Bernhard Holzer, CERN

Introduction to Transverse Beam Optics

II.) Twiss Parameters & Lattice Design

Z X Y( )

Bunch in a storage ring

Bernhard Holzer, CERN

Introduction to Transverse Beam Optics

... don't worry: it's still the "ideal world"

... Particle acceleration whithout emittance or beta function

)2/(sin

1*

Kr)8(

entZN(N

4222

0

42

i

Historical note:

Rutherford Scattering, 1911

Using radioactive particle sources:α-particles of some MeV energy

N(θ)

θ

ε & β

01

*ss

x

xM

x

x

Reminder of Part I

Solution of Trajectory Equations

)cos()sin(

)sin(1

)cos(

lKlKK

lKK

lKM foc

lKlKK

lKK

lKMdefoc

cosh(sinh(

sinh(1

cosh(

10

1 lMdrift

Equation of Motion:

… hor. plane:

… vert. Plane:

21K k

K k

0xKx

focusing lens

dipole magnet

defocusing lens

Transformation through a system of lattice elements

combine the single element solutions by multiplication of the matrices

*.....* * * *etotal QF D QD B nd DM M M M M M

x(s)

s

court. K. Wille

2 1

s2,s1( )*

s s

x xM

x x

0

typical values

in a strong

foc. machine:

x ≈ mm, x ≤ mrad

Question: what will happen, if the particle performs a second turn ?

x

... or a third one or ... 1010 turns

0

s

Astronomer Hill:

differential equation for motions with periodic focusing properties

„Hill‘s equation“

Example: particle motion with

periodic coefficient

equation of motion: ( ) ( ) ( ) 0x s k s x s

restoring force ≠ const, we expect a kind of quasi harmonic

k(s) = depending on the position s oscillation: amplitude & phase will depend

k(s+L) = k(s), periodic function on the position s in the ring.

The Beta Function

General solution of Hill´s equation:

( ) ( ) cos( ( ) )x s s s

β(s) periodic function given by focusing properties of the lattice ↔ quadrupoles

ε, Φ = integration constants determined by initial conditions

Inserting (i) into the equation of motion …

0

( )( )

sds

ss

Ψ(s) = „phase advance“ of the oscillation between point „0“ and „s“ in the lattice.

For one complete revolution: number of oscillations per turn „Tune“

1

2 ( )y

dsQ

so

( ) ( )s L s

(i)

8.) The Beam Emittance

General solution of Hill´s equation:

β(s) = periodic function given by focusing properties of the lattice

ε = constant, determined by initial conditions of the particle ensemble.

( ) ( )s L s

( ) ( ) cos ( )x s s s

( ) ( )cos ( ) sin ( )( )

x s s s ss

2 2( )* ( ) 2 ( ) ( ) ( ) ( ) ( )s x s s x s x s s x sx´

x

Liouville: in reasonable storage rings

area in phase space is constant.

A = π*ε=const

ε beam emittance = woozilycity of the particle ensemble, intrinsic beam parameter,

cannot be changed by the foc. properties.

Scientifiquely spoken: area covered in transverse x, x´ phase space … and it is constant !!!

2 2( )* ( ) 2 ( ) ( ) ( ) ( ) ( )s x s s x s x s s x s

Phase Space Ellipse

( ) ( ) cos ( )x s s sparticel trajectory:

max. Amplitude: )(ˆ sx x´ at that position …?

… put into and solve for x´)(ˆ sx

22 xx

/x

In the middle of a quadrupole β is maximum,

α = zero 0x

… and the ellipse is flat

*

* A high β-function means a large beam size and a small beam divergence.

… et vice versa !!!!

Phase Space Ellipse

… solve for x´2

2,1

xxx

2222

2 xxxxx

… and determine via:x 0dx

xd

x´

xx

x

shape and orientation of the phase space ellipse

depend on the Twiss parameters β α γ

)()()()()(2)()( 22sxssxsxssxs

2

1( ) ( )

2

1 ( )( )

( )

s s

ss

s

Emittance of the Particle Ensemble:

single particle trajectories, N ≈ 10 11 per bunch

))(cos()()( sssx

Gauß

Particle Distribution:

2

2

2

1

2)( x

x

x

eeN

x

particle at distance 1 σ from centre ↔ 68.3 % of all beam particles

)()(ˆ ssx

aperture requirements: r 0 = 10 * σLHC: mmmm 3.0180*10*5* 10

Emittance of the Particle Ensemble:

Example: HERA

beam parameters in the arc

)1(10*7

80)(

9 mrad

mx

mm75.0

Z X Y( )

particle bunch

9.) Transfer Matrix M… yes we had the topic already

( ) ( ) cos ( )x s s s

( ) ( )cos ( ) sin ( )( )

x s s s ss

general solution

of Hill´s equation

remember the trigonometrical gymnastics: sin(a + b) = … etc

( ) cos cos sin sins s sx s

( ) cos cos sin sin sin cos cos sins s s s s s

s

x s

starting at point s(0) = s0 , where we put Ψ(0) = 0

0

0

cos ,x

0 00 0

0

1sin ( )

xx

inserting above …

0 0 0 0

0

( ) cos sin sinss s s sx s x x

00 0 0 0

0

1( ) cos (1 )sin cos sins s s s s s s

ss

x s x x

which can be expressed ... for convenience ... in matrix form

0s

x xM

x x

0 0

0

0 0 0

0

cos sin sin

( )cos (1 )sincos sin

ss s s s

s s s ss s s

s

M

s

* we can calculate the single particle trajectories between two locations in the ring,

if we know the α β γ at these positions.

* and nothing but the α β γ at these positions.

* … ! * Äquivalenz der Matrizen

ψ turn = phase advance

per period

10.) Periodic Lattices

„This rather formidable looking

matrix simplifies considerably if

we consider one complete turn …“

turnsturnturns

turnsturnsturnsM

sincossin

sinsincos)(

0 0

0

0 0 0

0

cos sin sin

( )cos (1 )sincos sin

ss s s s

s s s ss s s

s

M

s

Tune: Phase advance per turn in units of 2π )(*

2

1

s

dsQ

Ls

s

turns

ds

)(

Delta Electron Storage Ring

transfer matrix for particle trajectories

as a function of the lattice parameters

Stability Criterion:

Question: what will happen, if we do not make too

many mistakes and your particle performs

one complete turn ?

Matrix for 1 turn:

cos sin sin

sin cos sin

turn s turn s turn

s turn turn s turn

M1 0

cos sin0 1

1 JMatrix for N turns:

1 cos sin 1 cos sinNNM J N J N

The motion for N turns remains bounded, if the elements of MN remain bounded

real 1cos 2)(MTrace

stability criterion …. proof for the disbelieving collegues !!

Matrix for 1 turn:cos sin sin

sin cos sin

turn s turn s turn

s turn turn s turn

M1 0

cos sin0 1

I JMatrix for 2 turns:

2211

2 sin*cos**sin*cos* JIJIM

21

2

212121

2 sinsincossin*sincos*coscos* JJIIJI

now …

II2

*10

01*JI

10

01** IJ

IJJI **

IJ10

01*

2

2

2

)sin(*)cos(* 2121

2JIM

)2sin(*)2cos(*2JIM

11.) Transformation of α, β, γ

consider two positions in the storage ring: s0 , s

0s s

x xM

x x

since ε = const (Liouville):

2 2

2 20 0 0 0 0 0 0

2

2

x xx x

x x x x

where …

SC

SCM

MMMMMM QFDriftBQDQF ...

Beta function in a storage ring

D

yx ,

1

0 s

x xM

x x

... remember W = CS´-SC´ = 1

0

0

x S x Sx

x C x Cx1 S S

MC C

express x0 , x´0 as a function of x, x´.

inserting into ε2 22x xx x

2 20 0 0( ) 2 ( )( ) ( )Cx C x S x Sx Cx C x S x Sx

sort via x, x´and compare the coefficients to get ....

0s s

x xM

x x ρ

sθ

z

s0

2 20 0 0( ) 2s C SC S

0 0 0( ) ( )s CC SC S C SS

2 20 0 0( ) 2s C S C S

in matrix notation:

2 20

0

2 20

2

2s

C SC S

CC SC CS SS

C S C S

!

1.) this expression is important

2.) given the twiss parameters α, β, γ at any point in the lattice we can transform them and

calculate their values at any other point in the ring.

3.) the transfer matrix is given by the focusing properties of the lattice elements,

the elements of M are just those that we used to calculate single particle trajectories.

4.) go back to point 1.)

