Measuring Atmospheric Parameters with SuperBeams Enrique Fernández Martínez Departamento de Física Teórica and IFT Universidad Autónoma de Madrid

Outline Introduction Oscillation parameters Experiments description Subleading effects in disappearance m 2 atm and the sign degeneracy 23 and the octant degeneracy The effects of 13 and T2K-1 bounds revised Conclusions

The oscillation parameters What we already know Solar sector Atm sector What we still dont know sin2 13 < 0.40 cp Mass hierarchy Octant of 23 M. C. González García hep-ph/ = 35º–55º 12 = 28º–38º

Fluxes T2K-1SPL flux from decay at flux from decay L=130Km Old SPL fluxes courtesy of Gilardoni New fluxes Campagne et al. hep-ex/ L=295Km T2K fluxes courtesy of J.J. Gómez Cadenas OA2º

Event Rates T2K-1 B1B2B3B4 No osc. N Signal N SPL - + No osc. N Signal N L=130Km 5yr exposure with a 22.5Kt water cerenkov detector for T2K-1 2yr + 8yr exposure with a 440Kt water cerenkov detector for the SPL L=295Km Statistics dominatedSystematics dominated 4 energy bins of 200MeV Between 0.4 – 1.2GeV

The importance of energy resolution 13 = 0º = 0º T2K-1SPL 90% CL contours 5% systematic error and backgrounds taken into account

The importance of energy resolution T2K-1SPL 13 = 0º = 0º GeVE27.0

The importance of energy resolution T2K-1SPL 13 = 0º = 0º GeVE25.0 GeVE27.0 E1 = GeV

The importance of energy resolution T2K-1SPL 13 = 0º = 0º GeVE25.0 GeVE27.0 E1 = GeV E2 = GeV

The importance of energy resolution T2K-1SPL 13 = 0º = 0º GeVE25.0 GeVE27.0 E1 = GeV E2 = GeV E3 = GeV

The importance of energy resolution T2K-1SPL 13 = 0º = 0º GeVE25.0 GeVE27.0 E1 = GeV E2 = GeV E3 = GeV E4 = GeV

The importance of energy resolution T2K-1SPL 13 = 0º = 0º GeVE25.0 GeVE27.0 E1 = GeV E2 = GeV E3 = GeV E4 = GeV

The disappearance channel Where sin2 2 cos ~ J sin 2 13 < L sol E m E. K. Akhmedov et al. hep-ph/ A. Donini et al. hep-ph/ E m atm

The sign degeneracy Input: 13 = 0º = 0º L L sol atm + O 2 2 sin 2 2sin L sol + O 2

The sign degeneracy L Ls L L sol atm sol atm + O 2 sin2 2 2 sin 2 2sin Input: 13 = 0º = 0º

The sign degeneracy Fit assuming inverted hierarchy Input: 13 = 0º = 0º L Ls L L sol atm sol atm + O 2 sin2 2 2 sin 2 2sin

The octant degeneracy Input: 13 = 0º = 0º L Ls L L sol atm sol atm + O 2 sin2 2 2 sin 2 2sin

The octant degeneracy Input: 13 = 0º = 0º L Ls L L sol atm sol atm + O 2 sin2 2 2 sin 2 2sin

The effect of 13 Input: 13 = 0º = 0º L Ls L L sol atm sol atm + O 2 sin2 2 2 sin 2 2sin

The effect of 13 Input: 13 = 0º = 0º Assuming 13 = 0º L Ls L L sol atm sol atm + O 2 sin2 2 2 sin 2 2sin s 2cos2sin sJ cos ~ 2 23

The effect of 13 Input: 13 = 0º = 0º Assuming 13 = 0º, 2º L Ls L L sol atm sol atm + O 2 sin2 2 2 sin 2 2sin s 2cos2sin sJ cos ~ 2 23

