Transcript From Raw Data to Physics Results

From
Raw Data
to
Physics Results
Grass
2009/08/07
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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.
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Trigger and veto
M. Sumihama Ph.D. thesis, 2003
-Tagging system
If there happened BCS, the TAG got fired.
3.5 eV
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
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Trigger and veto
- around the spectrometer
5
M. Sumihama Ph.D. thesis, 2003
Trigger 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
6
Trigger and veto
-Aerogel Cerenkov counter (AC)
• Main background event are the e+epairs 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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Trigger and veto
- Time-of flights (TOF)
• Time-of flights of charged particles are
measured by a TOF wall.
• This is one of trigger.
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M. Sumihama Ph.D. thesis, 2003
Trigger and veto
-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.
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M. Sumihama Ph.D. thesis, 2003
Trigger and Veto
Hadron trig. TAG UPveto TRG  AC TOF
 e e trig. TAG UPveto TRG TOF
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M. Sumihama Ph.D. thesis, 2003
Detectors :
TPC（time projection chamber）
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J.Y. Chen Ph.D. defence
Raw data
12
Return to original the physics events
vertex
track
13
J.Y. Chen Ph.D. defence
Return to original the physics events
And then … ?
14
J.Y. Chen Ph.D. defence
From Track to momentum
• If a particle in a magnetic field B tesla has charge Q coulombs
and velocity v m/s, the magnetic force is
F = BQv 2
v
m
 BQv
r
 mv  BQr
or p  BQr (units : Kg  m  s 1 )
•
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 = Energy
pc=BQrc
(units:Joules(J))
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http://lppp.lancs.ac.uk/motioninb/experiment.html
From Track to momentum
•
One electron volt, 1 eV = 1.6x10-19 J or, expressed another way, 1 J =
(1/1.6x10-19)eV. Therefore the units of the equation, above, can be
converted to eV as follows:
pc 
•
BQrc
1.6  10
19
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 = 1.6x1019
C. Therefore the equation above reduces to:
pc=Brc
•
•
(units : Electronvo lts (eV))
(units: eV)
By substituting in the value for c, on the right hand side, we get
pc=Br·3×108
(units: eV)
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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
pairs 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:
∠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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c2 = a2 + b2 - 2ab cos C
Find rest masses by dE/dx
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J.Y. Chen Ph.D. defence
What happen?
B
A
γ
p
C
pp
K+K-
• We got what is
B and C.
pK
π + π – π0
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What is A?
K-
-invariant mass
A
γ
p
K
K
A


K
K
 E   E   EA 
     
 PxK   PxK   Px A 
 K    K    A 
 Py   Py   Py 
 P K    P K    Pz A 
 z   z   
E
K
E
K
E
Px  Px
K
 Px
K
A
A
 (M
K+
Pp  (Px  Px )  (Py
(M A ) 2  (E A ) 2  (Pp A ) 2
A
K
K 2
)  (Pp )  (M
K 2
K 2
K
 Py
)  (Pz
K 2
K
)  (Pp )
K 2
K 2
 Pz
K 2
)
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Do we miss something?
• Conservation of Baryon Number
NBarion
γ+p → X → K+ + Kγ+p → φ → K+ + K0 +1 → 0 → 0 + 0
NBarion
Q
γ+p → X +Y → K+ + Kγ+p → φ +Y → K+ + K0 +1 → 0 +1 → 0 + 0 +1
0 +1 → 0 +1 → 0 + 0 +1
• There are something else …
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What is Y?
-missing mass
p
K
K
Y



p
K
K
E   M   E   E   EY 
  
  K   K   Y 
 0   0   Px   Px   Px 
 0    0    P K    P K    P Y 
  
  y   y   y 
 E   0   P K   P K   P Y 
  
  z   z   z 

γ+p → φ +p → K+ + KNBarion 0 +1 → 0 +1 → 0 + 0 +1
Q 0 +1 → 0 +1 → 0 + 0 +1
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M0 of proton from PDG =0.938 GeV
Use the cross section ratio to find the number of colours
 (e e  qf q f )
2
R 

N
z
c f
f
 (e  e-      )


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Result
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End
Thanks
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27
Lancaster Particle Physics Package (LPPP).
http://lppp.lancs.ac.uk/index.html
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