Vacuum Polarization. In QED, the bare charge of an electron is actually infinite!!! ''running coupling constant"

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1 Vacuum Polarization e e Thus, we never actually ever see a ''bare" charge, only an effective charge shielded by polarized virtual electron/positron pairs. A larger charge (or, equivalently, α) will be seen in interactions involving a high momemtum transfer as they probe closer to the central charge. ''running coupling constant" q In QED, the bare charge of an electron is actually infinite!!! Note: due to the field-energy near an infinite charge, the bare mass of the electron (E=mc 2 ) is also infinite, but the effective mass is brought back into line by the virtual pairs again!!

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3 Space-Time Symmetries F m 1 m 2 x 1 x 2 If the force does not change as a function of position, then force felt at x 2 : F = m 2 x 2 recoil felt at x 1 : F = m 1 x 1 subtracting: m 2 x 2 + m 1 x 1 = 0 d/dt [ m 2 x 2 + m 1 x 1 ] = 0 m 2 v 2 + m 1 v 1 = constant Translational Invariance Conservation of Linear Momentum

4 Consider a system with total energy 1 E = m x 2 + V 2 de dt = m x x + dv = m x x + x dx but dv/dx = F = mx dv dx dx dt assume this basic description also holds at other times (Newton s 2nd law) de dt = m x x m x x = 0 E = constant Time Invariance Conservation of Energy

5 Gauge Invariance in Electromagnetism: ''local" symmetry E = Φ - A A + χ(x,t) Φ Φ χ(x,t) t A t [Φ - χ(x,t) ] - [A + χ(x,t)] t t = Φ - A = E t B = A [A + χ(x,t)] = A = B Gauge Invariance Conservation of Charge

6 (Wigner, 1949) To see this, assume charge were not conserved So a charge could be created here by inputing energy E And destroyed here, with the output of some energy Eʹ PoP! q PoP! x 1 E = qφ(x 1 ) x 2 Eʹ = qφ(x 2 ) Thus we will have created an overall energy Eʹ E = q { Φ(x 2 ) Φ(x 1 ) } So, to preserve energy conservation, if Φ is allowed to vary as a function of position, charge must be conserved

7 Noether s Theorem Continuous Symmetries Conserved ''Currents" (Emmy Noether, 1917)

8 Gauge symmetry from another angle... Take the gauge transformation of a wavefunction to be where θ is an arbitrary ''phase-shift" as a function of space and time Ψ e iqθ Ψ Say we want the Schrodinger equation to be invariant under such a transformation clearly we re in trouble! Ψ = t i 2m 2 Ψ Consider the time-derivative for a simple plane wave: Ψ = Ae i(px-et) Ψ Ae i(px Et+qθ) Note that if we now introduce an electric field, the energy level gets shifted by qφ / t Ψ = i ( E + q θ/ t ) Ψ here s the problem! / t Ψ = i ( E + qφ + q θ/ t ) Ψ But we can transform Φ Φ θ/ t, thus cancelling the offending term! (a similar argument holds for the spatial derivative and the vector potential) Gauge invariance REQUIRES Electromagnetism!!

9 Another example... Special Relativity: Invariance with respect to reference frames moving at constant velocity global symmetry Generalize to allow velocity to vary arbitrarily at different points in space and time (i.e. acceleration) local gauge symmetry Require an interaction to make this work GRAVITY!

10 All known forces in nature are consequences of an underlying gauge symmetry!! or perhaps Gauge symmetries are found to result from all the known forces in nature!!

11 Pragmatism: Symmetries (and asymmetries) in nature are often clear and can thus be useful in leading to dynamical descriptions of fundamental processes True even for ''approximate" symmetries!

