Beam-beam effects. (an introduction) Werner Herr CERN. Slide 1. Why talk about beam-beam effects? Slide 2

Transcription

1 Beam-beam effects (an introduction) Werner Herr CERN Werner Herr, Beam-Beam effects, Zurich, 218 Why talk about beam-beam effects? Colliding beams in storage rings require a long lifetime to produce useful physics Two principles are important threats to this lifetime: 1. Magnet errors 2. Beam-beam interactions (unavoidable even in principle in colliders) For beams in collision the beam-beam interactions are the most significant source of nonlinear consequence Slide 1 Slide 2

2 Popular types of colliders: Particle types: 1. p p, p p 2. e+ e, e e (in the past..) 3. ion ion, p ion 4. lepton hadron 5. photon photon Main categories of colliders: 1. Circular colliders (mainly hadrons, low energy leptons) 2. Linear colliders (mainly leptons) Beam-beam collision beam-beam collision Typically: s.1% (or less) of particles have useful interactions % (or more) of particles are perturbed Slide 3 Slide 4

3 Some challenges (beam-beam related, incomplete): Linear colliders: 1. Luminosity, high energy 2. Beams are discarded after the collision (1 beam-beam interaction) 3. Strong focusing at collision point, beamstrahlung Circular colliders: 1. One or two rings (exotic: 4 beam collisions) 2. Beams are re-used - LHC: beam-beam interactions per production run (fill) challenge for the beam dynamics (many different types of beam-beam effects to be understood and controlled) Very different problems must be expected, already at early design phase We are interested in three main questions: What happens to a single particle? What happens to the whole beam? What happens to the machine? Unlike other effects, e.g. space charge or magnetic errors, the beam-beam effects do not become smaller with increasing energy, usually the effects become more important (so far it was always the case) Unfortunately, despite all progress not all aspects are well understood and a general theory does not exist. Slide 5 Slide 6

4 Beams are always a collection of (a large number of) charges Represent electromagnetic potential for other charges, resulting in: Forces on itself (space charge) Forces on opposing beam (beam-beam effects) Important for high density beams, i.e. high intensity and/or small beams: for high luminosity! Remember (Bruno): L= N 1N2 f nb 4πσxσy = N1N2 f nb 4π σxσy High luminosity is not good for beam-beam effects... Beam-beam effects are not good for high luminosity... 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 (213) Slide 7 Slide 8

5 One has to evaluate three main types of colliding beam machines: e.g. magnets: 1. Circular hadron colliders 2. Circular lepton colliders 3. Linear lepton colliders Although 1 and 2 result in very different problems, in the first part the treated is very similar Early warning: occasionally I shall make assumptions and/or simplifications, but only to keep the formulae readable, no compromise on the physics! A beam acts on particles like an electromagnetic lens, but unlike e.g. magnets: Very local force Very non-linear form of the forces, depending on particle distribution Does not represent simple form, i.e. well defined multipoles Can change distribution as result of interaction (time dependent forces..) Slide 9 Slide 1

6 Studying beam-beam effects - 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 2 years (but still a very active field of research) LHC is a wonderful epitome as it exhibits many of the features 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) In rest frame (denoted ) only electrostatic field: E, and B Derive potential U(x, y, z) fromρand Poisson equation: U(x, y, z)= 1 ǫρ(x, y, z) The electrostatic fields become: E = U(x, y, z) Slide 11 Slide 12

7 Transform into moving frame and calculate Lorentz force F on particle with charge q=z2 e E =E, E =γ E with : B = β E/c F= q( E+ β B) Example Gaussian distribution (not surprising): ρ(x, y, z)= σxσyσz ( NZ1e 3 exp 2π x2 2σ 2 x y2 2σ 2 y z2 2σ 2 z ) Simple example: Gaussian For 2D case the potential becomes (see proceedings): U(x, y,σx,σy)= NZ 1e 4πǫ exp( x2 2σ 2 x 2σ 2 y +q) (2σ 2 x+ q)(2σ 2 y+ q) +q y2 Once known, can derive E and B fields and therefore forces dq For arbitrary distribution (non-gaussian): difficult (or impossible, numerical solution required) Slide 13 Slide 14

8 Force for round Gaussian beams Assumption 1:σx=σy=σ, Z1= Z2 = 1 Assumption 2: very relativistic β 1 Only components Er and BΦ 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)= Ne2 (1+β 2 ) 2πǫ r [1 exp( r2 2σ ) 2 ] Compare with other forces (cheating in 1D): quadrupole force: F(x) x sextupole force: F(x) x 2 octupole force: F(x) x 3 beam-beam force: Fr(r)=± Ne2 (1+β 2 ) 2πǫ r [1 exp( r2 2σ 2 )... and we still assumeσx=σy=σ!! If not (usuallyσx σy in lepton colliders)? ] (remember : r = x 2 + y 2 ) Slide 15 Slide 16

9 ( ) E x = ne 2ǫ 2π(σ 2 x σ 2 y) Im erf x+iy 2(σ 2 x σ 2 y) e x2 2σ 2 x + y2 2σ 2 y erf xσy σx + iyσx σy 2(σ 2 x σ 2 y) ( ) E y = 2ǫ ne 2π(σ 2 x σ2 y ) Re erf x+ iy 2(σ 2 x σ2 y ) e x2 2σ 2 x + y2 2σ 2 y erf xσy σx + iyσx σy 2(σ 2 x σ2 y ) The function erf(t) is the complex error function [ erf(t) = e t2 1+ 2i t e z2 dz π ] The magnetic field components follow from: B y = β r E x /c and B x = β r E y /c Assumption/simplification: we shall continue with round beams... Fields are similar for Space Charge - what makes the difference? For the forces: - Space charge dependence on momentum: (1 β 2 ) Vanishes at high energy (almost always in lepton machines) - beam-beam: (1+β 2 ) instead of (1 β 2 ) Does not get smaller with increasing energy Space charge is "smooth", often integrable Beam-beam is extremely local, almost never integrable Totally different problems and effects! However: beam-beam only in colliders.. Slide 17 Slide 18

