Position Dependence Of The K Parameter In The Delta Undulator

Transcription

1 LCLS-TN-16-1 Position Dependence Of The K Parameter In The Delta Undulator Zachary Wolf SLAC January 19, 016 Abstract In order to understand the alignment tolerances of the Delta undulator, we must know how the K value depends on the transverse position of the beam. In this note we calculate the K value as a function of beam position for di erent polarization modes and row shift settings. The e ect of transverse eld rollo of the quadrants is included, and we will see that it plays a large role in the position dependence of K. In particular, it gives a gradient to K at the magnetic center in the elliptical modes, potentially making the undulator alignment requirements di cult to meet. 1 Introduction 1 The position dependence of the K parameter in the Delta undulator determines how accurately the undulator must be aligned in order to be resonant with other undulators. In an EPU the K parameter depends on position di erently for di erent quadrant row settings. In addition, the K parameter depends on the shape of the quadrant elds. In this note we explore these dependencies. The organization of the note is as follows. We rst calculate the magnetic elds in the Delta undulator. This is done in a rotated coordinate frame relative to the laboratory frame in which the undulator is used. Rotating the coordinates does not a ect the calculated K value and it simpli es the calculations. From the elds, we calculate the trajectory slopes which we use to calculate the slippage. The slope of the linear part of the slippage change with longitudinal position lets us calculate the K value of the undulator. The K value is calculated at di erent transverse positions and at di erent undulator row settings in order to study the position dependence of K. Magnetic Field In The Undulator.1 The Scalar Potential The scalar potential from a single magnet array satis es Laplace s equation. It varies sinusoidally down the array, it decreases away from the array, and it falls o as one moves away from the center line of the array in the transverse direction. In the coordinate system shown in gure 1, the scalar potential can be expressed as a Fourier sum of terms with a fundamental term of the form = 0 cos( s) exp ( r) cos ( (z z 0 )) (1) 1 Worupported in part by the DOE Contract DE-AC0-76SF This work was performed in support of the LCLS project at SLAC. 1

2 where 0 is a constant, = = u where u is the undulator period, z is the coordinate down the undulator, and z 0 gives the quadrant position along z. and determine the behavior of the potential in the transverse directions. The potential decreases as one moves radially away from the magnet and this is expressed by exp ( r). The potential also decreases as one moves to the side and this is given by the cos( s) dependence over a limited range of s. Since the Laplacian of the potential is zero, we have the constraint ks + kr ku = 0 () Figure 1: Coordinate system for the scalar potential of a single quadrant. Although the scalar potential is a sum of terms, we assume that the fundamental term dominates the K value. In a Fourier expansion of this type, typically the amplitude of the fundamental term is the largest. Furthermore, harmonics with a cos (n (z z 0 )) dependence give contributions to the transverse velocity of the beam that go as 1=n, and therefore the contributions to K from these harmonics go as 1=n. The smaller amplitudes of the higher harmonics combined with the 1=n dependence lead us to expect their contribution to K will be small. The scalar potential also has terms with di erent values (and di erent values as given by the constraint). Satisfactory ts of B r as a function of s above a magnet array can be done with only a few cos( s) terms. The fundamental term is the largest and has a value of determined by the e ective width of the magnet block. Additional terms have values which are not integer multiples of the fundamental, but are determined by the detailed magnet shape. These higher order terms are typically much smaller than the fundamental. The higher order terms may be useful to construct a precise eld behavior by shaping the magnet. But for now, we consider only the fundamental term which dominates K. We use the form of the potential given in equation 1 to calculate the magnetic scalar potential in the undulator by rotating the four quadrants and summing their rotated scalar potentials. The Delta undulator is oriented as shown on the left side of gure. In the laboratory, z is along the beam direction, y L is up, and x L makes a right handed system. For our calculations, it is more convenient to use the rotated system x, y, z, where x is in the direction from quadrant 3 to quadrant 1, y is in the direction from quadrant 4 to quadrant, and z is in the beam direction. Using equation 1, the scalar potential for each of the quadrants in the x, y, z system is 1 (x; y; z) = 0Q cos( y) exp ( x) cos ( (z z 01 )) (3) (x; y; z) = 0Q cos( x) exp ( y) cos ( (z z 0 )) (4) 3 (x; y; z) = 0Q cos( y) exp ( x) cos ( (z z 03 )) (5) 4 (x; y; z) = 0Q cos( x) exp ( y) cos ( (z z 04 )) (6)

