[math]\underline{\textbf{Navigation}}[/math]
[math]\vartriangleleft [/math]
[math]\triangle [/math]
[math]\vartriangleright [/math]
Differential Cross-Section
[math]\frac{d\sigma}{d\Omega}=\frac{1}{64\pi ^2 s}\frac{\mathbf p_{final}}{\mathbf p_{initial}} |\mathfrak{M} |^2[/math]
Working in the center of mass frame
[math]\mathbf p_{final}=\mathbf p_{initial}[/math]
Determining the scattering amplitude in the center of mass frame
[math]\mathfrak{M}=e^2 \left ( \frac{u-s}{t}+\frac{t-s}{u} \right )[/math]
[math]\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 )[/math]
[math]\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 )[/math]
[math]\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 )[/math]
[math]\mathfrak{M}^2=e^4 \left ( \frac{(t^2+s^2)}{u^2}+\frac{2s^2}{tu}+2-\frac{2us}{t^2}-\frac{2s}{t}-\frac{2ts}{u^2}-\frac{2s}{u}+\frac{(u^2+s^2)}{t^2}\right )[/math]
Using the fine structure constant ([math]with\ c=\hbar=\epsilon_0=1[/math])
[math]\alpha \equiv \frac{e^2}{4\pi}[/math]
[math]\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{2s}\left ( \frac{(t^2+s^2)}{u^2}+\frac{2s^2}{tu}+2-\frac{2us}{t^2}-\frac{2s}{t}-\frac{2ts}{u^2}-\frac{2s}{u}+\frac{(u^2+s^2)}{t^2}\right )[/math]
In the center of mass frame the Mandelstam variables are given by:
[math]s \equiv 4E^{*2}[/math]
[math]t \equiv -2p^{*2}(1-\cos{\theta})=-2p^{*2}\left (1-2\cos^2{\frac{\theta}{2}}+1 \right )=-4p^{*2} \left (1-2\cos^2{\frac{\theta}{2}} \right )=-4p^{*2}\sin^2{\frac{\theta}{2}}[/math]
[math]u \equiv -2p^{*2}(1+\cos{\theta})=-2p^{*2}\left (1+2\cos^2{\frac{\theta}{2}}-1 \right )=-4p^{*2}\cos^2{\frac{\theta}{2}}[/math]
Simplifying using the relationship
[math]\cos{\theta}=-1+\cos{\frac{\theta}{2}}[/math]
Calculating the parts:
[math]\frac{t^2}{u^2}=\frac{-2p^{*2}\left(1-\cos{\theta}\right\right)^2}{-2p^{*2}\left(1+\cos{\theta}\right\right)^2}=\frac{4p^{*2}\left(1-\cos{\theta}\right)^2}{4p^{*2}\left(1+\cos{\theta}\right)^2}=\tan^4{\frac{\theta}{2}}[/math]
[math]\underline{\textbf{Navigation}}[/math]
[math]\vartriangleleft [/math]
[math]\triangle [/math]
[math]\vartriangleright [/math]