Defining Occupancy - Revision history

Vanwdani at 21:38, 1 January 2019

2019-01-01T21:38:11Z

Vanwdani: /* Weighted Hits Occupancy */

2019-01-01T21:34:16Z

Weighted Hits Occupancy

Vanwdani: /* Weighted Hits Occupancy */

2019-01-01T21:33:51Z

Weighted Hits Occupancy

Vanwdani: /* Weighted Hits Occupancy */

2019-01-01T21:32:29Z

Weighted Hits Occupancy

Vanwdani: /* Weighted Hits Occupancy */

2019-01-01T21:30:06Z

Weighted Hits Occupancy

Vanwdani: /* Number of Incident Electrons */

2019-01-01T21:27:17Z

Number of Incident Electrons

Vanwdani: /* Lab Cross-Section */

2019-01-01T21:26:59Z

Lab Cross-Section

Vanwdani at 19:57, 29 December 2018

2018-12-29T19:57:59Z

Vanwdani: /* Weighted Hits Occupancy */

2018-12-29T19:56:16Z

Weighted Hits Occupancy

Vanwdani: /* Weighted Hits Occupancy */

2018-12-29T19:20:23Z

Weighted Hits Occupancy

@@ Line 3: / Line 3: @@
 [[Uniform_distribution_in_Energy_and_Theta_LUND_files|<math>\vartriangleleft </math>]]
-[[VanWasshenova_Thesis#Moller_Scattering|<math>\triangle </math>]]
+[[VanWasshenova_Thesis#Detector_Occupancy|<math>\triangle </math>]]
-[[Relativistic_Units|<math>\vartriangleright </math>]]
+[[Unweighted_Occupancy|<math>\vartriangleright </math>]]
 </center>
@@ Line 119: / Line 119: @@
 [[Uniform_distribution_in_Energy_and_Theta_LUND_files|<math>\vartriangleleft </math>]]
-[[VanWasshenova_Thesis#Moller_Scattering|<math>\triangle </math>]]
+[[VanWasshenova_Thesis#Detector_Occupancy|<math>\triangle </math>]]
-[[Relativistic_Units|<math>\vartriangleright </math>]]
+[[Unweighted_Occupancy|<math>\vartriangleright </math>]]
 </center>

← Older revision		Revision as of 21:34, 1 January 2019
Line 105:		Line 105:


−	<center><math>\int\limits_{\theta}^{\theta+d\theta} \int\limits_{0}^{0.2^{\circ}} \frac{\left(3+\left(\cos{\theta}\right)^2\right)^2}{\left(\sin{\theta}\right)^4}\sin{\theta} d\theta=\frac{9}{\pi}\left(\cos{\theta}-2\csc{\frac{\theta}{2}}^2+2\sec{\frac{\theta}{2}}^2\right)\~~big~~\|_{\theta}^{\theta+d\theta}</math></center>	+	<center><math>\int\limits_{\theta}^{\theta+d\theta} \int\limits_{0}^{0.2^{\circ}} \frac{\left(3+\left(\cos{\theta}\right)^2\right)^2}{\left(\sin{\theta}\right)^4}\sin{\theta} d\theta=\frac{9}{\pi}\left(\cos{\theta}-2\csc{\frac{\theta}{2}}^2+2\sec{\frac{\theta}{2}}^2\right)\Big\|_{\theta}^{\theta+d\theta}</math></center>

← Older revision		Revision as of 21:33, 1 January 2019
Line 105:		Line 105:


−	<center><math>\int\limits_{\theta}^{\theta+d\theta} \int\limits_{0}^{0.2^{\circ}} \frac{\left(3+\left(\cos{\theta}\right)^2\right)^2}{\left(\sin{\theta}\right)^4}\sin{\theta} d\theta=\frac{9}{\pi}\left(\cos{\theta}-2\csc{\frac{\theta}{2}}^2+2\sec{\frac{\theta}{2}}^2\right)\~~left~~\|_{\theta}^{\theta+d\theta}</math></center>	+	<center><math>\int\limits_{\theta}^{\theta+d\theta} \int\limits_{0}^{0.2^{\circ}} \frac{\left(3+\left(\cos{\theta}\right)^2\right)^2}{\left(\sin{\theta}\right)^4}\sin{\theta} d\theta=\frac{9}{\pi}\left(\cos{\theta}-2\csc{\frac{\theta}{2}}^2+2\sec{\frac{\theta}{2}}^2\right)\big\|_{\theta}^{\theta+d\theta}</math></center>

