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Particle Detection
In order to detect a particle it must interact with matter!
The most important processes are electromagnetic:
Energy loss due to ionization (e.g. charged track in drift chamber)
heavy particles (not electrons!)
electrons and positrons
Energy loss due to photon emission (electrons, positrons)
bremsstrahlung
Interaction of photons with matter (e.g. EM calorimetry)
photoelectric effect
Compton effect
pair production
Coulomb Scattering (multiple scattering)
Other electromagnetic processes
cerenkov light (e.g. RICH counters)
scintillation light (e.g. trigger and TOF systems)
transition radiation (e.g. particle id at high momentum)
Can calculate the above effects with a combo of classical E&M and QED.
In most cases calculate approximate results, exact calculations very difficult.
880.A20 Winter 2002
Richard Kass
Bethe-Bloch Formula for Energy Loss
Average energy loss for heavy charged particles
heavy= mincident>>me
Energy loss due to ionization and excitation
proton, k, , m
Early 1930’s Quantum mechanics (spin 0)
Valid for energies <100’s GeV and b >>za (z/137)
2
dE
2
2 Z z

 2N a re me c 
dx
A b2
Fundamental constants
re=classical radius of electron
me=mass of electron
Na=Avogadro’s number
c=speed of light
Incident particle
z=charge of incident particle
b=v/c of incident particle
g=(1-b2)-1/2
Wmax=max. energy transfer
in one collision
Wmax 
2me (cbg ) 2
1  me / M 1  ( bg )  (me / M )
880.A20 Winter 2002
2
2
 2me g 2 v 2Wmax
2
ln(
)

2
b


I2


=0.1535MeV-cm2/g
Absorber medium
I=mean ionization potential
Z= atomic number of absorber
A=atomic weight of absorber
=density of absorber
d=density correction
C=shell correction
 2me (cbg )2
Richard Kass
Corrected Bethe-Bloch Energy Loss
2
dE
2
2 Z z

 2N a re me c 
dx c
A b2
 2me g 2 v 2Wmax
C
2
ln(
)

2
b

d

2

I2
Z 

d=parameter which describes how transverse electric field of incident particle
is screened by the charge density of the electrons in the medium.
d2lng+z, with z a material dependent constant (e.g. Table 2.1of Leo)
C is the “shell” correction for the case where the velocity of the incident particle
is comparable (or less) to the orbital velocity of the bound electrons (bza ).
Typically, a small correction (see Table 2.1 of Leo)
Other corrections due to spin, higher order diagrams, etc are small, typically <1%
PDG
880.A20 Winter 2002
Richard Kass
Average Energy loss of heavy charged particle
The incident particle’s speed (bv/c) plays an important role in the amount
of energy lost while traversing a medium.


dE
1
 2 ln( bg )  2 b 2  K1
dx b

K1 a constant
1/b2 term dominates at low momentum (p)
b=momentum/energy
ln(bg) dominates at very high momenta (“relativistic rise”)
b term never very important (always <1)
p=0.1 GeV/c
p=1.0 GeV/c
880.A20 Winter 2002

K
p
b
0.58
0.20
0.11
1/b2
2.96
25.4
89.0
ln(bg)
-0.34
-1.60
-2.24

K
p
b
0.99
0.90
0.73
1/b2
1.02
1.24
1.88
ln(bg)
1.97
0.71
0.06
CLEO data
He-propane gas
pions
kaons
protons
Richard Kass
Energy loss of electrons and positrons
Calculation of electron and positron energy loss due to ionization and excitation
complicated due to:
spin ½ in initial and final state
small mass of electron/positron
identical particles in initial and final state for electrons
Form of equation for energy loss similar to “heavy particles”
dE
Z 1

 2N a re2 me c 2
dx c
A b2

 2 (   2)
C
ln(
)

F
(

)

d

2
 2( I 2 / m c 2 )
Z 
e

is kinetic energy of incident electron in units of mec2.  =Eke/ mec2 =g1
2
2
Note:
F ( ) e  1  b 2 
F ( ) e
 / 8  (2  1) ln 2
1  g 1
(2g  1) ln 2
 1  b 2  
 
