mass of the z boson - niels bohr institutexella/lecture_11feb2009.pdf · 2009-02-27 · • most...

The Physics plot at the end of your master thesis

Mass of the Z boson

• CM frame is Centre-of-Mass or Centre-of-Momentum– “Rest frame” for a system of particles– I.e. ∑pi=0 ( where p is the usual 3-vector)

• LAB frame – may be:– Rest frame of some initial particle (fixed target experiment), or– CM frame (pp interactions at LHC),or– Neither (asymmetric ee factory like Babar)

Particles and forces

• Lorentz invariant quantities exist for individual particles and systems.• Invariant mass of a system:

))((..1..1

2

∑∑==

==Ni

iNi

i ppps µ

µ

),(..1..1..1∑∑∑===

=NiNi

iNi

ipEpµ

2

..1

22

..1 ..1

=

=−

= ∑

∑∑

== Nii

Nii EpE

Nis

CM frame

within a frame Σpµi =constant

-> Conservation energy and momentum

Relativity : few kinematics reminders

Mass of Short-lived Particle

• From invariant mass of its decay products, e.g. 2-body:

– How to measure ma?

mc,Ec

ma

mb,Eb

θbc



Two-body Decay

• Initial invariant mass s = ma2

• Final invariant mass =

• If Eb, Ec >> mc , mc then Eb, Ec~ pb, pc

• So,

( ) ( )ppEE cbcb ++ −22

ppEEs cbcb .22 −=

( )θ abcba CosEEm −= 122

Which information is needed to make that physics plot?

Which detector can best provide that information?

y

z

x

E

B drift

chargedtrack

wire chamberto detectprojected tracks

gas volume with E& B fields

e-

e-e-

e-

e-What happens when a particle goes through a detector?

Muon lifetime

1. Physics principles of a particle interacting with detector material

2. Detectors best suited for measuring a basic particle property (mass, lifetime, energy,…)

3. Detection, collection and analysis of data to determine a basic particle property

In the lectures

In the laboratory

In this course you will learn :

Muon lifetime

Online resources for complementing the lectures Online resources for complementing the lectures • There are some good subject reviews available online from the

particle data group:– http://pdg.lbl.gov/2008/reviews/passagerpp.pdf

» passage of particles through matter– http://pdg.lbl.gov/2008/reviews/pardetrpp.pdf

» particle detectors– http://pdg.lbl.gov/2008/reviews/kinemarpp.pdf

» relativistic kinematics

• The Physics of Particle Detectors, D. Green, Cambridge ,2005.• Detector for Particle Radiation K. Kleinknecht, Cambridge, 1998.

It is in the library.• Techniques for Nuclear and Particle Physics Experiments, W.R.

Leo ,Springer-Verlag, 1987. Can be found in google books in largeparts. Else, I have a copy in my office you can browse.

Advised text for complementing the lectures Advised text for complementing the lectures

Teaching material for the course

• http://physics.web.cern.ch/Physics/ParticleDetector/BriefBook/

Physics principles of particle interaction with matterPhysics principles of particle interaction with matter

In this lecture you will refresh:- types of particles, forces, formalism needed to describe them - size of atoms and nuclei. (mostly not in Dan Green)

Goals of this lecture

Then you will learn about:-cross section , mean free path : quantitative description of a particle interaction with matter (Dan Green, pages 17-27. W.Leo, pages 17-20)

-charged particles interactions with matter (W.Leo, page 21 -60. Dan Green pages 89-125)


Which is it?

e.g. EM radiation/photons:Radio/microwave

Visible

X-ray/g-ray

Ener

gy

Wav

elen

gth

Particlebehaviourbecoming

more evident

FrequencyWavelength

EnergyMomentum

Wave Particle

• Alpha particle scattering from nuclei:– Rest mass of alpha = 3.7 GeV– Typical energy ~ 10 MeV– Can treat classically (fortunately for Rutherford!)

