DV MollerTrackRecon

Moller Lund

LUND file with Moller events (with origin of coordinates occurring at each event)

2       1       1       1       1       0       0.000563654     3.53715 0       6.2002
1 -1 1 11 0 0 0.69 -2.4999 10993.7998 10993.80 0.000511 0 0 0
2 -1 1 11 0 0 -0.69 2.4999 6.5852 7.08 0.000511 0 0 0

From a GEMC run WITH the Solenoid ced is used to obtain the information from the eg12_rec.ev file.

We take the phi angle from the Generated Event momentum as the initial phi angle. The obtain the final phi angle, we can look at the final position of the electron with in the drift chambers.

Examining the position from Timer Based Tracking, we can see that after rotations about first the y-axis, then the z-axis transforms from the detector frame of reference to the lab frame of reference.

Euler Angles

We can use the Euler angles to perform the rotations.

For the rotation about the y axis.

And the rotation about the z axis.

Transformation Matrix

The Euler angles can be applied using a transformation matrix

[math]\left( \begin{array}{ccc} \cos (\theta ) & 0 & -\sin (\theta ) \\ 0 & 1 & 0 \\ \sin (\theta ) & 0 & \cos (\theta ) \\ \end{array} \right).\left( \begin{array}{c} x \\ y \\ z \\ \end{array} \right)[/math]

[math]=\left( \begin{array}{c} x \cos (\theta )-z \sin (\theta ) \\ y \\ z \cos (\theta )+x \sin (\theta ) \\ \end{array} \right)[/math]

For event #29, in sector 3, the location of the first interaction is given by

Converting -25 degrees to radians, [math]\theta =-0.436332[/math] which is the rotation the detectors are rotated from the y axis.

[math]\left( \begin{array}{ccc} \cos (\theta ) & 0 & -\sin (\theta ) \\ 0 & 1 & 0 \\ \sin (\theta ) & 0 & \cos (\theta ) \\ \end{array} \right).\left( \begin{array}{c} -15.76 \\ 0 \\ 237.43 \\ \end{array} \right)[/math]

[math]=\left( \begin{array}{c} 86.0588 \\ 0. \\ 221.845 \\ \end{array} \right)[/math]

Finding [math]\phi =\frac{120\ 2 \pi }{360};[/math] since "sector -1" =3-1=2*60=120 degrees

[math]\left( \begin{array}{ccc} \cos (\phi ) & -\sin (\phi ) & 0 \\ \sin (\phi ) & \cos (\phi ) & 0 \\ 0 & 0 & 1 \\ \end{array} \right).\left( \begin{array}{c} 86.0588 \\ 0. \\ 221.845 \\ \end{array} \right)[/math]

[math]\left( \begin{array}{c} -43.0294 \\ 74.5291 \\ 221.845 \\ \end{array} \right)[/math]

This shows how the coordinates are transformed and explains the validity of using the TBTracking information to obtain a phi angle in the lab frame.

Phi shifts

gcard to generate electrons.

 <option name="BEAM_P"   value="e-, 6.0*GeV, 30.0*deg, 10*deg">
 <option name="SPREAD_P" value="5.5*GeV, 25*deg, 180*deg">

Cross-section

Calculations of 4-momentum components(Trial 1)

This trial did not take into account the initial electron energy loss as it traveled through the target material.

4-momentum components (Not accounting for initial energy loss to scattering)

Calculations of 4-momentum components(Trial 2)

Reconstructing Moller Events

Calculations of 4-momentum components (Trial 3)

Energy-Theta

Calculations of 4-momentum components (Trial 4)

Setup

Since we want to run for a evenly spaced energy range for Moller electrons, we will need to use some of the scattered electrons to help cover this range. A Moller scattering data file of 1E7 events has no Moller electrons with momentum over 5500 MeV. Since momentum is conserved, and the data is verified kinematicly verified, we cannot simply "switch" the data. This data can be altered to have a certain number of different phi values for each energy to match the Moller cross section. This data can then be written to a LUND file, and compared to the previous calculations which did not factor in loss of initial energy.