2
(  1)
8 g 
g2
Typo on P38 of Leo
2r2
b2 
14
10
4 
b2 
14
10
4 
 2 ln 2 




 23 
  2 ln 2 
 23 

2
3 
2
12 
(  2) (  2)
(  2) 
12 
(g  1) (g  1)
(g  1) 3 
For both electrons and positrons F() becomes a constant at very high incident energies.
Comparison of electrons and heavy particles (assume b=1):


2me c 2
dE

 2 ln(
)  A ln g  B
dx c 
I

880.A20 Winter 2002
electrons:
heavy
A
3
4
B
1.95
2
Richard Kass
Distribution of Energy Loss
Amount of energy lost from a charged particle going through material
can differ greatly from the average or mode (most probable)!
Measured energy loss of 3 GeV ’s and
2 GeV e’s through 90%Ar+10%CH4 gas
Landau and Vavilov energy
loss calculations
The long tail of the energy loss distribution makes particle ID using dE/dx difficult.
To use ionization loss (dE/dx) to do particle ID typically measure many samples and
calculate the average energy loss using only a fraction of the samples.
Energy loss of charged particles through a thin absorber (e.g. gas) very difficult
to calculate. Most famous calculation of thin sample dE/dx done by Landau.
880.A20 Winter 2002
Richard Kass
Landau model of energy loss for thin samples
Unfortunately, the central limit theorem is not applicable here so the energy
loss distribution is not gaussian.
The energy loss distribution can be dominated by a single large momentum
transfer collision. This violates one of the conditions for the CLT to be valid.
Conditions for Landau’s model:
1) mean energy loss in single collision < 0.01 Wmax, Wmax= max. energy transfer
2) individual energy transfers large enough that electrons can be considered free,
small energy transfers ignored.
3) velocity of incident particle remains same before and after a collision
f ( x,  )   (  ) / 
 ( ) 
1


 du exp( u ln u  u ) sin u
0
with  
1

[   (ln   ln   1  C )]
=energy loss in absorber
x=absorber thickness
= approximate average energy loss= 2Nare2mec2(Z/A)(z/b)2x
C=Euler’s constant (0.577…)
ln=ln[(1-b2)I2]-ln[2mec2b2)I2]+b2 ( is min. energy transfer allowed)
Must evaluate
integral
numerically
Can approximate f(x,) by:
f ( x,  ) 
1
1
exp(  (  e  ))
2
2
The most probable energy loss is mpln(/)+0.2+d]
Other (more sophisticated) descriptions exist for energy loss in thin samples by Symon,
Vavilov, Talman.
880.A20 Winter 2002
Richard Kass
Bremsstrahlung (breaking radiation)
Classically, a charged particle radiates energy when it is accelerated: dE/dt=(2/3)(e2/c3)a2
In QED we have to consider two diagrams where a real photon is radiated:
g
g
qf
qi
‘g’
Zf
Zi
qf
qi
+
‘g’
Zi
Zf
The cross section for a particle with mass mi to radiate a photon of E in a medium
with Z electrons is:
2
-2
d
Z ln E
 2
dE mi E
m behavior expected since classically
radiation  a2=(F/m)2
Until you get to energies of several hundred GeV bremsstrahlung is only important for
electrons and positrons:
(d/dE)|e/(d/dE)|m =(mm/me)237000
Recall ionization loss goes like the Z of the medium.
The ratio of energy loss due to radiation (brem.) and collisions (ionization) is for a particle
with energy E is:
(d/dE)| /(d/dE)| ZE/(600MeV)
rad
col
Define the critical energy, Ecrit, as the energy where (d/dE)|rad=(d/dE)|col.
Ecrit  600MeV for hydrogen
Ecrit  7.3MeV for lead
880.A20 Winter 2002
Another popular parameterization is
Ecrit  800MeV/(Z+1.2) (Leo P41)
Richard Kass
Bremsstrahlung (breaking radiation)
The energy loss due to radiation of an electron with energy E0 can be calculated:

dE
N
dx
Eg , max

0
Eg
d
dEg  NE0 rad with  rad  E01
dEg
N=atoms/cm3=Na/A
Eg,max=E0-mec2
Eg , max

0
Eg
d
dEg
dEg
(=density, A=atomic #)
The most interesting case for us is when the electron has several hundred MeV or
more, i.e. E0>>137mec2Z1/3. For this case we have:
 rad  4Z 2 re2a [ln( 183Z 1 / 3  1 / 18  f ( Z )]
Thus the total energy lost by an electron traveling dx due to radiation is:
dE

 4 Z 2 re2 a[ln( 183Z 1 / 3  1 / 18  f ( Z )] NE0
dx
We can rearrange the energy loss equation to read:

dE
 N rad dx
E
Since rad is independent of E we can integrate this equation to get:
E ( x )  E0 e  x / Lr
Lr is the radiation length, the distance the electron travels to lose all but 1/e of its original energy.
880.A20 Winter 2002
Richard Kass
Radiation Length (Lr)
The radiation length is a very important quantity describing energy loss of electrons
traveling through material. We will also see Lr when we discuss the mean free path for
pair production (i.e. ge+e-) and multiple scattering.
There are several expressions for Lr in the literature, differing in their complexity.
The simplest expression is:
Lr 1  4re2 aN a ln( 183Z 1 / 3 )( Z 2 / A)
Leo and the PDG have more complicated expressions:
Lr 1  4re2 aN a [ln( 183Z 1 / 3 )  f ( Z )]( Z ( Z  1) / A)
Lr 1  4re2 aN a [ Z 2 ( Lrad1  f ( Z ))  ZLrad 2 )]
Leo, P41
PDG
Lrad1 is approximately the “simplest expression” and Lrad2 uses 1194Z-2/3 instead of 183Z-1/3, f(z) is an infinite sum.
Both Leo and PDG give an expression that fits the data to a few %:
716.4 A
Lr 
( g  cm 2 )
Z ( Z  1) ln( 287 / Z )
The PDG lists the radiation length of lots of materials including:
Air: 30420cm, 36.66g/cm2
H2O: 36.1cm, 36.1g/cm2
Pb: 0.56cm, 6.37g/cm2
880.A20 Winter 2002
teflon: 15.8cm, 34.8g/cm2
CsI: 1.85cm, 8.39g/cm2
Be: 35.3cm, 65.2g/cm2
Leo also has a table of
radiation lengths on P42
but the PDG list is more
up to date and larger.
Richard Kass
Interaction of photons (g’s) with matter
There are three main contributions to photon interactions:
Photoelectric effect (Eg < few MeV)
Compton scattering
Pair production (dominates at energies > few MeV)
Contributions to photon interaction cross section for lead
including photoelectric effect (), rayleigh scattering (coh),
Compton scattering (incoh), photonuclear absorbtion (ph,n),
pair production off nucleus (Kn), and pair production
off electrons (Ke).
Rayleigh scattering (coh) is the classical physics process
where g’s are scattered by an atom as a whole. All electrons
in the atom contribute in a coherent fashion. The g’s energy
remains the same before and after the scattering.
A beam of g’s with initial intensity N0 passing through a medium is
attenuated in number (but not energy) according to:
dN=-mNdx or N(x)=N0e-mx
With m= linear attenuation coefficient which depends on the total interaction
cross section (total= coh+ incoh + +).
880.A20 Winter 2002
Richard Kass
Photoelectric effect
The photoelectric effect is an interaction where the incoming photon (energy Eghv)is
absorbed by an atom and an electron (energy=Ee) is ejected from the material:
Ee= Eg-BE
Here BE is the binding energy of the material (typically a few eV).
Discontinuities in photoelectric cross section due to discrete binding energies of atomic
electrons (L-edge, K-edge, etc).
Photoelectric effect dominates at low g energies (< MeV) and hence gives low energy e’s.
Exact cross section calculations are difficult due to atomic effects.
Cross section falls like Eg-7/2
Cross section grows like Z4 or Z5 for Eg> few MeV
Einstein wins Nobel prize in 1921 for his
work on explaining the photoelectric effect.
Energy of emitted electron depends on energy
of g and NOT intensity of g beam.
880.A20 Winter 2002
Richard Kass
Compton Scattering
Compton scattering is the interaction of a real g with an atomic electron.
gou
gin

