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Revision as of 21:45, 28 January 2009

15-1-09

Electron-Pion Contamination Estimate

The number of photons produced per unit path length of a particle with charge ze and per unit energy interval of the photons is proportional to the sine of the Cherenkov angle[1]

[math]\frac{d^2 N}{dEdx}=\frac{\alpha z^2}{\hbar c}sin ^2 \theta_c=\frac{\alpha z^2}{\hbar c}[1-\frac{1}{\beta^2 n^2 (E)}][/math]


[math]\frac{d^2 N}{d\lambda dx} = \frac{2 \pi \alpha z^2}{\lambda^2}[1-\frac{1}{\beta^2 n^2 (\lambda)}][/math]

[math]\beta=\frac{v}{c}=\frac{pc}{\sqrt{(pc)^2 + (mc^2)^2}}[/math]

after deriving the Taylor expansion of our function and considering only the first two terms, we get
[math]\frac{d^2 N}{dEdx}=\frac{\alpha z^2}{\hbar c}sin ^2 \theta_c=\frac{\alpha z^2}{\hbar c}[\beta^2 n^2 (E) - 1][/math]

The gas used in the CLAS Cerenkov counter is perfluorobutane [math]C_4 F_{10}[/math] with index of refraction equal to 1.00153.


  • For Electrons

The calculation of the number of photoelectrons emitted by electrons is shown below. Electron mass [math]m_e = 0.000511GeV[/math], [math]n=1.00153[/math] and [math]\beta=\frac{p}{E} = 1[/math], because mass of the electron is negligible and also [math]\frac{\alpha}{\hbar c}=370[ eV^{-1} cm^{-1}] [/math]

The Hall B cherenkov detector is [math]\sim \; 0.7[/math] m thick radiator. We assume the PMTs used to collect light have a constant quantum efficiency of 8% for photons with wavelength between 300 and 600 nm.

[math]\frac{dN}{dx} [/math] [math]= 2 \pi \alpha z^2 [{\beta^2 n^2 (\lambda)} - 1]\int_{300nm}^{600nm} \frac{1}{\lambda^2} d\lambda \times (0.08)=[/math]
[math]= 19 \times 10^{-9} [nm^{-1}][/math]

For the number of photoelectrons produced by the electrons we have the following value

[math] N = 19 \times 10^{-9} \times 0.7 m [\frac{10^9}{m}] = 13.3 [/math]
If we do the same calculation for pions we get as a result:

[math] N = 5.465 \times 10^{-9} \times 0.7 m [\frac{10^9}{m}] = 3.8255 [/math]


CLAS Cherenkov signal

Electrons

The cherenkov signal measured in CLAS for particles identified as electrons by the tracking algorithm is shown below. There are two distributions present. One distribution is centered around 1.5 PEs and the second distribution is at 8 PEs when two gaussians and a Landau distribution are combined and fit to the spectrum. As we will show below, the first peak is due to the misidentification of a negative pion as an electron.

PE Fit equation (Osipenko's CLAS Note 2004-20 File:CLAS Note-2004-020.pdf)
[math]N_{pe}= p_0 e^{-0.5 \left (\frac{x-p_1}{p_2} \right )^2} + p4\frac{1}{1-\left(\frac{x-p5}{p6}\right )} + p_6 e^{-0.5 \left (\frac{x-p_7}{p_8} \right )^2}[/math]
e_Momentum_vs_Number_of_Photoelectrons

The flag cut applied on the number of photoelectrons means that in CLAS detector instead of 5 superlayers were used 6 of them in track fit. As one can see from the histograms of the Number of photoelectrons, the cut on flag does not have effect on the peak around 1.5phe and decreases the number of entries by 37.17 %. The peak is due to a high energy pions(>2.5GeV), which have enough momentum to emit Cherenkov light and also because of the bad collection of light, there are a particular polar and azimuthal combination of angles where The Cherencov Detector cannot receive emitted light. .

