Difference between revisions of "DV RunGroupC Moller"

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<center>E<sub>cm</sub>=[(E<sub>1</sub>+E<sub>2</sub>)<sup>2</sup>-(p<sub>1</sub>+p<sub>2</sub>)<sup>2</sup>]<sup>1/2</sup></center>
 
<center>E<sub>cm</sub>=[(E<sub>1</sub>+E<sub>2</sub>)<sup>2</sup>-(p<sub>1</sub>+p<sub>2</sub>)<sup>2</sup>]<sup>1/2</sup></center>
  
E<sub>cm</sub>=[m<sub>1</sub><sup>2</sup>+m<sub>2</sub><sup>2</sup>+2E<sub>1</sub>E<sub>2</sub>(1-β<sub>1</sub>β<sub>2</sub>cosθ)]<sup>1/2</sup>
+
<center>E<sub>cm</sub>=[m<sub>1</sub><sup>2</sup>+m<sub>2</sub><sup>2</sup>+2E<sub>1</sub>E<sub>2</sub>(1-β<sub>1</sub>β<sub>2</sub>cosθ)]<sup>1/2</sup></center>
  
 
In the frame where one particle (m<sub>2</sub>) is at rest
 
In the frame where one particle (m<sub>2</sub>) is at rest
  
E<sub>cm</sub>=(m<sub>1</sub><sup>2</sup>+m<sub>2</sub><sup>2</sup>+2E<sub>1 lab</sub>m<sub>2</sub>)<sup>1/2</sup>
+
<center>E<sub>cm</sub>=(m<sub>1</sub><sup>2</sup>+m<sub>2</sub><sup>2</sup>+2E<sub>1 lab</sub>m<sub>2</sub>)<sup>1/2</sup></center>
  
 
This gives the momenta of the particles in the center of mass to have equal magnitude, but opposite directions
 
This gives the momenta of the particles in the center of mass to have equal magnitude, but opposite directions
  
p<sub>cm</sub>=p<sub>lab</sub>m<sub>2</sub>/E<sub>cm</sub>
+
<center>p<sub>cm</sub>=p<sub>lab</sub>m<sub>2</sub>/E<sub>cm</sub></center>
  
  

Revision as of 21:25, 23 April 2015

Simulating the Moller scattering background for EG1

Step 1

Determine the Moller background using an LH2 target to check the physics in GEANT4

Incident electron energy varies from 1-11 GeV.

LH2 target is a cylinder with a 1.5 cm diameter and 1 cm thickness.

  (Following dimensions listed on page 8 of File:PHY02-33.pdf)

Numbers Moller electrons per incident electron.

  While 2nd and 3rd generations are created, only 2 2nd generation daughter particles are created for 1E6 incident particles.  All knock on electrons are not counted.

Momentum and angular distributions. 11GeV

ElectronMomentumLabFrame.jpgMollerMomentumLabFrame.jpg

In the collision of two particles of mass m1 and m2, the total energy in the center of mass frame can be written

Ecm=[(E1+E2)2-(p1+p2)2]1/2
Ecm=[m12+m22+2E1E2(1-β1β2cosθ)]1/2

In the frame where one particle (m2) is at rest

Ecm=(m12+m22+2E1 labm2)1/2

This gives the momenta of the particles in the center of mass to have equal magnitude, but opposite directions

pcm=plabm2/Ecm


FinalMomCM.jpgMollerMomCM.jpg

FinalTheta.jpgMollerTheta.jpg

FinalThetaCM.jpgMollerThetaCM.jpg

Comparing experimental vs. theoretical for Møller differential cross section 11GeV

FinalThetaCM.jpgV3.jpg

Step 2

Replace the LH2 target with an NH3 target and compare with LH2 target.


Step 3

Determine impact of Solenoid magnet on Moller events

Papers used

A polarized target for the CLAS detectorFile:PHY02-33.pdf

An investigation of the spin structure of the proton in deep inelastic scattering of polarized muons on polarized protons File:1819.pdf

EG12