Beam-beam effects in circular colliders with strong emphasis on measurements, tools and methods (derived from a CAS lecture) 1. Linear colliders (single pass, no damping ...) 2. Lepton colliders (multipass, damping) 3. Hadron colliders (multipass, no damping)
Reading material [1 ]L. Evans, "Beam-beam interactions", in Proceedings CERN Accelerator school: Antiprotons for Colliding-beam Facilities, CERN, 1983. [2 ]W. Herr, "Beam-beam Interactions", in Proceedings CERN Accelerator School: Advanced Accelarator Physics Course, Trondheim, 2013, CERN-2014-009. [3 ]W. Herr, "Mathematical and Numerical Methods for Nonlinear Dynamis", in CERN Accelerator School, Trondheim, 2013, CERN-2014-009. [4 ]Proceedings "ICFA beam-beam workshop", 18. - 22. March 2013, ed. W. Herr, CERN-2014-004.
Beams in collision beam-beam collision 6 4 2 0 -2 -4 -6 -10 -5 0 5 10 s Typically: 0.001% (or less) of particles have useful interactions 99.999% (or more) of particles are just perturbed Note: typical numbers for hadron collisions, leptons a lot worse
Some challenges (beam-beam related, incomplete): Circular colliders: ≥ 5 · 10 10 beam-beam interactions per 1. Beams are re-used - LHC: production run (fill) challenge for the beam dynamics (many different types of beam-beam effects to be understood and controlled) 2. Must guarantee stability and beam quality for a long time 3. Particle distributions change as result of interaction (results in time dependent forces ..) 4. One critical perfomance parameter: high luminosity ! 5. ... Unfortunately, despite all progress not all aspects are well understood and a general theory does not exist [1, 2].
L = N 1 N 2 fn B N 1 N 2 fn B Luminosity: = 4 πσ x σ y 4 π · σ x σ y High luminosity is not good for beam-beam effects ... Beam-beam effects are not good for high luminosity ... Menu: Overview: which effects are important for present and future machines (LEP, PEP, Tevatron, RHIC, LHC, FCC, linear colliders, ...) Qualitative and physical picture of the effects Derivations in: Proceedings, Advanced CAS, Trondheim (2013) http://cern.ch/Werner.Herr/CAS2011_Chios/bb/bb1.pdf
Studying beam-beam effects - how to proceed - Need to know the forces - Apply concepts of non-linear dynamics - Apply concepts of multi-particle dynamics - Analytical models and simulation techniques well developed in the last 20 years (but still a very active field of research) LHC is a wonderful epitome as it exhibits many of the fea- tures revealing beam-beam problems
First step: Fields and Forces Need fields � E and � B of opposing beam with a particle distribution ρ ( x , y , z ) B ′ ≡ 0 In rest frame (denoted ′ ) only electrostatic field: � E ′ , and � Derive potential U ( x , y , z ) from ρ and Poisson equation: ∆ U ( x , y , z ) = − 1 ρ ( x , y , z ) ǫ 0 The electrostatic fields become: E ′ = − ∇ U ( x , y , z ) � Transform into moving frame to get � B and calculate Lorentz force
Example Gaussian distribution (a simplification !): � � − x 2 − y 2 − z 2 NZ 1 e ρ ( x , y , z ) = 3 exp √ 2 σ 2 2 σ 2 2 σ 2 σ x σ y σ z 2 π x y z For 2D case the potential becomes: y 2 x 2 exp( − x + q − y + q ) � ∞ 2 σ 2 2 σ 2 U ( x , y , σ x , σ y ) = NZ 1 e dq � 4 πǫ 0 (2 σ 2 x + q )(2 σ 2 y + q ) 0 Once known, can derive � E and � B fields and therefore forces For arbitrary distribution (non-Gaussian): difficult (or impossible), numerical solution required
Force for round Gaussian beams σ x = σ y = σ Simplification 1: round beams Simplification 2: very relativistic β ≈ 1 One finds: Only components E r and B Φ are non-zero Force has only radial component, i.e. depends only on distance r from bunch centre where: r 2 = x 2 + y 2 F r ( r ) = − Ne 2 (1 + β 2 ) 1 − exp( − r 2 � � 2 σ 2 ) 2 πǫ 0 · r For σ x � σ y the forces are more complicated:
� � σ y σ x + iy σ x + y 2 − x 2 x ne x + iy σ y 2 σ 2 2 σ 2 − e E x = Im erf erf x y � � � 2 ǫ 0 2 π ( σ 2 x − σ 2 2( σ 2 x − σ 2 2( σ 2 x − σ 2 y ) y ) y ) � � σ y σ x + iy σ x + y 2 − x 2 x ne x + iy σ y 2 σ 2 2 σ 2 − e E y = Re erf erf x y � � � 2 ǫ 0 2 π ( σ 2 x − σ 2 2( σ 2 x − σ 2 2( σ 2 x − σ 2 y ) y ) y ) The function erf(t) is the complex error function � t erf(t) = e − t 2 � � 1 + 2 i e z 2 dz √ π 0 The magnetic field components follow from: B y = − β r E x / c B x = β r E y / c and Assumption/simplification: we shall continue with round beams ...