Number of photoelectrons


E number of photoelectrons 27095 fits.gifE number of photoelectrons 27095 flag 10 fit with cut.gif


Table: Cherenkov fit values

Distributions amplitude mean width amplitude mean width
without cuts with cut(flag>10)
gauss(0) p0=2144+/-44.0 p1=5.342+/-0.343 p2=7.761+/-0.188 p0=1580+/-8.1 p1=3.75+/-0.06 p2=8.486+/-0.042
landau(3) p3=4.349e+04+/-2894 p4=1.049+/-0.026 p5=0.2197+/-0.0257 p3=8600+/-3648.7 p4=-3.861+/-1.414 p5=-4.88+/-1.41
gauss(6) p6=4960+/-270.6 p7=0.7345+/-0.0983 p8=0.8885+/-0.0594 p6=6219+/-54.2 p7=1.088+/-0.006 p8=0.6037+/-0.0052



When flag cut(flag>10 cut means that 6 superlayers were used in track fit) was applied the number of entries decreased by 37.17 % and the mean value for the number of photoelectrons is about 7-8. After 5<nphe<15 cut, the number of entries decreased by 66.63 %.The mean value of the nphe is ~9 which agrees with theory(mean value ~13).


Experiment B>0 without cuts flag>10 5<nphe<15 5<nphe<15 and flag>10
E momentum vs numb of photoelectrons 27095 exp without cuts 2.gif E momentum vs numb of photoelectrons 27095 exp with cuts flag 10 2.gif E momentum vs numb of photoelectrons 27095 exp with cuts ? nphe ?.gif E momentum vs numb of photoelectrons 27095 exp with cuts 5 nphe 15 flag 10.gif

Pions([math]\pi^-[/math])

[math]\pi^-[/math]_Momentum_vs_Number_of_Photons

After e_flag>10 cut, the number of entries decreased by 30.45 % and the mean value for the number of photons is ~9


Experiment B>0 without cuts e_flag>10 0<e_nphe<5 0<nphe<5 and e_flag>10
Pi momentum vs numb of photoelectrons 27095 exp without cuts 1 1.gif

Pi momentum vs numb of photoelectrons 27095 exp without cuts 2 1.gif Pi momentum vs numb of photoelectrons 27095 exp without cuts 3 1.gif

Pi momentum vs numb of photoelectrons 27095 exp with cuts flag 10 1 1.gif

Pi momentum vs numb of photoelectrons 27095 exp with cuts flag 10 2 1.gif Pi momentum vs numb of photoelectrons 27095 exp with cuts flag 10 3 1.gif

Pi momentum vs numb of photoelectrons 27095 exp with cuts ? nphe ?.gif Pi momentum vs numb of photoelectrons 27095 exp with cuts 0 nphe 5 flag 10.gif

Electron-pion contamination

Osipenko's CLAS Note 2004-20 File:CLAS Note-2004-020.pdf

EC_tot/P_vs_Number_of_Photoelectrons and EC_inner/P_vs_Number_of_Photoelectrons

Two types of cuts were applied on the distributions below, one on the energy deposited to the inner calorimeter [math]EC_{inner}/P\gt 0.08[/math] and another one on the total energy absorbed by the calorimeter [math]EC_{tot}/P\gt 0.2[/math], to improve the electron particle identification. In this case was used dst27095_05 file, the beam energy is 5.735 GeV and target NH3.

without cut [math]EC_{tot}/P[/math]_vs_nphe([math]EC_{tot}/P\gt 0.2[/math]) [math]EC_{inner}/P[/math]_vs_nphe ([math]EC_{inner}/P\gt 0.08[/math]) [math]EC_{tot}/P[/math]_vs_nphe([math]EC_{inner}/P\gt 0.08[/math])
Ec tot momentum vs numb of phe 27095 without cut.gif Ec tot momentum vs numb of phe 27095 with cut.gif Ec inner momentum vs numb of phe 27095 with cut ec inner momentum 0.08.gif Ec tot momentum vs numb of phe 27095 with cut ec inner momentum 0.08.gif


[math]EC_{inner}/P[/math]_vs_nphe([math]EC_{tot}/P\gt 0.2[/math]) [math]EC_{tot}/P[/math]_vs_nphe([math]EC_{tot}/P\gt 0.2[/math] and [math]EC_{inner}/P\gt 0.08[/math]) [math]EC_{inner}/P[/math]_vs_nphe ([math]EC_{tot}/P\gt 0.2[/math] and [math]EC_{inner}/P\gt 0.08[/math])
Ec inner momentum vs numb of phe 27095 with cut ec tot momentum 0.2.gif Ec tot momentum vs numb of phe 27095 with cut ec inner momentum 0.08 and ec tot momentum 0.2.gif Ec inner momentum vs numb of phe 27095 with cut ec tot momentum 0.2 and ec inner momentum0.08.gif