α-emitter

• Compton scattering of γ from electron:– Rest mass of γ = 0 eV– Rest mass of electron = 511 keV– Typical energy of γ ~ 10 MeV– Need to use both relativity and QM

γ -emitter γ

e



Relativity and Quantum Mechanics

• Relativity describes particle behaviour at– high speed ( close to speed of light)– I.e. high energy (compared with particle rest mass)

• Quantum mechanics describes behaviour of waves (or fields)– Probability interpretation for individual particles

• Often need both to analyse results of particle experiments

The “Fundamental Particles”• Quarks (matter particles)

– u,c,t d,s,b– We do not see free quarks, the particles actually observed are the

“traditional” particles such as protons, neutrons and pions.• Leptons (matter particles)

• e, µ , τ , νe , νµ , ντ• Gauge bosons (force particles)

– γ , W , Z, gluons ( only γ is observed directly )

• Most important particle properties from the detector point of view are:

– Mass– Charge (electric, “strong”, “weak”)

• Interactions ( EM, strong , weak )

– Lifetime


Tells us what happens to theparticle when it goes throughthe material

Tells us which particles we see in the detector and therefore use for understandingthe physics of a high energy collision


Relativity : few kinematics remindersMomentum: pµ = mνµ = ( γmc , γmν ) = ( E/c , p ) 4-vector

Lorentz 4-vectors are such that their “length” (magnitude) does not change under Lorentz transformation:

= constant = m2 c2pcEpps222 / −==

µ

µ

βγ 21/1 −=

c/υβ =

1== hcIn these units mass and energy and momentumare measured in eV = electronVolt

c=2.99 108 m s-1

h=h/2π= 6.58 10-22

MeV s

we have never observed quarksrunning aroundfree

we have only seen hadrons, madeof quarkstight togetherby the strongforce

2.6E-08 sec

139.5 MeV

139.9 MeV

stable

8.4E-17 sec

938.2 MeV

939.5 MeV

885 sec

+1

0

+1

0

Q: why lifetime so different?A: different massesB: different number of decay channelsC: different forces involved in the decayRaise hands for A,B,C



Decays via electroweakInteractionαEW = 10-11 αEM

Decays via electromagneticInteractionαEM = 1/137

Answer: Pi0 lifetime much shorter than pi+ due to different interaction strength:


e+

e-

q

q

π

p

K

J/ψ 2.6E-08 sec

139.5 MeV

139.9 MeV

stable

8.4E-17 sec

938.2 MeV

939.5 MeV

885 sec

+1

0

+1

0“stable”


Decay Width, Γ• The lifetime of particles can tell us about the strength of the interaction in

decay process (and about channels available)• Decay rate, W=1/τ (in rest frame) , τ = lifetime in rest frame• For short lived particles – reconstruct the mass, m, of the particle from

decay products.• Uncertainty principle:• Δt ~ τ , so τ/~hmE Δ=Δ

h~. tEΔΔ

• Define the decay width, Γ, to be the uncertainty in the mass, Δm

• For particle with several different modes of decay , can define partial widths, Γi

– Total width is the sum of all partial widths

τ/hh ==Γ W

Γ++Γ+Γ=Γ Ntot ....21


Relativity reminder: Time Dilation and Decay Distance

• Often measure particle lifetime by distance between creation and decayin lab.

• If mean life of particle is τ in its rest frame, in the lab frame the mean lifeis

• During this time it travels a distance βγcτ• Since p=βγm, … mean decay distance in lab

γττ ='

mcpc ττβγ =βγ 21/1 −=

c/υβ =

Decay length proportional to momentum


few cm


Stable particles• Can be used as beam particles or for “low-energy physics”• Decay prohibited by conservation laws

At high energy,90% of detectedparticles from an

hadronicinteraction arecharged pions!

Neutron and muon n: 3_1011m µ±: 6km

- Photon ( g ) - Neutrinos ( n )- Electron/positron- Proton/antiproton

Weakly decaying particlesmc /τDecay “parameter”

–Gives mean decay distance for 1GeV energy

1-10cmStrange baryons or “Hyperons”

50-200µmHeavy quark hadrons, τ lepton

π±,K±,K0L: 5-50mlight quark hadrons, mesons

cannot seedirectly


Very short-lived particles

• Detectable only by their decay products, eg. Z boson• Electromagnetic decays to photons or lepton pairs

– Includes π0 giving high-energy photons

• Strongly decaying “resonances”

π0: cτ/m= 180nm

Very massive fundamental particles

– W±,Z0

– top quark– Higgs boson ?– Super-symmetric particles, …

• Decay indiscriminately to lighter known (and possibly unknown) objects– leptons, quark “jets” (pions plus photons) etc.