q

t
Solve for energies and angles
using conservation of energy
and momentum
me c 2
cos q  1 
( Eg ,in  Eg ,out )
E
E
electron
g ,in g ,out
The result of the scattering is a “new” g with less energy and a different direction.
Eg ,in
Not the
Eg ,out 
with g  Eg ,in / me c 2
1  g (1  cos q)
usual g!
Kinetic Energy of Electron  T  Eg ,in  Eg ,out  Eg ,in
g (1  cos q)
1  g (1  cos q)
The Compton scattering cross section was one of the first (1929!) scattering cross
sections to be calculated using QED. The result is known as the Klein-Nishima
cross section.
re2
re2 E g ,out 2 E g ,out E g ,in
d
g 2 (1  cos q) 2
2

(1  cos q 
)
(
) (

 sin 2 q)
2
d 2[1  g (1  cos q]
1  g (1  cos q)
2 E g ,in
E g ,in E g ,out
At high energies, g>>1, photons are scattered mostly in the forward direction (q0)
r
At very low energies, g0, K-N reduces to the classical result: dd  2 (1  cos q)
2
e
880.A20 Winter 2002
2
Richard Kass
Compton Scattering
At high energies the total Compton scattering cross section can be approximated by:
8
3
 comp  ( re2 )( )(ln( 2g )  1 / 2)
3
8g
(8/3)re2=Thomson cross section
From classical E&M=0.67 barn
We can also calculate the recoil kinetic energy (T) spectrum of the electron:
re2
d
s2
s
2

(
2


(
s

)) with s  T / E g ,in
2 2
2
2
dT me c g
g (1  s )
(1  s )
g
This cross section is strongly peaked around
Tmax:
2g
Tmax  E g ,in
Tmax is known as the
Compton Edge
1  2g
Kinetic energy distribution
of Compton recoil electrons
880.A20 Winter 2002
Richard Kass
Pair Production (ge+e-)
This is a pure QED process.
A way of producing anti-matter (positrons).
ee+
gi
n
Z
g
e+
v
Z
Nucleus or electron
+ gi
n
Z
g
v
eZ
Threshold energy for
pair production in field
of nucleus is 2mec2, in
field of electron 4mec2.
Nucleus or electron
First calculations done by Bethe and Heitler using Born approximation (1934).
At high energies (Eg>>137mec2Z-1/3) the pair production cross sections is
constant.
pair =4Z2are2[7/9{ln(183Z-1/3)-f(Z)}-1/54]
Neglecting some small correction terms (like 1/54, 1/18) we find:
pair = (7/9)brem
The mean free path for pair production (pair) is related to the radiation length (Lr):
pairbeam
=(9/7)ofLrg’s with initial intensity N0 passing through
Consider again a mono-energetic
a medium. The number of photons in the beam decreases as:
N(x)=N0e-mx
The linear attenuation coefficient (m) is given by: m= (Na/A)(photo+ comp + pair).
For compound mixtures, m is given by Bragg’s rule: (m/)=w1(m1/1)++ wn(mn/n)
with wi the weight fraction of each element in the compound.
880.A20 Winter 2002
Richard Kass
Multiple Scattering
A charged particle traversing a medium is deflected by many small angle scatterings.
These scattering are due to the coulomb field of atoms and are assumed to be elastic.
In each scattering the energy of the particle is constant but the particle direction changes.
In the simplest model of multiple scattering we ignore large angle scatters.
In this approximation, the distribution of scattering angle qplane after traveling a distance x
through a material with radiation length =Lr is approximately gaussian:
dP(q plane )
dq plane
q plane
1

