Differential cross-section divergence

[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

[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.