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ANALISYS Of FINAL MANUAL SCANNING DATA U DORE june 3 204

ANALISYS Of FINAL MANUAL SCANNING DATA U DORE june 3 204 THE ANALYS HAS BEEN MADE USING ASCII FILES CONTAINING MANUAL SCANNING RESULTS 1) GOLDEN selection 0MU data ev 661 2) SATO selection 0MU ev 781

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ANALISYS Of FINAL MANUAL SCANNING DATA U DORE june 3 204

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  1. ANALISYS Of FINAL MANUAL SCANNING DATA U DORE june 3 204 THE ANALYS HAS BEEN MADE USING ASCII FILES CONTAINING MANUAL SCANNING RESULTS 1) GOLDEN selection 0MU data ev 661 2) SATO selection 0MU ev 781 3)GOLDEN selection CHARMS ev 2816 In THE following I will give some information and some personal considerations waiting for comments and CORRECTIONS

  2. The netscan tracks associated with the kink have been matched with the dt tracks Results in column 4 Notes At least one muon If several association are possible choose the bes So for 62/81 tracks is possible to have Momentum determination Charm kink will have positive momentum Tau decays negative

  3. COMPARING SATO and GOLDEN SELECTION We can measure efficiencies of the two selections WE HAVE SATO GOLDEN events Yes no 49 No yes 22 Yes yes 59 From these numbers we obtain eps(Sato)= 72% eps(Golden)= 54% THESE efficiencies are only the selection criteria ones To have total efficiency multiply by location and netscan Ones LUCA gives for netscan and selection 24% (??????)

  4. TOWARDS LIMITS • For events for which the momentum has been measured • We can look for candidates • AN EXAMPLE • Require • Good track quality GLBHTR FLAG >4 • p(chi2)>1% • Negative momentum P+3*s(P)<0 • PT <0.25 GEV • 5) no muons associated to the event • We obtain • SATO selection 1 event • GOLDEN “ 0 events • NEED efficiencies • Background estimate for golden selection • compare charms with 0 mu

  5. Mom distr of kink

  6. COMPARISON OF charms kinks with 0mu kinks IF we apply same cuts for kinks candidate (except the nomuon requirement) we have 11 EVENTS Charms have been obtained for 90kevents So for 20k 0c we expect 11*20/90 =2 events We assume that the number of wk is the same in cc and 0c events

  7. COMPARISON OF CHARM and 0MU samples BOTH golden selection we obtain fort the ratio 0mu/CHARMS FOR THE DIFFERENT TOPOLOGIES The RATIO IS TOPOLOGY independent AS EXPECTED

  8. Some COMMENTS • THE RATIO FOR KINKS • INDICATES THAT WK CONTRIBUTION • IS NEGLIGIBLE • THE RATIO INDICATES THAT • THE CONTRIBUTION OF cc • IN 0MU IS 0.17*90k =15 K • ALL THE 0MU ARE 25K • SO REAL NC IS 25-15=5K NC/CC=0.1 • where IS MY MISTAKE ?????????

  9. TOWARDS LIMITS We can think of using events wit no measurement of momentum GOLDEN SELCTION Make a cut at 50 mrad Golden 0mu 36 candidates Golden charms 223 “ THE EXPECTED NUMBER OF EVENTS will be 0.17*223 =38 so the effect will be 36-38=-2 Upper limit 90% CL 6 events THE ratio of sensitivity with the measured Momentum sample will 6/2.3*(0.32/0.68)=~ 1 0.32 effic for momentum reconstruction

  10. EXPERIMENTAL DETERMINATION OF WK BACKGROUND the charm sample has 478 kink events (461 good). of these 382 have a matching between the manual scan angular coeff and the ang coeff of one track as mesaured in the electronic detector (angular tolerance25 mrad) 203 have a good momentum measurement in the DT (rank bigger than 4) 155 have a probability of the momentum fit larger than 1%. there are 22 events with a negative momentum This number reduces to 11 if we require that -P+3.*s(P) still negative and Pt < 250 mev and and to 18 if we require teta(kink)>50 mrad (no Pt cut) the efficiency of momentum and charge determination is 155/478=0.32

  11. 12 COMPARISON wih Gianfranca paper if we assume that the number of negative and positive wk is the same then we would expect 18/0.32*2~~=110 wk in a situation in which all kinks are considered the charm sample is obtained in a sample of 100k CC events in the nubar sample of 2000 eventes (G Derosa) we would have 110*2/100=2.2events in the DE Rosa paper an expected background of 0.5+-0.3 is quoted Determination of interaction lenght We have in average 5 shower tracks so the total lenght f track scanned will be 5*100000*lmed*0.001*eps(kink)*eps(mom)=180m L =180/11=16.4m Assume lmed=4 mm (guess),Eps(kink)=0.3,eps(mom)=0.3 previous limit paper L=24.0+-8

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