Difference between revisions of "Neutron Polarimeter"

From New IAC Wiki
Jump to navigation Jump to search
Line 85: Line 85:
  
 
Some results are:
 
Some results are:
 +
 +
 +
{| border="1" cellpadding="20" cellspacing="0"
 +
|-
 +
!width=40|collimator diameter
 +
|detector distance
 +
|neutron energy
 +
|time of flight uncertainty
 +
|neutron <math>\beta<math>
 +
|neutron time of fligh
 +
|neutron absolute error
 +
|neutron relative error
 +
|photon absolute error
 +
|photon relatibe error
 +
|-
 +
|1 m||20 MeV||1 ns||4.79 cm||75 cm||7.49 cm
 +
|-
 +
|}
 +
 +
 +
 +
 +
 +
  
  

Revision as of 04:28, 17 June 2010

Go Back


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 calculation

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, neutron with time of flight uncertainly is 1 ns

The neutron's kinetic energy as function of the neutron's time of flight is:

[math]T_n = m (\gamma - 1) = m\left[ \frac{1}{\sqrt{1-\left(\frac{l}{c\ t}\right)^2}} - 1 \right][/math]

And it follows, that neutron's kinetic energy error as function of the neutron's time of flight error is:

Formula2.png

Also we need neutron time of flight as function of neutron kinetic energy:

[math]t:=\frac{l}{c\ \beta} = \frac{l\ E}{c\ p} =
         \frac{l\ (T+m)}{c\sqrt{T^2+2mT}} = 23\ ns[/math]

Some results are:


collimator diameter detector distance neutron energy time of flight uncertainty neutron [math]\beta\lt math\gt |neutron time of fligh |neutron absolute error |neutron relative error |photon absolute error |photon relatibe error |- |1 m||20 MeV||1 ns||4.79 cm||75 cm||7.49 cm |- |} So neutron uncertainty is: absolute:\lt br\gt \lt math\gt \delta T_n(\delta t = 1\ ns,\ t=23\ ns,\ l=1\ m) = 0.88\ MeV[/math]
relative:
[math]\frac{\delta T_n}{T_n} = \frac{0.88\ MeV}{10\ MeV} = 9%[/math]


Corresponding photon uncertainty is:

absolute:
[math]\delta T_{\gamma} = 2.051\cdot \delta T_n = 2.051\cdot 0.88\ MeV = 1.80\ MeV [/math]
relative:
[math]\frac{\delta T_{\gamma}}{T_{\gamma}} = \frac{1.80\ MeV}{(2.051\cdot 10 + 1.715)\ MeV} = 8%[/math]


Go Back