Introduction to Longitudinal Beam Dynamics

Transcription

1 Introduction to Longitudinal Beam Dynamics Andrea Latina (CERN) Credits: Elias Métral, Daniel Schulte (CERN) 1 st SLRI-CERN ASEAN Accelerator School August 28 - Setember 1, 2017

2 Table of contents Part 1: Introduction ; units ; basics Part 2: Transit-time factor ; momentum comaction ; transition energy Part 3: Longitudinal hase sace ; Synchronous article and synchronous oscillations ; longitudinal stability Part 4: Synchrotron frequency ; transfer matrix for an accelerator structure ; rinciles of bunch comression 2/65 A. Latina - Introduction to longitudinal beam dynamics

3 Purose of this course Discuss the oscillations of the articles in the Longitudinal lane of synchrotrons, called SYNCHROTRON OSCILLATIONS (similarly to the betatron oscillations in the transverse lanes), and derive the basic equations 3/65 A. Latina - Introduction to longitudinal beam dynamics

4 Some references 1. Mario Conte, William W. MacKay, An Introduction to the Physics of Particle Accelerators, Second Edition, World Scientific, Andrzej Wolski, Beam Dynamics in High Energy Particle Accelerators, Imerial College Press, Thomas P. Wangler, RF Linear Accelerators, 2nd Edition, Wiley, Shyh-Yuan Lee, Accelerator Physics, World Scientific, Helmut Wiedemann, Particle Accelerator Physics, Sringer, 4th Edition, The CERN Accelerator School (CAS) Proceedings, e.g. 1992, Jyväskylä, Finland (LINK) 4/65 A. Latina - Introduction to longitudinal beam dynamics

5 Part 1. Introduction ; units ; basics 5/65 A. Latina - Introduction to longitudinal beam dynamics

6 Phase sace view IN REAL SPACE IN PHASE SPACE Horizontal Vertical Longitudinal, bunched beam, below transition Longitudinal, unbunched beam, below transition Courtesy of A.W. Chao δ = Δ Longitudinal, bunched beam, above transition Longitudinal, unbunched beam, above transition 6/65 A. Latina - Introduction to longitudinal beam dynamics

7 Some movies (in hase sace) to have a better idea of what we will work on during this course and what you will be able to understand and do after this course... 7/65 A. Latina - Introduction to longitudinal beam dynamics

8 Useful units quantity symbol SI units ractical units force F N (newton) MeV/m electric field E V/m (volt/meter) MV/m electric otential Φ V (volt) MV electric charge q C e (e + charge) momentum P N s MeV/c energy E J (joule) MeV 8/65 A. Latina - Introduction to longitudinal beam dynamics

9 Acceleration by time-varying electric fields E r r s Let V RF be the amlitude of the RF voltage across the ga g The ar9cle crosses the ga at a distance r The energy gain is: ΔE = e g 2 g 2! E ( s, r, t)! ds g In the cavity ga, the electric field is suosed to be: [MeV] [n] (1 for electrons or rotons) E( s, r, t) = E1( s, r) E2( t) [MV/m] In general, E 2 (t) is a sinusoidal 9me varia9on with angular frequency w RF E ( t) = E sin Φ 2! ( t) where Φ( t) = t ω RF dt + Φ0 t 0 9/65 A. Latina - Introduction to longitudinal beam dynamics

10 Convention 1. For circular accelerators, the origin of time is taken at the zero crossing of the RF voltage with ositive sloe 2. For linear accelerators, the origin of time is taken at the ositive crest of the RF voltage Time t=0 chosen such that: 1 E 2 2 E 2 f = w RF t f = w RF t f 1 f 2 E ( t) = E sin 2! ( w t) RF E ( t) = E cos 2! ( w t) RF 10/65 A. Latina - Introduction to longitudinal beam dynamics