3 Figure : The left side of the gure shows the Delta undulator in its con guration in the tunnel where y L is up, z is in the beam direction, and x L makes a right handed system. For our calculations, we use the rotated coordinate system on the right, where x is along the line pointing from quadrant 3 to quadrant 1, y is along the line pointing from quadrant 4 to quadrant, and z makes a right handed system. where z 0i is the longitudinal shift of quadrant i, and 0Q is the amplitude of the scalar potential of all the identical quadrants on the axis of the undulator where x = 0 and y = 0. Quadrants 3 and 4 are loaded with opposite polarity magnets as quadrants 1 and in order to make a vertical eld planar undulator in the laboratory frame when all the rows are aligned. This accounts for the minus signs, 0Q, in the potentials for quadrants 3 and 4. The scalar potential for the undulator is the sum of the scalar potentials for the quadrants. Quadrants 1 and 3 both depend on x, and quadrants and 4 both depend on y. We rst add the scalar potentials for quadrants 1 and 3, and then add the scalar potentials for quadrants and 4, and then add the sums to get the scalar potential for the whole undulator. We will interpret this as forming the entire undulator from two crossed planar adjustable phase undulators. The scalar potential for the combination of quadrants 1 and 3 is given by Let = 0Q cos( y) exp ( x) cos ( (z z 01 )) 0Q cos( y) exp ( x) cos ( (z z 03 )) (7) z 01 = Z + z 03 = Z where Z = z 01 + z 03 is the average z-position of the quadrants, and is the z-shift between the quadrants. quadrants becomes = 0Q cos( y) sinh ( x) cos (8) (9) (10) = z 01 z 03 (11) With these de nitions, the scalar potential for the pair of + 0Q cos( y) cosh ( x) sin 3 cos ( (z Z )) sin ( (z Z )) (1)

4 This is the scalar potential for a planar adjustable phase undulator. The full range of amplitudes and phases are covered if the range of includes u ; u and the range of Z includes u ; u. Similarly, the scalar potential for the combination of quadrants and 4 is given by Let 4 = 0Q cos( x) exp ( y) cos ( (z z 0 )) 0Q cos( x) exp ( y) cos ( (z z 04 )) () z 0 = Z z 04 = Z 4 4 where Z 4 = z 0 + z 04 is the average z-position of the quadrants, and (14) (15) (16) 4 = z 0 z 04 (17) is the z-shift between the quadrants. With these de nitions, the scalar potential for the pair of quadrants becomes 4 4 = 0Q cos( x) sinh ( y) cos cos ( (z Z 4 )) 4 + 0Q cos( x) cosh ( y) sin sin ( (z Z 4 )) (18) This is again the potential for a planar adjustable phase undulator. The scalar potential for the undulator is the sum of the scalar potentials for the quadrant pairs. = + 4 (19) Performing the sum, we nd = 0Q cos( y) sinh ( x) cos cos ( (z Z )) + 0Q cos( y) cosh ( x) sin sin ( (z Z )) 4 + 0Q cos( x) sinh ( y) cos cos ( (z Z 4 )) 4 + 0Q cos( x) cosh ( y) sin sin ( (z Z 4 )) (0) By putting in the various values for Z, Z 4,, and 4, we get the scalar potential for the various undulator modes at di erent K values. Z. Wolf, "Variable Phase PPM Undulator Study", LCLS-TN-11-1, May,

5 . Magnetic Field In The Undulator Coordinate System The magnetic eld in the undulator is given by B = r. Taking the gradient we nd B x (x; y; z) = 0Q [ k r cos cos( y) cosh ( x) cos ( (z Z )) + k r sin cos( y) sinh ( x) sin ( (z Z )) 4 cos sin( x) sinh ( y) cos ( (z Z 4 )) sin 4 sin( x) cosh ( y) sin ( (z Z 4 ))] (1) B y (x; y; z) = 0Q [ cos sin( y) sinh ( x) cos ( (z Z )) sin sin( y) cosh ( x) sin ( (z Z )) + k r 4 cos cos( x) cosh ( y) cos ( (z Z 4 )) + k r 4 sin cos( x) sinh ( y) sin ( (z Z 4 ))] () B z (x; y; z) = 0Q [ cos + sin cos + sin 4 4 In order to simplify these formulas further, let cos( y) sinh ( x) sin ( (z Z )) cos( y) cosh ( x) cos ( (z Z )) cos( x) sinh ( y) sin ( (z Z 4 )) cos( x) cosh ( y) cos ( (z Z 4 ))] (3) B 0 = 0Q (4) B xc = B 0 cos (5) B xs = B 0 sin (6) 4 B yc = B 0 cos (7) 4 B ys = B 0 sin (8) 5

6 With these substitutions, the elds become B x (x; y; z) = B xc cos( y) cosh ( x) cos ( (z Z )) + B xs cos( y) sinh ( x) sin ( (z Z )) B yc sin( x) sinh ( y) cos ( (z Z 4 )) B ys sin( x) cosh ( y) sin ( (z Z 4 )) (9) B y (x; y; z) = B xc sin( y) sinh ( x) cos ( (z Z )) B xs sin( y) cosh ( x) sin ( (z Z )) + B yc cos( x) cosh ( y) cos ( (z Z 4 )) + B ys cos( x) sinh ( y) sin ( (z Z 4 )) (30) B z (x; y; z) = B xc cos( y) sinh ( x) sin ( (z Z )) We can change the origin of z such that If we let +B xs cos( y) cosh ( x) cos ( (z Z )) B yc cos( x) sinh ( y) sin ( (z Z 4 )) and drop the prime, we write the elds as a function of position as +B ys cos( x) cosh ( y) cos ( (z Z 4 )) (31) z = z 0 + Z (3) = (Z Z 4 ) (33) B x (x; y; z) = B xc cos( y) cosh ( x) cos ( z) +B xs cos( y) sinh ( x) sin ( z) B yc sin( x) sinh ( y) cos ( z + ) B ys sin( x) cosh ( y) sin ( z + ) (34) B y (x; y; z) = B xc sin( y) sinh ( x) cos ( z) B xs sin( y) cosh ( x) sin ( z) +B yc cos( x) cosh ( y) cos ( z + ) +B ys cos( x) sinh ( y) sin ( z + ) (35) 6

