Difference between revisions of "Neutron Polarimeter"

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= Four-vector Algebra =
 
= Four-vector Algebra =
  
Consider the two bode reaction <math>D(\gamma, n)p</math>:
+
Consider two body reaction <math>D(\gamma, n)p</math>:
  
 
[[File:Collision01.png | 450 px]]
 
[[File:Collision01.png | 450 px]]
  
  
Write down all four-vectors:
+
Write down four-momentum vectors before and after reaction:
  
 
  <math> p_{\gamma} = \left( T_{\gamma},\ T_{\gamma},\ 0,\ 0  \right) </math>  
 
  <math> p_{\gamma} = \left( T_{\gamma},\ T_{\gamma},\ 0,\ 0  \right) </math>  
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  <math> p_{p} = \left( E_p,\ \ p_p\cos \Theta_p,\ \ p_p\sin \Theta_p,\ \ 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:
+
Now apply the law of conservation of four-momentum vectors:
  
 
  <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>
 
  <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>
Line 45: Line 45:
 
Substituting the corresponding masses, we get finally:
 
Substituting the corresponding masses, we get finally:
  
  <math>T_{\gamma}\ [MeV] = 2.003\ T_n\ [MeV] + 1.577\ [MeV]\ \ \ \ [1]</math>
+
  <math>T_{\gamma}\ [MeV] = 2.003\ T_n\ [MeV] + 1.715\ [MeV]\ \ \ \ [1]</math>
  
 
and visa versa:
 
and visa versa:
  
  <math>T_n\ [MeV] = 0.499\cdot T_{\gamma}\ [MeV] - 0.787\ [MeV]\ \ \ \ [2]</math>
+
  <math>T_n\ [MeV] = 0.499\cdot T_{\gamma}\ [MeV] - 0.856\ [MeV]\ \ \ \ [2]</math>
  
  
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1) from formula [1] above we can predict the threshold of <math>^2D(n,\gamma)</math> reaction:
+
1) from formula [1] above we can predict the threshold of <math>^2D(\gamma, n)</math> reaction in  <math>\Theta_n = 90^o</math> direction:
  
  <math>E_{\gamma} = 1.577\ MeV</math>
+
  <math>E_{\gamma} = 1.715\ MeV</math>
 +
 
 +
comment: it's not true because by momentum conservation there are no neutron with zero kinetic energy.
  
 
2) from formula [1] above we can predict the incident photon energy based on the detected neutron energy (neutron polarimeter).
 
2) from formula [1] above we can predict the incident photon energy based on the detected neutron energy (neutron polarimeter).
 
[[File:Tgamma02.png | 1000 px]]
 
  
 
3) from formula [2] above we can predict the detected neutron energy based on the incident photon energy.
 
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>25\ MeV</math> we can detect neutrons up to  <math>11.62\ MeV</math>
  
   -  for the incident photons up to <math>44\ MeV</math> we can detect neutrons up to  <math>21.17\ MeV</math>
+
   -  for the incident photons up to <math>44\ MeV</math> we can detect neutrons up to  <math>21.10\ MeV</math>
  
 
4) we can do the error calculations.
 
4) we can do the error calculations.
  
=Example of error calculation (need to be updated to the formulas [1] and [2] above)=
+
=Example of error calculation=
  
==example 1==
+
==simple error calculation==
  
 
Say, we have, 10 MeV neutron with uncertainty 1 MeV,
 
Say, we have, 10 MeV neutron with uncertainty 1 MeV,
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  <math> \delta T_{\gamma} = 2.003\ \delta T_n = 2.003\times 1\ MeV = 2.003\ MeV </math>
 
  <math> \delta T_{\gamma} = 2.003\ \delta T_n = 2.003\times 1\ MeV = 2.003\ MeV </math>
  
==example 2==
+
==photon kinetic energy error <math>\delta T_{\gamma}</math> as function of TOF uncertainty <math>\delta t</math>==
 +
 
 +
In the calculations below I attempted to predict the uncertainty in photons energy based on uncertainty in neutrons time of flight.
  
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:
 
  
<math>\delta t = 1\ ns</math>
 
  
The neutron kinetic energy is:
+
The neutron kinetic energy as function of time of flight 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>
 
  <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:
+
By taking derivative of the expression above we can find the relative error for neutron energy:
  
 
  <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>
 
  <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>
  
 
+
In that formula for <math>\delta T_n</math> we need to know the neutron time of flight which is:
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)} =
 
  <math>t:=\frac{l}{c\ \beta_n} = \frac{l}{c\ (p_n/E_n)} =
Line 101: Line 100:
  