Size of atoms and nuclei

α Electromagnetic = e2/hc = α =1/137α Weak = 10-11 αElectromagnetic α Strong = g2/hc = 0.9

Binds atomRadioactive decaysBinds nucleus

Atomic scale and energy

M proton = 1 GeV , m=M electron = 0.5 MeV = 1/2000 M proton

nucleus = protons and neutrons, is at restE0 = lowest bound state in H atom = -melectron c2 α2 /2 = -13.6 eVEn=E0/n2

a0 = Bohr radius = 0.54 Å = λ/ α = h / αmc

λ = h / mc compton wave length electron=0.004Å1Å =0.1 nm

0

50

100

150

200

250

0 5 10 15 20 25 30 35 40

LiNa

K

Kr

He

NeAr

2nd period

3rd period 1st transitionseries

Radius (pm)

Atomic Number Z


Nuclear size Nucleus contains Z protons and A-Z neutronsA=atomic weight= number of nucleons

λ p= h/ mproton c = 0.2 fermi 1 fermi=10-15 m

If nucleus is imagined as containing A sphericalobjects of radius λ p packed in a volume V,thenV= 4/3 π aN

3 = A 4/3 π λ p 3

Nucleus radius = aN = A 1/3 λ p ~ 1.2 fermi for Hydrogen


aN / a0 = 0.5 Å / 1.2 fermi ~ 10-6 !! …. Lots of space ….

This could be also expected by using the analogue way: aN ~ λ αstrong ~ 0.22 fm

Cross section

Q: which of the three figures is correctly describing what happensto e in the material ? Raise hands for A,B,C

e e

Material of atomic weight/number A/Z

E,p

Length L

E,p

A

e

e

γ


E,p

Length L

E’,p’

B

Length L


E,p

ee

~E,p

e-e-e- e-

C

Cross section

We need to define a quantity which allows us to quantify the probability for a interaction of the electron with material in the detector with a given outcome.

This quantity will depend on the energy available for the interaction, on the materialA,Z of the detector, on the type of particle incoming, …..

Knowing quantitavely what is more probable or less probable to an electron to happenwhen crossing a given material, allows us to correctly choose the detector materialfor a specific particle property measurement.

This quantity is called the cross section

A : remember QM ! A, B, C are all correct, but they have different probability to occur.

Dimensionally, σ has the size of an area.One could interpret it as the geometrical cross sectional area of the targetintercepting the beam. The fraction of the flux intercepting the area of the targetwill interact, the rest will not. Eg. If the target is a sphere of radius a, then σ ~ π a2

Definition of differential cross section :Average fraction of beam particles scattered (Ns) in dΩ

TARGETUnit area

dΩ

Flux (F)=number of incident particles per unit area per unit time Beam broader than

target

Definition of cross section:σ = ∫ (dσ / dΩ) dΩDepends on energy available for the interaction

Cross section

per unit of time per unit of flux F of incoming beam particles per unit time and area

dσ/dΩ = 1/F dNs /dΩDepends on energy available for interaction and on angle of scattering

• In the most detailed definition, cross section incorporates:– Strength of underlying interaction (vertices)– Propagators for virtual exchange factors– Phase space factors (available energy)– Does not depend on rate of incoming particles.

Cross section

Name ”barn” comes from war time, during reasearch on the atomic bomb.Physicists scattering neutrons off Uranium nuclei were describing such nuclei“big as a barn”. The unit 10-28 m2 comes from the rough size of the uranium nucleus. They thought the term would hide any reference to the study ofthe nuclear structure

Called the “cross-section” because it has units of area.–Normally quoted in units of barns ( 10-28m2 )… or multiples eg. Nanobarns (nb), picobarns (pb)

Cross section

One can prove more formally the validity of this association of total cross sectionwith geometrical area using the quantum mechanical theory of scattering

(now blackboard, see also Dan Green, page 19-22 and Appendix B)

This “visual”interpretation, this “geometrical” cross section interpretation should betaken with caution, and the cross section value only “roughly” provides a measure ofthe physical dimensions of the target.