7 B z (x; y; z) = B xc cos( y) sinh ( x) sin ( z) 3 Trajectories In The Undulator 3.1 Equations Of Motion +B xs cos( y) cosh ( x) cos ( z) B yc cos( x) sinh ( y) sin ( z + ) +B ys cos( x) cosh ( y) cos ( z + ) (36) The trajectory of a charged particle beam in the undulator is determined by the Lorentz force law dp dt = q (v B) (37) where p = mv, = 1 1, and m is the particle rest mass. The energy of the particle is constant in the magnetic eld of the undulator, except for radiation losses which we neglect. With constant energy, is constant. The Lorentz force law becomes v = q (v B) (38) m With the substitution d=dt = v z d=, the changes in the individual velocity components with z are given by dv x dv y dv z = = = q (v y B z mv z v z B y ) (39) q (v z B x mv z v x B z ) (40) q (v x B y mv z v y B x ) (41) 3. Iterative Solution We solve these equations iteratively by expanding in powers of a small parameter. Let = qb 0 m c (4) be the small dimensionless expansion parameter. motion are dv x dv y dv z In terms of this parameter, the equations of = c B 0 v z (v y B z v z B y ) (43) = c B 0 v z (v z B x v x B z ) (44) = c B 0 v z (v x B y v y B x ) (45) At this point we could divide all velocities by c, divide all positions by 1=, and divide all magnetic elds by B 0 in order to eliminate the factor kuc B 0v z. We think of small quantities in terms of these dimensionless variables. In the interest of clarity, however, we keep the factor and continue to use mks units, so there is no confusion about the meaning of a quantity. 7

8 Expanding the particle position in terms of the expansion parameter, we have x = x (0) + x (1) + x () + (46) y = y (0) + y (1) + y () + (47) z = z (48) where the last line indicates that the z-position of the particle is the independent variable. numbers in parenthesis indicate the order of the expansion. The velocities are where The magnetic elds are given by The v x = v x(0) + v x(1) + v x() + (49) v y = v y(0) + v y(1) + v y() + (50) v z = v z(0) + v z(1) + v z() + (51) B x = B x j 0 x B x j 0 x (1) y B x j 0 y (1) + (5) B y = B y j 0 x B y j 0 x (1) y B y j 0 y (1) + (53) B z = B z j 0 x B z j 0 x (1) y B z j 0 y (1) + (54) B i j 0 = B i (x (0) ; y (0) ; z) j B i j 0 j (x (0) ; y (0) ; z) (56) where i = x; y; z, j = x; y. The transverse positions are found from the velocities as follows: v x = dx dt x(z) = x 0 + = x 0 + Z z 0 Z z = x 0 + v x(0) v z(0) z + 0 v x (z 0 1 ) v z (z 0 ) 0 v x(0) + v x(1) + 1 Z z 0 v x(1) v z(0) (57) 1 v z(1) + 0 v z(0) v z(0)! v x(0) v z(1) vz(0) 0 + (58) In these equations x 0 is the constant initial position where the beam enters the undulator. Similarly y(z) = y 0 + v Z! z y(0) v y(1) v y(0) v z(1) z + v z(0) 0 v z(0) vz(0) 0 + (59) We see that and x (1) = y (1) = x (0) = x 0 + v x(0) v z(0) z (60) y (0) = y 0 + v y(0) v z(0) z (61) Z z 0 Z z 0 v x(1) v z(0) v y(1) v z(0) v x(0) v z(1) v z(0) v y(0) v z(1) v z(0)!! 0 (6) 0 (63) 8

9 With the expansions in the small parameter, the equations of motion become dv x(0) + dv x(1) + dv x() + c = 1 v z(1) + B 0 v z(0) v z(0) f(v y(0) + v y(1) + )(B z j 0 x B z j 0 x (1) y B z j 0 y (1) + ) (v z(0) + v z(1) + )(B y j 0 x B y j 0 x (1) y B y j 0 y (1) + )g (64) dv y(0) + dv y(1) + dv y() + c = 1 v z(1) + B 0 v z(0) v z(0) f(v z(0) + v z(1) + )(B x j 0 x B x j 0 x (1) y B x j 0 y (1) + ) (v x(0) + v x(1) + )(B z j 0 x B z j 0 x (1) y B z j 0 y (1) + )g (65) dv z(0) + dv z(1) + dv z() + c = 1 v z(1) + B 0 v z(0) v z(0) f(v x(0) + v x(1) + )(B y j 0 x B y j 0 x (1) y B y j 0 y (1) + ) (v y(0) + v y(1) + )(B x j 0 x B x j 0 x (1) y B x j 0 y (1) + )g (66) We now proceed to solve these equations for zeroth, and rst orders Zeroth Order To zeroth order in, the equations of motion are The solutions can be written by inspection dv x(0) dv y(0) dv z(0) = 0 (67) = 0 (68) = 0 (69) v x(0) = v x0 (70) v y(0) = v y0 (71) v z(0) = v z0 (7) where the v x0, v y0, v z0 are the constant initial velocities. The corresponding zeroth order transverse positions are x (0) = x 0 + v x0 v z0 z (73) y (0) = y 0 + v y0 v z0 z (74) 9

