1. Low-Energy Ionization “Cooling” David Neuffer Fermilab

2. 2 Outline  Low-energy cooling-protons  ions and “beta-beams”  Low-energy cooling – muons  emittance exchange

3. 3 Combining Cooling and Heating:  Low-Z absorbers (H 2, Li, Be, …) to reduce multiple scattering  High Gradient RF  To cool before  -decay (2.2   s)  To keep beam bunched  Strong-Focusing at absorbers  To keep multiple scattering  less than beam divergence …  Quad focusing ?  Li lens focusing ?  Solenoid focusing?

4. 4 Low-energy “cooling” of ions  Ionization cooling of protons/ ions has been unattractive because nuclear reaction rate is competitive with energy-loss cooling rate  And other cooling methods are available  But can have some value if the goal is beam storage to obtain nuclear reactions  Absorber is also nuclear interaction medium  Y. Mori – neutron beam source –NIM paper  C. Rubbia, Ferrari, Kadi, Vlachoudis – source of ions for β-beams

5. 5 Cooling/Heating equations  Cooling equations are same as used for muons  mass = a m p, charge = z e  Some formulae may be inaccurate for small β=v/c  Add heating through nuclear interactions  Ionization/recombination should be included  For small β, longitudinal dE/dx heating is large  At β =0.1, g L = -1.64, ∑ g = 0.36  Coupling only with x cannot obtain damping in both x and z 0.01.03.05.0 

6. 6 μ Cooling Regimes  Efficient cooling requires:  Frictional Cooling (<1MeV/c) Σ g =~3  Ionization Cooling (~0.3GeV/c) Σ g =~2  Radiative Cooling (>1TeV/c) Σ g =~4  Low-ε t cooling Σ g =~2β 2 (longitudinal heating)

7. 7 Miscellaneous Cooling equations For small β: Better for larger mass?

8. 8 Example  ERIT-P-storage ring to obtain directed neutron beam (Mori-Okabe, FFAG05)  10 MeV protons  9 Be target for neutrons  σ ≈ 0.5 barns  β = v/c =0.145  Large δE heating  Baseline Absorber  5µ Be absorber  δE p =~36 keV/turn  Design Intensity  1000Hz, 6.5×10 10 p/cycle  100W primary beam  < 1.5 kW on foil –0.4 kW at n turns =1000

9. 9 ERIT results  With only production reaction, lifetime is 30000 turns  With baseline parameters, cannot cool both x and E  Optimal x-E exchange increases storage time from 1000 to 3000 turns (3850 turns = 1ms)  With x-y-E coupling, can cool 3-D with g i = 0.12  Cooling time would be ~5000 turns  With β  =0.2m,  δE rms = 0.4MeV,  ,N = 0.0004m (x rms =2.3cm)  x rms = 7.3cm at β  =2m (would need r =20cm arc apertures) (but 1ms refill time would make this unnecessary)

10. 10 3-D mixing lattice  For 3-D cooling, need to mix with both x and y  Possible is “Moebius” lattice (R. Talman)  Single turn includes x-y exchange transport –solenoid(s) or skew quads –in zero dispersion region for simplicity –Solenoid: BL = π Bρ –For 10 MeV p : BL = 1.44T-m  Complete period is 2 turns

11. 11 Emittance exchange parameters  with g L,0 = -1.63, η = 0.5m,  need G = 3.5 to get g L =0.12 –(L W =0.3m)   Beam Center Geometry of wedge absorber L W =1/G

12. 12 “Wedge” for thin foil  Obtain variable thickness by bent foil (Mori et al.)  Choose x 0 =0.15m, L W =0.3, δ o =5µ  then a=0.15, δ R =2.5  for g L,0 = -1.63, η = 0.5m  Barely Compatible with β * = 0.2m  Beam energy loss not too large?  <1kW Power on foil a=0.15 Beam

13. 13 Other heating terms:  Mixing of transverse heating with longitudinal could be larger effect: (Wang & Kim) At ERIT parameters: β x = 1.0m, β z = 16m, η=0.6m, Be absorber, dδ 2 /ds= 0.00032, d θ 2 /ds= 0.0133 only 5%, 1.5% changes … At β x = 0.2m, ~25% change …

14. 14 Beta beams for neutrinos (Zuchelli) ν-sources v-sources  Number of possible sources is limited (v sources easier)  Want:  lifetime ~1s  large ν-energy  ν and ν*  easily extracted atoms  Noble gases easier to extract  6 He 2 “easiest” E ν* =1.94MeV  18 Ne 10 for ν E ν =1.52MeV  Noble gas  ( 8 B, 8 Li) have E ν,ν * =~7MeV  Want 10 20 ν and ν* / year…

