Difference between revisions of "DV RunGroupC Moller"

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Replace the LH2 target with an NH3 target and compare with LH2 target.
 
Replace the LH2 target with an NH3 target and compare with LH2 target.
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[[File:FnlMomNH3.png]][[File:MolMomNH3.png]]
  
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[[File:FnlMomNH3CM.png]][[File:MolMomNH3CM.png]]
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[[File:FnlThetaNH3.png]][[File:MolThetaNH3.png]]
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[[File:FnlThetaNH3CM.png]][[File:MolThetaNH3CM.png]]
  
 
=Step 3=
 
=Step 3=

Revision as of 06:26, 2 May 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 distributions in the Lab Frame.

FnlMom.pngMolMom.png

Momentum distributions in the Center of Mass Frame.

Estimated Momentum Distribution

In the collision of two particles of mass m_1 and m_2, the total energy in the center of mass frame can be written


[math]E_{cm}=[(E_{1lab}+E_{2lab})^2-(\vec{p_{1lab}}+p_{2lab})^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]

For an incoming electron with momentum of 11GeV, we should find the momentum in the center of mass to be around 53 MeV which is confirmed in the the data/plots.


KEi Pxi Pyi Pzi xi yi z1 KEf Pxf Pyf Pzf xf yf zf KEm Pxm Pym Pzm xm ym zm
11000 0 0 11000.5 0 0 -510 10999.1 0.433025 -0.858867 10999.6 0 0 -509.276 0.905324 -0.433025 0.858867 0.905366 0 0 -509.276

Changing the code for the total Energy to [math]E=(p_1^2+m_1^2)^{1/2}[/math] in the lab frame gives

FnlMomCM new.pngMolMomCM new.png

Angular Distribution in the Lab Frame

FnlTheta.pngMolTheta.png


Angular Distribution in the Center of Mass Frame

FnlThetaCM.pngMolThetaCM.png

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

Using the equation from JLab to approximate Moller scattering (in the Center of Mass Frame)[1]

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


where [math]\alpha = \frac{1}{137}[/math]


Plugging in the values expected for a scattering electron:

[math]m = \frac{.511 MeV}{c^2}[/math]


[math]\alpha^2=\frac{1}{137^2}=5.33\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{ \alpha^2}{2mE}=\frac{.98\times 10^{-7}}{ MeV^{2}}=\frac{9.8\times 10^{-7}}{(1\times 10^6eV)^2}=\frac{9.8\times 10^{-7}}{1\times 10^{12}eV^2}=\frac{9.8\times 10^{-19}}{eV^2}=\frac{9.8\times 10^{-19}\times(1\times 10^{18})}{1\times 10^{12}eV^2\times(1\times 10^{18})}=\frac{.98}{(1\times 10^{9}eV)^2}\approx\frac{1}{(GeV)^2}[/math]

Using the conversion of


[math]\frac{1}{1GeV^{2}}=.3892 mb [/math]


We find that the differential cross section scale is [math]\frac{d\sigma}{d\Omega}\approx mb[/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 ρtarget is the density of the target material, ltarget is the length of the target, and iscattered 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]

FnlThetaCM.pngTheory.png


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


XSect.png

Step 2

Replace the LH2 target with an NH3 target and compare with LH2 target. FnlMomNH3.pngMolMomNH3.png


FnlMomNH3CM.pngMolMomNH3CM.png


FnlThetaNH3.pngMolThetaNH3.png


FnlThetaNH3CM.pngMolThetaNH3CM.png

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