10 where the x 0 and y 0 have been previously de ned to be the initial positions. At this point, we make a simplifying assumption. We assume that all particles have zero initial transverse velocity. In this case and 3.. First Order To rst order in, the equations of motion are dv x(1) dv y(1) dv z(1) With zero initial transverse velocity, these equations become = = = v x(0) = 0 (75) v y(0) = 0 (76) v z(0) = v z0 (77) x (0) = x 0 (78) y (0) = y 0 (79) c B 0 v z(0) fv y(0) B z j 0 v z(0) B y j 0 g (80) c B 0 v z(0) fv z(0) B x j 0 v x(0) B z j 0 g (81) c B 0 v z(0) fv x(0) B y j 0 v y(0) B x j 0 g (8) dv x(1) dv y(1) dv z(1) = c B 0 B y j 0 (83) = c B 0 B x j 0 (84) = 0 (85) Inserting the expressions for the elds in the undulator coordinate system, we have dv x(1) = c B 0 [ B xc sin( y 0 ) sinh ( x 0 ) cos ( z) B xs sin( y 0 ) cosh ( x 0 ) sin ( z) +B yc cos( x 0 ) cosh ( y 0 ) cos ( z + ) +B ys cos( x 0 ) sinh ( y 0 ) sin ( z + )] (86) dv y(1) = c B 0 [B xc cos( y 0 ) cosh ( x 0 ) cos ( z) +B xs cos( y 0 ) sinh ( x 0 ) sin ( z) B yc sin( x 0 ) sinh ( y 0 ) cos ( z + ) B ys sin( x 0 ) cosh ( y 0 ) sin ( z + ) (87) 10

11 dv z(1) = 0 (88) We integrate to nd the velocity components to rst order: v x(1) = c B 0 [ B xc sin( y 0 ) sinh ( x 0 ) sin ( z) +B xs sin( y 0 ) cosh ( x 0 ) cos ( z) +B yc cos( x 0 ) cosh ( y 0 ) sin ( z + ) B ys cos( x 0 ) sinh ( y 0 ) cos ( z + )] (89) v y(1) = c B 0 [B xc cos( y 0 ) cosh ( x 0 ) sin ( z) B xs cos( y 0 ) sinh ( x 0 ) cos ( z) B yc sin( x 0 ) sinh ( y 0 ) sin ( z + ) +B ys sin( x 0 ) cosh ( y 0 ) cos ( z + ) (90) v z(1) = 0 (91) 3.3 Solution To The Equations Of Motion The velocity of a particle in the undulator can now be given to rst order. Using v x = v x(0) + v x(1) + (9) v y = v y(0) + v y(1) + (93) v z = v z(0) + v z(1) + (94) with we get = qb 0 m c (95) v x = q m [ B xc sin( y 0 ) sinh ( x 0 ) sin ( z) +B xs sin( y 0 ) cosh ( x 0 ) cos ( z) +B yc cos( x 0 ) cosh ( y 0 ) sin ( z + ) B ys cos( x 0 ) sinh ( y 0 ) cos ( z + )] (96) 11

12 v y = q m [B xc cos( y 0 ) cosh ( x 0 ) sin ( z) B xs cos( y 0 ) sinh ( x 0 ) cos ( z) B yc sin( x 0 ) sinh ( y 0 ) sin ( z + ) +B ys sin( x 0 ) cosh ( y 0 ) cos ( z + )] (97) In order to simplify future equations, let v z = v z0 (98) K 0 = qb 0 m c (99) Then c v x = K 0 [ cos sin( y 0 ) sinh ( x 0 ) sin ( z) + k s sin sin( y 0 ) cosh ( x 0 ) cos ( z) + k r 4 cos cos( x 0 ) cosh ( y 0 ) sin ( z + ) sin 4 cos( x 0 ) sinh ( y 0 ) cos ( z + )] (100) 4 K Value c v y = K 0 [ k r cos cos( y 0 ) cosh ( x 0 ) sin ( z) sin cos( y 0 ) sinh ( x 0 ) cos ( z) 4 cos sin( x 0 ) sinh ( y 0 ) sin ( z + ) + sin 4 sin( x 0 ) cosh ( y 0 ) cos ( z + )] (101) 4.1 Slippage Slippage is the distance between the light wave and the electrons. The slippage is given by ds = (c v z )dt (10) = ( c v z 1) (103) We take the electron energy to be constant, so the Lorentz factor is constant. 1 = 1 v x c v y c v z c (104) 1

