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dissertation entitled

"Theory and Applications of Arbitrary-Order Achromats"

presented by

We ishi Wan

has been accepted towards fulfillment
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Ph. D . degree in Physics

 

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Date May 10, 1995

 

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Theory and Applications
of
Arbitrary-Order Achromats

By

Weishi Wan
A DISSERTATION

submitted to
Michigan State University
in partial fulfillment of the requirements
for the Degree of

DOCTOR OF PHILOSOPHY

Department of Physics and Astronomy

1995

ABSTRACT

An Analytical Theory of Arbitrary Order Achromats and Their Applications

By

Weishi Wan

An analytical theory of arbitrary order achromats for optical systems with mid-
plane symmetry is presented. Besides repetition of cells, mirror symmetry is used
to eliminate aberrations. Using mirror imaging around the x-y and x-z planes, we
obtain four kinds of cells: the forward cell (F), the reversed cell (R), the cell in which
the direction of bend is switched (S), and the cell where reversion and switching is
combined (C). Representing the linear part of the map by a matrix, and the nonlinear
part by a single Lie exponent, the symplectic symmetry is easily accounted for and

maps are easily manipulated.

It is shown that, independent of the choice and arrangement of such cells, there is
a certain minimum number of conditions for a given order; for example, this number
is five for the first order, four for the second order, fifteen for the third order, fifteen
for the fourth order, and thirty-nine for the fifth and sixth orders. It is shown that the
minimum number of cells necessary to reach this optimum level is four, and four of
the sixty-four possible four-cell symmetry arrangements are optimal systems. Various
third-, fourth- and fifth-order achromats are designed and potential applications are

discussed.

Copyright
VVEHSFHCVVAII

1995

To my wife .quiang and my parents.

ACKNOWLEDGMENTS

Throughout the five years of my graduate study at Michigan State University, I
have constantly received help from numerous individuals and organizations. I would

like to take this chance to express my gratitude to each of them.

First of all, I am grateful to my thesis advisor, Prof. Martin Berz, for exposing
me to the interesting topic of high-order achromats, for guiding my research work
through the past four years, for suggesting many important revisions of my thesis.
for teaching me critical and rigorous thinking, and, most importantly, for his faith
in science and his dedication to scientific research which inspired me to continue my
quest for a scientific career. My thanks go to Prof. Henry Blosser, Prof. Jerry Nolen.
Prof. Wayne Repko, Prof. Peter Schroeder, and Prof. Brad Sherrill, for kindly
serving on my advisory committee and for many helpful comments and suggestions

about my dissertation.

I would like to thank Prof. Karl L. Brown and Prof. Roger V. Servranckx for
various helpful discussions, Dr. Jorge More for providing me the optimization package
LMDIF, and Dr. Felix Marti for allowing me access to his private literature collection.
I would like to thank Don Lawton for drawing the wonderful pictures used for my
presentations. I also thank Prof. Julius Kovacs for supporting my first year of study

at MSU.

Special thanks goes to my former colleague, Dr. Georg Hoffstéitter, for many
stimulating discussions and a courageous act when my life was threatened. I would
like to thank other members of our group, Dr. Ralf Degenhardt, Kyoko Fuchi, Khodr
Shamseddine, and Meng Zhao, for discussions and friendship. I give my thanks to
.Iing Wang, Dr. Xiaoyu W11, and Bo Zhang for the exciting soccer games we played

together.

This work was supported in part by the (CS. National Science Foundation, Grant
No. PIIY 89-13815. and the Alfred P. Sloan Foundation. I express my deep appreci-

ation to them for their continuous support of basic scientific research.

I am deeply indebted to my wife .quiang for her endurance of my long working
days. especially on weekends: for her support of my commitment to scientific research
at a time when many are leaving the discipline: for her em‘ouragemeut; and for her
pressure on me when I was lazy at times. It is hard to imagine finishing my graduate
study without her. I am grateful to my parents for their wholehearted love, support.
and inspiration throughout my life. I am extremely sad, however. because my mother

is not able to share my happiness as I finish my graduate study.

vi

Contents

LIST OF TABLES

LIST OF FIGURES

1 Introduction
1.1 The Map Method in Beam Physics ...................

1.2 An Overview of Achromats ........................

2 The Lie Representation of Symplectic Maps
2.1 Lie Transformations and Symplecticity .................
2.2 Lie Factorizatious .............................

2.3 The Baker—('ampbell-Hausdorff Formula ................

3 Repetitive Achromat Theory
3.1 Midplane Symmetry ...........................
3.2 Second-Order Achromat Theory .....................
3.3 Third-Order Achromat Theory ......................

4 Arbitrary-Order Achromat Theory
4.1 Map Representations ...........................

4.2 Optimal Four-Cell Systems ........................
4.2.1 General Properties of k‘(.i(‘ll Systems ..............
1.2.2 Two— and Three-Cell Systems ..................
4.2.3 Four-Cell Systems .........................
4.2.4 The Optimal Four-Cell Systems .................

“1.3 Order-by-Order Solutions .........................

5 Applications
5.1 A Third-Order Achromat - FRSC ....................
5.1.1 The First-Order Layout ............ ~ .........

viii

111
112

5.1.2 The Second- and Third-Order Achromat ............

v1 ‘4‘ V‘
-- .3; ts;

A Third-Order Achromat - FRFR ....................
A Fourth-Order Achromat - FRFR.
A Fifth—Order Achromat - FRFR.

A Symplectic Properties of Matrices R and 5'

B The Proof of Equation (4.60)

LIST OF REFERENCES

viii

113
116
119
121

133

135

137

List of Tables

1.1

C}!
'1—

31
‘1

The number of abe rrations of orders I t05 for a system with midplane
symmetry. The interdependemy of aberrations comes from symplec-
ticity, which is a property of a Hamiltonian system. ..........

The FRSC third-order achromat: The COSY INFINITY output of the
first-order map of the forward cell. The first five columns are Taylor
coefficients of If. a], y], by, and If as functions of 1",, (1,, 11,-, [1,, t,,
and 61. and the sixth column contains the powers associated with the
coefficients on the same row. From left to right, the sub-columns stand
for .13. (1,, y,, 1),. h, and 15,-, respectively. .................

The FRSC third-order achromat: The field strengths and drift lengths
of the first-order layout. 11., is the first-order derivative of the field of
the bending magnet over the bending radius r (dimensionless). . . . .

The FRSC third-order achromat: The COSY output of the third-order
map. A third-order achromat is reached and only (t|6") (n = 1,2,3) is
not cancelled. (Any number smaller than ”3-11 is set to zero.) . . . .

The FRSC third-order achromat: Field strengths of the sextupoles and
the octupoles. The field index, 113, is the third-order derivative of the
field of the bending magnet over the bending radius (dimensionless). .

The FRFR third-order achromat: The field strengths of the quadrupoles
and the sextupoles. Since they are excited symmetrically, only half of
the multipoles are listed ..........................

The FRFR third-order achromat: The first-order map of half the ring.
(The zeroes are numbers smaller than lE-IS) The phase advances are
pl. = 11,, = 7r/2. Note that (tld) also vanishes ..............

The FRFR third-order achromat: The third-order two-turn map. (The
zeroes are numbers smaller than lE-S.) Note that time-of-fiight terms,
which depend on energy and mass, cannot be cancelled along with
(t|6") due to the fact that the magnetic field only distinguishes mag-
netic rigidity. which 1s a function of 6,. and 6m ..............

The FRFR third-order achromat: The field strengths of the octupoles.

The FR FR. fourth-order achromat: The field strengths of the quads and
the sextupoles. Only half of them are shown due to mirror symmetry.

The FRFR fourth-order achromat: The field strengths of the octupoles
and the decapoles. Here all the multipoles are very weak ........

ix

*1

Mil
111
11(1

117

I22
I23

5.11 The FR FR fifth-order achromat: The field strengths of the quads and
the sextupoles. Only half of them are shown due to mirror symmetry.

5.12 The FRFR. fifth-order achromat: The field strengths of the octupoles
and the decapoles. Note that the multipoles are extremely weak as a
result of good linear behavior. ......................

5.13 The FRFR. fifth-order achromat: The field strengths of the duode-
capoles. Note that the multipoles are extremely weak as a result of
good linear behavior. ...........................

130

List of Figures

1.1

1.2
1.3
1.4

3.1

3.3

4.1

CJI
H

91
[\D

The principle of a mirror symmetrical, first-order achromat. The con-
ditions are the cancellation of (al6), (alr), (.rla), (bly), and (ylb) in the
middle where the vertical line lies .....................

The Double Bend Achromat lattice of the APS at ANL. .......
The Triple Bend Achromat lattice of the ALS at LBL. ........

The layout and lattice functions of the Duke FEL storage ring, where
each arc is made up of 10 FODO cells, with phase advances per cell
chosen as 1,/',. = (3/10)27r and #1,, = (1/10)27r ...............

The geometric relationship among cells F, R, S, and C illustrated by
asymmetric boxes. ............................

Complex plane diagram for second-order geometric aberrations of a
four-cell repetitive system with phase advances 27r. ..........

Complex plane diagram for second—order geometric aberrations of a
three-cell repetitive system with phase advances 27r ...........

K. Browns four-cell, second-order achromat. Quadrupoles are used
to tune the system to phase advance 27r in both transfer planes, and
two families of sextupoles, SF and SD, are used to correct chromatic
second-order aberrations in the .r- and y-planes, respectively. To make
the sextupoles weak, they are placed such that or is larger at SF and
[3,, is larger at SD ............................. I .

Optimal four-cell systems and the first-order requirements to achieve
their optimum. ..............................

The FRSC third-order achromat: The first-order forward cell. Only the
bending magnet and the quadrupoles are shown. The phase advances
are 11; = 7r and 11,, = 7r/2. It also shows that at the end of the cell,
the dispersion is not corrected, but dispersive rays are parallel to the
on-energy rays that start with the same initial conditions. ......

The FRSC third-order achromat: The third-order .r-z beam envelope
where only the bending magnets are shown. The mirror symmetry
between cells F and R. and that between cells S and C is clearly shown
in the beam trajectories ..........................

8
9

37

37

38

98

1'13

114

Q1
_. C

v1
*1.

qt
U!

8;!
33

3;!
K1

5.9

5.12

The FRSC third—order achromat: The third-order lab layout. The
“S”-sl1aped geometry helps make it a compact system. ........

The FRSC third-order achromat: Remaining aberrations up to the
eighth order (scale: 30 [1111 X 20 11111). The vertical line represents the
final focal plane, where the deviations are around 10 pm. .......

The FRFR third-order achromat: The original layout of the Experi-
mental Storage Ring (ESR) at Darmstadt, Germany ..........

The FR FR third-order achromat: The first-order beam envelope of the
horizontal (.r-z) plane. The curve that does not coincide with the :-
axis in the straight section is the d—Iunction. The other curve is the
dispersive ray. The circumference is 108.36 m; the emittance is 12.571'
mm mrad; and the dispersion is 0.7% ...................

The FRFR third-order achromat: The 8th-order 1000-turn tracking
picture. The left and right columns display those of .r- and y-motion,
respectively; the top, middle. and bottom rows show those of 6 =
—0.1%. 0. and 0.1%, respectively. ....................

The FRFR third-order achromat: The multi-turn mass resolution as
a function of the number of turns. Due to the small emittance, the
resolution increases almost linearly with the number of turns ......

The FRFR fourth-order achromat: The layout, beam envelope and
dispersive ray. The phase advances are 11,. = M = 7r/2. The circum-
ference is 147.35 m; the emittance is 207r mm mrad; and the dispersion

is 0.6%. ..................................

The FRF R fourth-order achromat: Beam spots of different emittances.
The top two rows show that an achromat is reached after two turns.
The bottom row shows the remaining higher-order aberrations. . . . .

The FRF R fifth-order achromat: The layout, beam envelope and dis-
persive ray. The phase advances per cell are p, = 11,, = 1r/2. The
circumference is 266.64 m; the emittance is 301r mm mrad; and the
dispersion is 0.3%. ............................

The FRF R. fifth-order achromat: 1000-turn tracking of the .r-a motion
of on-energy particles. ..........................

The FRF R fifth-order achromat: Resolution vs numbers of turns at the
acceptance. The saturation comes from the accumulation of higher-
order aberrations over turns ........................

xii

118

119

126

127

p—u
Is;
’1.

128

129

Chapter 1

Introduction

1.1 The Map Method in Beam Physics

In the past six decades, accelerators and other beam optical systems have gone
through tremendous improvements [Wied93]. The energy that can be reached by
accelerators rose from a few MeV to 2 TeV [Law32, Dugan91, Finley91]; and the
beam spot size in the Stanford Linear Collider has been decreased to 75 nm x 1
pm [Schwar94]. All these improvements come from the detailed knowledge of the
motion of charged particles in electromagnetic fields, which is governed by a system

of first-order ordinary differential equations (ODEs)

-O

s: f(:2‘,t). (1.1)

&|&

t

Since in general the ODEs are nonlinear and complicated, computers become a
very useful tool in solving the ODEs numerically. Through the years, there have
been mainly two ways to do this: The ray tracing method and the map method.
The basic feature of the ray tracing method [Gordon59] is to send many particles
through a numerical integrator derived from the ODEs (for example, a Runge-Kutta
integrator), and obtain the final positions and angles of all particles. By studying

the distribution of the particles in the phase space (at a fixed position around the

1

reference orbit, called Poincare section), the dynamic behavior of the beam is studied.
Two examples for the different computer codes are RAYTRACE [Kowa185], which is
used for spectrographs, and TEAPOT [Tal87], which is used for repetitive systems,

especially synchrotrons.

Although ray tracing is conceptually simple, it is time-consuming and does not
readily provide the direct links between the final and initial coordinates. The map
method, on the other hand, focuses on finding the analytical relation between the fi-
nal and initial coordinates through solving the ODEs. Once these relations, functions
between the initial and final coordinates are obtained (the transfer map), all informa-
tion about the particle motion is known. Therefore, major efforts have been devoted
to find the map of the ODEs. It is customary and advantageous to use curvilinear
coordinates along a reference particle, such that other particles are always close to
the origin and the map is origin preserving. Hence, perturbation theory has been the

major tool for solving the ODEs.

Since all particles stay close to the origin, the nonlinearity in the map is weak.
This is why the map can be approximately represented by a truncated Taylor series.
With orders typically reaching ten, an accuracy in the range of 10 digits can be
achieved. The coefficients in the Taylor series, except Bay/0:13;, day/6a,, etc., are
called the aberrations. A intuitive way to obtain the aberrations is to numerically
differentiate the data of the ray-tracing output, like in the code MOTER [Thiess72].
where certain low-order derivatives can be extracted. Yet this method of obtaining
the derivatives is rather cumbersome and limited, due to the loss of accuracy resulting
from the numerical differentiation. This is the main reason for the development of

the map method.

Furthermore, for our convenience, the arc length of the reference particle, rather

than the time, is used as the independent variable. As a result, the phase space

variables are 2:, [1,, y, p,,, t, and E. Therefore, the ODEs are transformed to
d—‘i-s = flag). (1.2)

In order to keep the variables small and simultaneously canonical, a new set of
variables, x, a, y, b, At, and 6K, are used in computer codes like COSY INFINITY
[Ber293] and also in this thesis, where a = px/po, b = py/po, At = (t — to)v07/(l +7),
and 6x = (E1. — EkO)/Eko- Note that all quantities with subscript 0 are associ-
ated with the reference particle. In these coordinates, the map is denoted by M =

(man may my, m6, mt, m6), Where

 

 

 

 

( xi 1 f 3" l
a; a,-
i: = M i . (1.3)
Atf At;
1 5f A k 51' A

When man is written as a Taylor series, the coefficient of the term x? a§°y:"b§"At:'6:‘ is
represented by (zlzf‘a‘“y‘Vb‘°At“6“). The same rule applies to the other functions.
For example, (zlz) is the coefficient of term 2:,- in m, (617/31,), which is the magni-
fication; (a|x) is the coefficient of term 2:,- in ma, which is the defocusing power; and

(xlaz) is the coefficient of a? in m,, which is the second-order opening aberration.

For circular machines, the picture of aberrations can not describe the key aspects
of particle motion conveniently, because they are averaged out over many turns, with
only the ones with the same periodicity as the motion being important. Therefore,
instead of magnification, focusing power, and aberrations, the concepts such as tunes,
betatron functions, and resonances are used to describe the motion in a circular
machine. Generally speaking, the tunes (T) are defined as the remainder of the

number of periods of the motion in one plane over the number of turns, which is

a measure of the entire motion, including the nonlinearities [Berz92a]. To the first

order, the tune of the ith component is
T,- = arccos(Tr(L,-))/27r, (1.4)

where L,- is the linear matrix of one turn for the ith componant and T r(L,-) is the
trace of it. The tunes of the transverse motion are called the betatron tunes and the
tune of the longitudinal motion is called the synchrotron tune. In this thesis, only

betatron tunes (T x and Ty) are relevant. When the relation
1T3 + mTy = n (I, m, and n are integers.) (1.5)

holds, the motion is said to be on an (I + m)th order resonance. As shown in Section
3.3, there are always aberrations which grow exponentially under this resonance,
called the driving terms of the resonance. A motion which is on a resonance with
nonzero driving terms is unstable and the particle will eventually hit the wall and
get lost. Since there are infinitely many resonances, virtually every particle will be
lost after a long time. Practically speaking, particles are only stored in a machine for
a certain period, which means that mostly the low-order resonances are important.
because they cause the growth of certain lower-order aberrations, which are initially

bigger than higher-order aberrations.

After the development of the theory of the alternating-gradient synchrotron [Cour5. ‘.

1.

large synchrotrons took the center stage of high energy accelerators; the map method
has been developed with this. Currently there are dozens of various computer codes in
use, ranging from first-order codes like COMFORT [Wood83] and SYNCH [Garren75].
to second- and third-order codes based on explicit formulas, like TRANSPORT
[Brown73], MAD [Iselin85, Iselin88], DIMAD [Serv85], TRIO [Matsuo76], GIOS [WollS'ia .
and MARYLIE [Dragt85], to higher-order codes of the same approach, like COSY 5.0

[Berz87a], and finally to arbitrary-order codes that do not rely on explicit formulas,
like TLIE [Zeijt392], ZLIB [Yan90], and COSY INFINITY [Ber290, Ber293]. It is
worth noting that it is possible to compute transfer maps of an arbitrary order af-
ter the emergence of the differential algebraic (DA) techniques [Berz89]. Among the
high-order codes (beyond the third-order), COSY 5.0 is a fifth-order one, and the
rest are all DA codes. Of all the DA codes, COSY INFINITY is the first and, up to
now, probably the most general code. It has been shown through the history of the
code development that DA techniques are probably the only practical way to obtain
the transfer map of an arbitrary order. Therefore, they should be discussed in more

detail.

Without involving too much mathematics (see, for example, [Ber290, Ber292b]),
the differential algebraic techniques can be viewed as a way to solve the ODEs to
an arbitrary order in one attempt without loosing accuracy. The keys are, first, that
the arbitrary-order Taylor expansion of a large class of functions, whose variables
are Taylor series with constant parts, can be obtained through a finite number of
operations; second, that Taylor expansions of all functions are done simultaneously,
instead of the traditional way of obtaining higher-order solutions through lower-order

OIICS .

Since it is only necessary and possible to obtain a finite number of terms from
the Taylor series of a function, the truncation order n is always specified when Taylor
expansion is done, and terms of higher orders are neglected. Therefore, infinitely often
differentiable functions, including f of the beam optical systems, can be expanded
around any given point, and the expansion up to order n can be obtained by a finite
number of additions and multiplications, even when f is complicated. Together with
addition, multiplication, and differentiation of polynomials, expressed in the space of

polynomials up to order n and properly implemented in computers, the ODEs can

be solved with a DA-based numerical integrator where the values of the phase space
variables are replaced by the DA vectors containing the constants and derivatives of
those variables. At the end, the solution contains not only the final values but also
all derivatives up to a certain order, which are the aberrations in the transfer map.
It is worth noting that in the code COSY INFINITY, the integrator is a modified
eighth-order Runge-Kutta integrator with automatic step size control to ensure that

the accuracy of the solution is compatible with the specific order.

When the electromagnetic field in a beam optical system does not change lon—
gitudinally, there is a much quicker way to solve the ODEs using the so-called flow
operator. In this case, f in the ODEs does not depend on s, which leads to the result

that the transfer map is

M = exp((s — so)Lf-o)f, (1.6)
where L}- = f~ 6 is the flow operator, and f: (13;, 61,-, y;, b;, At;, 6,). Thus, the map
is obtained after only one step. Note that when f represents a Hamiltonian system,

the flow operator L f becomes a Lie operator similar to that discussed in Section 2.1.

The DA techniques offer a power tool to study beam optical systems, includ-
. ing tracking through a high-order map, map manipulations such as composition and
inversion, computing generating functions and Lie factorizations, studying parame-

ter dependence of certain quantities [Ber292c], and suppressing resonances through

normal forms [Ber292a, Ber292b, Ber293].

1.2 An Overview of Achromats

The search for achromats up to a certain order has generated substantial interest

for the past two decades. Here an achromat is defined as a beam optical system

 

Order 1 2 3 4 5
Aberrations 6 30 70 140 252
Independent aberrations 6 18 37 65 1 10

 

 

 

 

 

 

 

 

 

 

Table 1.1: The number of aberrations of orders 1 to 5 for a system with midplane
symmetry. The interdependency of aberrations comes from symplecticity, which is a
property of a Hamiltonian system.

whose map of the transverse motion is free of aberrations up to a certain order. The
advantage of achromats is that all aberrations of the transverse motion are cancelled,
as are all aberrations of the longitudinal motion except (t|6"); hence, an achromat
transports charged particles without distortion of the transverse motion. This is why
first- and second-order achromats have been so widely used in accelerators, storage
rings, and beam transport lines. Last but not least, this is also an interesting and

challenging problem from a purely theoretical point of view.

Since midplane symmetry has been employed in most of the beam optical systems,
cancelling half of the transverse aberrations, all achromat theories consider only sys-
tems with midplane symmetry. Table 1.1 lists the number of aberrations that have
to be cancelled for achromats up to the fifth order. It shows that the number of
independent aberrations grows so rapidly with increasing order that even up to only
the second order it is unrealistic to obtain an achromat by providing each aberration
with a knob. Therefore, the challenge is how to achieve an achromat with as few

knobs as possible.

Due to their simplicity, first-order achromats have been widely used, especially the
partial achromats where only dispersion is corrected. If we exclude the techniques of
dispersion matching and the dispersion suppressor (see, for example [Wied93]), there
are basically two ways of obtaining a first-order achromat, namely, through mirror

symmetry and repetition. The principle of a mirror symmetrical first-order achromat

 

 

 

 

 

 

 

 

 

 

Figure 1.1: The principle of a mirror symmetrical, first-order achromat. The condi-
tions are the cancellation of (al6), (a|:r), (:rIa), (bly), and (ylb) in the middle where
the vertical line lies.

is illustrated in Fig. 1.1. When (alx), (:r|a), (bly), (y|b) and (al6) vanish in the
middle, mirror symmetry entails that the total first—order map is f (here f stands for

the identity map).

Mirror symmetrical achromats are mainly used as building blocks of synchrotron
light sources [Jack87] due to the low equilibrium emittance achieved (see, for exam-
ple, [Wied93]). As examples, the lattices of the Advanced Photon Source (APS) at
Argonne National Laboratory (AN L), and the Advanced Light Source at Lawrence

Berkeley Laboratory (LBL) are shown in Figures 1.2 and 1.3, respectively [Murphy92].

As shown in Section 3.2, any repetitive system with integer tunes is a first-order
achromat. Since a repetitive first-order achromat (except a 3-cell system) cancels
all second-order geometric aberrations, it has been used for bending arcs of various
synchrotrons and storage rings, especially as a 180° bending arc of a racetrack lattice

[Serv83, Serv83, Litv93, Wu93a, Wu93b]. Fig. 1.4 presents the lattice of the Duke

ANL.APS

 

 

 

 

20 I

E mzozoza “2.12:

5050

15 20 25 30

10
DISTANCE [m]

Figure 1.2: The Double Bend Achromat lattice of the APS at AN L.

5

O

ANLAPS

 

 

‘!

s 10 1‘5 20 25 30
01er [m]

0

 

 

 

Figure 1.3: The Triple Bend Achromat lattice of the ALS at LBL.

10

 

 

 

at 3‘5
O

 

 

 

N
O

3 3 l0 '1 (m)
8
1IVTT'TTYTYV'TVYVVF'L
'. / "
k ‘
\
y,»
7
/
/

o
o}:

0

0

 

 

I (m)

Figure 1.4: The layout and lattice functions of the Duke FEL storage ring, where
each arc is made up of 10 FODO cells, with phase advances per cell chosen as 112,, =
(3/10)21r and 1b,, = (1/10)21r.

FEL (Free Electron Laser) storage ring.