11 Relativistic equations 2 E = mc normalized velocity energy v β = = c 1 1 γ 2 total E = E + kin E 0 rest kinetic γ = E E = m m total energy rest energy = v c = 1 2 β momentum E = mv = β = β γ m0 c c energy ev momentum ev c mass ev 2 c 2 c = E E γ = 1 + E kin E [ GeV/c] 0.3 B[ T] ρ [ m] 0 11/65 A. Latina - Introduction to longitudinal beam dynamics

12 1 normalized velocity electrons v β = = c 1 1 γ 2 Beta 0.5 rotons E_kinetic (MeV) γ = E E = m m total energy rest energy = v c = 1 2 β Gamma electrons rotons E_kinetic (MeV) 12/65 A. Latina - Introduction to longitudinal beam dynamics

13 First derivatives dβ = β 1 γ 3 dγ d 3 ( c ) = γ dβ E 0 2 ( 1 ) 3 dβ 2 dγ = β β Logarithmic derivatives d 2 β = β ( β γ ) dγ γ 2 d γ de γ de = = 2 γ 1 E γ + 1 E dβ ( γ ) β dγ = 2 1 γ kin kin 13/65 A. Latina - Introduction to longitudinal beam dynamics

14 Part 2. Synchrotrons Transit-time factor ; momentum comaction ; transition energy 14/65 A. Latina - Introduction to longitudinal beam dynamics

15 Synchrotron B R Synchronism condition T s = h T RF T s = L β = 2πR sc β sc h integer harmonic number E RF cavity RF generator 1. Note: both ω RF = 2π T RF and ω s = 2π T s increase with energy 2. Remember: to kee articles on the closed orbit, B must also increase with energy 15/65 A. Latina - Introduction to longitudinal beam dynamics

16 Synchrotrons B E Bending magnet R In reality, the orbit in a synchrotron is not a circle, straight sections are added for RF cavities, injection and extraction, etc.. injection extraction Usually the beam is re-accelerated in a linac (or a smaller synchrotron) before injection r E The bending radius r does not coincide to the machine radius R = L/2 16/65 A. Latina - Introduction to longitudinal beam dynamics

17 EPA (CERN) Electron Positron Accumulator LEAR (CERN) Low Energy Antiroton Ring CERN Geneva CERN Geneva CERN Geneva Examles of different roton and electron synchrotrons at CERN PS (CERN) Proton Synchrotron 17/65 A. Latina - Introduction to longitudinal beam dynamics CERN Geneva

18 Parameters for circular accelerators The basic rinciles, for the common circular accelerators, are based on the two relations: 1. The Lorentz equation: the orbit radius can be exressed as: R = P q B = m 0βγc q B 2. The synchronicity condition: The revolution frequency can be exressed as: f = 1 T s = βc 2πR = qb 2πm 0 γ According to the arameter we want to kee constant or let vary, one has different acceleration rinciles. They are summarized in the table below: Machine Energy (g) Velocity Field Orbit Frequency Cyclotron ~ 1 var. const. ~ v const. Synchrocyclotron var. var. B(r) ~ B(r)/g(t) Proton/Ion synchrotron var. var. ~ R ~! Electron synchrotron var. const. ~ R const. 18/65 A. Latina - Introduction to longitudinal beam dynamics

19 Transit-time factor The RF acceleration over a ga of length g is: E (s, r, t) = E 1 (s, r) E 2 (t) [V/m] and the energy gain E The transit-time factor T a is the ratio: g/2 E = q E (s, r, t) ds. g/2 T a = energy gained in the time-varying RF field or T a= E energy gained in a DC field of voltage V q V RF RF In the general case, g/2 ( ) s E = E 1 (s, r) cos ω RF + φ 0 ds g/2 v z g/2 V RF = E 1 (s, r) cos (φ 0 ) ds g/2 (energy gain) (DC voltage) and the transit-time factor is T a = 19/65 A. Latina - Introduction to longitudinal beam dynamics ( ) g/2 E g/2 E s 1 (s, r) cos ω RF ds v z = q V RF q g/2 g/2 E 1 (s, r) ds