13 So vz c = 1 1 Taking the square root, we nd for large and small v x =c and small v y =c v x c v y c (105) v z c = 1 1 v x c v y c (106) Taking the inverse, we nd So the slippage changes with z according to 4. De ne K c v z = v x c + v y c (107) ds = 1 + v x c + v y c (108) The slippage changes linearly with z with additional oscillatory terms. K value. ds = K So ds K = = c v x + v y We know c v x and c v y. We need to square them and nd the values averaged in z. 4.3 Calculate D E v c x From equation 100, the formula for c v x is c v x = K 0 [ 1 cos sin( y 0 ) sinh ( x 0 ) sin ( z) + k s sin sin( y 0 ) cosh ( x 0 ) cos ( z) + k r 4 cos cos( x 0 ) cosh ( y 0 ) sin ( z + ) sin 4 The linear term de nes the (109) (110) (111) cos( x 0 ) sinh ( y 0 ) cos ( z + )] (11) In order to simplify the calculations, we make the following de nitions for c 1 to c 4. c v x = c 1 sin ( z) + c cos ( z) + c 3 sin ( z + ) + c 4 cos ( z + ) (1)

14 Explicitly, we have Squaring c v x, we have c 1 = K 0 cos sin( y 0 ) sinh ( x 0 ) (114) c = K 0 sin sin( y 0 ) cosh ( x 0 ) (115) 4 c 3 = K 0 cos cos( x 0 ) cosh ( y 0 ) (116) c 4 = K 0 sin 4 cos( x 0 ) sinh ( y 0 ) (117) c v x = c 1 sin ( z) + c cos ( z) + c 3 sin ( z + ) + c 4 cos ( z + ) +c 1 c sin ( z) cos ( z) +c 1 c 3 sin ( z) sin ( z + ) +c 1 c 4 sin ( z) cos ( z + ) +c c 3 cos ( z) sin ( z + ) +c c 4 cos ( z) cos ( z + ) +c 3 c 4 sin ( z + ) cos ( z + ) (118) Taking the average value along z, we have 4.4 Calculate D E v c y v c x From equation 101, the formula for c v y is c v y = K 0 [ k r cos = 1 c 1 + c + c 3 + c c 1 c 3 cos() c 1 c 4 sin () +c c 3 sin () +c c 4 cos () +0 (119) cos( y 0 ) cosh ( x 0 ) sin ( z) sin cos( y 0 ) sinh ( x 0 ) cos ( z) 4 cos sin( x 0 ) sinh ( y 0 ) sin ( z + ) + sin 4 We make the following de nitions for d 1 to d 4. sin( x 0 ) cosh ( y 0 ) cos ( z + )] (10) 14

15 Explicitly, we have Squaring c v y, we have v y = d 1 sin ( z) + d cos ( z) + d 3 sin ( z + ) + d 4 cos ( z + ) d 1 = K 0 cos cos( y 0 ) cosh ( x 0 ) (11) d = K 0 sin cos( y 0 ) sinh ( x 0 ) (1) 4 d 3 = K 0 cos sin( x 0 ) sinh ( y 0 ) () d 4 = K 0 sin 4 sin( x 0 ) cosh ( y 0 ) (14) c v y = d 1 sin ( z) + d cos ( z) + d 3 sin ( z + ) + d 4 cos ( z + ) +d 1 d sin ( z) cos ( z) +d 1 d 3 sin ( z) sin ( z + ) +d 1 d 4 sin ( z) cos ( z + ) +d d 3 cos ( z) sin ( z + ) +d d cos ( z) cos ( z + ) +d 3 d 4 sin ( z + ) cos ( z + ) (15) Taking the average value along z, we have v c y = 1 d 1 + d + d 3 + d d 1 d 3 cos() d 1 d 4 sin () +d d 3 sin () +d d 4 cos () +0 (16) 4.5 Calculate K General Expression Since K = c v x + v y we can use equations 119 and 16 to write an expression for K K = c 1 + c + c 3 + c 4 + d 1 + d + d 3 + d 4 +(c 1 c 3 + c c 4 + d 1 d 3 + d d 4 ) cos() +( c 1 c 4 + c c 3 d 1 d 4 + d d 3 ) sin () (17) 15

16 Recalling from equations 114 to 117 and 11 to 14, the c i and d i in this expression for K have values c 1 = K 0 cos sin( y 0 ) sinh ( x 0 ) (18) c = K 0 sin sin( y 0 ) cosh ( x 0 ) (19) 4 c 3 = K 0 cos cos( x 0 ) cosh ( y 0 ) (0) c 4 = K 0 sin 4 cos( x 0 ) sinh ( y 0 ) (1) d 1 = K 0 cos cos( y 0 ) cosh ( x 0 ) () d = K 0 sin cos( y 0 ) sinh ( x 0 ) (3) 4 d 3 = K 0 cos sin( x 0 ) sinh ( y 0 ) (4) d 4 = K 0 sin 4 sin( x 0 ) cosh ( y 0 ) (5) These are the relations we seek. With these relations, we know K as a function of position for all polarization modes and row settings of the undulator Special Case 1, = 0 Consider the special case of two crossed planar undulators without eld rollo. In this case = 0 and =. The c and d parameters in equation 17 then become c 1 = 0 (6) c = 0 (7) 4 c 3 = K 0 cos cosh ( y 0 ) (8) c 4 = K 0 sin 4 Inserting these values in the expression for K we nd sinh ( y 0 ) (9) d 1 = K 0 cos cosh ( x 0 ) (140) d = K 0 sin sinh ( x 0 ) (141) d 3 = 0 (14) d 4 = 0 (143) K = c 3 + c 4 + d 1 + d (144) 16

