Lecture 3: Modeling Accelerators Fringe fields and Insertion devices. X. Huang USPAS, January 2015 Hampton, Virginia

Transcription

1 Lecture 3: Modeling Accelerators Fringe fields and Insertion devices X. Huang USPAS, January 05 Hampton, Virginia

2 Fringe field effects Dipole Quadrupole Outline Modeling of insertion devices Radiation damping and quantum excitation Damping, excitation, Ohmi envelope Longitudinal tracking with acceleration

3 By/Bymax Magnetic field profile and the hard-edge model Most accelerator modeling codes use the hard-edge model for magnet constant Hamiltonian. Real magnets always have a smooth transition at the edges fringe fields RADIA new Measured 00 fit meas 00 measured Z (m) SPEAR3 dipole B y ~z Hard edge model Z (m) Field or gradient is a constant equal to the average value inside the magnet body. The effective length is the integrated strength divided by the field or gradient. The hard-edge model is non-maxwellian! 3 B (T/m) all quads at 7 A in modeling SPEAR3 quadrupoles B ~z 60Q 50Q 34Q 5Q

4 Vector potential with fringe field With cylindrical symmetry, the vector potential of a normal multipole (n = for dipole, etc.) with fringe field (with Coulomb gauge A = 0) on a straight geometry is A r = A θ = cos nθ n! p=0 G n+p+ n,p+ s r p+n+, sin nθ n! p=0 G n+p+ n,p+ s r p+n+, cos nθ A s = G n! p=0 n,p s r p+n. With x = r cos θ and y = r sin θ and G n,p s = p n! 4 p n+p!p! d p G n,0 (s), and G ds p n,p+ s = dg n,p(s) ds Example: for dipole n =, assuming G,0 = B 0 Θ(z), then A x = x y B 0 Θ z + O(4), A y = xy B 0Θ (z) + O(4) 4 A z = B 0 Θ z x + 8 B 0Θ z x x + y + O(5). Correspondingly, the magnetic field is B x = B 0 Θ (z) xy 4, B y = B 0 Θ z 8 B 0Θ (z)(x + 3y ) B z = B 0 Θ z y. 4.

5 Approximate field profile with Enge function The longitudinal profile function can be approximated well with the Enge function Θ z = + e Where s 0 is at the effective edge, G is the full gap. n i=0 a i s s 0 G i B0 (T) model data SPEAR3 dipole Z (m) Effective edge at s 0 = m Gap G = 0.05 m. Coefficients: a=[ ] 5

6 Evaluation of fringe field effect for dipoles At the entrance edge: y and Y axises pointing out of page. x Z = D δ X 0 Z = 0 s Z = D Z Fields in (X, Y, Z) coordinates (to first order of X, Y, Z): B X = 0, B Y = B 0 Θ(Z), B Z = B 0 Θ Z Y. Suppose points, are boundaries of the fringe field, point 0 is the entrance face of the bend magnet. For the hard-edge model, the drift ends at the (x, y) plane at 0, followed by a sector dipole whose entrance face coincides with the plane. The difference between reality and the hard-edge model is captured by the following transformations: () At 0, in drift space, propagate particles to the (X, Y) plane. () In drift space, back propagate to (X, Y) plane at. (3) In actual fields, propagate to point. (4) In sector dipole, back propagate to point 0. (5) At 0, in sector dipole, back propagate particles to the original (x, y) plane. Coordinate transformations between (x, y, s) system and (X, Y, Z) system are performed at steps () and (5). The differences in phase space coordinates before and after the above procedure is the fringe field effects. 6

7 Edge focusing At the entrance edge: 0 M x = h tan β, 0 M y = h tan(β ψ ), At the exit edge 0 M x = h tan β, 0 M y = h tan(β ψ ), Definition of entrance and exit angles. (K. Brown, SLAC-75) Correction for the vertical plane due to finite extent of fringe field: ψ, = KhG( + sin β, )/ cos β, With G the full gap and K = B y z (B 0 B y z ) dz gb 0 7

8 Fringe field effect of quadrupoles The Hamiltonian for a quadrupole with fringe field From H + δ +δ Linear effect (for the entrance edge) p x + p y + +δ p xa x + p y a y a s, ignore δ, apply vector potential for quadrupole, get H = P x + P y + K s x y 4 K s x y xp x + yp y K s x 4 y 4 + O X 6, H = H 0 + H s, With the hard-edge model H 0 = P x + P y + K 0 x y, s s 0 P x + P y, s < s 0 and the perturbation term with H s = K(s)(x y ) K s = K s K 0, s s 0 K s, s < s 0 8

9 Linear effect as a thin element F Q Transfer matrix M = M D 0 F Q M Q (0 ) The generating function for map* F Q = e :f : is calculated to be f = I (xp x yp y ) With integral I defined as +D I = K(s)(s s 0 ) ds D The transfer matrix is diag(e I, e I, e I, e I ). Z = D 9 x 0 Z = 0 Z = D With D being the fringe length, roughly I = 6 K 0D At the exit edge, integral I has opposite sign. So the two edges tend to cancel. The cancellation is not complete for quadrupoles with finite length. The net effect includes a tune shift (always negative) K0 L I K0 L x y I x, y K0 K0 *The Lie map e :f: = +: f : + : f! : + is an operator, where f : g = [f, g], Poisson bracket. The Lie map for a constant Hamiltonian is e :LH:. s

