Introduction to µsr. Roberto De Renzi DiFeST, Department of Physics and Earth Sciences University of Parma Italy
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1 Introduction to µsr Roberto De Renzi DiFeST, Department of Physics and Earth Sciences University of Parma Italy Setup of the first spectrometer at ISIS, MuSR
2 Introduction to µsr Muon history The charged particles Anti-matter The neutrinos Parity violation How it works Production Spin polarization Transport Implantation Detection Few examples Summary 2
3 J.J. Thomson: the electron B e m J J Thomson Electrons orbit around B e Thomson measures the ratio : m a light (lepton) q<0 particle 1896 He suggests recipes for the best plum pudding
4 E. Rutherford: the nucleus particles through a gold foil scatter at large angles An even better cake! E Rutherford Bohr atom
5 E. Rutherford: the proton particles through N2 scatter hydrogen nuclei Let's call them protons! E Rutherford
6 P.A.M. Dirac, C. Anderson: the positron From relativity, quadratic energy form E =p c +m c E e- 4 0 Dirac predicted the electron sea e- E = p 2 c 2 +m 2 c 4 PAM Dirac e? E = p 2 c 2 +m 2 c 4 No, it's an anti-electron B CD Anderson
7 Antimatter Dirac's E<0 solutions: each particle has an antiparticle proton p p + p p antiproton electron e e antielectron = positron + e e B CD Anderson
8 J. Chadwick: the neutron particles from Po on Bo produce unknown radiation Po N2 n N Neutral particle with mass mnc2 = 938 MeV The name is = mpc2 my idea, yo! Nuclear mass = 1800 mec2 didn't add up! heavy (baryon) E Rutherford J Chadwick
9 Pauli suggested a neutral particle for β decays Beta decay at rest, if it were a 2-body decay products would have fixed energies There must be C 14 7 N +e N an additional neutral particle, the neutron! e W Pauli Instead they have an energy spectrum 3-body decay!?
10 Pauli suggested a neutrino Beta decay at rest, if it were a 2-body decay products would have fixed energies There must be C 14 7 N +e an additional neutral particle, the neutron! N e W Pauli Instead they have an energy spectrum After Chadwick's discovery of the neutron let's call it neutrino C N +e + ν e E Fermi
11 H. Yukawa: the meson Nuclei are made of n and p+. What force binds them with finite range? Coulomb force = exchange of photons e- hν p+ mν = 0 In analogy n meson p+ mc2 ~ 150 MeV Mass justifies screening, finite range V (r ) 1 r 1 V (r ) e r hr 2 mc V (q ) 12 q V (q ) m c q + 2 h 1935 H Yukawa Mev leptons mesons baryons
12 C. Anderson, S Neddermayer: mesotron Within two years a new particle with that mass (~) is found. C.D. Anderson calls it mesotron 100 < mc2 < 150 MeV with cosmic ray balloons VF Hess However its decay is a bit slow But I don't know that yet CD Anderson τ ~ 2 μs and it has spin S=1/ I measured it in 1941 BB Rossi
13 M. Conversi: not Yukawa's meson? Furthermore the mesotron does not interact strongly enough with matter. μ μ + μ μ M Conversi Fe C µ ± ± 0.05 µ ± ± 0.03 μ - + p + n + νμ
14 C. Powell G. Occhialini: Two particles, pion and muon μ e Three tracks in a photographic emulsions at Mt Chacaltaya (5600 m). C Powell The π is Yukawa's meson G Occhialini mπ = 140 MeV/c2 τπ = 26 ns S = 0 The μ is a lepton (a heavier electron) mμ = 106 MeV/c2 τμ = 2200 ns S = ½ μ π
15 C. Powell G. Occhialini: Two particles, pion and muon Three tracks in a photographic emulsions at Mt Chacaltaya (5600 m). C Powell μ e The π is Yukawa's meson G Occhialini τπ = 26 ns S = 0 W Pauli Hey, there's something missing here The μ is a lepton (a heavier electron) mμ = 106 MeV/c2 τμ = 2200 ns S = ½ μ π 1947 mπ = 140 MeV/c