15. 15 β-beam Scenario Neutrino Source Decay Ring Ion production: Target - Ion Conversion Ring Ion Driver 25MeV Li ? Decay ring B  = 400 Tm B = 5 T C = 3300 m L ss = 1100 m 8 B:  = 80 8 Li:  = 50 Acceleration to medium energy RCS 8z GeV/c Acceleration to final energy Fermilab Main Injector Experiment Ion acceleration Linac Beam preparation ECR pulsed Ion productionAccelerationNeutrino source

16. 16 β-beam Scenario (Rubbia et al.)  Produce Li and inject at 25 MeV  Charge exchange injection  nuclear interaction at gas jet target produces 8 Li or 8 B  Multiturn with cooling maximizes ion production  8 Li or 8 B is caught on stopper(W)  heated to reemit as gas  8 Li or 8 B gas is ion source for β- beam accelerate  Accelerate to Bρ = 400 T-m  Fermilab main injector  Stack in storage ring for:  8 B  8 Be + e + +ν or 8 Li  8 Be + e - + ν* neutrino source

17. 17 Cooling for β -beams (Rubbia et al.)  β-beam requires ions with appropriate nuclear decay  8 B  8 Be + e + + ν  8 Li  8 Be + e - + ν*  Ions are produced by nuclear interactions  6 Li + 3 He  8 B + n  7 Li + 2 H  8 Li + 1 H  Secondary ions must be collected and reaccelerated  Either heavy or light ion could be beam or target  Ref. 1 prefers heavy ion beam – ions are produced more forward (“reverse kinematics”)  Parameters can be chosen such that target “cools” beam  (losses and heating from nuclear interactions, however…)

18. 18 β-beams example: 6 Li + 3 He  8 B + n  Beam: 25MeV 6 Li +++  P Li =529.9 MeV/c Bρ = 0.59 T-m; v/c=0.09415  Absorber: 3 He  Z=2, A=3, I=31eV, z=3, a=6  dE/ds = 110.6 MeV/cm, L R =756cm –at (ρ He-3 = 0.09375 gm/cm 3 ) Liquid, (ρ He-3 = 0.134∙10 -3 P gm/cm3/atm in gas)  dE/ds= 1180 MeV/gm/cm 2, L R = 70.9 gm/cm 2  If g x = 0.123 (Σ g = 0.37), β ┴ =0.3m at absorber  ε N,eq = ~ 0.000046 m-rad  σ x,rms = 1.2 cm at β ┴ =0.3m,  σ x,rms = 3.14 cm at β ┴ =2.0m  σ E,eq is ~ 0.4 MeV  ln[ ]=5.68,  energy straggling underestimated?

19. 19 Cooling time/power : 6 Li + 3 He  8 B + n  Nuclear cross section for beam loss is 1 barn (10 -24 cm 2 ) or more  σ = 10 -24 cm 2, corresponds to ~5gm/cm 2 of 3 He  ~10 3-D cooling e-foldings …  Cross-section for 8 B production is ~10 mbarn  At best, 10 -2 of 6 Li is converted  Goal is 10 13 /s of 8 B production  then at least 10 15 Li 6 /s needed  Space charge limit is ~10 12 6 Li/ring  Cycle time is <10 -3 s  If C=10m, τ =355 ns, 2820 turns/ms  5/2820=1.773*10 -3 gm/cm 2 (0.019 cm @ liquid density …)  2.1 MeV/turn energy loss and regain required …(0.7MV rf)  0.944 MW cooling rf power…

20. 20 β-beams example: 7 Li + 2 H  8 Li + 1 H  Beam: 25MeV 7 Li +++  P Li =572 MeV/c Bρ = 0.66 T-m; v/c=0.0872  Absorber: 2 H  Z=1, A=2, I=19eV, z=3, a=7  dE/ds = 184.7 MeV/cm, L R =724cm –at (ρ H-2 = 0.169 gm/cm 3 ) Liquid, (ρ H-2 = 0.179∙10 -3 P gm/cm3/atm in gas)  dE/ds= 1090 MeV/gm/cm 2, L R = 70.9 gm/cm 2  If g x = 0.123 (Σ g = 0.37), β ┴ =0.3m at absorber  ε N,eq = ~ 0.000027 m-rad  σ x,rms = 1.0 cm at β ┴ =0.3m,  σ x,rms = 2.5 cm at β ┴ =2.0m  σ E,eq is ~ 0.36 MeV  ln[ ]=6.0,  Easier than 8 B example ?