•scattering state described by incoming plane wave (k) and an outgoing spherical wave with scattering amplitude A(θ) e ikz + A(θ) eikr /kr

• scattering amplitude A decomposed in sum over partial waves for all possible quantized valuesof angular momentum l (central forces, conserved l ) A=1/2i Σ (2l +1) (e iδl -1) Pl(cosθ)where δl is phase shift due to the scattering. If δl real, elastic scattering only, if δl is imaginary then absoprtion (inelastic) needsto be considered

• cross section is square of scattering amplitude σ E= (π/k2) Σ (2 l + 1) |1- ηl|2 σ I= (π/k2) Σ (2 l + 1) (1 - |ηl|)2

if scattering can also be absorptive, we need to distinguish between σ elastic (E) , σ inelastic (I), σ Total (T) = σ E + σ I

for pure elastic scattering σ T= σ E= (4 π/k2) Σ (2 l + 1) sin2 δl , σ I = 0if scattering off completely absorbing body , σ I = σ E , σ T = 2 σ Eoptical theorem (based on conservation energy) allows to say : σT < (4 π/k2) Σ (2 l + 1)

• assume incoming wave suffers no scattering when outside absorbing object of radiusa and is totally absorbed when it is inside radius a , then lmax = ka (L=r x p, L=hl, p=hk) , and if ka>>1

σ T < (4 π/k2) Σ (2 l + 1), Σ from 0 to lmax => σ T ~ 4 πa2

Cross section

σTOT is proportional to the geometrical cross section.It can exceed it due to quantum nature of scattering (shadowscattering): ΔpT ~h/a impulse given to wave in scattering off object offinite size aθ ~ Δ pT /p

Cross section

RoughlyConstantOver severalOrders of magnitude

1 Angstrom = 10-10 m ,1 fermi = 10-15 m,1 barn = 10-28 m2

Q: Try to “roughly” calculate the total cross section for scattering off an atom , and a nucleon, and compare it to measured data below . How good is your estimate?

Cross section

(go through results with transparencies)

Q: why is this lower?

A:σAtom = πa0 2 ~ 3.14 (0.54 Angstrom)2 = 3.14 0.54 2 10-20 m2 ~ 1. 10-20 m2 =1 108 10-28 m2 = 108 barn or 10-16 cm2

aN = A 1/3

( λp / α s ) ~ 1 fm if A=1σN = πaN 2 ~ 3.14 (1 fm)2 = 3.14 10-30 m2 ~ 3. 10-2 10-28 m2 =30 millibarn or 10-26 cm2

Please note that σAtom / σN = 10 10

When a particle enters material then probability to interact withatom (cloud of electrons around nucleus) is much higher thanto interact with the nucleus

Cross section

Photon has no e.m. nor strong charge so how will interact with proton?Photon has also a parton content, and that is how it can interact with proton. This requires anadditional factor αelectromagnetic for the photon to disassociate into a hadron and hence interactwith a nucleonα= 1/137, hence ~ two orders of magnitudelower cross section wrt pp

γ

p

The scattering we are most interested in when analyzing interactionsof particles with our detectors is the one where the target is nota single object (atom, nucleon), but several ones concentrated in thematerial of our detector

Mean free path

If target centers are uniformly distributed => N0 ρ /A , A=atomic weight, N0=avogadro n. = 6.0 1023mol-1

If the material is not too thick => dx ,the number of targets met by the particle in the material per unit perpendicular area is N0 ρ dx /A

If the beam is again characterized by flux F, then number of incidentparticles per unit area per unit time is F * Area of the target(a)

N = total number of particles scattered at all angles = F a N0 ρ dx σ /A

N/(Fa) = probability for scattering of one single particle in thickness dx = = N0 ρ σ dx /A = C dx 1/C= mean free path = average distance between scatterings

What is the probability of a particle to go through a thickness x of a materialwithout interacting?

Mean free path

P(x) = prob. of not having an interaction after distance xC dx = probability of interacting between x and x+dx

P(x+dx) = P(x) (1- C dx)

P(x) + ( dP/dx ) dx = P(x) - P(x) C dx

dP = - C P(x) dx

P(x) = const e-Cx

Asking P(0)=1 , const=1 => P(x) = e-Cx

Probability for a particle to survive a distance x is exponential in x

Mean free path = λ = ∫ x P(x) dx / ∫ P(x) dx = 1/ C = A / (N0 ρ σ)

Mean free path is the distance traversed by a particle without suffering a collision

Using the following tables (see next slide)

Q : what is the atomic mean free path in Argon (Ar) ? And the nuclear mean free path in Ar?

Q : how does it compare to the atomic and nuclear mean free path in Pb ?

