Difference between revisions of "Neutron Polarimeter"
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− | We can 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 - p_n cos\Theta_n)} =</math> | ||
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<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> 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= | ||
[[File:Kinetic_energy_0_900_MeV.jpeg]] [[File:Kinetic_energy_0_21_MeV.jpeg]] | [[File:Kinetic_energy_0_900_MeV.jpeg]] [[File:Kinetic_energy_0_21_MeV.jpeg]] | ||
− | + | ||
+ | =low energy approximation== | ||
As we can see from Fig.2 for low energy neutrons (0-21 MeV)<br> | As we can see from Fig.2 for low energy neutrons (0-21 MeV)<br> | ||
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<math> T_{\gamma} = 2.051\ T_n + 1.715 </math> | <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, | Say, we have, 10 MeV neutron with uncertainty 1 MeV, | ||
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<math> \delta T_{\gamma} = 2.051\ \delta T_n = 2.051\times 1\ MeV = 2.051\ MeV </math> | <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 | Say, the neutron's time of flight uncertainly is 1 ns |
Revision as of 16:46, 5 April 2011
Four-vector Algebra
Let's do the four-vector algebra:
By conservation of four-momentum:
Squaring both side of equation above and using the four-momentum invariants
we have:
Energy dependence of as function of neutron energy and neutron angle
We can easily solve the equation above with respect to incident photon energy:
Detector is located at , so
and visa versa
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:
So, the equation of the line is:
Finally for low energy neutrons (0-21 MeV):
example of error calculation
==example 1=+
Say, we have, 10 MeV neutron with uncertainty 1 MeV, the corresponding uncertainly for photons energy is:
example 2
Say, the neutron's time of flight uncertainly is 1 ns
The neutron's kinetic energy is:
And:
Also we need the neutron's time of flight as function of neutron's kinetic energy:
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:
Neutron kinetic energy errors are:
And photon energy errors are:
Below are some examples for different detector distance and neutron kinetic energy:
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 % |