10 Beam-beam kick: We are careless a and use (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 however there is a reason Kick ( r ): angle by which the particle is deflected during the passage Integration of force over the collision, i.e. time of passage t (assuming: m1=m2 and Z1= Z2= 1): with: Newton s law : r = 1 mcβγ Fr(r, s, t)= Ne2 (1+β 2 ) (2π) 3 ǫ rσs + t 2 t 2 [1 exp( r2 2σ 2 ) ] Fr(r, s, t)dt [exp( (s+vt)2 2σ 2 s ) ] Slide 19 Slide 2

11 Using the classical particle radius (implies Z1 = ± Z2): r= e 2 /4πǫmc 2 we have (radial kick and in normalized Cartesian coordinates): r = 2Nr ] r [1 exp( r 2 2σ 2 ) γ x = 2Nr γ x r 2 [1 exp( r2 2σ 2 ) ] y = 2Nr γ y r 2 [1 exp( r2 2σ 2 ) ] Beam-beam force/kick amplitude (units of beam size) kick beam-beam kick 1D For small amplitude: linear force (like quadrupole) For large amplitude: very non-linear force Particles in a bunch oscillate across the force Slide 21 Slide 22

12 What the particles "see": amplitude kick beam-beam kick 1D For small amplitudes: looks like a quadrupole, non-linear part not "seen" A quadrupole changes the tune, so must beam-beam Slope defines the "sign" and the "gradient" of the "quadrupole" What the particles "see": amplitude kick beam-beam kick 1D For large amplitude: particles move across the non-linear part amplitude dependent tune shift The particles "see" more than one slope (left and right of the peak)... Slide 23 Slide 24

13 Can we quantify the beam-beam strength? Tune shift may be a good indicator - We need the slope of force (kick r ) at zero amplitude (for the time being) - This defines: beam-beam parameterξ - For head-on interactions and round beams (β = β x = β y ): ξ = 4π δ( r β ) δr = N r o β 4πγσ 2 Can we quantify the beam-beam strength? In general for non-round beams (β x β y ): ξ x,y = N ro β x,y 2πγσx,y(σx+σy) Proportional to (linear) tune shift Qbb from beam-beam interaction: Qbb ±ξ Good: Reliable measure for strength of beam-beam interaction Bad: Does not take into account the non-linear part Ugly: Eventually we have to.. Slide 25 Slide 26

14 some examples: LEP - LHC LEP (e + e ) LHC (pp) (211) Beam sizes 2µm 4µm 3µm 3µm Intensity N /bunch /bunch Energy 1 GeV 35 GeV ǫx ǫy ( ) 2 nm.2 nm.5 nm.5 nm β x β y ( ) 1.25 m.5 m 1. m 1. m Crossing angle. 24µrad Beam-beam.7.8 parameter(ξ) (.37) Back to: ξx,y = N ro β x,y 2πγσx,y(σx+σy) β x,y σx,y(σx+σy) i.e.ξx ξy for flat beams (mostly leptons) but: ξx,y = β x,y σx,y if : 1 (σx+σy) = βx βy = ǫ x ǫy β x,y ǫ x,y = ξx = ξy 1 (σx+σy) For lepton machines this ratio is sometimes adjusted... ξx = ξy = ξ becomes a minimum Slide 27 Slide 28

15 Still, how big is the tune shift Take only the linear part Transformation matrix over the interaction becomes (like thin quadrupole - beam-beam is really thin! ) : f For small amplitudes linear force like a quadrupole with focal length f [ ] ξ 4π 1 f = x x = Nr γσ 2= β "Full turn matrix" without beam-beam: cos(2π(q)) β sin(2π(q)) 1 sin(2π(q)) cos(2π(q)) β Add "beam-beam thin lens", i.e. the (linear) beam-beam focusing: 1 β cos(2πq) β sin(2πq) sin(2πq) cos(2πq) 1 1 f 1 we allow for a change of the tune Q andβ: should become = cos(2π(q+ Q)) β sin(2π(q+ Q)) 1 sin(2π(q+ Q)) cos(2π(q+ Q)) β Slide 29 Slide 3

16 Solving this equation gives us: cos(2π(q+ Q))=cos(2πQ) β 2 f sin(2πq) and β β = sin(2πq)/ sin(2π(q+ Q)) [ β β = sin(2πq) sin(2π(q+ Q)) ] 1 = 1+4πξ cot(2πq) 4π 2 ξ 2 Both Q and β depend on ξ and tune Q β can become smaller or larger at interaction point This is called "Dynamic β" (for ξ = nothing changes) ξ= Q delta Q beam-beam tune shift versus tune ξ =.3 ξ=.2 ξ=.1 Strong dependence on Q for largerξ (dynamicβ) Q ξ only far from integer and half-integer Is this just a nuisance or can it be useful?? Slide 31 Slide 32

17 ξ= Q delta Q beam-beam tune shift versus tune ξ =.3 ξ=.2 ξ=.1 LEP working point (vertical plane): Q y decreased:.7 =.5 β decreased: 5 cm = cm (Luminosity!) LHC working point: Q ξ, Weaker effects onβ PEL s [m] CHL beta x [m] Dynamic Beta (IP5, LHC) proton-antiproton nominal proton-proton Dynamicβin LHC, computed for pp and p p (with standard LHC parameters) β :.55 m =.52 m Slide 33 Slide 34

18 Linear tune shift - two dimensions Start with standard working.311 Linear tune shift point.31 x Qy Qx Linear tune shift - two dimensions Start with standard working.311 Linear tune shift point.31 x Tune shift in both planes Qy.39 LHC (equally charged beams).38 Tune shift is negative x Qx Slide 35 Slide 36

19 Relative detuning Q/ ξ Tune measurement: linear optics 1.2 Tune distribution for linear optics Linear force all particles have the same tune Only one frequency (tune) visible Qx Next: include non-linear part Non-linear force: Amplitude detuning Tune depends on amplitude Different particles have different tunes Largest effect for small amplitudes (Calculations in proceedings...) amplitude in units of beam size α Detuning with amplitude round beams linear only [ withα= a we get: Q/ξ= 4 σ α 2 1 I ( α2 4 ) e α2 4 ] Slide 37 Slide 38

20 Non-Linear tune shift - two dimensions.311 Tune footprint for head on collision.31 X Qy Start with standard working.39 point Qx Non-Linear tune shift - two dimensions.311 Tune footprint for head on collision Start with standard working Qy.31 (,6) X (6,6) point.39 (6,) tudes No single tune in the beam: Tunes depend on x and y ampli (,) Tunes are "spread out" Point becomes a footprint Note: the coupling is evident! Qx Slide 39 Slide 4