exp[
]
2
q 0 2
2q 0
2
with q 0 
13.6MeV
z x / Lr (1  0.038 ln{x / Lr })
bpc
In the above equation b=v/c, and p=momentum of incident particle
The space angle q= qplane
The average scattering angle <qplane>=0, but the RMS scattering angle <qplane>1/2= q0
Some other quantities
of interest are given in
The PDG:
880.A20 Winter 2002
1 rms
1
q plane  q 0
3
3
1
1
rms
y rms

x
q

xq 0
plane
plane
3
3
1
1
rms
s rms
xq plane

xq 0
plane 
4 3
3
 rms
plane 
The variables s, y, q,are
correlated, e.g. yq=3/2
Richard Kass
Why we hate Multiple Scattering
Multiple scattering changes the trajectory of a charged particle.
This places a limit on how well we can measure the momentum of a charged particle
Trajectory of charged particle
(charge=z) in a magnetic field.
in transverse B field.
s=sagitta
L/2
r=radius of curvature
The apparent sagitta due to MS is:
s rms
plane

1
4 3
rms
Lq plane
L 13.6  103

Lq 0 
z L / Lr
4 3
4 3
pb
1
L2
L2
0.3L2 zB

The sagitta due to bending in B field is: s B  
8r 8 p
8 pc
0.3zB
GeV/c, m
GeV/c, m, Tesla
The momentum resolution dp/p is just the ratio of the two sigattas:
rms
dp s plane

p
sB
L 13.6  10 3
z L / Lr
L / Lr
p
b
4 3
3


52
.
3

10
0.3L2 zB
bLB
8p
Independent of p
As an example let L/Lr=1%, B=1T, L=0.5m then dp/p 0.01/b. Typical values
Thus for this example MS puts a limit of 1% on the momentum measurement
880.A20 Winter 2002
Richard Kass
Scintillation Devices
As a charged particle traverses a medium it excites the atoms (or molecules)
in the the medium. In certain materials called scintillators a small fraction
energy released when the atoms or molecules de-excite goes into light.
ENERGY IN LIGHT OUT
The use of materials that scintillate is one of the most common experimental
techniques in physics.
Used by Rutherford in his scattering experiments
Scintillation light can be used to:
Signal the presence of a charged particle
Measure energy since the amount of light is proportional to energy deposition
Measure the time it takes for a charged particle to travel a known distance
(“time of flight technique”)
There are lots of different types of materials that scintillate:
non-organic crystals (NaI, CsI)
organic crystals (Anthracene)
Organic plastics (see Table on next page)
Organic liquids (toluene, xylene)
880.A20 Winter 2002
Richard Kass
Scintillators
Properties of common plastic scintillators
Emission spectrum of NE102A
Plastic scintillator
violet
blue
Typical cost 1$/in2
A typical plastic
Scintillator system
880.A20 Winter 2002
Richard Kass
Photomultiplier tubes
light
e’s
violet blue green
Electric field accelerates electrons
Quantum efficiency of bialkali cathode vs wavelength
Electrons crash into dynodes  create more electrons
We need a way to convert the scintillation photons into an electrical signal.
Photons  photoelectric effect electrons
Use a photomultiplier tube to convert scintillation light into electrical current
Properties of phototubes:
In situations where a lot of light
very high gain, low noise current amplifier
is produced (>103 photons) a
6
gains 10 possible
photodiode can be used in place of a
possible to count single photons
phototube, e.g. CLEO’s calorimeter
Off the shelf item, buy from a company
wide variety to choose from (size, gain, sensitivity)
tube costs range from $102-$103
Sensitive to magnetic fields (shield against earth’s): use “mu-metal”
880.A20 Winter 2002
Richard Kass
Scintillation Counter Example
Some typical parameters for a plastic scintillation counter are:
energy loss in plastic scintillator:
scintillation efficiency of plastic:
collection efficiency (# photons reaching PMT):
quantum efficiency of PMT
2MeV/cm
1 photon/100 eV
0.1
0.25
What size electrical signal can we get from a plastic scintillator 1 cm thick?
A charged particle passing perpendicular through this counter:
deposits 2MeV which produces 2x104g’s
of which 2x103g’s reach PMT which produce 500 photo-electrons
Assume the PMT and related electronics have the following properties:
PMT gain=106 so 500 photo-electrons produces 5x108 electrons =8x10-11C
Assume charge is collected in 50nsec (5x10-8s)
current=dq/dt=(8x10-11 coulombs)/(5x10-8s)=1.6x10-3A
Assume this current goes through a 50  resistor
V=IR=(50  )(1.6x10-3A)=80mV (big enough to see with O’scope)
So a minimum ionizing particle produces an 80mV signal.
What is the efficiency of the counter? How often do we get no signal (zero PE’s)?
The prob. of getting n PE’s when on average expect <n> is a Poisson process:
 n n e n 
P ( n) 
n!
The prob. of getting 0 photons is e-<n> =e-500 0. So this counter is 100% efficient.
Note: a counter that is 90% efficient has <n>=2.3 PE’s
880.A20 Winter 2002
Richard Kass
Time of flight with Scintillators
Time of Flight (TOF) is a particle identification technique.
measure particle speed and momentum determine mass
t=x/v=x/(bc) with b=pc/E=pc/[(mc2)2+(pc)2]1/2
x(( mc) 2  p 2 )1/ 2
t
pc
Actually, we measure
the time it takes for the
particle to travel a
known distance.
Consider two particles with different masses but same momentum:
2
2
2
2
2
2
2
2
2
x
((
m
c
)