Although the concept of first-order achromats had been widely used in various
beam optical systems and accelerators for a long time, it was only in the 19708 that a
theory developed by K. Brown enabled the design of realistic second-order achromats

in a systematic and elegant way [Brown79, Brown82a].

The theory is based on the following observations. First, any system of 11 identical
cells (n > 1), with the overall first-order matrix equaling unity (I) in both transverse
planes, gives a first-order achromat. When n is not equal to three, it also cancels
all second-order geometric aberrations. Second, of all second-order chromatic aberra-
tions, only two are independent. Therefore, they can be corrected by two families of
sextupoles, each responsible for One of them in each transverse plane. These findings
make it possible to design a four-cell, second-order achromat with only one dipole,
two quadrupoles and two sextupoles per cell. Detailed studies on this theory will be

shown in Section 3.2.

Because of its simplicity, the second-order achromat concept has been applied to

11

the design of various beam optical systems such as the time-of-fiight mass spectrom-
eters, both single-pass(TOF I) [Wouter85, Wouter87] and multi-pass(ESR) [W01187b,
W01187c], the Arcs of the Stanford Linear Collider (SLC), the new facility at SLAC,
the Final Focus Test Beam [Brown85, Brown87a, Brown87b, Schwar94], and the MIT
South Hall Ring (SHR) [Flan289a, Flanz89b].

Since it is hard to generalize the second-order achromat theory to higher orders,
the first third-order achromat theory was developed by Dr. Alex Dragt based on
normal form theory and Lie algebra [Dragt87]. According to the theory, a system of
11 identical cells is a third-order achromat if the following conditions are met: (1) The
tunes of cells T, and T, are not full, half, third or quarter integer resonant, but nTx
and nTy are integers. (2) The two chromaticities and five independent third-order

aberrations are zero. Details of the theory will be discussed in Section 3.3.

Two examples of third-order achromats have been designed. The first design was
done by Dragt himself, containing thirty cells with T, = 1/5 and T3, = 1/6. Each
contains ten bends, two quads, two sextupoles and five octupoles. The whole system
forms a 180° bending arc. The second design was done by Neri [Neri91]. It is a
. seven-cell system with only one bend per cell, with the total bend also being 180°.
The tunes of a cell are T, = 1/7 and T3, = 2/ 7, which seems to violate the theory
1 because of the third-order resonance 2T: — Ty = 0. However, the achromaticity can
still be achieved because the driving terms are cancelled by midplane symmetry (see

Section 3.3). This approach greatly reduces the number of cells.

Similar to Brown’s theory, the Dragt theory cannot be immediately used to find
arbitrary order achromats in such a way that the number of the cells in a achromat
is independent of the order. The main reason is that for any given order, the tunes of
a cell have to be specially chosen such that most, if not all, of the resonances up to

one order higher are avoided. Thus the number of system cells has to be the smallest

12

F R
Figure 1.5: The geometric relationship among cells F, R, S, and C illustrated by
asymmetric boxes.

 

 

 

 

number possible that makes both tunes of the whole system integers, which depends
on the order and usually increases quickly. A second reason is that as the order
increases, the difficulty of obtaining an analytical formula increases rapidly because

of the complexity of the Baker-Campbell-Hausdorff formula.

Our approach for a general achromat theory does not use the normal form. method
and avoids the resonance concern by introducing mirror symmetry to cancel more
aberrations. With these considerations, we are able to study systems with arbitrary
numbers of cells and obtain solutions that are independent of the arrangements inside
a cell. Because of their simplicity, Lie transformations are used to represent symplectic
maps, but instead of an order-by-order factorization, we use a factorization formed
by a linear matrix and a single Lie operator, describing the linear and nonlinear parts
respectively. The introduction of mirror symmetry makes it possible for us to obtain
four total kinds of cells: the forward cell (F), the reversed cell (R), the switched cell
in which the direction of bend is switched (S), and the cell in which reversion and

switching is combined (C) (Fig. 1.5).

This thesis is organized as follows: The Lie representations of symplectic maps are
discussed in Chapter 2, including the definition of a Lie operator, various methods of
Lie factorization, and the Baker-Campbell-Hausdorff formula. As a comparison to our

achromat theory, the theories of repetitive achromats up to order three are presented

13

in Chapter 3. In Chapter 4, the analytical theory for arbitrary order achromats is
studied with a detailed proof provided for every theorem. Chapter 5 consists of four
example designs of achromats of orders three to five. Also presented are studies on
various aspects of the nonlinear dynamical behavior of those examples. Finally, a

summary concludes the thesis.

Chapter 2

The Lie Representation of
Symplectic Maps

In the following chapter the Lie representation of symplectic maps, developed by
Dragt and Firm [Dragt76, Dragt81] is outlined, proceeded by the introduction of Lie
transformations and symplecticity. Only the results closely related to the achromat

theory are presented.

2.1 Lie Transformations and Symplecticity

First, let us review the definition and basic properties of the Poisson bracket, since a

Lie transformation is defined based on the Poisson bracket. On a phase space R“

with variables (ql, ~-, qm, p1, ---, pm), a Poisson bracket of functions f and g is
defined as
"‘ afar W69 ~ ~ *1
, = _____=v .J.V, 2.1)

[f 9] gay. 8p.- 012.611.) f g (
where

.. 6 0 8 0

W =( f f f __f__

53,...,5qin.,5p—l-,...,apm)

l4

15

and J is an antisymmetric 2m x 2m matrix

~ 01
= . , 2.2
J ( _, O ) < 1
As an example, note that
[1,, [j] = Jgj, (2.3)

where f: (ql, ---, qm, p1, ---, pm) and J,,- is the (ij) element of the matrix j.

It is well known that Poisson brackets have the following properties:

Us+hb=UstflM, QA)
[f,tg] = t[f, g] where t is an arbitrary constant, (2.5)
UsM=dfldh+nfiM1 (am
lflhfifl+hdbflHdMUwH=0 @J)

With the Poisson bracket defined as the multiplication on R2”, the space of functions

on 722'" form a Lie algebra.

For any function f on 722'", a Lie operator : f : acting on another function g on

R2” is defined as

=f=g=UwI (ZS
The zero power of : f : is defined as

=f99=g (2%
and the square of : f : is

=f9g==UdfigW 910)

16

with higher powers being defined in the same way. Similar to the Poisson brackets,

Lie operators have the following properties:

=f=(g+h)==f=g+:f=h, (2.11)
: f: (tg) = t: f : g, where t is an arbitrary constant, (2.12)
:f = (M) = (1 f=g)h +9(= f = h), (2-13)
If==g=h—=g:=f=h==1f,~91=h- (2.14)

Note that, with the multiplication defined as
:f:x:g:=:f::g:—:g::f:, (2.15)

the Lie operators on 722'" form another Lie algebra.

A useful relation that : f :" obeys is the Leibniz rule

=f i" (9h) = i ( n )(I f ="‘ g)(= f 1“" h), (2-15)

m=0 m

where

n n!
( m ) = ml(n—m)l°

To prove it, the mathematical induction method is used. First, for n = 1, we have
=f=(9h) = (=f=g)h+g(=f=h),

which satisfies eq. (2.16). Second, for n — l, we assume eq. (2.16) holds. Third, for

n, we have

=f (gh) ——— =f = (. f eh»
f

l7
+Z(";1)(=1=mg>(=f:"-mh)
n n—l m 11—17;
= z;( _ )(=/: gm: h)
+E(”’1)(=f=mg)(=fz"-mh)

= Z(,’,‘,)(=f:mg>(=f="-mh),

which concludes the proof.

A Lie transformation associated with a function f on 722’" is defined as

exp(: f :) = i i : f z" . . (2.17)

1
”:0 n.

To avoid the subtle questions connected to the convergence of a Lie transformation,
where even the definition of the norm is not clear, we require that f is a polynomial of
orders 3 and up, expansion is always truncated at order n, and the functions, on which
the Lie transformation acts, are also polynomials. Since no infinite series is involved,
the conclusions are completely rigorous. On the other hand, due to the fact that n is
an arbitrary natural number, this approach does lose generality. Practically speaking,
this is always the case for the DA maps, which makes this treatment fit perfectly to
the implementation of the Lie transformation. Therefore, the new definition of a Lie

transformation is:

exp(: f :) =n i}, : f :1 . (2.18)

i=0 °

Lie transformations have the following properties:

eXp(= f =)(g + h) =n exp(: f =)9 + exp(: f =)h, (2-19)

exe(= f =)tg =7. teXp(= f :)g, (2.20)

18

exe(= f =)(9h) =7. (exp(: f =)9)(exp(= f Oh), (2-21)

exp(: f :)[g,h] =,, [exp(: f :)g,exp(: f :)h], (2.22)
The proof of eq. (2.21) makes use of the Leibniz rule (eq. 2.16):

(exp(: f :)9)(exp(= f :)h)
:1; (ill—,zf;lg) (i%:fzmh)

m=0 °

 

z“ (:12 fi‘f‘mgu—l ‘f‘m bl)

(=0 m=0 ' m)!
_ n 1 i I! . .m . .l-m h
.. (gfilng—mfl'f' N- )1
n 1 I f m 1_m
=7. (§E(;o(m)zf: ng: h))
=n (2:311! : :1 (gh) (Leibniz rule)
:7: exe(=f =)(gh)- 12-23)

Equation (2.22) can be obtained directly from eq. (2.21).

Now let us look at a Lie transformation acting on a polynomial

g(ql, . . . , qnhpl, . . . 1pm) —_—-_ Z afq1,---,fpmq;q1 . . . qjgmpflpl . . . pig": , (224)

i<11 ’°..’iQM tip] t""iPm

where 1,, + - - - +1.," + ip, + - - - + 1,," g n. Using eqs. (2.19) and (2.21) repeatedly, we
obtain

exp(:f=)g =. exp(=f:)( z qqmpp.)

iq! ooooo ‘qm ’ipl ’ooo"pm

=n Z (1qu ,m'ipm exp(: f :)(q;q1 . . . qigmplpl . . . pgm)

in vvvv iqm vim v"'vfpm

=1; 2 aiql,...,,pm(exp(: f :)q1)‘q1 .. - (exp(: f :)qm)‘?m

iql""’iqm’iP1""'iPm

19
(eXp(:f:)p1)‘Pt ---(exp(: f :)pm)ipm
=71. 9(exp(: f :)q1,~-,exp(: f:)qm1exp(: f I)p1,-",6Xp(1 f I)pm),

which leads to the following important theorem:

Theorem 2.1 [ff is a polynomial of order 3 or higher on 722’", and g is an arbitrary

polynomial, then we have

-o

exp(: f :)g =. g<exp<z f z) 1, ' (2.25)

where

eXP(= f :)f=n (exp(I f :)qt,---,exp(= f :)qn,exp(= f =)q1,---,eXp(= f :)qn). (226)

Theorem 2.1 is ready to be generalized to the case where g is a polynomial of a

map M. Thus, we have the following theorem:

Theorem 2.2 IfIFI is a map and g is an arbitrary polynomial on 722'", then we have

-0

exp(: f :)g(M) =n g(exp(: f :)M). (2.27)

Theorem 2.2 is probably even more important than Theorem 2.1 because in most
cases, the linear map of a system is unity. Therefore, it is the one which is used

directly and very frequently in Chapter 3 and 4.

Another useful concept is symplecticity. A 2m dimensional vector function M is

-o

symplectic when its Jacobian matrix Jac(M) satisfies

A

Jac(M) . J . Jac(M)t = i. (2.28)

20

This kind of function is important in that the transfer map of a Hamiltonian system is

always symplectic [Gold80, Arnold89]. Symplectic maps have the following properties:

[f(M).g(M)1 = (If, 91W), (229)
det(Jac(117)) = 1, (2.30)
Jac(M o N) . j..1ac(1v7 0 Nr = i, (2.31)
Jac(M-l).j-Jac(M-1)‘ = f, (2.32)

4 II
‘o

where [If and [Y are symplectic maps and denotes the composition of two func-

tions. Equations (2.29) and (2.31) can be proven using the chain rule of derivatives.
The proof of eq. (2.31) can be found in reference [Dragt81]. Since the Jacobian of
M“ is the inverse of that of M, we obtain
Jac(M‘1)-j - Jac([f/I'l)‘

= Jac(ll-lf)“1 - j. (Jac(M)-1f

= Jac(Mrl .j . (Jac(M)-1)t

= (Jac(M)‘ - —j . Jac(M))"l

= (_j)-1

= j,
which shows that M '1 is also symplectic (eq. 2.32). Equations (2.31) and (2.32)
indicate that, with the composition “0” as the multiplication, symplectic maps form

a group.

From eq. (2.21), we have

[exp(: f :)I,,exp(: f :)Ij] (2.33)
=.. exp(zf :)[Is 1,] (2.34)
=,, exp(: f 2)J," (2°35)

=71 Jija (2.36)

21

which leads to the following theorem:

Theorem 2.3 Ifa map M on 732'" has the form [Ff =,, exp(: f :)f, it is symplectic.

In the next section, it will be shown that any symplectic map can be represented

by either one or a series of Lie transformations up to an arbitrary order.

Next let us study the composition of exp(: f :)g(l) and map M. Note that this

operation plays a key role in the development of achromat theories.
Theorem 2.4 [fa map M on ”R” is symplectic, we have

(exp(: f =)g) 0 (M) :71 exp(: fUll) 09(43)- (2-37)
Ifa map [17! on 722'" is antisymplcctic, i.e., Jac(M) . j Jac(M)‘ = —j, we have

(exp(: f 09) 0 (117) :71 exp(— = HM) 09(44)- (238)

Proof.

First assume that [If is a symplectic map, which implies that a Poisson bracket is

an invariant under the transformation M (eq. 2.29), i.e.:

1f<M).g<M>1=-(u,g1><m

The rest of the proof is straightforward:

fl 1*

exp(: f(M) =)g(M)
=. go?) + [mam + 31111211112121.1212») + -- -
=. go?) + (1mm?) + 51-11(12). (maxim + - --
=. go?) + ([1,9])(11‘4) + $31. 1!. give?) + . - .

4

:11 (exp(: f :)g) 0 (4 ).

22

In the case of [If being antisymplectic, the proof is basically the same except that

the Poisson bracket changes sign under the transformation, i.e.:

lfUl-f) 91M )l= -(lf 911(M (239)
Therefore we obtain

exp<- =1<M1 0ng 1
— 9(MI-lfflf’l)911Ml+%lf(M)11f(M)19(M)ll+-~
ng(M)+(lf g]1(1+ Mat/11411911111115
=M..g( 1+ +(1f,gl)(M 1+ :(lf11f,gl])(M)+-~
=n (exp(: f :)g) 0 (M),

which concludes the proof.

2.2 Lie Factorizations

In this section, proofs are given for the fact that any symplectic map can be repre-
sented, up to a given order, by a product of Lie transformations, which is referred to
as a Lie factorization. Various methods of Lie factorization are presented, including

Dragt-Finn factorization [Dragt76] and the single-exponent factorization.

Theorem 2.5 Let [Ill be a symplectic map on R”. Then, to an arbitrary order 71,
there exists a matrix L and homogeneous polynomials f3, f4, - - -, fn+1 of orders 3, 4.

- and 71 +1, respectively, such that

.(o(LI"1 ((zexp =1oI"1 o(=(exp (=1. 1---I"1o o<=exp< 1.1101"). (2.401

23

Note that L is simply a matrix; Lf, on the other hand, is the linear map, which
is a set of functions. Similarly, exp(: f :) is just an operator while exp(: f :)fis the

map generated by it.
Proof:

The induction method is used.

(1) The second order:

Define M1 = (L'lf) o M and D71 = M1 — f According to eq. (2.31), M1 is
symplectic. Besides, note that M1 =1 f and A72 contain terms of orders two and up.
Suppose M1 =2 exp(: f3 :)f We have

M1 =2 f-l-[fmi] =2 —6f3°j+

which entails that

A

W, =2 —(M'1 —D.J‘ =2 —1V..J.
On the other hand, the symplecticity of M1 gives
Jac(M1)-j-Jac(M1)‘ = j,
=> (i + Jac(N1)) - j - (1" + Jac(M11‘) = 1‘,
=> j + Jac(lVl)j+ jJac(M1)‘ =1 j,
=> Jac(M)} =1 (Jac(fiflj)‘. (2.41)
Here, Jac(A-ll) . j - Jac(fil)‘ is eliminated because it contains terms of orders two and

up. According to the potential theorem [Meyer91], there exists a function f, which

satisfies 6 f = 9', if and only if § satisfies

2.2; _ 59.92
am,- _ 82:.“ (2.42)

Therefore, eq. (2.41) shows that f3 does exist.

24
(2) The (n — 1)st order:
Assume that the theorem holds, i.e.:
M =1.-. (11") 0 (exp(: f3 111 o - - = o (exm: f. :111.

(3) The nth order:

Define
MM; = (exp(: fn 0f)“ 0 - - - 0 (exp(: f3 :)l)"l o(L‘1i) o M.

Hence Mud =n__1 f. According to Theorem 2.3 and eq. (2.31), M -1 is symplectic.
Now define NM; = Mud — f. Note that [V -1 contains terms of orders n and up.

Suppose 111,4 =n exp(: fn+1:)f. Similar to the second-order case, we have

Mil-1 :n f‘l'lfrH-lan =11 “fifrH-l‘jo

Since Jac(Nn_1) contains terms of orders (n - 1) and up, the symplecticity of

-o

Mn_1 entails that
Jac(an_1)j =,._l (Jac(M.._.)j)t.
Hence fn+1 exists. Finally, it has been proven that
M =. (11") o (exp(: f3 =11") o(exp(:f.=11’1 o - - - 0 (exp(: 1... :11"),

which concludes the proof.

Using Theorem 2.4 repeatedly, eq. (2.40) can be written into

M =n exp(: fn+1 :)exp(: fn :)---exp(: f3 :)(L‘lf), (2.43)

where exp(: f; :) (i = 3, = - - ,n + 1) are operators acting on functions obtained from

previous Lie transformations. Please note the difference between exp(: f :) exp(: 9 :)f

25

and (exp(: f :)f)o( (exp g:).l) In the former expression, exp(: f :) acts on the
map exp(: g :)I; in latter expression, exp(: f z) and exp(: g :) act on the unity
map f separately, and the resulting maps exp(: f :)f and exp(: g :)f are composed

afterwards.

It is worth noting that M can also be expressed in the reverse order, where

M :7, (exp(:fn+1:)I-) 0 (exp(: fn :)l) o - - - 0 (exp(: f3 :)l) 0 (LL). (2.44)
The proof is the same except that M.- is defined as
M1: M 0 (II—1f)“ (=6XP( f3' )f) 1° ((‘exp (fi+1 )5 l

The next theorem shows that it is also possible to represent a symplectic map

using a single Lie transformation.

Theorem 2.6 Let M be a symplectic map on 722’". Then to an arbitrary order n,

there exists 'a matrix L and a polynomial H of orders 3 and 4, up to 12. +1, such that

=.. (Lf)o( ((=exp )Hrfl (2-45)

Proof:
First, define M1 = (L‘1)o M.
(1) The second order:

From the proof of the last theorem, we know that there exists a function f3 of

order 3 satisfying
M1 =2 exp(: f3 :)f.

(2) The third order:

26

Suppose there is a function f4 of order 4 such that

M: =3 eXP(= fa + f4 :)7

=3 1+ Us + 1.11 + $113 + f4.[f:= + 114.111

=3 I"+ (13,11 + gm. [£1.11] + (1.11

=3 exp(: f3 =11"+ (1.11

:3 exp(: f3 :)f+ Mf,.j. . (2.46)

The removal of f4 from [f3 + f4, [f3 + f4, ll] is due to the fact that f4 is of order 4;
hence [f3, [f4, ll] and [f4, [f3, Ll] give fourth-order terms.

Now define N: = M1 — exp(: f3 :)L, which contains terms of orders 3 and up.

Thus from eq. (2.46) we obtain

_. 45

6L: =3 —(A-’i -eXP(= f3 3)!) ' J
Since M‘. is symplectic, we have
Jac(Ml) -j-Jac(M1)‘ =2 j
(Jac(M=1+ 1ac(exp(= fa =1111 - J“ - (Jac(M11‘+ 1ac(exp(= f. 111‘) =. 1,
=> Jac(IVl) . J" . Jac(exp(: f3 :)I")* + Jac(exp(: f3 :)1") . J" - Jac(M)‘,
+1ac(exp(= fa =11"11 -J-1ac(exp(= f3 :11"1‘ =2 1‘,
=> Jac(M.) . j - Jac(exp(: f;., :)(")t + Jac(exp(: f3 :)1") . J“ . .1.-..c(.1)7.)t =2 (1,
(exp(: f3 :)1" is symplectic.)
=> Jac(N1)-j - it + L-j-Jac(N1)t =2 0,
(Jac(M) is of orders 2 and up.)
=> Jac(lVQj =2 (Jac(M.)J‘)‘.

Therefore f4 exists. Choosing H = f3 + f4, we have

-o

M :3 (LL) 0 (exp(: H :)I).

27
(3) The (n — 1)st order:
Assume that the theorem holds, i.e.
M ’=.-1 (11") o(e=<p(= 11:11"),

whereH=f3+f4+---+fn.

(4) The nth order:

Suppose there is a function fn+1 of order n + 1 such that

-o

M. =.. exp(: f3 + f. + ---+ f... :17

=,, exp(: f3 + f4 + - ° - + fn I)f+ <5an ' j (2-47)

Define NW4 = 11/71 — exp(: H ):L, which contains terms of orders n and up. Thus

from eq. (2.47) we obtain
6h =n -(M -6XP(= H 0f) - j
Similar to the third-order case, the symplecticity of 11?. gives
=> Jac(M.._.)j =,._1 (Jac(M,_.)j)*,
which implies that fn+1 exists. Choosing H = f3 + f; + + fn+1, we have
M =.. (Li) 0 (exp(= H Of).

which concludes the proof.

Note that the proofs for the above factorization theorems provide algorithms for
obtaining the Lie factorization of arbitrary orders. In practice, they are used by

Differential Algebraic codes such as COSY INFINITY [Ber293].

28

2.3 The Baker-Campbell-Hausdorfi‘ Formula

In this section we will study an important formula, the Baker-Campbell—Hausdorff
formula, which combines two Lie transformations into one. The complete proof of it
involves knowledge of Lie algebra that is beyond the scope of this thesis [Dynkin62,
Vara84]. Instead, we will present a partial proof to show that it holds to the third

order for Lie operators.

Theorem 2.7 Let A and B be two functions on 722'“. The following relation holds:

exp(: A :)exp(: B :)

:3 exp(: A + B + $114.81 + 11—,(1A. (A, 811 + (19. (13.14111 :1,

where A and B are polynomials of orders 3 and higher, and “ =3: means the truncation

of the polynomial of Lie operators at the third order.

Proof.

From eq. (2.14), we have
:A::B:—:B::A:=:[A,B]:.

Using this relation repeatedly, the rest of the proof is straightforward.

The left-hand side can be transformed to

exp(: A :) exp(: B :)

1 1
=3(1+:A:+—:A:2+-:A:3)(1+:B:+l:B:2+l:B:3).
2 6 2 6
1
=31+:A:+§:A:2+%:A:3+:B:+:A::B:+%:A:2:B:
l 1 1
-:B:2 -: :: :2 -: :3
+2 +2 A B +6 B

l
=31+:(A+B):+§(:A:2+:A::B:+:B::A:+:B:2+:[A,B]:)

29

+%(:A:3+3:A:2:B:+3:A::3:2+:B:3)
1 l

=31+:(A+B):+-2-:(A+B):2+§:[A,B]:

+%(:A:3+:AzzzB:+:A::B::A:+:A::[A,B]:+:B::A:2

+:A::[A,B]:+:[A,B]::A:+:A::B:2+:BzzAzzB:

+:[A,B]::B:+:B:2:A:+:[A,B]::B:+:B::[A,B]:+:B:3)

1

l l
=31+:(A+B):+-2—:(A+B):2+§:[A,B]:+6:(A+B):3

1

+-(2:A::[A,B]:+:[A,B]::A:+:B::[A,B]:+2:[A,B]::B:).

6

The right-hand side can be transformed to

exp(: A + B + éIA. B] + -1-1§([A, [A, 8]] + [B, [B,A]]) + - .. :)
:3 1+ : (A + B + %[A, B] + {IQ-((A, (A191) + [B, [B,A]])) :

1 1 1
+5:(A+B+§lA=Bl)32+g=(A+B)
1 1 1
=31+:(A+B):+§:(A+B):2+§:[A,B]:+6:(A+B):3

+£(;(A+B)::[A,B]:+: [A,B]::(A+B)1)
l

+1-2‘(3AiilA1Bli-31A13131A:+:B::[A,B]:—:[A,B]::B:)
=31+:(A+B):+%:(A+B):2+%;[A,B];+%:(A+B):3

1

+6(2:A:: [A,B]:+:[A,B] ::A:+:B:: [A,B]:+2:[A,B] 1133).

Altogether, the left-hand side equals the right-hand side, which ends the proof.

