Electron Polarimetry

Prakash

This is a schematic of trimmed down version of geometry fo the Geant4 simulation of Compton Polarimeter at Hall-A in Jefferson Lab.

Compton Scattering

Kinematics

  • Incident electron $e$ with energy $E$ and momentum $\vec p = (0,0,p)$ along $\vec z$ axis.
  • Incident photon $\gamma$ with energy $k$, incident angle $\alpha_e$ with respect to $\vec z$ and momentum $(0, -k \sin \alpha_e , - k \cos \alpha_e )$,
  • Scattered electrons $e'$ with energy $E'$, scattering angle $\theta_e$ with respect to $\vec z$ and momentum $\vec p\prime = (p\prime \sin \theta_c\sin\phi, p\prime\sin\theta_c\cos\phi, p\prime \cos\theta_c)$,
  • Scattered photons $\gamma\prime$ with energy $k\prime$, scattering angle $\theta_\gamma$ with respect to $\vec z$ and momentum $\vec k\prime = (k\prime \sin \theta_\gamma, k\prime \sin\theta_\gamma\cos\phi, k\prime\cos\theta_\gamma)$.

The scattered photon energy $k$’ is related to the scattered photon angle $\theta_\gamma$ by :

\begin{align*} k^{\prime}=k \frac{E+p \cos \alpha_c}{E+k-p \cos \theta_\gamma+k \cos \left(\alpha_c-\theta_\gamma\right)} . \end{align*}

For photon incident angle $\alpha_c=0$, this equation can be simplified, using $\gamma=E / m$, leading to

\begin{align*} \frac{k^\prime}{k} \simeq \frac{4 a \gamma^2}{1+a \theta_\gamma^2 \gamma^2} \end{align*}

where

\begin{align*} a=\frac{1}{1+\frac{4 k \gamma}{m}}=\frac{1}{1+\frac{4 k E}{m^2}} \end{align*}

The maximum scattered photon energy $k_{\text {max }}^{\prime}$, corresponding to the minimum scattered electron energy $E_{\min }^{\prime}$, is reached for $\theta_\gamma=0$.,

\begin{align*} k_{\max }^{\prime}=4 a k \gamma^2=4 a k \frac{E^2}{m^2}, \\ E_{\min }^{\prime}=E-k_{\max }^{\prime}+k=E-4 a k \frac{E^2}{m^2}+k \simeq E-4 a k \frac{E^2}{m^2}, \end{align*}

while the minimum scattered photon energy $k_{\min }^{\prime}$, corresponding to the maximum scattered electron energy $E_{\max }^{\prime}$, is for $\theta_\gamma=\pi$,

\begin{align*} k_{\min }^{\prime}=k \\ E_{\max }^{\prime}=E-k_{\min }^{\prime}+k=E . \end{align*}

The photon scattering angle at which $k^{\prime}=k_{\max }^{\prime} / 2$ is

\begin{align*} \theta_{\gamma 1 / 2}=\frac{m}{E \sqrt{a}}=\frac{1}{\gamma \sqrt{a}} \end{align*}

The scattered electron momentum $\mathbf{p}$ ’ is related to the scattered electron angle $\theta_e$ by a second order equation:

\begin{align*} p^{\prime 2}\left(C^2-B^2\right)-2 A B p^{\prime}+m^2 C^2-A^2=0 \\ p^{\prime}=\displaystyle \frac{A B \pm C \sqrt{\left(A^2-m^2\left(C^2-B^2\right)\right)}}{C^2-B^2} \end{align*}

where

\begin{align*} A= & m^2+E k+k p \cos \alpha_c \\ B= & p \cos \theta_e-k \cos \left(\theta_e-\alpha_c\right) \\ C= & E+k \end{align*}

The maximum electron angle is obtained for $A^2=m^2\left(C^2-B^2\right)$. For small photon incident angle and energy, one gets :

\begin{align*} \theta_e^{\max } \simeq 2 \frac{k}{m} \end{align*}

Calculations

  • Now the scattering angle for photon can be simplified to
\begin{align*} \theta_\gamma= \sqrt{\left( 4 \frac{k}{k^\prime} - \frac{1}{4a\gamma^2} \right)} \end{align*}

Updates Thu Jun 27 11:21:25 AM EDT 2024

  • 11:21> Wrote generator prototype. Keeps on segfaulting on SampleVertex. Using gdb to find that out.
  • 11:22>
  • 16:32> So basically can’t yet figure out. Reverted everythign back. Now can’t even run with “moller” generator.
  • 16:33> Don’t feel like working on this no more.

