Differential Cross-Section

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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}+\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}+\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]

Where the relationship

[math]\cos{\theta}=-1+\cos{\frac{\theta}{2}}[/math]

[math]\therefore E^2\equiv m^2+p^2 [/math]

[math]\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{8E^{*2}}\left( \frac{16p^{*4}\sin^4{\frac{\theta}{2}}+16E^{*4}}{16p^{*4}\cos^4{\frac{\theta}{2}}}-\frac{32E^{*4}}{4p^{*2}\sin^2{\frac{\theta}{2}}4p^{*2}\cos^2{\frac{\theta}{2}}}+\frac{16p^{*4}\cos^4{\frac{\theta}{2}}+16E^{*4}}{16p^{*4}\sin^4{\frac{\theta}{2}}}\right )[/math]

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