Multipole components in the RCS-BM Hideaki Hotchi Dec. 8, Tokai

Central orbit (cm) x y s ideal orbit orbit estimated by tracking 6.67mm  L=  =  m  x 24=  m for whole circumference)

Field distribution (181 MeV) along the actual central orbit B y (T) s (m)

Estimation of muptipole field components in the BM Assuming A x =A y =0 and  →∞, Assuming the mid-plane symmetry (no skew field) : a n =0, Case1 : estimation with the B y distribution on the medium plane Case2 : estimation with the B y distribution along a circle (radius=R) x y  Medium plane Central orbit

Estimation of multipole field components (case1) B y (T) s (m) ⅠⅡⅢⅣⅤⅥⅦⅧⅨ Ⅹ B y L (Tm) x (m) ⅠⅡ Ⅲ Ⅳ Ⅴ Ⅵ ⅦⅧ Ⅸ Ⅹ B y distribution along the central orbit The field area is divided into 10 pieces. B y L distribution for each region

Multipoles in the BM (case1) B y L (Tm) x (m) Ⅰ Ⅱ Ⅲ Ⅳ Ⅴ Ⅵ Ⅶ Ⅷ Ⅸ Ⅹ fitting curve s (m) K0K0 K 1 (m -1 ) K 2 (m -2 ) K 4 (m -4 ) K 6 (m -6 ) K 3 (m -3 ) K 5 (m -5 ) K 7 (m -7 ) ⅠⅡⅢⅣ ⅤⅥⅦ ⅧⅨⅩ

Estimation of multipole field components (case2) R=5.0 cm x y   (rad) B y (T) s=0 s=1.38 m s=1.75 m

Estimation of multipole field components (case2) - cont’d - B y (T) s (m) ⅠⅡⅢⅣⅤⅥⅦⅧⅨ Ⅹ B y L (Tm)  (rad) ⅠⅡ Ⅲ Ⅳ Ⅴ Ⅵ ⅦⅧ Ⅸ Ⅹ B y distribution along the central orbit The field area is divided into 10 pieces. B y L distribution for each region

Multipoles in the BM (case2) B y L (Tm)  (rad) Ⅰ Ⅱ Ⅲ Ⅳ Ⅴ Ⅵ Ⅶ Ⅷ Ⅸ Ⅹ Reconstructed curve (up to n=4) s (m) K0K0 K 1 (m -1 ) K 2 (m -2 ) K 4 (m -4 ) K 6 (m -6 ) K 3 (m -3 ) K 5 (m -5 ) K 7 (m -7 ) ⅠⅡⅢⅣ ⅤⅥⅦ ⅧⅨⅩ where

Comparison s (m) K0K0 K 1 (m -1 ) K 2 (m -2 ) K 4 (m -4 ) K 6 (m -6 ) K 3 (m -3 ) K 5 (m -5 ) K 7 (m -7 ) ⅠⅡⅢⅣ ⅤⅥⅦ ⅧⅨⅩ Blue - case1 Red - case2 A x, A y ≠0 in the end-field region !! The end field has a sextupole-like and octupole-like multipole field component. Assuming  →∞,

- cont’d- The parameters (b n, b 0 ”,b 1 ”) can be determined with the B y distribution on the medium plane. s=0.0 m s=0.8 m s=0.4 m s=1.2 m s=1.6 m s=0.2 m s=1.0 m s=0.6 m s=1.4 m s=1.8 m B y (T)  (rad) The B y distribution along a circle (r=5cm) can be reconstructed reasonably well using the parameters (bn, b0”,b1”) determined from the By distribution on the medium plane.

- cont’d- K 2 ( Ⅵ - Ⅹ ) =  m -2 for “case1” =  m -2 for “case2” K 3 ( Ⅵ - Ⅹ ) =  m -3 for “case1” =  m -3 for “case2” ○ K 2 and K 3 in the fringe field region ○ Driving term for “Q x  2Q y =  6” and “4Q x =27” The effects from b 0 ” and b 1 ” are canceled out due to a characteristic shape of the fringe field. G 1,-2,-6 ( Ⅵ - Ⅹ ) = (cos: , sin: ) for “case1” = (cos: , sin: ) for “case2” Q x  2Q y =  6 G 4,0,27 ( Ⅵ - Ⅹ ) = (cos:  , sin: ) for “case1” = (cos:  , sin: ) for “case2” 4Q x = (6.74,6.27)

Tracking by SAD - modeling - B y (T) s (m) ⅠⅡⅢⅣⅤⅥⅦⅧ - The bending field is considered as “step function”. - Multpole field components (K 1 ～ K 6 ) are introduced as “thin lens” at the center of each region. Ⅸ Ⅹ

Tracking by SAD - condition - -The start point of tracking is set at the entrance of the 1 st BM. -Initial condition of the beam particle : x=y, x’=y’=0. z=0.,  p/p=0., +0.5, +1.0% -Physical apertures are set for all the BEND, QUAD and SEXT. -Multipoles up to n=6 (14-pole) are introduced for tracking. -The field strength of quadrupole magnets is re-fitted after introducing multipole field components of the BM. -Q’s fringe : ON (f1=0.431) -Chromaticity correction : ON and OFF - Synchrotron oscillation : OFF -Number of turns : 1000

Tracking by SAD - results (1) - Q x (Q y ～ 6.19) Q x (Q y ～ 6.23) Q x (Q y ～ 6.27) Xmax=Ymax (cm)  p/p=0%  p/p=0.5%  p/p=1% w/o chromatic correction case1 case2 Sasha’s cal. (3D_BM & QFF)

Tracking by SAD - results (2) - Q x (Q y ～ 6.27) Xmax=Ymax (cm)  p/p=0%  p/p=0.5%  p/p=1% w/ chromatic correction case1 case2 Sasha’s cal. (3D_BM, QFF & SCC)

 mm mrad Qx Qy Mapping - Multipoles up to n=4 (case1) [calculations with n=<6 are underway.] - With chromatic correction - With synchrotron oscillation (assuming stationary bucket) - Start point of tracking : 1 st QDX - Initial condition of the beam particle:  x =  y, x=(  x /  x ) 1/2, x’=0, y=(  y /  x ) 1/2, y’=0, z=0,  p/p=0 and 0.5% - Number of turns : 5000 Q x -2Q y =-6 4Q x =27 Q x -4Q y =-18 Q x -Q y =0 5Q x =33 6Q x =39 ?  p/p=0%  p/p=0.5%