17 K = K0[cos 4 + cos cosh ( y 0 ) + sin 4 sinh ( y 0 ) cosh ( x 0 ) + sin sinh ( x 0 )] (145) Note that making the eld rollo small greatly simpli es the formula for K. The position dependence is second order near the magnetic center for all quadrant row positions. The polarization mode does not enter the equation. It is very desirable to shape the magnets so that is small Special Case, x 0 = 0; y 0 = 0 Consider the special case of being on the magnetic axis where x 0 = 0 and y 0 = 0. equations 18 to 5 become Inserting these values in equation 17 for K, we nd K = c 3 + d 1 = K 0 In this case, c 1 = 0 (146) c = 0 (147) 4 c 3 = K 0 cos (148) c 4 = 0 (149) d 1 = K 0 cos (150) d = 0 (151) d 3 = 0 (15) d 4 = 0 (153) kr cos + cos 4 (154) This equation shows that regardless of the value of, on the magnetic axis, the undulator has the ideal behavior where the K value is set by shifting the rows of the two crossed undulators. This is true for all polarization modes Special Case 3, = u, 4 = u This is the case when the undulator is "turned o ", that is, on the magnetic axis K = 0. case, In this = 4 = At general positions we have using equations 18 to 5 (155) (156) c 1 = 0 (157) c = K 0 sin( y 0 ) cosh ( x 0 ) (158) c 3 = 0 (159) c 4 = K 0 cos( x 0 ) sinh ( y 0 ) (160) 17

18 Inserting these values in equation 17 for K, we have d 1 = 0 (161) d = K 0 cos( y 0 ) sinh ( x 0 ) (16) d 3 = 0 (163) d 4 = K 0 sin( x 0 ) cosh ( y 0 ) (164) K = c + c 4 + d + d 4 +(c c 4 + d d 4 ) cos() (165) K = K 0f + + kr " ks sin ( y 0 ) cosh ( x 0 ) + ku ku kr cos ( x 0 ) sinh ( y 0 ) cos ( y 0 ) sinh ks ( x 0 ) + sin ( x 0 ) cosh ( y 0 ) # sin( y 0 ) cosh ( x 0 ) cos( x 0 ) sinh ( y 0 ) cos()g (166) cos( y 0 ) sinh ( x 0 ) sin( x 0 ) cosh ( y 0 ) Note that away from the magnetic axis, the undulator does not "turn o ". value. For the special case of = 0, It still has a nite K The undulator does not have K = 0 o axis even if = 0. K = K0 sinh ( y 0 ) + sinh ( x 0 ) (167) 5 K Values Near The Magnetic Center Typically the undulator is aligned so that the beam is close to the magnetic center. In this case we expand the c and d parameters of equations 18 to 5 to second order in x 0, x 0, y 0 and y 0. We get the following results. k c 1 = K s 0 x 0 y 0 cos (168) k s c = K 0 y 0 sin (169) 1 c 3 = K 0 1 x y0 4 cos (170) k r 4 c 4 = K 0 y 0 sin (171) 18

19 d 1 = K 0 1 d = K 0 k r x 0 sin k d 3 = K sk r 0 x 0 y 0 cos d 4 = K 0 k s x 0 sin 1 y0 + 1 x cos We can now nd K to second order in x 0, x 0, y 0 and y 0. Equation 17 gives K = c 1 + c + c 3 + c 4 + d 1 + d + d 3 + d 4 +(c 1 c 3 + c c 4 + d 1 d 3 + d d 4 ) cos() (17) (173) (174) (175) +( c 1 c 4 + c c 3 d 1 d 4 + d d 3 ) sin () (176) Inserting the second order position dependence, we get K k s 1 = K 0 y 0 sin + K 0 1 x y0 4 cos k r K 0 y 0 sin + K 0 1 y0 + 1 x 0 cos k r k s 4 + K 0 x 0 sin + K 0 x 0 sin 8 h k K i h i 9 s kr 0 x 0 y 0 cos k K r 0 cos k 4 u h >< k + K i h s 0 k + u y 0 sin k k u K i r 0 y 0 sin k 4 u >= h k + K r i h 0 cos k k u K i cos() s kr 0 x 0 y 0 cos k 4 u h >: k + K r 0 x 0 sin k i h k u K s 0 x 0 sin k 4 i u >; 8 h < k K s 0 k + u y 0 sin k i h K r 0 cos k 4 i 9 u h : k K r 0 cos k i h k u K s 0 x 0 sin k 4 i = sin () (177) u ; We proceed to simplify this expression. K = K 0 ku ksy 4 0 sin + K 0 ku kry 4 0 sin 4 + K 0 ku 8 >< + >: 8 < + : k 4 rx 0 sin K 0 ku K 0 k u K 0 ku K 0 k u K 0 ku K 0 ku + K 0 ku kr 1 ksx 0 + kry 0 cos 4 + K 0 ku kr 1 ksy 0 + krx 0 + K 0 ku ksx 4 0 sin 4 cos ku 4 9 ksk rx 0 y 0 cos k u ksk ry 0 sin k u sin ku 4 ksk rx 0 y 0 cos k u cos ku 4 ksk rx 0 sin k u sin ku 4 k s y 0 sin k s x 0 cos cos ku 4 sin ku 4 >= cos() cos >; 9 = sin () (178) ; 19

