Neutrino Interactions

Transcription

1 Neutrino Interactions Natasja Ybema Nathan Mol

2 Overview EM interaction Fermi s WI Parity violation Lefthandedness of neutrinos V-A interaction Cross sections of elastic scattering Quasi elastic scattering and confirmation V-A 2

3 Electron-proton scattering initial e j(e) γ q 2 e final p j(p) p 1 µ M fi jµ ( e) j ( p) 2 q 3

4 WI analogue to EM Example: Beta decay Currents in terms of Dirac spinors u: M fi e 2 (u p γ µ u p )( 1 q )(u eγ µ u 2 e ) Fermi s point interaction approximation: M fi G (u p γ µ u n ) (u e γ µ u ν ) n ν e p e 4

5 Limitations Good description, but... It doesn t explain two things - Parity violation - Left handedness of neutrino s 5

6 Helicity Left handed H=-1 Right handed H=1 h = s p s p s: Spin p: Momentum 6

7 Parity Violation Madame Wu Experiment Cooled 60 Co sample Nuclear spins align with magnetic field 60 Co 60 Ni * + e +ν e Real world: Spin pointing upwards Preferred electron emission direction is downwards 7

8 Parity Violation Madame Wu Experiment Cooled 60 Co sample Nuclear spins align with magnetic field 60 Co 60 Ni * + e +ν e Mirror world: Spin pointing downwards Preferred electron emission direction is still downwards Parity violation 8

9 Goldhaber experiment Experiment to determine the handedness of neutrinos 9

10 K capture Initial Situation 1 J z =1/ 2 e Eu σ e = 1 2 Situation 2 J z = 1/ 2 e Eu 10

11 K capture Initial Final Situation 1 J z =1/ 2 e Eu 152 Sm * ν σ e = 1 2 σ ν = 1 2 Situation 2 e - ν J z = 1/ Sm * 152 Eu 11

12 Spin Conservation Situation 1 J z =1/ 2 J z =1 1 2 = 1 2 J =1 Final 152 Sm * ν J z = 1+ 1 ν 2 = 1 2 J z = 1/ Sm * Situation 2 J =1 σ ν =

13 Helicity γ vs. helicity ν Situation 1 J z =1/ 2 Left-handed γ λ = -1 J =1 Final Left-handed ν 152 Sm * ν Situation 2 ν J z = 1/ 2 Right-handed γ 152 Sm * Right-handed ν λ = J =1

14 Measuring the γ helicity Analyzing magnet Lead shielding Scatterer Scintillator detector 14

15 Results Neutrino helicity = -1 Neutrinos are lefthanded 15

16 16

17 Dirac equations require gamma matrices! γ 0 = # 0 I " I 0 $ & γ k = %! # "# 0 σ k σ k 0 $ & %&! γ 5 = # I 0 " 0 I $ & %! σ 1 = # 0 1 " 1 0 $! & σ 2 = # 0 i % " i 0 $! & σ 3 = # 1 0 % " 0 1 $ & % Pauli spin matrices Gamma matrices are the transformation operators in the WI γ 5 = iγ 0 γ 1 γ 2 γ 3 17

18 Handedness of Leptons Apply 1 γ 5 operator to chiral states " 1 γ 5 = $ # I 0 0 I % " ' $ I 0 & # 0 I % " ' = $ 2I 0 & # 0 0 % ' & u ν & χ # = $! % φ " 1 2 (1 γ 5 )u ν = 1 2 (1+γ 5 )u ν = " $ # " $ # I and I %" ' & $ # %" ' & $ # χ ϕ χ ϕ % " ' = χ $ & # 0 % ' = " 0 $ & # ϕ % ' & % ' & 18

19 Handedness of Leptons Apply 1 γ 5 operator to chiral states " 1 γ 5 = $ # I 0 0 I % " ' $ I 0 & # 0 I % " ' = $ 2I 0 & # 0 0 % ' & u ν & χ # = $! % φ " 1 2 (1 γ 5 )u ν = 1 2 (1+γ 5 )u ν = " $ # " $ # I and I %" ' & $ # %" ' & $ # χ ϕ χ ϕ % " ' = χ $ & # 0 % ' = " 0 $ & # ϕ % ' & % ' & Lefthanded states 19

20 Possible operators constructed from gamma matrices O i 1 γ γ 5 µ iγ iγ 5 µ γ γ µ ν Scalar Pseudoscalar Vector Axial Tensor vector S P V A T Parity most general form of any matrix element M fi Ci ( u poiun ) ( ueoiuν ) i 20

21 Parity Violation in WI Add an extra term to take the parity violation into account M M M fi fi fi Ci ( u poiun ) ( ueoiu ) + Ci" i ( u O u ) ( u O ν p i n e iγ 5 ν i ' C( $ i 5 Ci ( u poiun ) % ueoi (1 + γ ) uν " i & Ci # From experiment (left-handedness): = G 2 i C i ( u p O u i 21 n! C i = C i ) [ u ] eo γ 5 (1 ) u i ν u )

22 V-A Interaction Left handed electrons require V A, for V: for A: 5 γ γ γ µ µ i O O i i = = ] ) (1 [ ] ) ( [ 2 } ] ) (1 [ ) ( ] ) (1 [ ) ( { ν µ µ ν µ µ ν µ µ γ γ γ γ γ γ γ γ γ γ γ γ u u u C C u G u i u u i u C u u u u C G M e n A V p e n p A e n p V fi + = + = 22

