The force vanishes at large distances; energy and angular momentum are conserved.
Incident beam: intensity (flux) \(I\) = particles crossing unit area normal to the beam per unit time.
Near the centre the orbit bends (attraction or repulsion) and then again becomes a straight line.
Differential cross section: \(D(\Omega)\,d\Omega=\dfrac{N(\Omega)}{I}\), \(N\) = particles scattered into \(d\Omega\) per unit time; dimensions of area.
Impact parameter \(b\): perpendicular distance between the centre and the incident line. \(E=\frac12mv_0^2\), \(l=mv_0b=b\sqrt{2mE}\).
\(E\) and \(b\) fix the scattering angle \(\Theta\); particles through the ring \(2\pi b\,db\) go into \(2\pi\sin\Theta\,d\Theta\): \[D(\Theta)=\frac{b}{\sin\Theta}\Big|\frac{db}{d\Theta}\Big|,\qquad \sigma=\int D\,d\Omega.\]
Orbit symmetric about periapsis: \(\Theta=\pi-2\psi\) and \[\Theta(b)=\pi-2\int_0^{u_m}\frac{b\,du}{\sqrt{1-\dfrac{V(u)}{E}-b^2u^2}},\qquad u=\frac1r.\]
Rutherford scattering
Coulomb force \(F=\dfrac{zz'e^2}{r^2}\): hyperbolic orbit, eccentricity \(\varepsilon=\sqrt{1+\big(\frac{2Eb}{zz'e^2}\big)^2}\), and \(\sin\frac\Theta2=\frac1\varepsilon\): \[b=\frac{zz'e^2}{2E}\cot\frac\Theta2,\qquad D(\Theta)=\Big(\frac{zz'e^2}{4E}\Big)^2\frac{1}{\sin^4(\Theta/2)}.\]
Total \(\sigma\) diverges (long-range force).
Example 15.1 Hard sphere
Classical scattering off a hard sphere of radius \(R\).
Solution
Reflection law: \(2\alpha+\Theta=\pi\), \(b=R\sin\alpha=R\cos\frac\Theta2\).
\(D=\dfrac{R\cos(\Theta/2)}{\sin\Theta}\cdot\dfrac R2\sin\dfrac\Theta2=\dfrac{R^2}{4}\) (isotropic).
\(\sigma=\dfrac{R^2}{4}\cdot4\pi=\pi R^2\) — the geometric cross section.
Example 15.2 Distance of closest approach
A particle with speed \(v_0\) and impact parameter \(b\) approaches a repulsive centre \(F=k/r^2\). Find the closest distance \(R\).
Solution
Example 15.3 Pion capture cross section
A pion of kinetic energy \(T\) hits a nucleus (charge \(Ze\), radius \(R\)) if it reaches \(r\le R\). Show \(\sigma=\pi R^2\big(1-\frac VT\big)\) for \(\pi^+\) and \(\pi R^2\big(1+\frac VT\big)\) for \(\pi^-\), \(V=\frac{Ze^2}{R}\).
Solution
Grazing orbit: \(\sqrt{2mT}\,b=mR\,v_R\) and \(T=\pm V+\frac12mv_R^2\).
\(b^2=R^2\,\dfrac{T\mp V}{T}\) \(\Rightarrow\sigma=\pi b^2=\pi R^2\Big(1\mp\dfrac VT\Big)\) (upper sign \(\pi^+\)).
Example 15.4 Counting scattered particles
\(\dfrac{d\sigma}{d\Omega}=a^2+b^2\cos^2\theta\), incident flux \(N\). Particles scattered per unit time?
Solution
