Difference between revisions of "GEANT Moller Simulations Comparison"

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(Created page with "https://wiki.iac.isu.edu/index.php/Converting_to_barns https://wiki.iac.isu.edu/index.php/Check_Differential_Cross-Section Converting the number of electrons scattered per angl…")
 
 
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<center><math>\textbf{Navigation}</math></center>
 +
 +
<center>
 +
[[Lorentz_Transformation_to_Lab_Frame|<math>\vartriangleleft </math>]]
 +
[[VanWasshenova_Thesis#Weighted_Isotropic_Distribution_in_Lab_Frame|<math>\triangle </math>]]
 +
[[CED_Verification_of_DC_Angle_Theta_and_Wire_Correspondance|<math>\vartriangleright </math>]]
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</center>
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https://wiki.iac.isu.edu/index.php/Converting_to_barns
 
https://wiki.iac.isu.edu/index.php/Converting_to_barns
  
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While this expression has no explicit dependancies on energy, the ratio is a function of the energy, as well as the physical makeup of the target.
 
While this expression has no explicit dependancies on energy, the ratio is a function of the energy, as well as the physical makeup of the target.
  
This gives, for LH2:
+
This gives,  
  
<center><math>\rho_{target}\times l_{target}=\frac{70.85 kg}{1 m^3}\times \frac{1 mole}{2.02 g} \times \frac{1000g}{1 kg} \times \frac{6\times10^{23} molecule}{1 mole} \times \frac{2\ atoms}{molecule}\times \frac{1m^3}{(100 cm)^3} \times \frac{1 cm}{ } \times \frac{10^{-24} cm^{2}}{barn} =4.2\times 10^{-2} barns^{-1}</math></center>
+
For 5 cm length of a LH2 target:
  
 +
<center><math>\rho_{target}\times l_{target}=\frac{70.85 kg}{1 m^3}\times \frac{1 mole}{2.02 g} \times \frac{1000g}{1 kg} \times \frac{6\times10^{23} molecule}{1 mole} \times \frac{2\ atoms}{molecule}\times \frac{1m^3}{(100 cm)^3} \times \frac{5 cm}{ } \times \frac{1 \times 10^{-24} cm^{2}}{barn} =.21 barns^{-1}</math></center>
  
 +
 +
----
 +
 +
For 1 cm length of a LH2 target:
 +
 +
<center><math>\rho_{target}\times l_{target}=\frac{70.85 kg}{1 m^3}\times \frac{1 mole}{2.02 g} \times \frac{1000g}{1 kg} \times \frac{6\times10^{23} molecule}{1 mole} \times \frac{2\ atoms}{molecule}\times \frac{1m^3}{(100 cm)^3} \times \frac{1 cm}{ } \times \frac{1 \times 10^{-24} cm^{2}}{barn} =.042 barns^{-1}</math></center>
  
 
From earlier simulations for random angle Phi, we know that the full range of Theta is limited depending on the target material.
 
From earlier simulations for random angle Phi, we know that the full range of Theta is limited depending on the target material.
  
  
<center>[[File:MollerThetaLab_4e7_LH2_11GeV.png]][[File:MollerThetaLab_4e7_NH3_11GeV.png]]</center>
+
<center>[[File:MollerThetaLab_4e7_LH2_11GeV.png]][[File:MollerThetaLab_4e7_LH2_11GeV_Detector.png]]</center>
  
  
<center>[[File:MollerThetaLab_4e7_LH2_11GeV_Detector.png]][[File:MollerThetaLab_4e7_NH3_11GeV_Detector.png]]</center>
+
<center>[[File:MolThetaLab4e8LH211GeV.png]][[File:MolThetaLab4e8LH211GeVDetector.png]]</center>
  
  
 +
<center>[[File:4e7_5cm_LH2_MolThetaLab.png]][[File:4e7_5cm_LH2_MolThetaLab_Detector.png]]</center>
 +
 +
----
 +
 +
For 4e7 incident electrons:
 
<center><math>\sigma = \frac{N_{events}}{N_{incident}\ \rho\ \ell}=\frac{975593}{40000000\ \cdot 4.2\times 10^{-2} barns^{-1}}=\frac{0.024}{4.2\times 10^{-2} barns^{-1}}=0.58 barns</math></center>
 
<center><math>\sigma = \frac{N_{events}}{N_{incident}\ \rho\ \ell}=\frac{975593}{40000000\ \cdot 4.2\times 10^{-2} barns^{-1}}=\frac{0.024}{4.2\times 10^{-2} barns^{-1}}=0.58 barns</math></center>
  