10 Nonlinear effect of quadrupole fringe field Apply the same approach to find the generating function for nonlinear effects (the map is e :f4: ). For the entrance edge (change sign for the exit edge). f 4 = K 0(x 3 p x + 3xy p x y 3 p y 3x yp y ) This cannot be symplectically integrated. However, the generating function for a skew quadrupole is integrable. f 4,skew = a 6 (x3 p y + y 3 p x ) exp : αy 3 p x : x = x αy 3, exp : αy 3 p x : p y = p y 3αy 3 p x, Similarly for the x 3 p y term. These are two kicks! Therefore, we can model the nonlinear effects of quadrupole fringe field with a symplectic map by rotating π, applying the skew quad fringe kick, and rotating 4 back. 0

11 Example Magnetic field B B B x y z 3 B[ y ( z) ''( z)(3x y y )] 3 B[ x ( z) ''( z)( x 3xy )] sgn( z) B '( z) xy SPEAR3 quadrupole fringe field Bf (normalized) Z (m) for 60Q I 3 6I m, m k k I a 0 0 xp f (rad) quadpass quadpass+matrix new quad pass field pass Case QuadLinearPass QuadLinearFPass Field Pass + Quad Field Element Quad Quad [Drift QuadField Drift] /9/05 x i (m) X. Huang, USPAS Jan 05

12 Wiggler and undulator Insertion device modeling y x z, s Wiggler parameter K = eb 0 kmc = 0.934B 0 T λ u [cm]. with k = π λ u. On the mid-plane, the magnetic field B y = B 0 cos kz Trajectory x z with amplitude A = = K k ρ γ = x 0 A cos kz λ u π.

13 Magnetic field in an ID The field is periodic in the longitudinal direction. Suppose the vertical field is B y = B x, y cos kz with symmetry B x, y = B( x, y) and B x, y = B(x, y). It can be shown that to satisfy B=0 and B=0, we need B x + B y k B = 0 The other field components B x = B x, y dy cos kz, x B z = ( k) B x, y dy cos kz. Symmetry leads to B x x = 0 = 0, B x y = 0 = 0 and B z y = 0 = 0. The Halbach wiggler field model B y = B 0 cosh k x x cosh k y y cos kz, B x = k x k y B 0 sinh k x x sinh k y y cos kz, B z = k k y B 0 cosh k x x sinh k y y cos kz. with k x + k y = k. 3

14 Effects of ideal ID on the beam The Hamiltonian for the Halbach model (to 4 th order) H = p x + p y + k 4k ρ x x + k y y + k k ρ x 4 x 4 + k 4 y y 4 + 3k x k x y sin ks kρ p x k x x + k y y k x p y xy. Assume k x = 0 (ideal planar wiggler with wide poles) H = p x + p y + y k sin ks 4ρ p ρ x y + k ρ y4 + O(X 5 ) For the ideal planar wiggler, the effect on beam is on the vertical plane. Linear effect Tune shift: Δν y = β L 8πρ + L 3 96πρ β where L is wiggler length, β is beta function at ID center (minimum) Nonlinear effects Octupole-like effect causing amplitude-dependent tune shift, B 0 L λ u 4

15 Effects of an imperfect wiggler Kick to the beam (by field integral on trajectory) Δx = B Bρ y x z, y, z dz = Bρ B y x, y, z dz + Bρ B y x Δx(z)dz Static dynamic Static field integrals (on straight path): First integral I y x = B y x, 0, z dz, Δx = I y /Bρ Second integral I y x = dz B y x, 0, z dz, Δx = I y /Bρ (similarly for I x and I x ) Quadrupole int. Skew quad int. Sextupole int. I q = B y dz, x I sq = B x dz, x 5 z Δν x = β xi q, Δν 4π y = β yi q 4π coupling I sext = B y x dz, nonlinear dynamics The field integrals can be obtained from measurements. These static effects can be modeled as multipole kicks.

16 Kick from the dynamic effect Δx = Bρ B y Δx(z)dz, x with Δx z = B cos kz, k ρ So the kick is Dynamic effects from field roll-off y x = B(x,0) x Δx = cos kz, (on the mid-plane) L B 0 B(x, 0) Bρ k x The dynamic kick effect is particularly severe for elliptically-polarized undulator (EPU) I. Blomqvist Field roll-off of the planar phase for an EPU Dynamic integral from the field roll-off 6

17 Modeling of effects of IDs on the beam There are methods of symplectic integration for s-dependent Hamiltonian when analytic forms of vector potential is available (not discussed here). A commonly used method for ID modeling is the kick map method: x, y (Δx, Δy ) The kicks can be derived from the potential (P. Elleaume) z z Ψ x, y = dz( B x x, y, z dz + By x, y, z dz ) with the kicks given by Δx =, Δy = Ψ(x,y) Bρ x Ψ(x,y) Bρ y, 4 x 0-3 hfield3, no shim x (mrad) 0 Comparison of the Elleaume kick map with numerical integration (Runge-Kutta) - Elleaume Numerical x (mm) 7

18 Example of ID field (EPU for MAX-II) Planar Circular Orbit for.5 GeV electrons I. Blomqvist 8

19 References. M. Basetti and C. Biscari, Part. Accel. 5, (996). Y. Cai and Y. Nosochkov, SLAC-PUB-8 (005) 3. K. Brown, SLAC-Report-75 (98) 4. E. Forest and Milutinovic, NIMA 69, (988) 5. J. Irwin and C.-x. Wang, Explicit soft fringe maps of a quadrupole, PAC 95 (995) 6. X. Huang, et al, Lattice modeling for SPEAR3, IPAC Lloyd Smith, Effect of wigglers and undulators on beam dynamics, ESG-Tech-note-4, (986) 8. P. Elleaume, A new approach to the electron beam dynamics in undulators and wigglers, EPAC 9, p. 66 (99) 9. J. Safranek, et al, Nonlinear dynamcis in a SPEAR wiggler, PRSTAB 5, 0070 (00) 9