Difference between revisions of "DV MollerTrackRecon"

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1 -1 1 11 0 0 0.69 -2.4999 10993.7998 10993.80 0.000511 0 0 0
 
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
 
2 -1 1 11 0 0 -0.69 2.4999 6.5852 7.08 0.000511 0 0 0
2      1      1      1      1      0.650745        0.00159988      3.52285 21.4844 17.5986
 
1 -1 1 11 0 0 -4.20 0.3862 10982.4004 10982.40 0.000511 -0.000456454 0.00077441 0
 
2 -1 1 11 0 0 3.45 0.8800 12.4191 12.92 0.000511 -0.000456454 0.00077441 0
 
2      1      1      1      1      0.650745        0.00159988      3.52285 21.4844 17.5986
 
1 -1 1 11 0 0 -4.20 0.3862 10982.4004 10982.40 0.000511 -0.000456454 0.00077441 0
 
2 -1 1 11 0 0 0.75 -1.2666 2.1155 2.58 0.000511 0 0 0
 
 
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Revision as of 18:47, 22 October 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.

Event29.png

Picture of ced output for event 29

Simulation

Event29 1.png

Reconstruction

No sol1.png

Simulation

Event29 2.png

Reconstruction

No sol2.png

Simulation

Event29 3.png

Reconstruction

No sol3.png

EulerAngles.png

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

Conversions.png


Converting -25 degrees to radians, [math]\theta =-0.436332[/math]


[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]


DV_RunGroupC_Moller#Moller_Track_Reconstruction