Interactions of Particles with Matter

Transkript

1 Interactions of Particles with Matter

2 Particle Detection Principle In order to detect a particle - it must interact with the material of the detector - transfer energy in some recognizable fashion i.e. The detection of particles happens via their energy loss in the material it traverses... Possibilities: Charged particles Hadrons Photons Neutrinos Energy loss by multiple reactions Ionization, Bremsstrahlung, Cherenkov... Nuclear interactions Photo/Compton effect, pair production Weak interactions Total energy loss via single interaction charged particles

3 Particle Interactions Examples Ionization: Pair production: Compton scattering: Charged Particle Positron Electron Electron γ γ Electron Nucleus x Electron γ γ Charged Particle Photon Electron Photon Photon Atom Electron Nucleus Positron Electron

4 Energy Loss by Ionization de/dx For now assume: Mc 2 mec 2 Charged Particle i.e. energy loss for heavy charged particles [de/dx for electrons more difficult...] γ Interaction dominated by elastic collisions with electrons... Electron Bethe-Bloch Formula de dx = Kz 2 Z A 1 β 2 [ 1 2 ln 2m ec 2 β 2 γ 2 T max I 2 β 2 δ(βγ) 2 ] 1/β 2 ln (const β 2 γ 2 )

5 Bethe-Bloch Classical Derivation Bohr 1913 Particle with charge ze and velocity v moves through a medium with electron density n. Electrons considered free and initially at rest. Momentum transfer: p = = F dt = ze 2 (x 2 + b 2 ) F dt dx dx = F dx v b x2 + b 1 2 v dx = ze2 b v Interaction of a heavy charged particle with an electron of an atom inside medium. p : x b 2 x 2 + b 2 averagesto zero Symmetry! = 2ze2 bv More elegant with Gauss law: [infinite cylinder; electron in center] E (2πb) dx =4π(ze) E dx = 2ze b and then... { F = ee dx p = e E v = 2ze2 bv

6 Bethe-Bloch Classical Derivation Bohr 1913 Energy transfer onto single electron for impact parameter b: E(b) = p2 2m e Consider cylindric barrel Ne = n (2πb) db dx $% "!!!!! "# }! Cylindric barrel with Ne electrons Energy loss per path length dx for distance between b and b+db in medium with electron density n: Energy loss! de(b) = p2 2πnb db dx = 4z2 e 4 2m e 2b 2 v 2 2πnb db dx = 4π nz2 e 4 m e m e v 2 db b dx Diverges for b 0; integration only for relevant range [bmin, bmax]: Bohr 1913 de dx = 4π nz2 e 4 m e v 2 bmax b min db b = 4π nz2 e 4 m e v 2 ln b max b min

7 Bethe-Bloch Classical Derivation Bohr 1913 Determination of relevant range [bmin, bmax]: [Arguments: bmin > λe, i.e. de Broglie wavelength; bmax < due to screening...] b min = λ e = h p = b max = γv ν e ; 2π γm e v γ = 1 1 β 2 Use Heisenberg uncertainty principle or that electron is located within de Broglie wavelength... Interaction time (b/v) must be much shorter than period of the electron (γ/νe) to guarantee relevant energy transfer... [adiabatic invariance] de dx = 4πz2 e 4 m e c 2 β 2 n ln m e c 2 β 2 γ 2 2π ν e Deviates by factor 2 from QM derivation Electron density: Effective Ionization potential: n = NA ρ Z/A!! I ~ h <νe>

8 Bethe-Bloch Formula [see e.g. PDG 2010] de dx = Kz 2 Z A 1 β 2 [ 1 2 ln 2m ec 2 β 2 γ 2 T max I 2 β 2 δ(βγ) 2 ] [ ρ] K = 4π NA re 2 me c 2 = MeV g -1 cm 2 NA = Tmax = 2mec 2 β 2 γ 2 /(1 + 2γ me/m + (me/m) 2 ) [Max. energy transfer in single collision] z : Charge of incident particle M : Mass of incident particle Z : Charge number of medium A : Atomic mass of medium [Avogardo's number] re = e 2 /4πε0 mec 2 = 2.8 fm [Classical electron radius] me= 511 kev [Electron mass] β = v/c [Velocity] γ = (1-β 2 ) -2 [Lorentz factor] I : Mean excitation energy of medium δ : Density correction [transv. extension of electric field] Validity: density.05 < βγ < 500 M > mμ

9 Energy Loss of Pions in Cu Minimum ionizing particles (MIP): βγ = 3-4 de/dx falls ~ β -2 ; kinematic factor [precise dependence: ~ β -5/3 ] de/dx rises ~ ln (βγ) 2 ; relativistic rise [rel. extension of transversal E-field] Saturation at large (βγ) due to density effect (correction δ) [polarization of medium] Units: MeV g -1 cm 2 βγ = 3-4 MIP looses ~ 13 MeV/cm [density of copper: 8.94 g/cm 3 ]