12 For ''pre-1974" hadrons, the following symmetries were also observed Q = I 3 + (B+S)/2 Gell-Mann - Nishijima Formula thus, define ''Hypercharge" as Y B + S Y Mesons Y Κ 0 (498) 1 Κ + (494) Κ (896) 1 Κ + (892) π (140) (135) π 0 η (547) ηʹ (958) π + (140) I 3 ρ (769) (769) ρ 0 ω (782) ϕ (1019) ρ + (769) I 3 Κ (494) ( Spin Parity ) -1 0 nonet Κ 0 (498) Note the presence of both particles and antiparticles Κ (892) -1 1 nonet Κ 0 (896)

13 Y Baryons Y n (940) 1 p (938) Δ (1232) Δ (1232) 1 Δ + (1232) Δ ++ (1232) Σ (1197) Σ 0 (1193) Λ (1116) Σ + (1189) I 3 Σ* (1387) Σ* ο (1384) Σ* + (1383) I 3 Ξ (1321) -1 Ξ 0 (1315) Ξ (1535) -1 Ξ ο (1532) ( Spin Parity ) 1/2 + octet Note antiparticles are not present Ω (1672) 3/2 + decuplet

14 Inelastic Scattering: Evidence for Compositeness

15 Consider a 3-component ''parton" model where the constituents have the following quantum numbers: Y Y 1 1 s d u -1 1 I 3-1 u d 1 I 3 s -1-1 ''quarks" ''anti-quarks"

16 Mesons are generally lighter than baryons, suggesting they contain fewer quarks Also, the presence of anti-particles in the meson nonets suggests they might be composed of equal numbers of quarks and anti-quarks (so all possible combinations would yield both particles and anti-particles) Further, if we assume quarks are fermions, the integer spins of mesons suggest quark-antiquark pairs We can add quarks and anti-quarks quantum numbers together graphically by appropriately shifting the coordinates of one ''triangle" with respect to the other: Y d s 1 u s d u -1 d d u u s s u d 1 I 3 s u -1 s d

17 Baryons: So try building 3-quark states Spin numbers of 1/2 and 3/2 suggest the superposition of 3 fermions Absence of anti-particles suggests there is not substantial anti-quark content (note that m(σ ) m(σ + ) so they are not anti-particles, and similarly for the Σ* group) Start with 2:

18 Baryons: Spin numbers of 1/2 and 3/2 suggest the superposition of 3 fermions Absence of anti-particles suggests there is not substantial anti-quark content (note that m(σ ) m(σ + ) so they are not anti-particles, and similarly for the Σ* group) So try building 3-quark states ddd ddu duu uuu Now add a 3rd: dds uds uus The baryon decuplet!! dss uss and the Ω sealed the Nobel prize sss

19 Baryons: So try building 3-quark states Spin numbers of 1/2 and 3/2 suggest the superposition of 3 fermions Absence of anti-particles suggests there is not substantial anti-quark content (note that m(σ ) m(σ + ) so they are not anti-particles, and similarly for the Σ* group) Start with 2:

20 Baryons: Spin numbers of 1/2 and 3/2 suggest the superposition of 3 fermions Absence of anti-particles suggests there is not substantial anti-quark content (note that m(σ ) m(σ + ) so they are not anti-particles, and similarly for the Σ* group) So try building 3-quark states ddd ddu duu uuu Now add a 3rd: dds uds uus The baryon decuplet!! dss uss and the Ω sealed the Nobel prize sss

21 But what about the octet? It must have something to do with spin... (in the decuplet they re all parallel, here one quark points the other way) We can ''chop off the corners" by artificially demanding that 3 identical quarks must point in the same direction Σ (1197) n (940) Ξ (1321) 1-1 Y p (938) Σ 0 (1193) Λ (1116) Ξ 0 (1315) Σ + (1189) I 3 J=1/2 But why 2 states in the middle? ddd ddu duu uuu ways of getting spin 1/2: u d s u d s u d s dds uds uus Σ 0 dss uss these ''look" pretty much the same as far as the strong force is concerned (Isospin) Λ sss J=3/2

22 Charge: d + d + u = 0 u = -2d Σ (1197) n (940) 1 Y p (938) Σ 0 (1193) Λ (1116) Σ + (1189) I 3 d + u + u = +1 d + 2(-2d) = +1-3d = +1 Ξ (1321) -1 Ξ 0 (1315) J=1/2 ddd ddu duu uuu d = -1/3 & u= +2/3 dds uds uus dss uss u + d + s = 0 s = -1/3 sss J=3/2