Q : Plot Xsection vs Atomic weight A. Does it behave as expected? (compare to Dan Green, page 24)

Remember that 1 mole of element = A (atomic weight) grams of that element

1 g/liter = 10 -3 g /cm 3

Mean free path

Mean free path

Ar A = 39.95 ρ = 1.782 g/liter σ nuclear = 0.868 barn = 868 mbarn σ atomic ~ 3 10 8 barn

N0 = 6 10 23 1 barn = 10 -24 cm 2 1g/lit = 10 -3 g/cm3

< λ atomic > = 39.95 ------------------------------------------------ ~ 1.2 10-4 cm 6 10 23 1.782 10-3 3 10 8 10 -24

< λ nuclear > = 39.95 ------------------------------------------------ ~ 4.3 104 cm 6 10 23 1.782 10-3 0.868 10 -24

A:

Mean free path

Pb A = 207.19 ρ = 11.35 g/cm3 σ nuclear = 2.9 barn σ atomic ~ 3 10 8 barn

N0 = 6 10 23 1 barn = 10 -24 cm 2

< λ atomic > = 207.19 ------------------------------------------------ ~ 1 10-7 cm 6 10 23 11.35 3 10 8 10 -24

< λ nuclear > = 207.19 ------------------------------------------------ ~ 10 cm = 0.1 m 6 10 23 11.35 2.9 10 -24

We start to see now why detectors which rely on atomic interactions for detection (e.g. tracking the path of a particle without stopping it) can be very compact and should preferibly be made of gas.

Detectors which rely on nuclear interactions for detection (e.g. calorimeters,Stopping the particle and collecting all its energy) in order to be compact should really be made in large fraction of heavy elements material (interleaved with moretransparent detectors to allow collection of energy signal)

Mean free path

1.0

0.01 10 100

A

σELσI

A->1, σEL /σ I -> 0, transparent

A-> large, σEL /σ I ->1Heavy nuclei are black,fully absorbing objects

Let us remember this whenwe discuss calorimeters

proton nucleusscattering

Cross section

Interaction of charged particles withmatter

The physical processes which allow us to detect particles aredifferent for charged or neutral particles.

Charged particles :When a charged particle passes through matter it will1. Loose energy2. Change direction

This is due to respectively, the following two processes1. Inelastic collisions with the atomic electrons of the material2. Elastic scattering off nuclei

1. and 2. occur many times per path length, and the effects on the particleof passage through matter is the cumulative result of this.1. happens much more often than 2.

Electromagnetic interactions

Charged particles interaction with matter

Electromagnetic interaction

Electromagnetic interactionStrong interaction

There are also other processes, which are more rare but nonetheless veryimportant for detection:

3. Emission of cherenkov or transition radiation 4. Nuclear reactions (inelastic scattering off nuclei)5. bremsstrahlung

Excitation (followed by de-excitation)Atomic electron is promoted to higher energy state by energy provided by particle. When it falls back to ground state, energy may be released as a photon

ChargedParticle

Free Electron

Electric Field

IonIonization Ionization

Atomic electron is knocked free from the atom.The remaining atom now has charge as well (it is an ion)The atom may also be left in an excited state and emit a photon.If you are a Solid State Physicist, the ionized atom is a “hole”.


By far 1. is the main responsible for energy loss of charged particles in matter.During this type of interaction energy is transferred to the atom eithervia excitation of the atom or ionization (electron kicked off) of the atom


BUT: the energy loss per interaction of type 1 is very small wrt to the totalkinetic energy of the particle . This of course depends on the material .

As we have seen, running through Lead is 100 times more “dangerous” than in Argon, so even in a relatively small piece of material one can actually see a significant difference of energy loss for particle.e.g. A 10 MeV proton will loose all its energy in 0.25 mm of copper (see later)

Sometimes electrons knocked off atoms due to ionization are so energeticthat they will their selves ionize -> delta-rays or knock-on electrons

To detect the path of a particle in modern particle physics detectors we mostoften use this process 1.

Drift chamber filled with ArCO2 gas



Track ina gaseouschamber

Gaseous chamber is innermost so onecan observe the track paths before they encountertoo much material (red areas in figure)

Process 1. Is statistical in nature, occurring with some quantum mechanicalprobability but the path in matter is generally large, so fluctuations in the total energy loss are small and one can define safely an

For particles heavier than the electron the loss of energy via ionization(process 1.) is more important than 3. or 5. (radiation)


Average energy loss per unit path length dE/dx (or stopping power)

At high energies (~1 GeV) , for electrons or positrons process 5. is actually very important as well as 1.

