Calculations of 4-momentum components

Setup

Since we want to run for a evenly spaced energy range for Moller electrons, we will need to use some of the scattered electrons to help cover this range. A Moller scattering data file of 4E8 events has no Moller electrons with momentum over 5500 MeV.

awk 'NR == 1 {line = $0; max = $15} NR >1 && $15 > max {line =$0; max =$15} END{print line}' MollerScattering_NH3_4e8incidnet.dat

11000    0     0     11000.5     0     0     -510     5502.94     -14.7812     -50.9086     5503.19     0     0     -505.319     5495.8     14.7812     50.9086     5496.06     0     0     -505.319

Since momentum is conserved, and the data is verified kinematicly, we cannot simply "switch" the data. This data can be altered to have a certain number of different phi values for each energy to match the Moller cross section. This data can then be written to a LUND file, and compared to the previous calculations which did not factor in loss of initial energy.

Prepare Data

Using the existing Moller scattering data from a GEANT simulation of 4E8 incident electrons, a file of just scattered momentum components can be constructed using:

awk '{print $9, $10, $11, $16, $17, $18}' MollerScattering_NH3_4e8incident.dat > Just_Scattered_Momentum.dat

Transfer to CM Frame

Reading in the data from the dat file, we use a C++ program to read the momentum components for the Scattered and Moller electrons into 4-momentum vectors defined as the Lab_final frame of reference.

Performing a Lorentz boost to a Center of Mass frame for the two 4-vectors from the Lab_final frame of reference, we move to a frame where the energies are equal and the momentum are equal but opposite.

Relativistic Kinematics

For Moller Electron energies above 500 MeV, in the Lab frame, histograms of momentum, and theta as well as a 2-D histogram of Energy vs. Theta for the Moller Electron in the CM frame will be filled.

Using the histogram for Theta in the CM frame, we can determine the relative number of events that occur at a given angle. This information will be used to keep the relative number of particles having the same Theta angle, but multiple Psi angles to evenly cover the detector area

Run for Necessary Amount to match Cross Section

Using the above plot for the target material, we can find the relative amount that each Theta angle should observe for this process which gives a known Moller differential cross section.

Theta (degrees)	Differential Cross-Section (barns)	Number of Events
90.000000	0.000017	5
91.000000	0.000017	5
92.000000	0.000017	5
93.000000	0.000017	5
94.000000	0.000017	5
95.000000	0.000017	5
96.000000	0.000017	5
97.000000	0.000017	5
98.000000	0.000017	5
99.000000	0.000018	6
100.000000	0.000018	6
101.000000	0.000018	6
102.000000	0.000019	7
103.000000	0.000019	7
104.000000	0.000019	7
105.000000	0.000020	8
106.000000	0.000020	8
107.000000	0.000021	9
108.000000	0.000022	10
109.000000	0.000022	10
110.000000	0.000023	11
111.000000	0.000024	12
112.000000	0.000025	13
113.000000	0.000025	13
114.000000	0.000026	14
115.000000	0.000028	16
116.000000	0.000029	17
117.000000	0.000030	18
118.000000	0.000031	19
119.000000	0.000033	21
120.000000	0.000035	23
121.000000	0.000036	24
122.000000	0.000038	26
123.000000	0.000040	28
124.000000	0.000043	31
125.000000	0.000045	33
126.000000	0.000048	36
127.000000	0.000051	39
128.000000	0.000054	42
129.000000	0.000058	46
130.000000	0.000062	50
131.000000	0.000067	55
132.000000	0.000072	60
133.000000	0.000077	65
134.000000	0.000083	71
135.000000	0.000090	78
136.000000	0.000097	85
137.000000	0.000106	94
138.000000	0.000115	103
139.000000	0.000126	114
140.000000	0.000138	126
141.000000	0.000152	140
142.000000	0.000167	155
143.000000	0.000185	173
144.000000	0.000205	193
145.000000	0.000228	216
146.000000	0.000254	242
147.000000	0.000285	273
148.000000	0.000321	309
149.000000	0.000362	350
150.000000	0.000411	399
151.000000	0.000468	456
152.000000	0.000537	525
153.000000	0.000618	606
154.000000	0.000716	704
155.000000	0.000834	822
156.000000	0.000978	966

We can set up conditional statements to check what range the Theta angle falls in, then by dividing

[math]\Delta \phi=\frac {2\pi}{number\ of\ events}[/math]

we should find the change in phi needed to give an evenly distributed distribution around the xy plane for a given Theta angle.

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