23 So having 2 states in the centre isn t strange... but why there aren t more states elsewhere?! u u s u u s i.e. why not and??? We can patch this up again by altering the previous artificial criterion to: The lowest energy state ''Any pair of similar quarks must is be in identical spin states" ( ) Not so crazy lowest energy states of simple, 2-particle systems tend to be ''s-wave" (symmetric under exchange) Σ (1197) ddd ddu duu uuu dds n (940) Ξ (1321) 1-1 Y p (938) Σ 0 (1193) Λ (1116) Ξ 0 (1315) uds Σ + (1189) I 3 J=1/2 uus What happened to the Pauli Exclusion Principle??? Why are there no groupings suggesting qq, qqq, qqqq, etc.?? What holds these things together anyway?? dss sss uss J=3/2

24 Pauli Exclusion Principle there must be another quantum number which further distinguishes the quarks (perhaps a sort of ''charge") Perhaps, like charge, it also helps hold things together! We see states containing up to 3 similar quarks this ''charge" needs to have at least 3 values (unlike normal charge!) Call this new charge ''colour," and label the possible values as Red, Green and Blue We need a new mediating boson to carry the force between colours (like the photon mediated the EM force between charges) call these ''gluons"

25 In p-n scattering, u and d quarks appear to swap places. But their colours must also swap (via gluon interactions). This suggests that an exchange-force is involved... But then we run into trouble while trying to conserve charge at an interaction vertex: R G (a quark-screw!) the only way out is to attribute colour to the gluons as well. For the above case, the gluon would have to carry away RG quantum numbers

26 We could explain only having the quark combinations seen if we only allowed ''colourless" quark states involving either colour-anticolour, all 3 colours (RGB), or all 3 anticolours. (hence the analogy with ''colour", since white light can be decomposed into either red, green & blue or their opposites - cyan, magenta & yellow) If the carriers of the force (the gluons) actually carry colour themselves, the field lines emanating from a single quark will interact: q q q ''flux tube" * formally still just a hypothesis (calculation is highly non-perturbative)

27 For this configuration, the field strength (flux of lines passing through a surface) does not fall off as 1/r 2 any more it will remain constant. The field energy will thus scale with the length of the string and so as L then E Clearly we can t allow this!! Can be stopped by terminating field line on another colour charge Ah! So only colourless states have finite energy! ''Confinement" PoP! q q q q q q q q q q ''fragmentation"

28 Getting very close to a quark: RB q q q q q So, on average, the colour is ''smeared" out into a sort of ''fuzzy ball" q RG Thus, the closer you get, the less colour charge you see enclosed within a Gaussian surface. So, on distance scales of ~1 fm, quarks move around each other freely ''Asymptotic Freedom"

29 Where Are The Coupling Constants Running??

30 1974

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32 ee+

33 Come The Revolution... (1974) SPEAR (Richter et al.)

34 e + π + Ψ π Ψʹ ( ) e

35 Brookhaven (Ting et al.)

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37 J/Ψ CC ''Charmonium" Not entirely unexpected had been some hints of needing another quark (Glashow, Bjorken,...) ( GIM mechanism) nicer parallel with the (then known) leptons: e µ ν e ν µ u d c s There are different energy states of Charmonium, in analogy to hydrogen atom:

38 Hydrogen Different mechanisms which separate energy levels: n basic Bohr energy level α 2 ''Fine Structure" spin-orbit coupling α 4 ''Hyperfine Structure" magnetic moment coupling (very small in hydrogen) α 4

39 Charmonium & Positronium Energy Levels Note in charmonium the various splittings are all of the same order so α S ~ 1 (best fit actually gives α S 0.2)

40 ''Naked" Charm:

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43 ''Meanwhile, back at the lab, Martin was about to make another unexpected discovery..." e + e µ + e + ''missing energy" lepton number violation?? e + e τ + + τ e + ν e + ν τ µ + + ν µ + ν τ m τ = 1.78 GeV!