20 Transit-time factor: an examle A simlified model is tyically used to comute it, where: E 1 (s, r) = V RF g E 2 (t) = cos (ω RF t + φ 0) [V/m] accelerating gradient, constant hase modulation In this case, E = q V RF g T a is the transit-time factor, defined as: g/2 g/2 ( ) s cos ω RF + φ 0 ds = q V RF T a v z T a = E q V RF If v z 0, and arallel to the electric field, T a = sin ω RFg 2vz ω RF g 2vz ( ) ωrf g sinc 2v z Note: T a is always 1 if g 0 then T a 1 (function sinc x := sin x x, is called cardinal sin ) 20/65 A. Latina - Introduction to longitudinal beam dynamics

21 Main RF arameters I. Voltage, hase, frequency In order to accelerate articles, longitudinal fields must be generated in the direction of the desired acceleration E s, t) = E ( s) E ( ) ( 1 2 t ω RF = 2 π f RF t E é ù 2( t) = E0 sin êë ò wrf dt + f0 t0 úû DE = e V RF T a sin f 0 Such electric fields are generated in RF cavities characterized by the voltage amlitude, the frequency and the hase II. Harmonic number T = h T Þ f = h rev RF RF f rev f rev f RF h = revolution frequency = frequency of the RF = harmonic number harmonic number in different machines: AA PS SPS LHC (AA) Antiroton Accumulator - (PS) Proton Synchrotron - (SPS) Suer Proton Synchrotron - (LHC) Large Hadron Collider 21/65 A. Latina - Introduction to longitudinal beam dynamics

22 Disersion 22/65 A. Latina - Introduction to longitudinal beam dynamics

23 Momentum comaction factor in a transort system In a article transort system, a nominal trajectory is defined for the nominal momentum. For a article with a momentum + the trajectory length can be different from the length L of the nominal trajectory. The momentum comaction factor is defined by the ratio: dl a = L d Therefore, for small momentum deviation, to first order it is: DL D = a L 23/65 A. Latina - Introduction to longitudinal beam dynamics

24 Examle: constant magnetic field dq ds = r dq ds + ( r ) dq 1 = x r ds x s + ds1 - ds = ds ( r + x) dq - r dq x D = = r dq r r x d By definition of disersion D x ds 1 α = 1 L 0 L D x (s) ρ(s) ds To first order, only the bending magnets contribute to a change of the trajectory length (! = in the straight sections) 24/65 A. Latina - Introduction to longitudinal beam dynamics

25 Momentum comaction in a ring In a circular accelerator, a nominal closed orbit is defined for the nominal momentum. For a article with a momentum deviation roduces an orbit length variation C with: For B = const. dc a = C d = DC D = a C The momentum comaction factor is defined by the ratio: dr R d and circumference C = 2 R 1 a = C ò C (average) radius of the closed orbit Dx( s) ds r( s) N.B.: in most circular machines, a is ositive Þ higher momentum means longer circumference 25/65 A. Latina - Introduction to longitudinal beam dynamics

26 Transition energy Proton (ion) circular machine with a ositive 1. Momentum larger than the nominal ( + D ) Þ longer orbit ( C+ D C ) 2. Momentum larger than the nominal ( + D ) Þ higher velocity ( v + D v ) What haens to the revolution frequency f = v/c? At low energy, v increases faster than C with momentum At high energy v c and remains almost constant There is an energy for which the velocity variation is comensated by the trajectory variation Þ transition energy Below transition: higher energy Þ higher revolution frequency Above transition: higher energy Þ lower revolution frequency 26/65 A. Latina - Introduction to longitudinal beam dynamics

