Old work

Moller Differential Cross Section

Using the equation from [1]

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

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

[math]\displaystyle{ \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]\displaystyle{ \frac{d\sigma}{d\Omega}\approx .5\times 10^{-3}mb=.5\mu b }[/math]

Different p₂¹ Values

Using the conversion of

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

[math]\displaystyle{ \sigma=\int d\sigma=\int \frac{d\sigma}{d\Omega_2'}d\Omega }[/math]

The range of the detector is considered to be [math]\displaystyle{ .10 \le \theta \le .87 }[/math],[math]\displaystyle{ -\pi \le \phi \le \pi }[/math]

[math]\displaystyle{ \sigma=\int_{ .611}^{2.531} \int_{-\pi}^{\pi} \frac{d\sigma}{d\Omega_2'}sin\theta \,d\theta \, d\phi }[/math]

[math]\displaystyle{ \sigma=2\pi \int_{.611}^{2.531} \frac{d\sigma}{d\Omega_2'} sin\theta \,d\theta }[/math]

[math]\displaystyle{ \sigma=2\pi (1.638)\frac{d\sigma}{d\Omega_2'} }[/math]

[math]\displaystyle{ \sigma=(10.294) \frac{d\sigma}{d\Omega_2'} }[/math]

Differential Cross Section Scale for Different p₂¹ Values

[math]\displaystyle{ p_{2}'(MeV) }[/math]	[math]\displaystyle{ \frac{d\sigma}{d\Omega_{2}^'}(eV^{-2}) }[/math]	[math]\displaystyle{ \frac{d\sigma}{d\Omega_{2}^'}(GeV^{-2}) }[/math]	[math]\displaystyle{ \frac{d\sigma}{d\Omega_{2}^'}(mb) }[/math]	[math]\displaystyle{ \frac{d\sigma}{d\Omega_{2}^'}(b) }[/math]	[math]\displaystyle{ \sigma(b) }[/math]
[math]\displaystyle{ 10000 }[/math]	[math]\displaystyle{ 9.357\times 10^{-11} }[/math]	[math]\displaystyle{ 9.357\times 10^{7} }[/math]	[math]\displaystyle{ 3.644\times 10^{7} }[/math]	[math]\displaystyle{ 3.644\times 10^{4} }[/math]	[math]\displaystyle{ 3.751\times 10^{5} }[/math]
[math]\displaystyle{ 5000 }[/math]	[math]\displaystyle{ 3.743\times 10^{-10} }[/math]	[math]\displaystyle{ 3.743\times 10^{8} }[/math]	[math]\displaystyle{ 1.458\times 10^{8} }[/math]	[math]\displaystyle{ 1.458\times 10^{5} }[/math]	[math]\displaystyle{ 1.501\times 10^{6} }[/math]
[math]\displaystyle{ 1000 }[/math]	[math]\displaystyle{ 9.357\times 10^{-9} }[/math]	[math]\displaystyle{ 9.357\times 10^{9} }[/math]	[math]\displaystyle{ 3.644\times 10^{9} }[/math]	[math]\displaystyle{ 3.644\times 10^{6} }[/math]	[math]\displaystyle{ 3.751\times 10^{7} }[/math]
[math]\displaystyle{ 500 }[/math]	[math]\displaystyle{ 3.743\times 10^{-8} }[/math]	[math]\displaystyle{ 3.743\times 10^{10} }[/math]	[math]\displaystyle{ 1.458\times 10^{10} }[/math]	[math]\displaystyle{ 1.458\times 10^{7} }[/math]	[math]\displaystyle{ 1.501\times 10^{8} }[/math]

CM to Lab Frame

We can substitute in for [math]\displaystyle{ \theta }[/math]

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

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

Using,

[math]\displaystyle{ sin(\theta^*)=sin(\theta_{2}^*)=\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right) }[/math]

[math]\displaystyle{ \frac{\partial ^2\sigma(E,\, \theta ,\, \phi)}{\partial E\, \partial \Omega} = \frac{\partial ^2\sigma^*(E^*,\, \theta^* ,\, \phi^*)}{\partial E^*\,\partial \Omega^*}\frac{p}{p^{*}} }[/math]

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

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

Now, using the trigometric identity,

[math]\displaystyle{ sin^2 t+cos^2 t=1\Longrightarrow cos^2(\theta^*)=1-sin^2(\theta^*) }[/math]

[math]\displaystyle{ \frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{*4}}{4E^{*2}p_{2}'^4}\frac{ (3+1-sin^2(\theta^*))^2}{sin^4 \left( \theta_{2}'\right)} }[/math]

[math]\displaystyle{ \frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{*4}}{4E^{*2}p_{2}'^4}\frac{ (4-sin(\theta^*)sin(\theta^*))^2}{sin^4 \left( \theta_{2}'\right)} }[/math]

[math]\displaystyle{ \frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{*4}}{4E^{*2}p_{2}'^4}\frac{ (4-\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right)\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right))^2}{sin^4 \left( \theta_{2}'\right)} }[/math]

[math]\displaystyle{ \frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{*4}}{4E^{*2}p_{2}'^4}\frac{ (4-\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right))^2}{sin^4 \left( \theta_{2}'\right)} }[/math]

[math]\displaystyle{ \frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 p_{2}^{*4}}{4E^{*2}p_{2}'^4}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{*4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)} }[/math]

Substituting,

[math]\displaystyle{ p_{2}^*=\sqrt{E_{2}^{*2}-m^2} }[/math]

[math]\displaystyle{ \frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 (\sqrt{E_{2}^{*2}-m^2})^4}{4E^{*2}p_{2}'^4}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{*4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)} }[/math]

[math]\displaystyle{ \frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 (E_{2}^{*2}-m^2)^2}{4E^{*2}p_{2}'^4}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{*4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)} }[/math]

Substituting in for m, E₂^*,and E^* [math]\displaystyle{ \alpha^2=5.3279\times 10^{-5} }[/math]

[math]\displaystyle{ \frac{d\sigma}{d\Omega '_1}=(\frac{ 5.3279\times 10^{-5}( ((53.015MeV)^{2}-(.511MeV)^2)^2}{4\times (106.031MeV)^{2}p_{2}'^4}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{*4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)} }[/math]

[math]\displaystyle{ \frac{d\sigma}{d\Omega '_1}=\frac{9.357\times 10^9eV^2}{p_{2}^{'4}}\frac{ (16-8\frac{p_{2}^{'2}}{p_{2}^{*2}}\ sin^2 \left( \theta_{2}'\right)+\frac{p_{2}^{'4}}{p_{2}^{*4}}\ sin^4 \left( \theta_{2}'\right))}{sin^4 \left( \theta_{2}'\right)} }[/math]