It is worth noting that only commutators appear in the right-hand side of the

B-C-H formula, which is remarkable in that, for any Lie algebra, the manipulation of

the corresponding Lie transformation does not require extra operations. Besides, the

B-C-H formula links the Lie operators with the Poisson brackets.

30

A direct result of the B-C-H formula is that we are able to find the inverse of

exp(: f :) easliy. Since [f, f] = 0, we obtain

eXp(I f I)exp(I -f I) = exp(: f + (-f) :) = exp(I 0 I) = 1-

Therefore, the inverse of exp(: f :) is exp(: — f z.) Futhermore, the inverse map of M,

-o

in the form of a single-factor Lie factorization, i.e., M = (L ) 0 (exp(: H :)L), is
M-1 = (exp(: —H :)1") 0(L’1T), (2.48)

which is shown below:

M-loM = (exp(:—H:)1")o(L-11")o(L1")o(exp(:H.)I")
= (Iexp(-—111I)I)o((Ierxp( :11
= exp(:H:)exp(:—-H :)I"=i‘

MoM“ = (Lf)°(exp(IHI)7)°(exp(I—H=)f)°(L"‘f)
= (Lf1o(exp(=H .1(exp=— H=o11"1 (II-‘11
= unoudn=i

Chapter 3

Repetitive Achromat Theory

In this chapter the theories of repetitive achromats up to order three will be dis-
cussed. This will give us an overview of what has been achieved and the remaining
difficulties. All achromat theories, including the arbitrary-order theory presented in
Chapter 4, deal with systems with midplane symmetry. The first section is devoted

to the definition and implications of midplane symmetry.

3.1 Midplane Symmetry

In a system with midplane symmetry, two particles that are symmetrical about the
midplane at the beginning stay symmetrical afterwards. In other words, let us con-
sider a beam optical system with transfer map M. Suppose a particle enters it at
(x;,a,-,y.-,b.-,t.~,6.-) and exits at (x ;,a f,yf,bf,t f,6f). Another particle that enters it at

(x;,a.~,—y.~,-b.-,t.-,6,-) will exit at (xf,af,—yf,—bf,tf,6f). Now the matrix P is defined

as
(100000\{z\
010000 a
~_00—1000 y
P" 00 0—100 b (3'1)
000010 t
\000001)\5)

 

 

 

 

31

32

Hence, midplane symmetry can be expressed as
wnofiow4n=hi am

Since we are only interested in symplectic maps, M can be represented by a linear

matrix and a Lie transformation (see Section 2.2):

q ...

M = (Li) 0 (exp(: H :)I).
Inserting into eq. (3.2) and using Theorems 2.4 and 2.2, we have
(111 2 (exp(: H 111 = (121) o (111 o(exp(: 11:11) 2 (11-111
=> (LL) 0 (exp(: H :)T) = exp(: H(P'1P) :)(P - L . P-ll).

For first-order matrices, midplane symmetry requires that
L=P-L-P'1, (3.3)

which implies that for higher orders

exp(: 11(1") =1 = exp(: 111(12—11) :1,

: mh=Hw4n an
Equations (3.3) and (3.4) determine that

((2121 (2121 o 1 (2111 (2111)
(a(|)x) (a(|)a) (Ill) ((11)) (aAt) (a(|)6)
L" o 0 (11:1 (1111 o 0 (3'5)
(1121 (1111 o o (1111 (1111
\(1121 (1121 o o (1111 (1111)

 

 

and

H = Z nggagygbg,g6$i’ai°yi”bibti'di", (3.6)

igiaivibic '6

33

where i, + in + iy + i1, + i, + i5 2 3 and iy + i1, is even. Here the independent variable

is the arc length of a reference trajectory s and x.-,a.-,y.-,b,-,t.-,6,- are canonical variables.

An achromat can be achieved only when acceleration is not present and syn-
chrotron radiation can be neglected. Therefore, the transfer map of such a system is
not only symplectic but also time-independent (static) and energy-conserving. These

put more constraints on the transfer map, and we now have

((2121 (2111 o o o (2111)
(2121 (2111 o o 3 (2111
0

 

 

0 0 (111(1) (ylb) 0
L = 3.7
o 0 (11.11 (1111 o ( )
(tII) (tla) 0 0 (tlt) (tl5)
\ 0 0 0 0 0 (6|6) )
and
H = Z nggagv,,;,xi’ai°yiybibbf‘, (3.8)
game...)
where i, + in + iy + i5 + i5 2 3 and i, + i), is even.
It is clear that, from eq. (3.8), we have
H = )3 009,101,911 (3.9)

i; :3

when an achromat is reached. Hence, only the terms (t|6") (n = 2,3, - - =) are left in

an achromat. Futhermore, symplecticity [Berz85] implies that

(tlz) = -($|$)(a|51=)+(a|$)(1|51=)1 (3-10)

(tla) = -($|a)(a|51=)+(ala)($|51=), ‘ (3-11)

which means that this is also true for the first order.

34

3.2 Second-Order Achromat Theory

Since a second-order achromat is always based on a first-order achromat, let us first
study how to obtain a repetitive first-order achromat. Consider a system consisting

of n identical cells (n > 1) and let L, be the x-matrix of one cell, which has the form

(1121 (2121 (2111 M L,
L, = ( (a(|)x) (ago) (all6) ) = ( 0 1 ). (3.12)

Therefore, the total x-matrix L7,: is

LT; = L:

(M: (Mn-1+Mn-2+I~+i)(:5)
0 1

( M" (M" — i)(M —1‘)-155). (3.13)

0 1
Equation (3.13) shows that the dispersion vanishes when M" = L, i.e., the phase
advance equals a multiple of 211‘. Together with the requirement that L, = L, a

first-order achromat is reached when the tunes of the whole system are integers.

Brown’s second-order achromat theory is built on a repetitive first-order achromat,

as described above. It consists of the following two theorems:

Theorem A: If a system contains N identical cells (N > 1 and N 75 3), all
second-order geometric aberrations vanish when, for a cell, both transverse planes

have the same non-integer tunes and the phase advance of the system is a multiple

of 21r.

Proof:

35

Here we adopt K. Brown’s original notation where the linear matrix is represented

by R and the second-order matrix is T. Therefore, the second-order map is
331,1 = Z Rijivjp + Z: Tijkxjodikp, (3.14)
i .1311

where xgp and xm are initial and final coordinates, respectively. According to pertur-
bation theory, high-order solutions can be obtained through the lower-order solution
and the inhomogeneous part of thelODEs. In the case of Brown’s theory, the ODEs
are expanded to the second order, where the the terms in the inhomogeneous part
are called the driving terms [Brown82b]. With all driving terms obtained, ngk can

be expressed as a function of Hg, which is
L o
T111 = [0 K.(s1(11..(s11"(11.1(s11mds, wlth (11 + m1 = 3, (3.111

and where K,(s) is the multiple strength at s. For geometric aberrations, Rij should

come from the geometric part of R only, which is

( cos 111(3) + 01(3) sin 1,!)(3) 3(3) sin 1/2(s) )
—7(s) sin 112(3) cos 1/1(3) — 0(3) sin w(s) '

As a result, T1,). can be written into
L o
11,-. = [o Fp(s)sm"(1,b(s)) cosm(1,b(s))ds. ~ (3.16)
Since sin"(z/)(s)) cosm(z/)(s)) (m +n = 3) gives only eii’f“) and efl‘f’“), the conditions
for all second-order geometric aberrations to vanish are

L , L .
/ Fpewds = 0 and / Fpeflwds = 0.
0 0

Due to the fact that the system consists of individual cells, the integral conditions

above become the following sums

N ~ . N 1.; .
z eriw" = O and Z Ffefl'd’“ = 0,

k=1 k=l

36

where
k 8 8
-k = /NL+ o+A Fp(s)€hw(3)-w"(’°))d3
fiL‘l-Jo-As
= A’ Fp(£L + .30 + §)eii(¢(%L+So+3)-¢k(so))d§
—As N
A3 _ -
= Fp(§)e*‘(“’(’”d§, (3.17)
-A3
- As _ -
F12 = / F,(.1)e*3'W<~'>)d.1. (3.18)
—As

Here, N, L, 30, and A3 are the number of cells, the length of the system, the position of
the center of the element considered, and the half-length of the element, respectively.
Repetition of the system is used to obtain eq. (3.17) and (3.18). Since 17",, and P; are
independent of 1:, eq. (3.17) is further reduced to

N . N .

2 e*"”“ = 0 and Z eim’“ = O.

k=l k=1
In conclusion, all second-order aberrations vanish when N aé 3, N 1b,,” = 2m,,.,y7r, and

mm, 75 2mN (m = 1, 2, -- ) (see Figs. 3.1 and 3.2).

The second theorem deals with the correction of second-order chromatic aberra-

tions left in a system satisfying Theorem A.

Theorem B: For a system that satisfies Theorem A and N > 3, a second-order
achromat is achieved when two families of sextupole components are adjusted so as
to make one chromatic aberration in each transverse plane vanish. In other words,

only two chromatic aberrations are independent.

The proof of this theorem can be found in reference [Carey81]. Another proof using
normal form theory will be given in Section 3.3 as part of the third-order achromat

theory. A typical four-cell, second-order achromat is shown in Fig. 3.3.

37

11‘" 611 l4 6311
3 1 3 1
—-¢ .— —-fi .—

 

 

 

 

Figure 3.1: Complex plane diagram for second-order geometric aberrations of a four-
cell repetitive system with phase advances 21r.

11..
F

 

 

 

 

Figure 3.2: Complex plane diagram for second-order geometric aberrations of a three-
cell repetitive system with phase advances 21r.

38

3' so 9 '° so SF °° so

OF OO OF OO OF OD OF OD

Figure 3.3: K. Brown’s four-cell, second-order achromat. Quadrupoles are used to
tune the system to phase advance 27r in both transfer planes, and two families of
sextupoles, SF and SD, are used to correct chromatic second-order aberrations in the
x- and y-planes, respectively. To make the sextupoles weak, they are placed such that
,8, is larger at SF and fly is larger at SD.

3.3 Third-Order Achromat Theory

In the mid 80’s, Dragt [Dragt87] developed a third-order achromat theory for repetI
itive systems based on the normal form theory [Dragt79]. Although the same result
can be obtained from other normal form algorithms, the Lie factorization has the ad-
vantage of explicitly showing the number of independent aberrations to be corrected.
In practice, however, it is difficult to implement the Lie normal form beyond the
fifth order; hence DA techniques have to be used, either combined with Lie algebraic

techniques [Forest89] or by themselves [Ber292a], to compute the normal form map.

The key idea of this theory is that when an achromat is achieved in the normal
form coordinates, it is achieved in any set of coordinates. Since the transfer map
in the normal form coordinates is much simpler than that in the original curvilinear
coordinates, the conditions for an achromat become much clearer. Consider an n-ccll
symplectic system with midplane symmetry. From Section 2.2, the transfer map of

one cell can be written into

-0

M =1. (11") e(exp(= 13 =1i‘1o(exp(= f. =111o---o(exp(: f... 111.

lo 01111

Whic

“he

can

39

To order 3, we have

=. (11") (211(13 11")( ((212: 1:11) (3.19)

Since t is not of interest, it is discarded. Therefore T = (x, a, y, b, 6). The normal form
transformation is done order by order [Forest89]. For the first order, there exists a

5 x 5 symplectic matrix

011 (112 015
021 022 025
A = (133 (134 . (3.20)
(143 044
1

which satisfies

A - L - A"1 = R, (3.21)
where
Ci“:
6-3.”:
R = 6"” . (3.22)
e"“9
1

Suppose s“, is the eigenvector of e’”‘FV. The fact that L is real entails that 3“, is the

eigenvector of e"“’19.

Now let us define N1 as the transfer map in the eigenvector coordinates, which

can be transformed to

4

N1 = (AT)oMo(A"’l)

= (A111 (11"1( ((exp 111o o(exp(= f1 1112 (11-11")

X011

l01

11).

40

= (111") o (11") o (AI-‘1") o (A!) 2 (exp(: fa :11") 0111-11")
o(AI"1 0 (exp(: f1 :11") 1 (21-11")
= (11111 ((212 1141111 (exp( (14(4-11111")
= (Rf12(exp(=g:=11"12(exp(=g.:11"). (3.23)
Note that f is the unity map in the eigenvector coordinates.

Next we define
N2 = (exp(: 03 :)T) 0 N1 0 (exp(: —G3 :)i). (3.24)

To the second order, we have

-.

N1 =2 (exp(: o. :11) 2 1111002121: -0, :11)

=2 (exp(: 03(111") 1111") 2 (exp(: 93 :11) 0 (exp(: -0, :11)

=2 exp(: -Ga:)exp(zga=12xp(=Ga(RI"1=1(RI"1

=2 exp(: ,3 - (a. - 01(1111)=1(111"1. (3.211
where use of the B-C-H formula has been made.

Since G3 and g3 are polynomials of order 3, in general they have the form

_ m n 3
G3 ‘— Z Gmgnxmyny36sx: 52331 1153,2355, (3.26)
mgngmyny‘la
_ m —n m -n i
m;n:myny36

where m3 + n, + my + ny + i5 = 3 and my + 12,, is even. Therefore, we have

93 — (Ga — CARI-l)

= Z (gmgngms,nyig _ (1 _ 61((mz—n:)#z+(my-ny)#y)) sznxmynyiJ)

mgnxmyny 35

mg —n: my -ny 36
3,, sin 3,, 3y 6

Z: (gmxngmynyio " (1 — Chiba-m) szngmynyi5) S?’§:'3;n”§:”5i6,(3.28)

mgngmyny36

41

where [I = (pm/1y), r'ri = (mmmy) and ii = (n3, ny). As a result, all the terms which
satisfy [I - (r'ri - ii) 75 2n7r can be removed. This is why normal form transformation

simplifies the transfer map.

Futhermore, there are two categories of terms that remain in the normal form

map. The first category satisfies fii = fi, i.e.,
m3 = nac and my 2 ny, (3.29)

which does not depend on 17. Since these terms cannot be removed intrinsically
by any normal form transformation, they are the minimum independent conditions
for obtaining an achromat. Note that they are also responsible for the amplitude-

dependent tune shifts.

The other category of terms remaining in the normal form map consists of the

nontrivial solutions to the equation
[£1055 — fi) = 2n7r, (3.30)

which are tune-dependent and can be removed by carefully choosing the tunes or by
having a certain symmetry. The tunes that give eq. (3.30) non-trivial solutions arr-
called resonances; and the terms associated with the solutions are called the driving

terms.

In the case of the second order, 2'11 and it must also satisfy
mz+n3+my+ny53 (3.3!)

Therefore, from eqs. (3.29) and (3.31), the tune-shift terms are g1100133336, gooulsysfi

and 90000363, and from eqs. (3.30) and (3.31) the resonance driving terms can also lw

Obtained.

42

Since the vector (x, a, y, b, 6) is real, from

33 x 3:, x
3,, a 33 a
s3, = A y we have A"1 3y = y is real.
3,, b 3,, b
6 6 6 6

Therefore, from the fact that all coefficients in f3 are real, we have
93(331 gr, 3y, 5y, 5) = f3(A‘1T) is real.
Thus we obtain

_ m: —n: my—ny 36
Z gmrnzmyny‘ts'sx 3:: 3y 3y 6
mgngmyny36

_ - , —m, n, -my 11;, i5
_ z: gmz‘nzmyfly‘osx 3:: 3y 3y 6
mgngmyny36

_ - . mg-ng my-ny 36
— E 9n,m,n,m,163x 3,, 3y 3y 6 .
mgngmyny‘iJ

Since this relation holds for any point, the coefficient of each term has to be equal

separately, which gives

gmgngmyny36 = gngmxnymyio- (3.32)

When m, = n, and my = ny, we have

gmgmgmymyig : gmgmgmymy36, (3.33)

which means that the coefficients of the tune-shift terms are real. Therefore, there are
only two independent chromatic aberrations left when driving terms do not appear,

which proves Theorem B from the last section.

Note that we have

gmgngm n i
Gmenemyn,;, = 1 _ 6,115,117.?) (3.34)

entai‘
well

is re

sud
the

TGSC

ME

18!

43

for terms that satisfy fl- (rfi — 77) # 2n7r. Thus, the relation

G . gnxmgnymyi6
ngmgnymybg — 1 - e_gfl.(fi-m)

 

gmxngmynyig
1 _ eifl‘bfi‘fl)

= Gmwxmynyg, (3.35)

 

entails that this part of G3 is real. Since the rest of G3 has no effect on A72, the
coefficients can be chosen as zeroes. Therefore, G3 is real. It also follows that G3(Rf)

is real.

Regarding the driving terms, one way to eliminate them is to choose the tunes
such that no resonances equal to or below order 3 occur; the other way is to choose
the tunes such that some resonances are avoided and the driving terms of the present

resonances are cancelled by midplane symmetry.

In the case of Brown’s second-order achromats, p3 = M = (q/p)21r, where q and p
are natural numbers, q < p, and q/ p is an irreducible fraction. Therefore, the driving

terms are the non-trivial solutions of

(m: — n3 + my — n,,) = n (n is an integer). (3.36)

"BIG

On the other hand, 2p 2 4 and m; + ndc + mg + 723, S 3 entail that

m, — n: + my — 72,, = —p, 0, p. (3.37)
l) E > 3:

Equation (3.37) can be further reduced to
m; — nx + my — ny = 0. (3.38)
Equation (3.38) has these non-trivial solutions

{m""“’ = _} and {mx‘nx = ‘1 (3.39)

my‘ny = my-ny

ll
p—a
‘0

44

which can be transformed to

m; — 1 mm = 0
n; = 0 n: = 1
my 2 O and my = 1 (3.40)
ny — 1 n3, - 0

These solutions entail that the driving terms are 3,3,,6 and 533,,6. Since midplane
symmetry cancels them, only tune-shift terms are left in such a system, which requires

at least four cells to be a first-order achromat.

2) E = 3:
From the requirements of midplane symmetry discussed above, eq. (3.37) can be

reduced to

mz - n3 + my — ny = -3, 3. (3.41)

With midplane symmetry taken into account, it has the non-trivial solutions

m, — n1. = —3, —1 m, - n, = 3, 1
{ m, _ n, = 0, _2 and { m, -72, = 0, 2 , (3.42)
which are equivalent to
m: — 0 m, — 0 m, — 3 m; = 1
n, = 3 n: = 1 n, = 0 n3 = 0
my 2 0 , my = 0 , my = 0 , and my = 2 . (3.43)
12,, - 0 n3’ — 2 71,, = 0 n1, - 0

3

Hence the driving terms are 32., 3,33, s, and sxsj. This shows that not all second-order

geometric aberrations vanish in a three-cell system, which agrees with Theorem A.

3)g=2:

Similar to case 2), we have

m3 — n1, + my — *ny = —2, 2, . (3.44)

45

which gives the non-trivial solutions

m, — n, = —2, 0 m,c — n; = 2, 0
{my—n3, = 0, —2 and {my—n, = 0, 2 (3'45)
Hence we have
m, = 0 m3 = O m, = 2 m, = 0
n, = 2 72,; = 0 123 = 0 n3, = 0
my = 0 ’ m3, = 0 ’ m3’ = 0 and my = 2 ’ (3'46)
111’ = 0 713’ = 2 ny = 0 ny = 0

which entail that the driving terms are 3:6, $26, 336 and 536. This shows that all
second-order geometric aberrations vanish, yet not all driving terms disappear. From
the proof of Theorem B given by D. Carey (see eq. (18) in [Carey81]), it is clear
that there are two independent second-order chromatic aberrations only when p 7E 1r.

In conclusion, normal form theory gives an alternative proof of K. Brown’s theorems.

Now let us move to the third order. From eq. (3.23) and (3.24), we have

17; = (exp(: G; :)i) 0 N1 0 (exp(: —G3 :)i) (3.47)
= (exp(: 03 :)i) o (Rf) 0 (exp(: g3 :)f) (3.48)
o(exp(: 94 :)f) 0 (exp(: —G3 :)f) (3.49)

=3 eXP(= -Ga =)exp(= 94 :)exp(: 93 :)exp(: 03(Rf) =)(Rf)
=3 eXp(= gs - Ga + CAR?) 0
exp(: 9. + guy. as] + [93, G3(Rf)] + [030213, 031) :)(Rf)
=3 exp(: g. + gag. as] + [93.031213] + [041213.03 =)
exp(: 9, -— Ga + 03021“) MRI“)

:3 (Rf) 0 (exp(: h3 :)f) 0 (exp(: 11., :)f), (3°50)

h3 = g.-G.+G.(Rf), (3.51)

. h. = g. + gag. as] + [93.041213] + [Gama 031). (3.52)

46

Now define N3 as
N3 = (exp(: H4 :)f) 0 A72 o(exp(:—H431.) (3.53)

To the third order, we have

-0

N3 =3 (exp(: H4(Rf) :)Rf) 0 (exp(: h;; :)f) 0 (exp(: (1., :)i) 0 (exp(: —H.; :)f)
:3 exp(: —H4 :)exp(: h4 :)exp(: ’23 :)exp(: H4(RI-) :)(Rf)

-_—.3 exp(: h3 + h., — (H4 — 114121)) :)(Ri), (3.54)
where use of the B-C-H formula has been made.

Similar to the second order, when resonances are avoided and/or driving terms
not present, only tune-shift terms are left, which are h22000(sx§z)2, hnuo(sz§x)sy.§y,
hoo220(3y§y)2, huoogsxs'xbz, hoongsysybz, and 11000046“. Therefore, the total pseudo-

Hamiltonian is

hT = 91100132§x6 + 9001118y5y5 + gooooatS3
+h22000(3x§1:)2 'l' h111103x5x3y§yh00220(3y§y)2
+h110023x§x62 + h001123y§y62 + (10000454- (3-55)
Since 93, 94, G3 and GAR]? are real, h is also real. Hence, hmmmvmyg, is real.
Therefore, 5 third-order knobs are needed to achieve a third-order achromat.

Define real coordinates

X,Y = (s.,,+§x,,)/2, (3.5m

PX,Py = (say—gravy, (3.57)

where X and PX are the real linear combinations of a: and a, and Y and Py are the

47

real linear combinations of y and b. The linear matrix in the new coordinates is

c05(u.-) sin(#x)
- Sin(#r) COSU‘I)
R = “350134) Similar) a (3°58)

-Sin()uy) cosy.)
1

where the linear motion is simply a rotation of X and PX, and Y and Py. Usually the
X -Px space is called the normal form space. Futhermore, the pseudo-Hamiltonian

has the form

h = g110m(X2 + P§)6 + 9001110”2 + P306 + 90000363
+h22000(X2 + 10;.)2 + hunger? + 19,?!)(2/2 + 13,3)hoom(Y2 + 1),?)2
Humor? + P})62 + hoomo/2 + P,Z)¢52 + hm46‘.
= 21);,(X2 + P§)6 + w;(Y2 + P§)6 + e63
+a(X2 + P}? + b(X’ + P}{)(Y2 + P§)c(Y’ + 13,?)2
+ch'(X2 + P962 + wflY2 + Pflb2 + f64. (3.59)
Here, w; and w; are first-order chromaticities; w: and w: are second-order chromatic-

ities; a,b and c are anharmonicities.

In summary, a third-order achromat is achieved when the total tunes of both trans-
verse planes are integers, all resonances of order 4 and below are avoided or all driving
terms of present resonances vanish, and all first- and second-order chromaticities and

anharmonicities are corrected.
When a third-order achromat is reached in the normal form coordinates, we have
N3 :3 (Rf) 0 (exp(: 663 4- f64 :.)I) (3.60)

Therefore, the total map of the n-cell achromat is

.9

N; :3 ((Rf) 0 (exp(: 663 + f64 :)n)"

48
:3 exp(n : 663 + f64 :)f (3.61)

Since the normal form transformation
A: (exp(: (H4: )c>(i) ((:exp (G3. )f)o (Ar) _ (3.62)

does not contain t, 1V; and [I commute. As a result, the total map in the original
coordinates is

.9 _. _. .-.

MT =3 AoNgoA-l
=3 (exp(zH4=)Bo(exp(=Gs=)f)o(Af)°1\7§
A11)°( ((exp: -Ga=)f)°(exp(=-H4=)f)
:3 exp(: —H4 ::)exp( —03 ::)exp(n 853+ f6“ )exp( 03 ::)exp( H, ~f)
=3 exp(n:653+ f54:)i' = 1V3. (3.63)

Indeed, MT is a third—order achromat, and the only aberrations left are (tl6‘), where

i=1,2,---,n

Let us come back to the examples designed by Dragt and N eri. Dragt’s example is
very straightforward, because all resonances of order 4 and below are avoided. N eri’s
example is more complicated because some resonances are present and their driving
terms are cancelled by midplane symmetry. In his example, the tunes are [ix/271' =

1/7 and py/21r = 2/7, which have the driving terms satisfying
m, — nx + 2(my — ny) = —7, 0, 7. (3.64)

It gives the non-trivial solutions

("h—n” = _1’ 2’ f’ (3.65)

1
my—ny = —3, —l, , 3
However, midplane symmetry requires that my+ny be even, which entails that my—ny

also be even. Hence, all the terms are cancelled.

