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

880.P20 Winter 2006 Richard Kass 1

Interaction of Particles with MatterIn 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 electrons/positrons!)electrons and positrons

Energy 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 (Coulomb 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 results, exact calculations very difficult.


Classical Formula for Energy LossAverage energy loss for a heavy charged particle: mass=M, charge=ze, velocity=vParticle loses energy in collisions with free atomic electrons: mass=m, charge=e, velocity=0Assume the electron does not move during the “collision” & M’s trajectory is unchangedCalculate the momentum impulse (I) to electron from M:

M, ze, v

b=impact parameter

dxEvedxdxdtEedtEeFdtI )/()/(

Next, use Gauss’s law to evaluate integral assuminginfinite cylinder surrounding M:

bv

zeIbzedxEzedxbEAdE

22/24)2(

The energy gained by the electron is E=p2/2me=I2/2me

The total energy lost moving a distance dx through a medium with electron density Ne fromcollisions with electrons in range b+db is:

b

db

mv

dxNez

m

dbdxbN

bv

ze

m

dbdxbNIdVEN

e

e

e

e

e

ee 2

422

22 4

2

)2()

2(

2

)2(

Therefore the energy loss going a distance dx is:

min

max2

42

2

42

ln44

/b

b

mv

Nez

b

db

mv

NezdxdE

e

e

e

e

But, what should be used for the max and min impact parameter?The minimum impact parameter is from a head on collisionAlthough E&M is long range, there must be cutoff on bmax, otherwise dE/dx Relate bmax to “orbital period” of electron, interaction short compared to periodMust do the calculation using QM using momentum transfer not impact parameter.


Incident particlez=charge of incident particle =v/c of incident particle=(1-2)-1/2

Wmax=max. energy transfer in one collision

22

max22

2

222 2)

2ln(2

I

Wvmz

A

ZcmrN

dx

dE eeea

Bethe-Bloch Formula for Energy Loss

Fundamental constantsre=classical radius of electronme=mass of electronNa=Avogadro’s numberc=speed of light

=0.1535MeV-cm2/g

Average energy loss for heavy charged particlesEnergy loss due to ionization and excitationEarly 1930’s Quantum mechanics (spin 0)Valid for energies <100’s GeV and >>z (z/137)

heavy= mincident>>me

proton, k, ,

2

22

2

max )(2)/()(1/1

)(2

cmMmMm

cmW e

ee

e

Absorber mediumI=mean ionization potentialZ= atomic number of absorberA=atomic weight of absorber=density of absorber=density correctionC=shell correction

min

max2

222

ln4

/b

bNcmrzdxdE eee

Note: the classical dE/dx formula contains many of the same features as the QM version: (z/)2, & ln[]


Corrected Bethe-Bloch Energy Loss

Z

C

I

Wvmz

A

ZcmrN

dx

dE eeea

c

22)2

ln(2 22

max22

2

222

=parameter which describes how transverse electric field of incident particle is screened by the charge density of the electrons in the medium.2ln+, with a material dependent constant (e.g. Table 2.1of Leo)C is the “shell” correction for the case where the velocity of the incident particleis comparable (or less) to the orbital velocity of the bound electrons (z ). Typically, a small correction (see Table 2.1 of Leo)

Other corrections due to spin, higher order diagrams, etc are small, typically <1%

PDGplots


Average Energy loss of heavy charged particleThe incident particle’s speed (v/c) plays an important role in the amount of energy lost while traversing a medium.

12

2 2)ln(1

Kdx

dE

K1 a constant

1/2 term dominates at low momentum (p)=momentum/energyln() dominates at very high momenta (“relativistic rise”)

term never very important (always 1) =p/m, =E/m, =E/p

K p

0.58 0.20 0.11

1/2 2.96 25.4 89.0

ln() -0.34 -1.60 -2.24

K p

0.99 0.90 0.73

1/2 1.02 1.24 1.88

ln() 1.97 0.71 0.06

p=0.1 GeV/c

p=1.0 GeV/c

Data from BaBar experiment


Energy loss of electrons and positronsCalculation of electron and positron energy loss due to ionization and excitationcomplicated due to:

spin ½ 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 particles”

Z

CF

cmIA

ZcmrN

dx

dE

eeea

c

2)())/(2

)2(ln(

12

22

2

222

is kinetic energy of incident electron in units of mec2. =Eke/ mec2 =-1

2

2

22

22 2ln)12(1

8

11

)1(

2ln)12(8/1)(

eF

Note:Typo on P38 of Leo 2r2

32

2

32

2

)1(

4

)1(

10

)1(

1423

122ln2

)2(

4

)2(

10

)2(

1423

122ln2)(

eF

For both electrons and positrons F() becomes a constant at very high incident energies.

