# Cross section of coulomb sacttering vs Rutherford

by oosgood
Tags: coulomb, cross, rutherford, sacttering
 P: 3,014 Let us suppose the impact factor is $b$, and the initial velocity is $v_\infty$. If the potential energy is spherically symmetric $V = V(r)$ and vanishes at infinity, then the law of conservation of angular momentum and energy give: $$m \, v_\infty \, b = m \, r^2 \, \dot{\phi}$$ $$E = \frac{m \, v^2_\infty}{2} = \frac{m}{2} \, \left(\dot{r}^2 + r^2 \, \dot{\phi}^2 \right) + V(r)$$ One can then eliminate $\dot{\phi}$ from the first equation and use it to go over from time dependence to angular dependance: $$\dot{r} = \frac{d r}{d \phi} \, \dot{\phi} = \frac{v_\infty \, b}{r^2} \, \frac{d r}{d \phi} = -v_\infty \, \frac{d}{d \phi} \left( \frac{b}{r} \right), \ z \equiv \frac{b}{r}$$ to get a 1st order differential equation for the trajectory of the particle: $$E = \frac{m \, v^2_\infty}{2} = \frac{m \, v^2_\infty}{2} \, \left[ \left( \frac{d z}{d \phi} \right)^2 + z^2 \right] + V \left( \frac{b}{z} \right)$$ $$\left( \frac{d z}{d \phi} \right)^2 = 1 - \frac{V(b/z)}{E} - z^2$$ The value for which the r.h.s. becomes zero determines the distance of closest approach $r_0(b, E) = b/z_0(b, E)$. If we measure the angle $\phi$ from that point, then r increases and z decreases with increasing angle. This means we need to take the negative root of the above equation and the variables separate: $$\phi = \int_{b/r}^{z_0(b, E)}{\frac{d z}{\sqrt{1 - \frac{V(b/z)}{E} - z^2}}}$$ As we take $r \rightarrow \infty$, the angle approaches the "scattering angle". Due to symmetry, we have: $$\theta(b, E) = \pi - 2 \, \int_{0}^{z_0(b, E)}{\frac{d z}{\sqrt{1 - \frac{V(b/z)}{E} - z^2}}}$$ where, again, $z_0$ is determined from: $$1 - \frac{V(b/z_0)}{E} - z^2_0 = 0$$ Once you know $\theta = \theta(b; E) \Rightarrow b = b(\theta; E)$, you can find the differential scattering cross section from: $$\sigma_d(\theta; E) = \frac{d\sigma}{d \Omega} = \frac{b(\theta; E)}{\sin \theta} \, \frac{d b}{d \theta}$$ More interesting for you is the inverse problem: Given $\sigma_d(\theta; E)$, can you find $V = V(r)$? This amounts to solving some intergral equations, and is rather involved, so I cannot present it right now. Once you have that, you can find the charge distribution from Poisson's equation: $$\nabla^2 V = \frac{1}{r^2} \, \frac{d}{d r} \left( r^2 \, \frac{d V}{d r} \right) = \frac{q\, \rho(r)}{\epsilon_0}$$ where q is the charge of the projectile.