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<center><math>\mathcal{L} \cdot t_{simulated}=\frac{N_{events}}{\sigma}</math></center>
 
<center><math>\mathcal{L} \cdot t_{simulated}=\frac{N_{events}}{\sigma}</math></center>
 
  
 
For a Luminosity of <math>\mathcal{L}=\frac{1.3\times 10^{11}}{barn\cdot s}</math>
 
For a Luminosity of <math>\mathcal{L}=\frac{1.3\times 10^{11}}{barn\cdot s}</math>
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<center><math>t=1.3\times 10^{-5}\ s</math></center>
 
<center><math>t=1.3\times 10^{-5}\ s</math></center>
 +
 +
----
 +
 +
For 4e8 incident electrons:
 +
<center><math>\sigma = \frac{N_{events}}{N_{incident}\ \rho\ \ell}=\frac{9757288}{400000000\ \cdot 4.2\times 10^{-2} barns^{-1}}=\frac{0.024}{4.2\times 10^{-2} barns^{-1}}=0.58 barns</math></center>
 +
 +
 +
 +
<center><math>\sigma=\frac{R_{events}}{\mathcal{L}} \Rightarrow \mathcal{L}=\frac{R_{events}}{\sigma}</math></center>
 +
 +
 +
 +
<center><math>\mathcal{L}=\frac{dN_{events}}{dt}\frac{1}{ \sigma}\Rightarrow \int_{0}^{t_{simulated}}\mathcal {L}\, dt= \int_{0}^{N_{events}}\frac{1}{\sigma}\, dN</math></center>
 +
 +
 +
 +
<center><math>\mathcal{L} \cdot t_{simulated}=\frac{N_{events}}{\sigma}</math></center>
 +
 +
For a Luminosity of <math>\mathcal{L}=\frac{1.3\times 10^{11}}{barn\cdot s}</math>
 +
 +
 +
 +
<center><math>\frac{1.3\times 10^{11}}{barn\cdot s} \cdot t_{simulated}=\frac{9757288}{.58\ barn}</math></center>
 +
 +
 +
 +
<center><math>t=1.3\times 10^{-4}\ s</math></center>
 +
 +
----
 +
 +
For 6e7 incident electrons with a 5cm long LH2 target:
 +
<center><math>\sigma = \frac{N_{events}}{N_{incident}\ \rho\ \ell}=\frac{4584834}{60000000\  \cdot .21barns^{-1}}=\frac{0.076}{.21 barns^{-1}}=.361 barns</math></center>
 +
 +
 +
 +
<center><math>\sigma=\frac{R_{events}}{\mathcal{L}} \Rightarrow \mathcal{L}=\frac{R_{events}}{\sigma}</math></center>
 +
 +
 +
 +
<center><math>\mathcal{L}=\frac{dN_{events}}{dt}\frac{1}{ \sigma}\Rightarrow \int_{0}^{t_{simulated}}\mathcal {L}\, dt= \int_{0}^{N_{events}}\frac{1}{\sigma}\, dN</math></center>
 +
 +
 +
 +
<center><math>\mathcal{L} \cdot t_{simulated}=\frac{N_{events}}{\sigma}</math></center>
 +
 +
For a Luminosity of <math>\mathcal{L}=\frac{1.32\times 10^{11}}{barn\cdot s}</math>
 +
 +
 +
 +
<center><math>\frac{1.32\times 10^{11}}{barn\cdot s} \cdot t_{simulated}=\frac{4584834}{.361\ barn}</math></center>
 +
 +
 +
 +
<center><math>t=9.62\times 10^{-5}\ s</math></center>
 +
 +
 +
----
 +
 +
For 6e6 incident electrons with a 5cm long LH2 target:
 +
<center><math>\sigma = \frac{N_{events}}{N_{incident}\ \rho\ \ell}=\frac{732603}{6000000\  \cdot .21barns^{-1}}=\frac{0.122}{.21 barns^{-1}}=.58 barns</math></center>
 +
 +
 +
 +
<center><math>\sigma=\frac{R_{events}}{\mathcal{L}} \Rightarrow \mathcal{L}=\frac{R_{events}}{\sigma}</math></center>
 +
 +
 +
 +
<center><math>\mathcal{L}=\frac{dN_{events}}{dt}\frac{1}{ \sigma}\Rightarrow \int_{0}^{t_{simulated}}\mathcal {L}\, dt= \int_{0}^{N_{events}}\frac{1}{\sigma}\, dN</math></center>
 +
 +
 +
 +
<center><math>\mathcal{L} \cdot t_{simulated}=\frac{N_{events}}{\sigma}</math></center>
 +
 +
For a Luminosity of <math>\mathcal{L}=\frac{1.32\times 10^{11}}{barn\cdot s}</math>
  
