Differential Cross-Section

[math]\displaystyle{ \underline{\textbf{Navigation}} }[/math]

[math]\displaystyle{ \vartriangleleft }[/math] [math]\displaystyle{ \triangle }[/math] [math]\displaystyle{ \vartriangleright }[/math]

Differential Cross-Section

[math]\displaystyle{ \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]\displaystyle{ \mathbf p_{final}=\mathbf p_{initial} }[/math]

Determining the scattering amplitude in the center of mass frame

[math]\displaystyle{ \mathfrak{M}=e^2 \left ( \frac{u-s}{t}+\frac{t-s}{u} \right ) }[/math]

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ with\ c=\hbar=\epsilon_0=1 }[/math])

[math]\displaystyle{ \alpha \equiv \frac{e^2}{4\pi} }[/math]

[math]\displaystyle{ \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]\displaystyle{ s \equiv 4E^{*2} }[/math]

[math]\displaystyle{ t \equiv -2p^{*2}(1-\cos{\theta}) }[/math]

[math]\displaystyle{ u \equiv -2p^{*2}(1+\cos{\theta}) }[/math]

Calculating the parts:

[math]\displaystyle{ \frac{t^2}{u^2}=\frac{4p^{*2}\left(1-\cos{\theta}\right)^2}{4p^{*2}\left(1+\cos{\theta}\right)^2}=\tan^4{\frac{\theta}{2}}=\frac{\left(1-\cos{\theta}\right)^4}{\sin^4{\theta}} }[/math]

[math]\displaystyle{ \frac{s^2}{u^2}=\frac{16E^{*4}}{4p^{*4}\left(1+\cos{\theta}\right)^2}=\frac{E^{*4}\sec^4{\frac{\theta}{2}}}{p^{*4}} }[/math]

[math]\displaystyle{ \frac{u^2}{t^2}=\frac{4p^{*2}\left(1+\cos{\theta}\right)^2}{4p^{*2}\left(1-\cos{\theta}\right)^2}=\cot^4{\frac{\theta}{2}}=\frac{\left(1+\cos{\theta}\right)^4}{\sin^4{\theta}} }[/math]

[math]\displaystyle{ \frac{s^2}{t^2}=\frac{16E^{*4}}{4p^{*4}\left(1-\cos{\theta}\right)^2}=\frac{E^{*4}\csc^4{\frac{\theta}{2}}}{p^{*4}} }[/math]

[math]\displaystyle{ \frac{2s^2}{tu}=\frac{32E^{*4}}{4p^{*4}\left(1+\cos{\theta}\right)\left(1-\cos{\theta}\right)}=\frac{8E^{*4}}{p^{*4}\sin^2{\theta}} }[/math]

[math]\displaystyle{ \frac{-2s}{t}=\frac{8E^{*2}}{2p^{*2}\left(1-\cos{\theta}\right)}=\frac{4E^{*2}}{p^{*2}\left(1-\cos{\theta}\right)}=\frac{4E^{*2}\left(1+\cos{\theta}\right)}{p^{*2}\left(1-\cos{\theta}\right)\left(1+\cos{\theta}\right)}=\frac{4E^{*2}\left(1+\cos{\theta}\right)}{p^{*2}\sin^2{\theta}} }[/math]

[math]\displaystyle{ \frac{-2s}{u}=\frac{8E^{*2}}{2p^{*2}\left(1+\cos{\theta}\right)}=\frac{4E^{*2}}{p^{*2}\left(1+\cos{\theta}\right)}=\frac{4E^{*2}\left(1-\cos{\theta}\right)}{p^{*2}\left(1+\cos{\theta}\right)\left(1-\cos{\theta}\right)}=\frac{4E^{*2}\left(1-\cos{\theta}\right)}{p^{*2}sin^2{\theta}} }[/math]

[math]\displaystyle{ \frac{-2ts}{u^2}=\frac{4p^{*2}\left(1-\cos{\theta}\right)4E^{*2}}{4p^{*2}\left(1+\cos{\theta}\right)^2}=\frac{E^{*2}\left(1-\cos{\theta}\right)\sec^4{\frac{\theta}{2}}}{p^{*2}} }[/math]

[math]\displaystyle{ \frac{-2us}{t^2}=\frac{4p^{*2}\left(1+\cos{\theta}\right)4E^{*2}}{4p^{*2}\left(1-\cos{\theta}\right)^2}=\frac{E^{*2}\left(1+\cos{\theta}\right)\csc^4{\frac{\theta}{2}}}{p^{*2}} }[/math]

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[math]\displaystyle{ \underline{\textbf{Navigation}} }[/math]

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