44 τ decays weakly and we would naively expect the lifetime to be roughly T τ (m µ /m τ ) 5 T µ = (0.106/1.78) 5 (2.2x10 6 s) = 1.6x10 12 s However The τ is heavy enough to decay into hadrons as well as muons and electrons (where as muon can only decay to electrons) τ W s Κ τ W d π u u ν τ ν τ Cabibo-mixing" (only counts as 1)

45 So there is a relative phase-space factor of roughly: µ-decay τ-decay electron electron muon quarks 1 versus quark colours!! T τ (1.6x10-12 s) (1/5) = 3.2x10 13 s Experiment: 3.0±0.1 x10 13 s

46 But is this real? fragmentation e + µ + e + q hadrons Consider: versus e µ e q hadrons Cross-sections for both processes should be basically the same except for an additional phase-space factor for the number of different quarks and different colour states that can be produced R σ (e + e hadrons) σ (e + e µ + µ ) = σ (e + e qq ) σ (e + e µ + µ ) For the CM energies we will look at, only the 5 lightest quarks can be produced. R = N C [ (q u /e) 2 + (q d /e) 2 + (q s /e) 2 + (q c /e) 2 + (q b /e) 2 ] = N C [ 4/9 + 1/9 + 1/9 + 4/9 + 1/9 ] = (11/9) N C (in fact, higher order corrections suggest a better estimate of R (11/9) N C (1+α S /π) )

47 N C = 3!!

48 π 0 More Evidence for Colour: u u u γ γ π 0 π 0 γ + γ In this case, the amplitudes add coherently and calculation yields: ( N 2 Γ (π 0 C ) α 2 m 3 π 2γ) = 3 64π 3 f 2 = 7.73 (N C /3) 2 π + u u u γ γ π 0 + u u u γ γ +... if there are other colours (''pion decay constant" f π = 92.4 MeV from charged pion decay rate) Experimentally Γ = 7.7 ± 0.6 ev N C = 2.99 ± 0.12

49 Parity Violation in Weak Interactions First suggested in 1956 by Lee & Yang based on review of kaon decay modes Directly observed by Wu et al. in 1957 from the decay 60 Co 60 Ni* + e + ν e γ (1.173 MeV) + γ (1.332 MeV) nuclear spins aligned by cooling to 0.01 o K in a magnetic field 60 Co e P 60 Co (degree of polarisation determined from the anisotropy of γ-rays) e Should be the same under parity transformation, but fewer electrons are actually seen going forward!

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51 Garwin, Lederman & Weinrich (1957) ν µ π+ e+ µ+ (polarised) ν µ ν e precess polarised muons

52 Also, in 1958, Goldhaber et al. measured the helicity of the neutrino: e Eu(J=0) 152 Sm*(J=1) + ν e 152 Sm(J=0) + γ events were chosen with the final states collinear γ and ν e travel in opposite directions, so helicity of the neutrino is found from that of the gamma all neutrinos are left-handed!

53 Kaons: K o = ds K o = sd (S = +1) (S = 1) But S is not conserved in weak interactions so K o -K o mixing can occur: K o d s W + u u W s d K o We can thus define two orthogonal mixtures: K 1o = 1/ 2 ( K o + K o ) K 2o = 1/ 2 ( K o K o ) Note: C P K 1o = + K 1o and C P K 2o = K 2o K 1 o π + π ; π o π o K 2 o π + π π o ; π o π o π o Allowed K 1 o π + π π o ; π o π o π o K 2 o π + π ; π o π o Forbidden

54 Experimentally, 2 kaon states are observed with different lifetimes: K S o π + π ; π o π o τ 9x10 11 s K L o π + π π ο ; π o π o π o ; π ± lepton ν (ν) τ 5x10 8 s ± So we associate K S o K 1 o and K L o K 2 o However, in 1964, Christenson, Cronin, Fitch & Turlay discovered K L o π + π (branching ratio ~ 2x10 3 )