Prepare Data

Using the existing Moller scattering data from a GEANT simulation of 4E7 incident electrons, a file of just scattered momentum components can be constructed using:

awk '{print $9, $10, $11, $16, $17, $18}' MollerScattering_NH3_Large.dat > Just_Scattered_Momentum.dat

Transfer to CM Frame

Center of Mass Frame

4-Momentum Invariants

Figure 1: Definition of variables in the Center of Mass Frame

Starting with the definition for the total relativistic energy:

[math]E^2\equiv p^2c^2+m^2c^4[/math]

[math]\Longrightarrow {E^2}-p^2c^2=(mc^2)^2[/math]

Since we can assume that the frame of reference is an inertial frame, it moves at a constant velocity, the mass should remain constant.

[math]\frac {d\vec p}{dt}=0\Rightarrow \frac{d(m\vec v)}{dt}=\frac{c\ dm}{dt}\Rightarrow \frac{dm}{dt}=0[/math]

[math] \therefore m=const[/math]

We can use 4-momenta vectors, i.e. [math]{\mathbf P}\equiv \left(\begin{matrix} E\\ p_x \\ p_y \\ p_z \end{matrix} \right)=\left(\begin{matrix} E\\ \vec p \end{matrix} \right)[/math] ,with c=1, to describe the variables in the CM Frame.

Using the fact that the scalar product of a 4-momenta with itself,

[math]{\mathbf P_1}\cdot {\mathbf P^1}=P_{\mu}g_{\mu \nu}P^{\nu}=\left(\begin{matrix} E\\ p_x \\ p_y \\ p_z \end{matrix} \right)\cdot \left( \begin{matrix}1 & 0 & 0 & 0\\0 & -1 & 0 & 0\\0 & 0 & -1 & 0\\0 &0 & 0 &-1\end{matrix} \right)\cdot \left(\begin{matrix} E & p_x & p_y & p_z \end{matrix} \right)[/math]

[math]{\mathbf P_1}\cdot {\mathbf P^1}=E_1E_1-\vec p_1\cdot \vec p_1 =m_{1}^2=s[/math]

is invariant.

Using this notation, the sum of two 4-momenta forms a 4-vector as well

[math]{\mathbf P_1}+ {\mathbf P_2}= \left( \begin{matrix}E_1+E_2\\\vec p_1 +\vec p_2 \end{matrix} \right)= {\mathbf P}[/math]

The length of this four-vector is an invariant as well

[math]{\mathbf P^2}=({\mathbf P_1}+{\mathbf P_2})^2=(E_1+E_2)^2-(\vec p_1 +\vec p_2 )^2=(m_1+m_2)^2=s[/math]

Equal masses

For incoming electrons moving only in the z-direction, we can write

[math]{\mathbf P_1}+ {\mathbf P_2}= \left( \begin{matrix}E_1+E_2\\ 0 \\ 0 \\ p_{1(z)}+p_{2(z)}\end{matrix} \right)={\mathbf P}[/math]

We can perform a Lorentz transformation to the Center of Mass frame, with zero total momentum

[math]\left( \begin{matrix}E^*_{1}+E^*_{2}\\ 0 \\ 0 \\ 0\end{matrix} \right)=\left(\begin{matrix}\gamma^* & 0 & 0 & -\beta^* \gamma^*\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta^* \gamma^* & 0 & 0 & \gamma^* \end{matrix} \right) . \left( \begin{matrix}E_{1}+E_{2}\\ 0 \\ 0 \\ p_{1(z)}+p_{2(z)}\end{matrix} \right)[/math]

Without knowing the values for gamma or beta, we can utalize the fact that lengths of the two 4-momenta are invariant

[math]s={\mathbf P^*}^2=(E^*_{1}+E^*_{2})^2-(\vec p\ ^*_{1}+\vec p\ ^*_{2})^2=(m_{1}^*+m_{2}^*)^2[/math]

[math]s={\mathbf P}^2=(E_{1}+E_{2})^2-(\vec p_{1}+\vec p_{2})^2=(m_{1}+m_{2})^2[/math]

This gives,

[math]\Longrightarrow (m_{1}^*+m_{2}^*)^2=(m_{1}+m_{2})^2[/math]

Using the fact that

[math]\begin{cases} m_{1}=m_{2} \\ m_{1}^*=m_{2}^* \end{cases}[/math]

since the rest mass energy of the electrons remains the same in inertial frames.

Substituting, we find

[math](m_{1}^*+m_{1}^*)^2=(m_{1}+m_{1})^2[/math]

[math]2m_{1}^*=2m_{1}[/math]

[math]m_{1}^*=m_{1}[/math]

[math]\Longrightarrow m_{1}=m^*_{1}\ ; m_{2}=m^*_{2}[/math]

This confirms that the mass remains constant between the frames of reference.

