Differential Cross-Section

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Differential Cross-Section

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

Working in the center of mass frame

$\mathbf p_{final}=\mathbf p_{initial}$

Determining the scattering amplitude in the center of mass frame

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

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

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

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

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

Using the fine structure constant ($with\ c=\hbar=\epsilon_0=1$)

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

$\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4s}\left ( \frac{(t^2+s^2)}{u^2}+\frac{2s^2}{tu}+2-\frac{2us}{t^2}-\frac{2s}{t}-\frac{2ts}{u^2}-\frac{2s}{u}+\frac{(u^2+s^2)}{t^2}\right )$

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

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

$t \equiv -2p^{*2}(1-\cos{\theta})$

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

Calculating the parts to have common denominators:

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

$\left(2\right)\qquad \qquad \frac{2s^2}{tu}=\frac{32E^{*4}}{4p^{*4}\left(1+\cos{\theta}\right)\left(1-\cos{\theta}\right)}=\frac{8E^{*4}}{p^{*4}\sin^2{\theta}}=\frac{8E^{*4}\sin^2{\theta}}{p^{*4}\sin^4}=\frac{8E^{*4}\left(1-\cos^2{\theta}\right)}{p^{*4}\sin^4{\theta}}$

$\left(3\right)\qquad \qquad \frac{t^2}{u^2}=\frac{4p^{*2}\left(1-\cos{\theta}\right)^2}{4p^{*2}\left(1+\cos{\theta}\right)^2}=\tan^4{\frac{\theta}{2}}=\frac{p^{*4}\left(1-\cos{\theta}\right)^4}{p^{*4}\sin^4{\theta}}=\frac{p^{*4}\left(\cos^4{\theta}-4\cos^3{\theta}+6\cos^2{\theta}-4\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}$

$\left(4\right)\qquad \qquad \frac{u^2}{t^2}=\frac{4p^{*2}\left(1+\cos{\theta}\right)^2}{4p^{*2}\left(1-\cos{\theta}\right)^2}=\cot^4{\frac{\theta}{2}}=\frac{p^{*4}\left(1+\cos{\theta}\right)^4}{p^{*4}\sin^4{\theta}}=\frac{p^{*4}\left(\cos^4{\theta}+4\cos^3{\theta}+6\cos^2{\theta}+4\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}$

$\left(5\right)\qquad \qquad \frac{s^2}{u^2}=\frac{16E^{*4}}{4p^{*4}\left(1+\cos{\theta}\right)^2}=\frac{E^{*4}\sec^4{\frac{\theta}{2}}}{p^{*4}}=\frac{4E^{*4}}{p^{*4}\left(1+\cos{\theta}\right)^2}=\frac{4E^{*4}\left(1-\cos{\theta}\right)^2}{p^{*4}\left(1+\cos{\theta}\right)^2\left(1-\cos{\theta}\right)^2}=\frac{4E^{*4}\left(\cos^2{\theta}-2\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}$

$\left(6\right)\qquad \qquad \frac{s^2}{t^2}=\frac{16E^{*4}}{4p^{*4}\left(1-\cos{\theta}\right)^2}=\frac{E^{*4}\csc^4{\frac{\theta}{2}}}{p^{*4}}=\frac{4E^{*4}}{p^{*4}\left(1-\cos{\theta}\right)^2}=\frac{4E^{*4}\left(1+\cos{\theta}\right)^2}{p^{*4}\left(1-\cos{\theta}\right)^2\left(1+\cos{\theta}\right)^2}=\frac{4E^{*4}\left(\cos^2{\theta}+2\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}$

$\left(7\right)\qquad \qquad \frac{-2s}{t}=\frac{8E^{*2}}{2p^{*2}\left(1-\cos{\theta}\right)}=\frac{4E^{*2}}{p^{*2}\left(1-\cos{\theta}\right)}=\frac{4E^{*2}\left(1+\cos{\theta}\right)}{p^{*2}\left(1-\cos{\theta}\right)\left(1+\cos{\theta}\right)}=\frac{4E^{*2}\left(1+\cos{\theta}\right)}{p^{*2}\sin^2{\theta}}=\frac{4E^{*2}p^{*2}sin^2{\theta}\left(1+\cos{\theta}\right)}{p^{*4}sin^4{\theta}}=\frac{4E^{*2}p^{*2}\left(1-cos^2{\theta}\right)\left(1+\cos{\theta}\right)}{p^{*4}sin^4{\theta}}=\frac{4E^{*2}p^{*2}\left(1+\cos{\theta}-\cos^2{\theta}-\cos^3{\theta}\right)}{p^{*4}sin^4{\theta}}$