49

As a summary of this chapter, let us look at the perspective of extending Dragt’s
theory to higher orders. If we follow Neri’s choice of resonance, it can be shown
that the minimum number of cells for a fourth-order achromat is 11, with T, = 1/11
and T, = 2/11. For fifth-order achromats, Neri’s approach fails because the driving
terms 3:33 and .9133 cannot be cancelled by midplane symmetry. To circumvent this
difficulty, different choices of the tunes have to be made. The next choices are T; 2
3T, or 3T, = Ty. The first one gives a smaller minimum number of cells, which is 17,
and the tunes are T: = 3/17 and T” = 1 / 17. These examples show that there is no
established pattern for the choices of the tunes or the number of the cells to design
an arbitrary-order achromat, and the driving terms for a resonance increase rapidly

with the order, as does the minimum number of cells. The theory presented in the

next chapter allows us to solve these difficulties.

Chapter 4

Arbitrary-Order Achromat
Theory

In the chapter that follows, an analytical theory of arbitrary-order achromats will be
developed. In Section 4.1, the maps of cells R, S, and C will be derived from the map
of the forward cell, and that of a four-cell system from those of single cells. Section
4.2 contains a classification of the systems with the best solutions. In the first part
(Section 4.2.3), it is shown that it is necessary to have at least four cells in a system
to achieve an efficient arbitrary-order achromat. Then the proof of the existence of
an optimal solution is given. It is further shown that 4 out of 64 four-cell systems give
the optimal solution while requiring a minimum number of linear constraints (Section
4.2.4). In Section 4.3, the four best systems are studied in detail to determine the
solutions for achromats order by order. First, a general solution for arbitrary-order
achromats is obtained, even though it is not the optimal solution we can obtain from
this theory, regarding the number of conditions that have to be satisfied. Then the

optimal solutions for achromats up to the sixth order are found.

50

51

4. 1 Map Representations

Let us consider a phase space consisting of 2m variables (ql, - - -, qm, p1, - . -, pm). Since
we do not take into account synchrotron radiation and acceleration, the transverse
motion of a beam optical system is described by a symplectic map. Therefore, its
transfer map M of order n can be represented by a matrix L and a polynomial H of

orders 3 up to n + 1 through Lie factorization (Theorem 2.6) via

M= nL(If)o( (exp (H :.)l) (41)
Furthermore, its inverse is (eq. 2.48)

M‘1= (exp H:)of) (L'lf). (4.2)

Next, let us define a “standard” and a “sub-standard” form of the maps. The

advantage of these forms will become evident later.
Definition 4.1 For a symplectic map M3, the standard form is defined as
M3 = exp(: H :)(MLl), (4.3)

where H is the pseudo-Hamiltonian of orders three and up, and ML is the linear

matrix. A representation of the form

M5 = [Hexp(: H,- :)] (MLI) (4.4)
is called a sub-standard form.

Note again the difference between eqs. (4.1) and (4.3), where, in the former equation.
exp(: H :) acts on f and the resulting map is then composed to the linear map, and, in

the latter equation, exp(: H :) acts on the linear map directly. Like the composition

52

“o”, Lie operators are are also associative, which implies the associativity of Lie
transformations. Apparently, use of the Baker-Campbell-Hausdorff formula allows

the transformation of a sub-standard form into a standard form.

From Theorem 2.1, M can be written in the standard form

-o

MF = exp(: H :)(Ll), (4.5)

which is called the forward map ll-lF.
To obtain the maps of R, S, and C cells, Theorem 2.4 has to be used repeatedly.

The reversed cell (R) is the one in which the order of the elements is reversed
from that of the forward cell. This means that if a particle enters the forward cell at
an initial point (1;, a,-, y;, b,-, 6;) and exits it at a final point (:61, a}, y}, bf, 6}), a
particle which enters the reversed cell at (1;, —a,, y,, —b;, 6}) will exit at (1);, —a,-,

y;, —b.-, 6;). This determines that the map of the reversed cell is

M“ = (121) o 12-1 0 (11-11“), (4.6)
where
1 0 0 0 0 a:
g 0 —1 o 0 o a '
RI = 0 0 1 0 o y (4 7)
0 0 O —1 0 b
0 0 0 0 1 6

Taking into account the fact that R is antisymplectic (see Appendix A), we can get

the standard form of MR:

-¢

M" = (Rf) o 113-1 0 (12-11")

= (Rf) 0 (exp(: —H :)1‘) o (1:11") 0 (12-11“)

(121) 0 (exp(: —H )1“) o (L’1R“i)

53

= (RI) 0 (exp(: H(L’1R‘1f) :)(L'IR’IB) (Theorem 2.4)
= exp(: H(L’1R"1f):)((Ri) o (L‘IR’ID) (Theorem 2.2)

= exp(: H(L‘IR"lf) :)(RL'IR-lf). (4.8)

The switched cell (S) is the mirror image of the forward cell about the y-z plane,

i.e.,
M5 = (51‘) o 101 0 (3-11‘), (4.9)
where
—1 O 0 0 0 x
.. 0 —1 0 0 O a
SI = O 0 1 0 0 31 (4.10)
0 0 0 1 0 b
0 0 0 0 1 6
Since the matrix S is symplectic (see Appendix A), we have
M5 = (Si) 0 M o (541.)
= (Si) 0 (LI) 0 (exp(: H :)f) o (541-)
= (3L1) 0 (exp(: 11(5-11‘) :)(s-lf))
= exp(: 11(5-11‘) :)(SLs-lf). (4.11)
The combined cell (C) is the switched and reversed cell, whose map is
M0 = (51) o (121) o 112-1 0 (12-11) 0 (5-11“). (4.12)

Due to the fact that matrix S R is antisymplectic, similar to the reversed cell, MC

can be brought into the standard form

12" = exp(: 11(L-11z-‘s-11") :)(SRL‘lR‘lS’lf). (4.13)

54

In summary, we have the maps of all four cell types listed below:

37" = exp(:H:)(L1‘), (4.14)
M“ = exp(:H(L-1R—1f):)(RL'1R"II-), (4.15)
M5 = exp(:H(S‘1l):)(SLS"l), (4.16)
M0 = exp(:H(L-112-15-11‘):)(SRL-IR-ls-lf). (4.17)

Since we have the maps of the different kinds of cells, we are ready to construct
the map of any given multi-cell system. As examples, the total maps of a few four-
cell systems are presented here. First, let us define a symbol denoting the map of a

multi-cell system.

Definition 4.2 Let C,- be the ith cell in a k-cell system, i.e., C.- can be F, R, S, or

C. The map of the total system is denoted by MC‘C""C*.

For example, MFRSC represents the map of a four-cell system consisting of the
forward cell, followed by the reversed cell, then by the switched cell and ending with
the combined cell. Next we will determine the sub-standard form for a variety of
four-cell combinations, which will be very useful later. As a proof of principle, we
show this process for system F RSC. In the transformation, we repeatedly make use
of eq. (2.1), Theorem 2.4, and the associativity of “0”. From the definition of system

FRSC, we can obtain its transfer map from the maps of single cells, which is
MFRSC=MCOMSOMROMF. (4.18)

Note that the order of the maps of the single cells is the reverse of that of the cells,

because the initial coordinates of the present cell are the final coordinates of the

MFRSC

previous one. Thus, is transformed to

Mme = [wt H(L—IR—ls-1f):)(SRL-ln-‘S“f)]

55

o(exp(: 11(5-11) :)(SLs-‘I‘n
o[exp(: H(L“R"F) :)(RL‘lR‘IN 0 [exp(: 11(1) :)(Li)]
exp(: 11(1) :){[exp(:H(L'1R-IS-1f) :)(SRL-‘R-ls-lm
oiexpc (us-‘1‘) :)(SLS-‘i'n
o(exp(: H(L‘1R"f) :)(RL-‘R-11)16(Lf)}
(Theorem 2.2)
exp(: Hu‘) :){[exp(= H(L"R“S’lf) :)(SRL-‘R-‘S-‘m
o[exp(: 11(5-11‘) :)(SLS“f)]
o[exp(: H(L"R“ . Lf) :)(RL-IR-1.L1")]}
(Theorem 2.4)
exp(: HU‘) :)exp(: H(L-‘R-‘ 41‘) =)
{[exp(: H(L’1R"S“f) :)(SRL“R“‘S“f)]
o[exp(: 11(5-‘13 .)(SLs-‘m 0 (1121:111-1 . L13}
(Theorem 2.2)
exp(: Ha“) :)exp(: («L-‘12“ .11) z)
{[exp(: H(L“R“S“i) :)(SRL-IR-ls-li'n
o[exp(: 11(5-1 .RL-1R-1 . Li') :)(SLS-1 . RL-IR-l . L “)1}
(Theorem 2.4)
exp(: 11(1) :)exp(: HW‘R‘1 ° L1) =)
exp(: 11(3-l -12L-112-1 - Lf) :){[exp(: H(L"R“S“f) :)
(.S‘RL“R"‘S"11-)]o(SLS‘1 . RL“R“ - Li')}
(Theorem 2.2)

exp(: H(l) :)exp(: H(L‘111’.‘l - Li) :)

56

exp(: H(S'1 - [BL-1R“1 - Li) :)
exp(: H(L“R“ -LS“ . RL"R‘1 - L1) :)

(SRL-‘R-1 . LS—1 . RL’IR“ -L ),

(Theorem 2.4)

which is now in the sub-standard form. In a similar way, the sub—standard forms of

the maps of the systems MFRFR, MFCSR, and MFCFC are obtained. Together with

MFRSC

AiFRSC

AZFRFR

AZFCSR

AZFCFC

, we have

-o

= exp(: H(f) .)exp(: H(L'1R-1 - L ):)

exp(: H(S’1-H’L’1R'1 - LL) 2)
exp(: H(L"‘R" . LS“‘ .RL’1R" . L1) :)

(SRL‘1R‘1 . LS-1 . RL'IR" - L1), (4.19)

- exp. HU) :)exp(. H(L‘1R‘1 - LL) :)

exp(: H( RL-IR- L1):)
exp(: H(L"‘R“ - L . RL‘IR“ . L1) :)

q

(BL-1R-1 . L - RL'IR" - LI), (4.20)

= exp(: H(f) )(e.xp H(L'1R"IS"'1 . LL) :)

exp(: HUM-IRAS"l . Li) :)
exp(: H(L"R" -SL . RL“R“S“ . L1) :)

(BL-IR-1 ~SL - 11L-112-15-l - L1), (4211

- exp(: H(l) :)exp(:H(L‘lR‘1.’§l”1 - L131)

exp(:H(SIifL'1R"lS"1 - L1) :)
exp(: H(L'1R"S" . L . S12L-112-15-1 . L1) :)

(SisL-IR-‘s-l . L - SRL‘1R‘15'“ . L1). (4.221

57

As shown in the next section, only those systems listed here are needed when
the solutions of arbitrary-order achromats are determined, because other systems are
not as efficient. What will also be shown here is the importance of the sub-standard
form where the optimal four-cell systems are decided. Finally, when solutions of
achromats are searched for among the four systems, the standard form of their maps

will be obtained from the sub-standard form using the B-C-H formula.

4.2 Optimal Four-Cell Systems

In this section, we first study general multi-cell systems and then all possible four-cell
systems using the maps obtained in the last section. The goal is to find the systems
that require the fewest conditions in order to be converted to achromats of a given
order. A few definitions have to be mentioned before the study can be started. They

are the keys to the proofs of the theorems discussed later in this section.

Like the previous theories, we consider only those systems with midplane symme-
try. Therefore, the transfer map of the forward cell can be represented in the form of
eq. (4.5) with its pseudo-Hamiltonian given by eq. (3.8), which we write as

H = Z €51“;ngti’ai°yi”bib6i6, (4.23)

441.9151,

where i, + in + iy + i1, + i5 _>_ 3, and i, + it, is even.

Definition 4.3 Define the polynomials A(H), B(H), C(H), and D(H) as the parts
of H which satisfy

A(H) = Z nggagygbgo$i’ai°yiybib6i6 (i3 + in is odd, ia + i1, is even), (4.24)
£94,431,
B(H) = Z nggagygbg6xi’a£“yi'bibdi‘ (it + in is odd, to + i1, is Odd), (4.25)

58

C(H) = Z Cannibalii‘aiwl’W5“ (ix-H}, is even, i, +i), is odd), (4.26)
magi“,

D(H) = Z: Ci,iai,ibi,$i’ai°yl"”"6“ (i, +ia is even, ia +2), is even). (4.27)
i,iaiyibi6

Definition 4.4 Define

11“ = H(f)=H(z,a,y,b), (4.28)
H“ = H(Rl)=H(a:,—a,y,—.b), (4.29)
H5 = H(Sl)=H(—x,—a,y,b), (4.30)
H0 = H(RSl)=H(—1:,a,y,—b). (4.31)

It is easy to show that

HF = A(H) + B(H) + C(H) + D(H), (4.32)
H” = A(H) — B(H) — C(H) + D(H), (4.33)
H5 = —A(H) — B(H) + C(H) + D(H), (4.34)
H0 = —A(H) + B(H) — C(H) + D(H). (4.35)

4.2.1 General Properties of k-Cell Systems

Consider a general system of 16 cells arranged using the above symmetry operations.
Using Theorems (2.2) and 2.4 repeatedly, its map can be brought to the sub-standard

form in a similar way as in eq. (4.19). The result has the form

4

M = exp(: HU) :)exp(: H(M(1)l) :) - - ~exp(: H(M(k‘1)f) :)(MTl), (4.36)
where MT is the linear matrix of the system and

( mg) in)? 0 0
mg? ml} 0 0' m

M“) = (2' = 1,2,---,k— 1) (4.37)

 

 

0 o 0
o o mg; m“ 0
l 0 o 1

59

is a midplane symmetrical matrix obtained from combinations of the linear matrices
of the previous cells and matrices R, R“, S, S", and C, C"1 depending on the

specific choices of the system. As a result, we have det(M(‘)) = 1.

Using the B-C-H formula, M can be transformed to a single Lie operator acting

on a linear map, which is
M =n exp(: H(f) + H(M(l)i) + - -- + H(M("'1)I-) + commutators :)(MTf), (4.38)

where the commutators are polynomials of order 4 and higher, and MT is the linear

matrix of the whole system.

For the time being, we restrict ourself to 2: H (M (‘71-), hoping that systems which
cancel the most number of terms in 2: H (M (ill), called the optimal systems, do the
same to the total pseudo-Hamiltonian. In Section 4.3, this will be shown to be true
at least up to the sixth order. Due to the fact that all third-order terms in the
total pseudo-Hamiltonian are contained in H3O) + H3(M(1)i) + - - - + H3(M(*‘1)i),
those not cancelled by symmetry must be cancelled by adjusting the second-order
elements. Therefore, 2H3(M(‘)l) is used to find the necessary conditions for the
optimal systems that cancel the most number of terms in 2 H3(M ('71-. ) (Theorems
4.2 and 4.3). Although the restriction of 2H3(M(‘)f) makes it more difficult to

prove Theorem 4.3, it makes the logical structure of the proof more straightforward.

Next let us show that there is no system that can cancel D(H(f)) + D(H(M(1)f))
+ . . . + D( H (M (k'llfi) just by symmetry without changing the nonlinear settings.

First, D(H(i)) is split into two parts:

Definition 4.5 D+(H(i)) is defined as the terms in D(H(f)) with all exponents on

x, a, y, b even, which has the form 2 Czn,,2n¢,2ny,2n,,g,xz"‘a2"“y2"9b2“ 6".

"3 sum My vnb 9‘6

D‘(H(f)) is defined as the terms in D(H(f)) with all exponents on 3:, a, y, b odd,

60

which has the form 2 C2ng+1,2n¢+1,2n,+1,2nb+1,i5$2nz+la2na+l y2fly+1b2nb+l6w.

nz.na.ny.n6.i3
Here n,, no, ny, n), are non-negative integers.

Note that D = D+ + D" because of eq. (4.27).

Theorem 4.1 For a given k-cell system, it is impossible to cancel any term from
+ H(l)) + D+(H(M(1)I-)) + + D+(H(M(k'1)l)) solely by the symmetrical ar-
rangements of the cells and the choices of special linear matrices under the assumption

that no relations are assumed among the Lie coefficients Cindy”).

Proof.

From eq. (4.36) and (4.37), the sum of H(M(‘)I-) is

H(f) + H(M(1)f) + ‘ ' ' ‘l' H(M(k—l)f) = Z CigioiyibiJ (xi’ai“y‘vb‘b6‘3

inimimihi;
+(m1(11)$ + mnzla + m115)6)33(m(11)$ + m(12)a +m(}5)6)ia
(m (3:331 + ml1)b)‘"(ml,1)y + m(1)b)"6“ + - - - +
(k- 1)

+(m11 :c + mgg‘lm + m)t'll6)“(mg'{ 1):: + m“ 1)a+ mat-'26)“

(m4: "9 +m "‘ "b)"v(m 5.227% + mfii"’b)‘"6“).
which entails that the sum of D+ is

D+(H(I)) + D+(H(M“>f)) + - - - + D+(H(M“°-"I)) =

2n 2n 2n 2n i
Z Cznxyzflaflflyflnhi); (I :a 6y 9b b6 6
11; Mom.) .7153;

+(m(11)x + m(12)a + m(15)6)2ns(m(11)$ + m(12)a + m(15)6)2na
("15329 + mglb)2"v(m(2y + mlllb)2"°6i5 + . . . +
+(mfi- 1):: + m(lc—1)a + m(k-1)6)2n,(mgli-1)z + m(k- -1) (1+ m§§'1’6)2"°

(my; ”9 + my; l’11)“("lilifny + mi'i‘llb)2"°5“) -

61

Since there are no connections demanded among 02",,2n¢.2nw2nb,g&, the vanishing of a

polynomial associated with Cznxgnmznwgnb’go requires that

k—l _ . . . . .
x“w“w“%““+Zlmflx+m9a+m§®hflmfix+m$a+m£®“°

i=1

("19:39 + méib)2"”(m1'§y + "1125)?“ = 0

for any point in the phase space. Due to the fact that all quantities that appear in
this polynomial are real numbers, they can not be cancelled regardless of the choices
for M (i). Therefore, these terms have to be cancelled separately when the achromats

are designed.

4.2.2 Two— and Three-Cell Systems

The next theorem shows that two- or three—cell systems cannot give optimal solutions

for achromats.

Theorem 4.2 Two- or Three-cell systems can not cancel A3(H(f)) + A3(H(M(1)i))
+ + Aa(H(M<*-1>I)). 8411(1)) + Bs(H(M“’f)) + + 84(H(M<*-l>i‘)) and
0.411(1)) + 03(H(M<”I)) + + 03(H(M<*-‘>I)).

Proof.

(1) Two-cell sflems:

 

The sum of H(M(‘)f) is

Hal ‘1‘ H(M(1)f) = 2 Camp“, (xi’ai“yi”bi°6i‘

i,,i¢,i,,ib,i5
-Hme+m9a+m9®fim9x+m9a+m$flr

(may + msbmmsy + manure) .

62

Cancelling the terms associated with C1.0.0.0.2 from A3(H(f)) + A3(H(M(1)i)) entails

that
6‘1,0,(),(),2(:r62 + (mllllh: + mglga + mlllsl6)62) = 0,

Since all coefficients are independent of each other, each term in the above equation

has to vanish separately, which gives the solution

”‘11 = "'1,
milz) = 07
m)? = O.

and cancelling the terms associated with C0.1.0.0.2 from B3(H(i)) + 83(H(M(1)i))

entails that

00.1.0.0.2(a52 + (mil)?! + "1112)“ + mils.)5)52) = 0,

which has the solution

mg) = — 11
m4? 0,

Considering the terms associated with C1434” from 03(H (13) + 03(H(M(1)f)), we

have
01,133,1(2306 + mill)mg12)xa6) = 2C1,1'o,o,1$a6,

which shows that C3(H(f)) + C3(H(M(1)f)) cannot be cancelled.

(2) Three-cell systems:

 

Similarly, the sum of H (M (01‘) is

H“) + H(M(1)f) + Hlei) = Z Ciziaiyibifi (xi‘ai°y‘vb‘°6ia

i3 9‘0 9"” sibvid

63
+(m§‘,’ :1: + m(12)a + ml,51)6)"(m(1) :1: + m(12)a + m21)6)‘“
(méé’y + mgi’b)‘2(m$3’y + "1533202622
+(mlfilx + m(22)a + ml25)6)"(m(21) a: + mlila + m(25)6)“
(m3 3 + 33,213)» (mté’y + 342693) .
Cancellation of the terms with 01,03,0’2 requires that
C'1,o,(),o,2(:c62 + (m)th + muga + ml1)6)62+(m£21)x + mma + ml25)6)62) = 0,

which entails that

1+ mm + ml? = 0 "1121)"- -(1+ mi?)

3.)? + 13%;? = . =3 3.9— _ -39) . (439)
1 2

m15)+ m(5 )_ _0 m(:)_ _ _ m(5) .

Now, let us look at those terms with 03,03,043, which have the form

x3+(m(11)x+m(1)a+m(l)5)3+(m(2)$+m(22)a+m(2)6)3
= 9:3 + (mmx + mgla)3 + 3(m(11)a: + mllz)a)2m95)6
+3<m “1’4- +m‘23 amx (2)2323 +m< (2)333
+(m (19:11: + muga )3 + 3(ml21):r + mingla)2 m15)6 (4.40)
+3633). + mli’aXm £19262 + (174239262
= (1 +(mi‘1’)2 + (mli’12)x2+ 3((m.‘.’)2m§‘3’+ (m.2.’)2m 2,)323
+3<m£"(m£‘3’)2 +mi2.’(m123’)2)m2+((mlz’)2+(1742392142
+ 3m(( £2.92") 3’+(m£‘i’)2 mfii’n 26+6<m£‘.’m£‘3’m£‘3’+mii’mi23’mt2’1m6
+ 3m(( $239244 “3’ +(mf?)2 mti’1a26+ 3(m.23’(m“3’)2 +m(2,>(mg§>)2)332
+3<m§‘3’(m£‘3’)2 +m£23’(-m £292 )332+((mgl,>)3+(mg25>)3)33. (4.41)
Similar to the case of the two-cell system, each term in eq. (4.41) has to vanish

separately.

64

Inserting mg) 2 —(1 + mi?) from eq. (4.39) into (1 +(m(111))3 + (mfi’)3) = 0 from
eq. (4.41), we have
1+ (mfii’r — <1 + 3399):) = 0.

=> m§:’(m§‘.’ + 1) = 0,

=> m)? = 0 or mill) = —1.
Since mg) = —(1 + mill), the solutions are
mW=0 m9=—1
(n- m m-

Inserting either of the two above solutions into (mgll))2m(112) + (mfi))2m§22) = 0 (eq.
4.41), and combining this with eq. (4.39), we have mg) = mg) = 0. Similarly, from
(mlel‘s’ + (mli’le? = 0. we obtain mi? = ml? = 0-

In summary, the solutions cancelling the terms with 01,039,; and 03,0333 are

m3=°

rfibz_h 0‘ 7fib=0m °
m3=m3=° m3=m3=°
mis =m15 =0 mis =m15 =0

Similarly, the solutions which cancel the terms with (70,139,; and Co,3,o,o,o are

m9=0 74?=-1
m3? = —1 33 3.12,) = 0
r49=m3=0 r4?=m3=o'
r42=m§=0 r42=m3=0

For all the combinations, there is at least one matrix whose determinant equals zero.
So there is no solution that cancels A3(H(f)) + A3(H(M(1)l)) + A3(H(M(2)i))
and Bg(H(i)) + B3(H(M(1)f)) + B3(H(M(2)f)) simultaneously, which concludes

the proof.

65

4.2.3 Four-Cell Systems

In this section we will show that certain four-cell systems do cancel A, B and C at

the same time.

3 3
Theorem 4.3 Given afour-cell system, the terms ZAn(H(M(‘)i)), ZBn(H(.M(‘)I-)),

i=0 i=0

3
and ZCn(H(M(‘)f)) (Mm) = I) are cancelled for all choices of n, if and only if
i=0 '

H(M(1)f), H(M(2)f), and H(M(3)f) equal a permutation ofHR, HS, and H0.

Proof.
(1) The sufficiency is obvious from eqs. (4.32)-(4.35).
(2) The necessity:

3 3 3
Since ZA3(H(M(‘)i)), 2B3(H(M(‘)f)), and 203(H(M(‘)1)) can only be can-
i=0 '=0 i=0
celled by symmetry and the first-order arrangements, the necessary conditions for the
vanishing of A3, 83, and C3, and hence for the vanishing of An, B", and C", are also

from symmetry and the linear map only. Therefore, A3, B3, and C3 are selected to

determine the necessary conditions.

