Difference between revisions of "Neutron Polarimeter"

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= Four-vector Algebra =
 
= Four-vector Algebra =
  
[[File:Collision.png || 380px]]
+
Consider two body reaction <math>D(\gamma, n)p</math>:
  
 +
[[File:Collision01.png | 450 px]]
  
Let's do the four-vector algebra:
+
 
 +
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>  
Line 14: Line 16:
 
  <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>  
  
By 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 22: Line 24:
 
  <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>
 
  <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=
  
= Case <math>\Theta = 90^o</math> =
 
  
 +
Detector is located at <math>\Theta_n = 90^o</math>, and the formula above is simplified:
  
Detector is located at <math>\Theta_n = 90^o</math>, so 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>
  
<math> m_p^2 = m_D^2 + m_n^2 +  2T_{\gamma}\cdot m_D  - 2m_D\cdot E_n - 2\ T_{\gamma}E_n </math>
 
  
 
We can easily solve the equation above with respect to incident photon energy:  
 
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}) </math>
+
  <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>
  
  
<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:
 
 
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>
 
  <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>:
+
Substituting the corresponding masses, we get finally:
  
  <math>T_{\gamma}\ [MeV] = 2.003\ T_n\ [MeV] + 1.577\ [MeV]</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\ T_{\gamma}\ [MeV] - 0.787\ [MeV]</math>
+
  <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.
  
Using the formulas above we can:
+
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.
  
1. Calculate the threshold for <math>^2D(n,\gamma)</math> reaction:
+
  -  for the incident photons up to <math>25\ MeV</math> we can detect neutrons up to  <math>11.62\ MeV</math>
  
  <math>E_{\gamma} = 1.577\ MeV</math>
+
  - for the incident photons up to <math>44\ MeV</math> we can detect neutrons up to  <math>21.10\ MeV</math>
  
2. Predict the incident photon energy based on the detected neutron energy.
+
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:
  
3. Predict the detected neutron energy based on the incident photon energy.
+
<math> \delta T_{\gamma} = 2.003\ \delta T_n = 2.003\times 1\ MeV = 2.003\ MeV </math>
  
  -  for the incident photons up to 25 MeV we have (0 - 11.69) MeV neutrons
+
==photon kinetic energy error <math>\delta T_{\gamma}</math> as function of TOF uncertainty <math>\delta t</math>==
  
  -  for the incident photons up to 44 MeV we have (0 - 21.17) MeV neutrons
+
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:
  
Also note:
+
<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 visa versa =
+
By taking derivative of the expression above we can find the relative error for neutron energy:
  
  <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>
+
<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>
  
=how it looks=
+
In that formula for <math>\delta T_n</math> we need to know the neutron time of flight which is:
  
[[File:Kinetic_energy_0_900_MeV.jpeg‎]] [[File:Kinetic_energy_0_21_MeV.jpeg‎]]
+
<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>
  
  
=low energy approximation==
+
And now we can calculate the relative error for photon energy using the formula derived before:
  
  As we can see from Fig.2 for low energy neutrons (0-21 MeV)<br>
+
  <math>\delta T_{\gamma} = 2.003\ \delta T_n</math>
energy dependence of incident photons is linear
 
  
Find that dependence. We have:<br>
 
<math> T_{\gamma}(0\ MeV) = 1.715360792\ MeV </math>
 
<math> T_{\gamma}(21\ MeV) = 44.78703086\ MeV </math><br>
 
So, the equation of the line is:<br>
 
<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):<br>
 
<math> T_{\gamma} = 2.051\ T_n + 1.715 </math>
 
  
=example of error calculation =
+
Say, the detector is 1.5 m away and neutron's time of flight uncertainty is:
  
==example 1=+
+
<math>\delta t = 1\ ns</math>
  
Say, we have, 10 MeV neutron with uncertainty 1 MeV,
+
And below are y numerical calculations and plot with error bars:
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>
+
[[File:table_1ns.png | 800 px]]
  
==example 2==
+
And in the plot below I have overlay my error calculations using the formulas above:
  
Say, the neutron's time of flight uncertainly is 1 ns
+
[[File:Tgamma_1ns_01.png | 900 px]]
  
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:
+
Now say, the detector is 1.5 m away and neutron's time of flight uncertainty 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>
+
  <math>\delta t = 3\ ns</math>
  
 +
[[File:table_3ns.png | 800 px]]
  
Also we need the neutron's time of flight as function of neutron's kinetic energy:
+
[[File:Tgamma_3ns_01.png | 900 px]]
  
<math>t:=\frac{l}{c\ \beta_n} = \frac{l}{c\ (p_n/E_n)} =
+
=<math>\Theta</math> dependence=
        \frac{l\ (T_n + m_n)}{c\sqrt{T_n^2+2m_nT_n}}</math>
 
  
 +
Starting from general kinematic relations:
  
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> 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>
  
<math>t(T_n = 10\ MeV) = 23\ ns</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>
  
Neutron kinetic energy errors are:
+
finally
  
  <math>\delta T_n(\delta t = 1\ ns) = 0.88\ 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{0.88\ 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>
+
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:
<math>\frac{\delta T_{\gamma}}{T_{\gamma}} = \frac{1.80\ MeV}{(2.051\cdot 10 + 1.715)\ MeV} = 8.1\ %</math>
 
  
  
 +
[[File:Tgamma_angle.png | 900 px]]
  
Below are some examples for different detector distance and neutron kinetic energy:
 
  
{| height="10" border="1" cellpadding="6" 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 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 %||||
 
|}
 
  
 +
And below are the photon energy as function of neutron angle and neutron energy:
  
  
 +
[[File:Tgamma_3D_01.png | 900 px]]
  
  
  
 
[http://wiki.iac.isu.edu/index.php/PhotoFission_with_Polarized_Photons_from_HRRL Go Back]
 
[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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