Marco Delmastro Experimental Particle Physics 1

particlephysicsExperimental

B. particle interactions and detector response

Abbildung

8.2: Entw

icklungeines

elektromagnetischen

Schauers(M

onteCarlo

Simulation)

•Nur

dieProzesse

γ+

K→

K+

e ++

e −

e+

K→

K+

e+

γ

werden

berucksichtigt(K

=Kern).

•Auf der

StreckeX

0verliert

dase −

durchBrem

sstrahlungdie

Halfte

seinerEnergie

E1

=

E02

•Das

Photon

materialisiert

nachX

0 , dieEnergie

vonPositron

undElektron

betragt

E±

=

E12

•Fur

E>

ϵtritt

keinEnergieverlust

durchIonisation/A

nregungauf.

•Fur

E≤

ϵverlieren

dieElektronen

Energie

nurdurch

Ionisation/Anregung.

FolgendeGroßen

sindbei der

Beschreibung

einesSchauers

vonInteresse:

•Zahl der

Teilchenim

Schauer

•Lage

desSchauerm

aximum

s

•Longitudinalverteilung

desSchauers

imRaum

•Transversale

Breite

desSchauers

Wir

messen

dielongitudinalen

Kom

ponentendes

Schauersin

Strahlungslangen:

t=

xX0

Nach

Durchlaufen

derSchichtdicke

t betragtin

unseremeinfachen

Modell die

Zahl derschnel-

lenTeilchen

N(t)

=2 t

,156

Simple shower model:[from Heitler]

Only two dominant interactions: Pair production and Bremsstrahlung ...

γ + Nucleus ➛ Nucleus + e+ + e−

[Photons absorbed via pair production]

e + Nucleus ➛ Nucleus + e + γ[Energy loss of electrons via Bremsstrahlung]

Electromagnetic Shower[Monte Carlo Simulation]

Shower development governed by X0 ...After a distance X0 electrons remain with only (1/e)th of their primary energy ...

Photon produces e+e−-pair after 9/7X0 ≈ X0 ...

Analytic Shower Model

Eγ = Ee ≈ E0/2

Simplification:

Ee ≈ E0/2

[Ee looses half the energy]

[Energy shared by e+/e–]

Assume:

E > Ec: no energy loss by ionization/excitation

E < Ec: energy loss only via ionization/excitation

Use

... with initial particle energy E0

A simple shower model

Marco Delmastro Experimental Particle Physics 2

t =x

X0

N(t) = 2t

E =E0

N(t)= E0 · 2�t

t = log2(E0/E)

N(E0

, Ec) = Nmax

= 2tmax =

E0

Ec

N(E0

, E1

) = 2t1 = 2 log2(E0/E1) =E

0

E1

N(E0, E1) / E0tmax

/ ln(E0/Ec)➛

Simple shower model:[continued]

Shower characterized by:

Number of particles in shower Location of shower maximum Longitudinal shower distribution Transverse shower distribution

Longitudinal components; measured in radiation length ...

Number of shower particlesafter depth t:

... use:

Energy per particleafter depth t:

Total number of shower particleswith energy E1:

Number of shower particlesat shower maximum:

Shower maximum at:

8.1 Electromagnetic calorimeters 231

8.1 Electromagnetic calorimeters

8.1.1 Electron–photon cascades

The dominating interaction processes for spectroscopy in the MeV energyrange are the photoelectric and Compton effect for photons and ionisa-tion and excitation for charged particles. At high energies (higher than100 MeV) electrons lose their energy almost exclusively by bremsstrahlungwhile photons lose their energy by electron–positron pair production [1](see Sect. 1.2).

The radiation losses of electrons with energy E can be described by thesimplified formula:

−(

dE

dx

)

rad=

E

X0, (8.1)

where X0 is the radiation length. The probability of electron–positronpair production by photons can be expressed as

dw

dx=

1λprod

e−x/λprod , λprod =97X0 . (8.2)

A convenient measure to consider shower development is the distancenormalised in radiation lengths, t = x/X0.

The most important properties of electron cascades can be understoodin a very simplified model [2, 3]. Let E0 be the energy of a photon incidenton a bulk of material (Fig. 8.1).

After one radiation length the photon produces an e+e− pair; electronsand positrons emit after another radiation length one bremsstrahlungphoton each, which again are transformed into electron–positron pairs. Letus assume that the energy is symmetrically shared between the particles at

0

E 0 /2

E0

E 0/4 E 0/8 E 0/16

1 2 3 4 5 6 7 8 t [X0]

Fig. 8.1. Sketch of a simple model for shower parametrisation.

Analytic Shower ModelSketch of simpleshower developmentA simple shower model

Marco Delmastro Experimental Particle Physics 3

A simple shower model

Marco Delmastro Experimental Particle Physics 4

� (1 + t0) · E0

EcX0 ⇥ E0

T =E0

Ec· X0 · F F < 1

T = X0

tmax�1X

µ=0

2µ + t0

· Nmax

· X0

= X0 · (2tmax � 1) + t0 · E0

EcX0

= X0

· (2log2E0/Ec � 1) + t

0

· E0

EcX

0

Analytic Shower Model

Simple shower model:[continued]

Longitudinal shower distribution increases only logarithmically with theprimary energy of the incident particle ... Some numbers: Ec ≈ 10 MeV, E0 = 1 GeV ➛ tmax = ln 100 ≈ 4.5; Nmax = 100 E0 = 100 GeV ➛ tmax = ln 10000 ≈ 9.2; Nmax =10000

Relevant for energy measurement (e.g. via scintillation light): total integrated track length of all charged particles ...

with t0: range of electron with energy Ec

[given in units of X0]

Energy proportionalto track length ...

As only electrons contribute ...

[ with ]

Particle detection• The number N of the particles which are created in shower is

proportional to the energy E of the original particle. Use this to show that the relative energy resolution is given by

Marco Delmastro Experimental Particle Physics 5

LAr calorimeter exercises• The electromagnetic calorimeter for the ATLAS detector is made from roughly 2 mm thick

layers of lead. Between the lead layers are 2 mm wide gaps filled with liquid Argon. Lead has a Z = 82, A = 206 and a density of 11.34 g/cm3. Liquid argon has a Z = 18, A = 40 and a density of 1.4 g/cm3. ü At η = 0 the depth of the ATLAS electromagnetic calorimeter is (about) 22 radiation

lengths X0. What would be the depth of the detector in cm if it was an homogeneous calorimeter (i.e. all made of liquid argon)? And if it was all made of lead?

ü An electron of 5 GeV is generating an electromagnetic shower. At what depth would the shower reach its maximum in liquid argon?

ü Compute the longitudinal depth of lead needed to contain 95% of the energy of a 10 GeVand a 100 GeV photons respectively.

ü How much energy does a minimum-ionizing-particle (mip) deposit in 22 X0 of liquid Argon, assuming:

ü How deep in cm is the real ATLAS electromagnetic calorimeter at η = 0, assuming a perfect succession of lead and liquid argon layers of the same thickness?

Marco Delmastro Experimental Particle Physics 6

Particle interactions• Compute the threshold energies an electron and a proton must possess

in water to emit Cherenkov radiationü Nwater = 1.3

Marco Delmastro Experimental Particle Physics 7