To prove the claim, the groups of solutions for the smallest decoupled sets of
equations are found first; the connections between two of them are found from the
equations containing the variables of the two groups, which form a second-level group.
and so on. Eventually, a necessary solution for all the equations is found, from which

the conclusion is drawn.
First observe that since all four cells contribute as a sum, cancelling the terms

associated with coefficients Ci,o,o,o,2, C0,),o,o,2, Cs,o,o,o,o, and Co,3,o,o,o requires that

3 . . .
71:62 + 2(ml'llx + mg'ga + m£26)62 = 0, (4.42)

i=1

w}.

66

362 + $333.12); + 33133 + "3126).? = o, (4.43)
i=1
3
1L23+Z:(m('l )r+m()a+m15)6)3- =,0 (4.44)
i=1
3
a3 + :(mgllx + mgzla +m (226? = 0. (4.45)
i=1

The fact that eqs. (4.42) and (4.44) are decoupled from eqs. (4.43) and (4.45) allows
us to solve the two groups separately. Let us first concentrate on eqs. (4.42) and
(4.44). Since the coefficient of each term in the equations has to vanish separately,

we obtain the necessary equations for the coefficients, which are

1+ m)? + m‘,” + m8’- _ 0, (1)
m)? + ml? + m)?— — O, (2)
mw+mm+mfl=, (m
1+(m12’)2 +(m123’)2+(m123’)2=o. (4)
(11212.92 m1232+(m1232)2m1232+ (m121212m123’=o, (5)
m1121m 12392 +m12321m 123212 +m13221m121’) =0, (6)
(m123’)2 + ("312392 + (m123’)2 = o, (7)
(m123’)2 m123’+(m12.’)2 m123’+(m 122)2m121’=o (8)
m1232m12 321721232 +m129m12 32m123’ +m1212m12 3’m1232— — o, (9)
(m123212m112+(m122)2m1?+ (m1‘1’)2m 122:0 o, (10)
m123’(m123’)2 +m123’(m 123212 +m112(m 12:21:03 (11)
11312326212392 +m123’(m12’)2 +m1232(m12:.’)2 =0, (12)
(m123’)2 + (m123’)2 + (m11’)2 = o, (13)

where eqs. (1)-(3) come from eq. (4.42), and eqs. (4)-(13) from eq. (4.44). Note that

all these equations are invariant under a permutation of the upper indices, as they

67

should be.

Since eqs. (1) and (4) are decoupled from the other ones and only contain mu,

let us first solve them. Equation (1) can be used to solve for m(31) to obtain
m9: —U+mm+mm) (fl

Inserting this into eq. (4), we have 1 + (mg?)3 + (mfi))3— (1 + mul) + m(2))3= —0,

which can be reduced to
(mW+nWWmW+1wfi?+U=0

Using eq. (*), we obtain three solutions for mm, which are

m11>= 372-112 {m112=—m112 m11>=—m11>
mll =_1 9 ,

, (4.46)

m“ — —l mm = —'1
As is to be expected from the permutational symmetry, the three solutions in eq.
(4.46) are permutations of each other. Without the loss of generality, we select one

of them, which induces one particular choice of the permutation, and all the other

solutions can be obtained at the end. Therefore, the representative solution of eqs.

(1) and (4) is

m11)_ __m(21)
13)_ -

(4.47)
m1] = "'1

i From now on, solutions of all other variable are expressed with respect to this repre-

sentative solution.

(3)

Now let us turn to eq. (3). from which m15 can be solved. We have

mg: -1mm+mm) 1w

Inserting this into eq. (13), we have (mg?)3 + (mg?)3— (m (15) + mu?)3 - —O, which can

be reduced to

m112m1121m112+m1121= o

68

From eq. (*’), the solutions of mm are

3 1 2
141=—ms mw=—ms mu=—ms
11) , 1 02 1s) -

(4.48)
m15 : 0 T7115 = 0

Note that the other three solutions are exactly the same as those shown in eq. (4.48),

respectively.

Combining eqs. (4.47) and (4.48), we obtain the the solutions for mm and mu;

satisfying eqs. (1), (3), (4), and (13), which are

3 2
PEEP-"2231222111” (m11=-m11m1:=°
9 ’
mm = ‘mii amn = ‘1 mm = —mu,m11 = ‘1
142=—ms)41=o 44
222 11) _ 12) 13) _ - ( ° )
mu - ‘mu amu — "1

Next, let us now decide how these solutions satisfy eqs. (8) and (11). Together
with the four equations above, they form a second-level group which contains purely

777.11 and m15.

Inserting the first solution from eq. (4.49) into (8) and (11), we have
{ 1m1:3;)2m1g+1—1)21—m125;) = o
1m11 )1m15 )2 + 1-1)1—m112)2 = 0

which has the solutions m)? = 0 or m)? = 1. In the case of mg) = O, we have m)?
= —m§§) = O, which gives
{ms=me=ms=o
1 2 3 -
mil) = -mi1)amil) = '"1
In the case of mg) = 1, we have mg) = —m§21) —1, which gives
2 3
mis) = “mi5)ami15) = 0 (4 50)
(1) _ (2) _ (3) _ ' -
mu - "lamu — 11mm - ‘1

Similarly, inserting the second solution from eq. (4.49) into (8) and (11) gives the

solutions
2 3 2
{ .. { .
mil) = ‘mi1)ami1) = *1 mi1)=1,mi1)= -1,mu = ‘1

69

Since the last solution from eq. (4.49) happens to satisfy eqs. (8) and (11), and

the solutions from eqs. (4.50) and (4.51) are special cases of eq. (4.49)’s solution,

those solutions satisfying the equations containing only mg) and mi? are

2 3 1 2 3
mils) = 0,m$5) = 0, m1.) = m1.) = -m)5),m)5) = 0 4 51
11) _ 12) 13) _ 1 0‘ (1) _ 12) (3) _ - ( - )
mu - -m11 3mm - ‘1 mu - —m11 1m11 - ‘1

Note that the first solution is actually a special case of the second one. But it is kept

as a separate solution for the convenience of the proof.

A little inspection shows that the equations containing only mm, which are eqs.
(2) and (7), are exactly the same as eqs. (3) and (13) for m15. Therefore, m1; has

the same solution as mm, which means that the solutions satisfying those equations

containing only mg) and mg) are exactly the same as those of the equations for mu
and m15, which are
1 2 3 1 2 3
miz) = miz) = miz) = 0 mi2) = "mi2)ami2) = 0 4 52
11) _ 12) 13) _ 02‘ 11) _ 12) 1s) _ - ( - l
mu — ‘mn )mu - "1 mu — “mu 3mm - "1

After checking with eq. (9), (10) and (12), we obtain the solutions for the whole

system of equations, which are

1 3 3
mifi = mi? = m); = 0 mi? = mi22)2= miaz) = 0
(1) mis = mis) = mis) = 0 v (2) mis) = ‘mi5)ami5) = 0 1

333112: —m123>,m11>=—1 m11>=-m11>,m11=-1
33112: —m112,m123>=o m112=-m112,m11>=o
2 3 2 3
232 "21121: 22223372232123 = ‘2 2 ‘22 "21121 = “2212122211? 2 ‘2
mm = _m111m11 = “1 mu = ‘mu )mn = "1

Again, the equations of mu, m2; and 771325, given by eqs. (4.43) and (4.45), are
exactly the same as those of mm, mm and m15. Since the solution for mu has been

fixed, all three solutions have to be taken account, which are

(1) _ (2) _ (3) _ (1) _ (2) _ (3) __
m21 -m21 -m21 -0 "'21 -m21 -m21 —0
"‘25 — ”‘25 = m25 1 "‘25 = "‘25 = "‘25 1

(ll) (1) _ (2) (3) = 0 (1) (2) (3) ._._. 0

"312.1: —m1=:>,m112= —1 33112 = —m1;>,m11 = —1

70

1 2 3
mil) = mii) = mil) = O

(1) (2) (3) 0

or "225 = "‘25 = "’25 =

(1) _ _
m22 — "m22 177122 - "

1 2 3 1 2 3
mil) = mifi = min) = 0 min) = mil) = mil) = 0

2 3 1 1 3 2
mis) = —mg5),mg5) = 0 1 mis) = —m(25)1mg5) = 0

(2) (3) (1) _ _

9

1 3 2
m22 = —m22 ,m22 - 1 min) = "mi2))mi12) = "1

1 2 3
mil) = mil) = mifi = 0

(1) (2) (3)

or m25 = —m25 1m25 = 0 i

7021 = —m21 1m21 = "‘21 - -m21 1m21 =

2 3 1 2 3
’ 1
m22 = "mzz ,m22 = ‘1 "‘22 = —m22 ,m22 = ‘1

11)_ 12) 13) 0

{ 12) 1:1) 11) 0 11)_ 1:1) 12) 0

"221 — —m21 ,m21 =
or m1? = "312,2 = m1? = 0 1
2 3
33322222 = —m13’,m§3’ = 1
2 3 1 1 3 2
mizl) = -migbmili) = 0 mi?) = “migbmizfi 2“ 0
(4’) mis) = “misbmig = 0 1 mis) = —m(25),mg5) = 0 1
2 3 1 1 3 2
772(22) = “mi2)ami2) = ‘1 "1(22) = —mg2),mg2) = ‘1
1 2 3
mil) = ‘milbmifi = 0
or m)? = —mg§),m§? = 0 -
1 2 3
111122 = —m13’,m13 = —1

So far there is an abundance of solutions. By using additional conditions connect-
ing matrix elements, their number will be reduced. Let us consider the terms with

0133,03, 02,19,043, and 0139,04. Cancelling them requires that

3 a o u n . .
ma2 + :(mg'lh: + mg'ga + m£26)(mg'1):c + mg'ga + mg26)2 = 0, . (4.53)
i=1
3 u . . . . o
+ Z1m1'322 + m1'32a + m1'126)21m112z + m1232a + m1‘26) = 11, 1151)

5:1

71
x05+232( (7711356 +m12)a+m(15)6)(m(21)(21+m 2)a+m(2:5)6)6 : 0. (4.55)
i=1

Next, all possible combinations of the two sets of solutions are considered, and
the final solutions are decided. We will start from the simplest case, and then move

to the more complicated ones.

Case (1) — (1’):

 

Since m” = m); = mg? = m)? = 0 (i = 1,2,3), the equations (4.53)-(4.55) are
simplified to
3 1') 1)
3702 + 22(7771'15’7)("'22221)2 = 01
i=1

3 . .
macs + :(mg'l):c)(mg'2)a)6 = 0,

where the coefficients are

1 2 2 3 3
1+m12121m12)2 +m1321m112)2 +m1121m132)2=—o, 14.56)
1 + (mii ))2m212) + (mii) )2m22)+ (mii))2 "1232)- - 01 (4-57)
1 + meglg) + mglmg) + mfilmg) = 0. (4.58)

Now let us look at the first solution from (1’), which gives

2 3
m(2)- —mg2) .
m::)- — —1

Inserting this into eqs. (4.56)-(4.58), we have

1+1—m1212)1-1)2+m12121m1212)2+1-1)1-m1212)2=0,
+1—m1122—)1 1+) 1m12i2)2m1212+ +1—1)21- m1212)=o

1+ 1—m1232)1— 1) + m1212m122+11)1—m1212)= 0,

72

which can be transformed to

1— m1'12 + m12121m1212)2 -1m1212)2 = o,
1 -(m121’)’+(m121’) ”11222—111132: 0,
1 + mg) + mgmg) + my =.0
After factorization, the equations are

2 2 -
11— m112)11—1m112)2)= o,
11-1m1212)2)11— m1212) = o,
2 2
(1+ mi1))(1+ 77222)) = 0-

Straightforward arithmetic reveals that the solutions are

mg) = —1 mg) = :l:1
12) 1 222' 12) -
777.22 It]. m22 = -1

In conclusion, the final results for the first solution are

{m1212=1m1212=—1,m12;2=—1 o1 { m11>=11,m12)=,111m<i>= =1
(1)
m

‘22 — -1 m122 = 1111111212: +1 ’ 11 = —1,m12;2 = —1,m1212- =1

Written in the form of matrices, the solutions of this case are

1 1) 1) —1 1) o —1 1) 1)
M122: 0 —1 o ,Mf22_-. 0 +1 0 ,Ml‘22= 1) $1 0
0 1) 1 o o 1 o 1) 1

‘01‘

4:1 0 0 i1 0 0 —1 0 0
M122: 0 —1 0 ,Ml‘22= 0 =1 0 ,Ml‘22= 0 1 0 .
o 01 1) 01 o 01

which correspond to the cases that (ll/[1(1),M(2),Ml(3)) = (R1,Cl,51), (R1,S1,C1),
(51, 121,01), or (R1,Sl,Cl). Therefore, the first solution of (1’) agrees with the as-
sertion of the theorem. Note that M1”) can only be S or 0, because mg) is fixed to
be —1. This is why Case (1) - (1’) gives four, rather than six, solutions. It can be

shown similarly that the second solution of (1’) draws the same conclusion.

73

In the case of the third solution, we have

2
{1111)— _ =11)

m3- = —1

Inserting it into eqs. (4.56)-(4.58) shows that eqs. (4.56) and (4.57) are automat-

ically satisfied, and only eq. (4.58) gives a nontrivial solution, which has the form 2

+ 2m( )m(12)= 0. The solution is m1§2m§9= —1. Together with mg) = —1, we have
(1) (1)_
m11m22— — 1
"2131 =

It can be shown that

mg) = —mg12) (see Appendix B). (4.60)

Thus, the solutions are

{ 1111:,2: :11,m§':;2— _ :1, "11:2: —1
m12’-= :t1,m$2)- = 2F1,m(22)- = —1

or in the matrix form
i1 0 0 3F1 0 0 —1 0

M111): 1) $1 0 M,122= o 11 o ,M,122= o — o, .
0 0 1 0 0 l 0 l

which correspond to the case that (Mlm,M1(2),Ml(3)) = (R1,Cl,51) or (01,121,331).

Or—to

Therefore, Case (1) — (1’) agrees with the assertion of the theorem.

 

Here mg) = mg}? = mg? = 0 (i = 1,2,3); thus the equations (4.53)-(4.55) become

:ta2 + Z(m§1)x)(mg2a + mg5)6)2— =0 (4.61)
i=1

a: 2a+2:(m§1):1:) 2(mg2)a+m(2 26)=0, (4.62)
1:]

xa6 + 2((m1121)(m11}a + m25)5)6= 0, (4.63)

i=1

74

which are equivalent to

1+m112(m1123) +m112(m1322> +m13 2(m13a2)2=o,
1 + (m112)3 m1? + (m1‘12)3 m1§2+ (m112)3m1322 = 0,
1 +m1112m1‘2+m1212m1322+m132m132-0,
m112(m112) +m1312<m1352) +m1312(m1352>3 =0,
(m112)3m112+ (m112)3m13;2+ (772131213 m1352— — o.

(1) (11+m(2)

3 3
mu m251m225) + mi1)m(5) =

1 1 1 2 2 2 3 3 3
mi1)mi2)m(25) + mi1)mi2)mi5) '1' mi1)m(22)mi5 )_ —0-

From the first three equations, we obtain the same solution as in Case (1) -- (1’). First
let us look at the case of (M111), M112) ,M13))= (R1, 5'1, Cl). Inserting this into the last

three equations, we have

m12‘52+m§§2+(m132)25 = 0, (1")
m1‘52—m1352—(m132)25 = o, (2")
—1?m1‘52+m (771(3))25 = 12- (3")

From (2”) and (3”), we have m25) — -,0 and from (1”) and (3”), we have m2? =

Therefore, the solution is mgs) = mg): m2? = O, which is means that this case is

reduced to Case (1) - (1’). The other three solutions from Case (1) — (1’) give the

same results, which is a consequence of the permutational symmetry.
Note that Cases (1) — (3’) and (1) — (4') are the same as Case (1) —- (2’).

Case (2) — (2’):

 

Here, mg— = m2: = 0 (i = 1,2,3); thus the equations (4.53)-(4.55) become

3
ma2 + Z(m1ll)x + m15)6)(m(2 a + m1'5)6)2 = 0,

i=1

75

.22 a + :(mmx + m(15)6)2 (m12 )a + m2 215) = 0,

3
3:06 + 2(m1111)x + 7711,3525me; a + m125)5= o.

Cancelling the terms a215, a152, and a6 requires that

1 2 2 3 3
m15’(m12‘22 )3 +m15’(m122)3 +m152(m122)3- =0,

1 1 2 2 3 3
(mid)2 miz) + (m115))2m22) ‘2' (m115))2mg2) = 0,

"Jump; + mugmg) + mo)", (3 )_ _ 0.

Since the diagonal elements are the same as those obtained in Case (1) — (1’), we
obtain mg5) = m5? = mgl— - 0. Therefore this case is simplified to Case (1) — (2’),
which gives the same solution as Case (1) — (1’). Note that Case (2) — (3’) and Case

(2) — (4’) are the same as Case (2) — (2’).

Case (3) — (3');

 

Here, m1? = m25- — 0 (i = 1,2,3); thus the equations (4.53)-(4.55) become

an2 + Z( mgr: + m(12a)(m2'1)x + m122a)3— =0, (4.64)
1:]

2:2 a + :(mul) a: + mggaf (mgx + m22a)= O, _ (4.65)

:ca6 + 2(m1111):r + m12a)(m(21)3: + magaw = O. (4.66)
1:!

Cancelling the terms a3 and a26 requires that

m132<m112>3+m1322(m112>3 +m1322(m13;2>3 =0,

( (1)2

2
m12)2 "3212) +( H

3 3
m12 )2m22)+ (m(12))2m122) = 0,

1 2 2 3 3
mi 2)mi2) ‘l' mi2)mi2) + mi2)mi2) = 0-

76

Inserting this in mg) = —m§22) and m?) = 0, we have

2
m£:3((m£‘3)3— (m £23)3>=0,
(ml‘23)3(m§323+m‘33)= 0.

1 1 2
mlz’(m$’ —m§2’) = 0,
which have the solutions

m£12)— — 0 or m22)- — mg): 0.

Let us consider these two solutions separately.
(a ) ms— — m£§3= o
(3)_

From solution (3’), we have m22 — -—1. Together with m(12) = 0, the equations

(4.64) and (4.66) become

1 1 1 2 2
3'02 + (mii)$ + m£2)a)(m21);1:)2 + (mi1) 2: + mi2 ) “)(m21)x)

+(mi31) ”("321)”: " “)2 = 09

2) 2 2
x+m"a )3(m£1’x)

32a +(m(111)a: + m(112)a)2(m211)2:) + (m(
+(mfi3x)3 (m £332: — a) = 0,

2a + (mi‘3z + m‘33a my.) >+ +(mfi3i3z + mli3a)(m£3.3m)
+(m( 33x)(mg33x_ a) = o.

Cancelling the terms xaz, 22a and ma requires that

1+ml313= , 1—(ml3i3) =0, 1—ml3i3x=o,

which have no solution.
(b) m9)- —

From solution (3), we have m§2)- — 0. Hence, this case is reduced to Case (1) — (3’).

77
Similarly, Cases (3) — (4’) and (4) — (4’) can be simplified to Case (1) — (4') and
Case (2) — (4’) respectively, which concludes the 2-motion.

Now let us study the y-b sub-matrices. Since it is proven that the x-a sub-
matrices are permutations of R1, 31, and C1, and permutational symmetry holds,

one solution out of six can be chosen without the loss of generality. Here we choose

(11493114133114.3333) = (31.51.01).

Cancelling the terms with 01,03,053 and C033,“) requires that

y3 + (méé’y + méi’bf — (m‘a'é’y + méi’bf — (mé‘é’y + méi’b)3 = 0, (1)
1 1 2 2 3 3
312 — ("333)?! + m34)b)2 — ("3393/ + mg4)b)2 + ("333)3/ + mis4)b)2 = 0- (2)

From (1) + (2), we obtain y2 — (mgy + mgb)2 = O, which has the solutions

2 2
{ m(33) = 1 or m5”) = —1 .
mg) = 0 mg? = 0

From (1) — (2), we obtain (mgy + mgl)b)2 - (mgy + mgi)b)2 = 0, which has the

solutions
1 3 1 3
{m3 ., {"131 = s:
"‘34 = m34 m 4 — “min

Altogether, we have the solutions

(1) _ (3) (2) _ (1) _ (3) (2) _ _
{ma—imammw—l or {ma—im”,m33— 1 (4.67)

m9) = iméihm‘a? = 0 m§13= imgihméi’ = 0

Cancelling the terms with 01,09,253 and 00,1335) requires that
b3 + (mSERy + m313b)3 — (m333y + m£i3b)3 — (mfi’y + m£i3b)3 = 0.
b3 - (mast + m£13b13 — (m333y + m£i3b)3 + (m333y + m£i3b)3 = 0,

which are exactly the same as those for m3 and m34. Therefore, the solutions of m43

and m.“ are

{m£§3=im£§3,m£§3=o or {m£;3=im£§3,m£333=0

3 2 -
ml? = imfii’mfi? = 1 mg}: = $7715.43ng = _1

(4.68)

Cancelling the terms with C133,”, and C041,”) requires that

31” + ("3313) y + m(1)b)(m413) y + mmb)
—(m §§3y>(m 3.3.—3b) (m3353y + m§i3b)<m$3a3y + m‘3.3b>— - o (5)
3112— (m (1ly + m(1)b)(m(1)y + m(1)b)

-+(m3333y)<m£i3b) (m ‘33y+m‘i3b)(m£§3y+m£33b)=0. (6)

From (5) + (6), we obtain yb— (mQmeflb) = O, which has the solutions

m8?— = m4? = 1 or m(333- — mfl— — —1.

From (5) — (6), we have

("333133/ + m(13b)(m133y + "33135) = (mmy + m(33b)(m333y + "313331),
which entails that

(1) (1)_ (3) (3)

m3 "7'43: "m33 m43 31

3 3 3 3
msms+msms=msms+msms,

l 3 3
rs.»32=msms

Inserting eq. (4.67) into the three equations above, we obtain mg) = :lzmg? and
(l) (3)

m4 44 = im44 , which shows that the total solutions are

(1) im(33 (2) 1 (1) (3) (2)

ms: 3fl33: 33:3333fl33: 3
mi?) _ imi31’m12) 0 0’ mi?) _ im 1725;), mi?) _ 0

m43=:l:m43,m43=0 m43=:l:m43,m43 =0

m3? = 47,2333, m3?- = 1171313: imli3, "1333: -1

Cancelling the terms with C0551,” requires that

yb + (méé3y + m§13b)(m£33y + mll’b)

+<msy><msb> + (m‘33y + m§i3bxm£33y + m£33b)— — o (7)

hm

1 .
Will

Du

M

M

Ii(

From (7) — (6), we obtain

3 3
(m§§)y)(m§i)b) + (mgy + m§3)b)(mia)y + "2335) = 0,

which entails that

méi’mli’ = 0.

1 + m‘a‘é’mli’ + méi’mfii? = 0,
3 3
mg4)m,(44) = 0-

Due to the fact that det(M(3)) 324 0, the solutions are

(3) -

mas — 0 77134 — O
0) mg) = O or 5) mg) = O
maimaa) =-1 méi’mli’= —1

Case a): mgi)m,(4§) = —1 implies that det(M(3)) = —det(M1(3)) - mgmg) = —1,

which is impossible.

Case b): Similar to the x-a-6 sub-matrix, it can be shown that mg? = —mg) (see

Appendix B). Taking into account midplane symmetry, this means that M1?) = R2,
Mr?) = 32, and M2”) = 02.

In summary, M (1) = R, M (2) = S, and M (3) = C, which concludes the proof.

Altogether, we have proven that there is only one way to cancel 2 A3(H (M (l) f )),
2B3(H(M(‘)f)), and 203(H(M(1)f)), which is that M“) (i = 1,2,3) is a permuta-
tion of R, S, and C.

Since this is the only solution, it is the best a four-cell system can do. If we
consider a system with more cells, there is a possibility that we can find solutions
cancelling Z D‘(H(M(‘)f)) as well. Due to the fact that D’(H(f)) is only a small
part of D(H(i)), a solution which cancels Z D‘(H(M(‘)I-)) will make only limited

80

improvements compared to the current solution. To illustrate this, there is only 1 term

in D§(H(l)) and D;(H(i)) as opposed to 15 terms in D3(H(l)) and D4(H(l)); there

-o -o

are only 5 terms in D;(H( )) and D;(H(f)), as opposed to 39 terms in Ds(H( ))
and 06(H(f)).

4.2.4 The Optimal Four-Cell Systems

Now the question is: What kinds of systems have the properties stated in Theorem
4.3, under what conditions, and which of them requires the least number of conditions.

The following theorem answers these questions.

Theorem 4.4 Among all (sixty-four) four-cell systems, there are only four which
reach the optimum asserted by Theorem 4.3 while imposing the minimum number of

constraints on the linear map. They are FRFR, FRSC, FCFC, and FCSR.

Proof.

