Neutron Polarimeter

Go Back

Four-vector Algebra

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]

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\ 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.

Using the formulas above we can:

1. Predict the threshold of [math]^2D(n,\gamma)[/math] reaction:

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

2. Predict the incident photon energy based on the detected neutron energy.

3. Predict the detected neutron energy based on the 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

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