Occupancy for Sector 1

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

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

A bash script to run the GEMC simulations is created. tcsh scripts to run root2evio on lds2 is called using sshpass. The lds2 scripts use sshfs The main script on lds3:

BUILD_GEMC_SIMULATION.sh

The 3 scripts on lds2:

first_commands.tcsh

second_commands.tcsh

last_commands.tcsh

LUND File Output

0.1 degree spacing in the Lab frame. CM Frame is not evenly spaced.

Applying the weight

Looking at the angles and the associated weight, we can find the sums

Once_Angles_and_weight=3399.930890560805437

Total_Angles_and_weight=1023379.198058736044914

Checking with Mathematica

"Integrating" with Cosine term

Finding the Cross Section

Performing a Riemann sum for [math]-30^{\circ} \lt \phi \lt 30^{\circ}[/math]

The cross section should be equal between both frames since the number of particles is an invariant. The differential cross section must differ between frames since the solid angle does vary.

[math]\sigma_{(CM)}=\sigma{(Lab)}[/math]

[math]\frac{d\sigma}{d\Omega}_{(CM)} d\Omega_{(CM)}=\frac{d\sigma}{d\Omega}_{(Lab)} d\Omega_{(Lab)}[/math]

[math]\frac{d\sigma}{d\Omega}_{(CM)} \sin \theta_{(CM)}\ d\theta_{(CM)}\ d\phi=\frac{d\sigma}{d\Omega}_{(Lab)} \sin \theta_{(Lab)}\ d\theta_{(Lab)}\ d\phi[/math]

[math]\rightarrow \frac{d\sigma}{d\Omega}_{(Lab)}=\frac{d\sigma}{d\Omega}_{(CM)} \frac{\sin \theta_{(CM)}\ d\theta_{(CM)}\ d\phi}{ \sin \theta_{(Lab)}\ d\theta_{(Lab)}\ d\phi}[/math]

[math]\rightarrow d\sigma_{(Lab)}=\frac{d\sigma}{d\Omega}_{(CM)} \frac{\sin \theta_{(CM)}\ d\theta_{(CM)}\ d\phi}{ \sin \theta_{(Lab)}\ d\theta_{(Lab)}\ d\phi}\sin \theta_{(Lab)} d\theta_{(Lab)}\ d\phi[/math]

Adjust for DC Sector 1 Limits

GEMC Cross Section

Only taking GEMC hits in Sector 1 with Track ID of the mother of the FP equal to zero:

[math]\frac{0.009731\ barn}{0.013924\ barn}=70\%[/math]Efficiency

Taking GEMC hits with ANY Track ID of the mother of the FP :

[math]\frac{0.012433\ barn}{0.013924\ barn}=90\%[/math]Efficiency

Using the Cross Section

If we make the assumption that the beam of incoming electrons is a flux over an area for a given time,

[math]N_{incident}=\Phi\ A_{beam}\ t_{run} \rightarrow dN_{incident}=\Phi\ dA_{beam}\ t_{run}\rightarrow\ \frac{dN_{incident}}{ dA_{beam}}=\Phi\ t_{run}[/math]

Using the definition of the differential cross section:

[math]\frac{d\sigma}{d\Omega}\equiv \frac{ \Biggl(\frac{dN_{scattered}}{d\Omega} \Biggr)}{\Biggl(\frac{dN_{incident}}{dA}\Biggr)}\rightarrow \frac{d\sigma}{d\Omega}\Biggl(\frac{dN_{incident}}{dA}\Biggr)=\Biggl(\frac{dN_{scattered}}{d\Omega} \Biggr)[/math]

Substituting using the flux

[math] \frac{d\sigma}{d\Omega}\Biggl(\frac{dN_{incident}}{dA}\Biggr)=\Biggl(\frac{dN_{scattered}}{d\Omega} \Biggr)\rightarrow \frac{d\sigma}{d\Omega}\Phi\ t_{run}=\Biggl(\frac{dN_{scattered}}{d\Omega} \Biggr)[/math]

[math]\rightarrow dN_{scattered}= \frac{d\sigma}{d\Omega}\Phi d\Omega= \frac{d\sigma}{d\Omega}\Phi\ t_{run}\ \sin \theta\ d\theta\ d\phi[/math]

Since the differential cross section is known in the Center of Mass frame of reference, but measurements are taken in the Lab Frame, a transformation must occur.

