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]
Different p21 Values
Using the conversion of
[math]\frac{1}{1GeV^2}=.3894 mb[/math]
[math]\sigma=\int d\sigma=\int \frac{d\sigma}{d\Omega_2'}d\Omega[/math]
The range of the detector is considered to be [math] .10 \le \theta \le .87[/math],[math]-\pi \le \phi \le \pi[/math]
[math]\sigma=\int_{ .611}^{2.531} \int_{-\pi}^{\pi} \frac{d\sigma}{d\Omega_2'}sin\theta \,d\theta \, d\phi [/math]
[math]\sigma=2\pi \int_{.611}^{2.531} \frac{d\sigma}{d\Omega_2'} sin\theta \,d\theta [/math]
[math]\sigma=2\pi (1.638)\frac{d\sigma}{d\Omega_2'} [/math]
[math]\sigma=(10.294) \frac{d\sigma}{d\Omega_2'} [/math]
Differential Cross Section Scale for Different p21 Values
| [math]p_{2}'(MeV)[/math]
|
[math]\frac{d\sigma}{d\Omega_{2}^'}(eV^{-2})[/math]
|
[math]\frac{d\sigma}{d\Omega_{2}^'}(GeV^{-2})[/math]
|
[math]\frac{d\sigma}{d\Omega_{2}^'}(mb)[/math]
|
[math]\frac{d\sigma}{d\Omega_{2}^'}(b)[/math]
|
[math]\sigma(b)[/math]
|
| [math]10000[/math]
|
[math]9.357\times 10^{-11}[/math]
|
[math]9.357\times 10^{7}[/math]
|
[math]3.644\times 10^{7}[/math]
|
[math]3.644\times 10^{4}[/math]
|
[math]3.751\times 10^{5}[/math]
|
| [math]5000 [/math]
|
[math]3.743\times 10^{-10}[/math]
|
[math]3.743\times 10^{8}[/math]
|
[math]1.458\times 10^{8}[/math]
|
[math]1.458\times 10^{5}[/math]
|
[math]1.501\times 10^{6}[/math]
|
| [math]1000 [/math]
|
[math]9.357\times 10^{-9}[/math]
|
[math]9.357\times 10^{9}[/math]
|
[math]3.644\times 10^{9}[/math]
|
[math]3.644\times 10^{6}[/math]
|
[math]3.751\times 10^{7}[/math]
|
| [math]500[/math]
|
[math]3.743\times 10^{-8}[/math]
|
[math]3.743\times 10^{10}[/math]
|
[math]1.458\times 10^{10}[/math]
|
[math]1.458\times 10^{7}[/math]
|
[math]1.501\times 10^{8}[/math]
|
CM to Lab Frame
We can substitute in for [math]\theta[/math]
[math]\frac{d\sigma}{d\Omega '_1}=\frac{ \alpha^2 }{4E^{*2}}\frac{ (3+cos^2\theta^*)^2}{sin^4\theta^*}[/math]
[math]\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]sin(\theta^*)=sin(\theta_{2}^*)=\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right)[/math]
| [math]\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]\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]\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]sin^2 t+cos^2 t=1\Longrightarrow cos^2(\theta^*)=1-sin^2(\theta^*)[/math]
[math]\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]\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]\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]\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]\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]p_{2}^*=\sqrt{E_{2}^{*2}-m^2}[/math]
[math]\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]\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]\alpha^2=5.3279\times 10^{-5}[/math]
[math]\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]\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]