Introduction to Accelerator Physics 2011 Mexican Particle Accelerator School

Transcription

1 Introduction to Accelerator Physics 20 Mexican Particle Accelerator School Lecture 3/7: Quadrupoles, Dipole Edge Focusing, Periodic Motion, Lattice Functions Todd Satogata (Jefferson Lab) Thursday, September 29, 20 T. Satogata / Fall 20 MePAS Intro to Accel Physics

2 MePAS Accelerator Physics Syllabus -2: Wednesday Relativity/EM review, coordinates, cyclotrons Weak focusing, transport matrices, dipole magnets, dispersion 3: Thursday Edge focusing, quadrupoles, accelerator lattices, start FODO 4: Friday Periodic lattices, FODO optics, emittance, phase space 5: Saturday Insertions, beta functions, tunes, dispersion, chromaticity 6: Monday Dispersion suppression, light source optics (DBA, TBA, TME) 7: Tuesday (Nonlinear dynamics), Putting it all together T. Satogata / Fall 20 MePAS Intro to Accel Physics 2

3 Parameterizing Particle Motion: Approximations coordinate system We have specified a coordinate system and made a few reasonable approximations: 0) No local currents (beam in a near-vacuum) ) Paraxial approximation: 2) Perturbative coordinates: 3) Transverse linear B field: 4) Negligible E field: T. Satogata / Fall 20 MePAS Intro to Accel Physics 3

4 Review Drift transport matrix: Dipole transport matrix without focusing: Dipole horizontal transport matrix including focusing and dispersion: T. Satogata / Fall 20 MePAS Intro to Accel Physics 4

5 Focusing Without Bending Quadrupole magnets have but No dipole field: design trajectory is straight Like taking in our previous analysis This is one reason why we changed our parameterizations from horizontal dipole vertical dipole normal quadrupole skew quadrupole T. Satogata / Fall 20 MePAS Intro to Accel Physics 5

6 d dθ = R d ds Quadrupole Equations of Motion d 2 x d nx =0 dθ2 2 y x + (Bρ) By dθ 2 + ny =0 n ρ B 0 (A2) By x =0 y x (Bρ) x + Kx =0 y Ky =0 K (Bρ) By x This is truly a simple harmonic oscillator when K is constant: for a quadrupole of length L focusing x By x T. Satogata / Fall 20 MePAS Intro to Accel Physics 6 y =0 [K] = [length] 2 x(l) cos(l K) x (L) K sin(l K) 0 0 y(l) = K sin(l K) cos(l K) cosh(l K) y K sinh(l K) (L) 0 0 K sinh(l K) cosh(l K) defocusing Thick quadrupole transport matrix Swap places when K goes to -K x 0 x 0 y 0 y 0

7 Thin Quadrupoles In most accelerator uses, we can take L->0 with KL constant Use small-angle approximation to rewrite as a thin quadrupole x(l) x 0 x (L) y(l) = KL 0 0 x y 0 y (L) 0 0 KL y0 This is just like a lens in classical optics with a focal length x(l) x 0 x (L) y(l) = /f 0 0 x y 0 y (L) 0 0 /f y0 Thin quadrupole transport matrix Swap places when K goes to -K f KL T. Satogata / Fall 20 MePAS Intro to Accel Physics 7

8 Picturing Drift and Quadrupole Motion KL=0.5 m -, f=2m KL=0. m -, f=20m KL=20 m -, f=0.05m KL=-0. m -, f=-20m T. Satogata / Fall 20 MePAS Intro to Accel Physics 8

9 Picturing Drift and Quadrupole Motion KL=0.5 m -, f=2m KL=0. m -, f=20m KL=20 m -, f=0.05m KL=-0. m -, f=-20m Thin Quadrupole Approximations T. Satogata / Fall 20 MePAS Intro to Accel Physics 9

10 Dipole Edge Focusing Quadrupoles are not the only place we get focusing! Recall our 3x3 sector dipole matrix Vertical motion is just a drift of length L = ρθ But this magnet is curved and therefore not easy to build In particular, the ends are tilted to be to design trajectory T. Satogata / Fall 20 MePAS Intro to Accel Physics 0

11 Sector and Rectangular Bends Sector bend (sbend) Beam design entry/exit angles are to end faces Simpler to conceptualize, but harder to build Rectangular bend (rbend) Beam design entry/exit angles are half of bend angle Easier to build, but must include effects of edge focusing T. Satogata / Fall 20 MePAS Intro to Accel Physics

12 Dipole End Angles +x displaced particle enters B field later than design trajectory particle Design trajectory particle -x displaced particle enters B field earlier than design trajectory particle Different transverse positions see different B field! Particles displaced by +x see B field later than design Particles displaced by x see B field earlier than design T. Satogata / Fall 20 MePAS Intro to Accel Physics 2

13 Dipole End Angles We treat general case of symmetric dipole end angles Superposition: looks like wedges on end of sector dipole Rectangular bends are a special case T. Satogata / Fall 20 MePAS Intro to Accel Physics 3

14 Kick from a Thin Wedge The edge focusing calculation requires the kick from a thin wedge What is L? (distance in wedge) Quadrupole-like defocusing term, linear in position T. Satogata / Fall 20 MePAS Intro to Accel Physics 4

15 Dipole Matrix with Ends The matrix of a dipole with thick ends is then Rectangular bend is special case where α=θ/2 T. Satogata / Fall 20 MePAS Intro to Accel Physics 5