10 Understanding Bethe-Bloch 1/β 2 -dependence: Remember: p = F dt = F dx v i.e. slower particles feel electric force of atomic electron for longer time... Relativistic rise for βγ > 4: High energy particle: transversal electric field increases due to Lorentz transform; Ey γey. Thus interaction cross section increases... particle at rest γ = 1 fast moving particle γ gross Corrections: low energy : shell corrections high energy : density corrections

11 Understanding Bethe-Bloch Density correction: Polarization effect... [density dependent] Shielding of electrical field far from particle path; effectively cuts of the long range contribution... More relevant at high γ... [Increased range of electric field; larger bmax;...] For high energies: δ/2 ln(ω/i)+lnβγ 1/2 Shell correction: Arises if particle velocity is close to orbital velocity of electrons, i.e. βc ~ ve. Assumption that electron is at rest breaks down... Capture process is possible... Density effect leads to saturation at high energy... Shell correction are in general small...

12 Energy Loss of Charged Particles Dependence on Mass A Charge Z of target nucleus Minimum ionization: ca. 1-2 MeV/g cm -2 [H2: 4 MeV/g cm -2 ] de/dx (MeV g 1 cm 2 ) H 2 liquid He gas Fe Sn Pb Al C βγ = p/mc Muon momentum (GeV/c)

13 Stopping Power at Minimum Ionization de/dx min (MeV g 1 cm 2 ) H 2 gas: 4.10 H 2 liquid: 3.97 Solids Gases ln(z) PDG H He Li Be B C NO Ne Fe Sn Z Stopping e 27.3: power Stopping at minimum ionization power for atthe minimum chemical elements. ionization The straight forline the is fitted chem for Z > 6. A simple functional dependence on Z is not to be expected, since < de/dx> also depends on other variables.

14 de/dx and Particle Identification TPC Signal [a.u.] Measured energy loss [ALICE TPC, 2009] Momentum [GeV] Bethe-Bloch Remember: de/dx depends on β!

15 The ALICE TPC

16 The ALICE TPC Simulation Particle tracks in the ALICE TPC

17 de/dx Fluctuations Bethe-Bloch describes mean energy loss; measurement via energy loss ΔE in a material of thickness Δx with E = N n=1 δe n δn: number of collisions δe : energy loss in a single collision rel. probability Below excitation threshold Excitations Ionization by distant collisions Ionization by close collisions production of δ-electrons Ionization loss δe distributed statistically... so-called Energy loss 'straggling' Complicated problem... Thin absorbers: Landau distribution Standard Gauss with mean energy loss E0 + tail towards high energies due to δ-electrons energy transfer δe see also Allison & Cobb [Ann. Rev. Nucl. Part. Sci. 30 (1980) 253.]

18 de/dx Fluctuations Landau Distribution /x (MeV g 1 cm 2 ) Normalized Energy loss probability f ( /x) p/x w 500 MeV pion in silicon 640 µm (149 mg/cm 2 ) 320 µm (74.7 mg/cm 2 ) 160 µm (37.4 mg/cm 2 ) 80 µm (18.7 mg/cm 2 ) Mean energy loss rate PDG 2010 Thicknesses x Energy loss Approximation: /x (ev/µm) e 27.7: Straggling functions in silicon for 500 MeV pions, norma f( /x) = 1 2π exp 1 2 /x a( /x)mip ξ + e /x a( /x) mip ξ for full form see e.g. Leo ξ: material constant

19 de/dx Fluctuations Landau Distribution C0'1> 8 1-; < 1DE*0-5+#)-1*+/0/1+#*?F B@9 <@A <@9 8@A 8@9 I)*)-+#,05.06,*+-. =0/53)-50$10#03>?1*+// % -% &'"&#: ;)# % -%5 $17$$<1&'"&#: ;)#!"#$"%&'"()*+(&,)-./0*1! "#$2+34 #"ρ1718j991µ; B<91µ; K91µ; PDG @A 9@8 8@9 89@9 899@9 8999@9 C%+#1G)#05)-10#03>?1DH0'F Bethe-Bloch de/dx, two examples of restricted energy loss, and the Landau most probable energy per unit 7.6: Bethe-Bloch,twoexamplesofrestrictedenergy thickness in silicon. The change of Δp/x with thickness x illustrates its a ln x + b dependence. Minimum ionization (de/dx min) is MeV g 1 cm 2. Radiative losses are excluded. The incident particles are muons.

