2006 Raw Survey Results - OpenRAW - Digital Image Preservation
From Raw Data to Physics Results
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Transcript of From Raw Data to Physics Results
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From Raw DataRaw Data
to Physics ResultsPhysics Results
Grass2009/08/07
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Data Analysis Chain• Have to collect data from many channels
on many sub-detectors (millions)• Decide to read out everything or throw
event away (Trigger)• Build the event (put info together)• Store the data• Analyze them• do the same with a simulation• Compare data and theory
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Trigger and veto
-Schematic view of the LEPS exp.Schematic view of the LEPS exp.
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TriggerTrigger and veto
-Tagging systemTagging system
SSD :Silicon strip detector
The precise hit position of recoil electron is measured by SSD layers.
PS : Plastic scintillator
If there are hits at PS associated with the hits in the SSD, we obtain the energy Ee’ with the hit position at SSD and obtain the photon energy by estimation.
We select the events finding only one hit in the region covered by the fired scintillators to reduce the background events
If there happened BCS, the TAG got fired.
3.5 eV
M. Sumihama Ph.D. thesis, 2003
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Trigger and veto
- around the spectrometer
M. Sumihama Ph.D. thesis, 2003
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TriggerTrigger and veto
-Trigger counter (TRG)
• The TRG is a plastic scitillation counter to identify the event signals from charged particles produced at the target.
• The trigger counter is used as reference counter to measure the time-of flight with the RF signal.
M. Sumihama Ph.D. thesis, 2003
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Trigger and vetoveto-Aerogel Cerenkov counter (AC)• Main background event are the e+e- p
airs producted at the target and at TRG in a measurement of hadronic reaction.
• When a particle with a velocity β>1/n passes through a transparent material with a refractive index n, Cerenkov lights are emitted.
--- n=1.03; β~0.97
M. Sumihama Ph.D. thesis, 2003
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TriggerTrigger and veto
- Time-of flights (TOF)• Time-of flights of charged particles are
measured by a TOF wall.• This is one of trigger.
M. Sumihama Ph.D. thesis, 2003
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Trigger and vetoveto
-Upstream-veto counter
• The photon beam partly converts to charged particles mainly by the e+e- pair production process in air, the residual gas or Al windows of the beam pipe.
• This counter is a plastic scintillator located at 4m upstream from the target.
M. Sumihama Ph.D. thesis, 2003
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Trigger and VetoTrigger and Veto
TOFTRGUPvetoTAGTOFACTRGUPvetoTAG
trig. ee trig. Hadron
-
M. Sumihama Ph.D. thesis, 2003
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Detectors :TPC ( time projection chambe
r )
J.Y. Chen Ph.D. defence
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Raw data
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Return to original the physics events
track
vertex
J.Y. Chen Ph.D. defence
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Return to original the physics events
And then … ?And then … ?J.Y. Chen Ph.D. defence
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From Track to momentum• If a particle in a magnetic field B tesla has charge Q coulombs an
d velocity v m/s, the magnetic force is
F = BQv
• The unit of Q is Coulombs (C), B is Tesla (T) and r is meters (m). If we multiply both sides of the equation by the speed of light, c = 3x108ms-1, then the units are now in Joules because:
Momentum x Speed = EnergyMomentum x Speed = Energy
pc=BQrc (units:Joules(J))
http://lppp.lancs.ac.uk/motioninb/experiment.html
):(units
1
2
smKgBQrporBQrmv
BQvrv
m
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From Track to momentum• One electron volt, 1 eV = 1.6x10-19 J or, expressed another way, 1 J = (1/1.6
x10-19)eV. Therefore the units of the equation, above, can be converted to eV as follows:
• Q is equal to the charge on the particle moving in the magnetic field. For this exercise Q is equal to the charge on one electron or proton = 1.6x10-19 C. Therefore the equation above reduces to:
pc=Brc (units: eV)
• By substituting in the value for c, on the right hand side, we get pc=Br·3×108 (units: eV)
•
(eV)) ltsElectronvo:(units . 191061
BQrc
pc
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From Track to momentum• or, because 1GeV = 1x109 eV
pc=0.3Br (units:GeV)
• Finally, by expressing the units in terms of c we obtain:
p=0.3Br (units:GeV/c)
• What we need to know are just B and r.• How could we know the radius?
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From Track to momentum-
How could we know the radius• AP = BP = CP = radii of the circle.• The machine applies Pythagoras theorem to pairs of the coordinates pai
rs to calculate AB, A and BC.• The cosine rule is then applied to ΔABC in order to calculate ∠ABC.• ΔBAP and ΔBCP are both isocoles. This can be used to show that:
c2 = a2 + b2 - 2ab cos C
• ∠ABC = ∠BCP + ∠ BAP. • Thus ∠APC = 360 - 2 ∠ABC• The cosine rule is now applied to ΔACP to find
the radius of the circle.
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Find rest masses by dE/dx
J.Y. Chen Ph.D. defence
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What happen?
• We got what is B and C.
A
B
C
γ p
pp
pKK+K-
π + π – π0
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What is A?
-invariant mass
Az
Ay
Ax
A
Kz
Ky
Kx
K
Kz
Ky
Kx
K
P
P
P
E
P
P
P
E
P
P
P
E
A K K
A
K-
K+
γ p
222
222
2222
)()()(
)()()(
)()()()(
Ap
AA
Kz
Kz
Ky
Ky
Kx
Kx
Ap
Kx
Kx
Ax
Kp
KKp
KKKA
PEM
PPPPPPP
PPP
PMPMEEE
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Do we miss something?• Conservation of Baryon Number
γ+p → X → K+ + K-
γ+p → φ → K+ + K-
NBarion 0 +1 → 0 → 0 + 0
• There are something else …γ+p → X +Y → K+ + K-
γ+p → φ +Y → K+ + K-
NBarion 0 +1 → 0 +1 → 0 + 0 +1
Q 0 +1 → 0 +1 → 0 + 0 +1
Wrong!!
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What is Y?-missing mass
Yz
Yy
Yx
Y
Kz
Ky
Kx
K
Kz
Ky
Kx
Kp
P
P
P
E
P
P
P
E
P
P
P
EM
E
EY K K p
0
0
0
0
0
M0 of proton from PDG =0.938 GeV
γ+p → φ +p → K+ + K- NBarion 0 +1 → 0 +1 → 0 + 0 +1 Q 0 +1 → 0 +1 → 0 + 0 +1
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Use the cross section ratio to find the number of colours
f fc-ff zN
qqR 2
)ee(
)e(e
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Result
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End
Thanks
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Lancaster Particle Physics Package (LPPP).http://lppp.lancs.ac.uk/index.html