1、880.P20 Winter 2006,Richard Kass,1,Interaction of Particles with Matter,In order to detect a particle it must interact with matter! The most important interaction processes are electromagnetic: Charged Particles:Energy loss due to ionization (e.g. charged track in drift chamber)heavy particles (not
2、electrons/positrons!)electrons and positronsEnergy loss due to photon emission (electrons, positrons) bremsstrahlung Photons:Interaction of photons with matter (e.g. EM calorimetry)photoelectric effectCompton effectpair production Other important electromagnetic processes:Multiple Scattering (Coulom
3、b scattering)scintillation light (e.g. energy, trigger and TOF systems)Cerenkov radiation (e.g. optical (DIRC/RICH), RF (Askaryan) 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 resul
4、ts, exact calculations very difficult.,880.P20 Winter 2006,Richard Kass,2,Classical Formula for Energy Loss,Average energy loss for a heavy charged particle: mass=M, charge=ze, velocity=v Particle loses energy in collisions with free atomic electrons: mass=m, charge=e, velocity=0 Assume the electron
5、 does not move during the “collision” & Ms trajectory is unchanged Calculate the momentum impulse (I) to electron from M:,M, ze, v,b=impact parameter,Next, use Gausss law to evaluate integral assuming infinite cylinder surrounding M:,The energy gained by the electron is DE=p2/2me=I2/2me The total en
6、ergy lost moving a distance dx through a medium with electron density Ne from collisions with electrons in range b+db is:,Therefore the energy loss going a distance dx is:,But, what should be used for the max and min impact parameter? The minimum impact parameter is from a head on collision Although
7、 E&M is long range, there must be cutoff on bmax, otherwise dE/dxRelate bmax to “orbital period” of electron, interaction short compared to period,Must do the calculation using QM using momentum transfer not impact parameter.,880.P20 Winter 2006,Richard Kass,3,Incident particle z=charge of incident
8、particle b=v/c of incident particle g=(1-b2)-1/2 Wmax=max. energy transfer in one collision,Bethe-Bloch Formula for Energy Loss,Fundamental constants re=classical radius of electron me=mass of electron Na=Avogadros number c=speed of light,=0.1535MeV-cm2/g,Average energy loss for heavy charged partic
9、les Energy loss due to ionization and excitation Early 1930s Quantum mechanics (spin 0) Valid for energies za (z/137),heavy= mincidentme proton, k, p, m,Absorber medium I=mean ionization potential Z= atomic number of absorber A=atomic weight of absorber r=density of absorber d=density correction C=s
10、hell correction,Note: the classical dE/dx formula contains many of the same features as the QM version: (z/b)2, & ln,880.P20 Winter 2006,Richard Kass,4,Corrected Bethe-Bloch Energy Loss,d=parameter which describes how transverse electric field of incident particleis screened by the charge density of
11、 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 (bza ). Typically, a small correction (see T
12、able 2.1 of Leo),Other corrections due to spin, higher order diagrams, etc are small, typically 1%,PDG plots,880.P20 Winter 2006,Richard Kass,5,Average Energy loss of heavy charged particle,The incident particles speed (b=v/c) plays an important role in the amount of energy lost while traversing a m
13、edium.,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)bg=p/m, g=E/m, b=E/p,p=0.1 GeV/c,p=1.0 GeV/c,Data from BaBar experiment,880.P20 Winter 2006,Richard Kass,6,Energy loss of el
14、ectrons and positrons,Calculation of electron and positron energy loss due to ionization and excitation complicated due to:spin in initial and final state small mass of electron/positronidentical particles in initial and final state for electrons Form of equation for energy loss similar to “heavy pa
15、rticles”,t is kinetic energy of incident electron in units of mec2. t =Eke/ mec2 =g-1,Note: Typo on P38 of Leo2r2t,For both electrons and positrons F(t) becomes a constant at very high incident energies.,Comparison of electrons and heavy particles (assume b=1):,A B electrons: 3 1.95 heavy 4 2,880.P2