21 Tune measurement: with beam-beam Non-linear force.16 Tune distribution for optics with beam beam Particles with different amplitudes have different tunes We get tune spectra Width is aboutξ Qx Slang: weak-strong and strong-strong Both beams are very strong (strong-strong): Both beam are affected and change during a beam-beam interaction: Beam 1 changes beam 2, beam 2 changes beam 1 beam-beam effects change every time the beams "meet" Examples: LHC, LEP, RHIC,... (FCC?) Evaluation of effects challenging (need to be self-consistent) One beam much stronger (weak-strong): Only the weak beam is affected and changed due to beam-beam interaction Examples: SPS collider, Tevatron,... Slide 41 Slide 42

22 weak-strong beam-beam collision s Counter-rotating beam unaffected and treated as a static field Equivalent to treat single particles, tracking etc. this is usually done to study single particle stability (so far only weak-strong was considered) weak-strong beam-beam collision 6 4 strong weak s Counter-rotating beam unaffected and treated as a static field Weak beam can be strongly perturbed Slide 43 Slide 44

23 strong-strong beam-beam collision s Both beams are (maybe heavily) distorted Size, shape, density, (losses?)... Always treat both beams - not particles (self-consistently) How to get it self-consistent, some examples: 1. Coupled Vlasov equations - Usually difficult to solve analytically, need perturbative treatment, but still... - Numerical solver 2. Simulation - Multi-particle tracking - Gives desired results, but requires computing resources and very careful analysis (numerical problems, intelligent choice of field solver (!),...) Slide 45 Slide 46

24 Single particle (very weak!) - incoherent effects Single particle dynamics: treat as a particle through a static electromagnetic lens Basically non-linear dynamics All bad single particle effects observed: - Unstable and/or irregular motion - Enhanced diffusion - Beam blow up - Bad lifetime, particle loss - Resonances Observations hadrons Non-linear motion can become chaotic - reduction of "dynamic aperture" - particle loss and bad lifetime Strong effects in the presence of noise or ripple Evaluation is done by simulation resonances... Slide 47 Slide 48

25 Slide 49 Slide = L = N1 N2 f nb 4πσ x σy -2 N2-1 What is observed? Beam-beam parameter should increase N for: N1 = N2 = N x 1 2 x 1 2 Horizontal Phase Space -1 Horizontal Phase Space Observations leptons Luminosity should increase N1 N2 Remember resonances - For some amplitudes they are on - Tune depends on amplitude - With (head-on) beam-beam (all on circles) - All particle have the same tune - Without beam-beam px px

26 Beam-beam observations: lepton colliders x x x x xx xxx x xx xxx const CESR PETRA PEP 3b α x xx x x xx x x x x x x xxx Current (A)/beam Luminosity [1 cm s ] ξ y α N const const α N N α N α N 2 α N α N 2 Does not show the expected behaviour (L N 2 ) Have to introduce the concept of "beam-beam limit" Beam-beam limit (schematic) Beam-beam parameter increases linearly with intensityξ N.1 linear increase but saturation above some inten-.8 sity, ξ = const. no more increase Luminosity increases with intensity squared L N 2 but above this intensity, L N This intensity: beam-beam limit Bunch Intensity Luminosity, tuneshift Beam-beam limit, schematic Luminosity beam-beam parameter ξ αn L αn 2 ξ = const L α N Slide 51 Slide 52

27 What is happening? we have ξy = Nrβy 2πγσy(σx + σy) (σx σy) rβy 2πγ(σx) N σy and L = N2 f nb 4πσxσy = N f n B 4πσx Above beam-beam limit:σy increases when N increases to keepξ constant equilibrium emittance! N σy Therefore:L N andξ constant ξlimit is NOT a universal constant! Difficult to predict, but hard to exceed.5 What is happening? Where does it come from? From synchrotron radiation: vertical plane damped, horizontal plane excited Horizontal beam size usually (much) larger Vertical beam-beam effect depends on horizontal (large) amplitude Coupling from horizontal to vertical plane Equilibrium between this excitation and damping determines the beam-beam limitξlimit Lesson: Keep the coupling small! Slide 53 Slide 54

28 Problems with hadron machines Difference: hadron (e.g. protons) machines have no or very little damping, beams usually round No equilibrium emittance - no hard beam-beam limit just gets worse and worse... Very hard to exceed.1 Losses or lifetime extremely hard to predict, a prediction within a factor 2 is pretty good.. Remember: = L= N 1N2 f nb 4πσxσy The next problem How to collide many bunches (for highl)?? Must avoid unwanted collisions!! Separation of the beams: Pretzel scheme (SPS,LEP,Tevatron) Bunch trains (LEP,PEP) Crossing angle (LHC) FCC? Slide 55 Slide 56

29 Separation: LHC Many equidistant bunches (288 per beam) Two beams already separated in two separate beam pipes except: The two beams have to exchange between inner and outer beam pipes Four experimental areas Need local separation Two horizontal and two vertical crossing angles Layout of LHC IP5 IP4 IP6 IP3 IP2 beam1 IP1 beam2 Symmetric crossing in experimental areas IP8 IP7 Slide 57 Slide 58

30 Two beams, 288 bunches each, every 25 ns In common chamber around experiments 12 m Over 12 m: about 3 parasitic interactions Crossing angles (example LHC) Long-range Head-on Particles experience distant (weak) forces Separation typically 6-12σ We get so-called long range interactions Slide 59 Slide 6

31 What is special about them? In addition to the head-on effects: Break symmetry between planes, stronger resonance excitation Mostly affect particles at large amplitudes Cause effects on closed orbit, tune, chromaticity,.. Special case: PACMAN effects Tune shift has opposite sign in plane of separation E.g. case with a horizontal crossing angle: Horizontal tune shift positive, vertical tune shift negative What do the particles "see": Why opposite tuneshift??? amplitude kick beam-beam kick 1D Local slope has opposite sign for large separation Opposite sign for focusing! Slide 61 Slide 62

