Neutron Polarimeter
Analysis of energy dependence [math]\displaystyle{ T_{\gamma}\left( T_n\right) }[/math]
four-vectors algebra
[math]\displaystyle{ E = T + m }[/math] [math]\displaystyle{ E = p^2 + m^2 }[/math]
writing four-vectors:
[math]\displaystyle{ p_{\gamma} = \left( T_{\gamma},\ T_{\gamma},\ 0,\ 0 \right) }[/math] [math]\displaystyle{ p_D = \left( m_D,\ 0,\ 0,\ 0 \right) }[/math] [math]\displaystyle{ p_{n} = \left( E_n,\ p_n\cos(\Theta_n),\ p_n\sin(\Theta_n),\ 0 \right) }[/math] [math]\displaystyle{ p_{p} = \left( E_p,\ p_p\cos(\Theta_p),\ p_p\sin(\Theta_p),\ 0 \right) }[/math]
Doing four-vector algebra:
[math]\displaystyle{ p^{\mu}_{\gamma} + p^{\mu}_D = p^{\mu}_p + p^{\mu}_n \Rightarrow }[/math]
[math]\displaystyle{ 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]\displaystyle{ m_p^2 - m_{\gamma}^2(=0) - m_D^2 - m_n^2 = }[/math]
[math]\displaystyle{ = 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]\displaystyle{ = 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]\displaystyle{ \Theta_n = 90^o }[/math], so
[math]\displaystyle{ 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]\displaystyle{ 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
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]\displaystyle{ T_{\gamma}(0\ MeV) = 1.715360792\ MeV }[/math] [math]\displaystyle{ T_{\gamma}(21\ MeV) = 44.78703086\ MeV }[/math]
So, the equation of the line is:
[math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ \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
After some works:
And it follows:
Now say we have 10 MeV neutron. The corresponding time of flight is:
[math]\displaystyle{ t:=\frac{l}{c\ \beta} = \frac{l\ E}{c\ p} =
\frac{l\ (T+m)}{c\sqrt{T^2+2mT}} = 23\ ns }[/math]
So neutron uncertainty is:
absolute:
[math]\displaystyle{ \delta T_n(\delta t = 1\ ns,\ t=23\ ns,\ l=1\ m) = 0.88\ MeV relative:<br> \frac{\delta T_n}{T_n} = 9% }[/math]
Corresponding photon uncertainty is:
absolute:
[math]\displaystyle{ \delta T_{gamma} relative:<br> <math>\frac{\delta T_{\gamma}}{T_{gamma}} = \frac{\delta T_n}{T_n} = 9% }[/math]
[math]\displaystyle{ \frac{\delta T_{\gamma}}{T_{gamma}} = \frac{\delta T_n}{T_n} = 9% }[/math]