27 Transition energy The frequency of a article revolving in a synchrotron is f = 1 T = βc C where T is the revolution eriod, and C = 2πR is the circumference orbit. Differentiating ln (f ) yields df f = dt T = dβ β dc C = ( 1 γ 2 α ) dp P The exression in arentheses is usually written as ( ) 1 η tr = γ 2 α = 1 γ 1 2 γtr 2 and is called the hase sli factor. Note that there is a transition, at η tr = 0, when γ = γ tr = 1 α. We identify two regimes: η tr > 0 : below transition, when E < E tr; for strong focusing synchrotrons and linacs (in linacs, α = 0) η tr < 0 : above transition, when E > E tr; for weak focusing machines and for strong focusing ones (or for electron machines, where E E tr) Having defined the transition energy E tr = m 0γ trc 2. 27/65 A. Latina - Introduction to longitudinal beam dynamics

28 Transition energy 28/65 A. Latina - Introduction to longitudinal beam dynamics

29 Equations related to synchrotrons d 2 = g tr dr db + R B d 2 2 df = g +g f db = g tr B 2 df f dr R é æ g tr ö + ê1 - ç êë è g ø B 2 df = g + - B f 2 2 dr ( g g ) R d 2 tr ù d ú ú û [ MeV/c] R [ m] B [ T] f g tr [ Hz] momentum orbit radius magnetic field rev. frequency transition energy 29/65 A. Latina - Introduction to longitudinal beam dynamics

30 I - Constant radius d R = 0 Beam maintained on the same orbit when energy varies d db = B d 2 = g df f If increases B increases f increases 30/65 A. Latina - Introduction to longitudinal beam dynamics

31 II - Constant energy d = 0 V RF = 0 Beam debunches d 2 = 0 = g tr dr db + R B d 2 2 df = 0 = g +g f dr R If B increases R decreases f increases 31/65 A. Latina - Introduction to longitudinal beam dynamics

32 III - Magnetic flat-to d B = 0 Beam bunched with constant magnetic field d 2 = g tr dr R db = 0 = g tr B 2 df f é æ g tr ö + ê1 - ç êë è g ø db B = 0 = γ df 2 + γ 2 2 ( γ tr ) dr f R 2 ù d ú ú û If increases R increases f increase decreases g < g tr g > g tr 32/65 A. Latina - Introduction to longitudinal beam dynamics

33 IV - Constant frequency d f = 0 Beam driven by an external oscillator d 2 = g dr R 2 db é æ g tr ö ù d = ê1 - ç ú B ê ú ë è g ø û db B = γ 2 2 γ tr ( ) dr R If increases R increases B decreases increase g < g tr g > g tr 33/65 A. Latina - Introduction to longitudinal beam dynamics

34 Summary of the four conditions Beam Parameter Variations Debunched D = 0 B Ý, R ß, f Ý Fixed orbit DR = 0 B Ý, Ý, f Ý Magnetic flat-to DB = 0 Ý, R Ý, f Ý (h > 0) f ß (h < 0) External oscillator Df = 0 B Ý, ß, R ß (h > 0) Ý, R Ý (h < 0) R B f momentum orbit radius magnetic field frequency 34/65 A. Latina - Introduction to longitudinal beam dynamics

35 Part 3. Longitudinal hase sace ; synchronous article and synchronous oscillations ; longitudinal stability 35/65 A. Latina - Introduction to longitudinal beam dynamics

36 Longitudinal hase sace D/ acceleration D/ move forward move backward reference deceleration f f The article trajectory in the hase sace (f, D/) describes its longitudinal motion Emittance: area of the hase sace including nearly all articles NB: if the emittance contour corresonds to a ossible orbit in hase sace, its shae does not change with time (matched beam) 36/65 A. Latina - Introduction to longitudinal beam dynamics

37 Synchronous article - no acceleration Simle case (no accel.): B = const. g < g tr Synchronous article: article that sees always the same hase (at each turn) in the RF cavity V RF e B w w = = g m h 0 RF ΔE = e ˆ V RF sinφ f 0 f 1 f = w RF t In order to kee the resonant condition, the article must kee a constant energy The hase of the synchronous article must therefore be f 0 = 0 (circular machines convention) Let s see what haens for a article with the same energy and a different hase (e.g., f 1 ) 37/65 A. Latina - Introduction to longitudinal beam dynamics

