Probability distribution function of the photon's scattering angle

Question

What is the normalized probability distribution function of the photon's scattering angle, $\theta$, in Compton scattering effect when a photon hits an electron?

Answer 1

The well-known cross section formula for Compton scattering is $$ \frac{d\sigma}{d\Omega} =\frac{\alpha^2(\hbar c)^2}{2s} \left(\frac{\omega'}{\omega}\right)^2 \left(\frac{\omega}{\omega'}+\frac{\omega'}{\omega}-\sin^2\theta\right) $$ where $$ \omega'=\frac{\omega}{1+\frac{\hbar\omega}{mc^2}(1-\cos\theta)} $$

We can integrate $d\sigma$ to obtain a cumulative distribution function. Let $I(\theta)$ be the following integral of $d\sigma$. (The $\sin\theta$ is due to $d\Omega=\sin\theta\,d\theta\,d\phi$) \begin{equation*} I(\theta)= \int \left(\frac{\omega'}{\omega}\right)^2 \left(\frac{\omega}{\omega'}+\frac{\omega'}{\omega}-\sin^2\theta\right) \sin\theta\,d\theta \end{equation*}

The solution is \begin{multline*} I(\theta)=-\frac{\cos\theta}{R^2} +\log\big(1+R(1-\cos\theta)\big)\left(\frac{1}{R}-\frac{2}{R^2}-\frac{2}{R^3}\right) \\ {}-\frac{1}{2R\big(1+R(1-\cos\theta)\big)^2} +\frac{1}{1+R(1-\cos\theta)}\left(-\frac{2}{R^2}-\frac{1}{R^3}\right) \end{multline*} where \begin{equation*} R=\frac{\hbar\omega}{mc^2} \end{equation*}

The cumulative distribution function is \begin{equation*} F(\theta)=\frac{I(\theta)-I(0)}{I(\pi)-I(0)}, \quad 0\le\theta\le\pi \end{equation*}

The probability of observing scattered photons in the interval $\theta_1$ to $\theta_2$ is \begin{equation*} P(\theta_1\le\theta\le\theta_2)=F(\theta_2)-F(\theta_1) \end{equation*}

Differentiate $F(\theta)$ with respect to $\theta$ to obtain a normalized probability density function. $$ f(\theta)=\frac{dF(\theta)}{d\theta} =\frac{1}{I(\pi)-I(0)} \left(\frac{\omega'}{\omega}\right)^2 \left(\frac{\omega}{\omega'}+\frac{\omega'}{\omega}-\sin^2\theta\right) \sin\theta $$

Answer 2

A differential cross section provides a probability distribution if it is multiplied with the flux $F$ of the incoming particles in the time interval $\Delta t$ (here photons):

$$dP = d\sigma \cdot F \Delta t $$

where $d\sigma$ is:

$$d\sigma = \frac{d\sigma_{KN}}{d\theta}(\theta) d\theta$$

So if you want to know the probability of a photon emitted between 0 and $\pi/6$ you just integrate the differential cross section over $\theta$:

$$P[0,\pi/6] =F \Delta t\int_0^{\pi/6} \frac{d\sigma_{KN}}{d\theta}(\theta)\,d\theta $$

and for the range $[\pi/6, \pi/3]$ in the same way. Due to angle dependence of the Klein-Nishina differential cross section you will get (most likely) a different result.

You can find the complete formula for the Klein-Nishina differential cross section $\frac{d\sigma_{KN}}{d\theta}$ for instance on Wikipedia.

EDIT

Integrating the differential cross section (times $F\Delta t$) from 0 to $\pi$ you might observe that the result much smaller than 1, this means that in most of the cases there no scattering reaction at all. On the other hand you could argue that making F or $\Delta t$ large enough that the probability becomes larger than 1. In case of very large flux, multi-photon reactions also have to be considered, i.e. the probability balance has to be rewritten. If $\Delta t$ is considered to be large, note that during all the time F has to be maintained on the (more or less) same value, which rarely occurs in real physical situations. In this respect physical equations have only to be used in their application limits and do not have the "absolute" character as mathematical assertions.

Therefore the above equations should be applied for rather small $F \Delta t$.

Answer 3

As per wiki photon scattering angle $\theta$ depends on electron scattering angle $\varphi$ (or vise versa) due to momentum conservation :

$$ {\displaystyle \cot \varphi =\left(1+{\frac {hf}{m_{e}c^{2}}}\right)\tan(\theta /2)~.} $$

As per $E(\theta,\varphi)$ chart after collision :

you can deduce an approximate outgoing angle $|\theta,\varphi|$ from given energies.

Answer 4

Note, I'm no expert for Compton scattering and it's been a while, but I think you look for the Klein-Nishina formula https://en.m.wikipedia.org/wiki/Klein%E2%80%93Nishina_formula. Unfortunately I cannot help much more, sorry...