Total Energy in CM

Setting the lengths of the 4-momenta equal to each other,

[math]{\mathbf P^*}^2={\mathbf P}^{'2}[/math]

we can use this for the collision of two particles of mass m. Since the total momentum is zero in the Center of Mass frame, we can express total energy in the center of mass frame as

[math](E^*_{1}+E^*_{2})^2-(\vec p\ ^*_{1}+\vec p\ ^*_{2})^2=s=(E_{1}^'+E_{2}^')^2-(\vec{p_{1}}^'+\vec {p_{2}}^')^2[/math]

[math](E^*)^2-(\vec p\ ^*)^2=(E_{1}^'+E_{2}^')^2-(\vec{p_{1}}^'+\vec {p_{2}}^')^2[/math]

[math](E^*)^2=(E_{1}^'+E_{2}^')^2-(\vec{p_{1}}^'+\vec {p_{2}}^')^2[/math]

[math]E^*=\sqrt{(E_{1}^'+E_{2}^')^2-(\vec{p_{1}}^'+\vec {p_{2}}^')^2}[/math]

[math]\left( \begin{matrix}E^*_{1}+E^*_{2}\\ p_{1(x)}^*+p_{2(x)}^* \\ p_{1(y)}^*+p_{2(y)}^* \\ p_{1(z)}^*+p_{2(z)}^*\end{matrix} \right)=\left(\begin{matrix}\gamma & 0 & 0 & -\beta \gamma\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta \gamma & 0 & 0 & \gamma \end{matrix} \right) . \left( \begin{matrix}E_{1}^'+E_{2}^'\\ p_{1(x)}^'+p_{2(x)}^' \\ p_{1(y)}^'+p_{2(y)}^' \\ p_{1(z)}^'+p_{2(z)}^'\end{matrix} \right)[/math]

[math]\left( \begin{matrix}E^*_{1}+E^*_{2}\\ 0 \\ 0 \\ 0\end{matrix} \right)=\left(\begin{matrix}\gamma & 0 & 0 & -\beta \gamma\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 &0 \\ -\beta \gamma & 0 & 0 & \gamma \end{matrix} \right) . \left( \begin{matrix}E^'\\ p_{x}^' \\ p_{y}^' \\ p_{z}^'\end{matrix} \right)[/math]

By the definition of the CM Frame we know

[math]\Longrightarrow\begin{cases} E_2^*=\gamma(E_{2}^'+m)-\beta \gamma p_{2(z)}^' \\ p^*_{x}=p^'_{x}=0 \\ p^*_{y}=p^'_{y}=0 \\ p^*_{2(z)}=-\beta \gamma(E_{2}^'+m)+\gamma p_{2(z)}^' \end{cases}[/math]

[math]\Longrightarrow \begin{cases} p_{1(x)}^'=-p_{2(x)}^' \\ p_{1(y)}^'=-p_{2(y)}^' \end{cases}[/math]

[math]\vec {p}\, ^*=\vec {p_1}^*+\vec {p_2}^*=0 \Longrightarrow \vec {p_1}^*=-\vec {p_2}^* [/math]

[math]E^*=\sqrt{(E_{1}^'+E_{2}^')^2-(\vec{p_{1}}^'+\vec {p_{2}}^')^2}[/math]

Using the relativistic definition of total energy:

[math]E^2 \equiv p^2+m^2[/math]

[math]E_1^*=\sqrt{p_1^{*2}+m^2}[/math]

[math]E_1^*=\sqrt{p_2^{*2}+m^2}=\sqrt{(-p_1)^{*2}+m^2}=\sqrt{p_1^{*2}+m^2}[/math]

[math]\therefore E^*=E_1^*+E_2^* \Longrightarrow E_1^*=E_2^*[/math]

Since the energies are equal, we use this fact to find the momenta

[math]|p_1^*|=|p_2^*|=\sqrt{E_1^{*2}-m^{*2}}[/math]

Moller electron Center of Mass Frame

Relativistically, the x and y components remain the same in the conversion from the Lab frame to the Center of Mass frame, since the direction of motion is only in the z direction.