32 25 ns X d sep Modified "kick" for horizontal separation d: x (x+d, y, r)= 2Nr γ (x+d) r 2 [1 exp( r2 2σ ) 2 ] (with: r 2 = (x+d) 2 + y 2 ) Red flag: to use this expression (e.g. in a simulation) there is a small complication (if interested ask offline, or discussion session) X d sep - Number of long range interactions depends on spacing and length of common part - In LHC 15 collisions on each side, 12 in total! - Effects depend on separation: Q N d 2 (for large enough d!) footprints?? 25 ns Slide 63 Slide 64

33 footprint from long range interactions - Tuneshift large for largest ampli-.312 tudes (where non-linearities are strong) Qy.311 (,) - We should expect problems at.31 (,2) small separation (2,2) - Footprint is very asymmetric.39 (2,) - Size proportional to 1 d Qx One observes a "folding" (may also appear for head-on + octupoles) For small separation, the size of the footprint can be large particle losses Tune footprint, head on and long range - Compare foot print for different.316 vertical separation contributions Qy Seem to be totally separated For horizontal and vertical separation go opposite directions Can we take advantage of that?? head on horizontal separation Qx (Note: a second head-on collision just doubles the size of the footprint) Slide 65 Slide 66

34 Two interaction points: Tune footprint, combined head on and long range.316 Two head-on footprints Qy One horizontal long range + One vertical long range Qx Alternating (i.e. one vertical and horizontal each), implemented in the LHC Seems to get some compensation, i.e. overall footprint decreased and symmetric Is that all? (see later) Small crossing angle small separation Stable region versus separation 14 in units of beam sizeσ Minimum separation for LHC: 1σ comfort zone separation d (sigma) For too small separation: particles may be lost and/or bad liftime stable region (sigma) Slide 67 Slide 68

35 A "little" headache: Beam separation d (in units ofσ) depends on: d β High luminosity smallβ small separation big problems since footprint size (and other effects) : 1 d 2 a very smallβ may be difficult to reach (also for other reasons). Possible solutions: - large crossing angle - but other detrimental effects - flat beams (i.e.β x β y ) - partial compensation (e.g. octupoles, wires, see later) More jargon: PACMAN bunches Trains not continuous: gaps for injection, extraction, dump of 3564 possible bunches Slide 69 Slide 7

36 Trains for beam 1 and beam 2 are symmetric: At interaction point: bunch 1 meets bunch 1, bunch 2 meets bunch 2, etc. hole 1 meets hole 1, hole 2 meets hole 2, etc. But not always... Long-range Head-on What we want... Slide 71 Slide 72

37 Long-range Head-on What we get... Long-range Head-on Some Bunches meet holes (at beginning and end of batch) Cannot be avoided Worst case: less than half of long range collisions (depends on collision scheme and gaps) Slide 73 Slide 74

38 When a bunch meets a "hole": Miss some long range interactions PACMAN bunches They see fewer unwanted interactions in total Different integrated beam-beam effect Long range effects for different bunches will be different: Different tune, chromaticity, orbit... May be more difficult to optimize Note: this case is specific LHC, but something similar happens in other machines in some form.. Example: tune along the train, two horizontal (H + H) crossings nominal pacman pacman Horizontal tune along bunch train (computed) Tune spread between bunches rather large (.25), effects add up for several crossings Too large ( Q.1) for 4 IPs Slide 75 Slide 76

39 Example: tune along the train, alternating (H + V) crossings Horizontal tune along bunch train (computed) Tune spread has disappeared due to compensation by alternating crossings (1 horizontal and 1 vertical) This is the real reason for alternating crossings in the LHC! Can be measured (without compensation): 2e+1 1.8e+1 1.6e+1 2e+9 1.4e+1 1.2e+1 1e+1 8e+9 6e+9 4e time (min) Integrated losses Integrated losses during scan in IP1 4 % 1 % (12 σ) 7 % 6 % 5 % 4 % (5 σ) 35 % 3 % (4 σ) Separation (crossing angle) slowly decreased from 12σto 4σ Bunches with more long range collisions suffer first Slide 77 Slide 78

40 ... along the train: 2.5e+1 2e+1 1.5e+1 1e+1 5e bunch number Integrated losses Integrated losses per bunch 16 long range interactions 7-8 long range interactions 7-8 long range interactions Losses follow exactly the number of long range interactions in LHC losses completely dominated by long range effects Interlude: Both, head-on and long range contribute to chromaticity Head-on (modulation ofβ-function): C HO x,y = Qx,y Wx,y, Wx,y = 1 βx,y βx,y δp Long range (modulation of separation): ( C LR x,y = Dx + Dy dx dy ) Q LR x,y Result: different bunches, different chromaticity Slide 79 Slide 8

41 Closed orbit effects Starting from the kick x for long range interactions: x (x+d, y, r)= 2Nr γ (x+d) r 2 [1 exp( r2 2σ 2 ) ] For well separated beams (d σ) the force (kick) has an amplitude independent contribution: a x = const. [ 1 x ( ) ] x 2 d d + O d X a THIS is the complication mentioned before... This constant and amplitude independent kick changes the orbit! Has been observed in LEP with bunch trains (and was bad) So should be evaluated by computation, however: Change of orbit change of separation change of orbit... Change of tune change of separation change of tune... All PACMAN bunches will be on different orbits In LEP: 8 bunches, in LHC: 288 bunches Requires the self-consistent computation of 5616 orbit! (first time done in 21, calculation took 45 minutes) Slide 81 Slide 82

42 PACMAN Orbit effects: calculation bunch number vertical offset (mum) vertical offset IP1 Vertical offset expected at collision points, sizeable with respect to beam size Predicted orbits from self-consistent computation Does it have anything to do with reality? Cannot be resolved with beam position measurement, but.. PACMAN Orbit effects: measurement file:///afs/cern.ch/user/z/zwe/desktop/png/bcid_vs_posy_pm_posyerr.png #1 Measured vertex centroid in ATLAS detector Very good agreement with computation Slide 83 Slide 84

43 Coherent beam-beam effect x When bunches are well separated: All particles in a bunch "see" the same kick Whole bunch sees a kick as an entity (coherent kick) The coherent kick of separated beams can excite coherent dipole oscillations All bunches couple because each bunch "sees" many opposing bunches: many coherent modes possible! When bunches are separated much less than oneσ(quasi head-on interactions): Remember orbit kick all particles "see" the same kick x = const. d 1 other parts { }} ( ) { x x 2 d + O d There is "one part" of the kick that is the same for all particles This part also excites dipolar oscillations There are "other parts" which are different for the particles These parts can change the particle distribution Slide 85 Slide 86

