Just for reference, this is the cross section for Compton scattering.

$$ \frac{d\sigma}{d\Omega} \propto\left(\frac{E'}{E}\right)^2\left(\frac{E'}{E}+\frac{E}{E'}-\sin^2\theta\right) $$

Currently I am integrating $d\sigma$ numerically to obtain scattering probabilities per angular bin. However, numerical integration would not be necessary if I had a cumulative distribution function.

Is there a closed form cumulative distribution function for Compton scattering?


1 Answer 1


Noting that $d\Omega=\sin\theta\,d\theta\,d\phi$, we have $$ d\sigma\propto \left(\frac{E'}{E}\right)^2 \left(\frac{E'}{E}+\frac{E}{E'}-\sin^2\theta\right)\sin\theta \,d\theta\,d\phi $$

Let $g(\theta)$ be the following antiderivative. $$ g'(\theta)= \left(\frac{E'}{E}\right)^2 \left(\frac{E'}{E}+\frac{E}{E'}-\sin^2\theta\right)\sin\theta $$

Then \begin{multline*} g(\theta)=-\frac{\cos\theta}{R^2} +\log\bigl(1+R(1-\cos\theta)\bigr)\left(\frac{1}{R}-\frac{2}{R^2}-\frac{2}{R^3}\right) \\ {}-\frac{1}{2R\bigl(1+R(1-\cos\theta)\bigr)^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{E}{mc^2} \end{equation*}

Then the cumulative distribution function $F(\theta)$ is $$ F(\theta)=\frac{g(\theta)-g(0)}{g(\pi)-g(0)},\quad0\le\theta\le\pi $$


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