We start with describing what happens to heavy charged particles (muons and higher masses)

M

M(Mc2γ , Mvγ)

(hω, hk)

ε = ε1 + iε2dielectric constantof material

Chargedparticle

Electromagnetic field of Interaction (photon)

in a material ω2 = (kc)2/ ε , ε1 =n2 refractive index , ε2= absorption coefficient

θc

<1

<1

>1

ε1

cm= c/√ε > c , transition radiation ifdiscontinuity in material crossed

<<1h ω > KeV

Excitation or ionizationsaturation at βγ ~ 1/√ρ

>0eV <h ω< KeV

cm= c/√ε < c, v can be > cm

then cherenkov radiation=0h ω ~eV

effectε2

Charged particles interaction with matter (Kleinknecht, page 9-13)

dE/dx = average energy loss per unit of path length

Bohr gave a classical calculation , and then Bethe-Bloch made the quantummechanical formulation

Bohr calculation :

b

e

M, ze

vx

Assume electronis free, at rest

Charged particles interaction with matter - massive particles

assume electronmoves very littleduring time of interaction withheavy particle

assume heavy particle undeviatedafter collision becauseM >> me

(following is in W.Leo, page 22. )

What is the energy gained by the electron duringThe interaction ?

I = ∫ F dt = e ∫ E⊥ dt = e ∫ E⊥ (dt/dx) dx = e ∫ E⊥ dx/v

only E perpendicular contributes due to symmetry


We now use gauss law integrating over a infinitely long cilinder centeredon particle trajectory and passing through the electron

∫ E⊥ 2π b dx = 4 π ze ∫ E⊥ dx = 2ze /b

I = 2ze2 / bv

E e =E gained by electron = I2/2me = 2 z2 e4 /( me v2 b2)

E lost by heavy particle to all electrons in b and b+db in thickness dx (givenNe density of electrons) is :

-dE = Ee Ne dV = 2 z2 e4 /( me v2 b2) Ne 2 π b db dx = 2 z2 e4 /( me v2 ) Ne 2 π (db/b) dx

Over which bmin and bmax should we integrate now?

bmax cannot be infinity, interaction has to happen during relatively short timefor our impulse calculation to be validbmin cannot be zero else our Ee goes to infinity



bmin can be obtained from a head-on collision where electron gets 1/2 me(2v)2

-> relativistically

2 z2 e4 /( me v2 b min2) = 2 γ2 me v2

bmax from time interaction takes place << time orbit around nucleus τ =1/ωFor our collision typical interaction time is t=b/v-> relativistically b max/(γ v) ≤ τ

So -dE /dx = 4 π z2e4 Ne ln bmax = me v2 bmin

4 π z2e4 Ne ln γ2 me v3 me v2 ze2ω

Bohr classical formula holdswell for alpha particles e.g. , butnot for lighter particles like p

v = β c is the incoming particle velocity

Bethe Bloch correct quantum calculation gives

-dE /dX = 2π N0 re2 me c2 ρZ z2 ln 2me c2 β2 Wmax - 2β2 -δ - C/Z

β2 A (1-β2)I

2π N0 re2 me c2 = 0.15 MeV cm2 /g


I = ionization potential = Z I0 = Z • 12 eV

δ= density effect correction : ~log(βγ)electric field of particle tends to polarize atoms along path. So electrons far from the path of the particle will be shielded from the full electric field density.Collisions with these electrons will therefore contribute less to the total energy loss. This effect is more important the more energetic particle is (classical formula : bmax becomes bigger, distant collisions more important).

C = shell correction : f(1/βγ)Effect which arises when v is comparable to the orbital velocity of electronIn atom

Wmax max energy transfer in 1 coll.

If β → 0 , -dE/dx term with 1/ β2 dominatesIf β → 1 , -dE/dx term with 1/ β2 almost constant and dE/dx increases slowly due to the logarithmic dependence. relativistic raise is cancelled by the density effect terms


the loss of energy per unit path length does not depend on the mass ofthe particle but only on its velocity

There is a minimum at βγ = P/Mc ~ 4 or roughly v~0.96 c.

Particles at this point are called Minimum Ionizing Particles (MIP)

Q : Try to draw qualitatively how the dE/dx curve goes as a function of momentum P of the incoming particle for protons and muons particlesin the region below/around the minimum ionizing energy.start by drawing one curve (e.g. for protons) giving some order of magnitudeon both axis, and then draw how you expect it to be for muons and other particlesIf you like.