Suppose the forward cell has a linear matrix L. For the time being, let us restrict

ourselves to the :c-a-6 block of L, which has the form

b n d —b —d1] + bn’

d 77’ and L;1 = —c a on — an’ . ' (4.69)
0 1 0 0 1

(1) Choices on the second cell.

ase a FF:

Recall that the standard form of a forward cell (eq. 4.5) is

-o

MF = exp(:H:)(Ll).

lo ti
5. l

81
Therefore, the map of the system FF is
M” = MP M3“
= ((:exp11(:)(1) ((:exp L:)(1))
= exp(: H(I3=)( (((:exp H(I3=) (L13) 0 (L13)

= exp(:H(1):)(exp:(L1):)(L-L1)

To reach the Optimum, L has to be H, S, or C. Since L is symplectic, it can only be

5. Hence, the linear matrix is

—100
L,_—_ 0—10,
001

which entails that the conditions of reaching the optimum are

Q

sac-as:
II II II || ||
t—‘OPOO

Therefore, five conditions have to be met to reach the optimum.

(Jase 1h FR:

From eq. (4.8), we have

M'RoMF

3
w
:5

ll

(exp(:H(L’lR’11-):) (RL 1R 11))0 o:(e(xp (:)H(1) (L1))

= exp(: 11(:)f) )(:exp (H(L"1R‘l - L1) :)(RL’IR'1 - L1).

Specifically, Lf‘Rfl - L1 can be obtained from eq. (4.69), which yields

d —b —dn + 1117' 1 0 O a b 17
LflRl'l -L1 = —c a cn—an’ 0 —1 0 c d 17’
0 0 1 0 0 l O 0 1

Snce

cond

Whi<

Hex
five

82

—2ac —(ad+bc) -2a17’
0 0 1

ad + bc 2bd 2bfl'
( ) (4.70)

Since L411"1 - L is antisymplectic, it can only be R or C, which leads to the following

conditions:
bd = 0
ac = 0
b", = 0 7
an’ = 0

which are equivalent to

b = 0 a : 0
c = O or d = 0 (4.71)
"I = 0 17! ___ 0

Hence, we obtained two solutions with three conditions, which in turn eliminate the

five-condition solution above. For further reference, they are listed below:

Solution A : Solution B :

a 0 17 and 0 b 17 (4.72)
L1 = 0 d 0 L1 = C 0 0 .

0 O 1 0 O 1
Case 1; FS:

Similar to FR, the map of FS is

-o

MFS = MSOMF
= exp( Holman-1311)((expzrwnxu‘)

.9

= exp(:H(1) ::)exp( H(S‘l-LI) :)(S’L‘1S‘1 .Lf).

JIUK

whit

Willi

For
the

83

Since 5' is symplectic, S ‘1 - L is also symplectic. Therefore, 3"1 - L can only be S,

which means that L = 1. Like Case 1a, the conditions are

‘
so

HOPOO

‘0

fiO‘Odfi

‘0

which is not an optimal solution.

Case 1d FC:

 

The transfer map of the system FC is

-o

MFG = MCOMF
= (exp(: H(L‘IR‘IS‘11) :)(RSL-IR-ls-ln) 0 (exp(: H(i’) :)(Lf))

= exp(: H(1) :)exp(: H(L'l [1’45"1 - L1) :)(R.S'L’1R'1.S“'1 - L1).

For this case, L‘IR'IS‘1 - L can be R or C, because it is antisymplectic. Since it has

)

the form

(1 —b —dr] + (717' —l 0 0 a
Li'lRflel - L1 = —c a C17 — an’ O 1 0 c
0 0 1 0 0 l 0

d

‘

b
d
0

r—Ié

2ac ad + bc 2617
0 0 1

—(ad+bc) —2bd —2d17
, (4.73)

the solutions for an optimal system are

b
c
’7

Illlll
cc
0
H

fl
9.9
”II
oo

a
II
o

84

The fact that these solutions require only three conditions makes them another set

of candidates, which are

Solution A : Solution B :

0 0 and 0 b 0 (4.74)
L1: d 7] L1: C O T], .
O 1 0 0 1

In conclusion, in order to impose minimum conditions, the second cell can only

‘

COG

be R or C.
(2) Choices for the third cell of the system FRX x

From Case 1b, we have

_9

MFR = exp(:H(1) :)exp(: H(L’IR'ILL) :)(RL‘IR—ILL).

Let W = (exp(: H(M(x)f) :)(L(x)f)) be the map of the third cell, which can be
F, R, c, or D. Thus, the total map of the three-cell system is
MFRX = Mx o MFR
= (exp(: H(M(><)I3 :)(L(><)I3)
o(exp(: H(I3 :)exp(: H(L-IR-ILI“) :)(RL-IR-IL“))
= exp(: H(f) :)exp(: H(L-‘R-‘m =)
«exp(: H(M(><)I‘) :)(L(><)f)) o (BL-‘R—‘Lfn
= exp(: H(i) :)exp(: H(L-‘R-‘m =)
exp(: H(M(x)RL“R"Lf) :)(L(x)RL“R"Lf).

For our convenience, let us define M (2)( x) as the linear matrix in the pseudo-Hamiltonian

for the third cell. For systems FRX, we have

M(2)(x) = M(><)-RL‘112‘1 . L = M(x) . LFR,

whe

I'd-

Wlllt

Whic

any

Whic

85

where LFR = I‘M-112‘l - L.
Easel; FRF:

Since M(F) = 1, we have M(2)(F) = 11’.L"1R"l - L. For the two solutions, the
x-a-b block M (2)(F ) are listed below.

In case of Solution A, we have

100100.100
M{2’(F)= 0—1 0 0—1 0 = 010,
001001 001

which does not reach the optimum because it does not agree with Theorem 4.3.

In case of Solution B, we have

100 —100 —100
Mf2)(F)= 0—10 010 = 0—10,
001 001 001

which is a possible solution, because it agrees with Theorem 4.3 and does not need

any more conditions.

Case 2b FRR:

From M(R) = L‘IR‘I, we obtain

MWR) = FIR" ~RL‘1R“ - L = L-1 ~L‘1R'1-L.

In case of Solution A, we have

d o —dn 1 o o d o —d17
Mf2)(R)= o a 0 0 —1 0 = o -—a o ,
0 0 1 o 01 o o 1

which does not reach the optimum unless d = 1 and 17 = 0. This case corresponds to

a five—condition solution, which is eliminated.

86

In case of Solution B, we have

0 —b 0 —1 0 0 0 —b O
Mlm(R)= -c 0 C17 0 1 0 = c 0 c1] ,
0 0 1 0 O 1 0 0 1

which does not reach the optimum because it does not agree with Theorem 4.3.
Case 2c FRS:
From M(S) = S“, we obtain

M(2)(S) = 5-1 ~RL'1R" - L.

In case of Solution A, we have

—100100 —100
Mf2’(5)= 0—1 0 010 = 0—10,
001001 001

which is a possible solution, because it agrees with Theorem 4.3.

In case of Solution B, we have

—100 —100 100
Mll2)(S)= 0—1 0 0—1 0 = 010,
001 001 001

which does not reach the optimum, because it does not agree with Theorem 4.3.

Case 2d FRC:

From M(C) = L'IR‘IS‘I, we obtain

M(2)(C) = L-IR-‘s-1 . RL‘IR‘I - L.

In case of Solution A, we have

dO—dn —100 100
Mf2)(C) = o a 0 0 —1 0 0 —1 o
001 001 001

87

—d 0 —dn
: 0 a O ,
0 0 1

which does not reach the optimum unless (I = 1 and n = 0. This case corresponds to

a five-condition solution, which is eliminated.

In case of Solution A, we have

O—bO —100 —10
Mf”(C) = —c 0 c1) 0 —1 0 01
001 001 00

0 b O
= -6 0 C11 9
0 O 1

which does not reach the optimum, because it does not agree with Theorem 4.3.
(3) Choices for the fourth cells of the systems FRFX and F RSx.
Case 3a FRFX:

Define LFRF = L - RL‘IR'1 - L. Similar to case (2), we have

M3) = M(x)LFRF.

Since solution B is the possible solution for this case, the linear matrix of the forward

From M(F) =1, we have

cell is

“1

FRFF:

b
O
0

COO
V-‘Oé

M(3) = L . RL‘IR" . L.

88

Since

0
RIL;IR;‘L,=( o —1
0

this system is not a solution (see Case 2b).
FRFR:

From M(R) = L'IR‘I, we have
M(3) = L’1R" . L . RL‘IR“ . L.

Therefore, the x-a—b block is

—100 —100 100
Mf3)(C)= 010 0—10 = 0—10,
001 001 001

which shows that this system is a solution, because it agrees with Theorem 4.3.

In summary, the total map of system FRF R is
MFRFR = exp(: HF :)exp(: HC :)exp(: HS :)exp(: H331. (4.75)
FRFS:

From M(S) = S“, we have

M“) = s-1 . L - RL-‘R-l . L.

Since

—1 O 0
RlLl-lRi-IIq: 0 —1 0 ,

0 0 1

which does not agree with Theorem 4.3, this system is not a solution.

FRFC:

89
From 111((') : L-l H—IH" . we have
111(3) 2 L‘lH—lhml ' L - H1141?" - L.

Therefore, the .r-a-b block is

_ 0 —b 0 —1 0 o o b 1, —1 0 0
11,1”(0) = —c 0 c1] 0 1 0 c 0 o 0 —1 0
0 0 1 0 0 1 0 0 1 0 0 1

which is not a solution. because it does not agree with Theorem 4.3.
Case 3|) FRSX:

Define LFRS = SLS" - 13114-11?-l - L. We have
M”) = M(x) - LFRS.

Since solution A is the possible solution for this case, the linear matrix of the forward

a 0 7]
L1 = 0 d 0 .
0 0 l

FRS F:

cell is

From M(F) = i, we have
111(3) = SLS“ . BL“ 11’“ . L.

Since

—1 0 0
.S'I-I'RlLrlR-I—I'le 0 —l 0 ,

O 0 1

which does not agree with Theorem 4.3. this system is not a solution.

90
FRSR:

From ARR) = I."I R". we have
.11“) = L“R" -sLs-' 3117‘s” -L.

Therefore, the .r-a-b block is

. (l 0 —(17] —l 0
11!)”(11) = 0 (1 11 11 1
0 11 1 0 0

which is not a solution. because it does not agree with Theorem 1.3.
FRSS:
From 131(5) 2 5". we have

111(3):.8'“ -.8'L.‘5"' -RL"R" ~L= LS“ ~RL'111'l oL,

which is the same as FRSF.
FRSC:

From AHC) = L‘1R‘IS", we have
111(3) = L'lR—1.S‘1..S'L»S_l -RL‘IR‘1 -L = L'IR’l -LS'1 - RL’IH—1 - L.

Therefore, the .r-a-o block is

, 1 0 0 —1 0 0 -1 0 0
.11}"’(C)= 0 —1 0 0 —1 0 = 0 1 0 ,
0 0 1 0 11 1 0 111

which shows that this system is a solution, because it agrees with Theorem 4.3.

Sll
In summary, the total map of system FRSC is
111,711“. 2 exp(: HF :)exp(: 11R :)exp(: IIS :)exp(: 11(1':)1. (4.76)
which satisfies the requirements.

(I) Choices for the third cell of system FCX X.

Define L”. 2 NHL" 11"5'" - L. For systems FC'X, we have

MWx) = M(x) - L""'.

('ase la F('F:

From 111(17) = 1. we have

M(WF) = SRL"R"S" . L.

In case of Solution A. we have

‘ —1 11 0 —1 0 0 1 0 0
Mf‘”(F)= 0 1 0 0 1 0 = 0 1 0 .
0 0 1 0 0 1 0 0 1

which does not reach the optimum, because it does not agree with Theorem 4.3.

In case of Solution B, we have

—1 0 0 1 0 0 —1 0 0
Mfz’w‘): 0 1 o 0 —1 0 = 0 —1 0 ,
0 0 1 0 0 1 0 0 1

which is a possible solution, because it agrees with Theorem 4.3.
Case vlb FUR:

From M(R) = L‘IR", we have

M‘”(R) = L"R" -sHL.-‘R-‘s-‘ . L.

Since

1 0 0 —1 0 O
R,"-.S‘,R,I.,"R,".S'f‘-L1= 0 —1 0 or 0 1 0 ,
0 0 l 0 0 1

which does not agree with Theorem 4.3, this is not a solution.
Case lc F(‘S:

From M(S) : 5“. WP have
1’tll”(.9')-_—.$"‘-.S'RL“R“.5'-l . L : RL“R“s-1L,

In case of Solution A. we have

, 1 0 0 -1 0
MV’LS'): 0 —1 0 0 1
0 0 1 0 0

which is a possible solution, because it agrees with Theorem 4.3.
In case of Solution B, we have

‘ 1001110 100
111:“(5'): 0—1 0 0—1 0 = 010 ,
001001 001

which does not reach the optimum, because it does not agree with Theorem 4.3.
Cas_e__4_d FCC:
From [M(C) = L"R".S", we have
M”)((.‘) = L"R“S“ - SRL"R"S" - L = L" . L“R“S" -L.

Since

L;‘R,".S‘,“ L, = (

OOt—

GHQ
F—‘oo
\—/
O
’1
/'_\
OO"
l
OHO
_‘oo
v

5):)
which does not agree with Theorem 1.3. this is not a. solution.

(:3) Choices for the fourth cell of the systems FCFX and FCSX.
Case 5a FCFX:

Define LIT," 2: L - NHL" 11’"H" - L. We have
M(‘W x) = m x) . I."""".

Since solution B is the possible solution for this case, the linear matrix of the forward

0 h 0
LI 2 c 0 I" .
0 0 l

FCFF:

cell is

From M(F) = 1. we have
1lvll3l(F) = L - SHL"R"S" - L.
Since

0
.qulLrlRl-lsl-l ° L1: ( 0 —l
0

which does not agree with Theorem 4.3, this system is not a solution.
FCFR:
From M(R) = L'lli", we have

M(WR) = L"R" - L - S'RL"R"S" - L.

Therefore, the .r-a-6 block is

0 —b by’ 1 0 0 0 b 0 —1 0 0
MPH?) = -c 0 0 0 —1 0 c 0 1/ 0 —1 0
' 0 0 1 0 0 1 0 0 1 0 0 1

94

l I) Zlm'
0 — l 0 .
0 0 l

which is not a solution. because it does not agree with Theorem 1.3.
FCFS:
From 141(5) 2 1?". we have

MWs) = s” . L . SHL" IT‘S“ - L.

Since

—1 0 0
.S'1R1Ll—IR1-ISI—I’ L1: 0 “l 0 ,
O 0 1

which does not agree with Theorem 4.3, this system is not a solution.
FCFC:
From M(C) = L" R’lb'", we have

1119111"): L-‘R-'s—‘ . L - .sRL-'R-'s-' . L.

Therefore, the .r-a-b block is

100 —100 —100
M,‘3’(C)—_- 0 —1 0 0 —1 0 = 0 1 0 ,
0 01 0 01 001

which shows that this system is a solution, because it agrees with Theorem 4.3.

In summary, the total map of system FCFC is

_o

11‘1”ch = exp(: 11F :)exp(: 113:)exp(:HS :)exp(: HC :)1. (4.77)

which satisfies the requirements.

Case 51) FCS X :

Define L”,8 :2 SL5" ..S'11’L"H"S" - L. We have

M”’( x) = .111 x) - If”.
Since solution A is the possible solution for this case, the linear matrix of the forward

cell is
(1 0 0
L1 = ( 0 (l 71' ) .
0 0 l

FCSF:
From M(F) = i. we have

111‘3’(F)=5‘L.S"' -s'HL-'H—‘s-‘ - L = .S'L- RL“R“S“ . L.

_ —1 0 0
RlLf'Rf'Sf‘le 0 —1 0 ,

which does not agree with Theorem 4.3, this system is not a solution.

Since

FCSR:
From M(R) = L‘IR", we have

11101112): L"R“-.S'L5"‘-S'RL"R“S“'-L -_— L-'R"..sL-RL"R-‘s-'-L.(4.7s)

Therefore, the .r-a-6 block is

. —1 0 0 —1 0 0 1 0 0
1111:”(11): 0 1 0 0 —1 0 = 0 -1 0 .
0 0 1 0 0 1 0 0 1

which shows that this system is a solution, because it agrees with Theorem 4.3.

In summary, the total map of system FCSR. is

1171170512 = exp(: HF:)exp(:HC :)0XP(1HSIl9XP(3HRill-i (4.79)

{)6
which satisfies the requirements.
FCSS:
From 31(5) 2 S". we have
111(3)(.S')=S" uSLS" '15'1‘114—lR-15-‘1'L': L - HL"H"S" - L,
which is the same as FCSF.
FCSC:

From M(C') : L" 11).".9'", we have
MONC) = L‘l 11)_13"-.SLS—'-.SHL"IT‘S—LL = L-lR-lL-RL'lR"S-l-L.(4.80)

Therefore. the .r-a-o block is

. (1 0 0 1 0 0 a 0 o —1 0 0
Mf'”((') = 0 (1 —ar]’ 0 —1 0 0 d 71’ 0 —-1 0
0 0 1 0 0 1 0 0 1 0 0 1

— l 0 0
= 0 1 —‘2(11/' .
0 0 l

which is not a solution, because it does not agree with Theorem 4.3, and this concludes

' the proof.

The next theorem gives the linear conditions of the y-b block for the optimal

systems.

Theorem 4.5 For the four systems obtained from the last theorem, the constraints
on the y-b block of the linear map are the vanishing of either the diagonal elements

or the off-diagonal elements.

Proof.

.07

Let. L; = ( (J f ) be the y-b block of L.
g h
(l) FRFR:
From eq. (4.20). the total map of system FRFR, is
ATFRFR = exp(: 11(1) :)exp(: H(L"R—l - L1) :)
exp(: I[(11’L“R“ . L1) :)
exp(: [ML-'1?"-L-RL"R"-L1):)1.

From the second cell, we have

_ _ _ h —f l 0 e f _ eh+fg 2fh
1,13,1.L.1—(_g )(0 _,)(g 1)-( _269 _(mm

which gives the following possible solutions.

, . , _ c = 0
Solution A. {h : 0

and Solution B: { g 2

In the case of solution A, we have

L3‘R;'L.=(‘(‘, ?)

_ _ . l 0 -1 0 —1 0
R2L2‘R'21'L2:( 0 —1)(o 1)" o 41’

L313? -L2°32LiiRil'L2: ((1) 11)),

which is a solution

In the case of solution B, we have

L;‘R;‘-L1=((‘, ff)

_ _ 1 o 1 o 10
RzL2‘Rz"L2=(0—1)(0—1):(01)’

1 0
L;‘R;‘-L2.RzL;'R;'.L2=( _1)

),(1.s1)

98

 

 

 

 

 

 

 

Systems Linear Conditions

F R S C ((1)6)z0. (.rla)=(a|.r)=0
F R F R (alb)=0.(.rl.r)=(a|a)=0
FCS R (.rlo)=0. (.r|a)=(a|.r)=0
F C F C (.rlo)=0, (;r|.r)=(a|a)=0

 

 

Figure 1.]: Optimal four-cell systems and the first-order requirements to achieve their
optimum.

which is also a solution.

Because of midplane symmetry.
.11me = exp(: HF :)exp(: HC' :)exp(: HS :)exp(: HR :)L (4.82)

in both cases.

(2) FRSC, FCI‘TT, FCSR

0 1
FCFC, and FCSR. have the same matrix M; (i = 1.2.3) as that of FRFR. So, the

Since 5, = < I 0 ) we have S2 = F; and C} = 3;. This implies that FRSC,

requirements 011 L2 should also be the same. which concludes the proof.

All four optimal systems are listed in Table 4.1.

4.3 Order-by-Order Solutions

Next we will study the nonlinear conditions for higher-order achromats of the optimal

systems. Of the four systems. there are only two different maps, which are

MFR“, = ATFFFF = exp(: 11F:)exp(: 11R:)exp(: HS :)exp(: HC :)L (4.83)
and
MFCSR : MFR”? : exp(: HF:)exp(:11C:)exp(: 115:)exp(: HR:)L. (4.84)

99

The lemma below shows that. one can be obtained from the other through a simple
transformation. Thus, we need to study only one map to find out the conditions for

achromats.

Lemma 4.1 By switching A and B. the maps .*lTFR5(' and IITFFFF are trans/owned

to A’IFRSF and AlFm'R. respectively.

Proof:

Under the transformation. we have

HF = A+B+('+D —» B+A+C+D=HF,
HR = :t—B—("+D —> B—.4—€+D=HC.
115 = -A—B+("+D —. —B—A+C+D=HS,
HC' = —A+B—(‘+D —. -B+A—-C+D=HR.

which entails that the transformation of the maps is

11117115le FF _.) [111T.‘sR.FRFFlq

which concludes the proof.

The next theorem gives a general solution for arbitrary-order achromats. It is not

necessarily the best possible, yet it holds for an arbitrary order.

Theorem 4.6 For a given, order n, the optimal systems FRFR, FRSC. FCSR, amt

FCFC are achromats if A =7, 0 or B 2,, 0, and D =n+1 0.

Proof.

The mathematical induction method is used.

100

Let us first consider systems FRSC and FCFC. From the proof of Theorem 4.4,

they have the same maps, which are

1¢T:exp(:1‘IF:)exp(: 11R: )exp(: 115:)exp(: [I( :)L.

Define 11,, = 2 f,- and f,- = A, + B,- + C, + D), where f,- is the sum of the ith order
1:3
terms in II.

(1) The second order:

Using the B-("l-I'I formula. 111 can be transformed to the “standard” form, which

M =2 exp(2”f=)e><1>(:HfdcwrH§S:)exp(=11§':)L
:2 exp( 1.: )cxptzf ”=21ex1=t 1;. . ):exp( 15' :11“
:2 exp(: .15 +1...” +./.4."+f.§' 2)?
:2 exp(: 403 :)L.
Therefore the second-order solution is D :3 o.
(‘2) The third order:
Similar to the second-order case. 113 is simplified to
17 =3 exp(: Hf=><=xptz Hf=1exp<= Hf=1exp1= Hf :)f
=3 exp(sz+.ff=)exp(zflf+.ffz)
exp(: f: +11“ :)exp(: 15’ +1.” :1!“
=1 exp(: ff + ff + Li" + .11" :1 exp(: .1; =)
exp( f3 “)expt [9 )cxmz ff :)L
23 exp(: 404:) )exp(: 13 + 11+ +3l13Furiil 1)
exp(: .1: + 114 7115“. 15*] =1 1!“

=3 exp(.10,+:1tf{ Lf1+1a f§l+[f3 +1511: +f§D =1)

101

where

[1511.5] + {13“.13'1 + 1.12:" + 135*. 1'3" + .13 '1
z [.13 + I33 +1"3+ D3. .313 — I33 — C3 + 0:1]
+[-.'l,3 — 83 +13 + D3. —A3 + 3.1- ('3 + 0.1]
+ 11.43 + 03. —.43 + 03]
= 2133 + C3143 + 031+ 21—133 + C3, —A3 + 031+ 14/133031
: .1[B3, A3] + i[('3. D3] + 8M3. D3]

= ~l[B3. 1‘3].

Altogether. the map has the form

4

1” =3 (‘Xp(l lD4 + 2[B3. (13].)i:

which entails that A 2:1 0 or B :3 O. and D :4 0 are solutions for a third-order

achromat.
(3) The nth order:

Let us assume that. A =,,_1 0 or B =n_1 0. and 1) =71. 0 are solutions for an

( 72. — l)th-0rder achromat.
a) A 273-1 0., D Zn 0.

In this case, we have

n—l

[if—1 = :(Bi'l'Ci).

i=3

n-l

Hf... = 21—8.- — 1:1.

1:3

n—l

”2;. = Elli—CL

1:3

which entails that.

11,511+".-. =11;7_,+11,f'_,:11 and [11F 11: ,1 =[H;:‘_,,11,E'_ 1:0.

77—1

Therefore M can he transformed to

-o

17 =3 exp(:/1,531 1131121111. 1333.115... 211311111331 :11
=. 33,31 11.5.1111. +1...» 11311 I111. +15+ff+1 2)
exp(: 11:; l+ 1,. +1... :)exp(. 11(_,+1,E + 3.1:)!“
=. exp1z-1D..1:1exp1:H,f_.+1,f‘-11exp1 Hf_.+ff:1
3331. 115.. + .11." =1exp1 Hf... + 15 :11"
=. exp(:-10.31 :1oxp1: 1.” +15 1111115.. + 15.11.51, + 151+ ~-- -1
exv1=1;?'+.1,f'+ $1115... +13" HE.1 + 15] + - -- :11”
=. 12111110331:1vx1>1:.1f+1f+ $11H513151+11£Hi1131
33111-113123511111.1511:118 11.11111”
=3 1~x111:-40.+1=1vxp12ff+1f+§1113151+11513RI1z1
3.31:1: 3+1: 31 111-; 1,E1+[1;?.131
=3 33.31.110.313. +11.11.5111+[1:11.51+[f.;151+[15.1311:11"
=3 3331340,... + 511515] + 11:11-11 — 1131151+ [15.1511 =11“
+31115115111+1.1,S"+1;f.1§'11:11:

‘)

:n exp(:‘an4-l +[/'1n~ _B3_('3]+[_/1n 31:3)3‘C3l

=n exp(: ‘10,.“

:11 exp(: ~an+1 + 2[B;1. An] :)f,

which shows that An, = 0 and 0.1+] = 0 are a solution for the nth order.

