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:

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

Using the relationship

$\cos{\theta}=-1+\cos{\frac{\theta}{2}}$

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

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

$\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{8E^{*2}} \frac{(16E^{*4}\sin^4{\frac{\theta}{2}}+16E^{*4}}{16E^{*4}\cos^4{\frac{\theta}{2}}}-\frac{324E^{*4}}{4E^{*2}\sin^2{\frac{\theta}{2}}4E^{*2}\cos^2{\frac{\theta}{2}}}+\frac{16E^{*4}\cos^4{\frac{\theta}{2}}+16E^{*4}}{16E^{*4}\sin^4{\frac{\theta}{2}}}\right )$