TRIUMF: Canada's national laboratory for particle and nuclear physics

Ellipse matching with doublet or triplet thin quadrupole lenses Tautz, M. F. Jun 1, 1968

Your browser doesn't seem to have a PDF viewer, please download the PDF to view this item.

Item Metadata

Download

Media
51833-TRI-1968-06.pdf [ 13.97MB ]
Metadata
JSON: 51833-1.0107796.json
JSON-LD: 51833-1.0107796-ld.json
RDF/XML (Pretty): 51833-1.0107796-rdf.xml
RDF/JSON: 51833-1.0107796-rdf.json
Turtle: 51833-1.0107796-turtle.txt
N-Triples: 51833-1.0107796-rdf-ntriples.txt
Original Record: 51833-1.0107796-source.json
Full Text
51833-1.0107796-fulltext.txt
Citation
51833-1.0107796.ris

Full Text

T R I - U N I V E R S I T Y M E S O N  F A C I L I T YTRIUMFELLIPSE MATCHING WITH DOUBLET OR TRIPLET THIN QUADRUPOLE LENSESbyMAURICE F. TAUTZUniversity of Victoria Physics DepartmenttUNIVERSITY OF ALBERTA SIMON FRASER UNIVERSITY UNIVERSITY OF VICTORIA UNIVERSITY OF BRITISH COLUMBIA TRI-68-6ifiifE R R A T Apage 10page 11page 14ELLIPSE MATCHING WITH DOUBLET OR TRIPLET THIN QUADRUPOLE LENSESM.F. Tautz TRI-68-6equation 1-5 Interchange a and g . equation 1-8 Read " Xt(T_1)t EL T_1X = e "line 7 Read " GB - A2 = detCT- 1 ) 11 det T_1 det Eq = yg - ct2 = 1 "G yRead " Xa - I (A - a ) -' Y G X X X Xpage 17 line 10 Read "into equations 11-13 and 11-17."ELLIPSE MATCHING WITH DOUBLET OR TRIPLET THIN QUADRUPOLE LENSESM. F. Tautz TRI-68-6Physics Department University of Victoria June 1, 1968TABLE of CONTENTSAppendixAppendixAppendix1. Introduction2. Doublet Matching3. Triplet Matching4. Computer ProgramI - Review of Basic ConceptsII - Solution of Thin Lens Doublet EquationsIII - Program Listing and Sample Output- 1 -1. Introduction:Matching based on the thin lens approximation can be used to determine starting values for digital computer search programs which attempt thick lens matching. The problem of matching with a symmetric* thin lens doublet has been solved by B. Chidley and S. Lokin.'*' That is, given the phase space ellipses representing the beam at the entrance and exit of such a system, they can find the two focal lengths and three drift lengths such that the resultant transfer matrix takes the input ellipses into the output ellipses. Now from a practical point of view there would seem to be two criticisms of Chidley and Lokin's treatment:1. The distance between the input and output ellipses is not specified as input data although this is usually a fixed quantity in a matching problem.2. The assumption of symmetry restricts the number of possible solutions.The following theory shows that these two objections can be overcome in a natural way.* The two thin lens focal lengths are equal in absolute value but of opposite sign.- 2 -2• Doublet Matching:The horizontal plane transfer matrix for a thin lens doublet is taken as:1 L3\ / 1 o\ /H =0 1 \l/f2 11 L 2 / 1 0 1 Li\ 0  1 j  yl/f! 1 / \0 11 + L2/fi + L 3 / f j + L3/f 2 L] + L2 + L 3 + L ^ / f i  + L!L3/f2+ L2L3/(f!f2) + L2L 3/f2 + L 1L3/f1 + L^Lg/(fjf2)2.11/f! + l/f2 + L2 /(f1f2) 1 + Li/fi + L x / f 2 + L 1L2 /(f1f2)\ + D2/f2The components for the verticle plane matrix V can be found by substitutingfl— * -fiin 2.1f2 *■ -f 2We assume that the particle beam is represented by a phase space ellipse and that the emittance is known. The shape and orientation of the ellipse can be specified by the two independent parameters a, y. 2The transfer matrix components for any system which takes an initial ellipse specified by a, y into a final ellipse specified by A, G are given by/ |  (cos <j) + A sin <f> ) |^(a-A)cos t|> + (1+aA) sin cj>J-v'Gy sin <j> (cos <f> - a sin <p)where is a free parameter. For more details see Appendix I.2.2- 3 -If