44 Simplest case - one bunch per beam: -mode π -mode TURN n TURN n+1 Coherent modes: two bunches are "locked" in a coherent oscillation, can be either: Two bunches oscillate "in phase": -mode Two bunches oscillate "out of phase":π-mode -mode has no tune shift and is stable π-mode has large tune shift and can be unstable Tune of the modes (schematic): Two separate modes visible -mode is at unperturbed tune beam-beam modes and tune spread π-mode is shifted by Y ξ π - mode Y is called: "Yokoya factor" Typically: 1 Y 1.35 This is due to the "other parts" a (otherwise Y = 1) a say goodbye to Gaussian distributions Q: what about the tune spread Q ξ? - mode Tune Slide 87 Slide 88

45 -mode is at the unperturbed tune = Q = beam-beam modes and tune spread π - mode π-mode is shifted by Y ξ = Q π 1.35 ξ Incoherent spread between [., 1.] ξ = Q i 1. ξ ξ -mode Tune This is bad news: Q π > Q i Strong-strong case:π-mode is shifted outside the tune spread No Landau damping possible for this mode, requires: Q π Q i What was measured: LEP.12 Q_FFT_TfsQ2_99_1_19:7_26_25 u 2: Qx Both modes clearly visible They are real! :.28 π :.34 Slide 89 Slide 9

46 What was measured: RHIC Blue Horizontal, single p bunch, at injection 1.E+1 1.E E E+7 1.E Tune Spectral power.4 Vx Beam-beam OFF Beam-beam ON Sx Courtesy W. Fischer (BNL) Spectral power as function of the tune, with and without beam-beam interaction With beam-beam interaction the two modes visible First we consider head-on collision of one bunch per beam a and b Particle distributionsψ a andψ b mutually changed by interaction (by the "other parts") Interaction depends on particle distributions - Beamψ a solution depends on beamψ b - Beamψ b solution depends on beamψ a Can one find a self-consistent solution? What is the equation of motion? For distribution function: Vlasov equation Slide 91 Slide 92

47 for beam a: ψ a t = q x p x ψ a x + ( p x t ) ψa p x ψ a t ψ a = q x p x x + ( + ) ρ b (x ; t) q x x+δ p (t) 4 πξ x p.v. dx x x } {{ } force from beam b on beam a ψ a p x ρ b (x; t)= ψ b (x, p x ; t)dp x The same thing for beam b, two coupled differential equations for beam distributionsψ a (x, p x ) andψ b (x, p x ) Normally cannot find exact solutions Need numerical solutions just the result (if interested, ask a lecturer): n What is shown: difference and sum of bunch centres <ψ a > <ψ b > π - mode <ψ a > + <ψ b > - mode oscillation amplitude Slide 93 Slide 94

48 Does it make any sense? w A Fast Fourier Transform of this motion: Tune shifts. ξ and 1.33 ξ Get the same frequencies as the simulation (but 1 5 times faster...) How to deal with the problems? Every "Coherent Motion" requires organized motion of many/all particles Requires a "high degree of symmetry" (between the beams) "High degree of symmetry" large Yokoya factor Y Possible countermeasure break the symmetry by: - Different bunch intensity (results in different tune shifts for the two beams) - Different nominal tunes of the two beams Y becomes smaller and smaller π-mode should move into the tune spread (and is damped) Slide 95 Slide 96

49 Beams with unequal tunes Q1 = Q2 Q1 Q Bunches with different tunes cannot maintain coherent motion Q 2 Q 1 =.2 Landau damping restored, noπ-mode visible Beams with unequal intensities I1 = I2 I1 I Bunches with different intensities cannot maintain coherent motion I 2.5 I 1 = Q.5 ξ =.187 (just enough!) Landau damping restored, noπ-mode visible Slide 97 Slide 98

50 These were coherent modes due to head-on interactions What about coherent modes due to long range interactions? Separation now much more thanσ, in LHC more than 1 σ! - Do they exist? yes... - Do we understand them? more or less... - Are they more complicated? YES a - many more modes... - Key problems: they are not damped by head-on incoherent spectrum b (even if they sit inside the incoherent spectrum!!!) a Deserves a dedicated lecture - offline discussions more appropriate b Deserves another dedicated lecture Can one compensate or reduce beam-beam effects? Find lenses to correct beam-beam effects Head on effects: Non-linear "electron lens" to reduce spread Tests in progress Long range effects: At very large distance: force is 1/r Same force as a wire! Tests in the LHC So far: mixed success with active compensation, passive compensation (better: reduction) more practical and (often) more beneficial Slide 99 Slide 1

51 Others: Möbius lattice - flat beams Idea: Reduced beam-beam problems for round beams Principle: Interchange horizontal and vertical plane each turn Effects: Round beams for leptons Some compensation effects for beam-beam interaction First test at CESR at Cornell Interlude... Some final words (circular hadron colliders): Is beam-beam always bad? Slide 11 Slide 12

52 Stability diagram and head-tail modes e Real Q Imag Q m =, Q = m =, Q = - 3 m =, Q = + 3 m = 1, Q = Stabilization with octupoles or colliding beams - (Head-on) collisions much more efficient than octupoles for Landau damping, in particular for high energies a Beam-beam collisions save the day (e.g. LHC) a see e.g. Proceedings CAS Trondheim (213), Some PhD theses at EPFL Is beam-beam always bad? No, but requires a careful compromise between: - Detrimental beam-beam effects (dynamic aperture, losses, instabilities,...) - Effects on optics (plenty..) - Effects on luminosity - Effects on Landau Damping properties and coherent instabilities - Requirements from particle physics experiments and machine protection - A few more... Now for something completely different.. Slide 13 Slide 14

53 Beam-beam effects in linear colliders Mainly (only?) e + e colliders Past collider: SLC (SLAC) Under consideration: CLIC, ILC Under construction: Special issues: Particles collide only once (dynamics)! Can accept to spoil the beam during collision Beam-beam considerations very different Particles radiate photons crossing the other beam, leading to an energy spread. Have to introduce "Differential Luminosity" Main effects to look at (here): Disruption (modification of beam size, mainly classical) Beamstrahlung (spread of CM energy, requires quantum treatment) High energy photons interact with EM fields Slide 15 Slide 16

