[NOTES/ME-06007]-no title

Recall that the unbounded orbit is symmetric about the line \(\phi_0\), where \(\phi_0\) corresponds to the point on the orbit for which the distance of the particle from the scattering centre is minimum.\\ \renewcommand{\epsdir}{/home/home1/WorkSpace/Live/me/eps/} \FigBelow{25,10}{90}{65}{AlphaScatt}{}\\ The scattering angle is related to the angle \(\Psi\) by \begin{equation} \theta = \pi -2\Psi \end{equation} The angle \(\Psi\) is seen to be the change in angle \(\phi\) as \(r\) increases from \(r_\text{min}\) to infinity. This information can be used to get the angle \(\Psi\) as an integral involving the potential \(V(r)\). Show that \begin{equation} \Psi = \int_{r_\text{min}}^ \infty \frac{(b/r^2)\, dr}{\sqrt{[1-(b^2/r^2)-(2V(r)/\mu v_\infty^2)]}}, \end{equation} where \(v_\infty\) is the velocity of the particle at infinity.