In conclusion. 11:30 and D=n+1 0 are a solution for an nth- order achromat.

103
h) B 2,.-. 0. D 2,. 0.

[n this case. we have

n—l

11* = 21.1. + (".).

71—l

11!: = 21.1.- — (31.

11,11 = 21—11. +1'.1.

n—l

Hf...‘ = Zl‘x’h-(Ul.

i=3

which implies that

11,5_.+11,f‘_ l=11R_

711

Therefore N? can be transformed to

+Hn_ 1=0 and [Hi l,H 2:1] =[HR

117 =1 exp( 11:.+1.‘.“" + =1exp1 115_.+1f+1..1:

exp( 11:-.5'1+1 +1.?11r1em1 HE..1+1 +1.11 :11

Zn 9XP(140n+1 I)?Xp(. Hilly—l +115 2)

e><p( ff+.;ff ‘1'?le -511fl+lf”1H 51-1)] )(eXp=

=11 exp(lz4Dn+1 :)exp(:Hn_1-l-ff1)€‘xp(1115.1"f'fyi2 +ff +ff

W1” 1fl+lff1H1L 1l+lfR+f51H§1l11

=11 exp(=4Dn+1+ :HH 1.1.ff+ f$+f§l+lff~flcj 1]

+111. 1.1.1+1.1 15.11:- 11+11f+ 15,11 _ .111 :11”
=.. 91111141)... + 3111;? 11,? + 1:? + 151+ 11.51151
+1.1f.1;.“1+1ff.1.1‘]+1ff + 15.1.51) :11"
=. 1.1111141)... + $311.15 + 1313511111 :11"

:n 9Xp(: 40714-1 + 2[I3n./13l 1),-O

104

which shows that B 2,, 0 and D 2,,“ 0 are also a solution for an nthoorder achromat.

According to Lemma 1.1. the systems FRFR and FCSR have the same solutions.
which concludes the proof.
Considering the number of conditions, the two solutions obtained above are equiv—

alent. This is shown by the following theorem.

Theorem 4.7 For a given order n. the number of monomials in. A is the some as

that in B.

Proof:

A general term from H is Ciriaiyibgfi.ri‘a‘“y‘yb'b6‘6, where (iy + i5) is even.

For terms from A,,, i, + in is odd and in + ii, is even. Specifically, when ix is odd.
2}, is even, ii, is even, and i,, is even; when if is even, in is odd, ii, is odd, and iy is odd.
Therefore .4” can be decomposed to

An = Z Z Z Z: Citiagyibgél'i’aiayiybib6i6

i5 iy in if
even even even odd

+ Z Z Z Z C,,,-,,-y.-,,,.era‘ay'ybwré (I:r + i, + 2', + 2:, + 2;, = n ..
Ii}, (y in it
odd odd odd even
For terms from B”, i1. + id is odd and in + ib is odd. Specifically, when it is odd.
in is even, ib is odd, and iy is odd; when i1. is even, in is odd, ib is even, and iy is even.
Therefore 8,, can be decomposed to

B" = Z Z Z Z Cinaiyrwxi’a‘w‘”bibéit

ib iy in z,t
even even odd even

105

+ Z Z Z Z (iiixzaz'yiml‘i’dial/“butt” (i1? +1}; + iy + it + is =71)-
17;, 27,, I}, 27,.
odd odd even odd

For any given iy and it. E Z = Z Z . which concludes the proof.
I}, 2', z}, (I,
even odd odd even

Since D cannot be cancelled by symmetry, we always have to kill it to get an
achromat. This means that the number of nonlinear conditions cannot be smaller
than the number of terms in D. So the best solution under this theory will be
that the number of nonlinear conditions equals that of the terms in D. The next
theorem shows that a best solution exists for up to the fourth order. According
to computational results, this solution is also valid for the fifth— and sixth-order
achromats. Before studying the theorem itself, a few observations necessary to the

proof are shown below.

Considering the commutator of two general terms from the Hamiltonian, we have

V ,i t , t i i .t’ I" I" it i!
[Ciriaiyibia-l ’0 “y ”b ”5 6» Cigigi’yigi’.“ ”1 “y ”b 66 “l
. . . . ' 'I _ ' 'I _ - “I - -r ' 'I
: (fixiaiytbtacizigigigi;((114; _ lal;)x1x+¢, laza-l-za ly1y+1ybab+zb615+16

+(I'ylig __ ibi;)Iix‘i-iltain-i-iayiy—iiL—lbib+i£_16i6+i3)
(1) If the two terms are from the same part of the Hamiltonian, we have

(i,+i;—1)+(ia+i;—1):(ir+ia)+(i;+i;)-2=even,
(2t.+zf;— 1)+(zfi.+z';,)= (i.+zfi.)+(z:;+z';,)—1=odd,
which entails that the commutator gives terms in C.
(2) If the first term is from A, i.e., tr + in is odd and in +2}, is even, and the second

term is from B, i.e. i; + if, is odd and if, + if, is odd, we have

(i,+i;—l)+(ia+i;—l):(ir+ia)+(i:.+i;)—2:even,

106
(2'. +11, — 1) +0. +17.) = (z. + m + (2". + it) - l = mm

which entails that [1. B] gives terms in D.

(3) If the first term is from A. i.e.. tr +1}, is odd and 17,, + I}, is even, and the second

term is from D. i.e.. i2. + if, is even and i; + if, is even. we have
~(n+2;—1>+(zf.+zf:,-1)=(i.+zfi.)+(i;+i;>-2=odd.
(27,. +13, — 1)+(z,+zf;,) = (2",, +i.)+(zfi;+zfig)— 1 =odd.
which entails that [.4. D] gives terms in B.
(4) If the first term is from B. i.e.. if +176 is odd and in +1), is odd, and the second
term is from D. i.e., i; + if, is even and i; + if, is even. we have
(i.+i;— l)+(i,+i; — 1) =(i,+ia)+(i;+i;)-2 =odd.

(z?.+zif,—1)+(ib+z':,)=(i.+ib)+(i;+i;)—1=even.

which entails that [B, D] gives terms in A.

(5) If the first term is from A. i.e.. i1. + in is odd and in + ib is even. and the second
term is from C, i.e.. i; + if, is even and z; + i’b is odd, we have
(i.+z?_',.— 1)+(i.+i’, — 1) =(2'.+i.)+(i;+ifi.)—2=odd.
(in +i; — 1) +(ib+i;) : (id-+45) +(i;+if,) — 1 =even.
which entails that [A, C] gives terms in A.

(6) If the first term is from B, i.e., ir +1}, is odd and ta + ib is odd, and the second

term is from C, i.e.. z"I + if, is even and i; +1}, is odd, we have

(i.+i;—1)+(i.+i;—1)=(i.+i.)+(i;+i;)-2=odd,

<i.+i;—1>+(i.+iz,)=<i.+zfi.)+<if.+i;>—1=odd.

which entails that [B, C] gives terms in B.

107
(7) If the first term is from D. i.e.. i1. + i,I is even and in + ib is even, and the
second term is from C'. i.e.. i_', + i; is even and i; + i; is odd. we have
(i_,. + i: — l) + (i,l +1.; — l) = (if +ia) +(i;+i;) — ‘2 = even,
(in, + if, — l) + (it +11) : (in +14.) + (i;+i§,) — l 2 even.

which entails that [D. ('l gives terms in D.

Theorem 4.8 For the optimal systems. achromats up to the fourth order can. be

obtained by cancelling I) in. the total map.

Proof:
(1) Systems FRSC. and FCFC:
a) The second order:
From the proof of Theorem 4.6, the map is
AT :2 exp(: Hf :)exp(: Hf :)exp(: H: :)exp(: H30 :)f
=2 exp(: 40:. :)I‘.
which shows that the second-order solution is D3 = O.
b) The third order:
Also from the proof of Theorem 4.6. the map is
AT :3 exp(: Hf :)exp(: Hf :)exp(: Hf :)exp(: Hf :)f
:3 exp(:/1D4 + 2[83. A3] :)ii
Since [83, Ag] belongs to D... a third-order achromat can be achieved by zeroing

4D4 + 2[B,3, A3] instead of cancelling D4 and A4 (or B4) separately. Therefore, the

best third—order solution is

00
Ci!
V

1
D4 = —;[B3,A3]. (4'

108

c) The fourth order:

Using the B-C-H formula. 117 can be transformed to

.17 =. exp(: It,” :)exp(: 11.51; )(xp( 11;: :)exp(: 11;" :)1"
=4 exp(: 1..” +1.” +1..” :)exp(: 13’" +1.” +1.." =1
exp(: 1;? +1.? +19," :) exp(: ff +1.? +15 :17
=4 MPH—105:)exp(:.li-f+f.’:)exm f, + ff :)
vxpt 1.; +1.. . 1mm: (1241:" :)f
z, (‘x1)(:1Dg:)eX[)(if3 +1, +f3 +f..+ :(1fzf..fl+[ff~ 4R1
+11f1f11+1fit1f11f 1*11+tff 11.. 1.111 1:)
exp(: 1.5“ +1.? +11; +1.? + 5(1159151 + [tiff]
+[1f.1.§'11+fi(11§',1f;;‘.1§'11+115.115.1511):)f
:4 exp(: 10.:) )exp(. ff + f, +ff' +f§'
+£11.151151 + 1.111.151 + 1151?] + 112.". 1.1 '1 + 1.1.5.1."1 + 115.151)
+fi<t11311111111 + 11.1. [1.1.1.111 + 11;". [1.3.1511 + [.153 115.1511)
+5111+11+ 1.1 + 11 + 151111.111+111.111H1111».
15+1f+1§+1f+fittff..f§‘1+tf§~ {H113 £111
+fi<11f+1f11§ +111:+1511+11§+1$11§+1$1§ +1511) =1!”
:4 exp(: 405 :)exp(: 41). + 2w... A3]
+fi<11f1fl+115151+11:.1f1+11:‘.1£1
+11.” + 1.5.1.? + 111+11f+ 1.5.1:." + 151)
+£11.15 «11.111. 111 + 111.11. 11.11 + 111)
+fi<11f. 11.11111 + 111.111.1111 + 115.115.1511 + 111.115.1311
+115 + 151.114.15.15 + 1311 +115? +1511? +151! + 13311) :11"

109

exp(: 10;, :)exp(: H :)f

:1

where
- 1 .. . ..
11:;mtm+m1nnn+wt+u1
+115. 1*] + 1.113.111 + [ff + 15.11;“ +1.11)

‘ 1.+11* 1.1.1 11*11+111F.11*1 11 +1111

117”? [f‘~fW1*11+1*111.H111 111.111..11'11+11111131111

mfl+htt+fih+fiflflfi+fibfiwflfi+fim

1
_. 3([143 + 83 «1—(‘75q A4 - B4 — C4 ‘1' D4]

+[—A3 -' B3 '1' (‘3. —/44 '1' B4 — C4 “‘1' D4] + ‘1[A39 _A4 '1' D4]

+[f44 '1' 84 +01 '1' D41 A3 — B3 — C3]

+1-A4 — 13.. + c". + D... —.4_. + 13.1 — 011+ 11A. + 04. —A3])

1
112.43. {—11.1 — 131+ ('1.-111 + Ba - C311

+1211... [113 + 83 + €3,113 — Ba - C311)
1 w q . Q
+fi<111 — 11.1.11.11*11+111 - 11.11.11.111)

—[.43 B. + (1.] + [83 + €3.11. + 0.] + [113. -—B.1 + C4]

+[—133+C3 —A4+D4]+1]43104l

_([‘2A3‘ [—A3 — 83 + C3, —A3 '1' B3 -' C3“

+[2.43, [A3 ‘1' B3 '1' C3» ‘43 _ B3 - C311)

1
(12133 '1‘ (73111143 '1' Ba '1' C31 A3 — B3 - C3”

E5
+[2(—B.1 + C3111‘A3 — 33 + (731-113 + 33 ‘ C3“)

= ~lB4431+ 3133 A4] + 2[C31D4l + 41/431041

+[.43. {—A3, B3 — C3” — [44341431 BS ‘1' C3“
"[33 + C3. [.43. B3 '1' (1'3” ‘1" [-83 '1” C3~ [“1439 B3 _ C3”

110

V‘ 1 1
= 2[B4. .43] + 2[B3. 44] — 2K 3. —;[B3. .43“ + -'l[.43. —§[Bg. .43“
2
5t [(3 [4% B3” + [Bib [.43. (bill
‘)

. 1 ..
: 2[B4, .43] + 1413:}. 4441+ §[('3. [4439 83]] + €[B39 [C39 f13ll'

—2[.43. [44% Bill _

Altogether, the map is

A7 :4 exp(: ng :)exp(: [I :)f
l 2 . ..
:4 (‘Xp(2 405 + 2484, 13] + 2[83. A4] + §[('3. [.43. 8311+ §[Bg. [C3, .43” .)1.

SlllCG‘ ..).[[34. .43] + 2[Bf;. 41]+%[(43.[.43.I33]]+%[83.[Cl3. 143]] bGlOIIgS tO 05, the l)(‘h‘l

solution for the fourth order is
1 2
Dr, = —(2[B4. A3] ‘l” 2[B3. .44] 'l' §[("3. [A3, B3” ‘l' ;[B3. [C3, 14.3”) (4.86)

(2) Systems FCSR and FRFR:

The best solution for a fourth-order achromat can be obtained by switching A and

B, which is
D13 2 O, (4.87)
1
D4 = — .2[/43. B3]. (4.88)
l 2
05 : —(2[‘449 331+ 2[443~ 83] + :i[(w3~ [839 143]] + €[4439 [C39 831]), (4'8()l

which concludes the proof.

Computer results show that this property holds for the fifth and sixth orders also.

As the conclusion for this chapter. the following conjecture is presented:

Conjecture 4.1 For the optimal systems. achromats up to an arbitrary order can. be

obtained by cancelling D in. the total map.

Chapter 5

Applications

When a beam optical system is actually designed, one of the design codes has to be
used (see Section 1.1). In the case of high-order achromats, DA techniques are needed,
because when the order is as high as 5, the map and Lie factorization can be computed
efficiently only through DA techniques. The proofs in Theorems 2.5 and 2.6 actually
provide the DA algorithms for computing various Lie factorizations. Since the Lie
factorizations are always obtained from the map, the B-C-H formula is not necessary.
which greatly simplifies the complexity of the processes. Due to the fact that no
second-order polynomials appear in the Lie exponents. this algorithm works for an
arbitrary order as well. Throughout the design processes, COSY INFINITY is used.
Which contains all the tools important to beam optical design. Therefore, everything
Can be done in one shot, including map computation, Lie coefficients extraction.
fitting, tracking, and resolution calculation. First, the DA map of the desired order
is computed. Second, relevant Lie coefficients are extracted from the Lie exponent
Obtained from the map. Third, fitting routines are used to cancel the Lie coefficients
Which cannot be cancelled by symmetry. In our case, sometimes the package LMDIF

from Argonne National Laboratory was used, other times a matrix inverter was used.

In the next sections, several designs of third-, fourth-, and fifth-order achromats

are presented. All of them are direct results of the arbitrary-order achromat theory

111

1112

presented in Chapter .1; and the circular systems are designed in such a way that
Brown’s second-order achromat theory can be applied, where two second-order knobs
are saved. One of them. a circular system (see Section 5.2), is not only a third-
order achromat. but is also energy-isochronous to the third order, which is an ideal
multi-pass time-of—flight mass spectrometer. Another design, a single—pass third-order
achromat (see Section 5.1). presents a compact system which can be transformed to
a. single-pass time-of-flight mass spectrometer. Two more designs push the orders to

4 and 5, which further verified the arbitrary-order achromat theory.

5.1 A Third-Order Achromat - FRSC
5.1.1 The First-Order Layout

Our design is aimed at a system for 200 MeV protons [Wan93a]. The first-order for-
ward cell contains a l‘ZO-degree inhomogeneous bending magnet and four quadrupoles
( Fig. 5.1). The compact layout shows the potential application of high-order achro-
Inats for single-pass time-of-flight spectrometers. The forward cell is symmetric
around the midpoint, which entails that (:rlr) = (ala) and (y|y) = (blb). So we
need to fit only 5 conditions, instead of 7, in order to obtain a map of the form shown
in Table 5.1. During the process of fitting, the two drifts between quadrupoles and
dipole are fixed; the variables are the field index of the dipole, the two field strengths
of the quadrupoles, and the drifts before and between the quadrupoles. Table 5.2
Shows the parameters chosen as our starting conditions for higher order optimization.

The long drifts are spaces where higher-order elements will be placed.

l13

 

 

 

 

 

 

 

 

 

 

.0:
F!
—U

Q

 

Figure 5.1: The FRSC third-order achromat: The first-order forward cell. Only the
bending magnet and the quadrupoles are shown. The phase advances are p, = 1r and
[1,, = 7r/‘2. It also shows that at the end of the cell, the dispersion is not corrected.
but dispersive rays are parallel to the on-energy rays that start with the same initial
conditions.

5.1.2 The Second- and Third-Order Achromat

To make this system a second-order achromat, ten sextupoles were inserted symmet-
rically with respect to the dipole (but not symmetrically excited). The sextupoles
provide us the knobs to correct second-order aberrations. The required values are

rather weak (Table 5.4), which indicates that the first-order layout gives weak higher-

“1.000000 0.00000008+00 0.00000003+00 0.00000008+00 0.24833078-15 100000
0.00000008+00 -1.000000 0.00000008+00 0.00000003+00 -7.386987 010000
0.00000005+00 0.00000003+00 0.1387779E-15 1.000000 0.00000003+00 001000
0.00000008+00 0.00000008+00 -1.000000 0.15265573-15 0.00000008+00 000100
0.00000008+00 0.00000008+00 0.00000003+00 0.00000003+00 1.000000 000010
7.386987 -0.2775558B-15 0.00000008+00 0.00000003+00 1.733828 000001

Table 5.1: The FRSC third-order achromat: The COSY INFINITY output of the
first-order map of the forward cell. The first five columns are Taylor coefficients of
If, a], y], bf, and if as functions of ff, (1;, y;, b,, t;, and 6;, and the sixth column
contains the powers associated with the coefficients on the same row. From left to
right, the sub-columns stand for 93,-, a,, y,, b,, t.-, and 6,, respectively.

114

 

 

Element Field/ Length Gradient

Drift 1 0.871354(m)

Quadrupole 1 1.86498(kG) 0.372997(kG/cm)
Drift 2 0.624612(m)

Quadrupole 2 —2.67177(kC) -0.534353(kG/cm)
Drift 3 2.0(m)

Dipole 8.6(kG)

Field index(nl) 0.398455

 

 

 

 

 

Table 5.2: The FRSC third-order achromat: The field strengths and drift lengths
of the first-order layout. nl is the first-order derivative of the field of the bending

magnet over the bending radius r (dimensionless).

order aberrations, and the newly introduced sextupoles will not produce strong third-
order aberrations either. The map of the four-cell system was then computed. As
the theory predicted, it is free of all second-order aberrations. Note that this sys-
tem was designed before the analytical theory was developed. The requirement of

ten sextupoles was obtained from a computer theorem-proving program discussed in

references [Wan92, Wan93a].

 

Figure 5.2: The F RSC third-order achromat: The third-order x-z beam envelope
where only the bending magnets are shown. The mirror symmetry between cells F
and R and that between cells S and C is clearly shown in the beam trajectories.

115

Figure 5.3: The FRSC third-order achromat: The third-order lab layout. The “S"-

shaped geometry helps make it a compact system.

The last step is to correct third-order aberrations. All third-order elements are
superimposed in the existing dipoles, quadrupoles, and sextupoles because this con—
figuration tends to require weaker octupole fields. Specifically, l4 octupoles are placed
inside other multipoles and an octupole component is added to the inhomogeneous
dipole field through the curvature of the pole faces. So there are 15 variables for 15
conditions. The results show that very weak octupoles can meet all the conditions
(Table 5.4). Figures 5.2 and 5.3 show the beam envelope and the lab coordinate
layout of the whole system. Regarding the area it occupies, this is a rather compact
system. Table 5.3 presents the third-order map, which shows that the only terms left
that are nonzero are the dependents of time-of—fiight on energy spread up to the third

order. This makes the system an effective time-of—flight spectrograph.

Using an emittance of 1 mm mrad and a momentum spread of 1%, Figure 5.4 shows
the eighth-order beam around the final focal point. The sum of the aberrations at

the focal point is about 10 pm horizontally and 3 pm vertically, which is quite small.

116

1.000000 0.00000008+00 0.00000008+00 0.00000008+00 0.00000003+00 100000
0.00000005+00 1.000000 0.00000003+00 0.00000003+00 0.00000008+00 010000
0.00000003+00 0.00000008+00 1.000000 0.00000008+00 0.00000008+00 001000
0.00000008+00 0.00000003+00 0.00000005+00 1.000000 0.00000003+00 000100
0.00000008+00 0.00000003+00 0.00000003+00 0.00000005+00 1.000000 000010
0.00000003+00 0.00000008+00 0.00000005+00 0.00000003+00 6.935312 000001
0.00000003+00 0.00000008+00 0.00000008+00 0.00000008+00 -21.18904 000002
0.00000008+00 0.00000008+00 0.00000003+00 0.0000000E+OO 59.36542 000003

 

Table 5.3: The FRSC third-order achromat: The COSY output of the third-order
map. A third-order achromat is reached and only (t|6") (n = 1,2,3) is not cancelled.
(Any number smaller than lE-ll is set to zero.)

5.2 A Third-Order Achromat - FRFR

A third-order achromat based on the first—order layout of the Experimental Storage
Ring (ESR) at Germany is designed to convert it into a time-of—flight mass spec-
trometer without using electron cooling, so that the masses of short-lived radioactive

nuclei can be directly measured.

The ESR ring contains six dipoles, twenty quadrupoles, and eight sextupoles, as
well as RF cavities, beam cooling devices, and the injection-extraction system. Two
long, straight sections divide it into two identical parts, each of which is symmetric
about its center. (Figure 5.5) [FranzST]. It is much easier to take half rather than
a quarter of the ring as the forward cell. Consequently, the other half should be the

reversed cell, and an achromat corresponds to two turns of the ring.

Since this is a FRFR system, the first-order matrix of the forward cell has to
satisfy (:rlx) :2 (ala) = (115) = 0 (Table 4.1). In this design, the quadrupoles are
excited symmetrically, which ensures that (III) = (ala) and (yly) = (blb) and reduces
the number of first-order knobs from 5 to 3. In order to make a system a mass
spectrometer, (tlri) has be cancelled, too, which adds one more condition and hence

the total number of conditions is 4. After the fine tuning of the free knob (the third

 

Strengths of the Multi]

:)oles (Aperture 10 cm)

 

Sextupoles

Octupoles

 

 

~0.218686E-02
-0.484701E-03
0.304526 E-02
0.15272lE-02
0.147310E-02
-0.270373E~02
-0.644784E-03
0.187468E-02

~0.621716E-01
-0.l21175E-01
0.761314E-01
0.381804E-01
0.368275E-01
—0.675933E-01
-0.161196E-01
0.468670E-01

-0.359389E-03
0.180765E-03
-0.506326E-04
0.854076E-04
-0.794297E-04
-O.592687E-04
0.615846E-04
-0.61 380513-04
0.320186E-03
-0.886455E-03
0.287362E-02
-0.177754E-02

Gradient (kG/cmz) Field (kG) Gradient (kG/cma) Field (kG)
-0.254023E-03 -0.635057E-02 -0.673151E-03 -0.84l438E-01
0.14342413-02 0.358559E-01 0.11024113-02 0.137801

-0.449236E-01
0.225956E-01
-0.632907E-02
0.106759E-01
-0.992872E-02
-0.740859E~02
0.769808 E-02
-0.767256E-02
0.400233E-01
-O.110807
0.359203
-0.222193

 

 

 

 

 

-1.12893(n3)

 

Table 5.4: The F RSC third-order achromat: Field strengths of the sextupoles and
the octupoles. The field index, 713, is the third-order derivative of the field of the
bending magnet over the bending radius (dimensionless).

quadrupole), the best-behaved first-order solution was found and the horizontal beam
envelope is shown in Figure 5.6. The strengths of the quadrupoles are displayed in

Table 5.5.

