Old work

Moller Differential Cross Section

Using the equation from [1]

[math]\frac{d\sigma}{d\Omega '_1}=\frac{ e^4 }{8E^{*2}}\left \{\frac{1+cos^4(\frac{\theta^*}{2})}{sin^4(\frac{\theta^*}{2})}+\frac{1+sin^4(\frac{\theta^*}{2})}{cos^4(\frac{\theta^*}{2})}+\frac{2}{sin^2(\frac{\theta^*}{2})cos^2(\frac{\theta^*}{2})} \right \}[/math]

[math]where\ \alpha=\frac{e^2}{\hbar c}\quad with\quad \hbar = c =1\ and\ \theta^*=\theta^*_1=\theta^*_2[/math]

This can be simplified to the form

[math]\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 }{4E^{*2}}\frac{ (3+cos^2\theta^*)^2}{sin^4\theta^*}[/math]

Plugging in the values expected for 2 scattering electrons:

[math]\alpha ^2=5.3279\times 10^{-5}[/math]

[math]E^*\approx 106.031 MeV[/math]

Using unit analysis on the term outside the parantheses, we find that the differential cross section for an electron at this momentum should be around

[math]\frac{5.3279\times 10^{-5}}{4\times 1.124\times 10^{16}eV^2}=1.18\times 10^{-21} eV^{-2}=\frac{1.18\times 10^{-21}}{1eV^2}\times \frac{1\times 10^{18} }{1\times 10^{18}}=\frac{.0012}{GeV^2}[/math]

Using the conversion of

[math]\frac{1}{1GeV^2}=.3894 mb[/math]

[math]\frac{.0012}{1GeV^2}=\frac{.0012}{1}\frac{1}{1GeV^2}=.0012\times .3894 mb=.467\times 10^{-3}mb[/math]

We find that the differential cross section scale is [math]\frac{d\sigma}{d\Omega}\approx .5\times 10^{-3}mb=.5\mu b[/math]