21. 21 Cooling time/power- 7 Li + 2 H  8 Li + 1 H  Nuclear cross section for beam loss is 1 barn (10 -24 cm 2 ) or more  σ = 10 -24 cm 2, corresponds to ~3.3gm/cm 2 of 2 H  ~9 3-D cooling e-foldings …  Cross-section for 8 Li production is ~100 mbarn  10 -1 of 7 Li is converted ?? 10 × better than 8 B neutrinos  Goal is 10 13 /s of 8 Li production  then at least 10 14 Li 7 /s needed  Space charge limit is ~10 12 7 Li/ring  Cycle time can be up to 10 -2 s, but use 10 -3 s  If C=10m, τ =355 ns, 2820 turns  3.3/2820=1.2*10 -3 gm/cm 2 (0.007 cm @ liquid density …)  1.3 MeV/turn energy loss and regain required …(0.43MV rf)  0.06 MW cooling rf power…

22. 22 Space charge limitation  At N = 10 12, B F =0.2, β = 0.094, z=3, a=6, ε N,rms = 0.000046: δν ≈ 0.2  tolerable ??  Space charge sets limit on number of particles in beam

23. 23 Inverse vs. Direct Kinematics  6 Li 3 beam + 3 He 2 target  25 MeV Li  gas-jet target  tapered nozzle for wedge?  product is ~ 8 B  8 to 23 MeV  Stop in Tantalum absorber  penetrates more (5 to 12μ)  <14.5°  Beam divergence ~6° (2.5σ)  Boil out to vapor  neutralize to vapor, extract, ionize, accelerate  100% efficiency …  3 He 2 beam + 6 Li 3 target  12.5 MeV He beam  Li foil target (or liquid waterfall?)  fold to get wedge …  product is ~ 8 B  ~1 to 8 MeV  Stop in Tantalum absorber  penetrates less (2 to 5μ)  <30°  Beam divergence is ~10° (2.5σ)  Boil out to vapor

24. 24 X-sections, kinematics

25. 25 Rubbia et al. contains errors  Longitudinal emittance growth 2× larger than his  synchrotron oscillations reduce energy spread growth rate but not emittance growth rate  Emittance exchange with only x does not give 3-D cooling – need x-y coupling and balancing of cooling rates  Increases equilibrium emittance, beam size  increase needed for space charge, however  Includes z 2 factor in dE/dx but not in multiple scattering, …

26. 26 Low-Energy “cooling”-emittance exchange  dP μ /ds varies as ~1/β 3  “Cooling” distance becomes very short: for liquid H at P μ = 10MeV/c  Focusing can get quite strong:  Solenoid:  β  =0.002m at 30T, 10MeV/c  ε N,eq = 1.5×10 -4 cm at 10MeV/c  Small enough for “low-emittance” collider PµPµ L cool 200 MeV/c 10 0.1 cm 100 cm

27. 27 Emittance exchange: solenoid focusing  Solenoid focusing(30T)    0.002m  Momentum (30→10 MeV/c)  L ≈ 5cm  R < 1cm  Liquid Hydrogen (or gas) PµPµ L cool 200 MeV/c 10 0.1cm Use gas H 2 if cooling length too short  ε N,eq = 1.0×10 -4 cm at 10MeV/c, 50T -Will need rf to change  p to  z

28. 28 Gas absorbers  With gas absorbers can adjust energy loss rate  avoid beam loss

29. 29     Collider Parameters

30. 30 Summary

31. 31 β-beams alternate: 6 Li + 3 He  8 B + n  Beam: 12.5MeV 3 He ++  P Li =264 MeV/c Bρ = 0.44 T-m; v/c=0.094  Absorber: 6 Li  Z=3, A=6, I=43eV, z=2, a=3  dE/ds = 170 MeV/cm, L R =155cm –at (ρ Li-6 = 0.46 gm/cm 3 )  Space charge 2  smaller  If g x = 0.123 (Σ g = 0.37), β ┴ =0.3m at absorber  ε N,eq = ~ 0.000133m-rad  σ x,rms = 2.0 cm at β ┴ =0.3m,  σ x,rms = 5.3 cm at β ┴ =2.0m  σ E,eq is ~ 0.3 MeV  ln[ ]=5.34