55 beam collimator lead-glass cuts out photons 30 GeV protons K S +K L K L 18 m steel target magnets sweeps out charged particles CM of π + π pair )θ K L beam direction

56 K So = 1/ 1+ ε 2 ( K 1o ε K 2o ) K Lo = 1/ 1+ ε 2 ( ε K 1o + K 2o ) where ε small complex number parameterizing the size of the CP violation (experimentally, ε 2.3x10 3 ) What does this mean?? Reason for antimatter assymmetry?? Perhaps we can learn more from studying CP violation in other particle systems...

57 Matter-Antimatter Asymmetry Revisited: Sakarov Conditions (1967)!!! (GUTs) 1) Baryon Number Violation allows baryons and anti-baryons to appear and disappear independently of each other 2) CP Violation so the rate of appearance/disappearance of baryons is different from anti-baryons 3) Non-Equilibrium Conditions since equilibrium would then tend to ''average-out" any asymmetry Establishes Asymmetry Locks In Asymmetry

58 Recall that the ''matrix element" for scattering from a Yukawa potential is Ψ f V Ψ o = g 2 /(q 2 +M 2 ) In the Fermi theory of β decay, this is what essentially becomes G F or, more precisely, G F / 2 = g 2 /(q 2 +M 2 ) = 4πα W /(q 2 +M 2 ) σ G F 2 and the relatively small value of G F characterizes the fact that the weak interaction is so weak We can get this small value either by making α W small or by making M large So what if we construct things so α W = α??? UNIFICATION!! Assuming M q 2, M = 4π 2 α / G F α = 1/137 G F = 10 5 GeV 2 M ~ 100 GeV

59 p u u d hadrons ν e, ν µ, ν τ W - p u u d hadrons e-, µ-, τ- Stochastic Cooling Electron Cooling

60 Recall that the ''matrix element" for scattering from a Yukawa potential is Ψ f V Ψ o = g 2 /(q 2 +M 2 ) In the Fermi theory of β decay, this is what essentially becomes G F or, more precisely, G F / 2 = g 2 /(q 2 +M 2 ) = 4πα W /(q 2 +M 2 ) σ G F 2 and the relatively small value of G F characterizes the fact that the weak interaction is so weak We can get this small value either by making α W small or by making M large So what if we construct things so α W = α??? UNIFICATION!! Assuming M q 2, M = 4π 2 α / G F α = 1/137 G F = 10 5 GeV 2 M ~ 100 GeV CERN, 1983 M W = 80 GeV!!

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62 A Brief Theoretical Interlude (electroweak theory... at pace!!)

63 But how can this be the ''same" force when the W s are charged and the photon certainly isn t!? Is there a way we can ''bind up" the W s along with a neutral exchange particle to form a ''triplet" state (i.e. like the pions)?? Well, like with the pions, we seem to have a sort of ''Weak" Isospin since the weak force appears to see the following left-handed doublets ν e e µ τ dʹ sʹ bʹ ( ) ( ν ) µ ( ν ) τ ( u ) ( c ) ( t ) L L L L L L as essentially two different spin states: I W (3) = ± 1/2 (like p-n symmetry) Thus, in the process ν e W + e The W + must carry away +1 units of I W (3) so let s symbolically denote W + and, similarly, W If I W = 1 for the W s then, similar to the π o, there is also a neutral state: W o = 1/ 2 ( ) (which completes the triplet)

64 There is, however, another orthogonal state: 1/ 2 ( + ) If we ascribe this to the photon, then perhaps we might expect to see weak ''neutral currents" associated with the exchange of a W o with a similar mass to the W ± so we d have a nice ''single package" which describes EM and weak forces! Hold on... any simple symmetry is obviously very badly broken the photon is massless and the W s are certainly not! The photon is also blind to weak isospin and also couples to right-handed leptons & quarks as well Assume the symmetry was initially perfect and all states were massless Then postulate that there exists some overall (non-zero) ''field" which couples to particles and gives them additional virtual loop diagrams : but in the limit of zero momentum transfer (rest mass), so represent as (kind of like an ''aether" which produces a sort of ''drag") Higgs Mechanism

