Most of what is known about the forces between particles, from the nucleus Rutherford found to the quarks inside protons, was learned by scattering: fire a beam at a target and count what comes out in each direction. The count is summarised by the differential cross section $d\sigma/d\Omega$: the number scattered per unit time into a small solid angle, divided by the incident flux, the number crossing unit area per unit time. It has the dimensions of an area — the area of beam the target turns into that direction — and over all directions it gives the total cross section $\sigma$. For classical particles it follows from the trajectories: a particle arriving with impact parameter $b$ leaves at an angle $\theta(b)$. In quantum mechanics there are no trajectories, and the problem becomes a stationary wave: a plane wave coming in, and an outgoing spherical wave carrying the scattered particles away.

Throw tennis balls blindfold at an unseen statue and note where they fly off: from enough throws you can piece together the statue’s shape. Scattering experiments read the shape of a potential the same way, from where the particles go.

Classical particles bouncing off a hard sphere of radius $a$, arriving from the left at different impact parameters $b$: the nearer the axis, the further back they are thrown, with $b=a\cos(\theta/2)$. Every particle with $b<a$ is scattered, so the classical cross section is the disc .

A beam of flux $j$, particles per unit area per unit time, falls on a target, and $dN/dt$ particles per unit time leave into the solid angle $d\Omega$ about the direction $(\theta,\phi)$:

• $d\sigma/d\Omega$ is an area per unit solid angle; a common unit is the barn, $10^{-28}$ m$^{2}$.
• Classically a particle with impact parameter leaves at the angle , and for a central force .

A particle reflects off the sphere like light off a mirror, so the angle of incidence fixes the scattering angle, and the impact parameter fixes the angle of incidence. The particles between $b$ and $b+db$, all round the ring, land between $\theta$ and $\theta+d\theta$; dividing the ring’s area by the solid angle gives a constant, whose integral is the sphere’s shadow.

Incidence at the angle $\alpha$ to the normal, and mirror reflection.

Eliminate $\alpha=(\pi-\theta)/2$.

The ring of impact parameters and the ring of directions it lands in.

$\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$.

The same in every direction, over the full $4\pi$: exactly the area the sphere blocks.

The cross section is the area of beam a target removes: for a classical hard sphere, every particle aimed within its radius.

The rate of particles scattered into a small solid angle, divided by the incident flux and by the solid angle.

A rate divided by a flux, particles per second over particles per second per square metre, is an area.

For a hard sphere, $b=$ a $\cos(\theta/2)$: a head-on particle bounces straight back, a grazing one passes almost undeflected.

Great work! You now know how to:

• define the differential and total cross sections
• derive the classical cross section of a hard sphere
• set up the stationary quantum scattering problem