38 Synchronous oscillations - no acceleration f 1 - The article is accelerated - Below transition, an increase in energy means an increase in revolution frequency - The article arrives earlier tends toward f 0 V RF f 2 f 0 f 1 f = w RF t f 2 - The article is decelerated - decrease in energy - decrease in revolution frequency - The article arrives later tends toward f 0 38/65 A. Latina - Introduction to longitudinal beam dynamics

39 Synchronous oscillations - hase sace - no acceleration V RF f 2 f 0 f 1 φ = ω RF t Phase sace icture D f 39/65 A. Latina - Introduction to longitudinal beam dynamics

40 Synchronous article - acceleration Case with acceleration B increasing g < g tr V RF f s t f = w RF sinf DE = evˆrf The hase of the synchronous article is now f s > 0 (circular machines convention) The synchronous article accelerates, and the magnetic field is increased accordingly to kee the constant radius R g v m0 R = The RF frequency is increased as well in order to kee the resonant condition eb e B w w = = g m h 0 RF 40/65 A. Latina - Introduction to longitudinal beam dynamics

41 Synchronous oscillations - acceleration V RF 1 2 f s t f = w RF 41/65 A. Latina - Introduction to longitudinal beam dynamics

42 Synchronous oscillations - hase sace - acceleration V RF f t f = w RF f s The symmetry of the case with B = const. is lost D f < f < - s f s stable region f unstable region searatrix 42/65 A. Latina - Introduction to longitudinal beam dynamics

43 Part 4. Synchrotron frequency ; transfer matrix for an accelerator structure ; rinciles of bunch comression 43/65 A. Latina - Introduction to longitudinal beam dynamics

44 RF acceleration of the synchronous article To accelerate the synchronous article we need to have φ s > 0. The energy gain is then which corresonds to an increment of P by E = q ˆV RF sin φ s, P = E βc. Let s see how this relates with the rate of increase of B. The stability of the closed orbit wants that ρ = const in Bρ = P q then Over one turn db dt = 1 1 dp ρ q dt Ḃ = P t 1 qρ P = ( E) turn βc t = T rev = 2πR βc Thus 44/65 A. Latina - Introduction to longitudinal beam dynamics Ḃ = 1 q ( E) turn 2πR ρ [T/s]

45 RF acceleration of the synchronous article 1. Knowing the rate of increase of Ḃ, one can calculate the magnetic field rate needed for a given RF voltage and synchronous hase: Ḃ = ˆV RF sin φ s, 2πR ρ [T/s] as ( E) turn = q ˆV RF sin φ s 2. Knowing the magnetic field variation and the RF voltage, one can calculate the value of the synchronous hase: sin φ s = 2πRρ ( Ḃ ˆV RF φ s = arcsin 2πRρ ) Ḃ ˆV RF 3. Note that, knowing the rate of increase of Ḃ, one can calculate the necessary energy gain er turn: ( E) turn = qρ 2πR Ḃ [MeV] 45/65 A. Latina - Introduction to longitudinal beam dynamics

46 RF acceleration of the synchronous article Knowing the magnetic field B, one can calculate the required RF frequency 1. The RF frequency must satisfy: 2. We know that B (t) ρ = P (t) q m (t) β (t) c = q ω RF = h ω s thus β (t) = q m(t)c B (t) ρ γ (t) β (t) = q m 0 c B (t) ρ 3. From ω s (t) = 2π β(t)c 2πR write 4. Calling B 0 = m 0 c qρ, = β(t)c R, and from m (t) = m0γ (t) = m0 1 + γ (t) 2 β (t) 2, we can ω s (t) = c R ω s (t) = q ρ B (t) m 0 R 1 + γ (t) 2 β (t) 2 B (t) /B 0 = c B (t) /B γ (t) 2 β (t) 2 R 1 + (B (t) /B 0) 2 One obtains: f RF = hc 2πR B (t) /B (B (t) /B 0) 2 46/65 A. Latina - Introduction to longitudinal beam dynamics

