1 / 29

Neutrino Phenomenology

Neutrino Phenomenology. Boris Kayser Scottish Summer School August 11, 2006 +. At least 4 mass eigenstates, hence at least 4 flavors. Are There Sterile Neutrinos?. Rapid neutrino oscillation reported by LSND —.  1eV 2. in contrast to.  m 2 atm = 2.7 x 10 –3 eV 2. >.

tamarat
Télécharger la présentation

Neutrino Phenomenology

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Neutrino Phenomenology Boris Kayser Scottish Summer School August 11, 2006 +

  2. At least 4 mass eigenstates, hence at least 4 flavors. Are There Sterile Neutrinos? Rapid neutrino oscillation reported by LSND —  1eV2 in contrast to m2atm = 2.7 x 10–3 eV2 > m2sol = 8 x 10–5 eV2 Measured (Z) only 3 different active neutrinos. At least 1sterile neutrino.

  3. Is the so-far unconfirmed oscillation reported by LSND genuine? MiniBooNE aims to definitively answer this question.

  4. What Is the Pattern of Mixing? • How large is the small mixing angle 13? We know only that sin213 < 0.032 (at 2). The theoretical prediction of 13is not sharp: Present bound ( ) Albright & Chen sin213

  5. The Central Role of 13 If sin213 > (0.0025 – 0.0050), we can study both of these issues with intense but conventional  and  beams. Both CP violation and our ability to tell whether the spectrum is normal or inverted depend on 13. Determining 13 is an important stepping-stone.

  6. How 13 MayBe Measured sin213 3 m2atm (Mass)2 2 } m2sol 1 sin213 = Ue32 is the small e piece of 3. 3 is at one end of m2atm. We need an experiment with L/E sensitive to m2atm (L/E ~ 500 km/GeV), and involving e.

  7. Complementary Approaches Reactor Experiments Reactore disappearance while traveling L ~ 1.5 km. This process depends on 13 alone: P(e Disappearance) = = sin2213 sin2[1.27m2atm(eV2)L(km)/E(GeV)]

  8. Accelerator Experiments Accelerator  e while traveling L > Several hundred km. This process depends on 13, 23, on whether the spectrum is normal or inverted,and on whether CP is violated through the phase.

  9. Neglecting matter effects (to keep the formula from getting too complicated): (—) The accelerator long-baseline e appearance experiment measures — The plus (minus) sign is for neutrinos (antineutrinos).

  10. The Mass Spectrum: or ? Generically, grand unified models (GUTS) favor — GUTS relate the Leptons to the Quarks. is un-quark-like, and would probably involve a lepton symmetry with no quark analogue.

  11. Exploit the fact that, in matter, ( ) e e W ( ) e e raises the effective mass of e, and lowers that of e. How To Determine If The Spectrum Is Normal Or Inverted This changes both the spectrum and the mixing angles.

  12. E6 GeV Matter effects grow with energy E. At E ~ 1 GeV, matter effects  sin2 2M = sin2 213 [ 1 + S ] . Sign[m2( ) - m2( )] At oscillation maximum, P(e) >1 ; P(e)<1 ; ~ (—) (—) { Note fake CP violation. In addition, { ( ) PHi E(e) >1 ; PLo E(e) <1 ; Mena, Minakata, Nunokawa, Parke

  13. CP Violation and the Matter-Antimatter Asymmetry of the Universe

  14. Leptonic CP Violation • Is there leptonic CP, or is CP special to quarks? • Is leptonic CP, throughLeptogenesis, the origin of theMatter-antimatterasymmetry of the universe?

  15. Look forP( )  P( ) How To Search for Leptonic CP “   ”is a different process from   even when i= i e  - e-  Source Detector “ e  ” + e+  Source Detector

  16. CPT: P() = P()  P() = P() No CP violation in a disappearance experiment. But if is present, P(e)  P(e): Note that all mixing angles must be nonzero for CP.

  17. Separating CP From the Matter Effect Genuine CP and the matter effect both lead to a difference between  and  oscillation. But genuine CP and the matter effect depend quite differently from each other on L and E. To disentangle them, one may make oscillation measurements at different L and/or E.

  18. What Physics Is Behind Neutrino Mass?

  19. The See-Saw Mechanism — A Summary — This assumes that a neutrino has botha Majorana mass term mRRcRand a Dirac mass term mDLR. No SM principle prevents mRfrom being extremely large. But we expect mD to be of the same order as the masses of the quarks and charged leptons. Thus, we assume that mR >> mD.

  20.    When We have 4 mass-degenerate states: This collection of 4 states is a Diracneutrino plus its antineutrino.

  21.  When= We have only 2 mass-degenerate states: This collection of 2 states is a Majorananeutrino.

  22. N mN ~ mR – Splitting due to mR Dirac neutrino  m ~ mD2 / mR – What Happens In the See-Saw? • The Majorana mass term splits a Dirac neutrino into two Majorana neutrinos. Note that mmN  mD2  mq or l2. See-Saw Relation

  23. { Familiar light neutrino  } Very heavy neutrino N The See-Saw Relation

  24. Predictions of the See-Saw • Each i =i (Majorana neutrinos) • The light neutrinos have heavy partners N • How heavy?? • mN ~ ––––– ~ –––––– ~ 1015 GeV • Near the GUT scale. – m2top m2top m0.05 eV Coincidence??

  25. A Possible Consequence of the See-Saw — Leptogenesis The heavy see-saw partners N would have been made in the hot Big Bang. Then, being very heavy, they would have decayed. The see-saw model predicts — Nl- + … and Nl+ + … If there was CP in these leptonic processes, then unequal numbers of leptons and antileptons would have been produced. Perhaps this was the origin of today’s matter-antimatter asymmetry.

  26. Enjoy The Rest Of The School!

  27. Backup Slides

  28. Here atm lies between the (very nearly equal) 31 and 32. This measurement determines sin2223, but if 23  45°, there aretwo solutions for 23: 23 and 90° – 23. A reactor experiment may be able to resolve this ambiguity. • What is the atmospheric mixing angle 23?

  29. Assumes sin2223 = .95  .01 23 Sensitive to sin2213 = 0.01 (McConnel, Shaevitz)

More Related