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What Is a Muon ?

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What Is a Muon ? What is a muon ? Rob Veenhof PDG Review ● The PDG lists: – family – name – spin – mass – life-time – decay modes Families ● Some particles mainly act as glue: – photons – W, Z – gluons – gravitons ● Other particles are building blocks: – quarks + ● always glued together into hadrons (e.g. ) – leptons ● free or in bound states – neutrinos ● free Leptons & Neutrinos ● There are 3 leptons (families), each associated with a neutrino. ● All of these have their own anti-particle: - - - particles e e + + + e e anti-particles ● Leptons and neutrinos come and go in pairs: + + + + ( u d W ) + - + - J / ( c c ) - - - - - e e ( W e e ) + - + - e e ( e e Z ) Leptons & Neutrinos (cont'd) ● Some key properties of leptons: – They do not interact directly with hadrons, they do so via photons or via the very heavy W and Z. – Leptons that want to change family, need a W and 2 neutrinos ... not very efficient ! – Leptons have an appreciable mass and are charged, neutrinos have a tiny mass and are not charged – Thanks to their charge, leptons can exchange energy (photons) with anything that has charge. – They are currently not known to be composite. Muons ● Muons are 207 times heavier than electrons, 17 times lighter than 's. ● Muons are lighter than all hadrons, hence they can only decay into electrons. ● The Earth (and we !) are made of quarks, with which muons don't interact directly, and elec- trons, an interaction with which requires a W + a neutrino + an antineutrino. Muons interact mostly electromagnetically (via photons) with normal matter. This doesn't destroy the muon, the basis of muon identification ! History of muons ● Muons were discovered in the 1930's as a component of cosmic radiation which pene- trates matter more than electrons and has an ionisation rate between electrons and protons: – B. Rossi, Z. Phys. 82 (1933) 151 – J. C. Street et al., Phys. Rev. 47 (1935) 891 – C. D. Anderson and S. H. Neddermeyer, Phys. Rev. 50 (1936) 263 – J. C. Street and E. C. Stevenson, Phys. Rev 52 (1937) 1003 – (and several more) ● The discovery is sometimes attributed solely to C.D. Anderson (1936) but it is the positron for which he received the 1936 Nobel prize. The first muon experiment ... Trigger counters Lead absorber Trigger counter Cloud chamber, filter, 0.35 T magnetic field Veto counter [J. C. Street and E. C. Stevenson, Phys. Rev 52 (1937) 1003] Cosmic radiation ● Outside the atmosphere, cosmic radiation consists of protons and light nuclei. ● On entering the atmosphere, they form (mostly) pions. ● The pions decay into muons, neutrinos and photons. ● Some muons decay into electrons and neutrinos. ● We're irradiated by about 200 muons/sec.m² ! Exercise ● Given: – the energy of incident cosmics as shown – pion and muon life time: c =7.8 m c =660 m ● Assuming the initial interaction is at 15 km, which particles do we see here ? Solution ● A reasonable incident energy could be 10 GeV ● Assume that this produces a 6 GeV pion ● The pion will travel d = c ( =E / m ) =6 GeV/0.140 MeV×7.8 m =330 m ● and decay into a 5 GeV muon which travels d = c =5GeV/0.106 MeV×660 m =31 km ● So, we see overwhelmingly muons – pions are found in the upper atmosphere. But ... ● The atmosphere isn't empty ... – will our 5 GeV muon, produced at 15 km altitude, reach us at all or will it be absorbed before ? ● Assuming it reaches us ... – will it still have an energy of 5 GeV ? – will it perhaps traverse the Earth ? Energy loss ● In fact, muons lose about 2 GeV of their energy while traversing the atmosphere. ● Which isn't all that different from what happens in NA60 ! Absorber vs Atmosphere ● Composition of the NA60 absorber: Material Density Thickness Material RL # RL [g/cm³] [cm] [g.cm²] [cm] [-] BeO 2.81 41.0 115.21 14.1 2.9 Al2O3 3.52 25.4 89.41 7.4 3.5 C 1.93 440.0 849.20 18.8 23.4 Fe 7.87 40.0 314.80 1.8 22.7 Total 1368.62 52.5 ● Weight of the Earth atmosphere: – 1 atm = 101325 N/m² (1 N = 1 kg 9.80665 m/sec²) – equivalent weight: 1033 g/cm² ● (Note: there is more to it than only density !) How straight does the muon go ? ● The muon loses energy by numerous elastic photon exchanges with charged particles (p,e) in the medium. ● These exchanges also lead to a change in its direction, known as multiple scattering. ● For practical purposes, one estimates the angle with the formula, RL is the radiation length of the material: 13.6 MeV length = p RL How to read this ? Emperical normalisation Square root: adding sigma's ! 13.6 MeV length = p RL 1/momentum: the mean free path Radiation length: natural unit rises ~linearly with the energy for electromagnetic interactions Examples ● Earth atmosphere: – air has a radiation length of 36.66 g/cm², – the atmospheric thickness is 1033 g/cm², – the atmosphere represents 28 radiation lengths, – a 5 GeV muon will be scattered by: 13.6 MeV = 28=15 mrad 5GeV – how much is this in meters over 10 km ? ● NA60: – the absorber represents 52.5 radiation lengths, which leads to 37 % extra scattering – how much is this over 10 m ? Muons in NA60 ● Muons are not produced directly in hadronic interactions. They result from the decays of hadrons such as (at 450 GeV in pBe): tot =130−140 m b Br ≈1, c =7.8 m ≈0.5 m b + - −4 tot =7−15 m b Br =3.1×10 ≈0.0035 m b + - −5 tot ≈9−12 m b Br =4.55×10 ≈0.0005 m b + - −4 tot ≈0.7 m b Br =2.85×10 ≈0.0002 m b ● Muons are only a tiny fraction of what is produced in the target ! ● The overwhelming majority of the muons result from the uninteresting decays of pions ... Spin and Helicity ● A particle with mass and spin J can have a component of the spin in the direction of motion, or helicity, of -J, -J+1, ... J-1, J. ● Massless particles only have helicity -J and +J. ● Light particles (i.e. mass much smaller than the energy) like to have helicity -J and +J. ● Spin components, i.e. helicities, are conserved. ● Weak interactions feel helicity ! ● Strong and EM interactions do not. Angular effects of spin ● Question: does spin affect the angular distribution of the decay , knowing that the spins are : 0 , : 1 , : 1 ? 2 2 ● Answer: no, the decay is isotropic because the pion has spin 0. - + ● Question: does spin affect the decay , given that the has a spin of 1 ? ● Answer: yes, the helicity of the can be +1, 0 or -1 and in case it is +1 or -1, the muon will preferentially go in the direction of the Branching ratios and Spin ● Spin is a little more subtle than that, though: Br =99.98770 % , Br e e =0.01230 % ● The reason is that weak interactions couple, in the massless limit, only to left-handed particles and right-handed anti-particles: + + J = 0 Neutrinos are almost massless, So, the muon is right-handed hence they are right-handed. too, which it does not like ! Homework ● The helicity (spin in the direction of motion) is preferentially -½ for + and +½ for - , real photons can have +1 and -1 ● Why is the following decay far more probable with muons than with electrons ? ? Spin: 0 Spin: 1 Spin: ½ Summary ● Muons are charged, massive, spin ½ leptons. ● They are the main component of the cosmic radiation at the Earth surface, but they are secondaries, and rare, in hadronic interactions. ● Owing to their higher mass than electrons, some decays work more easily with them, for the better ( + -) or the worse ( ). ● Muons penetrate material very easily, which makes them easy to identify. They lose some energy traversing materials and scatter a bit..
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