47 Examle: PS At the CERN Proton Synchrotron machine, one has: R =100 m!b = 2.4 T/s 100 dioles with l eff = m. The harmonic number is 20 Calculate: 1. The energy gain er turn 2. The minimum RF voltage needed 3. The RF frequency when B = 1.23 T (at extraction) 47/65 A. Latina - Introduction to longitudinal beam dynamics

48 RF acceleration for non synchronous articles Parameter definition (subscrit s stands for synchronous article): f = f s + Df f = f + Df s = s + D E = E + DE q = q + Dq s s revolution frequency RF hase Momentum Energy Azimuth angle ds = R dq t ( ) = ω ( τ ) θ t t 0 dτ 48/65 A. Latina - Introduction to longitudinal beam dynamics

49 Azimuthal angle vs RF hase v V RF VˆRF Dq q s R f f s Df t f = w RF Dq > 0 Þ Df < 0 Since f RF = h f rev Df = -h Dq Over one turn q varies by 2 f varies by 2 h 49/65 A. Latina - Introduction to longitudinal beam dynamics

50 Parameters vs φ 1. Angular frequency t ( ) = ω ( τ ) θ t t 0 dτ d Dw = dt 1 d = - h dt 1 d = - h dt 1 df = - h dt ( Dq ) ( Df ) ( f -f ) s d dt f s = 0 by definition Dw = - 1 df h dt 50/65 A. Latina - Introduction to longitudinal beam dynamics

51 Parameters vs φ 2. Momentum 3. Energy D D = = w w w w h d d ø ö ç è æ - = D = D t h s s s s d d 1 f w h h w w t h s s d df w h - = D R v E = w = D D v E = d d t h R E s d df h = - D 51/65 A. Latina - Introduction to longitudinal beam dynamics

52 Derivation of the equation of motion Energy gain after the RF cavity Average increase er time unit ( DE ) = e sinf turn VˆRF e w R ( D ) = sinf turn VˆRF ( D ) T rev turn e = 2 R Vˆ RF sinf R = e V ˆ sinf 2! valid for any article! RF ( R! - R! ) = e Vˆ ( sinf sinf ) 2 - s s RF s 52/65 A. Latina - Introduction to longitudinal beam dynamics

53 Derivation of the equation of motion After some develoment (see J. Le Duff, in Proceedings CAS 1992, CERN 94-01) An aroximated version of the above is d æ DE ö 2 = e Vˆ RF - dt ç w è s ø ( D) d dt e Vˆ RF = 2 R s ( sinf sinf ) ( sinf - sinf ) s s Which, together with the reviously found equation df wsh = - D dt Describes the motion of the non-synchronous article in the longitudinal hase sace ( D, f ) s 53/65 A. Latina - Introduction to longitudinal beam dynamics

54 Solution of the equation of motion for small amlitudes After some develoment (see J. Le Duff, in Proceedings CAS 1992, CERN 94-01) For h cos > 0 f s the motion around the synchronous article is a stable oscillation: ( ) ( ) Δφ = Δφ max sin Ω sync t +φ 0 Δ = Δ max cos Ω sync t +φ 0 Δ max = Ω sync with B Δφ max and Ω 2 0 = Ω 2 sync = e V ˆ RF η hc 2 cosφ s 2π R 2 s E s h B = - s b sc R! max ="-! s s 54/65 A. Latina - Introduction to longitudinal beam dynamics

55 Synchrotron angular frequency and synchrotron tune for small amlitudes Ω sync = ω s e ˆ V RF h 2π β 2 E s η cosφ s Ω sync = 2 π f sync ω s = 2 π f s Number of synchrotron oscillations er turn: Q sync = Ω sync ω s = e ˆ V RF h 2π β 2 E s η cosφ s synchrotron tune 55/65 A. Latina - Introduction to longitudinal beam dynamics