16 What About Vertical Edge Focusing? Side view N Overhead view, α>0 S Field lines go from y to +y for a positively charged particle B x <0 for y>0; B x >0 for y<0 Net focusing! Field goes like sin(α) get cos(α) from integral length Quadrupole-like focusing y = ( B xy sin α/l)(l/ cos α) (Bρ) = tan α ρ y T. Satogata / Fall 20 MePAS Intro to Accel Physics 6

17 ========== Almost There ========== T. Satogata / Fall 20 MePAS Intro to Accel Physics 7

18 Matrix Example: Strong Focusing Consider a doublet of thin quadrupoles separated by drift L f D f F M doublet = 0 f D L 0 0 f F = L f doublet f D f F f F f D f D = f F = f = L f doublet f 2 There is net focusing given by this alternating gradient system A fundamental point of optics, and of accelerator strong focusing = L f F f D f F L f F f D L + L f D T. Satogata / Fall 20 MePAS Intro to Accel Physics 8

19 Strong Focusing: Another View f F f D M doublet = 0 f D incoming paraxial ray L 0 x x 0 f F For this to be focusing, x must have opposite sign of x Equal strength doublet is net focusing under condition that each lens s focal length is greater than distance between them = x0 = M doublet 0 L f F f D f F L f F f D f F = f D x < 0 BUT x>0iff f F >L = L f F f D f F L f F f D L + L f D x 0 T. Satogata / Fall 20 MePAS Intro to Accel Physics 9

20 Strong Focusing Homework f D f F The previous argument also works when the defocusing quadrupole comes before the focusing quadrupole Homework: Calculate the net focusing condition for this system Since quadrupoles focus in one plane and defocus in the other, alternating quadrupoles continuously produces a system that is overall net focusing and stable Horizontal Vertical F O D O FODO lattice: Periodic! T. Satogata / Fall 20 MePAS Intro to Accel Physics 20

21 More Math: Hill s Equation Let s go back to our equations of motion for x + Kx =0 y Ky =0 K (Bρ) What happens when we let the focusing K vary with s? Also assume K is periodic in s with some periodicity C x + K(s)x =0 K(s) (Bρ) By (s) x This periodicity can be one revolution around the accelerator or as small as one repeated cell of the layout (Such as a FODO cell in the previous slide) R By x K(s + C) =K(s) The simple harmonic oscillator equation with a periodically varying spring constant K(s) is known as Hill s Equation T. Satogata / Fall 20 MePAS Intro to Accel Physics 2

22 Hill s Equation Solution Ansatz x + K(s)x =0 K (Bρ) By x Solution is a quasi-periodic harmonic oscillator x(s) =Aw(s) cos[φ(s)+φ 0 ] (s) where w(s) is periodic in C but the phase φ is not!! Substitute this educated guess ( ansatz ) to find x = Aw cos[φ + φ 0 ] Awφ sin[φ + φ 0 ] x = A(w wφ 2 ) cos[φ + φ 0 ] A(2w φ + wφ )sin[φ + φ 0 ] x + K(s)x = A(2w φ + wφ )sin(φ + φ 0 )+A(w wφ 2 + Kw) cos(φ + φ 0 )=0 For w(s) and φ(s) to be independent of φ 0, coefficients of sin and cos terms must vanish identically T. Satogata / Fall 20 MePAS Intro to Accel Physics 22

23 Courant-Snyder Parameters 2ww φ + w 2 φ =(w 2 φ ) =0 φ = k w(s) 2 w (k 2 /w 3 )+Kw =0 w 3 (w + Kw)=k 2 Notice that in both equations w 2 k so we can scale this out and define a new set of functions, Courant-Snyder Parameters or Twiss Parameters β(s) w2 (s) φ = ds φ(s) = k β(s) β(s) α(s) 2 β (s) γ(s) +α(s)2 β(s) Kβ = γ + α β(s), α(s), γ(s) are all periodic in C φ(s)isnot periodic in C T. Satogata / Fall 20 MePAS Intro to Accel Physics 23

24 Towards The Matrix Solution φ = β(s) What is the matrix for this Hill s Equation solution? x x x(s) =A β(s) cos φ(s)+b β(s)sinφ(s) x (s) = β(s) {[B α(s)a] cos φ(s) [A + α(s)b]sinφ(s)} A = x 0 β(s) B = β(s) [β(s)x 0 + α(s)x 0 ] This all looks pretty familiar and pretty tedious We have done this many times so we skip to the solution s 0 +C = cos φc + α(s)sin φ C γ(s)sin φ C φ C = s0 +C s 0 β(s)sin φ C x cos φ C α(s)sin φ C x ds β(s) s 0 T. Satogata / Fall 20 MePAS Intro to Accel Physics 24

25 x x s 0 +C = Interesting Observations cos φc + α(s)sin φ C γ(s)sin φ C φ C = s0 +C s 0 β(s)sin φ C x cos φ C α(s)sin φ C x ds β(s) s 0 φ C is independent of s: betatron phase advance again Determinant of matrix M is still! (Check!) Still looks like a rotation and some scaling M can be written down in a beautiful and deep way M = I cos φ C + J sin φ C I = 0 0 J = α(s) γ(s) β(s) α(s) J 2 = I M = e J(s) φ C and remember x(s) =A β(s) cos[φ(s)+φ 0 ] T. Satogata / Fall 20 MePAS Intro to Accel Physics 25

26 ========== Once Again ========== T. Satogata / Fall 20 MePAS Intro to Accel Physics 26