Difference between revisions of "Uniform distribution in Energy and Theta LUND files"

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Starting with the Energy, and a Phi angle of 10 degrees, the 4-momentum vector can be kinematically determined:
 
Starting with the Energy, and a Phi angle of 10 degrees, the 4-momentum vector can be kinematically determined:
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<center><math>E=\sqrt{p^2+m^2}\Rightarrow p=\sqrt{E^2-m^2} \Rightarrow p=\sqrt{100MeV^2-.2611MeV^2}=9.9869MeV/c</math></center>
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Using the fact that phi is defined in the xy plane to be:

Revision as of 16:24, 23 May 2016

File:LUND Spread.C

The LUND file is created by creating an isotropic distribution of particles within the energy range of 2MeV-5.5GeV as is found through GEANT simulation. These particles are also uniformly distributed through the angle theta with respect to the beam line in the range 5-40 degrees. This is done at a set angle phi (10 degrees) with respect to the perpendicular components with respect to the beam line.

Init Mol E Lab.pngInit Mol Theta Lab.png


Init Mol Mom Lab.png

Center of Mass for Stationary Target

For an incoming electron of 11GeV striking a stationary electron we would expect:

Screen Shot 2016-05-12 at 6.42.46 PM.png


Boosting to the Center of Mass Frame:

Screen Shot 2016-05-12 at 6.44.44 PM.png

Init e Mom CM.pngInit Mol Mom CM.png


Init e Theta CM.pngInit Mol Theta CM.png

Phase space Limiting Particles

Since the angle phi has been constrained to remain constant, the x and y components of the momentum will increase in the positive first quadrant. This implies that the z component of the momentum must decrease by the relation:

[math]p^2=p_x^2+p_y^2+p_z^2[/math]

In the Center of Mass frame, this becomes:

[math]p_x^{*2}+p_y^{*2} = p^{*2}-p_z^{*2}[/math]


Since the momentum in the CM frame is a constant, this implies that pz must decrease. For phi=10 degrees:

MolPxPyLab.pngMolPxPyCM.png

This is repeated for rotations of 60 degrees in phi in the Lab frame.


We can use the variable rapidity:

[math]y \equiv \frac {1}{2} \ln \left(\frac{E+p_z}{E-p_z}\right)[/math]

where

[math] P^+ \equiv E+p_z[/math]
[math] P^- \equiv E-p_z[/math]

this implies that as

[math]p_z \rightarrow 0 \Rrightarrow \frac{E+p_z}{E-p_z} \rightarrow 1 \Rrightarrow \ln 1 \rightarrow 0 \Rrightarrow y=0[/math]


For forward travel in the light cone:

[math]p_z \rightarrow E \Rrightarrow \frac{E+p_z}{E-p_z} \rightarrow \infin \Rrightarrow \ln \infin \rightarrow \infin \Rrightarrow y \rightarrow \infin [/math]


For backward travel in the light cone:

[math]p_z \rightarrow -E \Rrightarrow \frac{E+p_z}{E-p_z} \rightarrow 0 \Rrightarrow \ln 0 \rightarrow -\infin \Rrightarrow y \rightarrow -\infin [/math]


For a particle that transforms from the Lab frame to the CM frame where the particle is not within the light cone:

[math]p_x^2+p_y^2=52.589054^2+9.272868^2=53.400MeV \gt 53.015 MeV (E) \therefore p_z \rightarrow imaginary[/math]

These particles are outside the light cone and are more timelike, thus not visible in normal space. This will reduce the number of particles that will be detected.

MolEThetaRapidityCM.png

Using Initial Condition based on After Collision in Lab Frame

Starting with the Energy, and a Phi angle of 10 degrees, the 4-momentum vector can be kinematically determined:

[math]E=\sqrt{p^2+m^2}\Rightarrow p=\sqrt{E^2-m^2} \Rightarrow p=\sqrt{100MeV^2-.2611MeV^2}=9.9869MeV/c[/math]

Using the fact that phi is defined in the xy plane to be: