Differential cross-section divergence

1. Jul 24, 2008

lonetomato

Hello, I was wondering if anyone could explain the troubling divergence here of the differential cross-section for rutherford scattering for $$\theta = 0$$. I know it must have something to do with the fact that the em force extends to infinity, which makes sense to me for the $$total$$ cross section.. Why would the differential be 0 for $$\theta = 0$$ and not over all angles?

$$\frac{d\sigma}{d\Omega} \theta = -\frac{b}{\sin \theta}\frac{db}{d\theta}$$

Thanks..
Anna

2. Jul 24, 2008

humanino

Hi, and welcome on PF,

the formula
$$\frac{d\sigma}{d\Omega} = \left|\frac{b}{\sin \theta}\left(\frac{db}{d\theta}\right)\right|$$
is not the Rutherford formula. It is hard sphere scattering, and is merely geometrical.

Rutherford formula is
$$\frac{d\sigma}{d\Omega} = \left(\frac{q_1q_2}{4E\sin^2 (\theta/2)}\right)^2$$

I have no clue whether you are refering to the optical theorem which relates the total cross section to the forward amplitude.