Geometry of the ring:

centrifugal force = Lorentz force

2

* *mv

e v B

p = momentum of the particle,

ρ = curvature radius

Bρ= beam rigidity

* /mv

e B p

* /B p e

Example: heavy ion storage ring TSR

8 dipole magnets of equal bending strength

12.) Lattice Design: „… how to build a storage ring“

High energy accelerators circular machines

somewhere in the lattice we need a number of dipole magnets,

that are bending the design orbit to a closed ring

Circular Orbit:

„… defining the geometry“

*

*

ds dl

B dl

B

2 2 **

Bdl pBdl

B q

The angle swept out in one revolution

must be 2π, so

field map of a storage ring dipole magnet

ρ

α

ds

… for a full circle

Nota bene: 410

B

Bis usually required !!

7000 GeV Proton storage ring

dipole magnets N = 1232

l = 15 m

q = +1 e

Tesla

es

mm

eVB

epBlNdlB

3.8

103151232

1070002

/2

8

9

Example LHC:

00

0 0

'( ) *cos( * ) *sin( * )

'( ) * *sin( * ) ' *cos( * )

yy s y K s K s

K

y s y K K s y K s

„ Focusing forces … single particle trajectories“

'' * 0y K y

ρ

Solution for a focusing magnet

y21/K k

K k

hor. plane

vert. plane

qp

B

/

1dipole magnet

qp

gk

/quadrupole magnet

Example: HERA Ring:

Bending radius: ρ = 580 m

Quadrupol Gradient: g = 110 T/m

k = 33.64*10-3 /m2

1/ρ2 = 2.97 *10-6 /m2

For estimates in large accelerators the weak focusing term 1/ρ2 can

in general be neglected

The Twiss parameters α, β, γ can be transformed through the lattice via the

matrix elements defined above.

2 2

2 2

0

2

' ' ' ' *

' 2 ' ' 'S

C SC S

CC SC S C SS

C S C S

Question: „ What does that mean ???? “

Most simple example: drift space

10

1 l

'S'C

SCM

0

1*

' 0 1 'l

x l x

x x0 0

0

( ) * '

'( ) '

x l x l x

x l x

particle coordinates

transformation of twiss parameters:

2

0

1 2

0 1 *

0 0 1l

l l

l2

0 0 0( ) 2 * *s l l

Stability ...?

( ) 1 1 2trace M

A periodic solution doesn‘t

exist in a lattice built exclusively

out of drift spaces.

13.) The FoDo-Lattice

A magnet structure consisting of focusing and defocusing quadrupole lenses in

alternating order with nothing in between.

(Nothing = elements that can be neglected on first sight: drift, bending magnets,

RF structures ... and especially experiments...)

Starting point for the calculation: in the middle of a focusing quadrupole

Phase advance per cell μ = 45°,

calculate the twiss parameters for a periodic solution

Periodic solution of a FoDo Cell

Output of the optics program:

4521250 *.

x

y

Nr Type Length Strength βx

αx

ψx

βy

αy

ψy

m 1/m2 m 1/2π m 1/2π

0 IP 0,000 0,000 11,611 0,000 0,000 5,295 0,000 0,000

1 QFH 0,250 -0,541 11,228 1,514 0,004 5,488 -0,781 0,007

2 QD 3,251 0,541 5,488 -0,781 0,070 11,228 1,514 0,066

3 QFH 6,002 -0,541 11,611 0,000 0,125 5,295 0,000 0,125

4 IP 6,002 0,000 11,611 0,000 0,125 5,295 0,000 0,125

QX= 0,125 QY= 0,125

L

QF QFQD

Can we understand, what the optics code is doing?

1cos( * ) sin( * )

,

sin( * ) cos( * )

q q

QF

q q

K l K lKM

K K l K l

1

0 1Drift

d

lM

strength and length of the FoDo elements K = +/- 0.54102 m-2

lq = 0.5 m

ld = 2.5 m

* * * *FoDo qfh ld qd ld qfhM M M M M M

0.707 8.206

0.061 0.707FoDoM

Putting the numbers in and multiplying out ...

The matrix for the complete cell is obtained by multiplication of the element matrices

matrices

The transfer matrix for 1 period gives us all the information that we need !

1.) is the motion stable? ( ) 1.415FoDotrace M < 2

2.) Phase advance per cell

cos sin sin( )

sin cos( ) sinM s

1cos( ) * ( ) 0.707

2trace M

1cos( * ( )) 45

2arc trace M

3.) hor β-function

(1,2)11.611

sin( )

Mm

4.) hor α-function

(1,1) cos( )0

sin( )

M

Can we do it a little bit easier ?

