Difference between revisions of "DV MollerTrackRecon"
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==Kinematic agreement== | ==Kinematic agreement== | ||
− | Finding the correct kinematic values starting from knowing the momentum of the Moller electron in the Lab frame, | + | Finding the correct kinematic values starting from knowing the momentum of the Moller (m) electron in the Lab frame, |
− | <center>Using <math>\theta=\arccos(\frac{P_{z Lab}}{P_{Lab}})</math></center> | + | <center>Using <math>\theta=\arccos(\frac{P_{z(m) Lab}}{P_{(m)Lab}})</math></center> |
{| class="wikitable" align="center" | {| class="wikitable" align="center" | ||
− | | style="background: gray" | <math>\Longrightarrow {P_{z Lab}=P_{Lab}\cos(\theta)}</math> | + | | style="background: gray" | <math>\Longrightarrow {P_{z(m) Lab}=P_{(m)Lab}\cos(\theta)}</math> |
|} | |} | ||
− | <center>Similarly, <math>\phi=\arccos(\frac{P_{x Lab}}{P_{xy Lab}})</math></center> | + | <center>Similarly, <math>\phi=\arccos(\frac{P_{x(m) Lab}}{P_{xy(m) Lab}})</math></center> |
− | <center>where <math>P_{xy Lab}=\sqrt{P_{x Lab}^2+P_{ | + | <center>where <math>P_{xy Lab}=\sqrt{P_{x(m) Lab}^2+P_{y(m)Lab}^2}</math></center> |
− | <center><math>P_{xy Lab}=(P_{x Lab}^2+P_{y Lab}^2)^2</math></center> | + | <center><math>P_{xy(m) Lab}=(P_{x(m) Lab}^2+P_{y(m) Lab}^2)^2</math></center> |
− | <center>and using <math>P_{Lab}^2=P_{x Lab}^2+P_{y Lab}^2+P_{z Lab}^2</math></center> | + | <center>and using <math>P_{(m)Lab}^2=P_{x(m) Lab}^2+P_{y(m) Lab}^2+P_{z(m) Lab}^2</math></center> |
− | <center>this gives <math> | + | <center>this gives <math>P_{(m)Lab}^2=P_{xy(m) Lab}^2+P_{z(m) Lab}^2</math></center> |
− | <center><math>\Longrightarrow P_{Lab}^2-P_{z Lab}^2=P_{xy Lab}^2</math></center> | + | <center><math>\Longrightarrow P_{(m)Lab}^2-P_{z(m) Lab}^2=P_{xy(m) Lab}^2</math></center> |
− | <center><math>\Longrightarrow P_{xy Lab}=\sqrt{P_{Lab}^2-P_{z Lab}^2}</math></center> | + | <center><math>\Longrightarrow P_{xy(m) Lab}=\sqrt{P_{(m)Lab}^2-P_{z(m) Lab}^2}</math></center> |
− | <center>which gives<math>\phi = \arccos(\frac{P_{x Lab}}{\sqrt{P_{Lab}^2-P_{z Lab}^2}})</math></center> | + | <center>which gives<math>\phi = \arccos(\frac{P_{x(m) Lab}}{\sqrt{P_{(m)Lab}^2-P_{z(m) Lab}^2}})</math></center> |
{| class="wikitable" align="center" | {| class="wikitable" align="center" | ||
− | | style="background: gray" | <math>\Longrightarrow P_{x Lab}=\sqrt{P_{Lab}^2-P_{z Lab}^2} \cos(\phi)</math> | + | | style="background: gray" | <math>\Longrightarrow P_{x(m) Lab}=\sqrt{P_{(m)Lab}^2-P_{z(m) Lab}^2} \cos(\phi)</math> |
|} | |} | ||
− | <center>Similarly, using <math>P_{Lab}^2=P_{x Lab}^2+P_{y Lab}^2+P_{z Lab}^2</math></center> | + | <center>Similarly, using <math>P_{(m)Lab}^2=P_{x(m) Lab}^2+P_{y(m) Lab}^2+P_{z(m) Lab}^2</math></center> |
− | <center><math>\Longrightarrow P_{Lab}^2-P_{x Lab}^2-P_{z Lab}^2=P_{y Lab}^2</math></center> | + | <center><math>\Longrightarrow P_{(m)Lab}^2-P_{x(m) Lab}^2-P_{z(m) Lab}^2=P_{y(m) Lab}^2</math></center> |
{| class="wikitable" align="center" | {| class="wikitable" align="center" | ||
− | | style="background: gray" | <math>P_{y Lab}=\sqrt{P_{Lab}^2-P_{x Lab}^2-P_{z Lab}^2}</math> | + | | style="background: gray" | <math>P_{y(m) Lab}=\sqrt{P_{(m)Lab}^2-P_{x(m) Lab}^2-P_{z(m) Lab}^2}</math> |
|} | |} | ||
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{| class="wikitable" align="center" | {| class="wikitable" align="center" | ||
− | | style="background: gray" | <math>P_{x CM}\Leftrightarrow P_{x Lab}</math> | + | | style="background: gray" | <math>P_{x(m) CM}\Leftrightarrow P_{x(m) Lab}</math> |
|} | |} | ||
{| class="wikitable" align="center" | {| class="wikitable" align="center" | ||
− | | style="background: gray" | <math>P_{y CM}\Leftrightarrow P_{y Lab}</math> | + | | style="background: gray" | <math>P_{y(m) CM}\Leftrightarrow P_{y(m) Lab}</math> |
|} | |} | ||
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{| class="wikitable" align="center" | {| class="wikitable" align="center" | ||
− | | style="background: gray" | <math>P_{z CM}=\sqrt{P_{CM}^2-P_{x CM}^2-P_{y CM}^2}</math> | + | | style="background: gray" | <math>P_{z(m) CM}=\sqrt{P_{(m)CM}^2-P_{x(m) CM}^2-P_{y(m) CM}^2}</math> |
|} | |} | ||
[[DV_RunGroupC_Moller#Moller_Track_Reconstruction]] | [[DV_RunGroupC_Moller#Moller_Track_Reconstruction]] |
Revision as of 16:17, 23 December 2015
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
For event #29, in sector 3, the location of the first interaction is given by
Converting -25 degrees to radians,
which is the rotation the detectors are rotated from the y axis.
Finding
since "sector -1" =3-1=2*60=120 degrees
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
Center of Mass Energy
Using the definition
We can use 4-momenta vectors, i.e.
we can use the fact that the scalar product of a 4-momenta with itself,
is invariant
Using this, the sum of two 4-momenta forms a 4-vector as well
The length of this four-vector is an invariant as well
For incoming electrons moving only in the z-direction, we can write
We can perform a Lorentz transformation to the Center of Mass frame, with zero total momentum
Without knowing the values for gamma or beta, we can use the fact that lengths of the two 4-momenta are invariant
Setting these equal to each other, we can use this for the collision of two particles of mass m1 and m2. Since the total momentum is zero in the Center of Mass frame, we can express total energy in the center of mass frame as
Using the relations
In the frame where one particle (m2) is at rest
which implies,
where
in MeV
Inspecting the Lorentz transformation to the Center of Mass frame:
For the case of a stationary electron, this simplifies to:
which gives,
Solving for , with
Similarly, solving for
by substituting in
Using the fact that
This gives the momenta of the particles in the center of mass to have equal magnitude, but opposite directions
Using the relation
where M is the total mass
Using the fact that
Kinematic agreement
Finding the correct kinematic values starting from knowing the momentum of the Moller (m) electron in the Lab 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.