BA

I

cm

dx

dE e

c

ln)2

ln(22

Comparison of electrons and heavy particles (assume =1):

A Belectrons: 3 1.95heavy 4 2


Distribution of Energy LossAmount of energy lost from a charged particle going through materialcan differ greatly from the average or mode (most probable)!

Measured energy loss of 3 GeV ’s and2 GeV e’s through 90%Ar+10%CH4 gas

Landau (L) and Vavilov energyloss 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.


Landau model of energy loss for thin samplesUnfortunately, 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

)]1ln(ln[1

withsin)lnexp(1

)(

/)(),(

0

Cuuuudu

xf

=energy loss in absorber x=absorber thickness= approximate average energy loss= 2Nare

2mec2(Z/A)(z/)2xC=Euler’s constant (0.577…)ln=ln[(1-2)I2]-ln[2mec22)I2]+2 ( is min. energy transfer allowed)

The most probable energy loss is mpln(/)+0.2+]

Other (more sophisticated) descriptions exist for energy loss in thin samples by Symon, Vavilov, Talman.

Must evaluate integral numerically

Can approximate f(x,) by:

))(2

1exp(

2

1),(

exf

The energy loss pdf, f(x, ), is very complicated:


Bremsstrahlung (breaking radiation)Classically, a charged particle radiates energy when it is accelerated: dE/dt=(2/3)(e2/c3)a2

ZiZf

qiqf

‘’

ZiZf

qiqf

‘’

+

In QED we have to consider two diagrams where a real photon is radiated:

The cross section for a particle with mass mi to radiate a photon of E in a medium with Z electrons is:

E

E

m

Z

dE

d

i

ln2

2

Until you get to energies of several hundred GeV bremsstrahlung is only important for electrons and positrons:

(d/dE)|e/(d/dE)| =(m/me)237000

m-2 behavior expected since classicallyradiation a2=(F/m)2

Recall 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:

(d/dE)|rad/(d/dE)|col(Z+1.2) E/800 MeV

Define the critical energy, Ecrit, as the energy where (d/dE)|rad=(d/dE)|col.

pdg


Bremsstrahlung (breaking radiation)The energy loss due to radiation of an electron with energy E can be calculated:

dE

dE

dEENEdE

dE

dEN

dx

dEE

radrad

E

max,max,

0

1max,max

0

with

N=atoms/cm3=Na/A (=density, A=atomic #)E,max=E-mec2

The most interesting case for us is when the electron has several hundred MeV or more, i.e. E>>137mec2Z1/3. For this case we rad is practically independent of energy and E,max=E:

)](18/1183[ln(4 3/122 ZfZrZ erad

NEZfZrZdx

dEe )](18/1183[ln(4 3/122

Thus the total energy lost by an electron traveling dx due to radiation is:

We can rearrange the energy loss equation to read:

dxNE

dErad

Since rad is independent of E we can integrate this equation to get:rLxeExE /

0)(

Lr is the radiation length, Lr is the distance the electron travels to lose all but 1/e of its original energy.


Radiation Length (Lr)The radiation length is a very important quantity describing energy loss of electronstraveling through material. We will also see Lr when we discuss the mean free path forpair production (i.e. e+e-) and multiple scattering.

There are several expressions for Lr in the literature, differing in their complexity.The simplest expression is:

)/)(183ln(4 23/121 AZZNrL aer

Leo and the PDG have more complicated expressions:

)/)1()](()183[ln(4 3/121 AZZZfZNrL aer Leo, P41

)]))(([4 21221

radradaer ZLZfLZNrL 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 %:

)()/287ln()1(

4.716 2

cmgZZZ

ALr

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: 0.56cm, 6.37g/cm2 Be: 35.3cm, 65.2g/cm2

Leo also has a table ofradiation lengths on P42but the PDG list is more up to date and larger.


Interaction of Photons (’s) with Matter

There are three main contributions to photon interactions:Photoelectric effect (E < few MeV)Compton scatteringPair production (dominates at energies > few MeV)

Contributions to photon interaction cross section for leadincluding 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 processwhere ’s are scattered by an atom as a whole. All electronsin the atom contribute in a coherent fashion. The ’s energyremains the same before and after the scattering.