54 Assumption for classical treatment: Basically same arguments like beam-beam: No acceleration by longitudinal fields Lorentz contraction: beam-beam longitudinal positions must overlap High energy: interaction within bunch can be ignored: O( E γ 2 ) Only E fields needed: E + v B 2 E (remember lectures on relativity) - Basic formula: Luminosity in linear colliders From : L= N2 f nb 4πσxσy to : L= N2 frep nb 4πσxσy - Replace Revolution Frequency f by Repetition Rate frep. - And introduce effective beam sizesσx,σy : L= N2 frep nb 4πσx σy Slide 17 Slide 18

55 Final focusing in linear colliders β 1/2 [m 1/2 ] D [m] Diagnostics Energy collimation Transverse collimation Final Focus system βx βy 1/2 1/2 Longitudinal location [km] Dx (Courtesy R. Tomas) "Final Focus" and "Beam delivery System" At the end of the beam line only! Smallerβ in linear colliders (1 mm.2 mm!!) Why "effective beam sizes? Using the enhancement factor HD: L= N2 frep nb 4πσx σy L= H D N 2 frep nb 4πσx σy Enhancement factor HD takes into account reduction of nominal beam size by the disruptive field (pinch effect) Related to so-called disruption parameterd: Slide 19 Slide 11

56 For weak disruptiond 1and round beams: H D = π D+O(D2 ) For strong disruption and flat beams: computer simulation necessary, maybe can get some scaling Disruption plays a key role! Pinch effect - disruption beam-beam collision s Slide 111 Slide 112

57 Pinch effect - disruption beam-beam collision s Additional focusing by opposing beams Equation of motion (e.g. of the electrons): beam-beam collision z1 z s Coordinates of bunches (overlap of bunch centres at t = ) z2 = z1 + 2t Slide 113 Slide 114

58 Assuming N the number of particles in a beam,γthe "energy" (in terms of electron rest mass) and ρl(z) = ρ(x, y, z, t)dxdy the longitudinal density, the equation of motion for an electron is d 2 x dt 2 + 4Nr e γ ρl(z2) Φ x = ( 1 Φ = 2π ρt(x, y, z2, t) is the Poisson equation fbb = x x = Nr e γσ 2 ) andρt(x, y, z, t) is the transverse distribution which is obviously ρt(x, y, z, t) = ρ(x, y) ρl(z) For a round beamσ = σ x = σ y a solution is Φ(x, y) = r 1 e r2 /2σ 2 r dr with : r 2 = x 2 + y 2 The solutions for "flat" or "elliptic" beams can be found in the literature For the latter and near the centre of a Gaussian (what else) beam, we get Φ(x, y) = then, (see above) d 2 x + 4Nr e dt 2 γ x 2 2σ x (σ x +σ y ) ρ l (z) + x σ x (σ x +σ y ) y 2 2σ y (σ x +σ y ) = giving the beam-beam deflection angle (and focal length) x (rather) ẋ = 2Nr e γ x σ x (σ x +σ y ) = 1 f bb Slide 115 Slide 116

59 Disruption parameter Using the bunch lengthσz and the focal length due to the beam-beam interaction, one can define and write Dx,y = 2Nr ( e σz 1 γ σx,y(σx + σy) fbb = x x = Nr e γσ 2 The disruption parameter is defined as the ratio of the r.m.s. bunch length to the focal length of the interaction ) Equation of motion: Using this definition and for a uniform longitudinal density the final equation of motion becomes (near the axis, as said above) d 2 x dt 2 + D 3 σ 2 } {{ z } x = only valid in the range: 3σz 2 ω 2 < t < 3σ z 2 This gives rise to sinusoidal oscillations and for the number of oscillations in this range we get: 3D 2π Slide 117 Slide 118

60 Slide 119 Kink instability... Observed and relevant in (plasmas): Fusion reactors Astrophysics (sun, jets, pulsars,..) Treat as (relativistic) fluid... Assume again the beam as an elliptical cylinder a (σy σx) When the beams collide with an offset, e.g. y = (y1 y2) the equations of motion are (without derivation, see books on hydrodynamics): ( t ( t + s s )2 )2 y1= ω 2 (y1 y2) y2=+ω 2 (y 1 y2) Of course the right hand side of the equations represent the forces a warning: there are many different models Slide 12

61 The solution is an oscillation y1,2 = a1,2 e (ks ωt) where (ω k)2 ω 2 ω 2 ω 2 (ω+k) 2 ω 2 a 1 a2 = then ω 2 = k 2 +ω 2 ± 4ω 2 k2 +ω 4 The exponential growth if k < 2ω is the so-called "kink instability": In this mode the bunch (cylinder) wobbles along the longitudinal direction Side note: when instead the diameter changes, it is called "sausage" mode However: If the offset is much larger than the vertical beam size, i.e. σy y σx then we can approximate the equation of motion by: d 2 y dt 2 + π 6 σy Dy σ 2 z = The solution is obviously a parabola and the beam centres can overlap in time if y π 4 σ ydy i.e. for large disruption parameters the is no large loss of luminosity... Slide 121 Slide 122

62 Beamstrahlung Disruption at interaction point is basically a strong "bending" Results in strong synchrotron radiation: beamstrahlung This causes (unwanted): Spread of centre-of-mass energy Pair creation and detector background Again: luminosity is not the only important parameter Beamstrahlung... Bremsstrahlung/synchrotron radiation in the electro-magnetic fields of opposing bunch leads to emission of photons Fields are extremely strong: can be in kt range... Special: photons can produce pairs of leptons (and quarks!), i.e. QED and QCD background Slide 123 Slide 124

63 Beamstrahlung... For a relativistic flat beam (see before!) Ey(y, z) = q N 2 2πǫ σx er f ( y 2σ y ) ρ(z) assuming a Gaussian longitudinal profileρ(z): Ey = q N 4πǫ σxσz for the magnetic field we get: very small high fields B E y c Some numbers: N = e/b σ x = 7 nm σ y =.7 nm σ z = 35µm E y 1 12 V m B>3kT Slide 125 Slide 126