16、0 Winter 2006,Richard Kass,7,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 ps and 2 GeV es through 90%Ar+10%CH4 gas,Landau (L) and Vavilov energy loss calcula
17、tions,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 andcalculate the average energy loss using only a fraction of the samples.,Energy loss of charged particles through a thin abso
18、rber (e.g. gas) very difficultto calculate. Most famous calculation of thin sample dE/dx done by Landau.,880.P20 Winter 2006,Richard Kass,8,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. Th
19、e 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 Landaus model:1) mean energy loss in single collision 0.01 Wmax, Wmax= max. energy transfer2) individual energy transfers large enoug
20、h that electrons can be considered free,small energy transfers ignored.3) velocity of incident particle remains same before and after a collision,D=energy loss in absorber x=absorber thickness x= approximate average energy loss= 2pNare2mec2r(Z/A)(z/b)2x C=Eulers constant (0.577) lne=ln(1-b2)I2-ln2me
21、c2b2)I2+b2 (e is min. energy transfer allowed),The most probable energy loss is Dmpxln(x/e)+0.2+d,Other (more sophisticated) descriptions exist for energy loss in thin samples by Symon, Vavilov, Talman.,Must evaluate integral numerically,Can approximate f(x,D) by:,The energy loss pdf, f(x, D), is ve
22、ry complicated:,880.P20 Winter 2006,Richard Kass,9,Bremsstrahlung (breaking radiation),Classically, a charged particle radiates energy when it is accelerated: dE/dt=(2/3)(e2/c3)a2,Zi,Zf,qi,qf,g,g,Zi,Zf,qi,qf,g,g,+,In QED we have to consider two diagrams where a real photon is radiated:,The cross sec
23、tion for a particle with mass mi to radiate a photon of E in a medium with Z electrons is:,Until you get to energies of several hundred GeV bremsstrahlung is only important for electrons and positrons:,(ds/dE)|e/(ds/dE)|m =(mm/me)237000,m-2 behavior expected since classically radiation a2=(F/m)2,Rec
24、all ionization loss goes like the Z of the medium.,The ratio of energy loss due to radiation (brem.) & collisions (ionization)for an electron with energy E is:,(ds/dE)|rad/(ds/dE)|col(Z+1.2) E/800 MeV,Define the critical energy, Ecrit, as the energy where (ds/dE)|rad=(ds/dE)|col.,pdg,880.P20 Winter
25、2006,Richard Kass,10,Bremsstrahlung (breaking radiation),The energy loss due to radiation of an electron with energy E can be calculated:,N=atoms/cm3=rNa/A (r=density, A=atomic #) Eg,max=E-mec2,The most interesting case for us is when the electron has several hundred MeV or more, i.e. E137mec2Z1/3.
26、For this case we frad is practically independent of energy and Eg,max=E:,Thus the total energy lost by an electron traveling dx due to radiation is:,We can rearrange the energy loss equation to read:,Since frad is independent of E we can integrate this equation to get:,Lr is the radiation length, Lr
27、 is the distance the electron travels to lose all but 1/e of its original energy.,880.P20 Winter 2006,Richard Kass,11,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 f
28、ree 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:,Leo and the PDG have more complicated expressions:,Leo, P41,PDG,Lrad1 is approximately the “simplest expression” and L
29、rad2 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 %:,The PDG lists the radiation length of lots of materials including:,Air: 30420cm, 36.66g/cm2 teflon: 15.8cm, 34.8g/cm2 H2O: 36.1cm, 36.1g/cm2 CsI: 1.85cm, 8.39g/cm2 Pb:
30、 0.56cm, 6.37g/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.,880.P20 Winter 2006,Richard Kass,12,Interaction of Photons (gs) with Matter,There are three main contributions to photon interactions:Photoelectric effect (Eg few
31、 MeV),Contributions to photon interaction cross section for lead including photoelectric effect (t), rayleigh scattering (scoh), Compton scattering (sincoh), photonuclear absorbtion (sph,n), pair production off nucleus (Kn), and pair production off electrons (Ke). Rayleigh scattering (scoh) is the c