Use h=c=1 units for p and E

Bethe Bloch correct quantum calculation gives

-dE /dX = 2π N0 re2 me c2 ρZ z2 ln 2me c2 β2 - β2 -δ - C/Z

A β2 (1-β2)I

2π N0 re2 me c2 = 0.15 MeV cm2 /g

v = β c is the incoming particle velocityI = ionization potential = Z I0 = Z • 12 eV

the loss of energy per unit path length does not depend on the mass ofthe particle but only on its velocity

If β → 0 , -dE/dx term with 1/ β2 dominatesIf β → 1 , -dE/dx term with 1/ β2 almost constant and dE/dx increases slowly due to the logarithmic dependence. relativistic raise is cancelled by the density effect terms

There is a minimum at βγ = P/Mc ~ 4 or roughly v~0.96 c. Particles at this point are called Minimum Ionizing Particles (MIP)

Q:


2me c2 = 1 MeV

Neglect these

Minimum at

P ~ 4 * 1 GeV protonP ~ 4 * 0.1 GeV muon(c=h=1 here)

Value of dE/dx at minimumFor proton =

P=M βγ 4 GeV = (1 GeV) βγ 4 √1-β2 = 1 β β=0.97

-dE /dX = 0.15 ln(16 x 1 103/1) (0.97)2

β2/(1- β2 )= β2γ2 = (P/M)2

-dE /dX ~ 1.5 MeV see next slide


same P , m1<m2 -> β1> β2 -> dE/dx smaller

S(E)

= -d

E/d

x

S(E) = stopping power = - dE/dxdepends on energy of the particle and type of material The stopping power (and hence the density of ionization along the path)increases towards the end of the path and reaches a maximum, theBragg peak, shortly before the energy drops to zero

Bragg curve

This behaviour is important in radiation therapy


Energy loss for electronsFor electrons one needs to modify the Bethe Bloch formula for two reasons:

Bethe Bloch correct for electrons gives

-dE /dX = 2π N0 re2 me c2 Z z2 ln 2me v2 γ2 - 1

A I

2π N0 re2 me c2/A = 0.15 MeV cm2 /g

β ~1

It is basically a constant since electron β ~1 already above 100 MeV …

Charged particles interaction with matter - electrons

1. Their mass is the same as the target (electron in atom) the assumption that the incident particle remains undeflected is invalid

2. For electrons the collisions are among identical particles, the calculation must take into account this fact.

electron


Electron orbeta X-Ray

Target Nucleus(Heavy metal)

X-rays are produced when a charged particles (electrons or betas) aredecelerated by a strong electrostatic field, such as that found near the nucleiof heavy metals (tungsten, lead).

Electrons and positrons are the only particles - in current energies in particlephysics - where radiation contributes substantially to the loss of energy

Bremsstrahlung emission:

Radiation emission by charged particle while being accelrated or decelerated in the coulombfield of a nucleus.The emission probability σ ~ (ze Ze )2 / mparticle

2 (=constant) so for a muon (m=200 me)this process is 1/40 000 times less important


process is basically the same asphoton pair production, crosssection behaves in a similar way

Depends on E but also on Z ofmaterial, and for high E becomesa constant

For detailed calculation, if interested, see W.Leo pg 35 onwards, or Google BooksPrinciples of Radiation Interaction in Matter and Detection By Claude Leroypage 83 onwards

Given emission probability is a constant (not dep on energy), can we guess how dE/dx goes with E ? Remember radioactive decay dN/dt …

-dE/dx = 4 Z2 re2 α N0 ρ ln 183 E = 1 E

A Z 1/3 X0

Where X0 is the radiation length

radiation length is the thickness of material which reduces the electron energy by a factor 1/e


At high energies, the energy loss by ionization can be neglected and

dE/E = - dx/ X0

So that the mean energy <E> of an electron with energy E0 after havinggone through a thickness X is

<E> = E0 e-X/X0

Comparison of ionization and bremsstralhung energy loss for electrons :

dE/dXB ~ Z2 re2 α E ~ Z2 (α λelectron=projectile)2 α E <-electron is projectile ( see

emission prob. formula)dE/dXI ~ Z re

2 mec2 ~ Z (α λelectron)2 / (λelectron) <-electron is target

What is the critical E at which : R =( dE/dXB )/ (dE/dXI ) = 1


Ec~ Z α2 λelectron / Z2 α3 λ(projectile)

2 = ~ (m (projectile)/ m electron ) (m (projectile)c2 / Z α)

Ec ~ 1200 m ec2/ Z [MeV] if projectile=electron

(comes from factors and π’s lying around in formulas and divided by α)

Ec is 8 MeV for electrons in lead is 300 GeV for muons in lead

Landau Distribution

The energy loss distribution The formula for dE/dx so far considered provides an average

The energy loss distributionhowever is not a gaussian. It depends on the path lengthin matter .

