In Taylor, read sections 14.2 - 14.5 for today, and 14.5 to 14.6 for Friday.

Since we can't measure the trajectories of projectile and target well enough to predict if a collision will occur in subatomic processes, we generate a probability. The apparent size of the target is called the cross section.

The number of collisions that occur must be proportional to the number of incident projectiles, the number of targets, and the collision cross section.
To determine the number of collisions (scatters), we assume that the targets don't overlap, and that the incident projectiles (beam) intercepts a finite area of the target assembly, that is, the target assembly (the sum of all the individual target atoms, nuclei, or whatever) is much larger than the beam.
Now draw some area around the region where a projectile is incident on the target assembly.
This area can be of any convenient shape, say it is a square of area A = 1 mm².
The number of targets in this square is the density of targets times the area, N_{tar} = n_{tar}A where N is a number and n is a density (number per unit area).
Since the targets don't overlap, the cross sectional area where a collision can occur is σN_{tar} = n_{tar}σA.
The probability that a given incident particle scatters equals the fraction of the time that it lands within the cross sectional area of a target, which is the fraction of the area A occupied by targets, or P_{sc} = σN_{tar}/A = n_{tar}σ.
The total number of scatters is the total number of incident projectiles times the probabiltiy for one of them to scatter,

N_{sc} = N_{inc}P_{sc} = N_{inc}n_{tar}σ.

By dividing both sides by Δt, we can find the rate of scatterings in terms of the rate of incoming projectiles, often an interesting quantity,

R_{sc} = R_{inc}n_{tar}σ.

A spacecraft is being sent through the asteroid belt on a straight, unguided trajectory. If the typical radius of asteroids is 3m and their density is 1/km² (totally fabricated numbers), and the spacecraft is 1m in radius, what is the probability that the spacecraft will hit an asteroid?

The typical cross section for a collision between the spacecraft and an asteroid is σ = π(R_{ast} + R_{cft})² = 50m².
The density of targets is n_{tar} = 1/km² = 10^{-6}/m².
The probability of a collision is P_{sc} = n_{tar}σ = (10^{-6}/m²)(50m²) = 5×10^{-5}.

In nuclear and particle physics, the targets are rather small, and m² is an awkward unit for cross section.
The unit commonly used is the barn where 1 barn = 1b = 10^{-28}m².
The cross section of an atomic nucleus is approximately 1 barn.
Many processes have much smaller cross sections, millibarns (mb), microbarns (μb), picrobarns (pb), and femtobarns (fb) are often encountered.

As used in the example above, the cross section for the scattering of two hard spheres of radii R_{1} and R_{2} has cross section σ = π(R_{1} + R_{2})².

The definition for cross section was for an elastic scattering, a process where the objects that emerge from the collision are the same as the initial projectile and target. But there are many other types of processes in which the projectile, target, or both are changed by the collision. For example, we can imagine scattering electrons off an atom in its ground state. If the final particles are just an electron and the atom, still in its ground state, then the scattering was elastic. But many other processes are also possible:

- the electron can excite the atom to an excited state,
- the electron can pass near the nucleus and produce a photon,
- the electron can knock another electron from the atom, leaving behind a positive ion and two or more outgoing electrons, or
- the electron can be absorbed by the atom, leaving just a negative ion.

We can assign a cross section to each process, and sum them to get the total cross section for a collision: σ_{tot} = σ_{elastic} + σ_{photon} + σ_{ion} + σ_{cap}.
The number of captures is the given by N_{cap} = N_{inc}n_{tar}σ_{cap}, and the total number of collisions of any sort is N_{tot} = N_{inc}n_{tar}σ_{tot}.

In the definition of the scattering cross section, we counted any scattering. It is possible to be more specific, and ask how many scattering occur in a particular direction.

ΔOmega; = A/r²

dΩ = sinθ dθ dφ

© 2007 Robert Harr