← Older revision		Revision as of 21:32, 1 January 2019
Line 105:		Line 105:


−	<center><math>\int\limits_{\theta}^{\theta+d\theta} \int\limits_{0}^{0.2^{\circ}} \frac{\left(3+\left(\cos{\theta}\right)^2\right)^2}{\left(\sin{\theta}\right)^4}\sin{\theta} d\theta=\frac{9}{\pi}\left(\cos{\theta}-2\csc{\frac{\theta}{2}}^2+2\sec{\frac{\theta}{2}}^2\right)\left \|_{\theta}^{\theta+d\theta}</math></center>	+	<center><math>\int\limits_{\theta}^{\theta+d\theta} \int\limits_{0}^{0.2^{\circ}} \frac{\left(3+\left(\cos{\theta}\right)^2\right)^2}{\left(\sin{\theta}\right)^4}\sin{\theta} d\theta=\frac{9}{\pi}\left(\cos{\theta}-2\csc{\frac{\theta}{2}}^2+2\sec{\frac{\theta}{2}}^2\right)\left\|_{\theta}^{\theta+d\theta}</math></center>



−	Here we will use a ~~indefinite~~ integral to find the cross-section in the lab. For a constant <math>\Delta \theta_{lab}=0.1^{\circ}</math> this corresponds to a <math>\Delta \theta_{CM}</math> that will become smaller as <math>\theta_{lab}</math> increases. <math>\Delta \phi=0.2^{\circ}</math> in both frames due to it being Lorentz invariant.	+	Here we will use a definite integral to find the cross-section in the lab. For a constant <math>\Delta \theta_{lab}=0.1^{\circ}</math> this corresponds to a <math>\Delta \theta_{CM}</math> that will become smaller as <math>\theta_{lab}</math> increases. <math>\Delta \phi=0.2^{\circ}</math> in both frames due to it being Lorentz invariant.

← Older revision		Revision as of 21:30, 1 January 2019
Line 105:		Line 105:


−	<center><math>\int\limits_{\theta}^{\theta+d\theta} \int\limits_{0}^{0.2^{\circ}} \frac{\left(3+\left(\cos{\theta}\right)^2\right)^2}{\left(\sin{\theta}\right)^4}\sin{\theta} d\theta=\frac{9}{\pi}\left(\cos{\theta}-2\csc{\frac{\theta}{2}}^2+2\sec{\frac{\theta}{2}}^2\right)</math></center>	+	<center><math>\int\limits_{\theta}^{\theta+d\theta} \int\limits_{0}^{0.2^{\circ}} \frac{\left(3+\left(\cos{\theta}\right)^2\right)^2}{\left(\sin{\theta}\right)^4}\sin{\theta} d\theta=\frac{9}{\pi}\left(\cos{\theta}-2\csc{\frac{\theta}{2}}^2+2\sec{\frac{\theta}{2}}^2\right)\left \|_{\theta}^{\theta+d\theta}</math></center>

← Older revision		Revision as of 21:27, 1 January 2019
Line 110:		Line 110:

	Here we will use a indefinite integral to find the cross-section in the lab. For a constant <math>\Delta \theta_{lab}=0.1^{\circ}</math> this corresponds to a <math>\Delta \theta_{CM}</math> that will become smaller as <math>\theta_{lab}</math> increases. <math>\Delta \phi=0.2^{\circ}</math> in both frames due to it being Lorentz invariant.		Here we will use a indefinite integral to find the cross-section in the lab. For a constant <math>\Delta \theta_{lab}=0.1^{\circ}</math> this corresponds to a <math>\Delta \theta_{CM}</math> that will become smaller as <math>\theta_{lab}</math> increases. <math>\Delta \phi=0.2^{\circ}</math> in both frames due to it being Lorentz invariant.
−
−	~~===Number of Incident Electrons===~~
−	~~Solving for the number of incident electrons that would be required to produce 1 scattered electron (event),~~
−
−	~~<center><math>0.000000020919615=\frac{1\,e^{-}}{N_{in}\rho l}\rightarrow N_{in}=226,549,890.755\,e^{-}</math> for a 5cm long LH2 target.</center>~~
−
−
−	~~Near the maximum of the DC, this corresponds to more events.~~
−
−	~~<pre>Labθ=39.8 Labdσ=0.000000341148416 ΔLabθ=0.1 CMθ=178.674400001996105 CMdσ=0.000000341148416 ΔCMθ=0.00008</pre>~~
−
−	~~<center><math>0.000000341148416=\frac{N_{out}}{N_{in}\rho l}\rightarrow N_{out}=16.3\,e^{-}</math></center>~~