What does it look like?(see Kleinknecht, page 16 -18)

Energy loss distribution peaking at E=0


σ= tot. cross section for collision of moving particle and atomic electronsσNx = probability of interaction in length x

I divide x in n intervals such that α = probability of one collision = σNx /n <<1

Then our energy loss distribution in x/n is

F(x/n, E) = (1 - σNx /n ) δ(E) + N x/n dσ/dE

Energy loss in two steps : 2x/n

F(2x/n, E) = ∫ 0E f(x/n, ε) f(x/n, E-ε) dε


Iterating n times, we do the whole thickness x

The result depends on the form of dσ/dE . Taking a simplified modeldσ/dE = σ δ(E-E*)Where we assume to have only one energy transfer value possible = E*, thenF(x/n, E) = (1 - α ) δ(E) + α δ(E-E*)

And for n collisions in the n layers

F(x, E) = Σ ν=0n n α ν (1 - α )n - ν δ(E- ν E*)

ν

If n large and α small then F(x, E) = Σ (βν/ν!) e-β δ(E- ν E*) -> poisson distrβ= nα = σNx for n very large, we get a gaussian

with a real formula for the diff cross section dσ /dEe.g. rutherford scattering , ~1/E2 for large energy transfers and small transverse parameters => There is some (small) prob. for large energy transfers …

Assume energy loss in each collision is small compared to initial E of particle (velocity unchanged). Then total energy lost is sum of each small loss.

Landau Distribution

Average valueMean value

There is a sizeable probability of large energy transfers to the electrons(e.g. delta electrons) . In case of electrons, bremsstralhung alsocontributes to the tails.With this included, the distribution develops a long asymmetric tail at high energies, the Landau tail.

These two quantitiescharacterize the distribution

Mean valueof energylost in depth L= <dE/dx >* L Different!


Now -> 2. elastic scattering off nuclei:


Charged particles :When a charged particle passes through matter it will1. Loose energy2. Change direction

This is due to respectively, the following two processes1. Inelastic collisions with the atomic electrons of the material2. Elastic scattering off nuclei

1. and 2. occur many times per path length, and the effects on the particleof passage through matter is the cumulative result of this.1. happens much more often than 2.

√

√

General very little energy is lost since masses of the nuclei of most materialsare generaly very large compared to the incident particle

The incident particle will mainly be deflected from its initial trajectory.This is the phenomenon of Multiple Scattering, and it is one of the importantlimitations in the precision with which we can determine the direction ofa particle in our high energy physics detectors

Multiple scattering


High angle scattering less probableFor high E, inelasticcoll. with nuclei becomesimportant

Gaussian distribution of many smallScatterings wrt to the initial direction

Tail due to only fewLarge angle scatterings(1/theta^4 )

θ plane



Const. happensto be this value,nothing deeper.X0 is not characterizing this phenomenon

2

Details about the labs

Details about the labs

Please, make groups of 2 (not more people) There will be 6 labs you can try, probably will have time to try all (lab is Wednesday aft.)

Keep a log of what you are doing, starting with a few notes about what youunderstood about the goals of the lab, and then reporting all your measurements(with error associated !) or figures etc.. And finally giving some conclusions.Maximum 3 pages, hand written is very fine (please write legible)

The log file will not be used for grading! It is more for us teachers to understand whatwe have not explained clearly, or if something additional needs to be taught during class. It is a valuable feedback to us.

During the last week of the course, we devote 3 hours to groups presentations.You will choose one lab among those you have done, and present your results.The log file is helpful also in this sense, to remember what you have done in each lab

Please return lab log file the Wednesday of the week after, 13:00 (before starting new lab)

mass of the z boson - niels bohr institutexella/lecture_11feb2009.pdf · 2009-02-27 · • most...

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