Difference between revisions of "Neutron Polarimeter"

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The main problem here we need the connection between the neutron kinetic energy and time of flight
 
The main problem here we need the connection between the neutron kinetic energy and time of flight
  
After some relativistic algebra we can derive:
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After some work:
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 +
[[File:formula0.png]]
 +
 
 +
and it follows that:
  
 
[[File:formula1.png]]
 
[[File:formula1.png]]
 +
  
  
  
 
[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]
 
[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]

Revision as of 21:15, 16 June 2010

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Analysis of energy dependence [math]T_{\gamma}\left( T_n\right)[/math]

four-vectors algebra

Collision.png
[math] E = T + m[/math]
[math] E = p^2 + m^2[/math]
writing four-vectors:
[math] p_{\gamma} = \left( T_{\gamma},\ T_{\gamma},\ 0,\ 0 \right) [/math] [math] p_D = \left( m_D,\ 0,\ 0,\ 0 \right) [/math] [math] p_{n} = \left( E_n,\ p_n\cos(\Theta_n),\ p_n\sin(\Theta_n),\ 0 \right) [/math] [math] p_{p} = \left( E_p,\ p_p\cos(\Theta_p),\ p_p\sin(\Theta_p),\ 0 \right) [/math]


Doing four-vector algebra:
[math] p^{\mu}_{\gamma} + p^{\mu}_D = p^{\mu}_p + p^{\mu}_n \Rightarrow [/math]
[math] p^{\mu\ 2}_p = \left(p^{\mu}_{\gamma} + p^{\mu}_D - p^{\mu}_n\right)^2 = p^{\mu\ 2}_{\gamma} + p^{\mu\ 2}_D + p^{\mu\ 2}_n + 2\ p^{\mu}_{\gamma}\ p^{\mu}_D - 2\ p^{\mu}_{\gamma}\ p^{\mu}_n - 2\ p^{\mu}_D\ p^{\mu}_n [/math]
[math] m_p^2 - m_{\gamma}^2(=0) - m_D^2 - m_n^2 = [/math]
[math] = 2\ T_{\gamma}\ m_D - 2\left( T_{\gamma}\ E_n - T_{\gamma}\ p_n\cos(\Theta_n)\right) - 2\ m_D\ E_n [/math]
[math] = 2\ T_{\gamma}\left( m_D - E_n + p_n\cos(\Theta_n) \right) - 2\ m_D\ E_n [/math]

Detector is located at [math]\Theta_n = 90^o[/math], so

[math] T_{\gamma} = \frac {2\ m_D\ E_n + m_p^2 - m_D^2 - m_n^2} {2\left( m_D - E_n \right)} =
                     \frac {2\ m_D\ (T_n + m_n) + m_p^2 - m_D^2 - m_n^2} {2\left( m_D - (T_n + m_n) \right)}[/math]

and visa versa

 [math] T_n = \frac {2\ T_{\gamma}\ m_D + m_D^2 + m_n^2 - m_p^2} {2\left( T_{\gamma} + m_D \right)} - m_n[/math]

how it looks

Kinetic energy 0 900 MeV.jpeg Kinetic energy 0 21 MeV.jpeg

low energy approximation

As we can see from Fig.2 for low energy neutrons (0-21 MeV)
energy dependence of incident photons is linear
Find that dependence. We have:
[math] T_{\gamma}(0\ MeV) = 1.715360792\ MeV [/math] [math] T_{\gamma}(21\ MeV) = 44.78703086\ MeV [/math]
So, the equation of the line is:
[math] T_{\gamma} = \frac{T_{\gamma}(21\ MeV) - T_{\gamma}(0\ MeV)}{21\ MeV - 0\ MeV}\ T_n + T_{\gamma}(0\ MeV) [/math]
Finally for low energy neutrons (0-21 MeV):
[math] T_{\gamma} = 2.051\ T_n + 1.715 [/math]

example of error analysis

example 1

Say, we have, 10 MeV neutron with uncertainty 1 MeV, the corresponding uncertainly for photons energy is:

[math] \delta T_{\gamma} = 2.051\ \delta T_n = 2.051\times 1\ MeV = 2.051\ MeV [/math]

example 2

Say, we have, 1 meter away detector with 1 ns time of flight neutron uncertainly

The main problem here we need the connection between the neutron kinetic energy and time of flight

After some work:

Formula0.png

and it follows that:

Formula1.png



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