20 Ordering the terms in powers of x 0, x 0, y 0 and y 0, we nd K = K 0 ku f kr cos + cos 4 +ksk y0 sin k u cos ku 4 r sin ku 4 +k 4 sy 0 sin +k 4 ry 0 cos 4 +krk sy 0 cos +krx 4 0 sin k sk r 6 4 x 0 cos k sin () u + x 0 cos 4 + k 4ry 0 sin 4 + k 4rx 0 cos + k 4sx 0 sin 4 x 0 y 0 cos k u cos ku 4 +y0 sin k u sin ku 4 +x 0 y 0 cos k u cos ku 4 +x 0 sin k u sin ku 4 Rearranging the second order terms and simplifying further, we nd K = K 0 ku f kr cos + cos 4 +ksk y0 sin k u cos ku 4 r x 0 cos k u sin ku 4 + ksy krx 4 0 sin + ksx kry 4 0 sin k sk r sin () + kr k rx 0 + ksy kr k sx 0 + kry 0 cos cos ku 4 x 0 y 0 cos k u + x 0 + y0 sin ku sin ku cos()g (179) cos 4 cos()g (180) This is our desired result. Note that when 6= 0 and 6=, there is a rst order gradient in the K value. The gradient of K near the magnetic center is problematic for aligning the undulator. To rst order in x 0, x 0, y 0 and y 0, K can be expressed as s K = K 0 kr cos k + cos u 4 " 1 + k r cos k s + cos 4 y0 sin k u cos ku 4 x 0 cos k u sin ku 4 # sin () (181) 0

21 The gradient in K at the magnetic center is 0 = 0 = K 0 q cos k s q cos + cos 4 k s + cos 4 We will calculate the size of these terms below. 6 Numerical Calculations 6.1 Parameter Values cos In the calculations below, the following parameter values are used: 4 sin sin () (18) sin 4 cos sin () (183) u = :03 m (184) = u = 196 1/m (185) = 70 1/m (186) = 186 1/m (187) B 0 = :61 T (188) K 0 = 1:8 (189) The value for was measured in the Delta undulator 3. The value for came from the constraint kr = ks + ku. The peak quadrant eld B 0 was calculated from the peak measured eld of 1: T in linear polarization vertical eld mode. The peak eld in this mode B peak is given by 4 0Q = p k and B 0 is given by 0Q, so B 0 = B u peak pkr. The value for K 0 was calculated from B 0 using equation Primary Modes From equation 180 we see that the linear terms near the magnetic center are largest when = = and they are smallest when = 0 or. We now consider these cases. = 0 gives the linear polarization vertical magnetic eld mode, = gives the linear polarization horizontal magnetic eld mode, = = gives the circular polarization right hand magnetic eld mode, and = = gives the circular polarization left hand magnetic eld mode. In these primary modes = 4 and the K value is set by the and 4 row shifts Linear Polarization Vertical Magnetic Field Figure 3 shows how the K value changes with x 0, the horizontal position in the undulator frame, when y 0 = 0. For this plot, the rows were shifted so that = 4 = :008 m, and = 0. This is the linear polarization vertical eld mode. On the magnetic axis, K is given by s kr K = K 0 cos k + cos u 4 (190) 3 Yurii Levashov, private communication. 4 Z. Wolf, H.-D. Nuhn, "Setting The K Value And Polarization Mode Of The Delta Undulator", LCLS-TN-14-, September,

22 The maximum value is obtained when cos k u = 1 and cos ku 4 = 1, and is K max = K 0 kr p (191) When = 4 = :008 m, cos p p = 1= and cos ku 4 = 1=, so kr K = K 0 (19) and K K max = 1 p (193) Figure 3 has several curves. The blue curve is calculated from equation 17. The green dashed curve is an independent calculation starting with the expressions for the magnetic eld components and numerically integrating to get the transverse velocities and from them to calculate K. This provides a check on the calculations in this note. The red curve is from equation 180. It gives an approximate expression for K near the magnetic center. The black line is a linear t to the curves at the magnetic center. It uses equation 190 to nd the K value at the magnetic center and equation 18 to nd the slope. The values of K at the magnetic center, K(0; 0), and the slope at the magnetic center, dk=dx(0; 0), are shown in the gure. The red curve deviates from the blue curve away from x 0 = 0, y 0 = 0. We expect the second order approximation to be valid for x 0, x 0, y 0 and y 0 much less than 1. Since <, we expect the approximation to be valid for or x 0 1 (194) x 0 5 mm (195) For x 0 = 0:5 mm, x 0 = 0:1, ( x 0 ) = 0:01, and the missing third order terms should contribute about K 0:001 = :003 to K. This is roughly in line with the deviations of the red line that we see. We can zoom in on the region near x 0 = 0 in order to estimate the alignment tolerance of the undulator. This is done in gure 4. From the blue curve, K changes by about 10 4 relative to its value at the magnetic center when x 0 is approximately 70 microns from the magnetic center. This is in line with the hyperbolic cosine position dependence of a planar undulator, and is in fact very close to the value in the LCLS undulators which have a similar period (30 mm). The plot of K vs y 0 at x 0 = 0 is shown in gure 5. The gure is identical to the plot of K vs x 0 at y 0 = 0, as we expect from the symmetry of the undulator. 6.. Linear Polarization Horizontal Magnetic Field Figure 6 shows how the K value changes with horizontal position x 0 when y 0 = 0. For this plot, the rows were shifted such that = 4 = :008 m, and =. This is the linear polarization horizontal eld mode. K changes with position in this mode in the same way as in the linear polarization vertical eld mode, as one would expect from symmetry Circular Polarization Right Hand Magnetic Field Figure 7 shows how the K value changes with horizontal position x 0 when y 0 = 0. For this plot, the rows were shifted such that = 4 = :008 m, and = =. This is the circular polarization right hand eld mode. Note the gradient at the magnetic center.