32、lassical physics process where gs are scattered by an atom as a whole. All electrons in the atom contribute in a coherent fashion. The gs energy remains the same before and after the scattering.,A beam of gs with initial intensity N0 passing through a medium is attenuated in number (but not energy)
33、according to:dN=-mNdx or N(x)=N0e-mx With m= linear attenuation coefficient which depends on the total interaction cross section (stotal= scoh+ sincoh + +).,880.P20 Winter 2006,Richard Kass,13,Photoelectric effect,The photoelectric effect is an interaction where the incoming photon (energy Eg=hv) is
34、 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
35、dominates at low g energies ( 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.P20 Winter 2006,Richard Kass,14,Compton Scattering,Compton scattering is the interaction of
36、a real g with an atomic electron.,q,f,gin,gout,electron,The result of the scattering is a “new” g with less energy and a different direction.,Solve for energies and angles using conservation of energy and momentum,The Compton scattering cross section was one of the first (1929!) scattering cross sec
37、tions to be calculated using QED. The result is known as the Klein-Nishima cross section.,At high energies, g1, photons are scattered mostly in the forward direction (q=0) At very low energies, g0, K-N reduces to the classical result:,Not the usual g!,880.P20 Winter 2006,Richard Kass,15,Compton Scat
38、tering,At high energies the total Compton scattering cross section can be approximated by:,(8/3)pre2=Thomson cross section From classical E&M=0.67 barn,We can also calculate the recoil kinetic energy (T) spectrum of the electron:,This cross section is strongly peaked around Tmax:,Tmax is known as th
39、e Compton Edge,Kinetic energy distribution of Compton recoil electrons,880.P20 Winter 2006,Richard Kass,16,Pair Production (ge+e-),This is a pure QED process. A way of producing anti-matter (positrons).,gin,gv,e-,e+,Nucleus or electron,Z,Z,gin,gv,e+,e-,Nucleus or electron,Z,Z,+,First calculations do
40、ne by Bethe and Heitler using Born approximation (1934).,Threshold energy for pair production in field of nucleus is 2mec2, in field of electron 4mec2.,At high energies (Eg137mec2Z-1/3) the pair production cross sections is constant.spair =4Z2are27/9ln(183Z-1/3)-f(Z)-1/54 Neglecting some small corre
41、ction terms (like 1/54, 1/18) we find:spair = (7/9)sbrem The mean free path for pair production (lpair) is related to the radiation length (Lr):lpair=(9/7) Lr,Consider again a mono-energetic beam of gs with initial intensity N0 passing through a medium. The number of photons in the beam decreases as
42、:N(x)=N0e-mx The linear attenuation coefficient (m) is given by: m= (Nar/A)(sphoto+ scomp + spair).For compound mixtures, m is given by Braggs rule: (m/r)=w1(m1/r1)+ wn(mn/rn)with wi the weight fraction of each element in the compound.,880.P20 Winter 2006,Richard Kass,17,Multiple Scattering,A charge
43、d 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 scatterin
44、g 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:,with,In the above equation b=v/c, and p=momentum of incident particle,The space angle q= qplane2,T
45、he average scattering angle =0, but the RMS scattering angle 1/2= q0,Some other quantities of interest are given in The PDG:,The variables s, y, q, y are correlated, e.g. ryq=3/2,880.P20 Winter 2006,Richard Kass,18,Why We Hate Multiple Scattering,Multiple scattering changes the trajectory of a charg
46、ed particle. This places a limit on how well we can measure the momentum of a charged particle(charge=z) in a magnetic field.,s=sagitta,r=radius of curvature,L/2,Trajectory of charged particlein transverse B field.,The apparent sagitta due to MS is:,The sagitta due to bending in B field is:,GeV/c, m
47、,GeV/c, m, Tesla,The momentum resolution dp/p is just the ratio of the two sigattas:,Independent of p,As an example let L/Lr=1%, B=1T, L=0.5m then dp/p 0.01/b.,Thus for this example MS puts a limit of 1% on the momentum measurement,Typical values,note: r2=(L/2)2+(r-s)2 r2=L2/4+r2+s2-2rs s=L2/(8r)+s2/(2r) sL2/(8r),
