# How is the probability density for the deflection angle calculated from the differential cross-section?

In a scattering process, the probability density for the deflection angle of the incident particle is proportional to the differential cross-section of that scattering process. Given the differential cross-section, how is the probability density for the deflection angle of the particle calculated?

For example, the differential cross-section for a Moller scattering process (collision of two electrons) is given in https://en.wikipedia.org/wiki/Møller_scattering (end of the article). What would be the probability that the electron is deflected by an angle between θ and 90 degrees, after colliding another electron?

Well, you know that $$p(x) \propto \frac{d\sigma}{dx}$$ from which it is easy to find $$p(x) = \frac{1}{\sigma} \frac{d\sigma}{dx}$$ which follows from $$\int p(dx) dx = 1$$.

• is dσ/dx the differential cross section here? What are σ and x? – Ali Lavasani May 29 '19 at 8:08
• it;s the differential cross section with respect to some variable $x$. The $x$ should be some information about the phase space of the final state particles. In your case, you would probably want $x \to \theta$, the angle between outgoing electron and beam axis (in com frame) – innisfree May 29 '19 at 9:40

A probability density can be obtained by integrating the differential cross section over $$d\Omega$$.

For Moller scattering we have $$\frac{d\sigma}{d\Omega} =\frac{\alpha^2}{8E^2} \left( \frac{1+\cos^4(\theta/2)}{\sin^4(\theta/2)} +\frac{8}{\sin^2\theta} +\frac{1+\sin^4(\theta/2)}{\cos^4(\theta/2)} \right)$$

Let $$I(\xi)=2\pi\int_\alpha^\xi\frac{d\sigma}{d\Omega}\,\sin\theta\,d\theta, \quad\alpha\le\xi\le\pi-\alpha$$

for some $$\alpha>0$$. The support range is restricted because $$d\sigma$$ is undefined at $$\theta=0$$ and $$\theta=\pi$$.

The cumulative distribution function is $$F(\theta)=\frac{I(\theta)}{I(\pi-\alpha)}, \quad\alpha\le\theta\le\pi-\alpha$$

The probability density is $$f(\theta)=\frac{dF(\theta)}{d\theta} =\frac{2\pi}{I(\pi-\alpha)} \left(\frac{d\sigma}{d\Omega}\right) \sin\theta, \quad\alpha\le\theta\le\pi-\alpha$$

Here is a graph of $$f(\theta)$$ for $$\alpha=\pi/180$$. 