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 referring to the optical theorem which relates the total cross section to the forward amplitude.

 

[sir-pinski]

Dear Anna,

The divergence in the differential cross-section for Rutherford scattering at \theta = 0 can be explained by the behavior of the electromagnetic force at very small angles. As you mentioned, the force extends to infinity, which means that at \theta = 0, the particles are extremely close to each other and experience a very strong repulsive force. This results in a very high scattering cross-section, which is represented by the singularity in the differential cross-section formula.

However, as the angle increases, the particles are further apart and the force decreases, leading to a decrease in the cross-section. This is reflected in the negative sign in the formula you provided, which represents the decreasing trend of the cross-section as the angle increases.

It is important to note that this divergence at \theta = 0 is a mathematical artifact and does not represent a physical phenomenon. In reality, the particles cannot be brought infinitely close together, and the singularity would be smoothed out by the finite size of the particles.

I hope this helps to clarify the issue. Let me know if you have any further questions.