65 Further suppose that this field is blind to weak isospin and, thus, allows for it s violation. This would allow the neutral weak isospin states to mix like with the mesons (the W ± are charged and cannot mix) We will call the ''pure," unmixed states W o and Γ And we will call the physical, mixed states Z o and γ

66 Think about mathematically introducing this Higgs coupling by applying some ''mass-squared" operator to the initial states (since mass always enters as the square in the propagator) M 2 W ± = W ± G + W G W W ± M 2 W ο = W ο G W G W + W ο G W G Γ + Γ M 2 G Γ G Γ G Γ G W Γ = Γ + Γ + W o where the right-most terms represent the weak isospin - violating terms Assume couplings to W s are all the same (G W ) but coupling to Γ may be different (G Γ ) For the W ± the mass would then simply be given by M W 2 = G W 2 (where G 2 contains the coupling plus a few other factors) For the latter 2 equations, we can think of M 2 as an operator which yields the mass-squared, M 2, for the coupled state: M 2 W o = G W 2 W o + G W G Γ Γ M 2 Γ = G Γ 2 Γ + G W G Γ W ο

67 G From the second of these: W G Γ Γ = (M 2 -G W o Γ2 ) G 2 Substituting into the first: M 2 W o = G 2 W W o W G 2 Γ + W o (M 2 -G Γ2 ) M 4 M 2 G Γ 2 = M 2 G W 2 G W 2 G Γ 2 + G W 2 G Γ 2 M 2 ( M 2 G Γ 2 G W2 ) = 0 M 2 = 0 or M 2 = G W 2 + G Γ 2 Thus, associate M γ 2 = 0 and M Z 2 = G W 2 + G Γ 2 Note also that M Z > Μ W

68 We can parameterize the γ as a mixture of W o and Γ as follows: γ Γ sinθ W Wo cosθ W θ W ''Weinberg Angle" Thus, applying M 2 : M 2 γ = M 2 (Γ sinθ W W o cosθ W ) = 0 0 = ( G Γ 2 Γ + G W G Γ Wο ) sinθ W (G W 2 W o G W G Γ Γ) cosθ W Coefficient of W o G W G Γ sinθ W G W 2 cosθ W = 0 Coefficient of Γ G Γ 2 sinθ W G W G Γ cosθ W = 0 tan θ W = G Γ / G W ''unification condition" ''anomaly condition" M Z 2 /M W 2 = (G W 2 + G Γ2 )/G W 2 = 1/cos 2 θ W M Z = M W /cosθ W ΣQ l + 3ΣQ q = 0 (leptons) (quarks) is satisfied separately for each generation

69 Neutral Current Event (Gargamelle Bubble Chamber, CERN, 1973) ν µ π p

70 From comparing neutral and charged current rates sin 2 θ W = M W = 80 GeV M Z = 91 GeV (predicted) Z e + e M Z = 91 GeV (observed!!)

71 While we re here... pre-abba weak doublet = So, consider the coupling to the Z 0 : ( u ) d = ( u ) d cosθ C + s sinθ C u u Z 0 + (d cosθ C + s sinθ C ) (d cosθ C + s sinθ C ) Z 0 Probability product of wave functions: uu + (dd cos 2 θ C + ss sin 2 θ C ) + (sd + ds ) sinθ C cosθ C ΔS = 0 ΔS = 1 Flavour-Changing Neutral Currents never seen!