The forces will result in a deflection: "beam-beam kick" We are careless a and use ( r , r ′ , x , x ′ , y , y ′ ) as coordinates We need the deflections (kicks ∆ x ′ , ∆ y ′ ) of the particles: r Incoming particle (from left) deflected by force from opposite beam (from right) Deflection depends on the distance r to the centre of the fields a One should always use canonical variables, but here x, x’ more convenient
After a short calculation (integration along bunch): Using the classical particle radius: r 0 = e 2 / 4 πǫ 0 mc 2 we have (radial kick and in Cartesian coordinates): � � 1 − exp( − r 2 ∆ r ′ = − 2 Nr 0 · r r 2 · 2 σ 2 ) γ � � 1 − exp( − r 2 ∆ x ′ = − 2 Nr 0 · x r 2 · 2 σ 2 ) γ � � 1 − exp( − r 2 ∆ y ′ = − 2 Nr 0 · y r 2 · 2 σ 2 ) γ
Form of the kick (as function of amplitude) beam-beam kick 1D 1 0.5 kick 0 -0.5 -1 -8 -6 -4 -2 0 2 4 6 8 amplitude (units of beam size) For small amplitudes: linear force (like "beam-beam quadrupole") For large amplitudes: very non-linear force
Can one quantify the beam-beam strength ? Tune shift by the "beam-beam quadrupole" may be a good indicator - Use the slope of quadrupole force (kick ∆ r ′ ) at zero amplitude - This defines: beam-beam parameter ξ - For head-on interactions (general case, non round beams): N · r o · β ∗ x , y ξ x , y = 2 πγσ x , y ( σ x + σ y ) Note: it is independent of β ∗ , important for (circular) colliders
some examples: LEP - LHC (most recent) LEP (e + e − ) LHC (pp) (2017) Beam sizes ≈ 200 µ m · 4 µ m ≈ 11 µ m · 11 µ m 4.0 · 10 11 /bunch 1.40 · 10 11 /bunch Intensity N Energy 100 GeV 6500 GeV ǫ x · ǫ y ( ≈ ) 20 nm · 0.2 nm 0.4 nm · 0.4 nm β ∗ x · β ∗ y (nominal) ( ≈ ) 1.25 m · 0.05 m 0.30 m · 0.30 m Crossing angle 0.0 340 µ rad Beam-beam 0.0700 0.0070 parameter( ξ ) (0.0037) Unlike often assumed: Linear tune shift ∆ Q bb from beam-beam interaction proportional, but not equal to ξ
Still, how big is the tune shift for a given ξ Take only the linear part ("beam-beam quadrupole") and add to the lattice Transformation matrix of a thin quadrupole (beam-beam is really thin): 1 0 1 1 − f For small amplitudes linear force like a quadrupole with focal length f � ξ · 4 π � f = ∆ x ′ 1 = Nr 0 γσ 2 = β ∗ x
Use the "Full Turn Matrix" without beam-beam: β ∗ sin(2 π ( Q )) cos(2 π ( Q )) − 1 β ∗ sin(2 π ( Q )) cos(2 π ( Q )) Add "beam-beam thin lens", i.e. the (linear) beam-beam focusing: β ∗ cos(2 π Q ) 0 sin(2 π Q ) 1 0 ◦ − 1 1 sin(2 π Q ) cos(2 π Q ) 1 β ∗ − f 0 allow for a change of the tune Q and β in the resulting matrix: β ∗ sin(2 π ( Q +∆ Q )) cos(2 π ( Q +∆ Q )) should become − 1 β ∗ sin(2 π ( Q +∆ Q )) cos(2 π ( Q +∆ Q ))
Solving this equation gives us (like a "tuning quadrupole"): cos(2 π ( Q + ∆ Q )) = cos(2 π Q ) − β ∗ β ∗ 0 2 f sin(2 π Q ) = sin(2 π Q ) / sin(2 π ( Q + ∆ Q )) and β ∗ 0 � � β ∗ sin(2 π Q ) 1 √ 0 = = β ∗ sin(2 π ( Q +∆ Q )) 1 + 4 πξ cot(2 π Q ) − 4 π 2 ξ 2 At a "real" quadrupole: β at quadrupole changes slightly, (usually assumed constant) At beam-beam interaction: Both ∆ Q and β depend also on ξ and tune Q (must not be ignored) β can become significantly smaller or larger at interaction point This is called "Dynamic β "
C P H L E L beam-beam tune shift versus tune 0.1 0.08 0.06 delta Q 0.04 ξ = 0.03 ξ= 0.02 0.02 ξ= 0.01 ξ= 0.0045 0 0 0.1 0.2 0.3 0.4 0.5 Q LEP working point, close to integer (vertical plane): ∆ Q y decreased: 0.07 = ⇒ ≈ 0.04 !!! β ∗ decreased: 5 cm = ⇒ ≈ 2.5 - 2.8 cm (Luminosity !) LHC working point, far from integer: Weaker effects on β ∗ ξ , ∆ Q ≈
Dynamic Beta (IP5, LHC) 0.7 0.65 proton-antiproton beta x [m] 0.6 nominal 0.55 proton-proton 0.5 0.45 13329 13329.1 13329.2 13329.3 13329.4 13329.5 s [m] Dynamic β in LHC, computed for pp and p ¯ p (with standard LHC parameters) β ∗ : 0.55 m = ⇒ 0.52 m
beam-beam linear tune shift in working diagramm Start with standard working Linear tune shift 0.311 point, no beam-beam x 0.31 With beam-beam: Tune shift in Qy both planes 0.309 LHC (equally charged beams) 0.308 Tune shift is negative (pp) 0.307 x Whole beam moves to new tune 0.306 0.305 0.275 0.276 0.277 0.278 0.279 0.28 0.281 Qx LHC is/was round (in most hadron colliders) equal tune change in both planes Usually not the case for leptons
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