The effect of 13 Input: 13 = 0º = 0º Assuming 13 = 0º, 2º, 4º L Ls L L sol atm sol atm + O 2 sin2 2 2 sin 2 2sin s 2cos2sin sJ cos ~ 2 23

The effect of 13 Input: 13 = 0º = 0º Assuming 13 = 0º, 2º, 4º, 6º L Ls L L sol atm sol atm + O 2 sin2 2 2 sin 2 2sin s 2cos2sin sJ cos ~ 2 23

The effect of 13 Input: 13 = 0º = 0º Assuming 13 = 0º, 2º, 4º, 6º, 8º L Ls L L sol atm sol atm + O 2 sin2 2 2 sin 2 2sin s 2cos2sin sJ cos ~ 2 23

The effect of 13 Input: 13 = 0º = 0º Assuming 13 = 0º, 2º, 4º, 6º, 8º, 10º L Ls L L sol atm sol atm + O 2 sin2 2 2 sin 2 2sin s 2cos2sin sJ cos ~ 2 23

The effect of Input: 13 = 8º = 0º Assuming = 0º L Ls L L sol atm sol atm + O 2 sin2 2 2 sin 2 2sin s 2cos2sin sJ cos ~ 2 23

The effect of Input: 13 = 8º = 0º Assuming = 0º, 90º L Ls L L sol atm sol atm + O 2 sin2 2 2 sin 2 2sin s 2cos2sin sJ cos ~ 2 23

The effect of Input: 13 = 8º = 0º Assuming = 0º, 90º, 180º L Ls L L sol atm sol atm + O 2 sin2 2 2 sin 2 2sin s 2cos2sin sJ cos ~ 2 23

T2K-1 errors revised m 2 = (2.43 – 2.60)·10 -3 eV 2 sin 2 2 > 0.97 tan 2 = 0.73 – 1.39 m 2 = (2.50 ± 0.06)·10 -3 eV 2 sin 2 2 > 0.98 Y. Itow et al. hep-ex/ m 2 = (2.45 – 2.56)·10 -3 eV 2 sin 2 2 > 0.98 tan 2 = 0.76 – 1.31 Present: m 2 = (1.7 – 3.5)·10 -3 eV 2 sin 2 2 > 0.9 tan 2 = 0.53 – 2.04 m 2 = (-2.63 – -2.49)·10 -3 eV 2 Input: 13 = 0º 23 = 45º = 0º 90% CL

T2K-1 errors revised m 2 = (2.42 – 2.61)·10 -3 eV 2 sin 2 2 = 0.94 – 0.99 tan 2 = 0.62 – 0.85, 1.21 – 1.66 m 2 = (2.44 – 2.58)·10 -3 eV 2 sin 2 2 = 0.95 – 0.99 tan 2 = 0.63 – 0.81, 1.24 – 1.58 Present: m 2 = (1.7 – 3.5)·10 -3 eV 2 sin 2 2 > 0.9 tan 2 = 0.53 – 2.04 m 2 = (-2.64 – -2.47)·10 -3 eV 2 90% CL Input: 13 = 0º 23 = 40º = 0º

Conclusions The measurement of 13 and will rely heavily on an improvement of the measure of 23 and m 2 23 Precision measurements of 23 and m 2 23 need energy resolution and events above and below the oscillation peak The errors on the 23 and m 2 23 are somewhat larger due to the dependence of the disappearance signal on 13, and the mass hierarchy This dependence can be exploited combined with the appearance channel to solve degeneracies

T2K-2 T2K-1 5% systematic error

T2K-2 T2K-1 2% systematic error

No background and no systematic Systematic 0% No Background Systematic 5% With Background Errors dominated by statistics

10% systematic Systematic 10%Systematic 5% Errors dominated by statistics

Energy Resolution Red histogram for true QE events Figure taken from Y. Itow et al. hep-ex/

Double energy resolution 4 bins of 200MeV 8 bins of 100MeV

Double energy resolution 4 bins of 200MeV 8 bins of 100MeV