we use the subscripts x and y to denote quantities in the horizontal and vertical planes respectively, then we get (by comparing equations 2.1 and 2 .2) the eight equations given below.Hi i = 1 + + L3/fi + + L2L3/(fif2) - J r  (cos<J)x + A^ sincJj^ ) 2.3Jx-FH22 = 1  + L i / f i  + L i / f 2 + 22/^2  L 1L2 / ( f i f 2) ~ (cosc(>x - a^sincfi^) 2.4x-FH21 = i/fi + 1/f 2 + ^2 / (f 1^2) ~ - y v x sin<*’x 2’^V n  = 1 - L2/fi - Ij3/f 1 - L 3/f2 + L2L 3/(fif2) = (cos<()y + Aysin<)>y) 2.6GV22 = 1 - Li/fi - Lx/f2 - L2/f2 + LiL2/(fif2) = 1 ^ -  (cos4>y - otysin<}>y) 2.7V2i = - 1/f 1 - l/f2 + IJ2/(f 1 f2> = sin<i)y 2,8H12 = Li + L2 + L 3 + LiL2 /fi + LiL3/f2 +L2L 3/f2 + LiL3/f 1 + LiL2L3/ (f if 2) =sj x :(a - A ) cos<)> +X X X(1 + a A ) sincfi x x  x2.9V12 = L i  +  L2 + L 3 -  L i L 2 / f i  -  L i L 3 / f 2 -L2L3/f2 - LiL3/fi + LiL2L 3/(fif3) =J y  y( a  -  A ) cos<j) +y y y(1 + a A ) sin<f> 1 2.10y y y)- 4 -Since det H = det V = 1, only six of the above equations are independent.We thus have six independent equations in the seven unknowns Lj, L2 , L3, f 1 j f2 > (J)X > 41 • Chidley and Lokin assumed the system was symmetric, i.e. they added the seventh equation fj = - f2 and were then able to solve the seven independent equations in seven unknowns. However, a less restrictive seventh equation may be taken. If we use^1 + ^2 + L3 = D 2.11where D (the system length) is assumed known, then equations 2.3 through 2.11 can also be solved. This is done by showing that solving this set of nine equations can be reduced to finding the real roots of an eight degree polynomial in sin<j) . The mathematical details are given in Appendix II. There are up to eight solutions depending on the number of distinct real roots of the polynomial. Some of these solution systems will have to be discarded as they contain negative drift lengths.- 5 -3. Triplet Matching:In order to accomplish thin lens triplet matching, one mustfind the transfer matrix analagous to 2.1 and then equate transfermatrix components as in equations 2.3 through 2.10. However, we nowhave six independent equations in the nine unknowns: Lj, L2, L3 , Li*,f 1 , f2, f 3 , <}>x , <f)y. Clearly several assumptions could be made. Onemight add again equation 2.11 withfl = f3 = - f2and try to solve for a "symmetric" triplet system. This has been done and a program has been written to find such systems. However, again one could raise the objection that the assumption of symmetry is too strong and rules out many good solutions. A more satisfactory method is to keep equation 2.11 but to leave the system with two free para­meters. These are arbitrarily taken as the first drift length and the focal length of the first thin lens. By allowing these to vary through a range of values we can use the doublet matching theory previously described to match to the rest of the system. The computer program "MATCH" uses the above procedure to do thin lens triplet matching. At present the first drift length is varied through values of Li = D/Iwhere I = 4, 5, --- 10, and the first thin lens focal lengths aregiven values of fi = ± J where J = 1, 2, —  4 meters. These valueswere chosen arbitrarily from the range of reasonable values.