Differential Cross-Section

$\frac{d\sigma}{d\Omega}=\frac{1}{64\pi ^2 s}\frac{\mathbf p_{final}}{\mathbf p_{initial}} |\mathfrak{M} |^2$

$\mathfrak{M}=e^2 \left ( \frac{u-s}{t}+\frac{t-s}{u} \right )$

$\mathfrak{M}^2=e^4 \left ( \frac{u-s}{t}+\frac{t-s}{u} \right )\left ( \frac{u-s}{t}+\frac{t-s}{u} \right )$

$\mathfrak{M}^2=e^4 \left ( \frac{(u-s)^2}{t^2}+\frac{(t-s)^2}{u^2} +2\frac{(u-s)}{t}\frac{(t-s)}{u}\right )$

$\mathfrak{M}^2=e^4 \left ( \frac{(u^2-2us+s^2)}{t^2}+\frac{(t^2-2ts+s^2)}{u^2} +2\frac{(ut-st+s^2-us)}{tu}\right )$

$\mathfrak{M}^2=e^4 \left ( \frac{(t^2+s^2)}{u^2}-\frac{2s^2}{tu}+\frac{(u^2+s^2)}{t^2}\right )$

Using the fine structure constant

$\alpha \equiv \frac{e^2}{4\pi}$

$\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{2s}\left ( \frac{(t^2+s^2)}{u^2}-\frac{2s^2}{tu}+\frac{(u^2+s^2)}{t^2}\right )$

In the center of mass frame the Mandelstam variables are given by:

$s \equiv 4E^{*2}$

$t \equiv -2E^{*2}(1-\cos{\theta})=-2E^{*2}\left (1-2\cos^2{\frac{\theta}{2}}+1 \right )=-4E^{*2} \left (1-2\cos^2{\frac{\theta}{2}} \right )=-4E^{*2}\sin^2{\frac{\theta}{2}}$

$u \equiv -2E^{*2}(1+\cos{\theta})$

$\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{8E^{*2}}\left ( \frac{(4E^{*4}(1-\cos{\theta})^2+16E^{*4})}{4E^{*4}(1+\cos{\theta})^2}-\frac{32E^{*2}}{4E^{*4}(1+\cos{\theta})(1-\cos{\theta})}+\frac{(4E^{*4}(1+\cos{\theta})^2+16E^{*4})}{4E^{*4}(1-\cos{\theta})^2}\right )$

$\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{8E^{*2}4E^{*4}}\left ( \frac{(4E^{*4}(1-\cos{\theta})^2+16E^{*4})}{(1+\cos{\theta})^2}-\frac{32E^{*2}}{(1+\cos{\theta})(1-\cos{\theta})}+\frac{(4E^{*4}(1+\cos{\theta})^2+16E^{*4})}{(1-\cos{\theta})^2}\right )$