Difference between revisions of "Neutron Polarimeter"

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Say, we have 10 MeV neutron, 1.5 m away detector, and neutron's time of flight uncertainty is <math>1\ ns</math>. Then the neutron time of flight is:
+
Say, we have 10 MeV neutron, 1.5 m away detector, and neutron's time of flight uncertainty is <math>1\ ns</math>.
 +
 
 +
 
 +
 
 +
Using the formulas above we can find:
  
 
  <math>t(T_n = 10\ MeV) = 35\ ns</math>   
 
  <math>t(T_n = 10\ MeV) = 35\ ns</math>   
 
Using the formulas above the errors become:
 
  
 
  <math>\delta T_n(\delta t = 1\ ns) = 0.59\ MeV</math>
 
  <math>\delta T_n(\delta t = 1\ ns) = 0.59\ MeV</math>
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Below are some other calculations for different neutron energy based on time flight uncertainty <math>1\ ns</math>:
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Below are some calculations for different neutron energy based on time flight uncertainty <math>1\ ns</math>:
  
{| height="10" border="1" cellpadding="10" cellspacing="0"
+
{| height="10" border="1" cellpadding="12" cellspacing="1"
 
|-
 
|-
 
|<math>\delta t_n</math>
 
|<math>\delta t_n</math>
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|<math>\delta T_{\gamma}</math>
 
|<math>\delta T_{\gamma}</math>
 
|<math>\frac{\delta T_{\gamma}}{T_{\gamma}}</math>
 
|<math>\frac{\delta T_{\gamma}}{T_{\gamma}}</math>
 +
|-
 +
|1 ns||1.5 m||0.1 MeV||0.015||343 ns||0.0006 MeV||0.6 %||0.0012 MeV||0.07 %
 
|-
 
|-
 
|1 ns||1.5 m||0.5 MeV||0.033||153 ns||0.006 MeV||1.3 %||0.013 MeV||0.5 %
 
|1 ns||1.5 m||0.5 MeV||0.033||153 ns||0.006 MeV||1.3 %||0.013 MeV||0.5 %
 
|-
 
|-
|1 ns||1.5 m||1.0 MeV||0.046||108 ns||0.02 MeV||1.8 %||0.04 MeV||1.0 %
+
|1 ns||1.5 m||1.0 MeV||0.046||108 ns||0.02 MeV||1.9 %||0.04 MeV||1.0 %
 
|-
 
|-
|1 ns||1.5 m||2.0 MeV||0.065||77 ns||0.05 MeV||2.6 %||0.10 MeV||1.9 %
+
|1 ns||1.5 m||2.0 MeV||0.065||77 ns||0.05 MeV||2.6 %||0.11 MeV||1.9 %
 
|-
 
|-
 
|1 ns||1.5 m||5.0 MeV||0.103||49 ns||0.21 MeV||4.1 %||0.41 MeV||3.6 %
 
|1 ns||1.5 m||5.0 MeV||0.103||49 ns||0.21 MeV||4.1 %||0.41 MeV||3.6 %
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|1 ns||1.5 m||10.0 MeV||0.145||35 ns||0.59 MeV||5.9 %||1.18 MeV||5.5 %
 
|1 ns||1.5 m||10.0 MeV||0.145||35 ns||0.59 MeV||5.9 %||1.18 MeV||5.5 %
 
|-
 
|-
|1 ns||1.5 m||20.0 MeV||0.203||25 ns||1.77 MeV||8.4 %||3.36 MeV||8.1 %
+
|1 ns||1.5 m||20.0 MeV||0.203||25 ns||1.68 MeV||8.4 %||3.36 MeV||8.1 %
 
|}
 
|}
  

Revision as of 03:18, 6 April 2011

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Four-vector Algebra

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

Collision01.png


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 in [math]\Theta_n = 90^o[/math] direction:

[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).

Tgamma02.png

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 checked)

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 uncertainty in photons energy based on uncertainly of neutrons time of flight.


Say, the neutron's time of flight uncertainty is:

[math]\delta t = 1\ ns[/math]

The neutron 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]

By taking derivative of the expression above we can find the relative neutron energy error:

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


Also we need to know the neutron time of flight as function of the neutron 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, 1.5 m away detector, and neutron's time of flight uncertainty is [math]1\ ns[/math].


Using the formulas above we can find:

[math]t(T_n = 10\ MeV) = 35\ ns[/math]   
[math]\delta T_n(\delta t = 1\ ns) = 0.59\ MeV[/math]
[math]\frac{\delta T_n}{T_n} = \frac{0.59\ MeV}{10\ MeV} = 5.9\ %[/math] 
[math]\delta T_{\gamma} = 2.003\cdot \delta T_n = 2.003\cdot 0.59\ MeV = 1.18\ MeV [/math] 
[math]\frac{\delta T_{\gamma}}{T_{\gamma}} = \frac{1.18\ MeV}{(2.003\cdot 10 + 1.577)\ MeV} = 5.5\ %[/math]


Below are some calculations for different neutron energy based on time flight uncertainty [math]1\ ns[/math]:

[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 1.5 m 0.1 MeV 0.015 343 ns 0.0006 MeV 0.6 % 0.0012 MeV 0.07 %
1 ns 1.5 m 0.5 MeV 0.033 153 ns 0.006 MeV 1.3 % 0.013 MeV 0.5 %
1 ns 1.5 m 1.0 MeV 0.046 108 ns 0.02 MeV 1.9 % 0.04 MeV 1.0 %
1 ns 1.5 m 2.0 MeV 0.065 77 ns 0.05 MeV 2.6 % 0.11 MeV 1.9 %
1 ns 1.5 m 5.0 MeV 0.103 49 ns 0.21 MeV 4.1 % 0.41 MeV 3.6 %
1 ns 1.5 m 10.0 MeV 0.145 35 ns 0.59 MeV 5.9 % 1.18 MeV 5.5 %
1 ns 1.5 m 20.0 MeV 0.203 25 ns 1.68 MeV 8.4 % 3.36 MeV 8.1 %




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