64 Critical field B c : Schwinger limit: for higher fields the field becomes non-linear: High enough energy: pair production fromγγ collisions The critical electric field is Ec = m2 e c 3 2π q h V/m and the corresponding critical magnetic field: Bc = E c = T Bc means onset of quantum regime... Beamstrahlung Parameter Y The parameter Y is defined as: Ec ωc Y 2 3 E = 2 3 E where E is the electron energy before radiating and Ec the photon energy corresponding to a bending radiusρ: Ec = 2 cγ 3 3 ρ The bending radiusρof the trajectory crossing a Gaussian bunch is not constant. Slide 127 Slide 128

65 Beamstrahlung Parameter Y Averaging E c over the bunch distribution one obtains (ignoring the pinching): Y=γ < E+B> 5 B c 6 reγ 2 N ασ z (σ x +σ y ) with: α = fine structure constant = Important numbers (again: Y is not constant during the collision): Y max 2 re 2γ N ασ z (σ x +σ y ) < Y> 5 12 Y max Note: bunch lengthσ z becomes a very important parameter! What about Y? Since Bc relevant for quantum regime: Low field, small Y: classical regime Typical storage ring (e.g. LEP): fields smaller than 1T (LEP.1 T) Y 1 High field, large Y: quantum regime Beamstrahlung: for large B (from collision) and highγ Y 1 (deep quantum regime) Slide 129 Slide 13

66 Number of photons and Energy Loss Average number of γ per electron (relevant for energy loss/spread): Nγ = 2.54 ασ zy λcγ U (Y) where U(Y) 1 (1+Y 2/3 ) 1/2 Note:λc is the (reduced) Compton wavelength: λc = me c = m) Number of photons and Energy Loss Average energy loss per electron: δe= E E 1.24 ασ z< Y> λcγ < Y> U1(< Y>) where U1(Y) 1 (1+(1.5Y) 2/3 ) 2 δe characterizes the spread in the centre-of-mass energy of the electrons/positrons Note: for Y the fractional energy loss can exceed 1 and the quantum correction U1(Y) is needed Slide 131 Slide 132

67 Functions U, U1 and Y U1 1 U.1 Y*U1.1 U Y Note: U is approximately 1. for not too large Y Y U 1 approximately constant for large Y δe! Some comments: You may have seen these formulae in a different form, just remember: re = m ǫ = Farad m 1 c m α = Slide 133 Slide 134

68 Differential Luminosity Beamstrahlung changes the energy spectrum of electrons (positrons), in particular because they can emit more than one photon: n γ > 1. Energy spectra for the two beams areψ(x 1, t) andψ(x 2, t). We define t = when the bunches first meet and a longitudinal variable z = at the front of each bunch. One writes for a "differential luminosity" the convolution integral: d 3 L(x 1, x 2, ) dx 1 dx 2 dz = 2 l l/2 dtψ(x 1, t)ψ(x 2, ) where l is the total bunch length and l 2 the total collision "time Note: the first "z" slice of beam 1 always sees an unperturbed (fresh) beam 2 As time (and z) evolve, a slice in z of beam 1 sees at a time t = z/2: d 3 L(x1, x2, z) dx1 dx2 dz = 2 l l/2 dtψ(x1, t) ψ(x2, z/2) after adding the z-slices for beam 1: d 2 L(x1, x2) dx1 dx2 = 4 l l/2 ψ(x1, t) dt l/2 ψ(x2, z) dz Slide 135 Slide 136

69 Effective particle energy The centre-of-mass energy for two colliding particles is always expressed as s, the square of the energy: s = (E1 + E2) 2, i.e. in the usual case s = 4E and normalized to the reference energy E it becomes s = 4 It is not difficult to show that for the effective energy one has: se f f x1x2 Provided a model forψfor the two beams is available, one can compute (probably numerically) the differential luminosity as a function of s, i.e. of the effective centre-of-mass energy: dl(s) ds = L 1 s 1 δ(s x1 x2) ψ(x1)ψ(x2) dx1 dx2 The differential luminosity d 2 L(x1, x2) dx1 dx2 = 4 l l/2 ψ(x1, t) dt l/2 ψ(x2, z) dz will normally have a sharp peak at the nominal energies and strong contributions from particles with full (or almost full) energy and particles with energies degraded by the beamstrahlung. Slide 137 Slide 138

70 How to keep beamstrahlung and side effects small? Some options: Shorter bunches... Flat beams... For flat bunches with a ratio R = σx/σy we can reduce δe R/(1+R) 2. For a fixed areaσxσy the luminosity would not be reduced Not mentioned: "Flip-flop", special feature for strong-strong interactions Asymmetric beams Coasting beams Synchrobetatron coupling Monochromatization... and many more Slide 139 Slide 14

71 Slide 141 BACKUP SLIDES 1 S X x S Y x S alternating x NO long range 1/3 long range 2/3 long range 3/3 long range Slide 142

72 Slide 143 Slide 144 S X x S Y x S alternating x beam beam eigenmodes head on : Ψ1 long range : Ψ2

73 Long range modes, 1σ x Fourierspectrumofcoherentmodes,fromHFMM Slide 145 Slide 146

74 Long range modes, 6σ x Fourierspectrumofcoherentmodes,fromHFMM Long range modes, 1σ x Fourierspectrumofcoherentmodes,fromHFMM Slide 147 Slide 148

75 Long range modes, 6σ x Fourierspectrumofcoherentmodes,fromHFMM Results: n oscillation amplitude 1 -<x(n)> 2 1 +<x(n)> n ε x εx ( 1) εx ( 2) Two beams have slightly different tunes Modes are damped, but emittance grows Slide 149 Slide 15

76 Multiple interaction points: Important: phase advance between interactions LHC has two independent rings Options to cancel coherent interaction at IPs: Different working points= Phase advance redistribution (precision!) Integer tunesplit Small tune split and same working point Phase advance redistribution n <y(n)> V 6.1 V 6.1p Local phase change of.2 Slide 151 Slide 152

77 2.2 2 Phase advance redistribution n (ε x +ε y )/2 V 6.1 V 6.1p Local phase change of.2 Coherentmodespectrum,softGaussianapproach Slide 153 Slide 154