  
===Comparison of GEANT Simulation to Whitney Data===
 
  
 +
<center><math>\frac{1.32\times 10^{11}}{barn\cdot s} \cdot t_{simulated}=\frac{732603}{.58\ barn}</math></center>
  
{| class="wikitable" align="center" border=1
 
|+ '''Rates'''
 
|-
 
| Simulation || GEANT4 ||  Whitney
 
|-
 
| Event Cross-Section (nb) || .61 || 0.079
 
|-
 
|Moller Cross-Section (nb) || 0.58||0.075
 
|-
 
| Length of Target (cm)  || 1 || 5
 
|-
 
| t_{simulated} (s) || 1.3E-5 ||9.54E-5
 
|-
 
| N_{events} || 1026940 || 1001700
 
|-
 
| N_{Moller} || 975593 ||951138
 
|-
 
| N_{incident} || 4E7 || 6E7
 
|}
 
  
  
Moller events occur for about 2.5% of the incident electrons on a LH2 target.  We can assume the number of Moller events that occur within the DC range to be around 30% of the total Moller events occuring for the number of incident electrons for LH2 as well.  Since the differential cross-section over the angel theta is proportional to the differtial cross-section over wire number we can dividing the Moller differential cross-section by the product of the density and length of the target material
+
<center><math>t=9.57\times 10^{-6}\ s</math></center>

Latest revision as of 14:22, 29 March 2018

[math]\textbf{Navigation}[/math]

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\vartriangleright [/math]

https://wiki.iac.isu.edu/index.php/Converting_to_barns

https://wiki.iac.isu.edu/index.php/Check_Differential_Cross-Section

Converting the number of electrons scattered per angle theta to barns, we can use the relation

[math]\mathcal{L}=\frac{R_{scattered}}{\sigma} = \Phi_{beam}\ \rho\ \ell [/math]


[math]\sigma=\frac{R_{scattered}}{\mathcal{L}} =\frac{R_{scattered}}{\Phi_{beam}\ \rho\ \ell}= \frac{R_{scattered}}{R_{incident}\ \rho\ \ell}=\frac{N_{scattered}}{\Delta t}\cdot \frac{\Delta t}{N_{incident}}\cdot \frac{1}{\rho \ell}[/math]

If the time is taken to be the same for the amount scattered as for the amount incident (the time simulated), this can be viewed as the probability of one incident electron producing a Moller event.


[math]\sigma=\frac{N_{scattered}}{N_{incident}}\frac{1}{\rho \ell}[/math]


While this expression has no explicit dependancies on energy, the ratio is a function of the energy, as well as the physical makeup of the target.

This gives,

For 5 cm length of a LH2 target:

[math]\rho_{target}\times l_{target}=\frac{70.85 kg}{1 m^3}\times \frac{1 mole}{2.02 g} \times \frac{1000g}{1 kg} \times \frac{6\times10^{23} molecule}{1 mole} \times \frac{2\ atoms}{molecule}\times \frac{1m^3}{(100 cm)^3} \times \frac{5 cm}{ } \times \frac{1 \times 10^{-24} cm^{2}}{barn} =.21 barns^{-1}[/math]



For 1 cm length of a LH2 target:

[math]\rho_{target}\times l_{target}=\frac{70.85 kg}{1 m^3}\times \frac{1 mole}{2.02 g} \times \frac{1000g}{1 kg} \times \frac{6\times10^{23} molecule}{1 mole} \times \frac{2\ atoms}{molecule}\times \frac{1m^3}{(100 cm)^3} \times \frac{1 cm}{ } \times \frac{1 \times 10^{-24} cm^{2}}{barn} =.042 barns^{-1}[/math]

From earlier simulations for random angle Phi, we know that the full range of Theta is limited depending on the target material.


