1. Alain Blondel Neutrino Factory scenarios I will endeavour to address some principle design issues related to the physics use of high intensity muon beams. Some ideas will be given. 1. intense muon beams in neutrino factory complex. (CERN model) 2. neutrino factory parameters.

2. Alain Blondel High Intensity Low Energy muon beams 1. Thin inner target in proton accumulator advantages: very efficient use of proton beam, point source difficulties: - can target take the heat? - creates high-radiation area inside ring 20 - 120 PSI already has 1 MW DC beam of 590 MeV protons with 5% I target for muons. How can one do 1000 times better? 2. Or Use full DC SPL 24 MW with thin muon target DC beams (   e ,   eee ) 20 + solenoid collection (1/.16) 2 =40 + better experiments ? Pulsed beam (     e -  1. Use proton beam from buncher 2. Use muons at the end of cooling channel! --> need now conceptual design of target station and muon beams

3. Alain Blondel Thoughts for muon targets in neutrino factory complex    1. Use SPL pulsed beam (3ms at 50 Hz) and thin transmission target    2. Use beam stored in accumulator and inner target    2. Use cooled muon beam ?    1. Use bunched proton beam (train of 2.3  s, 12 bunches of 10 ns each at 40 MHz)

4. Alain Blondel

5. From neutrino factory to Higgs collider        h (115) Upgrade to 57.5 GeV Separate   &  , add transfer lines More cooling +  E /E reduction Muon collider: a small…. but dfficult ring

6. Alain Blondel Neutrino Factory parameters Energy Intensity ring geometry running mu+ and mu- at the same time

7. Alain Blondel Energy The sensitivity to  13 is a steep function of energy: -- signal event rate increases like E -- background in wrong sign muons decreases fast with E The sensitivity to CP violation is not, energies >30 GeV do not contribute significantly to the sensitivity *-> The e   channel is essential threshold at 3.5 GeV and full efficiency at 15 GeV The matter resonance is essential (12 GeV) => E  >~ 20 GeV…

8. Alain Blondel Detector zIron calorimeter zMagnetized yCharge discrimination yB = 1 T zR = 10 m, L = 20 m zFiducial mass = 40 kT Baseline 3500 Km 732 Km 3.5 x 10 7 1.2 x 10 6 5.9 x 10 7 2.4 x 10 6 1.1 x 10 5 1.0 x 10 5  CC e CC  signal (sin 2  13 =0.01) Events for 1 year Also: L Arg detector: magnetized ICARUS Wrong sign muons, electrons, taus and NC evts *-> CF e signal at J-PARC =40

9. Alain Blondel CP violation Matter effect must be subtracted. spectrum of matter effect and CP violation is different  It is important to subtract in bins of measured energy.  knowledge of spectrum is essential here! 5-10 GeV 10-20 GeV 20-30 GeV 30-40 GeV 40-50 GeV 40 kton L M D 50 GeV nufact 5 yrs 2.10 20  /yr In fact, 20-30 GeV is enough! Best distance is 2500-3500 km

10. Alain Blondel Intensity of muons in Neutrino factory storage ring As high as possible i.e. >10 14 per second

11. Alain Blondel muon polarization: triangle or bow-tie? For a race track or triangle decay ring, in which polarization precesses it is possible to calibrate beam energy and energy spread with g-2 precession measurement A bow-tie has been suggested to avoid this spin precession and depolarization (net bend is zero, so muon polarization does not precess either) This has several inconvenients: -- P is different for the three straights in triangle (who shall be pleased?) -- P cannot be reversed -- E and  E) can no longer be measured -- in order to know the flux to 0.1% on must know P to 0.1% and this is hard! end of the bow tie.

12. Alain Blondel CERN baseline scenario

13. Alain Blondel -- Neutrino Factory -- CERN layout    e + e   _ interacts giving   oscillates e     interacts giving    WRONG SIGN MUON 10 16 p/ s 1.2 10 14  s =1.2 10 21  yr 3 10 20 e  yr 3 10 20   yr 0.9 10 21  yr

14. Alain Blondel Angular divergence If the muons have transverse momentum comparable to that of muon decay (50 MeV) the neutrino beam will be seriously degraded this corresponds to  = 0.5 m  /E 

15. Alain Blondel

16. Divergence of 0.1/  keeps 95% of the original flux. Straight section with this property were designed for the US Study II and by Keil Divergence MUST be measured. A gas Cerenkov device to measure the beam emittance was devised by Piteira. Various efects were considered (optical aberrations, heating of gas, multiple scattering, etc…and concluded that the divergence is easier to measure the bigger it is so that this should not be a problem)

17. Alain Blondel Main parameters to MONITOR 1. Total number of muons circulating in the ring, BCT, near detector for purely leptonic processes 2. muon beam polarisation, polarimeter 3. muon beam energy and energy spread, race-track or triangle. NO BOW-TIE! +polarimeter 4. muon beam angle and angular divergence. straight section design +beam divergence monitors e.g. Cerenkov 5. Theory of  decay, including radiative effects OK Yes, we believe that the neutrino flux can be monitored to 10 -3 IF + design of accelerator foresees sufficient diagnostics. + quite a lot of work to do to design and simulate these diagnostics. Conclusions I

18. Alain Blondel Injecting two polarities RF phase rotation and linear cooling channel work for both signs of muons (and rings also with adequate separate injection/ejection lines) this opens the possibility to double the flux by running mu+ and mu- at the same time, but the detector must be able to identify them by timing. The detectors can presumably identify by timing the source of events at the level of 100ns (number to be refined) i.e. 30 m. mu+ and mu- microbunches from 200 MHz RF (2.5 ns) cannot be separated. This requires some design gymnastics for the injection in the storage ring and its geometry.

19. Alain Blondel     l l-l- l+l+ ex: race track geometry: constraint: ¦l - - l + ¦ > l +  where  is the precision of the experiments time tag Muons of both signs circulate in opposite directions in the same ring. The two straight sections point to the same far detector(s). OK There is one inconvenient with this: the fact that there are two decay lines implies two near detectors. In addition this does not work for the triangle. this can be solved by dog bone or two rings with one or more common straights 

20. Alain Blondel     l l-l- l+l+ L this requires more arcs and possibly more tunnel I am sure part of this can be solved (rings could be on top of each other) 's

21. Alain Blondel