Counting Cosmic Rays through the passage of matter

By Edwin Antillon.

Energy loss of charged particles(collision loss)

Inelastic collisions with atomic electrons of the material and elastic scattering due to nuclei. Collision with atomic electrons are far more probable and it’s the main component of energy loss by collisions.

(radiation loss)Radiation loss by Bremsstrahlung radiation occurs from deceleration influenced by the presence of the electric field from the nucleus. Only for electrons is this radiation substantial since it is proportional to m –

2 . Cherenkov radiation can also be experienced by particles travelling faster than light on that same medium, however cherenkov radistion is small as compared to collision loss.

(hard component)Muons mostly lose energy through collision loss by ionizing atoms on their path due to their large mass.

(soft component)Electrons experience both radiation and collision loss.

• If there are more than one scattering target, we need the number of scattering centers per unit volume. (N)

• Density( )At.Weight (A) = # moles / unit volume

• Then, N = Na A

Collision Energy Loss and Stopping Power

dd

E, 1F

d Nsd

Ns averagenumberof particlesscattered perunittime

Ntotd Nsd FluxN Area x d

d F N areax FareaNAAx

Prob.ofinteractioninx withalltheelectronsNtotFA

Z xNA ZAdEdx

2 NAre2mec

2ZA

z2

2ln2me2 v2 WmaxI2

22 2CZE Prob.ofinteractioninxEavginteraction

dEdx

NA ZA Eint.

• The energy dissipated depends on the product of the density x thickness = “Mass thickness”.

• Equal mass thickness has the same effect on same energy.

• Energy loss due to Cherenkov radiation is already included in the stopping power formula as seen in the logarithmic and and dependence.

Bethe-Bloch Formula:

1dEdx

2 NAre2me c

2 ZA

z2

2ln2me2 v2 WmaxI2

22 2CZ

Stopping Momentum vs Mass Thickness

• For a charged particle of given energy, the minimum amount of mass thickness neccesary to stop the particle (Range) can be calculated.

• Stopping momentum as a function of Range are plotted on the right (due to Rossi)

• The lack of overlap for the two metals shown comes from the fact that the Z/A ratio for aluminum is about 17% bigger than Lead.

Flux vs. Energy• Muons result from the

interaction of primary Cosmic Rays with the atmosphere and their subsequent decay of produced particles.

• An integral distribution spectrum of muons as a function of energy at sea level is shown to the right (due to Sandstrom)

• Our estimation of the minimum energy obtained from a given mass thickness, provides the expected flux of particles.

Electron / Positron loss

• Electrons/Positrons lose energy additionally by Bremsstrahlung radiation.

• The critical energy corresponds to the energy where radiation and collision losses are equal.

• Ec=1600 mec2/ Z

(60 MeV for Al)• Electron above this energy can

give rise to Electromagnetic showers.

Electromagnetic showers• A high energetic photon

produces pair production (e-+ pair) and this pair in turn radiates energetic photons via bremsstrahlung on average after one radiation length. (1 rad. length (t) =9cm for aluminum, .56 cm for Lead)

• So number of particles produced is N ~ 2 t and the initial energy decreases as ~ Eo/2 t

• Since the cascade stops at the critical energy Ec, then the max. number of particles produced N~ Eo / Ec.

A closer approach• A closer fit on the number of particles

produces is given by Leo [Eq. 2.125]

• From the flux rates below a fit was done to find the dependece on Energy.

> 10 Mev 30/ (m^2 s sr)

>100 Mev 6 / (m^2 s sr)

>1000 Mev 0.2 / (m^2 s sr)

Shower development has to be taken into account

N Notaebt

where No 5.51 EoGeVZba1a 1b 0.634 0.0021Z

a 1.77 0.52ln Eo : ForZ 13

2.0Z340

.664 Z340Log: forZ 26

Where N is the number of shower particles at a depth t

Totalnumberof particlesdetectedEmin.

EM E

NEE

Theory and Experiment.Variable Mininum Maximun

Mass thickness (R) Composite density ~ Aluminum density

r dx = (2.75 g/cm3 ) ( 45 cm) = 124 g/cm2

r dx = (2.75 g/cm3 ) ( 100 cm) = 275 g/cm2

Radiation Length (Al) 1 Lrad = 8.9 cm

Lrad = 5 Lrad = 11

Mass thickness /mc

2

117 g cm-2 /108 eV 260 g cm-2 /108 eV

Momentum (p) 3.1 mc = 330 MeV /c 5.6 mc

= 590

MeV /c

Kinetic Energy (KE)

1. 50 mc2 = 160 MeV 2. 20 mc

2 = 230

MeV

Integral muon flux = 122 / (m2 s) = 105 / (m2 s)

Integral electron flux

= 3 / (m2 s) = 1 / (m2 s)

Total Flux = 126 / (m2 s) = 118 / (m2 s)

Counts = total Flux * area (=3.92E.2 m2)

= 300 per min. = 280 per min.

Observed rates = 270 counts per min = 270 counts per min.

Position Expected

counts/ m2 s.

Observed

Roof 180 +_ 5% 148 +_12

4th floor 116 _+ 10 114+- 11

2nd Floor 90_+ 5% 104 _+ 10

Basement 90 +- 9.5

Position Muon Electron Total observed

Roof 130 50 180

4th Floor 118-126