DV RunGroupC Moller

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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_1+E_2)^2-(p_1+p_2)^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 lab}m_2)^{1/2}[/math]

where [math]E_{1 ab}=KE_{1 lab}+m_1[/math] in MeV


The velocity of the center of mass in the lab frame is


[math]\beta_{cm}=\frac{p_{lab}}{(E_{1 lab}+m_2)}[/math]


where plab≡p1 lab and


[math]\gamma_{cm}\frac{(E_{1 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_{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.

Simulation Verification

Sample output from GEANT4 simulation:


KEi Pxi Pyi Pzi xi yi zi 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


Running a GEANT simulation just to put this set into a .dat file, then using the .dat file through the Moller_OG.C file in ROOT, we find in the Lab Frame:


Fnl4Mom.P 10999.599651
Fnl4Mom.E 10999.610609
Mol4Mom.P 1.320928
Mol4Mom.E 1.416324
Fnl4Mom.Theta(degrees) 0.005013
Fnl4Mom.Theta(radians) 0.000087
Mol4Mom.Theta(degrees) 46.756514
Mol4Mom.Theta(radians) 0.815641
Fnl4Mom.P 52.455386
Fnl4Mom.E 54.704969
Mol4Mom.P 52.455386
Mol4Mom.E 52.457875
Fnl4Mom.Theta(degrees) 1.051202
Fnl4Mom.Theta(radians) 0.018338
Mol4Mom.Theta(degrees) 179.040096
Mol4Mom.Theta(radians) 3.123255


FnlMomCM.pngMolMomCM.png

Explaining graph distribution

Trying to explain the graphs, we look at the data files for strange KE, hence weirder momentums


KEi Pxi Pyi Pzi xi yi zi KEf Pxf Pyf Pzf xf yf zf KEm Pxm Pym Pzm xm ym zm
11000 0 0 11000.5 0 0 -510 9855.29 -31.7166 -6.48546 9855.75 0 0 -501.365 1144.53 31.7166 6.48546 1144.59 0 0 -501.365
11000 0 0 11000.5 0 0 -510 10999.0996 -0.152974 -0.761885 10999.7 0 0 -502.19 0.590904 0.152974 0.761885 0.590932 0 0 -502.191986



Using the equations derived earlier,

[math]E_{cm}=(m_1^2+m_2^2+2E_{1 lab}m_2)^{1/2}[/math]


[math]p_{cm}=\frac{p_{lab}m_2}{E_{cm}}[/math]

For the KEf=9855.29 MeV, we find that

[math]E=KE_f+.511\times 10^6=9855.29\times 10^6+.511\times 10^6=9.855\times 10^9 eV[/math]


[math]p_{lab}=((-31.72\times 10^6)^2+(-6.48\times 10^6)^2+(9855.75\times 10^6)^2)^{1/2}=9.855\times 10^9 \frac{eV}{c}[/math]


[math]E_{cm}=(m_1^2+m_2^2+2E_{1 lab}m_2)^{1/2}=((0.511\times 10^6)^2+(0.511\times 10^6)^2+2\times (9.855\times 10^3)\times 0.511\times 10^6)^{1/2}[/math]


this gives

[math]E_{cm}=1.00\times 10^8 eV[/math]


[math]p_{cm}=\frac{p_{lab}m_2}{E_{cm}}=\frac{(9.855\times 10^9)\times (.511\times 10^6)}{1\times 10^8}=50 \frac{MeV}{c}[/math]

This explains the lower bound part of the graph in that some of the initial particles transfer a large amount of energy to the Moller electron.


For the second set of data, simulation finds the boost to be

Pcm Pxcm Pycm Pzcm Ecm
58.379341 -0.152974 -0.761885 58.374168 37.904019

Using the approximation used above doesn't give the Energy in the Center of Mass as 58 MeV, it yields 53 MeV. Using the other form,

[math]E_{cm}=(E_1+E_2)^2-(p_1+p_2)^2[/math]


[math]E_{cm}=((10999.0996\times 10^6+.511\times 10^6)+(0.590904\times 10^6+.511\times 10^6))^2-(1.09997\times 10^{10}+9.762298\times 10^5)^2[/math]


[math]E_{cm}=1.210157\times 10^{20}-1.210149\times 10^{20}[/math]


[math]E_{cm}=8\times 10^{14}[/math]


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 Halzen and Martin (p121) to approximate Moller scattering (in the Center of Mass Frame)

[math]\frac{d\sigma}{d\Omega}=\frac{m^2 \alpha^2}{16 p^4}\left(\frac{1}{\sin^4 \frac{\theta}{2}}+\frac{1}{\cos^4 \frac{\theta}{2}}-\frac{1}{\sin^2 \frac{\theta}{2}\cos^2 \frac{\theta}{2}}\right)[/math]


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


Plugging in the values expected for a scattering electron:

[math]m^2 = \frac{(.000511 GeV)^2}{c^4}=2.6\times 10^{-7} GeV^2/c^4[/math]


[math]\alpha^2=\frac{1}{137^2}=7.3\times 10^{-3}[/math]


[math]p^4=\frac{(.053 GeV)^4}{c^4}=7.9\times 10^{-6} GeV^4/c^4[/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{m^2 \alpha^2}{16 p^4}=\frac{1.5\times 10^{-5}}{ GeV^{2}}[/math]

Using the conversion of


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


We find that the differential cross section is [math]5.8\times 10^{-6} mb=5.8 nb[/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.pngDiffCrossSect.png

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