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

Cross-section Area

Calculations of 4-momentum components

DV_Calculations_of_4-momentum_components

Summary of 4-momentum calculations

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

x=POSITIVE

y=NEGATIVE

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

x=POSITIVE

y=POSITIVE

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

x=NEGATIVE

y=NEGATIVE

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

x=NEGATIVE

y=POSITIVE

4 momentum calculations for different frames of reference

Moller electron Lab Frame	Moller electron Center of Mass Frame	Electron Center of Mass Frame	Electron Lab Frame
[math]p_{2}'\equiv INPUT[/math]	[math]p_{2}^*\approx 53.013 MeV[/math]	[math]p_{1}^*\approx 53.013 MeV[/math]	[math]p_{1}'=\sqrt{E_{1}^{'\ 2}-(.511 MeV)^2}[/math]
[math]\theta_{2}'\equiv INPUT[/math]	[math]\theta_{2}^*=\arcsin \left(\frac{p_{2}'}{p_{2}^*}\ sin \left( \theta_{2}'\right) \right)[/math]	[math]\theta_{1}^*=\theta_{2}^*+\pi[/math]	[math]\theta_{1}'=\arcsin \left(\frac{p_{1}^*}{p_{1}'}\ sin \left( \theta_{1}^*\right) \right)[/math]
[math]E_{2}'=\sqrt{p_{2}^{'\ 2}+(.511 MeV)^2}[/math]	[math]E_{2}^*\approx 53.013 MeV[/math]	[math]E_{1}^*\approx 53.013 MeV[/math]	[math]E_{1}'\equiv E'-E_{2}'[/math]
[math]p_{2(x)}'=\sqrt{p_{2}^{'\ 2}-p_{2(z)}^{'\ 2}} cos(\phi '_2)[/math]	[math]p_{2(x)}^*\equiv p_{2(x)}'[/math]	[math]p_{1(x)}^*\equiv-p_{2(x)}^*[/math]	[math]p_{1(x)}'\equiv p_{1(x)}^*[/math]
[math]p_{2(y)}'=\sqrt{p_{2}^{'\ 2}-p_{2(x)}^{'\ 2}-p_{2(z)}^{'\ 2}}[/math]	[math]p_{2(y)}^*\equiv p_{2(y)}'[/math]	[math]p_{1(y)}^*\equiv -p_{2(y)}^*[/math]	[math]p_{1(y)}'\equiv p_{2(y)}^*[/math]
[math]p_{2(z)}'\equiv p_{2}'\ cos(\theta'_2)[/math]	[math]p_{2(z)}^*=\sqrt{p_{2}^{*\ 2}-p_{2(x)}^{*\ 2}-p_{2(y)}^{*\ 2}}[/math]	[math]p_{1(z)}^*\equiv -p_{2(z)}^*[/math]	[math]p_{1(z)}'=\sqrt{p_{1}^{'\ 2}-p_{(1(x)}^{'\ 2}-p_{1(y)}^{'\ 2}}[/math]

[math]sin \left( \theta_{LAB}\right)=\left(\frac{p_{x(m) Lab}}{p_{(m)Lab}}\right)[/math]

[math]\Longrightarrow p_{(m)Lab}\ sin \left( \theta_{LAB}\right)=p_{x(m) Lab}[/math]

[math]\Longrightarrow p_{(m)CM}\ sin \left( \theta_{CM}\right)=p_{x(m) CM}[/math]

[math]p_{x(m) LAB}\equiv p_{x(m) CM}[/math]

[math]\Longrightarrow p_{(m)CM}\ sin \left( \theta_{CM}\right)=p_{(m)Lab}\ sin \left( \theta_{LAB}\right)[/math]

[math]\Longrightarrow sin \left( \theta_{CM}\right)=\frac{p_{(m)Lab}}{p_{(m)CM}}\ sin \left( \theta_{LAB}\right)[/math]

[math](E_{(e^-) CM}+E_{(m) CM})^2-(\vec p_{(e^-) CM}+\vec p_{(m) CM})^2=s=(E_{(e^-) Lab}+E_{(m) Lab})^2-(\vec{p_{(e^-) Lab}}+\vec p_{(m) Lab})^2[/math]

where previously it was shown

[math]E_{(m) CM}=E_{(e^-)CM}[/math]

[math] \vec p_{(m) CM}=-\vec p_{(e^-)CM}[/math]

[math] \vec p_{(m) Lab}=0[/math]

[math] E_{(m) Lab}\equiv m_{(m)Lab}[/math]

[math]m_{(m)Lab}=m_{(e^-)Lab}\equiv m[/math]

[math](2E_{(m) CM})^2=(E_{(e^-) Lab}+m)^2-(\vec{p_{(e^-) Lab}})^2[/math]

[math]2E_{(m) CM}^2=E_{(e^-) Lab}^2+2E_{(e^-) Lab}m+m^2-p_{(e^-) Lab}^2[/math]

Using the fact,

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

[math]2E_{(m) CM}^2=p_{(e^-) Lab}^2+m^2+2E_{(e^-) Lab}m+m^2-p_{(e^-) Lab}^2[/math]

[math]2E_{(m) CM}^2=m^2+2E_{(e^-) Lab}m+m^2[/math]

[math]2E_{(m) CM}^2=2m^2+2E_{(e^-) Lab}m[/math]

[math]2E_{(m) CM}^2=2m(m+E_{(e^-) Lab})[/math]

[math]\Longrightarrow E_{(m) CM}=\sqrt{\frac{m(m+E_{(e^-) Lab})}{2}}[/math]

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 \left( \frac{\theta}{2}} \right){sin^4 \left( \frac{\theta}{2}} \right) \right)+\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]

where [math]\alpha=\frac{e^2}{\hbar c}\quad with\quad \hbar = c =1[/math]

This can be simplified to the form

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

Plugging in the values expected for a scattering electron:

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

[math]E\approx 53 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 2.809\times 10^{15}eV^2}=4.74\times 10^{-21} eV^{-2}=\frac{4.74\times 10^{-21}}{1eV^2}\times \frac{1\times 10^{18} }{1\times 10^{18}}=\frac{.0047}{GeV^2}[/math]

Using the conversion of

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

We find that the differential cross section scale is [math]\frac{d\sigma}{d\Omega}\approx 2\times 10^{-3}mb=2\mu b[/math]

Converting the number of electrons to barns,

[math]L=\frac{i_{scattered}}{\sigma} \approx i_{scattered}\times \rho_{target}\times l_{target}[/math]

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

[math]L=\frac{70.85 kg}{1 m^3}\times \frac{1 mole}{2.02 g} \times \frac{1000g}{1 kg} \times \frac{6\times10^{23} atoms}{1 mole} \times \frac{1cm}{100 cm} \times \frac{1 m}{ } \times \frac{10^{-23} m^2}{barn} =2.10\times 10^{-2} barns[/math]

[math]\frac{1}{L \times 4\times 10^7}=1.19\times 10^{-6} barns[/math]

Combining these plots, and rescaling the Final Theta in the Center of Mass for micro-barns, we find

DV_RunGroupC_Moller#Moller_Track_Reconstruction