# Integration boundaries in formula of central-force at classical scattering development

In "Classical Mechanics" book by Goldstein, Poole & Safko they develop the formulas of scattering to due to central-force problem.

In page 88, after receiving the expression for the differential cross section

$$\sigma(\theta) =\frac{s}{sin \theta} |\frac{ds}{d\theta}|$$

They said that the relation between $s$(impact parameter), $\theta$ (deflection angle) can be achieved from the central-force problem formulas.

General central-force formula: (3.36)

Questions:

1. How they have chosen such integration boundaries?

2. How did they use the formula: how the angles and radius match, for example: why does $r=\infty$ corresponds to $\theta=\pi$?

These are polar coordinates, so the scattered particle moves from $r=\infty, \theta=\pi$ (far left) via the point of closest approach ($r=r_m, \theta=\pi-\psi$) to $r=\infty, \theta$ (roughly 45 degrees in your image up to the right). You start from the differential equation (up to constant factors):

$$d\theta=\pm\frac{dr}{r^2\sqrt{E-U(r)}}$$

The sign takes care of the fact that $\theta$ is always increasing while $r$ first decreases and the increases.

If you integrate this from the starting point of the travjectory to some point $(r,\theta)$ along the trajectory, you get:

$$\int_{\pi}^\theta d\theta=\int_\infty^r\frac{dr}{r^2\sqrt{E-U(r)}}$$

which you can rewrite as (exchanging the boundaries of integration gives a minus sign):

$$\theta = \pi - \int^\infty_r\frac{dr}{r^2\sqrt{E-U(r)}}$$

The distance of closest approach, $r=r_m$ is reached at $\theta=\pi-\psi$, which leads to:

$$\psi = \int^\infty_{r_m}\frac{dr}{r^2\sqrt{E-U(r)}}$$