Fourth International Accelerator School for Linear Colliders Beijing, China 7-18 September 2009 Lecture A3 Beam Delivery System and Beam-Beam Effects Part 2 Olivier Napoly CEA-Saclay, France Preliminary Version 17 September 2009 Collider Parameters, Physical Constants and Notations E , energy and p , momentum c = 299 792 458 m/s Bρ , magnetic rigidity (Bρ = p/e) e = 1.602 177 33 10-19 C 2 L , luminosity mec = 510 999 eV = 4π 10-7 N A-2 permeability , bunch charge µ0 Q 2 ε0 = 1/µ0c permittivity N , number of particles in the bunch (N = Q/e) -15 re = 2.818 10 m e 2 nb , number of bunches in the train 2 with e cm = 4πε 0 re frep , pulse repetition rate εx and εy , horizontal and vertical emittances * * σx and σy , rms horizontal and vertical beam sizes at the IP σz , bunch length α,β,γ , Twiss parameters μ , tune ψ , phase advance, Beam Delivery System 2 Beam Delivery System The Beam delivery system is the final part of the linear collider which transports the high energy beam from the high energy linac to the collision point (Interaction Point = IP). ILC BDS Most Important Functions: Final Focus: Focus the beams at the interaction point to achieve very small beam sizes. Collimation: Remove any large amplitude particles (beam halo). Tuning: Ensure that the small beams collide optimally at the IP. Matching: Precise beam emittance measurement and coupling correction. Diagnostics: Measure the key physics parameters such as energy and polarization. Extraction: Safely extract the beams after collision to the high-power beam dumps. Beam Delivery System 3 ILC Beam Delivery System Layout (RDR) Linac Exit Beam Diagnostics Switch Yard Sacrificial β-collimator collimators E-collimator Final Focus Tune-up & Emergency extraction Tune-up 14mr IR dump Muon wall grid: 100m*1m Main dump Extraction Beam Delivery System 4 Contents: Four (+ Two) Outstanding Questions ? Part 1 Q1: do we need a Final Focus System ? Luminosity, Emittance, Q2: do we need High Field Quadrupole Magnets ? Quadrupole Magnets, Multipoles, Superconducting Quadrupoles Part 2 Q3: do we need Flat Beams ? Beam-beam forces, Beamstrahlung, e+e- Pairs, Fast Feedback Q4: do we need Corrections Systems ? Beam Optics, Achromat, Emittance Growth Beam Delivery System 5 Contents: Fundamentals in Beam Physics Part 1 Luminosity Emittance Magnetism (continued) Part 2 Beam-Beam Effects Beam Optics Synchrotron Radiation Beam Delivery System 6 Magnetic Measurements Magnet fields are measured to checks: 1. the field integral of the main component (Bdl for a dipole, Gdl for a quadrupole) 2. , or quality)the field harmonics content (~ field homogeneity These measurement are usually performed by a magnetic probe located at a Reference Radius Rref the magnet bore.along Hence, magnet experts expansion:use the following n-1 + ∞ ⎛ z ⎞ ⎜ ⎟ (B ) z = ∑ () n b − i n a⎜ ⎟ zfor< Ri n=1 ⎝ Rref ⎠ with a +)(sometimes (n- 1 ) B ( 0 ) n−1 so that b− i a = R forn ≥ 1 n n (n − 1 )! ref Beam Delivery System 7 Quadrupole Magnets: the Engineer Standpoint Given a Central Axis Oz , we rajectoryTwith the Reference which coincides introduce an “Engineering Symmetry” RS(θ ) which allows categorize the Magnets into “Engineering Multipoles”. y (θ ) Notations: R R(θ ) is the rotation with angle θ around Oz . zˆ θ • S is the symmetry with respect to the Oxy plane. O x RRSSR()()S θ ≡ θ o= o ()θ of the two.is the combination y y Note that : B 1) For a transverse B field (pseudo-vector) r r ⋅SBB = − zˆ zˆ B • • electrical current 2) For a longitudinal J O O r r ⊗ J x • J x ⋅S JJ = − 3) S RRR( )S θ o (S θ )= (θ 2 ) Beam Delivery System 8 Symmetric Current Distributions y electrical current a distribution of longitudinal Consider ,J(rθ) . ,J(rθ) is invariant with respect to If a current distribution RS zˆ • ()(,)(,)RS θ0J⋅ rθ = Jθ r O x then: y 1) the integrated current through the plane is zero, since ,J(rθ) (,)()(,)J rθ +R Jθ ⋅ rθ (,)(,) J rθ − Jθ + rθ J(,) rθ = S 0 = 0 zˆ + 2 2 • O x 2) 2 of is a sub-multiple 2 , since − θ0 π (,)J () rθ RR= S θ ()(,)0 o S θ0 ⋅J rθ = R (2)(,)θ0J⋅ rθ = Jθ r− θ0 (, 2) 2π ⇒ either orθ = 2 (J , ) rθ ≡ ( J ) r zero function constant θ of . 0 n Beam Delivery System 9 Multipolar Current Distributions e introduce the following categories of current distributions WJ(r,θ) : Monopole (n=0): J ( 0 () ) r distributionsindependent , azimuthally Dipole ( =1), such that: ( 1 ) ( 1 ) n ()(,)(,)RSπJ⋅ r θ = Jθ r Quadrupole (n=2): ( 2 ) ( 2 ) (RS 2π )J⋅ ( r ,θ =) Jθ ( r , ) Sextupole (n=3): ( 3 ) ( 3 ) (RS 3π )J⋅ ( r ,θ =) Jθ ( r , ) Octupole ( =4): ( 4 ) ( 4 ) n (RS 4π )J⋅ ( r ,θ =) Jθ ( r , ) Decapole ( =5): ( 5 ) ( 5 ) n (RS 5π )J⋅ ( r ,θ =) Jθ ( r , ) …. pole ( ): ()n ()n 2n- n ()(,)(,)RSnπ ⋅ J rθ = Jθ r ( 0 ) ( 0 ) : Note that , for the Monopole( R )S⋅θJ ( r ) = − J ( ) r θ ∀ Beam Delivery System 10 Decomposition of Current Distributions The space J over the sum can be decomposed current distributions of longitudinal of symmetric subspaces as follows. 1. Since any current , one can decompose RRR( )S π o (S π )= (π 2= ) 1 into symmetric (distribution n=1) and antisymmetric .t. r.components wRS(π): =JJ(R +( )S π ) ⋅JJ 2(R +S −π (J ) ⋅ ) 2 =JJJ(n= 1 +(R ) ( +π ) ) ⋅ 2 (n= 1 ) =JJ + π is invariantJπ under the rotationwhere R()π 2. ( R S 2 π Since )o R S ( π 2 = R ) theπ current ( ) distrib- , one can decompose utionJπ into symmetric (n=2) and antisymmetric .t. r.components wRS(π/2): =JJπ(R + π (S π 2 ) ⋅JJπ )(R + 2 π (S −π 2 )Jπ ) ⋅ 2 (n= 2 ) =JJ +(Rπ ( +π 2 )Jπ ⋅ ) 2 =JJ(n= 2 ) + etc, etc,… π 2 leadingJJJJJJ to= : (1)n= ⊕ (2)( n= ⊕( (4)n= n ⊕= ...))) (8)+ (n= 0) + octupole + quadrupole dipole + …. + monopole 16-pole + Beam Delivery System 11 Natural Harmonics Question: Where are the Sextupoles, and Decapoles, … ? Answer: Sextupoles (n=3) and Decapoles (n=5) are Dipole (n=20) Natural Harmonics: ( 3n= ) (n= 3 ) 3 (n= 3 ) ( 3n= ) (n= 3 ) R(S π⋅ 3JJ ) = ⇒ (SR π⋅ 3 )J =RS()π JJ ⋅ = ( 5n= ) (n= 5 ) 5 (n= 5 ) ( 5n= ) (n= 5 ) R(S π⋅ 5JJ ) = ⇒ S(R π⋅ 5 )J =RS()π JJ ⋅ = Dodecapoles (n=6) and 20-poles (n=10) are Quadrupole (n=21) Natural Harmonics: (π⋅ 6(JJ ) 6n= =) (n= 6 ) ⇒(3 π ⋅ 6 )J(n= 6 = ) (π 2(JJ ⋅ ) 6n= ) = (n= 6 ) RS SR RS (π ⋅ 10(JJ ) 10n= =) (n= 10 ⇒ ) (5 π 10⋅(J ) n= 10 = ) (π 2(JJ ⋅ ) 10n= ) = (n= 10 ) RS SR RS In general, 2n-poles with n =(2m+1)2p are 2.2p-pole (n=2p) Natural Harmonics: p R ()πn⋅()() Jn = J n ⇒ 2Rm+() 1 πn⋅() Jn =R(π 2⋅JJ )()()n = n S S S . In other words, using the example of Dipoles, the ensemble of Dipole (RS(π)- invariant) current distributions includes that of Sextupole (RS(π/3)-invariant) ones. As a consequence, Dipole magnets built from a Dipole invariant current distribution most likely contains Sextupole, Decapole, etc… components . Other harmonics are excluded, unless induced by non-invariant fabrication errors. Beam Delivery System 12 Quadrupoles Have Four Poles A Quadrupole magnet can be designed as follows: 1. Define a current distribution J(r,θ) over a 90° wide sector : θ ∈[θ0,θ 0+ π 2] nd 2. Rotate it by the symmetry RS(π/2) over the 2 sector θ ∈[θ0 + π ,2 θ0 + π ] y y J (1)(r,θ) J (2)(r,θ) J (1)(r,θ) − − − + + + RS(π/2) x x rd 3. Repeat the RS(π/2) operation to the 3 sector θ ∈[θ0 +,π θ0 + 3π ] 2 4. Repeat the R (π/2) operation to the 4th sector θ ∈[θ3+ π 2θ ,+ π ] 2 S y 0 0 J (2)(r,θ) J (1)(r, ) − θ − + + x + + (4) J (3)(r,θ) − − J (r,θ) Beam Delivery System 13 The Simplest Monopole Magnet current line transporting a The simplest Monopole magnet is given by longitudinal total current I ^z . r By rotation symmetry B : = B () θ ˆ r . 2π μ I sAmpère’From law : 0 . rdθ B()= r μ0 ⇒ I() B = r ∫0 2π r y B • I zˆ x μ I The complex magnetic fieldB= B+ iBis then given by B() z= 0 y x 2π z Nota Bene 1: The magnetic field inside of the conductor is an ideal focusing field ! Nota Bene 2: do not exist on earth ! Point-like Magnetic Monopoles Beam Delivery System 14 Magnet ‘Dipole’A The simplest Dipole Field configuration is given by a pair of linear conductors: y μ I μ I B(n= 1() ) z= 0 − 0 2 (πz− R0 )π 2z+ ( R0 ) μ0IR 0 −I +I = − • + 2 2 x π()z− R0 zˆ B R0 for :can be expanded This expressionz< R0 m μIR μ I μ I ∞ ⎛ z 2 ⎞ B(n= 1() ) z= 0 0 = − 0 = − 0 ⎜ ⎟ 2 2 2 2 ∑⎜ 2 ⎟ (πz− ) R0 Rπ 0 (− 1 z0 R ) π R0 m=0 ⎝ R0 ⎠ (to the mathematician expressions Comparing page 71) one gets : μ0 I μ0 I 0=m → : Dipoleb1= − ; 2=m →Decapole : b5= − 5 π R0 π R0 μ I μ I Sextupole:0 ; (2m+1)-pole : 0 1=m → b3= − 3 m → b2m+= 1 − 2m+ 1 π R0 π R0 Nota Bene: Magnet still has zero curvature radius ! This ‘Dipole’ Beam Delivery System 15 Quadrupole MagnetA of conductors:is given by a quadruplet Field configuration The simplest Quadrupole (n= 2 ) μ0 I μ0 I μ0 I μ0 I B() z= − iπ 2 + iπ − 3i π 2 2(πz )2(− R0 πz− R0 e )2(πz− R0 e πz )2(− R0 e ) m μ I ∞ ⎛ z ⎞ 0 (1)2−i m + π (1)22−i m + π −i m + (1)32π z< R = − ∑⎜ ⎟( 1−e+ e − e ) 0 2π R0 m=0 ⎝ R0 ⎠ m y μ I ∞ ⎛ z ⎞ − 0 m+1 m+1 m + 1 I = − ∑⎜( 1−⎟ ( − )i + ( − 1 )i − − ) 2π R0 m=0 ⎝ R0 ⎠ m +I +I ∞ ⎛ 4⎞ ⎧ ifm = k +4+ 1 • + μ0 I z x = − ∑⎜ ⎟ × ⎨ zˆ 2π R0 m=0 R0 ifm ≠ k +4 1 ⎝ ⎠ ⎩ R0 −I − (to the mathematician expressions Comparing page 71) one gets : 2μ0 I 2μ0 I 1=m → b Dodecapole:2= ; − 2 =m 5 → Quadrupole: b6= − 6 π R0 π R0 Beam Delivery System 16 Better Quadrupole MagnetA An improved Quadrupole Field configuration is given by an octuplet of conductors: y −−I −I − +I + + +I α R • 0 zˆ x + + +I The calculation proceeds like the preceding one +I leading to: −−I −I − m ∞ (n= 2 ) μ I 4⎛ z cos(⎞ ⎧ m 1+ )α m=if k + 4 1 B()= z − 0 ⎜ ⎟ × ⎨ ∑⎜ ⎟ z< R0 π R0 m=0 ⎝ R0 ⎠ ⎩0 ifm ≠ k +4 1 (to the mathematician expressions Comparing page 71) one gets : 2μ0 I 1=m → Quadrupole:b2= −cos(2 α ) ; π R0 2μ I π Dodecapole:0 cancels for α = 5=m → b6= − 6 cos(6α ) π R0 12 Beam Delivery System 17 Sector Quadrupole MagnetA is realised with 4 finite sectors of This quadrupole current density J .