Moller events WITH Solenoid
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
![Field effects.png](/./images/5/59/Field_effects.png)
Cross-section Area
The scalar product of two 4-momenta [math]p_1\cdot p_2=E_1E_2-\textbf p_1\cdot p_2[/math]
In the collision of two particles of mass m1 and m2, the total energy in the center of mass frame can be written
[math]E_{cm}=[(E_{1 lab}+E_{2 lab})^2-(\vec{p_{1 lab}}+p_{2 lab})^2]^{1/2}[/math]
[math]E_{cm}=[m_1^2+m_2^2+2E_1E_2(1-\beta_1\beta_2cos\theta)]^{1/2}[/math]
where θ is the angle between the particles.
In the frame where one particle (m2) is at rest
[math]E_{cm}=(m_1^2+m_2^2+2E_{1_{initial} lab}m_2)^{1/2}[/math]
where [math]E_{1_{initial} lab}=KE_{1_{initial} lab}+m_1[/math] in MeV
The velocity of the center of mass in the lab frame is
[math]\beta_{cm}=\frac{p_{1_{initial}lab}}{(E_{1_{initial} lab}+m_2)}[/math]
[math]\gamma_{cm}\frac{(E_{1_{initial} lab}+m_2)}{E_{cm}}[/math]
This gives the momenta of the particles in the center of mass to have equal magnitude, but opposite directions
[math]p_{cm}=\frac{p_{1_{initial}lab}m_2}{E_{cm}}[/math]
Finding the correct kimematic values starting from knowing the momentum of the Moller electron in the Lab frame,
Using [math]\theta=\arccos(\frac{P_z}{P})[/math]
[math]\Longrightarrow {P_z=P\cos(\theta)}[/math]
Similarly, [math]\phi=\arccos(\frac{P_x}{P_{xy}})[/math]
where [math]P_{xy}=\sqrt{P_x^2+P_y^2}[/math]
[math]P_{xy}=(P_x^2+P_y^2)^2[/math]
and using [math]P^2=P_x^2+P_y^2+P_z^2[/math]
this gives [math]P^2=P_{xy}^2+P_z^2[/math]
[math]\Longrightarrow P^2-P_z^2=P_{xy}^2[/math]
[math]\Longrightarrow P_{xy}=\sqrt{P^2-P_z^2}[/math]
which gives[math]\phi = \arccos(\frac{P_x}{\sqrt{P^2-P_z^2}})[/math]
[math]\Longrightarrow P_x=\sqrt{P^2-P_z^2} \cos(\phi)[/math]
Similarly, using [math]P^2=P_x^2+P_y^2+P_z^2[/math]
[math]\Longrightarrow P^2-P_x^2-P_z^2=P_y^2[/math]
Yields [math]P_y=\sqrt{P^2-P_x^2-P_z^2}[/math]
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_{x Lab}\Leftrightarrow P_{x CM}[/math]
[math]P_{y Lab}\Leftrightarrow P_{y CM}[/math]
DV_RunGroupC_Moller#Moller_Track_Reconstruction