Difference between revisions of "Defining Occupancy"

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<center><math>\sigma \frac{N_{scattered}}{\Phi \rho l}\Rightarrow N_{evt}=\sigma N_{incident} \rho l</math></center>
 
<center><math>\sigma \frac{N_{scattered}}{\Phi \rho l}\Rightarrow N_{evt}=\sigma N_{incident} \rho l</math></center>
  
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we can define a differential scattering cross-section, <math>d\sigma(\theta')/d{\mit\Omega}'</math>, in the laboratory frame, where <math>d{\mit\Omega}'=2\pi\,\sin\psi\,d\theta'</math> is an element of solid angle in this frame. Thus, <math>(d\sigma(\psi)/d{\mit\Omega}')\,d{\mit\Omega}'</math> is the effective cross-sectional area in the laboratory frame for scattering into the range of scattering angles <math>\psi$ to $\psi+d\psi</math>. Likewise, <math>(d\sigma(\theta)/d{\mit\Omega})\,d{\mit\Omega}</math>is the effective cross-sectional area in the center of mass frame for scattering into the range of scattering angles <math>\theta </math> to <math>\theta+d\theta</math>. Note that <matH>d{\mit\Omega} =2\pi\,\sin\theta\,d\theta</math>. However, a cross-sectional area is not changed when we transform between different inertial frames. Hence, we can write
  
  

Revision as of 15:44, 23 July 2018

General Occupancy

The occupancy measures the number of particles interactions per a detector cell per an event. For the CLAS12 drift chamber, there are 112 wires on each layer, with 12 layers within a region, giving 1344 cells. This can simply be defined as the "Unweighted Occupancy" for the CLAS12 DC and follows the equation:


Unweighted CLAS12 DC OccupancyNhitsNevtsNcells


where


NhitsThe number of DC wires intersected by primary and secondary events throughout the drift chamber in Region 1


Nevtϕ×Prob(interacting)


ϕ Number of incident particles on the face of drift chamber per cm2


Ncells112wireslayer×12layersRegion


CLAS12 DC Occupancy

The registering of a "hit" takes a finite time in which the detector and its associated electronics are not able to register an additional signal if it occurs. This time window is known as the "dead time" during which only limited events are registered. For Region 1:

Δt 250 ns: The time needed for events to be read by the electronics within Region 1

Since the events are simulated outside the dead time constraints of the DC, we can factor in the number of event windows that occur by dividing the dead time window per region by the time that would have been required to produce the number of incident electrons given a known current.


tsim Time of simulation = NincidentI(A)1A1C1s1.602E19 C1 e


When applying the Moller differential cross-section as a weight, this gives the CLAS12 occupancy as:


CLAS12 DC occupancy NhitsNevtΔttsim1112112

Weighted Hits Occupancy

Using the definition of the cross-section:


σNscatteredLt=NscatteredΦρlt


where the flux is defined as:


ΦNumber of es


Making some assumptions that the flux can be taken over an same time range as the time found in the cross-section, which allows


ΦNincident


For a LH2 target of length 5cm.


ρtarget×ltarget=70.85kg1m3×1mole2.02g×1000g1kg×6.022×1023molecules LH21mole2atoms1 molecule LH2×1m3(100cm)3×5cm=2.11×1023cm2×124cm2barn=0.211barns1


Additionally, for Moller Scattering, we can assume that almost 100% of the incident electrons occur as events.

Nincident=Nevt

This allows us to rewrite the cross-section expression as,

σNscatteredΦρlNevt=σNincidentρl


we can define a differential scattering cross-section, dσ(θ)/dΩ, in the laboratory frame, where dΩ=2πsinψdθ is an element of solid angle in this frame. Thus, (dσ(ψ)/dΩ)dΩ is the effective cross-sectional area in the laboratory frame for scattering into the range of scattering angles ψ$to$ψ+dψ. Likewise, (dσ(θ)/dΩ)dΩis the effective cross-sectional area in the center of mass frame for scattering into the range of scattering angles θ to θ+dθ. Note that dΩ=2πsinθdθ. However, a cross-sectional area is not changed when we transform between different inertial frames. Hence, we can write


Since the expression for the differential cross-section for Moller Scattering is well know, we can solve for the minimum angle detected by the DC (5 degrees in Theta).

σ(θ=5)=DdσdΩdΩ=Nscatteredsolid angleNincidentarea