p
)
x
((
m
c
)

p
)
x
(
m

m
1
2
1
2)
t12  t22 


( pc) 2
( pc) 2
p2
t12  t22  (t1  t2 )(t1  t2 )
x
x 2 (m12  m22 )
t1  t2 
(t1  t2 ) p 2
For high momentum (e.g. p>1 GeV/c for ’s):
t1+t2=2t and x/tc
x(m12  m22 )
x(m12  m22 )
t1  t2 
 1667
psec/meter
2
2
2cp
p
880.A20 Winter 2002
Richard Kass
Time of Flight with Scintillators
x(m12  m22 )
x(m12  m22 )
t  t1  t2 
 1667
psec/meter
2
2
2cp
p
As an example, assume m1=m (140MeV) , m2=mk (494MeV), and x=10m
t=3.8 nsec for p=1 GeV
t=0.95 nsec for p=2GeV
Scintillator+phototubes are capable of measuring such small time differences
Time resolution of a “good” TOF system is 150ps (0.15 ns)
In colliding beam experiments, 0.5 <x< 1 m small x puts a limit of t.
For x=1 m, p=1 GeV K/ separation t0 psec < 3  separation
x =1 meter
1.4 GeV/c ’s and K’s
No pulse
height correction
with pulse
height correction
880.A20 Winter 2002
Richard Kass
Cerenkov Light
The Cerenkov effect occurs when the velocity of a charged particle traveling through
a dielectric medium exceeds the speed of light in the medium.
Index of refraction (n) = (speed of light in vacuum)/(speed of light in medium)
Will get Cerenkov light when:
speed of particle
v
1
 b
speed of light in medium c
n
For water n=1.33, will get Cerenkov light if v > 2.25x1010 cm/s
Huyghen’s wavefronts
radiation
(c/n)t
q
bct
1
Angle of Cerenkov Radiation: cos q 
nb
880.A20 Winter 2002
No radiation
In a time t wavefront moves (c/n)t
but particle moves bct.
Richard Kass
Threshold Momentum for Cerenkov Radiation
Example: Threshold momentum for Cerenkov light:
1
n
1
1
gt 