According to the analytical theory, four independent sextupoles are required to
obtain a second-order achromat. However, because of the fact that for the first order,
cell R is identical to cell F, a simplification is possible based on Brown’s theory of
second-order achromats[Brown79, Carey81]. As shown in Section 3.2, a second-order
achromat can be achieved by placing two families of sextupoles in the dispersive region
of every cell and correcting one second-order chromatic aberration in each transverse
plane. This is the second reason why the forward cell is made symmetric itself and

(yly) is cancelled instead of (ylb) and (1713/). Even though in principle a second-order

118

 

Figure 5.4: The FRSC third-order achromat: Remaining aberrations up to the eighth
order (scale: 30 pm x 20 pm). The vertical line represents the final focal plane, where
the deviations are around 10 pm.

achromat can be achieved with two sextupoles per cell (half ring), the constraint that
the second half be the reverse of the first requires that the sextupoles be split into
symmetrically excited pairs. Again, one more pair is inserted to kill the term (tlé’)
(Table 5.6). Sixteen octupoles are introduced to correct all the remaining third-order
aberrations, including (1163). The positions of some of the multipoles are carefully
chosen to minimize the required field strengths, which results in a realistic setting

(Table 5.8).

Since our goal is to make ESR a multi-pass time-of—flight mass spectrometer, its
dynamic aperture has to be studied in detail. The dynamic aperture was obtained
by tracking the system of a number of turns using its 8th-order one turn map. For
particles of a momentum spread of i0.25% to survive 100 turns, the horizontal and

vertical apertures are roughly 1 7r mm mrad.

The mass resolution was determined in a statistical way: First, a large number
of particles (1000) with a certain mass deviation inside a certain phase space area

are sent through the 8th-order one-turn map n times and the n-turn time-of-flight

 

Figure 5.5: The F RF R third-order achromat: The original layout of the Experimental
Storage Ring (ESR) at Darmstadt, Germany

of each particle is computed. Second, the random errors of the detector are taken
into account, which in this design is chosen as 100 ps maximum, and the predicted
mass deviation of each particle is calculated. Finally, the difference between the
predicted and initial mass deviation of each particle is obtained and the n-turn mass
resolution is determined by calculating the inverse of the average differences. The

mass resolution's dependence on the number of turns is presented in Figure 5.8 .

5.3 A Fourth-Order Achromat - FRFR

A fourth-order achromat was also designed using the best solution obtained in Section
4.3. It is a storage ring with mirror symmetry. Since cells S and C are not allowed,
the only choice is FRFR. The first half of the ring forms the forward cell (F) and
the second half is the reversed cell (R). So, like the last example, the achromat is
achieved after two turns. In the process of the first-order design, the four elements
of (:rlzr), (x16), (ylb) and (bly) were fitted to zero. Since the first-order layout of each

cell is symmetric about its own center, (ala) equals (:rlzr), which implies that (ala)

/ \
4 ’I ’ ‘7 ~
\g/ \t\~/

 

Figure 5.6: The F RF R third-order achromat: The first-order beam envelope of the
horizontal (.r-z) plane. The curve that does not coincide with the z-axis in the straight
section is the [i-function. The other curve is the dispersive ray. The circumference is
108.36 m; the emittance is 12.57r mm mrad; and the dispersion is 0.7%.

is also cancelled. Ten quadrupoles are used in each cell, which give five free knobs
(Figure 5.9). Four of them are used for cancelling the four above terms with one left

for optimizing the first-order solution.

There are four sextupoles in each cell to correct second-order aberrations, all of
which have to be placed inside the dispersive region because all second-order aber-
rations left are chromatic ones. This is also true for the decapoles, but not for the
octupoles. By placing the fifteen octupoles and fifteen decapoles more or less equidis-

tant, a fourth-order achromat requiring very weak multipoles is achieved (Tables 5.9

and 5.10).

To study the influence of the remaining aberrations, the final spots of beams
starting from a square lattice are plotted and shown in Fig. 5.10. It clearly shows
that the achromat is reached at the end of the second turn, because up to the fourth
order there are still significant distortions after one turn. It also shows that the

higher-order remaining aberrations grow exponentially with the emittance.

 

 

 

 

Strengths of the Multipoles (Aperture 10 cm)
Quadrupoles Sextupoles
Gradient (kG/cm) Field (kG) Gradient (kG/cm2) Field (kG)

0.624520 -3. 12260 -0.208441E-01 -0.521102
0.742699 3.71350 0.106122E-01 0.265306
0.212000 -1.06000 -0.260702E-02 —0.65l754E-01
0.389363 1.94 682

-0.323636 —1.61818

 

 

 

 

 

 

Table 5.5: The FR FR third-order achromat: The field strengths of the quadrupoles
and the sextupoles. Since they are excited symmetrically, only half of the multipoles
are listed.

0.45619143-14 0.2306961 0.00000008+00 0.00000008+00 0.00000008+00 1000000
-4.334708 0.41633368-14 0.00000008+00 0.00000003+00 0.00000008+00 0100000
0.00000008+00 0.00000005+00-0.14432908-14 0.5119279 0.00000003+00 0010000
0.00000008+00 0.00000008+00 -1.953400 -0.181799OE-14 0.00000003+00 0001000
0.00000003+00 0.00000008+00 0.00000003+00 0.00000003+00 1.000000 0000100
0.00000008+00 0.00000008+00 0.00000002+00 0.00000008+00 0.16792128-14 0000010
0.00000005+00 0.00000008+00 0.00000005+00 0.00000005+00 -4.187160 0000001

 

Table 5.6: The FRFR third—order achromat: The first-order map of half the ring.
(The zeroes are numbers smaller than 113-15) The phase advances are ya, = py = 7r/2.
Note that (116) also vanishes

5.4 A Fifth-Order Achromat - FRFR

- As a proof of principle, a circular fifth-order achromat is designed, which proves
Theorem 4.8 and adds a case supporting Conjecture 4.1. The first-order layout should
avoid large changes in the beta functions in order to minimize nonlinear aberrations;
furthermore, there should be enough room for the insertion of correction multipoles.
Another consideration is that, if possible, the number of first-order conditions should

be further reduced through symmetry arrangements inside a cell.

The result of these considerations is a ring shown in Figure 5.11, which consists of

sixteen FODO cells plus two dispersion correction sections, each of which includes two

1.000000 0.00000008+00 0.00000008+00 0.00000003+00 0.00000003+00 1000000
0.00000008+00 1.000000 0.00000008+00 0.00000008+00 0.00000008+00 0100000
0.00000008+00 0.00000008+00 1.000000 0.00000003+00 0.00000003+00 0010000
0.00000003+00 0.00000003+00 0.00000008+00 1.000000 0.00000003+00 0001000
0.00000003+00 0.00000003+00 0.00000003+00 0.00000003+00 1.000000 0000100
0.00000003+00 0.00000005+00 0.00000008+00 0.00000008+00 -16.74864 0000001
0.00000008+00 0.00000008+00 0.00000008+00 0.00000005+00 22.80066 0000011
0.00000003+00 0.00000008+00 0.0000000E+00 0.00000005+00-0.2394999 0000002
0.00000003+00 0.00000008+00 0.00000008+00 0.00000003+00 ~23.27966 0000021
0.00000003+00 0.00000008+00 0.00000008+00 0.00000003+00 -3.467360 0000012
0.00000003+00 0.00000005+00 0.00000008+00 0.00000008+00 0.4997533 0000003

 

Table 5.7: The FRF R third-order achromat: The third-order two-turn map. (The
zeroes are numbers smaller than 1E-8.) Note that time-of-flight terms, which depend
on energy and mass, cannot be cancelled along with (t|6") due to the fact that the
magnetic field only distinguishes magnetic rigidity, which is a function of 6,, and 6m.

quadrupoles. The left half is the forward cell (F) and the right half is the reversed cell
(R). Achromaticity is achieved after two turns. The forward cell itself consists of two
parts, one of which is the reverse of the other. This guarantees that (:rlx) = (ala) and
(gly) = (blb). All four FODO cells within one part of a cell are identical except that
the last one has an extra quadrupole for dispersion correction. Hence there are three
knobs for the first-order design which can zero out (1311:), (a|a), (yly), (blb), (:rl6), and
(a|6) at the same time. Figure 5.11 shows that the beam travels around the ring in a
very uniform manner, avoiding large ray excursions and beta functions. As described
in the last example, a second-order achromat is achieved by symmetrically placing

and exciting two pairs of sextupoles in each half.

After the investment in a careful first- and second-order layout, the third-, fourth-.
and fifth-order corrections actually turn out to be conceptually straightforward, even
though they are computationally more demanding. In the whole process of nonlinear
optimization, only two aspects seem to be worth considering. First, the required
multipole strengths strongly depend on the average distance between multiples of

the same order. In order to keep their strength limited, it is important to have the

123

 

Strengths of the Octupoles (Aperture 10 cm)

 

Gradient (kG/Cm31

Field (kG)

Gradient (kG/cm3)

Field (kG)

 

-0.128089E-02
0.180353E-02
0.275862 E-02

-0.429976E—02
0.344373 E-02
0.153106E—03

-0.263437E-02
0.904441E-02

 

 

-0.160111
0.225441
0.344828

-0.5374 70
0.430466
0.191382E-01

—0.329296
1.13055

4;

 

-0.882350E-02
0.581219E-03
0.394693 E-02
0.706907E-02
0.453311E-02

-0.114925E-01
0.308289E—02
0.148141E-01

 

-1. 10294
0.726523 E-Ol
0.493366
0.883634
0.566639

-1.43656
0.385361
1.85176

 

Table 5.8: The FRFR third-order achromat: The field strengths of the octupoles.

 

Strengths of the Multipoles (Aperture 10cm)

 

Quadrupoles

Sextupoles

 

Gradient (kG/cm)

Field (kG)

Gradient (kG/cmz)

Field (kG)

 

0.277788
-0.273114

0.881770E-01
—0.282065E—01
~0.361-197E-01

1.38894
-1.36557

0.440885
-0. 141032
0. 180748

-0.114852E—02
-0. 184780E-03

0.101651E-02
-0.941644E-03

-0.287130E-01
-0.461950E-02

0.254127E-01
-O.235411E-01

 

 

 

 

 

 

 

Table 5.9: The FRFR fourth—order achromat: The field strengths of the quads and
the sextupoles. Only half of them are shown due to mirror symmetry.

dimension of the total size of the ring and the dispersive region sufficiently large, and
distribute roughly uniformly multipoles of the same order. Second, all the decapoles
have to be placed in regions with sufficient dispersion, because all the fourth-order
aberrations that remain after third-order corrections are chromatic aberrations. The
combination of these considerations results in weak multipole strengths for third-.

fourth-, and fifth-order corrections. (Tables 5.12 and 5.13).

The 1000-turn dynamic aperture for both horizontal and vertical motions are
studied using the 11th-order map. For particles of a momentum spread of $0.570 to

survive 1000 turns, the dynamic aperture is at least 100 7r mm mrad both horizontally

12/1

 

 

 

 

 

 

 

 

Strengths of the Multipoles (Aperture 10cm)
Octupoles Decapoles
Gradient (kG/cm3) Field (kG) Gradient (kG/cm“) Field (kG)
0.865187E-04 0.108148E-01 -0.184214E-03 -0.115134
-0.730041E—04 -0.912551E—02 0.993688E-04 0.621055E-01
0.118558E-03 0.148198E-01 0.267539E-03 0.167212
-0.101019E-03 -0.126274E-01 0.353274E—03 0.220796
0.474184E-04 0.592730E-02 -0.577517E-03 -0.360948
-0.663923E-05 -0.829904E-03 -0.153650E-03 -0.9603l4E-01
-0.132363E-04 -0. 165454 E-02 0.764958E-03 0.478099
0.466092E-04 0.582615E-02 -0.433794 E—03 -0.271121
-0.849342E-04 -0.106168E-01 0.405756E-03 0.253597
-0.515629E-05 -0.644536E-03 -0.843814E-03 -0.527384
0.280953E—04 0.351 191 E-02 0.538596 E-03 0.336623
0.125900E-03 0.157375E-01 0.369327E-03 0.230830
-0.423299E-03 -0.529124E-01 -0.128477E-03 -0.802978E-01
0.245393E-03 0.306741E-01 -0.177871E-03 -0.111169
0.796716E-04 0.995896E-02 0.207354E-03 0.129596

 

 

Table 5.10: The FRFR fourth-order achromat: The field strengths of the octupoles
and the decapoles. Here all the multipoles are very weak.

and vertically. It is much larger than the acceptance of the ring, which is about 30 ~
40 7r mm mrad. As an example, Figure 5.12 shows the horizontal motion of on-energy
particles up to 1000 turns. The absence of linear effects, as well as any aberration
up to order five, leads to a behavior that is entirely determined by nonlinearities of

order six and higher.

The time-of-flight energy resolution of this ring is determined in a statistical man-
ner similar to the example discussed in Section 5.2 except that the one-turn map
is of the ninth-order. The dependence of the resolution on the number of turns is

presented in Figure 5.13 .

 

 

 

 

Strengths of the Multipoles (Aperture 10 cm)
Quadrupoles Sextupoles
Gradient (kG/cm) Field (kG) Gradient (kG/cmz) Field (kG)
-0.162869 -0.814344 -0.718659E-03 -0.179665E-01
0.134119 0.670597 0.364420E—03 0.91 105013-02
-0.131803 0.659013

 

 

 

 

 

 

Table 5.11: The FRFR fifth-order achromat: The field strengths of the quads and

the sextupoles. Only half of them are shown due to mirror symmetry.

2 mrad l2 mrad
. 1 cm 1 cm
— L 3'1."- L
I2 mrad '2 mrad
l- 1 cm -' -' '. 1 cm

 

 

 

I2 mrad I2 mrad
_ a; 1 cm _ . 9" 1 cm

Figure 5.7: The FRF R third-order achromat: The 8th-order 1000-turn tracking pic-
ture. The left and right columns display those of :r- and y-motion, respectively; the
top, middle, and bottom rows show those of 6 = —0.1%, 0, and 0.1%, respectively.

 

40000 —

30000

1 1 1 1 1 1 Y

20000 r—

10000 *-

 

 

 

-(
O 111111L1411111LJJJllLLLJl

0 10 20 30 4O 50

 

Figure 5.8: The FRFR third-order achromat: The multi-turn mass resolution as a
function of the number of turns. Due to the small emittance, the resolution increases
almost linearly with the number of turns.

 

Figure 5.9: The FRFR fourth-order achromat: The layout, beam envelope and dis-
persive ray. The phase advances are yr = 11,, = 7r/2. The circumference is 147.35 m;
the emittance is 207r mm mrad; and the dispersion is 0.6%.

2 s
8 I:I I III I IT I VI I] I I1 q H .- II I l I I I I l I I I I I I I I q
>- L ' ° \ ; _ >— _ . t ‘ .
I. _I L? _I
1: ‘ r - ' . : 2: ‘ - I it:
.. .7 1— . .-
0 L. . . . o——.: 1 turn 0 L; c o . .)—-
t 2 4th order I 9' j
; . 0 . 0 . : ~ : . O ‘ f.:,
'1 .— —. '2 n L.
I o 0 O ' 7 d : ’ ' I
1.1 Ill 1 I 111 I I II I I 1 1 1 I 1 14 I I I I I 11111
'5 E
; FI I I l I I I I I I I I I I I I.I . ; .- I.I I I I I I I I I I I.I l I I. -
- q 1— u
1_ .4 2 _ __
'- o "' l- o "1
O L .._- 2 turns 0 L. ...:
: 2 4th order : :
‘1:— —_ '2 :— ‘5
.- 11 11111 I l 14 1111 IJ - I. I I 1511 I I l I I I J l 1 I 1 q
E E
8 I I III I I I I I I I I l I I I 3 I I I I III I l I I I I l I I I L.
>_ .- g o o O I :1 )- _ i . . . ‘ :i
1.— -: 2 .— —:
'- 0 o o o O u '- ! O I Q ‘ .1
O :——o o o o o——: 2 turns 0 L‘ o o . '—:
I 2 5th order I 2
: o o o o I : : ! O o 0 g j
‘1.— —. '2 .— ‘1
: - - - : : 1 - - - I *
11111 1 I 1 l I I 1 II I II 1 1 111 11 I '11 1 I] I 1 I

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

'1

0

1

X (cm)

-2

0

2 X (cm)

AE/E (X): 0.6: Aa. Ab (mrad): 2.0 AE/E (7.): 0.6: Aa. Ab (mrad): 1.0

Figure 5.10: The FRFR fourth-order achromat: Beam spots of different emittances.
The top two rows show that an achromat is reached after two turns. The bottom row
shows the remaining higher-order aberrations.

 

Figure 5.11: The FRF R fifth-order achromat: The layout, beam envelope and dis-
persive ray. The phase advances per cell are pr = W = 7r/2. The circumference is
266.64 m; the emittance is 307r mm mrad; and the dispersion is 0.3%.

15 mrad

' ‘ \\,\'\'-. m

Figure 5.12: The FRFR fifth-order achromat: 1000-turn tracking of the .r-a motion

of on-energy particles.

 

Strengths of the Multi]

soles (Aperture 10 cm)

 

Octupoles

Decapoles

 

Gradient (kG/cm3)

Field (kG)

Gradient (kG/cm“)

Field (kG)

 

-0.996975E-06
-0.246999E-05
0.204723 E-05
-0. 135901 E-05
0.951498E-06
—0.228548E-04
0.177119E-04
-0. 158309 E-04
0.420261E-05
0.871498E-07
0.377365 E-06
0.533332 E-05
0.321821E-05
0.191867E-05
—0.130343E-05

 

 

-0.124622E-03
-0.308749E~03
0.255903E-03
-0.169876E-03
0.118937E-03
-0.285685E702
0.221399Ev02
-0.197886E-02
0.525326E-03
0.108937E-04
0.471706E-04
0.666665 E-03
0.402276E-03
0.239833E-03
-0.162929E-03

-0.391808E-06
0.239260E-06
-0.346336E-07
-0.413315E-07
0.100518E-06
-0.501265E—07
-0.953086E-07
0.51 125613—06
-0.305803E-07
-0.775351E-07
0.506782E-08
0.153783E-07
—0.152854E-07
0.159598E-06
-0.317045E-06

 

 

-0.244880E-03
0.149538E-03
-0.216460E—04
-0.258322E-04
0.628240E—04
—0.3l3291E-04
-0.595678E-04
0.319535E-03
-0.191127E-04
-0.484594E-04
0.316738E-05
0.961144E-05
-0.955335E~05
0.997489E—04
-0.198153E-03

 

Table 5.12: The F RF R fifth-order achromat: The field strengths of the octupoles and
the decapoles. Note that the multipoles are extremely weak as a result of good linear

behavior.

 

20000
15000
10000

5000

 

 

i 1 I 1 1 L 1 I J 1 1 I

1144—111L1J

J .I 1

 

 

Figure 5.13: The FRF R fifth-order achromat: Resolution vs numbers of turns at the
acceptance. The saturation comes from the accumulation of higher-order aberrations

over turns.

60 80 100

 

130

 

Strengths of the Duodecapoles (Aperture 10 cm)

 

 

-0.602391E-07
0.115200E-06
-0.129574E-06
0.167172E-06
-0.146698E-06
0.109038E-07
-0.897166E-07
0.905100E-07
0.422171 E-07
-0.119032E-06
0.812032E-07
0.859254 E-07
0.143652E-06
-0.192421E-06
0.231 12213-06
-0.729862E-07
-0.102382E-06
-0.913997E-07

 

 

-0.376494E-04
0.720003 E-04
-0.809839E-04
0.104483E-03
—0.916861E-04
0.681489E-05
-0.560728E-04
0.565687E-04
0.263857E-04
-0.743948E-04
0.507520E-04
-0.537034E-04
0.897825E-04
-0.120263E-03
0.144451E-03
-0.456164E-04
-0.639889E-04
-0.571248E-04

-0.392296E-06
0.426602E-06
-0.251765E-06
0.101758E-06
-0.81297l E-07
0.113277E-06
-0.423092E-07
-0.733480E-07
0.173217E-07
0.970192E-07
0.745327E-07
-0.158631E-06
0.230450E-06
~0.172798E-06
0.923330E-07
0.126337E-06
—0.256941E-06

 

 

Gradient (kG/cms) Field (kG) Gradient (kG/cms) Field (kG)
0.2605261‘3-06 0.162829E-03 0.143366E-06 0.896036E-04
-0.141949E-06 -0.887180E-04 0.111585E-06 0.697405E-O4

-0.245185E-03
0.266626E-03
-0.157353E-03
0.635989E-04
-0.508107E-04
0.707979E-04
-0.264433E-04
-0.458425E-04
0.108261E-04
0.606370E-04
0.465829E-04
-0.991446E-04
0.144031E-03
-0.107999E-03
0.577081E-04
0.789607E-04
-0.160588E-03

 

Table 5.13: The F RFR fifth-order achromat: The field strengths of the duodecapoles.

Note that the multipoles are extremely weak as a result of good linear behavior.

 

Summary

The theoretical and experimental development of achromats is reviewed in detailed
studies of Brown's second-order achromat theory and Dragt’s third-order achromat.
theory. It is shown that Dragt‘s theory provides a complete proof of Brown‘s theory

in a different way, thus demonstrating that Dragt's theory is more general.

The second- and third-order theories are extended to an arbitrary-order theory
where detailed proofs of all conclusions are given. As opposed to repetition, this
theory explores the role of mirror symmetry in building an achromatic system. It is
shown that two- and three-cell systems are not the best choices for making achromats.
because they require more conditions than all efficient four-cell systems. On the other
hand, systems with five or more cells cannot give solutions that are distinctively better
than those of four-cell systems. Therefore, four-cell systems are the best choices
for building arbitrary-order achromats. For four-cell systems, the best solution is
found. Four four—cell systems are found optimal for solutions because they require
the smallest number of linear conditions. A general solution for four-cell arbitrary—
order achromats based on the optimal systems is presented; it is proved analytically
up to the fourth order and computational results suggest that it is valid up to the

sixth order. This is close to the best solution that can be obtained from this theory.

Four examples of achromats of the third, fourth, and fifth orders are presented.
An “S”-shaped third-order achromat shows the possibility for use as a single-pass

time-of-flight spectrometer. A circular third-order isochronous achromat can be used

131

132

as a multi—pass time-of—flight mass spectrometer for studying very short-lived nuclei.
A fourth-order and a fifth-order achromats verify the analytical theory to the sixth

order.

In conclusion, an understanding of arbitrary-order, mirror symmetrical achromats,

particularly four-cell systems, has been developed.

Appendix A

Symplectic Properties of Matrices
R and S

Theorem A.l R is antisymplectic and S is symplectic.

Proof.

To be consistant with the definition of J in eq. (2.2), we have

GOD'—

1000 0010 1000
A,_ 0100 0001 0100
[UR-[00404000 00—10
000—1 0—100 000—1
_ 10 01 10
‘ 0—1 —10 0—1
_ 10 0—1
‘ 0—1 —10
0—1
_(,0)__.,

1 33

which shows that R is antisymplectic.

Similarly, we have

134

—1 0 0 0
S = 0 1 0 0
0 0 —1 0
0 0 0 1
, —1 0
Now let us define a = ( 0 1 ). Since
—1 0 0 0 0
5.2125, : O l 0 0 0
0 0 —l 0 —1
0 0 0 l 0

||
A
I
q~ O
O QM
V
11
A
|
N O

S is symplectic.

coer—

OOHO

GOO—-

OOF'O

FOOD

Appendix B

The Proof of Equation (4.60)

Recall eq. (4.59), we have

marms = —1
m)? = —l .

Next we show that eq. (4.60) holds, i.e.,

"111’ = —-rr»$L’- (131)
Since 771.111,) : ”"1? and "712-12) : -mfzg)‘
Inglllrngizl = —1 means that mfilmg) = _1. (8.2)

To keep the generality, we have to consider the other cases because when the spe-
cific knowledge of the odering the cells are needed, as shown below, the permutational
symmetry is broken. Therefore, the conclusion drown from eq. (B2) is that only one
out of the last three cells satisfies m1? = my; 2 —1 (i = 1,2,3).

From eq. (4.70) and (4.73) in Section 4.2.4, the second cell has to be either R or

C when milling? = —1.

First. let the second cell be R. From eq. (4.70), we have

ad+ be 26d 2120’
1’ 1,“) 2 L713;1 - L1 = —2ac —(ad+ be) —207}’ .
0 0 l

135

136

Under the condition that 711.112) 2 711.121,) 2 mils) : 771.1215) 2 0. we have

1 0 0 —1 0 0
111,“): 0 —1 0 or 0 1 0 ,
0 0 1 0 0 1

which means that 1111111) 2 —m.g.12). If the second cell is C, the result is the same as for

the case of R.
Similarly. the second cell has to be either F or S when mill) 2 m3? = -—1.
When it is F, we have

—1 0 0
L,=M,“’= 0 —1 0 .

0 0 1

This implies that the third cell must satisfy megl 2 —1, Thus for R and C, A1112)

is
' —1 0 0
a) Ml” = Lf‘Rl'1-L1L1: 0 1 0 , and
0 0 1
‘ 1 0 0
1)) 111,1” 2 L;‘R;‘S;‘ - L,Ll = 0 —1 0 ,
0 0 1
respectively. This shows that in)? = —m.(2§). If the second cell is S, the result is the

same as the case above, except that L1 = I.

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