23 Matrix element for beta decay Experimental evidence for C A C V = 1 M fi = G 2 ( u p γ µ (1 γ 5 ) u n ) [ u e γ µ (1 γ 5 ) u ν ] Inspiration for neutrino scattering 23

24 (Anti-)neutrino-electron elastic scattering ( ) ν + e α ( ) ν + e α Can be split into two scenarios: 1) 2) ( ) ν ( ) ν e+ e ν ( ) e+ e µ,τ + e ν ( ) µ,τ + e 24

25 (Anti-)neutrino-electron elastic scattering ( ) ν + e α ( ) ν + e α Can be split into two scenarios: 1) 2) ( ) ν ( ) ν e+ e ν ( ) e+ e µ,τ + e ν ( ) µ,τ + e 25

26 Electronneutrino-electron scattering initial final initial final Charged current Neutral current 26

27 Lagrangian Charged current L eff (ν e e ν e e ) = G F {# $ 2 ν eγ µ (1 γ 5 )e%# & $ eγ (1 γ 5 µ )ν % e & + " # ν eγ µ (1 γ 5 )ν e $ " % eγ (g l # µ V Neutral current g Al γ 5 )e$ % } g-factors are obtained by the electroweak unification s W = sinθ W 27

28 (Anti-)neutrino-electron elastic scattering ( ) ν + e α ( ) ν + e α Can be split into two scenarios: 1) 2) ( ) ν ( ) ν e+ e ν ( ) e+ e µ,τ + e ν ( ) µ,τ + e 28

29 Neutral current scattering initial final Neutral current L eff (ν α e ν α e ) = G F # 2 ν αγ µ (1 γ 5 )ν %# $ α & eγ l $ µ (g V g Al γ 5 )e% & (α=µ,τ ) 29

30 Fierz Transformation The electronneutrino Lagrangian can be transformed using the Fierz transformation into: L eff (ν e e ν e e ) = G F # 2 ν eγ µ (1 γ 5 )ν %# $ e & $ eγ µ ((1+ g l V ) (1+ g l A )γ 5 )e% & This looks similar to the neutral current expression: L eff (ν α e ν α e ) = G F # 2 ν αγ µ (1 γ 5 )ν %# $ α & eγ l $ µ (g V g Al γ 5 )e% & (α=µ,τ ) 30

31 Kinematics From energy and momentum conservation: Electron kinetic energy: initial E νi final e - p ef T e for θ = 0 : ν p νi the maximum kinetic energy of the electron: p ei e - θ p νf ν E νf By rewriting this equation one obtains the minimum (threshold) energy of the neutrino 31

32 Threshold energy Kamiokande detector requires: T e thr = 4.5MeV E ν min = E ν thr = 4.74MeV 32

33 Differential Cross Sections Obtained from the Feynman diagrams: with the momentum transfer squared with g-factors and combine to Integrate over all possible angles to get the total cross section 33

34 Total Cross Section without threshold 34

35 Total Cross Section with threshold 35

36 Total Cross Sections LH RH LH RH } }?? where s = (E νi + E ei ) 2 is the total squared center-of-mass energy 36

37 Total Cross Sections LH RH LH RH } } CC and NC Only NC where s = (E νi + E ei ) 2 is the total squared center-of-mass energy 37

38 Weinberg Angle Using the total cross sections to calculate the Weinberg angle: sin 2 θ W = 0, ± 0, (Global average) Recall: g-factors depend on the Weinberg angle CHARM II Experiment (CERN 1990) 38

39 Quasi Elastic Scattering ν µ + e ν e + µ (Inverse muon decay) The threshold for a scattering process of type can be calculated by taking into account that This leads to the energy threshold: s 2 2 & # 2 Eν ma + ma $ mx! X " = % Why is the energy threshold of the inverse muon decay orders of magnitudes bigger then the energy threshold of the elastic scattering? 39

40 Question Why is the energy threshold of the inverse muon decay E ν thr = 4.74MeV orders of magnitudes bigger then the energy threshold of the elastic scattering? E th =10. 92GeV ν 40

41 Answer Electron mass: 0,511 MeV Muon mass: 105,658 MeV 41

42 Confirmation V-A prediction Measured cross section: σ/e ν =(16.51±0.93) cm 2 /GeV Prediction of V-A Standard Model: σ/e ν = cm 2 /GeV 42

43 Summary Neutrinos are left-handed (and anti-neutrinos righthanded) Cross sections depend on neutrino-flavor and handedness V-A theory is confirmed 43

44 References Carlo Giunti and Chung W. Kim, Fundamentals of Neutrino Physics and Astrophysics Oxford University Press 2007, ISBN Burcham & Jobes, Nuclear and Particle Physics, Longman Scientific & Technical B. R. Martin, Nuclear Particle Physics, John Wiley and sons 2009 H. Löhner, lectures, Student seminars on subatomic physics, Stefan A. Gärtner, Neutrino Helicity measurement, University of Saskatchewan

45 Questions: How can one determine the neutrino helicity? Why are there different cross sections for different neutrino flavors? 45

46 46

47 47

48 Thank you for your attention 48