bt 
n
1  b t2
n2 1 bt n2 1
b tg t 
1
n2 1

1
(n  1)(n  1)
Medium
helium
CO2
H2O
glass
d=n-1
3.3x10-5
4.3x10-4
0.33
0.46-0.75
For gases it is convenient to let d=n-1. Then we have:
b tg t 
1
d (d  2)
The momentum (pt) at which we get Cerenkov radiation is:
pt  mb tg t 
m
d (d  2)
For a gas d+2  so the threshold momentum can be approximated by:
pt  mb tg t 
m
2d
For helium d=3.3x10-5 so we find the following thresholds:
electrons 63 MeV/c
pions
17 GeV/c
880.A20 Winter 2002
kaons
protons
61 GeV/c
115GeV/c
Richard Kass
gt
123
34
1.52
1.37-1.22
Number of photons from Cerenkov Radiation
From classical electrodynamics (Frank&Tamm 1937, Nobel Prize 1958) we
find the following for the energy loss per wavelength () per dx for charge=1, bn>1:
dE
2aE
1
 2 [1  2
]
dxd

b n ( ) 2
For example see Jackson section 13.5
With a=fine structure constant, n() the index of refraction which in general depends
on the wavelength () of light.
We can re-write the above in terms of the number of photons (N) using:
dN=dE/E
dE
2aE
1
dN
2a
1
 2 [1  2
]


[
1

]
2
2
2
2
dxd

b n( )
dxd

b n(  )
We can simplify the above by considering a region were n() is a constant=n:
dN
2a
1
dN
2a 2
1
2
2

[
1

]


sin q
1 2
 1  cos q  sin q 
2
2
2
2
2
dxd


b
n
(

)
dxd


b n ( )
We can calculate the number of photons/dx by integrating over the wavelengths that
can be detected by our phototube (1, 2):
2
dN
d
1
1
 2asin 2 q  2  2asin 2 q[  ]
dx

1  2
1
880.A20 Winter 2002
Note: if we are using a phototube
with a photocathode efficiency that
varies as a function of  then we have:
2
dN
f ( )d
2
 2a sin q 
dx
2
1
Richard Kass
Number of photons from Cerenkov Radiation
For a typical phototube the range of wavelengths (1, 2) is (350nm, 500nm).
dN
1 1
1 1 105
2
2
 2a sin q [  ]  2a sin q [
 ]
 390 sin 2 q photons/cm
dx
1 2
3.5 5 cm
We can simplify using:
sin 2 q  1  cos 2 q  1 
1
( bn  1)( bn  1)
(n  1)( n  1)


b 1
b 2n2
b 2n2
n2
For a highly relativistic particle going through a gas the above reduces to:
sin 2 q 
( bn  1)( bn  1)
(n  1)( n  1)

 2(n  1)
2 2
2
b

1
b

1, gas
b n
n
dN
 780(n  1) photons/cm
dx
GAS
For He we find: 2-3 photons/meter (not a lot!)
For CO2 we find: ~33 photons/meter
For H2O we find: ~34000 photons/meter
Photons are preferentially emitted
at small ’s (blue)
For most Cerenkov counters the photon yield is limited (small) due to space
limitations, the index of refraction of the medium, and the phototube
quantum efficiency.
880.A20 Winter 2002
Richard Kass
Types of Cerenkov Counters
There are three different types of Cerenkov counters used to identify particles.
Listed in order of their sophistication they are:
Threshold counter (on/off device)
Differential counter (makes use of the angle of the Cerenkov radiation)
Ring imaging counter (makes use of the “cone” of light)
Each of the above counter is designed to work in a certain momentum range.
Threshold counter:
Identify the particle(s) which give off light.
Can use to separate electrons from heavier particles (, K, p) since electrons
will give off light at a much lower momentum (e.g. 68 MeV/c vs 17 GeV/c for He)
Problems with device:
above a certain momentum several particles will give light.
usually threshold counters use gas which implies low light levels (n-1 small)
low light levels leads to inefficiency, e.g. <ng>=3, the prob. of zero photons: P(0)=e-3=5%!
Phototubes must be shielded from magnetic fields above a few tenths of a gauss.
880.A20 Winter 2002
Richard Kass
Types of Cerenkov Counters
Differential Cerenkov Counter:
Makes use of the angle of Cerenkov radiation and only samples light at certain angles.
For fixed momentum cosq is a function of mass:
m2  p2
1
1
cos q 


