What does it mean that the Rutherford's cross section is infinite?

Question

I'm studying elastic scattering and I read that the Rutherford's differential cross section is defined as:

$$\left( \frac{d \sigma}{d \Omega} \right)_R = \frac{Z^2}{4} r_o^2 z^2 \frac{(m_ec / \beta p)^2}{\sin^4(\theta/2)}$$

And the total cross section is defined as:

$$ \sigma_T = \int_0^{2 \pi} d \phi \int_0^\pi \frac{d \sigma}{d \Omega} \sin \theta d \theta$$

For the Rutherford's cross section the total cross section equals infinity ($(\sigma_T)_R = \infty$).

What does this mean physically? The cross section is the probability of interaction per unit of surface. Does this mean that we are considering as an interaction even when the particle continues its path as if nothing happens (i.e. $\theta=0$)?

Answer 1

That the total cross section is infinite just means that every charged particle that passes by the (bare) nucleus is scattered to some extent. This is a consequence of the Coulomb potential being long range. Classically, it is sufficient for the potential to be nonzero for all radii in order to have an infinite total cross section. Quantum mechanically, if the potential approaches zero rapidly enough, the total cross section will be finite. However, the Coulomb potential goes to zero slowly at infinity, and the resulting Rutherford cross section (which is correct both classically and quantum mechanically) has an infinite total cross section.