Neutron Polarimeter

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Four-vector Algebra

Consider the two bode reaction [math]D(\gamma, n)p[/math]:

Write down all 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] 

Now apply the 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]

Detector located at [math]\Theta = 90^o[/math] case

Detector is located at [math]\Theta_n = 90^o[/math], and the formula above is simplified:

[math] m_p^2 = m_D^2 + m_n^2 +  2\ T_{\gamma}\cdot m_D  - 2\ m_D\cdot E_n - 2\ T_{\gamma}E_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)}
                  = \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 the formula above is become:

[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:

[math]T_{\gamma}\ [MeV] = 2.003\ T_n\ [MeV] + 1.577\ [MeV]\ \ \ \ [1][/math]

and visa versa:

[math]T_n\ [MeV] = 0.499\cdot T_{\gamma}\ [MeV] - 0.787\ [MeV]\ \ \ \ [2][/math]

Here I derived the formula [2] just inversing the formula [1]. I can as well start from exact solution above, solve this equation with respect to neutron energy, do the non-relativistic approximation and get exactly the same formula [2]. But anyway we ended up with two useful non-relativistic formulas we can analyze now:

1) from formula [1] above we can predict the threshold of [math]^2D(n,\gamma)[/math] reaction:

[math]E_{\gamma} = 1.577\ MeV[/math]

2) from formula [1] above we can predict the incident photon energy based on the detected neutron energy (neutron polarimeter).

3) from formula [2] above we can predict the detected neutron energy based on the incident photon energy.

 -  for the incident photons up to [math]25\ MeV[/math] we can detect neutrons up to  [math]11.69\ MeV[/math]

 -  for the incident photons up to [math]44\ MeV[/math] we can detect neutrons up to  [math]21.17\ MeV[/math]

4) we can do the error calculations.

Example of error calculation (need to be updated to the formulas [1] and [2] above)

example 1

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

[math] \delta T_{\gamma} = 2.003\ \delta T_n = 2.003\times 1\ MeV = 2.003\ MeV [/math]

example 2

In the calculations below I attempted to predict the uncertainly in photons energy based on uncertainly of neutrons time of flight. 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]

So the relative error in neutron energy is:

[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]

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 %

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