[math]p^*_{2(x)}\Leftrightarrow p_{2(x)}'[/math]

[math]p^*_{2(y)}\Leftrightarrow p_{2(y)}'[/math]

[math]p^*_{2(z)}=\sqrt {(p^*_2)^2-(p^*_{2(x)})^2-(p^*_{2(y)})^2}[/math]

Figure 2: Definition of Moller electron variables in the Lab Frame in the x-z plane.

[math]\theta '_2\equiv \arccos \left(\frac{p^'_{2(z)}}{p^'_{2}}\right)[/math]

Following the same geometry as for the Lab frame,

[math]\Longrightarrow \theta ^*_2=\arccos \left(\frac{p^*_{2(z)}}{p^*_{2}}\right)[/math]

Electron Center of Mass Frame

Relativistically, the x, y, and z components have the same magnitude, but opposite direction, in the conversion from the Moller electron's Center of Mass frame to the electron's Center of Mass frame.

[math]p^*_{2(x)}= -p^*_{1(x)}[/math]

[math]p^*_{2(y)}= -p^*_{1(y)}[/math]

[math]p^*_{2(z)}=-p^*_{1(z)}[/math]

where previously it was shown

[math]|p^*_{1}|\equiv |p^*_{2}|[/math]

[math]E^*_{1}\equiv E^*_{2}[/math]

[math]\theta_{1}^*=\pi-\theta_{1}^*[/math]

Determing Angles

Figure 3: Definition of Moller electron variables in the Lab Frame in the x-y plane.

[math]\phi '_2\equiv \arccos \left( \frac{p^'_{2(x) Lab}}{p^'_{2(xy)}} \right)[/math]

where [math]p_{2(xy)}^'=\sqrt{(p_{2(x)}^')^2+(p^'_{2(y)})^2}[/math]

[math](p^'_{2(xy)})^2=(p^'_{2(x)})^2+(p^'_{2(y)})^2[/math]

and using [math]p^2=p_{(x)}^2+p_{(y)}^2+p_{(z)}^2[/math]

this gives [math](p^'_{2})^2=(p^'_{2(xy)})^2+(p^'_{2(z)})^2[/math]

[math]\Longrightarrow (p'_{2})^2-(p'_{2(z)})^2 = (p'_{2(xy)})^2[/math]

[math]\Longrightarrow p_{2(xy)}^'=\sqrt{(p^'_{2})^2-(p^'_{2(z)})^2}[/math]

which gives[math]\phi '_2 = \arccos \left( \frac{p_{2(x)}'}{\sqrt{p_{2}^{'\ 2}-p_{2(z)}^{'\ 2}}}\right)[/math]

[math]\Longrightarrow p_{2(x)}'=\sqrt{p_{2}^{'\ 2}-p_{2(z)}^{'\ 2}} \cos(\phi)[/math]

Similarly, using [math]p_{2}^2=p_{2(x)}^2+p_{2(y)}^2+p_{2(z)}^2[/math]

[math]\Longrightarrow p_{2}^{'\ 2}-p_{2(x)}^{'\ 2}-p_{2(z)}^{'\ 2}=p_{2(y)}^{'\ 2}[/math]

[math]p_{2(y)}'=\sqrt{p_{2}^{'\ 2}-p_{2(x)}^{'\ 2}-p_{2(z)}^{'\ 2}}[/math]

[math]p_{x}[/math] and [math]p_{y}[/math] results based on [math]\phi[/math]

Checking on the sign from the cosine results for [math]\phi '_2[/math]

We have the limiting range that [math]\phi[/math] must fall within:

[math]-\pi \le \phi '_2 \le \pi\ Radians[/math]

Examining the signs of the components which make up the angle [math]\phi[/math] in the 4 quadrants which make up the xy plane:

[math]For\ 0 \ge \phi '_2 \ge \frac{-\pi}{2}\ Radians[/math]

px=POSITIVE

py=NEGATIVE

[math]For\ 0 \le \phi '_2 \le \frac{\pi}{2}\ Radians[/math]

px=POSITIVE

py=POSITIVE

[math]For\ \frac{-\pi}{2} \ge \phi '_2 \ge -\pi\ Radians[/math]

px=NEGATIVE

py=NEGATIVE

[math]For\ \frac{\pi}{2} \le \phi '_2 \le \pi\ Radians[/math]

px=NEGATIVE

py=POSITIVE

Partial Check

Partial Check

Alter Phi Angles

Using the fact that

[math]\cos{\phi} \equiv \frac{p_x}{\sqrt{p^2-p_z^2}}[/math]

