Difference between revisions of "Cos(Theta) between two correlated neutrons. Unpolarized case."

From New IAC Wiki
Jump to navigation Jump to search
Line 1: Line 1:
 
Below is the energy spectrum of two correlated neutrons in LAB frame. 10 million events. Sampled up to 10 MeV.
 
Below is the energy spectrum of two correlated neutrons in LAB frame. 10 million events. Sampled up to 10 MeV.
  
[[File:Sum 2n energy correlated.png | 900px]]
+
[[File:Sum 2n energy correlated.png | 700px]]
  
  
Line 7: Line 7:
 
Below is <math>cos(\Theta)</math> between two correlated neutron. No energy cut was made. Integrated over all energy spectrum.
 
Below is <math>cos(\Theta)</math> between two correlated neutron. No energy cut was made. Integrated over all energy spectrum.
  
[[File:Cos theta 2n correlated.png | 900 px]]
+
[[File:Cos theta 2n correlated.png | 800 px]]
  
  
 +
And with <math>E_n > 2\ MeV</math> energy cut.
  
Below is <math>cos(\Theta)</math> between two correlated neutron. Energy cut was made for each neutrons <math>E_n > 2\ MeV</math>. Integrated over all energy spectrum.
+
[[File:Cos theta 2n correlated energy cut.png | 800 px]]
 
 
[[File:Cos theta 2n correlated energy cut.png | 900 px]]
 
  
  
Line 23: Line 22:
  
  
We need bis statistics!!!
+
Looks good, but we need big statistics!

Revision as of 17:00, 10 June 2011

Below is the energy spectrum of two correlated neutrons in LAB frame. 10 million events. Sampled up to 10 MeV.

Sum 2n energy correlated.png


Below is [math]cos(\Theta)[/math] between two correlated neutron. No energy cut was made. Integrated over all energy spectrum.

Cos theta 2n correlated.png


And with [math]E_n \gt 2\ MeV[/math] energy cut.

Cos theta 2n correlated energy cut.png


[math]Asymmetry = \frac{counts\ cos(\Theta)_{nn} \lt -0.9}{counts\ cos(\Theta)_{nn} \gt 0.9} = frac{209,578}{3,219} = 65.11[/math]
[math]Energy\ Cut\ Yield \frac{counts\ with\ cut}{counts\ without\ cut} = \frac{1,227,505}{10,000,000} = 0.12277[/math]
[math]Asymmetry\ Yield \frac{(counts\ parallel) + (counts\ antiparticle)}{event\ without\ cut} = \frac{209,578 + 3,219}{10,000,000} = 0.02128[/math]


Looks good, but we need big statistics!