- 6 -4. Computer Program:The computer program "MATCH" has the following input-output conventions. The first data card contains two numbers read in format (12, F 7.5). The fixed point number is a control parameter, MR, which specifies the matching routine as one of the following:MR = 1 doublet and triplet matchingMR = 2 doublet matching onlyMR = 3 triplet matching onlyThe second number on the card is the requested system length in meters. The next data card reads 8 numbers in format (8 F 10.5). These are the values of a and 8 for the input and output ellipses. They are read inthe following order: a , 0 , A , B , a , g , A , B ,  where capitalsx x x x y y y yrepresent the ellipse parameters at the exit of the system. The units3are as in "TRIUMF" , i.e. cm. rad/100. Any number of matching jobsmay be run consecutively with each one specified by the two data cardsjust described. The last card in the data deck must be a blank card.The output of the program consists of a print out of the input data followed by a list of drift lengths and focal lengths (both in meters) for solutions which have been found. Solutions with negative drift lengths have been discarded.- 7 -REFERENCES1. Chidley, B. and S. Lokin, Thin Lens Doublet Notes, PrivateCommunication received from Chalk River, September 8 , 1966.2. Steffen, K. G., High Energy Beam Optics, Interscience, London (1965).3. Tautz, M. F., M. Sc. Thesis, March, 1968.ACKNOWLEDGEMENTThe author wishes to thank Dr. R. M. Pearce, Dr. D. Lobb, and Dr. L. P. Robertson, whose comments regarding this work were much appreciated.Appendix IReview of Basic ConceptsWe use orthogonal Cartesian co-ordinates wherex is the horizontal co-ordinate, y is the vertical co-ordinate, and z is the beam direction.The i ^  beam handling element (drift space or quadrupole magnet) canbe represented to first order by a 2 x 2 transfer matrix M.. This1matrix operates on input displacement-divergence vectors and the re­sulting vectors describe the particle motion at the exit of the element. We have= M1-1where primes denoteA similar equation holds in the yz-plane. The transfer matrix T for an entire system is obtained by computing the product, in the correct order, of the individual element matrices; i.e.,NT = n M. i-1 1where N is the number of elements in the system. The drift space transfer matrix in both planes is given bywhere L is the length of field free region.- 9 -For a quadrupole short compared to its focal length we have, to good approximation,The focal length is positive if the lens is defocussing and negative if it is focussing. A quadrupole lens which is focussing in one plane is defocussing in the other.*We note that the determinant of any transfer matrix equals unity since this is true for the individual element matrices.A beam of particles may be represented by a region indisplacement-divergence phase space which is enclosed by a central3ellipse. This ellipse may be written for the xz-plane aswhere f is the focal length.yx2 + 2axx' + $x'2 = z 1-2where the emittance e= ellipse areair 1-3and y3 - a2 = 1 1-4A similar equation can be written for the yz-plane.* It is assumed that the magnet poles are oriented with respect to the co-ordinate system in the usual way so that the x and y motions are independent.- 10 -When the beam travels through a transport system with transfer matrix T the ellipse coefficients are transformed according to/M /-V o) \- Tx!T2i T11T22 + Tl2T 21 “ T12T2 2 \-  2 TU T12 T i 22~ 2 t 21T h  t 22TilT 2 1\ f a  \ie1 V /1-5The transformation above is easily proved. For transfer matrix T we haveX = TX/xwhere X =vx'XX = oX1-6and we can write equation 1-2 asX E X = e o o o y awhere E = oa 6and the superscript t denotes the transpose operation.1-7Then substituting for IL in equation 1-7 gives1-8where e does not change since det T = 1.