Neutron Polarimeter

From New IAC Wiki
Jump to navigation Jump to search

Go Back


Four-vector Algebra

Collision.png


Let's do the four-vector algebra:

[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] 

By conservation of four-momentum:

[math] p^{\mu}_{\gamma} + p^{\mu}_D = p^{\mu}_p + p^{\mu}_n\ \ \Rightarrow \ \  p^{\mu}_p = p^{\mu}_{\gamma} + p^{\mu}_D - p^{\mu}_n[/math]

Squaring both side of equation above and using the four-momentum invariants [math](p^{\mu})^2 = m^2[/math] we have:

[math] m_p^2 = m_D^2 + m_n^2 +  2T_{\gamma}\cdot m_D  - 2m_D\cdot E_n - 2\left( T_{\gamma}E_n - T_{\gamma}p_n\cos\Theta_n\right) [/math]


Energy dependence of [math]E_\gamma[/math] as function of neutron energy [math]E_n[/math] and neutron angle [math]\Theta_n[/math]

We can easily solve the equation above with respect to incident photon energy:

[math]T_{\gamma} = \frac {m_p^2 - m_n^2 - m_D^2 + 2\ m_D E_n}{2\ (m_D - E_n + p_n cos\Theta_n)} [/math]

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

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

For non-relativistic neutrons [math]T_n \ll m_n = 939.5\ MeV[/math], and we can neglect the [math]T_n[/math] in denominator, so

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


Substituting the corresponding masses, we get finally for the case of non-relativistic neutrons detected at the angle [math]\Theta_n = 90^o[/math]:

[math]T_{\gamma}\ [MeV] = 2.003\ T_n\ [MeV] + 1.577\ [MeV][/math]

and visa versa:

[math]T_n\ [MeV] = 0.499\ T_{\gamma}\ [MeV] - 0.787\ [MeV][/math]


Using the formulas above we can:

1. Find the incident photon energy based on detected neutron energy.


2. The threshold for [math]^2D(n,\gamma)[/math] reaction is [math]E_{\gamma} = 1.577\ MeV[/math]


3. Predict the detected neutron energy based of incident photon energy.


  For the incident photons up to 25 MeV we have (0 - 11.69) MeV neutrons


  For the incident photons up to 44 MeV we have (0 - 21.17) MeV neutrons



Also note:

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, the neutron's time of flight uncertainly is 1 ns

The neutron's kinetic energy is:

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

And:

[math]\delta T_n \left(\delta t\right) = -\ \frac{m\ l^2 \delta t}{\left(1-\left(\frac{l}{c\ t}\right)^2\right)^{3/2}c^2 t^3}[/math]


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

[math]t:=\frac{l}{c\ \beta_n} = \frac{l}{c\ (p_n/E_n)} =
         \frac{l\ (T_n + m_n)}{c\sqrt{T_n^2+2m_nT_n}}[/math]


Say, we have 10 MeV neutron, 1 m away detector, and neutron's time of flight uncertainty is 1 ns. Then the neutron time of flight is:


[math]t(T_n = 10\ MeV) = 23\ ns[/math]   


Neutron kinetic energy errors are:

[math]\delta T_n(\delta t = 1\ ns) = 0.88\ MeV[/math]
[math]\frac{\delta T_n}{T_n} = \frac{0.88\ MeV}{10\ MeV} = 8.8\ %[/math] 


And photon energy errors are:

[math]\delta T_{\gamma} = 2.051\cdot \delta T_n = 2.051\cdot 0.88\ MeV = 1.81\ MeV [/math] 
[math]\frac{\delta T_{\gamma}}{T_{\gamma}} = \frac{1.80\ MeV}{(2.051\cdot 10 + 1.715)\ MeV} = 8.1\ %[/math]


Below are some examples for different detector distance and neutron kinetic energy:

[math]\delta t_n[/math] [math]l[/math] [math]T_n[/math] [math]\beta_n[/math] [math]t_n[/math] [math]\delta T_n[/math] [math]\frac{\delta T_n}{T_n}[/math] [math]\delta T_{\gamma}[/math] [math]\frac{\delta T_{\gamma}}{T_{\gamma}}[/math]
1 ns 1 m 5 MeV 0.103 32 ns 0.31 MeV 6.2 % 0.64 MeV 5.3 %
1 ns 1 m 10 MeV 0.145 23 ns 0.88 MeV 8.8 % 1.81 MeV 8.1 %
1 ns 1 m 20 MeV 0.203 16 ns 2.51 MeV 12.6 % 5.16 MeV 12.1 %
1 ns 1 m 0.5 MeV 0.033 102 ns 0.010 MeV 1.9 %
1 ns 1 m 1 MeV 0.046 72 ns 0.028 MeV 2.8 %
1 ns 1 m 2 MeV 0.065 51 ns 0.078 MeV 3.9 %
1 ns 1 m 4 MeV 0.092 36 ns 0.22 MeV 5.5 %




Go Back