← Older revision		Revision as of 21:26, 1 January 2019
Line 109:		Line 109:


−	Here we will use a indefinite integral to find the cross-section in the lab. For a constant <math>\Delta \theta_{lab}=0.1^{\circ}</math> this corresponds to a <math>\Delta \theta_{CM}</math> that will become smaller as <math>\theta_{lab}</math> increases. <math>\Delta \phi=0.2^{\circ}</math> in both frames due to it being Lorentz invariant. ~~Using the Lorentz boost vectors in the LUND file generator we find:~~	+	Here we will use a indefinite integral to find the cross-section in the lab. For a constant <math>\Delta \theta_{lab}=0.1^{\circ}</math> this corresponds to a <math>\Delta \theta_{CM}</math> that will become smaller as <math>\theta_{lab}</math> increases. <math>\Delta \phi=0.2^{\circ}</math> in both frames due to it being Lorentz invariant.
−
−	~~<pre>Labθ=5.0 Labdσ=0.000000020919615 ΔLabθ=0.1 CMθ=167.426100001950005 CMdσ=0.000000020919615 ΔCMθ=0.00429</pre>~~

	===Number of Incident Electrons===		===Number of Incident Electrons===

@@ Line 1: / Line 1: @@
 =General Occupancy=
 The occupancy measures the number of particles interactions per a detector cell per an event.  For the CLAS12 drift chamber, there are 112 wires on each layer, with 12 layers within a region, giving 1344 cells.  This can simply be defined as the "Unweighted Occupancy" for the CLAS12 DC and follows the equation:
@@ Line 115: / Line 124: @@
 <center><math>0.000000341148416=\frac{N_{out}}{N_{in}\rho l}\rightarrow N_{out}=16.3\,e^{-}</math></center>

← Older revision		Revision as of 19:56, 29 December 2018
Line 100:		Line 100:


−	Here we will use a indefinite integral to find the cross-section in the lab. For a constant <math>\Delta \theta_{lab}=0.1^{\circ}</math> this corresponds to a <math>\Delta \theta_{CM}</math> that will become smaller as <math>\theta_{lab}</math> increases. <math>\Delta \phi=0.2^{/circ}</math> in both frames due to it being Lorentz invariant. Using the Lorentz boost vectors in the LUND file generator we find:	+	Here we will use a indefinite integral to find the cross-section in the lab. For a constant <math>\Delta \theta_{lab}=0.1^{\circ}</math> this corresponds to a <math>\Delta \theta_{CM}</math> that will become smaller as <math>\theta_{lab}</math> increases. <math>\Delta \phi=0.2^{\circ}</math> in both frames due to it being Lorentz invariant. Using the Lorentz boost vectors in the LUND file generator we find:

	<pre>Labθ=5.0 Labdσ=0.000000020919615 ΔLabθ=0.1 CMθ=167.426100001950005 CMdσ=0.000000020919615 ΔCMθ=0.00429</pre>		<pre>Labθ=5.0 Labdσ=0.000000020919615 ΔLabθ=0.1 CMθ=167.426100001950005 CMdσ=0.000000020919615 ΔCMθ=0.00429</pre>

← Older revision		Revision as of 19:20, 29 December 2018
Line 96:		Line 96:


−	<center><math>\~~int_~~{\theta}^{\theta+d\theta} \~~int_~~{0}^{0.2^{\circ}} \frac{\left(3+\left(\cos{\theta}\right)^2\right)^2}{\left(\sin{\theta}\right)^4}\sin{\theta} d\theta=~~\left(~~\frac{9}{\pi}\cos{\theta}-2\csc{\frac{\theta}{2}}^2+2\sec{\frac{\theta}{2}}^2\right)</math></center>	+	<center><math>\int\limits_{\theta}^{\theta+d\theta} \int\limits_{0}^{0.2^{\circ}} \frac{\left(3+\left(\cos{\theta}\right)^2\right)^2}{\left(\sin{\theta}\right)^4}\sin{\theta} d\theta=\frac{9}{\pi}\left(\cos{\theta}-2\csc{\frac{\theta}{2}}^2+2\sec{\frac{\theta}{2}}^2\right)</math></center>