  
 +
And now we can calculate the relative error for photon energy using the formula derived before:
 +
 +
<math>\delta T_{\gamma} = 2.003\ \delta T_n</math>
 +
 +
 +
 +
Say, the detector is 1.5 m away and neutron's time of flight uncertainty is:
 +
 +
<math>\delta t = 1\ ns</math>
 +
 +
And below are y numerical calculations and plot with error bars:
 +
 +
[[File:table_1ns.png | 800 px]]
 +
 +
And in the plot below I have overlay my error calculations using the formulas above:
 +
 +
[[File:Tgamma_1ns_01.png | 900 px]]
 +
 +
 +
 +
Now say, the detector is 1.5 m away and neutron's time of flight uncertainty is:
 +
 +
<math>\delta t = 3\ ns</math>
 +
 +
[[File:table_3ns.png | 800 px]]
 +
 +
[[File:Tgamma_3ns_01.png | 900 px]]
 +
 +
=<math>\Theta</math>  dependence=
 +
 +
Starting from general kinematic relations:
 +
 +
<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>
 +
 +
we can solve the equation above with respect to <math>T_{\gamma}</math> as function of neutron energy <math>T_n</math> and neutron angle <math>\Theta_n</math>
 +
 +
<math>T_{\gamma} = \frac{m_p^2-m_n^2-m_D^2+2m_D E_n}{2(m_D-E_n-p_n\cos\Theta_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>. Then the neutron time of flight is:
+
and using
  
  <math>t(T_n = 10\ MeV) = 35\ ns</math>  
+
  <math>E_n = T_n+m_n</math>
 +
<math>p_n = \sqrt{E_n^2-m_n^2} = \sqrt{T_n^2+2T_n m_n}</math>
  
Using the formulas above the errors become:
+
finally
  
  <math>\delta T_n(\delta t = 1\ ns) = 0.59\ MeV</math>
+
  <math>T_{\gamma} = \frac{m_p^2-m_n^2-m_D^2+2m_D (T_n+m_n)}{2\left(m_D-m_n-T_n-\sqrt{T_n^2+2T_n m_n}\cos\Theta_n\right)}</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>
+
Let's do some plots using the equation derived above. Below is presented the photon energy as function of neutron angle for several values of neutron energy:
  
  
 +
[[File:Tgamma_angle.png | 900 px]]
  
Below are some other calculations for different neutron energy based on time flight uncertainty <math>1\ ns</math>:
 
  
{| height="10" border="1" cellpadding="8" cellspacing="0"
 
|-
 
|<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||0.5 MeV||0.033||153 ns||0.01 MeV||1.3 %||0.01 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||2.0 MeV||0.065||77 ns||0.05 MeV||2.6 %||0.10 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||1.77 MeV||8.4 %||3.36 MeV||8.1 %
 
|}
 
  
 +
And below are the photon energy as function of neutron angle and neutron energy:
  
  
 +
[[File:Tgamma_3D_01.png | 900 px]]
  
  
  
 
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[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]

Latest revision as of 05:17, 27 February 2013

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Four-vector Algebra

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

Collision01.png


Write down four-momentum vectors before and after reaction:

[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 law of conservation of four-momentum vectors:

[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] + 1.715\ [MeV]\ \ \ \ [1][/math]

and visa versa:

[math]T_n\ [MeV] = 0.499\cdot T_{\gamma}\ [MeV] - 0.856\ [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(\gamma, n)[/math] reaction in [math]\Theta_n = 90^o[/math] direction:

[math]E_{\gamma} = 1.715\ MeV[/math]

comment: it's not true because by momentum conservation there are no neutron with zero kinetic energy.

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]11.62\ MeV[/math]
 -  for the incident photons up to [math]44\ MeV[/math] we can detect neutrons up to  [math]21.10\ MeV[/math]

4) we can do the error calculations.

Example of error calculation

simple error calculation

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]

photon kinetic energy error [math]\delta T_{\gamma}[/math] as function of TOF uncertainty [math]\delta t[/math]

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


The neutron kinetic energy as function of time of flight 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 error for neutron energy:

[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]

In that formula for [math]\delta T_n[/math] we need to know the neutron time of flight which is:

[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]


And now we can calculate the relative error for photon energy using the formula derived before:

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


Say, the detector is 1.5 m away and neutron's time of flight uncertainty is:

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

And below are y numerical calculations and plot with error bars:

Table 1ns.png

And in the plot below I have overlay my error calculations using the formulas above:

Tgamma 1ns 01.png


Now say, the detector is 1.5 m away and neutron's time of flight uncertainty is:

[math]\delta t = 3\ ns[/math]

Table 3ns.png

Tgamma 3ns 01.png

[math]\Theta[/math] dependence

Starting from general kinematic relations:

[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]

we can solve the equation above with respect to [math]T_{\gamma}[/math] as function of neutron energy [math]T_n[/math] and neutron angle [math]\Theta_n[/math]

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

and using

[math]E_n = T_n+m_n[/math]
[math]p_n = \sqrt{E_n^2-m_n^2} = \sqrt{T_n^2+2T_n m_n}[/math]

finally

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


Let's do some plots using the equation derived above. Below is presented the photon energy as function of neutron angle for several values of neutron energy:


Tgamma angle.png


And below are the photon energy as function of neutron angle and neutron energy:


Tgamma 3D 01.png


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