Defining Occupancy

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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 Occupancy$\equiv \frac{N_{hits}}{N_{evts}N_{cells}}$

where

$N_{hits}\equiv$The number of DC wires intersected by primary and secondary events throughout the drift chamber in Region 1

$N_{evt} \equiv \phi \times Prob(interacting)$

$\phi \equiv$ Number of incident particles on the face of drift chamber per cm$^2$

$N_{cells} \equiv 112 \frac{wires}{layer} \times 12 \frac{layers}{Region}$

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:

$\Delta t \equiv$ 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.

$t_{sim} \equiv$ Time of simulation = $\frac{N_{incident}}{I(A)}\frac{1A}{1C}\frac{1s}{}\frac{1.602E-19\ C}{1\ e^{-}}$

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

CLAS12 DC occupancy $\equiv \frac{N_{hits}}{N_{evt}}\frac{\Delta t}{t_{sim}}\frac{1}{112}\frac{1}{12}$

Weighted Hits Occupancy

Lab Cross-Section

Using the definition of the cross-section:

$\sigma \equiv \frac{N_{scattered}}{\mathcal L t}=\frac{N_{scattered}}{\Phi \rho l t}$

where the flux is defined as:

$\Phi \equiv \frac{Number\ of\ e^{-}}{s}$

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

$\Phi \approx N_{incident}$

For a LH2 target of length 5cm.

$\rho_{target}\times l_{target}=\frac{70.85 kg}{1 m^3}\times \frac{1 mole}{2.02 g} \times \frac{1000g}{1 kg} \times \frac{6.022\times10^{23} molecules\ LH_2}{1 mole} \frac{2 atoms}{1\ molecule\ LH_2} \times \frac{1m^3}{(100 cm)^3} \times \frac{5 cm}{ }=\frac{2.11\times 10^{23}}{cm^2} \times \frac{1^{-24} cm^{2}}{barn}=0.211 barns^{-1}$

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

$N_{scattered}=N_{evt}$

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

$\sigma \frac{N_{scattered}}{\Phi \rho l}\Rightarrow N_{evt}=\sigma N_{incident} \rho l$

we can define a differential scattering cross-section, $\frac{d\sigma(\theta')}{d{\mit\Omega}'}$, in the laboratory frame, where $d{\mit\Omega}'=\sin\theta'\,d\theta'd\phi$ is an element of solid angle in this frame. Therefore, $\frac{d\sigma(\theta)}{d{\mit\Omega}'}\,d{\mit\Omega}'$ is the effective cross-sectional area in the laboratory frame for scattering into the range of scattering angles $\theta$ to $\theta+d\theta$ and transverse angles $\phi$ to $\phi+d\phi$. Likewise, $\frac{d\sigma(\theta)}{d{\mit\Omega}}\,d{\mit\Omega}$is the effective cross-sectional area in the CM frame for scattering into the range of scattering angles $\theta$ to $\theta+d\theta$ and transverse angles $\phi$ to $\phi+d\phi$. By relativity, only the scattering angles are Lorentz contracted in the direction of motion, leaving the transverse angles invariant between frames of reference. Additionally, an effective cross-section with corresponding cross-sectional area is not changed when we transform between different inertial frames. The number of scattered particles and incident particles are invariant between frames, while the cross-sectional areas are also measuring the same relative space.

$\frac{d\sigma(\theta')}{d{\mit\Omega}'} \,d{\mit\Omega}' = \frac{d\sigma(\theta)}{d{\mit\Omega}}\,d{\mit\Omega}$

$d\sigma(\theta') =\frac{d\sigma(\theta)}{d{\mit\Omega}}\, \frac{d{\mit\Omega}}{d{\mit\Omega}'}$

$d\sigma(\theta') =\frac{d\sigma(\theta)}{d{\mit\Omega}}\, \sin\theta d\theta d\phi$

Since the expression for the differential cross-section for Moller Scattering is well known in the CM, we can solve for the minimum angle detected by the DC (.55 degrees in Theta) in the lab frame.

$\sigma(\theta'=0.55^{\circ})=\frac{d\sigma(\theta)}{d{\mit\Omega}}\, \sin\theta\, d\theta\, d\phi$

$\int\limits_{\theta}^{\theta+d\theta} \int\limits_{0}^{0.2^{\circ}} \frac{\left(3+\left(\cos{\theta}\right)^2\right)^2}{\left(\sin{\theta}\right)^4}\sin{\theta} d\theta=\frac{9}{\pi}\left(\cos{\theta}-2\csc{\frac{\theta}{2}}^2+2\sec{\frac{\theta}{2}}^2\right)\Big|_{\theta}^{\theta+d\theta}$

Here we will use a definite integral to find the cross-section in the lab. For a constant $\Delta \theta_{lab}=0.1^{\circ}$ this corresponds to a $\Delta \theta_{CM}$ that will become smaller as $\theta_{lab}$ increases. $\Delta \phi=0.2^{\circ}$ in both frames due to it being Lorentz invariant.

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