We can: … the „thin lens approximation“

Matrix of a focusing quadrupole magnet:

1cos( * ) sin( * )

sin( * ) cos( * )

QF

K l K lKM

K K l K l

If the focal length f is much larger than the length of the quadrupole magnet,

1Q

Q

f lkl

the transfer matrix can be aproximated using

1 0

1 1M

f

, 0q qkl const l

FoDo in thin lens approximation

lD

Calculate the matrix for a half cell, starting in the middle of a foc. quadrupole:

/ 2 / 2* *halfCell QD lD QFM M M M

2

1

1

DD

halfCell

D D

ll

fM

l l

f f

/ 2,

2

Dl L

f f

1 0 1 01

* *1 11 10 1

D

halfCell

lM

f f

for the second half cell set f -f

L

FoDo in thin lens approximation

2 2

1 1

*

1 1

D DD D

D D D D

l ll l

f fM

l l l l

f f f f

Matrix for the complete FoDo cell:

2

2

2 2

3 2 2

21 2 (1 )

2( ) 1 2

D DD

D D D

l ll

f fM

l l l

f f f

Now we know, that the phase advance is related to the transfer matrix by

2 2

2 2

4 21 1cos ( ) *(2 ) 122

D Dl ltrace M

f f

After some beer and with a little bit of trigonometric gymnastics

2 2 2cos( ) cos ( ) sin ( / 2) 1 2sin ( )2 2

x xx x

we can calculate the phase advance as a function of the FoDo parameter …

22

2

2cos( ) 1 2sin ( / 2) 1

sin( / 2) /2

D

CellD

l

f

Ll f

f

Example:

45-degree Cell

LCell = lQF + lD + lQD +lD = 0.5m+2.5m+0.5m+2.5m = 6m

1/f = k*lQ = 0.5m*0.541 m-2 = 0.27 m-1

sin( / 2) 0.4054

CellL

f

47.8

11.4m

45

11.6m

Remember:

Exact calculation yields:

sin( / 2)4

CellL

f

Stability in a FoDo structure

2

2

2 2

3 2 2

21 2 (1 )

2( ) 1 2

D DD

FoDo

D D D

l ll

f fM

l l l

f f f

Stability requires:

4

cellLf

For stability the focal length has to be larger than a quarter of the cell length !!

SPS Lattice

2~4

2)(2

2

f

lMTrace d

2)(MTrace

Transformation Matrix in Terms of the Twiss parameters

Transformation of the coordinate vector (x,x´) in a lattice

General solution of the equation of motion

Transformation of the coordinate vector (x,x´) expressed

as a function of the twiss parameters

ρ

x

s1

s2

0

0

2,1')('

)(

x

xM

sx

sxss

))(cos(*)(*)( sssx

))(sin())(cos()(*)(

)(' ssss

sx

)sin(cossin)1(cos)(

sin)sin(cos

12212

2

1

21

12211221

122112112

1

2

21M

Transfer matrix for half a FoDo cell:

In the middle of a foc (defoc) quadrupole of the FoDo we allways have α = 0,

and the half cell will lead us from βmax to βmin

Compare to the twiss

parameter form of M

lD

LL

1 2

f

l

f

l

lf

l

MDD

DD

halfcell

~1~

~1

2

)sin(cossin)1(cos)(

sin)sin(cos

12212

2

1

21

12211221

122112112

1

2

21M

2cos

2sin

1

2sin

2cos

'' SC

SCM

Solving for βmax and βmin and remembering that ….

1 sinˆ 1 /' 2

1 / 1 sin2

D

D

l fS

C l f

22

2

ˆ'

sin2

DlSf

C

sin2 4

Dl L

ff

(1 sin )2ˆ

sin

L

(1 sin )2

sin

L

The maximum and minimum values of

the β-function are solely determined by

the phase advance and the length of the cell.

Longer cells lead to larger β

Z X Y( )

typical shape of a proton

bunch in the HERA FoDo Cell

!

!

14.) Scaling of the Twiss Paramerters

Conclusion:

* „the arc“ of a storage ring is usually built out of a periodic sequence

of single magnet elements eg. FoDo sections

* a first guess of the main parameters of the beam in the arc is

obtained by the settings of the quadrupole lenses in this section

* we can get an estimate of the beam parameters using a selection

of „rules of thumb“

Usually the real beam properties will not differ too much from these estimates

and we will have a nice storage ring and a beautifull beam and everybody is

happy around.

And then someone comes and spoils it all by saying

something stupid like installing a tiny little piece of detector

in our machine …