[math]\Longrightarrow \sqrt{p^2-p_z^2}=\frac{p_x}{\cos{\phi}}=constant[/math]

We can simply use the expression

[math]\frac{p_x}{\cos{\phi}}=\frac{p_x'}{cos{\left(\phi+\delta \phi\right)}}[/math]

[math]\Longrightarrow p_x'=\frac{p_x \times \cos{\left(\phi+\delta \phi\right)}}{\cos{\phi}}[/math]

Then, using

[math]\sqrt{p^2-p_z^2}=\sqrt{p_x^2+p_y^2}[/math]

[math]\Longrightarrow p_y'=\sqrt{p^2-p_z^2-p_x^{'2}}[/math]

Run for Necessary Amount to match Cross Section

Using the above plot for the target material, we can find the relative amount that each Theta angle should observe for this process which gives a known Moller differential cross section.

Theta (degrees)	Number of events
35	130
30	120
25	115
20	100
15	90
12	70
10	60
8	40

We can set up conditional statements to check what range the Theta angle falls in, then by dividing

[math]\Delta \phi=\frac {2\pi}{number\ of\ events}[/math]

we should find the change in phi needed to give an evenly distributed distribution around the xy plane for a given Theta angle.

Starting with a data file of momentum components constructed using awk as described above

A program was written to rotate the phi angle as described above. The changing x and y components for this distribution can be seen with

Lastly a LUND file was written that was 643360680 lines in length, which equates to 214453560 entries. This was divided into 8579 file parts of 75000 each. The first set from the original data set is shown below. To make sure the full 2 pi is covered, the rotation starts in the 1st quadrant.

split -a 4 -d -l 75000 Extra_Phi.LUND Phi_Parts_

---

Back to Recon

Differential Cross Section

Variables used in Elastic Scattering

Variables Used in Elastic Scattering

Scattering Cross Section

Scattering Cross Section

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]

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]

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]

Substituting for Moller range and energies

Converting the number of electrons to barns,

[math]\frac{d\sigma}{d\Omega_{2}'}=\frac{dN}{\mathcal L d\Omega}[/math]

[math]\Longrightarrow dN=\frac{d \sigma}{d \Omega} \mathcal L[/math]

[math]where \mathcal L=i\, \rho I[/math]

where ρ_target is the density of the target material, l_target is the length of the target, and N is the number of incident particles scattered.

[math]\mathcal L=\frac{.86g}{1 cm^3}\times \frac{(100cm)^3}{1m^3} \times \frac{1 kg}{1000 g}\times \left[ \left(.75 \frac{1 mole}{1.01 g} \times \frac{1000g}{1 kg} \right)+\left(.25 \frac{1 mole}{14.01 g} \times \frac{1000g}{1 kg} \right)\right] \times \frac{6.022\times10^{23}particles}{1 mole} \times \frac{1cm}{100 cm} \times \frac{1 m}{ } \times \frac{10^{-28} m^2}{barn} [/math]

[math]\mathcal L=\frac{860kg}{1 m^3}\times \left[ \left(\frac{742.574 mole}{kg} \right)+\left(\frac{17.844 mole}{1 kg} \right)\right] \times \frac{6.022\times 10^{-7}\ particles\cdot m^3}{1 mole\cdot barn} [/math]

[math]\mathcal L=\frac{5.170\times 10^{-4}kg\cdot particles}{1 mole\cdot barn} \left(\frac{760.418 mole}{kg} \right)[/math]

[math]\mathcal L=\frac{.394\ particles}{1 barn} [/math]

[math]\Longrightarrow N=\sigma \frac{.394\ particles}{1 barn}[/math]

Number of electrons from Moller electron Momentum

[math]p_2^{'}(MeV/c)[/math]	[math]\sigma(b)[/math]	[math]Number of electrons[/math]
[math]\equiv 10000[/math]	[math]3.751\times 10^{5}[/math]	[math]\approx 1.478\times 10^5[/math]
[math]\equiv 5000 [/math]	[math]1.501\times 10^{6}[/math]	[math]\approx5.914\times 10^5[/math]
[math]\equiv 1000 [/math]	[math]3.751\times 10^{7}[/math]	[math]\approx 1.478\times 10^7[/math]
[math]\equiv 500[/math]	[math]1.501\times 10^{8}[/math]	[math]\approx 5.914\times 10^7[/math]