$\left(8\right)\qquad \qquad \frac{-2s}{u}=\frac{8E^{*2}}{2p^{*2}\left(1+\cos{\theta}\right)}=\frac{4E^{*2}}{p^{*2}\left(1+\cos{\theta}\right)}=\frac{4E^{*2}\left(1-\cos{\theta}\right)}{p^{*2}\left(1+\cos{\theta}\right)\left(1-\cos{\theta}\right)}=\frac{4E^{*2}\left(1-\cos{\theta}\right)}{p^{*2}sin^2{\theta}}=\frac{4E^{*2}p^{*2}sin^2{\theta}\left(1-\cos{\theta}\right)}{p^{*4}sin^4{\theta}}=\frac{4E^{*2}p^{*2}\left(1-cos^2{\theta}\right)\left(1-\cos{\theta}\right)}{p^{*4}sin^4{\theta}}=\frac{4E^{*2}p^{*2}\left(1-\cos{\theta}-\cos^2{\theta}+\cos^3{\theta}\right)}{p^{*4}sin^4{\theta}}$

$\left(9\right)\qquad \qquad \frac{-2ts}{u^2}=\frac{4p^{*2}\left(1-\cos{\theta}\right)4E^{*2}}{4p^{*2}\left(1+\cos{\theta}\right)^2}=\frac{4E^{*2}\left(1-\cos{\theta}\right)\sec^4{\frac{\theta}{2}}}{p^{*2}}=\frac{4E^{*2}p^{*2}\left(1-\cos{\theta}\right)}{p^{*4}\left(1+\cos{\theta}\right)^2}=\frac{4E^{*2}p^{*2}\left(1-\cos{\theta}\right)\left(1-\cos{\theta}\right)^2}{p^{*4}\left(1+\cos{\theta}\right)^2\left(1-\cos{\theta}\right)^2}=\frac{4E^{*2}p^{*2}\left(-\cos^3{\theta}+3\cos^2{\theta}-3\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}$

$\left(10\right)\qquad \qquad \frac{-2us}{t^2}=\frac{4p^{*2}\left(1+\cos{\theta}\right)4E^{*2}}{4p^{*2}\left(1-\cos{\theta}\right)^2}=\frac{4E^{*2}\left(1+\cos{\theta}\right)\csc^4{\frac{\theta}{2}}}{p^{*2}}=\frac{4E^{*2}p^{*2}\left(1+\cos{\theta}\right)}{p^{*4}\left(1-\cos{\theta}\right)^2}=\frac{4E^{*2}p^{*2}\left(1+\cos{\theta}\right)\left(1+\cos{\theta}\right)^2}{p^{*4}\left(1-\cos{\theta}\right)^2\left(1+\cos{\theta}\right)^2}=\frac{4E^{*2}p^{*2}\left(\cos^3{\theta}+3\cos^2{\theta}+3\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}$

Combing like terms further,

$\left(1,3/4\right)\rightarrow \qquad \qquad 4p^{*4}\cos^4{\theta}+8p^{*4}\cos^2{\theta}+4p^{*4}$

$\left(2,6/5\right) \rightarrow \qquad \qquad 16E^{*4}$

$\left(8/7,10/9\right) \rightarrow \qquad \qquad 16E^{*2}p^{*2}+16E^{*2}p^{*2}\cos^2{\theta}$

Expressing this as the differential cross-section

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

$\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{sp^{*4}\sin^4{\theta}}\left( p^{*4}\cos^4{\theta}+2p^{*4}\cos^2{\theta}+p^{*4}+4E^{*2}p^{*2}+4E^{*2}p^{*2}\cos^2{\theta}+4E^{*4}\right)$

In the Ultra-relativistic limit as $E \approx p$

$\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4E^{*2}\sin^4{\theta}}\left( \cos^4{\theta}+6\cos^2{\theta}+9\right)=\frac{\alpha ^2}{4E^{*2}\sin^4{\theta}}\left(3+\cos{\theta}\right^2$

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