Presentazione di PowerPoint - PowerPoint PPT Presentation

1 / 12
About This Presentation
Title:

Presentazione di PowerPoint

Description:

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 ... – PowerPoint PPT presentation

Number of Views:10
Avg rating:3.0/5.0
Slides: 13
Provided by: ubald
Category:

less

Transcript and Presenter's Notes

Title: Presentazione di PowerPoint


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
(No Transcript)
3
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
4
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 (??????)
5
  • TOWARDS LIMITS
  • For events for which the momentum has been
    measured
  • We can look for candidates
  • AN EXAMPLE
  • Require
  • Good track quality GLBHTR FLAG gt4
  • p(chi2)gt1
  • Negative momentum P3s(P)lt0
  • PT lt0.25 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

6
Mom distr of kink
7
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 1120/90 2 events We assume that the
number of wk is the same in cc and 0c events
8
COMPARISON OF CHARM and 0MU samples BOTH
golden selection we obtain fort the ratio
0mu/CHARMS FOR THE DIFFERENT TOPOLOGIES
C1 0.17-0.02
V2 0.17-0.01
C3 0.14-0.02
V4 V4 0.18-0.03
total total 0.16-0.01

The RATIO IS TOPOLOGY independent AS EXPECTED
9
Some COMMENTS
  • THE RATIO FOR KINKS
  • INDICATES THAT WK CONTRIBUTION
  • IS NEGLIGIBLE
  • THE RATIO INDICATES THAT
  • THE CONTRIBUTION OF cc
  • IN 0MU IS 0.1790k 15 K
  • ALL THE 0MU ARE 25K
  • SO REAL NC IS 25-155K NC/CC0.1
  • where IS MY MISTAKE ?????????

10
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 0.17223 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
11
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 tolerance 25 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 -P3.s(P) still negative and
Pt lt 250 mev and and to 18 if we require
teta(kink)gt50 mrad (no Pt cut) the efficiency
of momentum and charge determination is
155/4780.32
12
12
COMPARISON wih Gianfranca paper if we assume
that the number of negative and positive wk is
the same then we would expect
18/0.322110 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 1102/1002.2events 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 5100000lmed0.001eps(kin
k)eps(mom)180m L 180/1116.4m Assume lmed4 mm
(guess),Eps(kink)0.3,eps(mom)0.3 previous
limit paper L24.0-8
Write a Comment
User Comments (0)
About PowerShow.com