New Ideas for 6 D Ionization Cooling R B Palmer Oxford, UK 5/11/05 • Muon Collider requirements • Transverse Ionization Cooling Theory • Longitudinal Emittance Cooling Theory • Final Reverse Emittance Exchange • New Ideas on How to do it • Conclusion • More New ideas if we have time 1
Why a Muon Collider 40 TeV SSC FNAL pp (2 TeV) 14 TeV LHC pp (1.5 TeV) SC ILC ee (.5-.8 TeV) MuMu (3 TeV) BNL 10 km 2
3 TeV Collider requirements from Snowmass 98 Average bending field T 5.2 Assume 10 33 cm − 2 70 Luminosity E cm N µ N b × f P µ β ⊥ = σ z dp/p emit ⊥ ∆ ν ǫ 6 10 − 12 m TeV 10 12 Hz MW mm % mm 3 2 2 × 15 28 3 0.16 .05 .044 170 dp = 1 . 5 10 4 0 . 003 0 . 16 = 7 . 2 10 − 2 ǫ � = β v γ σ z m p 100 ǫ 6 = ǫ � ( ǫ ⊥ ) 2 = 7 . 2 10 − 2 × (50 10 − 6 ) 2 = 180 10 − 12 m 3 Initial ǫ 6 ≈ 2 ( . 02) 2 ≈ 10 − 3 m • Cooling Required ≈ 1/5000,000 • Final Longitudinal Emmittance = 72,000 (pi mm mrad) • Final Transvers Emittance = 50 (pi mm mrad) 3
What is Emittance ? normalized emittance = Phase Space Area π m c If x and p x are both Gaussian and uncorrelated, then the area is that of an upright ellipse, and: ǫ ⊥ = π σ p ⊥ σ x = ( γβ v ) σ θ σ x ( π m rad ) π mc π σ p � σ z = ( γβ v ) σ p ǫ � = p σ z ( π m rad ) π mc ǫ 6 = ǫ 2 ( π m ) 3 ⊥ ǫ � Note that the π , added to the dimension, is a reminder that the emittance is phase space/ π 4
x’ What is Beta ⊥ ( Twiss ) of Beam x Upright phase ellipse in x ′ vs x , width = σ x β ⊥ = height σ θ low σ x and large σ θ low β Strong focus → → � 1 � � � σ x = � ǫ ⊥ β ⊥ � � � β v γ � ǫ ⊥ 1 � � σ θ = � � � � β ⊥ β v γ � 5
Transverse Cooling p � less p � restored ✘ ✿ ✘✘✘ p ⊥ less p ⊥ still less ✟ ✯ ✟✟✟✟✟✟✟✟✟✟✟✟✟✟ ✟ ✟✟✟ Material Acceleration Rate of Cooling without scattering dǫ dp = J x,y ǫ x,y p For the moment the ”partition functions” J x,y = 1 Explanation later 6
Minimum (Equilibrium) Emittance β ⊥ ǫ x,y ( min ) = C ( mat, E ) β v J x,y 1 J x,y = 1 C ( mat, E ) ∝ L R dγ/ds At minimum of dE/dx ( ≈ 300 MeV/c) material density dE/dx L R C o A o kg/m 3 MeV/m m % % Liquid H 2 71 28.7 8.65 0.38 1.36 Li 530 87.5 1.55 0.69 1.31 Be 1850 295 0.353 0.89 1.28 C 2260 394 0.47 1.58 1.25 Al 2700 436 0.089 2.48 1.23 • Hydrogen much the best material • Coefficient A o is for longitudinal cooling - explanation to come 7
An Aside: Beam Divergence Angles β ⊥ ǫ x,y ( min ) = C ( mat, E ) β v J x,y ǫ ⊥ � � � σ θ = � � � β β v γ � so for a beam in equilibrium � C ( mat, E ) � � � σ θ = independent of emittance � � � β 2 v γ � � for 75 % of maximum cooling rate, an aperture at 3 σ , and β 2 v γ = 2 the required angular acceptance A of the system must be � √ � C ( mat, E ) � � A = 3 4 � � � β 2 v γ � � Material H2 Li Be C Al Ang Acceptance (RaD) 0.25 .35 .4 .54 .66 These are very large angular acceptances ! 8
How to get low beta (strong focus) ? • Strong Solenoid – Practical limit is 10 T – Expensive 9
• Lithium Lens – For uniform i then perfect lens I ∝ A ∝ r 2 Bending ∝ B ∝ I/r ∝ r – Maximum current limited by breaking containment tube – Pressure ∝ Surface Field – Current lenses get up to near 10 T 10
Compare Solenoids and Li Lenses min emittance (pi mm mrad) 2000 Practical 1000 500 200 Solenoid 100 50 Required Li Lens 20 2 3 4 10 20 30 Mag Field (T) – Even a 20 T Solenoid will not get required emittance – Existing Li Lenses (10T) will not reach it – 30 T Li Lens ok, but not developed and probably impossible from cavitation 11
• At Multiple foci e.g. Mice cells • Beta of order 1/3 average beta for moderate B (3-6 T) • Harder as B rises because of coil thickness • Hard to get emittances < 400 pi mm mrad 12
Longitudinal Cooling ? • At mom ≪ 200 MeV/c dp/p is increased (heating) • At mom ≫ 200 MeV/c dp/p is weakly reduced (cooling) • We Use ≈ 200 MeV/c negligible heating or cooling 4 relative(dE/dx) 3 Partition function J z : dǫ z dp 2 = J z ǫ z p J z ≈ 0 1 10.0 10 2 10 3 6 dimensional emittance change: Muon Energy (MeV) dǫ 6 dp = J 6 ǫ 6 p where J 6 = J x + J y + J z ≈ 2 13