16 So what is missing? W Pauli Hey, there's something missing there μ e μ π Linear momentum conservation!
17 Pion and muon, both weak decays νe π π+ μ + + νe μ νe μ e + + μ e + ν μ + νe Matter Antimatter + e e =e p p =p n n + ππ 0= π 0 π =π + μ μ =μ νμ
18 Recognitions Incidentally: Nobel prize for 1906 J.J. Thomson 1908 E. Rutherford 1933 P.A.M Dirac and E. Schrödinger 1935 J. Chadwick 1936 C.D. Anderson, V.F. Hess 1938 E. Fermi 1945 W. Pauli 1949 H. Yukawa 1950 C.F. Powell Physics Chemistry Physics Physics Physics Physics Physics Physics Physics 18
19 What we know today π=u d π0 = u u d d c t g up charm top gluon d s b down strange bottom photon e μ τ W νe νμ ντ W boson Z0 Unified forces Baryons Mesons u Quarks n=ddu Leptons p =uud Z0 boson Three families
20 So we now have everything We can produce pions + p + n n + n + π + and they produce muons + + π μ + νμ However μsr needs another ingredient to work... 20
21 Lee and Yang: parity violation In certain interactions (e.g. magnetic) parity is broken i.e. the mirror image does not exist in nature TD Lee CN Yang Weak interactions violate parity
22 Parity violation CB Wu Madame CB Wu demonstrated that weak interactions violate parity Only right-handed anti-neutrinos and left-handed neutrinos exist in nature Anisotropic decay TD Lee and CN Yang got the 1957 Nobel prize
23 Parity violation Also Garwin Lederman & Weinrich showed that weak interactions violate parity + + μ e + ν μ + νe Nµ = Ne = = 0 N µ =
24 Let's sum up Accelerate protons to Ek > 280 MeV ~ 2mπc2, to impinge on a target p+ n 24
25 Pion production Accelerate protons to > 280 MeV and impinge them on a target n π n + p +n n + n + π Lots of pions + 25
26 Pion decay Pions that decay at rest on the surface of the target π τ π =26 ns 26
27 Parity violation Remember! The pion is S=0 Two body decay Sν=½ π+ μ + + νμ τ π =26 ns π Sµ=½ 100% spin polarized muon beams thanks to parity violation 27
28 Energy and momentum m 2π +p 2π= m 2μ +p 2μ + m 2ν + p 2ν (c = 1) p π p p μ = p ν =p Energy and momentum conservation m π = m 2μ +p 2 +p I.e. Hence 2 π m 2μ +m 2π E μ= = MeV/c 2 2m π 2 μ m m p= = 29.8 MeV/c 2m π And the muon kinetic energy is 2 v p 29.8 β= = = c E μ E μ,k = m +p m μ= 4.12 MeV/c 2 μ 2
29 Muon beam transport Quadrupole Dipole Qu ad ru p ole pa ir brings muons to stop in a sample (mostly at an interstitial site) 29
30 Let's introduce Muonium Mu = µ+ + e- In matter it most often binds to other ions forming covalent bonds 1s Mu O Bound state, light isotope of H Paramagnetic Diamagnetic 30
31 Thermalization 4 MeV Electron scattering (ionization) s 2-3 kev Muonium formation (e- capture/loss + collisions) s 200 ev few ev 31
32 Muon decay Average lifetime τμ = 2.2 μs + + μ e + ν e + ν μ Ne t τμ e+ 32
33 Muon decay Three body decay μ+ e + + νe + ν μ Takes place like this: νe ν μ μ+ (by parity violation this does not take place) e+ μ+ e+ νe ν μ Emax ~ ½mµc2 ~ 50 MeV 33
34 Muon decay Three body decay μ+ e + + νe + ν μ or like this: νe μ+ e+ Emin = 0 ν μ 34
35 Energy distribution in the muon decay Positron distribution P (x, θ)=1+a (x )cos θ with asymmetry A ( x )= 2 x x and 2x2 probability of emission E (x )= 3 2 x 40
36 Asymmetry of the muon decay Probability of e+ emission Sµ Sµ E =E max average over all energies P (θ) 1+cos θ 1 P (θ) 1+ cos θ 3 41
37 No spin dynamics Sµ B F Asymmetry 42
38 Spin dynamics: precession Magnetic moment Semiclassical dynamics m=γ ℏ S ℏ z^ ds =m B loc dt m(0) z^ B = B z^ θ y^ x^ x^ [ cos ωt sin θ m (t )=m sin ω t sin θ cos θ ] [ ] [ sin ω t sin θ sin ωt sin θ ω m cos ω t sin θ =m ( γ B loc ) cos ω t sin θ 0 0 Larmor 43 ]
39 Transverse field µsr No spin dynamics Spin precession at the Larmor frequency ω= γ B loc γ =135.5 MHz/T 2π B 44
40 Local field ω= γ B loc m Larmor frequency magnetic moment e.g. in a magnetic material dipolar Fermi contact B loc α= D αi β m βi + A c m α1 i α, β=x, y, z 45
41 MuSR nowadays µ 46
42 PSI GPS: another workhorse µ y x z 47
43 Example: Antiferromagnetic YBa2Cu3O6+x 48
44 The first magnet ever: Fe3O4 A spinel ferrimagnet with the metal-insulator Verwey transition 49
45 Examples MnSi helimagnet site determined by DFT 50
46 Where? TRIUMF J-PARC ISIS PSI 51
47 Summary Muon properties S γ m τ ½ MHz/T MeV µs γe/206.8 (*) me 3.18 γp mp/9 γ=g e 2m B θ Sμ * The anomalous electron g (QED corrections) is (5) cfr. the anomalous muon g (1) 52
48 Bibliography Particle hystory survey D. Griffith, John Wiley, New York, 1987, Ch. 1 μsr A. Schenck, Adam Hilger, Bristol 1986 A. Yaouanc, P. Dalmas de Reotier, Oxford Univ. Press, p S.J. Blundell Contemporary Physics 40, 175 (1999) Some private notes at 53
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