* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Experiments
Survey
Document related concepts
Quantum electrodynamics wikipedia , lookup
Renormalization wikipedia , lookup
Density of states wikipedia , lookup
Conservation of energy wikipedia , lookup
Cross section (physics) wikipedia , lookup
Photon polarization wikipedia , lookup
Elementary particle wikipedia , lookup
Nuclear physics wikipedia , lookup
History of subatomic physics wikipedia , lookup
Matter wave wikipedia , lookup
Atomic theory wikipedia , lookup
Photoelectric effect wikipedia , lookup
Wave–particle duality wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Transcript
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 2N 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 2N 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. d2lng+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 (bza ). 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 (bv/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 2N 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 2r2 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= 2Nare2mec2(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 mpln(/)+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)237000 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 E01 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. ge+e-) and multiple scattering. There are several expressions for Lr in the literature, differing in their complexity. The simplest expression is: Lr 1 4re2 aN a ln( 183Z 1 / 3 )( Z 2 / A) Leo and the PDG have more complicated expressions: Lr 1 4re2 aN a [ln( 183Z 1 / 3 ) f ( Z )]( Z ( Z 1) / A) Lr 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 Eghv)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 (q0) r At very low energies, g0, 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 (ge+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 <qplane>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 13.6 103 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/tc 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 t0 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 2aE 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 2aE 1 dN 2a 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 2a 1 dN 2a 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 2asin 2 q 2 2asin 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 2a 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 2a sin q [ ] 2a 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