20 Energy Loss at Small Energies Shell corrections to correct for atomic binding Higher order corrections relevant only at low energies Bloch corrections (higher orders) Barkas correction (pos. vs. neg. charge) With these corrections BB yields 1% accuracy down to βγ =0.05. Stopping power [MeV cm 2 /g] Lindhard- Scharff Nuclear losses µ Anderson- Ziegler ~ 1/β Bethe-B Minimum Minim ionization ionizat βγ For βγ < 0.05 there are only phenomenological fitting formulae available [MeV/c] Muon momentum M

21 Mean Particle Range Integrate over energy loss from E down to Pb Fe C PDG 2010 R = Example: 0 E de de/dx Proton with p = 1 GeV Target: lead with ρ = g/cm 3 R/M = 200 g cm -2 GeV -1 R = 200/11.34/1 cm ~ 20 cm R/M (g cm 2 GeV 1 ) βγ = p/mc H 2 liquid He gas Muon momentum (GeV/c) Pion momentum (GeV/c) Proton momentum (GeV/c)

22 Mean Particle Range Integrate over energy loss from E down to 0 R = 0 E de de/dx Example: Proton with p = 1 GeV Target: lead with ρ = g/cm 3 R/M = 200 g cm -2 GeV -1 R = 200/11.34/1 cm ~ 20 cm

23 Particle Energy Deposit βγ > 3.5: de dx βγ < 3.5: de dx de dx de dx min min Applications: Tumor therapy Possibility to precisely deposit dose at well defined depth by Ebeam variation [see Journal Club]

24 Heidelberg Ion-Beam Therapy Center (HIT)

25 Energy Loss of Electrons Bethe-Bloch formula needs modification Incident and target electron have same mass me Scattering of identical, undistinguishable particles de dx el. = K Z A 1 β 2 ln m eβ 2 c 2 γ 2 T 2I 2 + F (γ) [T: kinetic energy of electron] Wmax = ½T Remark: different energy loss for electrons and positrons at low energy as positrons are not identical with electrons; different treatment...

26 Bremsstrahlung Bremsstrahlung arises if particles are accelerated in Coulomb field of nucleus * A JKL de dx =4αN A z 2 Z 2 A 1 e 2 2 4π 0 mc 2 E ln 183 Z 1 3 E m 2 i.e. energy loss proportional to 1/m 2 main relevance for electrons or ultra-relativistic muons Consider electrons: de dx =4αN A de dx = E with X 0 = X 0 Z 2 A r2 e E ln 183 Z 1 3 E = E 0 e x/x 0 A 4αN A Z 2 r 2 e ln 183 [Radiation length in g/cm 2 ] Z 1 3 After passage of one X0 electron has lost all but (1/e) th of its energy [i.e. 63%]

27 Bremsstrahlung Critical Energy Critical energy: de dx (E c) = de Brems dx (E c) Ion 200 de dx Tot = de dx Ion + de dx Brems Approximation: E Gas c E Sol/Liq c = = 710 MeV Z MeV Z de/dx X 0 (MeV) Copper X 0 = g cm 2 E c = MeV Rossi: Ionization per X 0 = electron energy Brems E Total Ionization Exact bremsstrahlung Example Copper: Ec 610/30 MeV 20 MeV Brems = ionization Electron energy (MeV) Figure 27.12: Two definitions of the critical energy.

28 Bremsstrahlung Energy Spectrum Cross Section [high energy approximation] dσ dk = 1 k A X 0 N A y + y2 [Main dependence: dσ/dk ~ 1/k] Formula accurate for a large y range... except near y = 1 and near y = 0; deviations greater for higher electron energies. [see PDG for further details] Example Pb: (X 0 N A /A) ydσ LPM /dy GeV 10 GeV 1 TeV 10 TeV 100 TeV Bremsstrahlung 1 PeV 10 PeV y = k/e The normalized bremsstrahlung cross section Normalized Bremsstrahlung cross section k dσ/dk in lead versus fractional photon energy k = Eγ Ee = 010 GeV: Suppression for k <.23 MeV [y = ] Ee = 100 GeV: Suppression for k < 2.3 GeV [y = 0.023]

29 Total Energy Loss of Electrons from PDG 2010 Positrons Lead (Z = 82) 0.20 Møller e e e e 1 de (X 1 E dx 0 ) Electrons Møller (e ) Ionization Bremsstrahlung (cm 2 g 1 ) e + e + Bhabha (e + ) 0.05 e Bhabha e + e γ 0 1 Positron annihilation E (MeV) re 27.10: Fractional energy loss per radiation length in lead as Fractional energy loss per radiation length in lead γ as a function of electron or positron energy Annihilation

30 Energy Loss Summary Plot for Muons Stopping power [MeV cm 2 /g] Lindhard- Scharff Nuclear losses µ Anderson- Ziegler Bethe-Bloch µ + on Cu Radiative effects reach 1% [GeV/c] Muon momentum Radiative Radiative losses Without δ βγ [MeV/c] Minimum ionization Fig. 27.1: Stopping power (= E µc [TeV/c] ) forpositivemuonsincopperasa PDG 2010