Past exam of the particle physics course of the University of Oxford / 2022 / 4 / b / i

For azimuthally symmetric Rutherford scattering, integrate over the solid angle, $d\Omega =2\pi \sin \theta d\theta$. Substituting $u=\sin (\theta /2)$ gives $\sin \theta d\theta =4udu$, soThe small lower cutoff dominates when ${\theta }_{\max }\gg {\theta }_{\min }$ and ${\theta }_{\min }\ll 1$. Using $\sin ({\theta }_{\min }/2)\simeq {\theta }_{\min }/2$, the leading total cross section isA nonzero cutoff is essential: the unscreened forward cross section otherwise diverges.