What is Cross-section? — Physics

Definition 15.1 University Physics — Year 3 · Chapter 15 — Scattering Theory

Send a uniform beam of flux $\Phi$ (particles per unit area per second) onto one target particle. The rate of projectiles scattered into the solid angle $\dd\Omega$ around the direction $(\theta, \varphi)$ defines the differential cross-section,

\dd\dot N = \Phi\,\frac{\dd\sigma}{\dd\Omega}\,\dd\Omega ,

and its integral $\sigma = \int(\dd\sigma/\dd\Omega)\,\dd\Omega$ is the total cross-section: the area the target effectively presents. For a thin slab with $n$ targets per unit volume, a beam attenuates as $\eu^{-n\sigma x}$ : cross-sections are measured by counting what emerges. Units: nuclear physics uses the barn, $1\,\mathrm{b} = 10^{-28}\,\mathrm{m}^{2}$ — about a uranium nucleus’s geometric size, “as big as a barn” to a neutron.

The scattering experiment: a plane wave in, a spherical wave out, a detector at angle counting. The angular pattern of the counts is the physics of the target. — The scattering experiment: a plane wave in, a spherical wave out, a detector at angle $\theta$ counting. The angular pattern of the counts is the physics of the target.

Examples

Example 15.4 (From Yukawa to Rutherford)

For the screened Coulomb potential $V = \dfrac{Q_1Q_2}{4\pi \varepsilon_0 r}\,\eu^{-r/a}$ the Born integral is elementary (Exercise 15.5) and, as the screening radius $a \to \infty$ ,

\frac{\dd\sigma}{\dd\Omega} = \Big(\frac{Q_1Q_2}{16\pi\varepsilon_0 E}\Big)^2 \frac{1}{\sin^4(\theta/2)} :

the Rutherford cross-section. By a coincidence unique to $1/r$ , the classical calculation, the Born approximation and the exact quantum answer all agree — which is why Rutherford, computing classically in 1911, got the right formula and, from its verification flash by flash, the nucleus (Problem 15.1).

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