23 Figure 3: This gure shows how K varies with x 0 when y 0 = 0 in linear polarization vertical magnetic eld mode. Figure 8 shows how K varies near the magnetic center. The slope of the K change with x 0 is very large, 161 1/m. A misalignment of the undulator by 100 microns changes K by 0:016, or 0:6%. This is a very large change for a modest misalignment. Figure 9 shows how K varies with y 0 when x 0 = 0. The sign of the change of K with y 0 is opposite to the sign of the change of K with x 0, but the magnitude is the same, as expected from equation 180 with = Circular Polarization Left Hand Magnetic Field Figure 10 shows how the K value changes with horizontal position x 0 when y 0 = 0. For this plot, the rows were shifted such that = 4 = :008 m, and = =. This is the circular polarization left hand eld mode. Again note the gradient at the magnetic center. The variation of K with position in this mode is very similar to the case of right hand circular polarization, as one would expect from symmetry. 3

24 Figure 4: Relative deviation of K from its value at the magnetic center in a small region near the magnetic center. Figure 5: K variation with y 0 when x 0 = 0 in linear polarization vertical magnetic eld mode. 4

25 Figure 6: This gure shows how K varies with x 0 when y 0 = 0 in linear polarization horizontal magnetic eld mode. Figure 7: This gure shows how K varies with x 0 when y 0 = 0 in circular polarization right hand magnetic eld mode. 5

26 Figure 8: K variation with x 0 when y 0 = 0 in circular polarization right hand magnetic eld mode. It is a zoom view near the magnetic center. Figure 9: This gure shows how K varies with y 0 when x 0 = 0 in circular polarization right hand magnetic eld mode. 6

27 Figure 10: Circular polarization left hand magnetic eld mode K vs x 0 with y 0 = 0. 7

28 7 Laboratory Coordinate System The calculations for the Delta undulator are most easily done in the undulator reference frame. This was used in all the calculations up to this point. For comparing the calculations to measurements, however, it is useful to also work in the laboratory reference frame. The relation between frames is x 0L = 1 p (x 0 y 0 ) (196) y 0L = 1 p (x 0 + y 0 ) (197) or equivalently x 0 = 1 p (x 0L + y 0L ) (198) y 0 = 1 p ( x 0L + y 0L ) (199) The subscript L indicates the laboratory frame, and the unsubscripted quantities refer to the undulator frame. Consider equation 181 giving to rst order in x 0, x 0, y 0 and y 0 the expression for the K value near the magnetic center and reproduced below. " K = K 0 s k r cos + cos 4 k 1 + s y0 sin k u cos ku 4 kr cos k u + cos k 4 u x 0 cos k u sin ku 4 Expressing this formula in laboratory coordinates, we nd " 1 + k r cos K = K 0 k s which can be expanded to give s k r + cos 4 K = K 0 s k r cos + cos 4 " p 1 ( x 0L + y 0L ) sin k u cos ku 4 1 p (x 0L + y 0L ) cos cos + cos 4 sin ku 4 # sin () # # sin () (00) x 0L f K 0 q k r sin ksk r cos k u + cos k 4 u cos 4 + cos 4 sin sin ()g +y 0L f K 0 q k r sin ksk r cos k u + cos k 4 u cos cos sin sin ()g (01)

29 The gradients in K at the magnetic center in the laboratory 0L = K 0 sin k s q cos k u + cos k 4 u 4 cos + cos 4 sin sin 0L = K 0 sin k s q cos k u + cos k 4 u 4 cos cos In the primary modes, = 4 =. In 0L = K 0 k s sin 4 sin sin () (03) sin 0L = 0 (05) In the linear modes, sin () = 0 and there are no gradients. In the circular modes, however, sin () = 1 for right and left handed magnetic elds, respectively. In this case for the circular = K 0 k s 0L (06) Inserting the parameter values used previously, namely 0L = 0 (07) = 196 1/m (08) = 186 1/m (09) K 0 = 1:8 (10) sin = p 0L = 7 0L = 0 () where the plus sign is for right handed magnetic elds and the minus sign is for left handed magnetic elds. This is a very large gradient in x 0L. A misalignment of 100 microns in x L causes K to change by :07 for this setting. Note that smaller values of give a smaller gradient. A misalignment in y L does not produce a change in K. 9

30 8 Conclusion The general expressions for the position dependence of K were derived in this note. We found that it is highly desirable to have as small as possible. With > 0, there are rst order gradients in K with position, and the K value will be di cult to set by conventional magnetic measurement and alignment techniques. Acknowledgements I am grateful to Heinz-Dieter Nuhn for many discussions about this work. 30