56 Phase sace searatrix and article trajectories (Bucket) searatrices: Below transition Above transition ϕ / Ω ϕ / Ω ϕ [ ] - ϕ [ ] φ s = 0 φ s = 30 φ s = 60 φ s = 85 φ s π φ s φ s =180 φ s =150 φ s =120 φ s = 95 Equation of the searatrix 56/65 A. Latina - Introduction to longitudinal beam dynamics

57 Phase sace searatrix and article trajectories Particle trajectories: Below transition φ s = 0 φ s = 30 ϕ / Ω ϕ / Ω ϕ [ ] ϕ [ ] 57/65 A. Latina - Introduction to longitudinal beam dynamics

58 Longitudinal article tracking The motion of the articles can be tracked turn by turn using the recurrence relation (between turn n and turn n+1) ΔE n+1 = ΔE n + e ˆ V RF φ n+1 = φ n 2 π h η β s 2 E s ΔE n+1 [ sin φ n sinφ s ] 58/65 A. Latina - Introduction to longitudinal beam dynamics

59 Transfer matrix for an accelerating structure (linacs) Let s find the full 6 6 transfer matrix: x x y y z P P 0 = s?? ?? ?? ?? ?? ?? x x y y z P P 0 0 To satisfy the Maxwell s equations, we need to slit this matrix in three arts: x x x ( ) ( ) ( ) x y y exit body entrance y = end field (acceleration) end field y z z P P 0 L P P 0 0 where the end fields are thin elements. 59/65 A. Latina - Introduction to longitudinal beam dynamics

60 Transfer matrix for an accelerating structure: body Assume a homogeneous longitudinal RF electric field in the structure, G; assume a structure of length L. The energy gain is E = q G cos (ω RF t + φ 0 ) L }{{} Ĝ, effective gradient Let s define the normalized energy gain δ = E E 0 = q Ĝ L E 0. The longitudinal art is: therefore since E 0 E 0 + E 1 1+δ. ( P P 0 z (s) = z (0) ) E 0 (s) = E 0 + E ( P ( ) 1 0 M z = δ P 0 ) (0) 60/65 A. Latina - Introduction to longitudinal beam dynamics

61 Transfer matrix for an accelerating structure: body The transverse angles can be found recalling the conservation of the transverse momentum x (s) = x P 0 (0) P 0 + q Ĝ x E 0 (0) s/vz E 0 + q Ĝ s The transverse ositions can be found integrating the above exression over s, from 0 to L: ( ) log 1 + q Ĝ L E 0 x (L) = x (0) + x (0) q Ĝ E 0 Using the normalized energy gain, δ, one obtains: ( ) 1 L log(1+δ) M x,y = δ δ 61/65 A. Latina - Introduction to longitudinal beam dynamics

62 Transfer matrix for an accelerating structure: end fields Recall the Gauss law: Φ total = Q ɛ 0 = 0 where Φ total is the total flux of the electric field, and Q is the total charge inside the cylinder, in our case Q = 0. The total flux is Φ total = Φ + Φ, with Φ = G πr 2 Φ = G 2πr L Since Φ total = 0, then Φ = Φ and G = G r 2 L The transferred transverse energy is E = q G L = q G r and the entrance transverse kick is 2 then: r = E = q G r = δ r E 0 2E 0 2L (the exit kick must be divided by (1 + δ)) This alies to both x and y: which corresonds to a focusing effect. 62/65 A. Latina - Introduction to longitudinal beam dynamics M x,y = ( ) 1 0 δ 2L 1

63 Transfer matrix for an accelerating structure: summary The full 6 6 transfer matrix is: M = δ 2L(1+δ) δ 2L(1+δ) L log(1+δ) δ δ L log(1+δ) δ δ δ δ 2L δ 2L where δ is the the normalized energy gain δ = E/E 0. 63/65 A. Latina - Introduction to longitudinal beam dynamics

64 Princiles of bunch comression: chicane comression 64/65 A. Latina - Introduction to longitudinal beam dynamics

65 ...The End! Thank you for your attention! 65/65 A. Latina - Introduction to longitudinal beam dynamics