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

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Below is the energy spectrum of two correlated neutrons in LAB frame. 10 million events. Sampled up to 10 MeV.
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[https://wiki.iac.isu.edu/index.php/2n_correlations_in_Photofission go back]
  
[[File:Sum 2n energy correlated.png | 900px]]
 
  
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Follow are some Monte Carlo simulations of 2n correlations. One neutron are from one FF, two neutrons per fission. No accidental neutron was considered, all neutrons are correlated. Total 10M events was simulated.  Every neutron was sampled up to 10 MeV. The polarization of gammas are zero.
  
  
Below is <math>cos(\Theta)</math> between two correlated neutron. No energy cut was made. Integrated over all energy spectrum.
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==Below is the energy spectrum of two correlated neutrons in LAB frame.==
  
[[File:Cos theta 2n correlated.png | 900 px]]
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[[File:Sum 2n energy correlated.png | 700px]]
  
  
  
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.
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==Below is <math>cos(\Theta)</math> between two correlated neutrons. Integrated over all energy spectrum without any energy cut.==
  
[[File:Cos theta 2n correlated energy cut.png | 900 px]]
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[[File:Cos theta 2n correlated.png | 700 px]]
  
  
:<math>Asymmetry = \frac{2n\ \downdownarrows}{2n\ \downpdownarrows}</math>
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Even for the case above (without energy cut) the asymmetry between neutrons emitted anti-parallel and parallel is large and about 9.
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==But let's do the energy cut.==
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[[File:Cos theta 2n correlated energy cut.png | 700 px]]
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The calculated asymmetry from the last plot is:
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:<math>\mbox{Asymmetry} = \frac{\mbox{counts}\ \leftrightarrows}{\mbox{counts}\ \leftleftarrows} = \frac{209,578}{3,219} = 65.10655</math>
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here:
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*<math>\mbox{counts}\ \leftrightarrows</math> are all events with angle between two neutrons is <math>\Theta_{nn} \approx (180\pm25)^o</math>
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*<math>\mbox{counts}\ \leftleftarrows</math> are all events with angle between two neutrons is  <math>\Theta_{nn} \approx (0\pm25)^o</math>
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The yields are:
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:<math>\mbox{Energy}\ \mbox{Cut}\ \mbox{Yield} = \frac{\mbox{counts}\ \mbox{with}\ \mbox{cut}}{\mbox{total}\ \mbox{counts}} = \frac{1,227,505}{10,000,000} = 0.12277</math>
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:<math>\mbox{Asymmetry}\ \mbox{Yield} = \frac{(\mbox{counts}\ \leftrightarrows) + (\mbox{counts}\ \leftleftarrows)}{\mbox{total}\ \mbox{counts}} = \frac{209,578 + 3,219}{10,000,000} = 0.02128</math>
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Looks good, but we need big statistics. We can increase statistics by decreasing the energy cut but we will reduce the asymmetry.

Latest revision as of 17:48, 14 June 2011

go back


Follow are some Monte Carlo simulations of 2n correlations. One neutron are from one FF, two neutrons per fission. No accidental neutron was considered, all neutrons are correlated. Total 10M events was simulated. Every neutron was sampled up to 10 MeV. The polarization of gammas are zero.


Below is the energy spectrum of two correlated neutrons in LAB frame.

Sum 2n energy correlated.png


Below is [math]cos(\Theta)[/math] between two correlated neutrons. Integrated over all energy spectrum without any energy cut.

Cos theta 2n correlated.png


Even for the case above (without energy cut) the asymmetry between neutrons emitted anti-parallel and parallel is large and about 9.


But let's do the energy cut.

Cos theta 2n correlated energy cut.png


The calculated asymmetry from the last plot is:

[math]\mbox{Asymmetry} = \frac{\mbox{counts}\ \leftrightarrows}{\mbox{counts}\ \leftleftarrows} = \frac{209,578}{3,219} = 65.10655[/math]

here:

  • [math]\mbox{counts}\ \leftrightarrows[/math] are all events with angle between two neutrons is [math]\Theta_{nn} \approx (180\pm25)^o[/math]
  • [math]\mbox{counts}\ \leftleftarrows[/math] are all events with angle between two neutrons is [math]\Theta_{nn} \approx (0\pm25)^o[/math]


The yields are:

[math]\mbox{Energy}\ \mbox{Cut}\ \mbox{Yield} = \frac{\mbox{counts}\ \mbox{with}\ \mbox{cut}}{\mbox{total}\ \mbox{counts}} = \frac{1,227,505}{10,000,000} = 0.12277[/math]
[math]\mbox{Asymmetry}\ \mbox{Yield} = \frac{(\mbox{counts}\ \leftrightarrows) + (\mbox{counts}\ \leftleftarrows)}{\mbox{total}\ \mbox{counts}} = \frac{209,578 + 3,219}{10,000,000} = 0.02128[/math]


Looks good, but we need big statistics. We can increase statistics by decreasing the energy cut but we will reduce the asymmetry.