[math]\rightarrow dN_{scattered}= \frac{d\sigma}{d\Omega_{Lab}}\Phi\ t\ \sin \theta_{Lab}\ d\theta_{Lab}\ d\phi_{Lab}[/math]

[math]\frac{d\sigma}{d\Omega_{Lab}}\sin \theta_{Lab}\ d\theta_{Lab}\ d\phi_{Lab}=\frac{d\sigma}{d\Omega_{CM}}\sin \theta_{CM}\ d\theta_{CM}\ d\phi_{CM}[/math]

[math]\frac{d\sigma}{d\Omega_{Lab}}=\frac{d\sigma}{d\Omega_{CM}}\frac{\sin \theta_{CM}\ d\theta_{CM}\ d\phi_{CM}}{\sin \theta_{Lab}\ d\theta_{Lab}\ d\phi_{Lab}}[/math]

[math]\rightarrow dN_{scattered}=\frac{d\sigma}{d\Omega_{CM}}\frac{\sin \theta_{CM}\ d\theta_{CM}\ d\phi_{CM}}{\sin \theta_{Lab}\ d\theta_{Lab}\ d\phi_{Lab}}\Phi\ t_{run}\ \sin \theta_{Lab}\ d\theta_{Lab}\ d\phi_{Lab}[/math]

If we divide both sides by time

[math]\rightarrow \frac{dN_{scattered}}{t_{run}}=\frac{d\sigma}{d\Omega_{CM}}\frac{\sin \theta_{CM}\ d\theta_{CM}\ d\phi_{CM}}{\sin \theta_{Lab}\ d\theta_{Lab}\ d\phi_{Lab}}\Phi \sin \theta_{Lab}\ d\theta_{Lab}\ d\phi_{Lab}[/math]

[math]\rightarrow \frac{1}{t_{run}}\int dN_{scattered}=\frac{N_{scattered}}{t_{run}}=\iint\limits_{\Omega}\frac{d\sigma}{d\Omega_{CM}}\frac{\sin \theta_{CM}\ d\theta_{CM}\ d\phi_{CM}}{\sin \theta_{Lab}\ d\theta_{Lab}\ d\phi_{Lab}}\Phi \sin \theta_{Lab}\ d\theta_{Lab}\ d\phi_{Lab}[/math]

Expressing the differential cross section as a function of the Center of Mass scattering angle Theta, since it does not depend on Phi or the radius

[math]\frac{d\sigma}{d\Omega_{CM}}=\sigma(\theta_{CM})[/math]

We can approximate the integral over the solid angle by using a left handed Riemann sum. A right handed sum would produce an overestimate.

[math]\rightarrow \frac{1}{t_{run}}\int dN_{scattered}=\frac{N_{scattered}}{t_{run}}=\sum_{\theta}\sum_{\phi}\sigma(\theta_{CM})\frac{\sin \theta_{CM}\ \Delta\theta_{CM}\ \Delta\phi_{CM}}{\sin \theta_{Lab}\ \Delta\theta_{Lab}\ \Delta\phi_{Lab}}\Phi \sin \theta_{Lab}\ \Delta\theta_{Lab}\ \Delta\phi_{Lab}[/math]

Occupancy

LH2_NOSol_0Tor_11GeV_IsotropicPhi_v2_6_ShieldOut

Run

./BUILD_GEMC_SIMULATION.sh 

DVMacro

Clas12Mon

Create hipo file

Move hipo file to clas12mon folder

mv LH2_NOSol_0Tor_11GeV_IsotropicPhi_v2_6_ShieldOut.hipo ~/clas12mon

Run monitor program

./README

Load hipo file

"Press H for hipo"
"Press play"
"Switch to 

For [math]5^{\circ}\gt \theta\lt 40^{\circ}\ -30^{\circ}\gt \phi\lt 30^{\circ}[/math]

FOR DC Limits

Calculating

[math]\sigma_{Moller\ (\theta_{lab} = 5^{\circ}-40^{\circ})}=0.86\ barn[/math]

[math]t_{simulated}=\frac{N_{events}}{\sigma_{events} \Phi \rho \ell}=\frac{96105\ barn \cdot s}{7.87\times 10^{-2} \cdot 1.33\times 10^{11}\ barn}=9.3\times 10^{-6} s[/math]

[math]N_0=\Delta t \cdot R_{events}=\Delta t \cdot \frac{N_{events}}{t_{simulated}}=250\times 10^{-9}\ s \cdot \frac{98181}{9.3\times 10^{-6}\ s}=2639[/math]

[math]Occupancy=\frac{N_{hits}}{N_0}=\frac{N_{hits}}{\Delta t \cdot R_{events}}=\frac{t_{simulated}\cdot N_{hits}}{N_{events}\cdot \Delta t}=[/math]