 + E2 - E3 - E4) δ3(-p3 - p4)

Relation to Perkins notation

Consider a two-body to two-body scattering process

A B --> C D
in which a parallel beam of particles A, moving with speed vi, impinges on a target containing particles of type B. For instance, A could be a beam of protons, pions, electrons, or neutrinos, and B is some material containing nuclei with protons and neutrons. (When the typical interaction energies exceed the binding energy of protons and neutrons in nuclei (about 10MeV) then we can consider the nucleus as a collection of quasi-free protons and neutrons.) The outgoing particles are called C and D. If the scattering is elastic then C and D are the same as A and B. If the scattering is inelastic then C and D are different from A and B (different types of particles; particles of the same type are indistinguishable).

Consider the target to be of thickness dx, and to have nB B particles per unit volume. The flux of incoming particles (particles/cm2/sec) is Φ = nAvi where nA is the density of A particles in the beam and vi 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 nB dx. The scattering rate is the rate that A particles will hit this area

f s nB dx
The reaction rate per target particle is
W = f s


Copyright © Robert Harr 2003