- 11 -The new ellipse coefficients are then given byG A A B= (T 1)tE T 1 1-9owhich we recognize as the transformation law of a second rank covariant tensor. Multiplying out the matrices in 1-9 and equating components gives immediately equation 1-5.Taking the determinant of equation 1-9 yieldsGB - A2 = det(T_1)t det T_1 = det Eq = y3 - a2 = 1and we see that equation 1-4 still holds. We note that at any point on the beam line only three of the ellipse parameters are independent.If we substitute the components of the transfer matrix given by 2.2 into equation 1-5 we obtain an identity.Appendix IISolution of This Lens Doublet EquationsTo solve equations 2.3 through 2.11 we proceed as follows eq. 2.5 - 2.82(1/f! + l/f2) = PSM - V21 eq. 2.3 + 2.4 - 2.6 - 2.72(1/f! + l/f2) D = Hn  + H22 - (Vn  + V22).. (H23 - V21) D = Hj i + H22 - (Vn + V22)eq. 2.5 + 2.82L2/(f^f2) = H21 + V21eq. 2.3 + 2.62 + 2L2L3/(f! f 2) = H u  + VnH n  f  V11 - 2H2 1 + v21eq. 2.4 + 2.7S A STCT2/(f^f2) = H22 + V22H22 + V22 - 2. .  Li =Jl H21 + V21eq. 2.9 + 2.102D + 2IqL2L3/(f^f2) = P C2 + Vj2II-lII-2II-3II-4II-5II-611-1II-8II-911-10Hi2 + V12 - 2D LlL3 H2i + V2i- 13 -From equations II-6 , II-8 , 11-10:Hi 2 + V 12 - 2D H22 + V22 - 2 Hn  + V n  - 211-11H i  2 +  v 12 Y X H i  V21 H21 +  V21Multiplying out and using det H = det V = 1 we get^21^12 d" H2 iVi2 — H22Vn  - V22HU  — 4H2 iD + 4Hi i + 4H22 — 6 = 0 11—12Now if we substitute for the components of H and V (using equations 2.3 to 2.10) we get for equation II-3:whereYi a + Y o b + Y a c + Y k d = 0  11-13a = cos d> b = sin d> and a2 + b2 = 1 11-14x xc = cos d> d = sin <j> and c2 + d2 = 1 11-15y y^=k+fiY2 = A h r -  -  a  —  + d/g yX G X V X XI X  I XYu = - A - 2 - + a - 2 -  ~ d/g y y j G y yx/yy y y11-16- 14 -The same sort of substitutions in eq. 11-12 yieldsa(Xid + Xi+c + X5) + b(X2d + X3c + X6) + X7 = 0 11-17whereXi  =+  D' f f  (ak - x> -X X2+  Dx y T/  +  D  5  D  Jv x  	  + D  x x/  +  5  52 / D 5(1+a A )+a A /—*^ 55 5  D  5   D  +   D  5+ /  / x x+  Dy y           R ID  D  5  D  D  +  5  DG  /x 3 = / 7T~~ (A - a ) + a , „2 / v /  D  D  v   /  +y y 5  DG y  D  55  I D  +  D  5v   	  H N G N2 5  D     D  5/  +   D  +  5  D    D  5X5 =Xc =v  xx 	   2  D   	    5 5    + 5  5    / 511-18X7 = - 6Solving equations 11-13 and 11-17 for a and b gives-(X2d + X3c + Xg)(Y3c + Y^d) + Y2X7a =b =det(Xxd + X^c + X5)(Y3c + Y^d) - Y ^det11-1911-20wheredet = Yj (X2d + X3c + X6) - Y2(Xid + X^c + X5) 11-21- 15 -Substituting equations 11-19 and 11-20 into 11-14 we eventually get:c4Zi + 1 - + d4Z3 + d^Z^ + cddZ5 + cd^Zg + c2d2Z7 + czdZ8 + cd2Zg + c2Ziq ^2Zii + cdZ32 + £^13 + dZjt^ . + Z15 = 0where11-222 2 2Zj = - Y 3 (X3 +  X 4 )Z2 = -  2 Y 3 (X3X 6 +  X ^ X 5)2 2 2Z 3 = - Y 4 (X2 +  X :)2Zh = - 2Y 4 (X6 X 2 +  XgXi)2 2 2Z 5 = -  2 (Y3 (X2 X 3 +  X ^ )  +  Y 3Y k ( X 3 +  X k ) )2 2 2Z 6 = - 2(Yl+(X2 X 3 +  X ^ )  +  Y 3Y 4 (X2 + X 3))2 2 2 2 2 2Z 7 = -  ( Y 3 (X 2 +  X 3)  +  Y^ ( X 3 +  X , , )  +  AYgY^ ( X 2X 3 +  X ^ ) )2Z 8 = -  2 ( Y 3 (X 2X 8 +  X i X s )  +  2Y3Y k ( X 3X 6 +  X ^ X 5 ) )2 11-23Z 9 = - 2(Y 1+(X5 X 3 +  X 5 X 4 ) +  2 Y 3 Ytt (X2 X 6 +  X ^ g ) )Z 1 0 = - Y 3 (Xg+X5)+X 3Y i + X ^ 2 - 2 Y 1 Y 2 X 3X 1++ 2 Y 2 X 7 X 3Y 3+ 2 Y 1 X 7 X 4 Y 32 2 2 2 2 2 2Z n  = - Y Lt(X6+ X 5 )+Y 1X 2+ Y 2 X 1 - 2 Y 1 Y 2 X 2 X 1+ 2 Y 2 X 7 X 2 Y [++ 2 Y 1 X 7 X 1 Y lt2 2 2 2 Z 12 = 2(-Y 3Y 1+(X6+ X 5 )+Y 1X 2 X 3+ Y 2 X 1 X 1+- Y 1 Y 2 (X2 X 1++ X 3X 1)+ Y 2 X 7 (X2Y 3+ X 3Y 1+)+Y 1 X 7 ( X ^ + X ^ )  )\ 32 1 Z = - +5 65 27 - Z5 (5 YX - 3- Z = 5 25 Y7 5 65 ( ) 7 - Z5 ^5 6- 27 - 35 ^5 Y- 2 )2 2Z 11+ = 2 ( Y 1X2X 6+ Y 2X 1X 5- Y 1Y2 (X 2X 5+X5X 1) + Y 2X 7X 6Y ^ + Y 1X 7X 5Y ^ )Z 15 = Yi(Xg -  X 7 ) +  Y 2 (X8 -  X 7) -  2Y 1 Y 2 X 6X 5- 16 -We have from equation 11-15 that c = ±/l - d2.Substituting this in equation 11-22 and isolating the radical gives:±/l - d2 (Wx + W2d2+ W 3d3 + W^d ) = -(W5 + Wgd 4- W?d2 + Wgd3+ Wgd4) whereWi = Z2 +  Z i 3W2 = Z 5 +  Z i 2W3 =  Zg -  Z 2Wg = Zg - z 5W 5 = z x +  z 1 0 +  z 1 5W6 = Zg + Z1LhW7 = - 2ZX + Z7 - Z10 + ZnFinally