Updates Fri Jun 28 03:33:03 PM EDT 2024

  • 15:33> So progress, I hadn’t initialized the kinematic variables correctly so there were buncha nan’s and that was causing the issue. Have to get the kinematic variables right.
  • 15:33>

Exploration

  classDiagram
G4VUserPrimaryGeneratorAction <|-- comptonPrimaryGeneratorAction
comptonPrimaryGeneratorAction: G4ParticleGun
comptonPrimaryGeneratorAction: comptonBeamTarget
comptonPrimaryGeneratorAction: comptonEvent
comptonPrimaryGeneratorAction: map{stdstring,comptonVEventGen}

comptonBeamTarget: comptonMultScat

comptonEvent: ProduceNewParticle()
comptonEvent: comptonBeamTarget

comptonVEventGen: G4ParticleGun
comptonVEventGen: comptonBeamTarget
comptonVEventGen: SamplePhysics()

comptonVerted: G4double fBeamEnargy
comptonVerted: G4double fRadiationLength
comptonVerted: G4Material fMaterial

Now what does comptonPrimaryGeneratorAction do inside GeneratePrimaries()?

  • Calls GenerateEvent() method of comptonVEventGen
  • for each fPartType.size() in the event
    • fParticleGun->SetParticleDefinition()
    • fParticleGun->SetParticleEnergy();
    • fParticleGun->SetParticlePosition();
    • fParticleGun->SetParticleMomentumDirection();
  • Gets nthrown from rundata
  • Gets Samplign type from VEventGen That’s it.

Geometry Notes

Bend angle at each dipole  (beam height /  D1-D2 z pos)
root [19] 215.4/5395.7
(double) 0.039920678


Bend angle for compton edge (roughly)
root [20] 215.4/5395.7*11./8.
(double) 0.054890932


Distance edet to center D4
root [22] 38.47*25.4
(double) 977.13800

Distance D3 to edet
root [23] (5395-977.13)
(double) 4417.8700

y-pos of CE at edet
root [24] 215.4/5395.7*11./8.*(5395.-977.14)
(double) 242.50045

height above primary beamline of CE at edet
root [25] 215.4/5395.7*11./8.*(5395.-977.14) - 215.4
(double) 27.100451

Location of unscattered beam at e-det
root [26] 215.4/5395.7*(5395.-977.14)
(double) 176.36396

Distance CE from unscattered at edet
root [27] 242-176.
(double) 66.000000

root [29] (2.01-1.87)/2.
(double) 0.070000000
root [30] (2.01-1.87)/2.*25.4
(double) 1.7780000

root [31] 1.87*25.4 + 1.8
(double) 49.298000

thickness of top plate on weldment box
root [32] 0.25*25.4
(double) 6.3500000

root [33] 0.25*25.4 - 6.35
(double) 0.0000000

(I think?) height of DS opening in weldment box relative to beam height
root [34] 1.87*25.4 + 1.8 -6.35
(double) 42.948000

  • Center to D3 = 1645.92

  • D3 to that pipe = 212.44-38.47-32.30-34.84 = 106.83*25.4 = 2713.482

  • Center to left end of that pipe ~ 2713.482+1645.92 = 4359.402

  • Length of that pipe: 46.20*25.4 = 1173.48

  • Height of that pipe: 7.42*25.4 = 188.468

  • Width of that pipe: 66.50

  • Above the beamline: 1.271*25.4 = 32.2834

  • Absolute y position of the beam: 8.484*25.4 = 215.4936

  • Center y relative to beamline: (2.650 - 1.271)*25.4 = 35.0266 # WRONG

  • Center y relative to beamline: (2.650 - 1.271)*25.4 = 35.0266 # WRONG

  • Center y absolute origin: 35.0266+215.4936 = 250.5202

  • The left tube cut: x: 2.600*25.4 = 66.04

  • The left tube cut: y: (2.600+2.650)*25.4 = 133.35

  • Off center of the lft cut ((2.600+2.650)/2-2.650)*25.4 = 0.635

  • The right tube cut: x: 3.130*25.4 = 79.502

  • The right tube cut: y: 5.800*25.4 = 147.32

  • That end plate joint radius: 7.970/2*25.4 = 101.219

Power Deposition.

  • Simulate with compton generator.
  • Look at power deposition.
  • Find ionization threshold of the diamond.

    • How to find ionization threshold?

    • Compare electrons after and before the diamond.

      There is slightly more electron but how do I actually know the ionization energy. Look athe mother track and find the hit.e may be?

  • Then calculate the rate of sync photon and total energy deposited by synchrotron photon greater than ionization threshold.