A beam of ’s with initial intensity N0 passing through a medium is attenuated in number (but not energy) according to: dN=-Ndx or N(x)=N0e-x

With = linear attenuation coefficient which depends on the total interaction cross section (total= coh+ incoh + +).


Photoelectric effectThe photoelectric effect is an interaction where the incoming photon (energy Ehv) is absorbed by an atom and an electron (energy=Ee) is ejected from the material:

Ee= E-BEHere 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 energies (< MeV) and hence gives low energy e’s.Exact cross section calculations are difficult due to atomic effects. Cross section falls like E

-7/2

Cross section grows like Z4 or Z5 for E> few MeV

Einstein wins Nobel prize in 1921 for hiswork on explaining the photoelectric effect.Energy of emitted electron depends on energyof and NOT intensity of beam.


Compton ScatteringCompton scattering is the interaction of a real with an atomic electron.

in

ou

t

electron

The result of the scattering is a “new” with less energy and a different direction.

2,

,, /

)cos1(1cmEwith

EE ein

inout

)cos1(1

)cos1(ElectronofEnergyKinetic ,,,

inoutin EEET

)(1cos ,,,,

2

outinoutin

e EEEE

cm

Solve for energies and anglesusing conservation of energy and momentum

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.

)sin()(2

))cos1(1

)cos1(cos1(

]cos1(1[22

,

,

,

,2

,

,222

22

2

out

in

in

out

in

outee

E

E

E

E

E

Err

d

d

At high energies, >>1, photons are scattered mostly in the forward direction ()At very low energies, 0, K-N reduces to the classical result: )cos1(

22

2

er

d

d

Not theusual !


Compton ScatteringAt high energies the total Compton scattering cross section can be approximated by:

)2/1)2)(ln(8

3)(

3

8( 2

ecomp r (8/3)re

2=Thomson cross section From classical E&M=0.67 barn

We can also calculate the recoil kinetic energy (T) spectrum of the electron:

ine

e ETsss

s

s

s

cm

r

dT

d,22

2

22

2

/with))2

()1()1(

2(

This cross section is strongly peaked around Tmax:

21

2,max inET

Tmax is known as the Compton Edge

Kinetic energy distributionof Compton recoil electrons


Pair Production (e+e-)This is a pure QED process.A way of producing anti-matter (positrons).

i

nv

e-

e+

Nucleus or electronZ Z

i

nv

e+

e-

Nucleus or electronZ Z

+

First calculations done by Bethe and Heitler using Born approximation (1934).

Threshold energy forpair production in fieldof nucleus is 2mec2, infield of electron 4mec2.

At high energies (E>>137mec2Z-1/3) the pair production cross sections is constant.pair =4Z2re

2[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):pair=(9/7) LrConsider again a mono-energetic beam of ’s with initial intensity N0 passing through a medium. The number of photons in the beam decreases as: N(x)=N0e-x

The linear attenuation coefficient () is given by: = (Na/A)(photo+ comp + pair). For compound mixtures, is given by Bragg’s rule: ()=w1()++ wn(nn) with wi the weight fraction of each element in the compound.


Multiple ScatteringA 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 plane after traveling a distance x through a material with radiation length =Lr is approximately gaussian:

]2

exp[2

1)(20

2

0

plane

plane

plane

d

dP })/ln{038.01(/

6.130 rr LxLxz

pc

MeV

with

In the above equation =v/c, and p=momentum of incident particle

The space angle = plane

The average scattering angle <plane>=0, but the RMS scattering angle <plane>1/2= 0

Some other quantities of interest are given inThe PDG:

0

0

0

3

1

34

13

1

3

13

1

3

1

xxs

xxy

rmsplane

rmsplane

rmsplane

rmsplane

rmsplane

rmsplane

The variables s, y, are

correlated, e.g. y=3/2


Why We Hate Multiple ScatteringMultiple scattering changes the trajectory of a charged 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 particle in transverse B field.

rrmsplane

rmsplane LLz

p

LLLs /

106.13

3434

1

34

1 3

0

The apparent sagitta due to MS is:

The sagitta due to bending in B field is:pc

zBL

zB

pL

r

LsB 8

3.0

3.088

222

GeV/c, m

GeV/c, m, Tesla

The momentum resolution p/p is just the ratio of the two sigattas:

LB

LL

p

zBL

LLzp

L

s

s

p

p rr

B

rmsplane

/103.52

8

3.0

/106.13

34 32

3

Independent of p

As an example let L/Lr=1%, B=1T, L=0.5m then p/p 0.01/. 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+s2-2rss=L2/(8r)+s2/(2r)sL2/(8r)

Documents

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