nb n( p / E )
np
Differential cerenkov counters typically on work over a fixed momentum range
(good for beam monitors, e.g. measure  or K content of beam).
Problems with differential Cerenkov counters:
Optics are usually complicated.
Have problems in magnetic fields since phototubes must be shielded from B-fields
above a few tenths of a gauss.
Not all light will make it to phototube
880.A20 Winter 2002
Richard Kass
Ring Imaging Cerenkov Counters (RICH)
RICH counters use the cone of the Cerenkov light.
The ½ angle (q) of the cone is given by:
2
2 

1
1
1  m  p 
cosq 
 q  cos


nb
np


q
r
L
The radius of the cone is: r=Ltanq, with L the distance to the where the ring is imaged.
For a particle with p=1GeV/c, L=1 m, and LiF as the medium (n=1.392) we find:

K
P
q(deg)
43.5
36.7
9.95
r(m)
0.95
0.75
0.18
Great /K/p separation!
Thus by measuring p and r we can identify what type of particle we have.
Problems with RICH:
optics very complicated (projections are not usually circles)
readout system very complicated (e.g. wire chamber readout, 105-106 channels)
elaborate gas system
photon yield usually small (10-20), only a few points on “circle”
880.A20 Winter 2002
Richard Kass
CLEO’s Ring imaging Cerenkov Counter
The figures below show the CLEO III RICH structure. The radiator is LiF, 1 cm thick,
followed by a 15.7 cm expansion volume and photon detector consisting of a wire chamber
filled with a mixture of TEA and CH4 gas. TEA is photosensitive. The resulting photoelectrons
are multiplied by the HV on the wires and the resulting signals are sensed by a rectangular
array of pads coupled with highly sensitive electronics.
880.A20 Winter 2002
Richard Kass
CLEO’s Ring imaging Cerenkov Counter
Lithium Floride (LiF) radiator
Assembled radiators.
They are guarded by
Ray Mountain. Without
Ray “living”at the factory
that produced the LiF
radiators we would still
be waiting for the order
to be completed.
Assembled
photodetectors
A photodetector:
CaF2 window+cathode pads
880.A20 Winter 2002
Richard Kass
Performance of CLEO’s RICH
Number of detected
photons on 5 GeV
electrons
D*’s without/with
RICH information
Preliminary data
on /K separation
A track in the
RICH
880.A20 Winter 2002
Richard Kass
SuperK
SuperK is a water RICH. It uses phototubes to measure the Cerenkov ring.
Phototubes give time and pulse height information
481 MeV muon neutrino produces 394 MeV muon which later decays at rest into 52 MeV electron.
The ring fit to the muon is outlined. Electron ring is seen in yellow-green in lower right corner.
This is perspective projection with 110 degrees opening angle, looking from a corner of the Super-K
detector (not from the event vertex). Color corresponds to time PMT was hit by Cerenkov photon from
the ring. Color scale is time from 830 to 1816 ns with 15.9 ns step. In the charge weighted time histogram to
the right two peaks are clearly seen, one from the muon, and second one from the delayed electron
from the muon decay. Size of PMT corresponds to amount of light seen by the PMT.
From: http://www.ps.uci.edu/~tomba/sk/tscan/pictures.html
For water n=1.33
For b=1 particle cosq=1/1.33, q=41o
SuperK has: 50 ktons of H2O
Inner PMTS: 1748 (top and bottom) and 7650 (barrel)
outer PMTs: 302 (top), 308 (bottom) and 1275(barrel)
880.A20 Winter 2002
From SuperK site
Richard Kass