Difference between revisions of "Weight"

From New IAC Wiki
Jump to navigation Jump to search
 
Line 88: Line 88:
 
<center><math>\textbf{\underline{Navigation}}</math>
 
<center><math>\textbf{\underline{Navigation}}</math>
  
[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\vartriangleleft </math>]]
+
[[Lorentz_Transformation_to_Lab_Frame|<math>\vartriangleleft </math>]]
[[VanWasshenova_Thesis#Determining_wire-theta_correspondence|<math>\triangle </math>]]
+
[[VanWasshenova_Thesis#Weighted_Isotropic_Distribution_in_Lab_Frame|<math>\triangle </math>]]
 
[[CED_Verification_of_DC_Angle_Theta_and_Wire_Correspondance|<math>\vartriangleright </math>]]
 
[[CED_Verification_of_DC_Angle_Theta_and_Wire_Correspondance|<math>\vartriangleright </math>]]
  
 
</center>
 
</center>

Latest revision as of 15:01, 30 May 2017

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

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


4.1.5 Weight

Using the theoretical differential cross section from previous

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


[math]\alpha ^2=5.3279\times 10^{-5}[/math]


[math]E\approx 53.013 MeV[/math]

We can take the Moller electron distribution of Theta in the Center of Mass frame, and multiply each given angle Theta by the expected differential cross section.

MolThetaCM spread.png


This causes the Moller Theta distribution in the Center of Mass frame to directly follow the theoretical differential cross section.

MolThetaCMWeightedTheory LH2 11GeV.png


The Lab frame distribution of Theta can also be weighted similarly. However, instead of having it be a differential cross section, we can find the necessary number of particles.

[math]\frac{d\sigma}{d\Omega} = \frac{\left(\frac{number\ of\ particles\ scattered/second}{d\Omega}\right)}{\left(\frac{number\ of\ incoming\ particles/second}{cm^2}\right)}=\frac{dN}{\mathcal L\, d\Omega} =differential\ scattering\ cross\ section[/math]


[math]\frac{d\sigma}{d\Omega} =\frac{dN}{\mathcal L\, d\Omega} =\frac{dN}{\Phi \rho \ell\, d\Omega}[/math]


[math]\Rightarrow \int \frac{d\sigma}{d\Omega} \rho\ \ell \Phi d\Omega=\int dN[/math]



Checking versus the given Luminosity for the experiment


[math]\mathcal L = \frac{1.33 \times 10^{35}}{cm^2\cdot s} \times \frac{10^{-24} cm^2}{barn}=1.33\times 10^{11} barns^{-1}s^{-1}[/math]


[math]\mathcal L \int \frac{d\sigma}{d\Omega} d\Omega = 1.33 \times 10^{11} barns^{-1} \times 2\pi\ \int \frac{d\sigma}{d\Omega}\ \sin{\theta} d\theta=N[/math]


Limiting the range of Theta to within 5 to 40 degrees in the Lab frame:

DetectorRangeLab.png


We can find the corresponding angular range in the CM frame:

DetectorRangeCM.png


Integrating the differential cross-section over the solid angle:

DetectorIntegrationXSect LH2 11GeV.png
IntegralDiffXSect->Integral()

4.97493824519086629e+04 barns


Multiplying by the Luminosity, we find:

[math]4.97\times 10^{4} barns \times\frac{1.35\times 10^{11}}{barns\cdot s}=6.71\times 10^{15}\frac{electrons}{s}[/math]





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

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