[NOTES/CM-06008] Rutherford Scattering

For a particle moving in coulomb potential
\begin{equation}
V(r) = k/r , \qquad E > 0,
\end{equation}
the scattering angle is given by
\begin{equation}
\cos\Big(\frac{\theta}{2}\Big)= \frac{-1}{\epsilon},
\end{equation}
where for the Coulomb potential \( \epsilon = \sqrt{\Big(1+ \frac{2 E L^{2}}{\mu k^{2}}}\Big)\) is the eccentricity of the orbit. Therefore
\begin{eqnarray} \cot^{2}\Big(\frac{\theta}{2}\Big) = \epsilon^{2} - 1 = \frac{2 E L^{2}}{\mu k^{2}}\label{EQ03}
\end{eqnarray} where $L$ is the angular momentum and $E$ = Energy in the centre of mass frame. If \(v\) is the velocity of the incident beam, we have \begin{equation} L= mv\rho = \sqrt{2mE}\rho. \end{equation}
Knowing the Scattering angle as a function of \(L\) and \(E\) from \eqref{EQ03}, we may express it in terms of the impact parameter $\rho$ and $E$ \begin{equation} \cot^{2}\Big(\frac{\theta}{2}\Big) = \Big(\frac{2E}{\mu k^{2}}\Big) ( 2 \mu E\rho^2) \Longrightarrow \cot\Big(\frac{\theta}{2}) = \frac{2\rho E}{k}. \end{equation}
From above equation we have
\begin{equation} \rho = \Big(\frac{k}{2E}\Big) \cot\Big(\frac{\theta}{2}\Big). \end{equation}
Therefore,
\begin{equation}
\frac{d \rho}{d \theta} = \frac{1}{2} \Big( \frac{k}{2 E} \Big)
\csc^{2} \Big(\frac{\theta}{2}\Big)
\end{equation}
\begin{eqnarray}
\sigma(\theta)
&=& \frac{\rho}{\sin\theta} \Big|\frac{d \rho}{d \theta}\Big|
= \left( \frac{k \cot(\frac{\theta}{2})}{2E \sin\theta} \right) \times \left(
\frac{k}{4E \sin^{2}(\frac{\theta}{2})} \right) \\
&=& \frac{1}{4} \left( \frac{k}{2E} \right)^{2} \frac{1}{\sin^{4}
\left(\frac{\theta}{2} \right)}
\end{eqnarray}
If we complete total cross section we would get
$$\sigma_{t} = \infty.$$
The total cross section turning out to be infinity is a reflection of the fact
that the range of Coulomb potential is infinite.