From the EC_tot/P_vs_Number_of_Photoelectrons histogram one can see that the released energy fraction([math]EC_{tot}/P[/math]) at ~1.5 nphe peak is much smaller than it should be for electrons. In conclusion, the ~1.5 nphe peak is produced by the tail of negatively charged particles(pions). To eliminate negatively charged pions [math]EC_{tot}/P\gt 0.2[/math] cut is applied on Calorimeter. After the cut was applied the number of entries decreased by ~33.47%.

E numb of photoelectrons with cuts 27095 ec inner p 0.08 and ec tot p 0.2 file dst27095 fitted with normalized gaussian and landau.gif

Gausn 0 fitting Histogram.gifGausn 6 fitting Histogram.gifLandaun 3 fitting Histogram.gif

Pion contamination in the electron sample (without cuts) =
[math] = \frac {Integral(gauss(0))}{Integral(gauss(0) + landau(3) + gauss(6))} =[/math]
[math] = \frac{2.126 \times 10^6}{ 2.126 \times 10^6 + 2.589 \times 10^5 + 4.659 \times 10^5} = [/math]
[math] = 0.74575 [/math]

with cut([math]EC_{tot}/P\gt 0.2[/math]) with cut([math]EC_{inner}/P\gt 0.08[/math]) with cuts([math]EC_{inner}/P\gt 0.08[/math] and [math]EC_{tot}/P\gt 0.2[/math])
E numb of photoelectrons with cuts 27095 ec tot p 0.2.gif E numb of photoelectrons with cuts 27095 ec inner p 0.08.gif E numb of photoelectrons with cuts 27095 ec inner p 0.08 and ec tot p 0.2.gif
Distributions amplitude mean width amplitude mean width amplitude mean width
with cut([math]EC_{tot}/P\gt 0.2[/math]) with cut([math]EC_{inner}/P\gt 0.08[/math]) with cuts([math]EC_{inner}/P\gt 0.08[/math] and [math]EC_{tot}/P\gt 0.2[/math])
gauss(0) p0=1919+/-17.8 p1=3.593+/-0.233 p2=8.269+/-0.134 p0=2025+/-9.4 p1=3.44+/-0.05 p2=8.442+/-0.037 p0=1732+/-8.4 p1=3.886+/-0.050 p2=8.069+/-0.037
landau(3) p3=8600+/-1.4 p4=-3.063+/-1.414 p5=-7.138+/-0.000 p3=8600+/-3648.7 p4=-3.092+/-1.414 p5=-10.91+/-0.000 p3=8600+/-5160.0 p4=-2.821+/-1.414 p5=-7.986+/-1.414
gauss(6) p6=5568+/-69.1 p7=1.164+/-0.006 p8=0.543+/-0.008 p6=5773+/-76.6 p7=1.162+/-0.008 p8=0.5562+/-0.0052 p6=4301+/51.3 p7=1.189+/-0.008 p8=0.5299+/-0.0059


E numb of photoelectrons ec inner 0.08 ec total 0.2 e momentum 3GeV.gif

e_numb_of_photoelectrons with the following cuts [math]EC_{inner}/P\gt 0.08[/math], [math]EC_{tot}/P\gt 0.2[/math]. To eliminate the photons produced by the negatively charged pions was used the cut on the momentum e_momentum<3 GeV. Because the high energy pions are able to produce photons, which are misidentified with photoelectrons.


Electrons nphe with cuts ecinner 0.08p ectotal 0.2p emomentum 3 nphe 2.5 file dst27095 with gauss fit.gif Electrons nphe with cuts ecinner 0.08p ectotal 0.2p emomentum 3 nphe 2.5 file dst27095 with gauss landau fit.gif Electrons nphe with cuts ecinner 0.06p ectotal 0.24p emomentum 3 nphe 0.5 file dst27095 with gauss fit.gif Electrons nphe with cuts ecinner 0.06p ectotal 0.24p emomentum 3 nphe 0.5 file dst27095 with gauss landau fit.gif


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