78 Coherentmodes,HybridFastMultipoleMethod BACKUP SLIDES Slide 155 Slide 156

79 Tools and methods: Analysis and purpose: Evaluate stability of solution Calculate frequency spectra of oscillations Tools: Identify discrete spectral lines of oscillations Numerical integration of Vlasov equation Multi-particle and multi-bunch tracking Perturbation theory Numerical integration Vlasov equation is Partial Differential Equation Aim: find distribution and its time evolution Use: numerical integration with Finite Difference Methods Basic concept: Replace derivative by finite differences Represent continuous function ψ(x, t) by two-dimensional grid u n j (t n, x j) Slide 157 Slide 158

80 Example 1: δu u δt =λδ2 δx 2, u(x, t=)=u j = f (x) becomes : u n+1 j u n j t =λ un j+1 2un j + un j 1 x 2 = u n+1 j = u n j+ λ t x 2 (un j+1 2u n j+ u n j 1) spacial step x: x j= j x time step t for t: tn= n t initial distribution: u j = f (x j) Example 2: δ 2 u u δt 2 =λ2δ2 δx 2, u(x, t=)=u j= f (x) becomes : u n+1 j 2u n j + un j 1 t 2 =λ 2 un j+1 2un j + un j 1 x 2 = u n+1 j = 2(1 Exercise try to solve: ( λ t x )2 )u n j+ ( λ t x )2 δ 2 u δt 2 = u 4δ2 δx2, < x<1, u(x, )=sin(πx) (u n j+1+ u n j 1) u n 1 j Slide 159 Slide 16

81 Strategy for numerical integration Use finite difference scheme Back to our problem which looks like (per beam): ψ(x, px; t) t = A(x; t) ψ(x, p x; t) px B(px) ψ(x, p x; t) x In each substep integrate in one direction (operator splitting): ψ(x, px; t) t ψ(x, px; t) t = A(x; t) ψ(x, p x; t) px = B(px) ψ(x, p x; t) x, Discretiseψ(x, px; t) on the grid (i,j,n): U n i j (n=t/ t) Grid calculations: Each substep equation is of the type: δu with: f (u)=a(x) u δt=λ δ f (u) δpx u n i j is the discretisation at time t=n t of the densityψ(x, p x; t) for x=i x and y= j px Use the Lax-Wendroff scheme which looks like (e.g. first half-step in j-direction): u n+1/2 i j = u n i j A(x; t) 2 t (u n i p j+1 u n i j 1)+ 1 x 2 ( A(x; t) t p x Exercise try to derive (hint: Taylor expansion of u in t) )2 (u n i j+1 2u n i j+ u n i j 1) Slide 161 Slide 162

82 Grid calculations: Putting it all together with f= A(x)u and a similar half-step for g= B(px)u (now going the i-direction) one gets u n i j 1 un+ 2 i j ui j n+1 Small complication: in presence of discontinuities this method may generate oscillations Remedy: introduce artificial viscosity Strategy for simulations Represent bunches by macro particles (1 4 ) Track each particle individually around machine At interaction points evaluate force from other beam on each particle (that is where space charge effects are treated similar) In principle: for each particle in beam a calculate the integral overρ b (x, t) In practice not possible, unless one makes assumptions (e.g. Gaussian beams etc.), need other techniques Slide 163 Slide 164

83 FIELD COMPUTATION Solve Poisson equation for potentialφ(x, y) with charge distributionρ(x, y): ( ) 2 2 x2+ y 2 Φ(x, y)= 2πρ(x, y) formally possible with Green s function: Φ(x, y)= G(x x, y y )ρ(x, y)dx dy and (for open boundary): G(x x, y y )= 1 2 ln[(x x ) 2 + (y y ) 2 ] Techniques for field (force) calculation Soft Gaussian approximation: Assume Gaussian distribution with varying centre and width (fast but not precise, but o.k. for incoherent studies) Particle-particle methods: (precise but slow, typical: N = 1 4 ) Particle-mesh methods: evaluate field on a finite mesh (precise, but slow for separated beams) Hybrid Fast Multipole Methods (HFMM): recent method, precise and much better for separated beams and beam halos Slide 165 Slide 166

84 PARTICLE-PARTICLE METHODS (PP) Simple: accumulate forces by finding the force F(i,j) between particle i and particle j Problem: computational cost is O(Np) 2 For our problems typically Np 1 4 Used sometimes in astrophysics For Np < 1 3 and for close range dynamics good PARTICLE-MESH METHODS (PM) Approximate force as field quantity on a mesh. Differential operators are replaced by finite difference approximations. Particles (i.e. charges) are assigned to nearby mesh points (various methods). Problems: Computational cost is O(Ngln(Ng)) Bad to study close encounters Not ideal when mesh is largely empty (e.g. long range interactions) Slide 167 Slide 168

85 PARTICLE-MESH METHODS Main steps: Assign charges to mesh points (NG, TSC, CIC) Solve field equation on the mesh (many variants) Calculate force from mesh defined potential Interpolate force on grid (Ng Ng) to find force on particle PARTICLE ASSIGNMENT METHODS NGP: (Nearest Grid Point), densities at mesh points are assigned by the total amount of charge surrounding the grid point, divided by the cell volume. Drawback are discontinuous forces. CIC: (Cloud in cell), involve 2 K nearest neighbours, (K = dimension of the problem), give continuous forces. TSC: (Triangular Shaped Cloud), use assignment interpolation function that is piecewise quadratic. Slide 169 Slide 17

86 IN PRACTICE... Must look at: Stability Noise reduction (short scale fluctuations due to granularity) Number of particles Size of grid cells... A MORE RECENT APPROACH Fast Multipole Method (FMM) Derived from particle-particle methods, i.e. particles are not on a grid Tree code: treat far-field and short-field effects separately Relies on composing multipole expansions Computing cost: between O(Np) and O(Npln(Np)) Slide 171 Slide 172

87 Fast Multipole Method Well-known multipole expansion at point P for k point charges qi: Φ( r) = l= with : M m l = l m= l k i=1 Ml m r l+1 Ym l (θ,φ) qia l i Y m l (αi, βi) q1 q5 q2 a q3 q4 r P Order determined by desired accuracy Slide 173 Slide 174

88 FMM PROCEDURE Hierarchical spatial decomposition into small cells and sub-cells (e.g. quad-tree) Multipole expansion for each sub-cell Expansions in cells are combined to represent effect of larger and larger groups of particles A calculus is defined to relocate and combine multipole expansion Far-field effects are combined with near-field effects to give potential (and field) at every particle QUAD-TREE DIVISION Slide 175 Slide 176