72 Postulate 2 doublets: ( u ) d = ( u ) d cosθ C + s sinθ C (Glashow, Iliopolis & Maiani: GIM mechanism) & ( c ) s = ( c ) s cosθ C d sinθ C u (d cosθ C + s sinθ C ) Z 0 + Z 0 u (d cosθ C + s sinθ C ) c (s cosθ C d sinθ C ) + Z 0 + Z 0 c (s cosθ C d sinθ C ) uu + cc + (dd+ss)cos 2 θ C + (ss+dd) sin 2 θ C ) + (sd + ds - sd - sd) sinθ C cosθ C ΔS = 0 ΔS = 1

73 Blam! ( dp ) Rate W = Γ 0 dn Transition formation ''rate" of initial state (recall Γ = /τ) prob for decay to particular final state given the total number of available states = Γ 0 ( dp )( de ) de dn dp 1 Γ = f de 2π (E-E 0 ) 2 + Γ 2 /4 σ = But recall that W = v B σ / V π Γ 0 Γ f q 2 (E-E 0 ) 2 + Γ 2 /4 ( ) 1 ( V q 2 dq dω ) = (2π) 3 de dn de ( V q 2 ) 1 2π 2 v 1

74 Thus, for the production of Z 0 near resonance and the subsequent decay to some final state ''X" : σ (e + e X) = 12π M Z 2 Γ(Ζ 0 e + e ) Γ(Ζ 0 X) E 2 (E 2 M Z2 ) 2 + M Z2 Γ Ζ 2 CM [ ] CM since Γ(Ζ 0 e + e ) can be related by time-reversal to Γ(e + e Ζ 0 ) Peak of resonance M Z Height of resonance product of branching ratios Br(Ζ 0 e + e ) Br(Ζ 0 X) = Γ(Ζ 0 e + e ) Γ(Ζ 0 X) Γ Ζ Γ Ζ

75 Results: M Z = ± GeV Γ(Ζ 0 hadrons) = ± GeV Γ Z = ± GeV Γ(Ζ 0 l + l ) = ± GeV (3 x ) = !! So what s left??? ''Invisible modes" Neutrinos!! (limit for light, ''active" neutrinos)

76 An End To The Generation Game??? (not necessarily a bad thing!)

77 ''π In The Sky" p Blam! γ γ π 0 µ + π + π µ N µ N e 2 e + e ν e ν µ ν µ ν e ν µ ν µ

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82 Neutrinos oscillate! or maybe decay... ν µ s disappear in an energy-dependant way They have mass!!

83 The Standard Solar Model (p-p chain) p+p 2 H+e + +ν e p+e - +p 2 H+ν e 99.7% 0.23% 2 H+p 3 He+γ 84.92% ~10-5 % 15.08% 3 He+ 3 He α+2p 3 He+p α+e + +ν e 3 He+α 7 Be+γ 15.07% 0.01% 7 Be+e - 7 Li+γ+ν e 7 Be+p 8 B+γ 7 Li+p α+α 8 B 2α+e + +ν e

84 Homestake Gold Mine (South Dakota) ν e + 37 Cl = 37 Ar + e - (Ray Davis)

85 What s SNU? 1 SNU = captures per second per target nucleus

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87 ν Reactions in SNO CC ν e + d p + p + - e - ν e only - Gives ν e energy spectrum well - Weak direction sensitivity 1-1/3cos(θ) NC ν + d p + n + x ν x - Measures total 8 B ν flux from Sun - Equal cross section for all ν types ES ν x e ν x e - Low Statistics - Mainly sensitive to ν e,, some sensitivity to ν µ and ν τ - Strong direction sensitivity

88 The Basic Maths: In terms of the Standard Solar model, we see: Φ SNO = Φ(ν e ) = 0.35 If this is due to flavour conversion, SuperK would see: Φ SK (predicted) = (1 0.35) (1/6.5) = 0.45 electron neutrinos non-electron neutrinos relative sensitivity of ES reaction to neutral current Φ SK (measurement) = 0.451!!!

89 SSM Φ total (unconstrained CC spectrum) June 2001 (indirect) April 2002 (direct) Sept 2003 (salt - unconstrained) May 2008 (indep. NCD measurement)

90 SK ΦES SNO ΦES SNO ΦCC SNO ΦNCD Φ SNO NC SNO ΦSalt Φ SSM

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94 MINOS (θ 23 )

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