Emittance Exchange High dp/p ✲ ✲ Low ǫ n ✛ ✂✂ ❇ ❇ Low dp/p ✂ ❇ ✂ ❇ ✂ ❇ High ǫ n ✂ ❇ ✂ ❇ ✂ ❇ ✛ ✂ ❇ Material Magnet • dp/p (and Longitudinal emittance) reduced • But σ y (and transverse emittance) increased • Transverse cooling from mean loss in absorber • ”Emittance Exchange” + Transverse Cooling = 6 D cooling J x = ( J x ) o + ∆ J x J y = ( J y ) o + ∆ J y J z = ( J z ) o + ∆ J z ∆ J x + ∆ J y + ∆ J z + = 0 14
e.g. If cooling only by wedges Rate of Cooling without straggling y Wedge ∆ J z (wedge) = = D h Beam where D = dy/ ( dp/p ) is the Dispersion h s given a finite J z we get a minimum (equilibrium) dp/p: � σ p γ 1 − β 2 1 � � v = A o � � � β 2 p 2 J z � v The values of A o were given in the above table For Hydrogen, A o ≈ 1.36 %, but it is almost the same for other materials 15
For J z = 2 / 3 minimum (equilibrium) dp/p 15.0 (%) 12.5 10.0 σ p /p 7.5 5.0 2.5 0.0 .2 5 2 5 0.1 1.0 10.0 mom (GeV/c) Minimum at 2.5% around 200 MeV/c ǫ z = σ p × β v γ σ z p σ z , the bunch length depends on the RF gradient and frequency less at higher frequency 16
e.g. 6 D cooling in ”RFOFO” Ring with Wedges • Lattice similar to MICE • Bending gives dispersion • Wedge absorbers: Cooling also in longitudinal • Many turns in Ring gives more cooling at lower cost Injection/Extraction Vertical Kicker Alternating 3T Solenoids Tilted for Bending B y 201 MHz rf 12 MV/m Hydrogen Absorbers 33 m Circ • Could be converted to Helicoil • No Injection/extraction • Better performance by tapering • But more expensive 17
performance emit 1 /emit 2 × transmission = 188 10 2 n/n o (%) ǫ ⊥ ( π mm ) ǫ � ( π mm ) ǫ 6 ( π cm 3 ) n/n o 0.36 10.0 ǫ � 41.1 to 2.4 1/17 ǫ ⊥ 12.1 to 2.17 1/5.6 1.0 0.1 ǫ 6 6.4 to 0.012 1/533 10 − 2 0 5 10 15 20 turns Final Long Emittance 2400 (pi mm mrad) Second ring at 400 MHz → ≈ 1400 (pi mm mrad) c.f. 7200 (pi mm mrad) Req for Collider 18
Conclusion as of 2 years ago • Longitudinal emittance can be Achieved • Transverse emittance not Achieved – ≈ 800 pi mm mrad Possible with 10 T Solenoid – ≈ 400 pi mm mrad Possible with Lattice – ≈ 200 pi mm mrad Possible with Lithium lend – Required = 50 pi mm mrad • Final Reverse Emittance Exchange Proposed with wedges But is found to be hard in practice See below 19
New idea Li Jet ? • No containing tube to break • Use Magnetic field to stabilize (and form ?) jet • Jet larger at nozzle to avoid damage • Ends in indestructible pool Is this crazy ? 20
Old solution: Reverse Emittance Exchange at End • Assume 10 T Li Lens for transverse emittance • Assume RFOFO Ring for Longitudinal emittance • This is not quite fair, but reasonable Required Achievable Achievable/Req 50 10 − 6 200 10 − 6 Transverse 4.0 70 10 − 3 1.5 10 − 3 Longitudinal 1/50 180 10 − 12 60 10 − 12 6 D 1/3 • Required 6D emittance seems achievable • Longitudinal emittance even too small ! • But Transverse emitance too Large Suggests Final Reverse emittance Exchange 1. Wedges with wrong Dispersion Old Method 2. By use of septa (potato slicer) New idea 3. Very Low energy in Li Lens New idea 21
1) Wedges with wrong Dispersion (Old Idea) High dp/p ✛ OUT ✛ Low ǫ n ✲ ❇ ✂ ❇ ✂ Low dp/p ❇ ✂ IN ❇ ✂ High ǫ n ❇ ✂ ❇ ✂ ❇ ✂ ✲ ❇❇ ✂ ✂ Material Magnet Require 4 times smaller equilib transverse emittance thus J x,y = ( J x,y ) o × 4 = 1 × 4 = 4 and J z = 2 − 2 × J x,y = 2 − 8 = − 6 • Required transverse emittance achieved, but • Required longitudinal emittance lost 22
2) Potato Slicer (New idea) • This can be done at any momentum • Gaussian shapes of beams, and septa, lead to dilution • Realization may be hard Needs study, but must work at some level 23
3) Li Lens at very low Energy Remember: β ⊥ ǫ x,y ( min ) = C ( mat, E ) β v J x,y 1 J x,y = 1 C ( mat, E ) ∝ L R dγ/ds 4 relative(dE/dx) 3 • dE/dx × 4 at 10 MeV 2 • C(mat,E) = 1/4 10 MeV • Equilib. emittance × 1/4 1 = 50 (pi mm mrad) 10.0 10 2 10 3 • Now meets trans. requirement Muon Energy (MeV) 24
Effect on Longitudinal emittance Partition Functions J 3 J 6 2 J x , J y 1 J z 0 -1 -2 10 100 1000 Muon Energy (MeV) • Long. Emittance will rise from J z = − 1 • But J 6 remains positive • So 6D emittance should not rise • Effectively: Reverse Emittance Exchange Looks good, but needs study 25
Schematic of Collider 26
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