squaring equation 11-24 yields:Vj +  V 2d +  V 3 d 2 +  V g d 3 +  Vgd 4 +  V 6 d 5 +  V 7 d 6 +  V 8 d 7 +  V 9 d 8 = 0 where2 2V 1 = W1 -  W 5V2 = 2 ( W 1W2 - W 5W6)2 2 2V 3 = 2 (W1W3 — WgW7) — W 3 + W2 - WgVg = 2 (W^Wg d- W 2 W 3 — W^ W 2 — W 5W 3 — W 6W 7)2 2 2V 5 = 2 (W2Wg — W ^ W 3 — WgWg — WgWg) — W 2 +  W 3 — W 7Vg = 2 (W3Wg — W^Wg — W 2W 3 — WgWg — W 7 Wg)2 2 2V 7 = — 2 (W2Wg +  W 7 Wg) — W 3 +  Wg — WgVg = - 2(W3Wg +  WgWg)2 2 Vg = — Wg — Wg11-2411-2611-27- 17 -tti thOnce an i real root, cL, of this 8 degree polynomial is known,then the corresponding c^, a , and can easily be determined. Wecan find ± c^ immediately from equation 11-15. Thus Y = ± Y3c + Y^d^is known. Putting this in equation 11-13 and using equation 11-14 gives:± YT/l - b? + Yob. + Y = 0 with the solution2 2 2 2 2 i2YY2 ± [4Y2Y 2 - 4(Yi +  Y 2 )(Y - Y x)] ^  b =      5  1 T - 2 82 (Yf + Y2)and from equation 11-14 we get ± a^.The ambiguities in sign can be resolved by testing each possibility by substitution of a_^ , b^, c^, d^ into equations 11-13 and 11-27.When values of a^, b^, c^, d^ satisfying those equations have been determined, we can find the values of the transfer matrix components from equations 2.3 to 2.10. Then L3 and can be found from equations II-6 and II-8 respectively. Substitution of these values in equation 2.11 yields L2.We can now solve for the two focal lengths as shown below: eq. 2.3 - 2.6(L2 + L3) /f! + L 3/f2 = 1/2 (Hn  - Vn ) 11-29- 18 -eq. 2.4 - 2.7Lj/fj + (Lj + L2 )/f2 = 1/2 (H22 — V2 2) 11—30(L]_ + L2)(Hn  - Vn ) - L3 (H22 - V22).'. 1/f x = -----------------------------------  11-312 Det- lq (Hi 1 - Vn) + (L2 + L 3)(H2 2 ~ V2 2)1/f 2 = -------— ---------------------------- 11-322 Detwhere Det = (L2 + L3)(Lj + L2) — IqL3The matrix components for H and V should be recalculated for a final check that the drift lengths and focal lengths found do satisfy equations 2.3 to 2.10. In theory the problem has now been completely solved for a thin lens doublet.To find the real roots of equation 11-26 it was noted that d = sin (J> and hence the abscissa is limited to from -1 to +1. Values of the polynomial can be calculated by the computer for every .01 in this range and the roots located by detecting changes in sign of the function. This method works well and a sample problem is shown in Appendix III.It might happen in using this method that some real roots may be missed because of the finite step size. To eliminate this possibility a more sophisticated method for determining the roots of a polynomial (e.g. I.B.M. System/360 Scientific Subroutine Package "POLRT") could be used.

Cite

Citation Scheme:

        

Citations by CSL (citeproc-js)

Usage Statistics

Share

Embed

Customize your widget with the following options, then copy and paste the code below into the HTML of your page to embed this item in your website.
                        
                            <div id="ubcOpenCollectionsWidgetDisplay">
                            <script id="ubcOpenCollectionsWidget"
                            src="{[{embed.src}]}"
                            data-item="{[{embed.item}]}"
                            data-collection="{[{embed.collection}]}"
                            data-metadata="{[{embed.showMetadata}]}"
                            data-width="{[{embed.width}]}"
                            async >
                            </script>
                            </div>
                        
                    
IIIF logo Our image viewer uses the IIIF 2.0 standard. To load this item in other compatible viewers, use this url:
http://iiif.library.ubc.ca/presentation/dsp.51833.1-0107796/manifest

Comment

Related Items