MollerThetaLab 4e7 LH2 11GeV.pngMollerThetaLab 4e7 LH2 11GeV Detector.png


MolThetaLab4e8LH211GeV.pngMolThetaLab4e8LH211GeVDetector.png


4e7 5cm LH2 MolThetaLab.png4e7 5cm LH2 MolThetaLab Detector.png

For 4e7 incident electrons:

[math]\sigma = \frac{N_{events}}{N_{incident}\ \rho\ \ell}=\frac{975593}{40000000\ \cdot 4.2\times 10^{-2} barns^{-1}}=\frac{0.024}{4.2\times 10^{-2} barns^{-1}}=0.58 barns[/math]


[math]\sigma=\frac{R_{events}}{\mathcal{L}} \Rightarrow \mathcal{L}=\frac{R_{events}}{\sigma}[/math]


[math]\mathcal{L}=\frac{dN_{events}}{dt}\frac{1}{ \sigma}\Rightarrow \int_{0}^{t_{simulated}}\mathcal {L}\, dt= \int_{0}^{N_{events}}\frac{1}{\sigma}\, dN[/math]


[math]\mathcal{L} \cdot t_{simulated}=\frac{N_{events}}{\sigma}[/math]

For a Luminosity of [math]\mathcal{L}=\frac{1.3\times 10^{11}}{barn\cdot s}[/math]


[math]\frac{1.3\times 10^{11}}{barn\cdot s} \cdot t_{simulated}=\frac{975593}{.58\ barn}[/math]


[math]t=1.3\times 10^{-5}\ s[/math]

For 4e8 incident electrons:

[math]\sigma = \frac{N_{events}}{N_{incident}\ \rho\ \ell}=\frac{9757288}{400000000\ \cdot 4.2\times 10^{-2} barns^{-1}}=\frac{0.024}{4.2\times 10^{-2} barns^{-1}}=0.58 barns[/math]


[math]\sigma=\frac{R_{events}}{\mathcal{L}} \Rightarrow \mathcal{L}=\frac{R_{events}}{\sigma}[/math]


[math]\mathcal{L}=\frac{dN_{events}}{dt}\frac{1}{ \sigma}\Rightarrow \int_{0}^{t_{simulated}}\mathcal {L}\, dt= \int_{0}^{N_{events}}\frac{1}{\sigma}\, dN[/math]


[math]\mathcal{L} \cdot t_{simulated}=\frac{N_{events}}{\sigma}[/math]

For a Luminosity of [math]\mathcal{L}=\frac{1.3\times 10^{11}}{barn\cdot s}[/math]


[math]\frac{1.3\times 10^{11}}{barn\cdot s} \cdot t_{simulated}=\frac{9757288}{.58\ barn}[/math]


[math]t=1.3\times 10^{-4}\ s[/math]

For 6e7 incident electrons with a 5cm long LH2 target:

[math]\sigma = \frac{N_{events}}{N_{incident}\ \rho\ \ell}=\frac{4584834}{60000000\ \cdot .21barns^{-1}}=\frac{0.076}{.21 barns^{-1}}=.361 barns[/math]


[math]\sigma=\frac{R_{events}}{\mathcal{L}} \Rightarrow \mathcal{L}=\frac{R_{events}}{\sigma}[/math]


[math]\mathcal{L}=\frac{dN_{events}}{dt}\frac{1}{ \sigma}\Rightarrow \int_{0}^{t_{simulated}}\mathcal {L}\, dt= \int_{0}^{N_{events}}\frac{1}{\sigma}\, dN[/math]


[math]\mathcal{L} \cdot t_{simulated}=\frac{N_{events}}{\sigma}[/math]

For a Luminosity of [math]\mathcal{L}=\frac{1.32\times 10^{11}}{barn\cdot s}[/math]


[math]\frac{1.32\times 10^{11}}{barn\cdot s} \cdot t_{simulated}=\frac{4584834}{.361\ barn}[/math]


[math]t=9.62\times 10^{-5}\ s[/math]



For 6e6 incident electrons with a 5cm long LH2 target:

[math]\sigma = \frac{N_{events}}{N_{incident}\ \rho\ \ell}=\frac{732603}{6000000\ \cdot .21barns^{-1}}=\frac{0.122}{.21 barns^{-1}}=.58 barns[/math]


[math]\sigma=\frac{R_{events}}{\mathcal{L}} \Rightarrow \mathcal{L}=\frac{R_{events}}{\sigma}[/math]


[math]\mathcal{L}=\frac{dN_{events}}{dt}\frac{1}{ \sigma}\Rightarrow \int_{0}^{t_{simulated}}\mathcal {L}\, dt= \int_{0}^{N_{events}}\frac{1}{\sigma}\, dN[/math]


[math]\mathcal{L} \cdot t_{simulated}=\frac{N_{events}}{\sigma}[/math]

For a Luminosity of [math]\mathcal{L}=\frac{1.32\times 10^{11}}{barn\cdot s}[/math]


[math]\frac{1.32\times 10^{11}}{barn\cdot s} \cdot t_{simulated}=\frac{732603}{.58\ barn}[/math]


[math]t=9.57\times 10^{-6}\ s[/math]