III.) Twiss Parameters & Lattice Design...Layout of a storage ring lattice calculate the...

Introduction to Transverse Beam OpticsIntroduction to Transverse Beam OpticsBernhard Holzer, CERN Bernhard Holzer, CERN

yx ,ββββ

D

III.) Twiss Parameters & Lattice Design III.) Twiss Parameters & Lattice Design The The „„ not so ideal world not so ideal world ““

Transformation through a lattice:

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

Reminder: Particle TrajectoriesReminder: Particle Trajectories

−=

)cos()sin(

)sin(1

)cos(

lKlKK

lKK

lKM foc

=lKlKK

lKK

lKM defoc

cosh(sinh(

sinh(1

cosh(

=

10

1 lM drift

Equation of Motion:

… hor. plane:

… vert. Plane:

21= −K kρ

=K k

0=+′′ xKx●

ρ

x

s1s2

0)(*)()(

1)(

2=

−+′′ sxsks

sxρ

Hill equation

General solution of Hill s equation:

( ) ( ) cos( ( ) )= ⋅ +x s s sε β ψ φ

ββββ(s) = periodic functiongiven by focusing properties

εεεε = constant

( ) ( )s L sβ β+ =

HERA number of stored particles:

revb

mAN *

e=

100180

τ

b

* Cb * sN * * *

s . * Cb

− −

−=

3 6

19

100 10 21 10180 1 6 10

bN . *=107 3 10

Reminder: Reminder: ββββββββ function & function & Beam Emittance Beam Emittance εεεεεεεε

Emittance of the Particle Ensemble:Emittance of the Particle Ensemble:

single particle trajectories, N ≈≈≈≈ 10 11 per bunch

))(cos()()( φβε +Ψ⋅= sssx

GaußParticle Distribution:

2

2

2

1

2)( x

x

x

eeN

x σσσσ

σσσσππππρρρρ

−

⋅⋅=

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

)()(ˆ ssx βε=

aperture requirements: r 0 = 10 * σσσσLHC: mmmm 3.0180*10*5* 10 === −βεσ

Phase Space EllipsePhase Space Ellipse

shape and orientation of the phase space ellipse depend on the Twiss parameters ββββ αααα γγγγ

)()()()()(2)()( 22 sxssxsxssxs ′+′+= ββββααααγγγγεεεε

x´

xεβ

εα β−εγ

εα γ−

●

●

●

●

●

●

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

2 1

s2,s1( )*

= ′ ′ s s

x xM

x x

cos sin sin( )

sin cos( ) sinM s

µ α µ β µγ µ µ α µ+

= → − −

ββββββββ

αααααααα

μμμμμμμμ

Calculation of the Twiss Parameters Calculation of the Twiss Parameters

Transformation Matrix & Twiss Parameters:Transformation Matrix & Twiss Parameters:

−−+

=turnsturnturns

turnsturnsturnsMψψψψααααψψψψψψψψγγγγ

ψψψψββββψψψψααααψψψψsincossin

sinsincos)(

transfer matrix for particle trajectoriesas a function of the lattice parameters

−+−−

+=→

)sin(cossin)1(cos)(

sin)sin(cos

122122

1

21

12211221

1221121121

2

21

ψψψψααααψψψψββββββββ

ββββββββψψψψααααααααψψψψαααααααα

ψψψψββββββββψψψψααααψψψψββββββββ

M

transfer matrix for periodic structures... to determine the Twiss babies

2 20

0

2 20

2

2

− ′ ′ ′ ′= − + − ⋅

′ ′ ′ ′− s

C SC S

CC SC CS SS

C S C S

ββα αγ γ

transfer matrix for Twiss parametersbetween two lattice locations

elementary cell of a storage ring

xβ

yβ

...**** 21 QDDBendDQFtotal MMMMMSC

SCM =

′′=

(1 sin )2ˆ

sin

Lµ

βµ

+=

(1 sin )2

sin

Lµ

βµ

∨ −=

Z X, Y,( )

Layout of a storage ring latticeLayout of a storage ring lattice

calculate the periodic solution

determine the beta functions in the cell

to get the size of a particle bunch

15.) Beam dimension:15.) Beam dimension:

Optimisation of the FoDo Phase advance:Optimisation of the FoDo Phase advance:Optimisation of the FoDo Phase advance:Optimisation of the FoDo Phase advance:Optimisation of the FoDo Phase advance:Optimisation of the FoDo Phase advance:Optimisation of the FoDo Phase advance:Optimisation of the FoDo Phase advance:

In both planes a gaussian particle distributionis assumed, given by the beam emittance εεεε and the ββββ-function

2x x y yr ε β ε β= +

σ εβ=

In general proton beams are „round“in the sense that

x yε ε≈So for highest aperture we have to minimise the ββββ-functionin both planes:

typical beam envelope, vacuum chamber and pole shape in a foc. Quadrupole lens in HERA

HERA beam size

Optimising the FoDo phase advance Optimising the FoDo phase advance 2x x y yr ε β ε β= +

search for the phase advance μμμμ that results in a minimum of the sum of the beta’s

(1 sin )* (1 sin )*2 2ˆ

sin sin

L Lµ µ

β βµ µ

∨ + −+ = +

2( ) 0sind L

d µµ=

2*cos 0 90

sin

L µ µµ

= → = °

2ˆsin

Lβ βµ

∨

+ =

0 30 60 90 120 150 18010

12

14

16

18

2020

10

βgesµ( )

1800 µ

electron beams are usually flat, εy ≈ 2 - 10 % εx

���� optimise only ββββhor

(1 sin )2ˆ( ) 0 76

sin

Ld d

d d

µ

β µµ µ µ

+= = → ≈ °

red curve:ββββmax blue curve:ββββminminminmin

as a function of the phase advance μμμμ

Electrons are differentElectrons are different

1 36.8 72.6 108.4 144.2 1800

6

12

18

24

3030

0

βmax µ( )

βmin µ( )

1801 µ

... the most complicated insertion:the drift space

Question to the audience:what will happen to the beam parameters αααα, ββββ, γγγγ if westop focusing for a while …?

2 2

2 2

0

2

' ' ' ' *

' 2 ' ' 'S

C SC S

CC SC S C SS

C S C S

β βα αγ γ

− = − + −

− 1

' ' 0 1

C S sM

C S

= =

„0“ refers to the position of the last lattice element

„s“ refers to the position in the drift

20 0 0

0 0

0

( ) 2

( )

( )

s s s

s s

s

β β α γα α γγ γ

= − += −=

yx ,ββββ

D

The The „„ not so ideal world not so ideal world ““

16.) Insertions16.) Insertions

ββββββββ--------Function in a Drift:Function in a Drift:Function in a Drift:Function in a Drift:Function in a Drift:Function in a Drift:Function in a Drift:Function in a Drift:

let‘s assume we are at a symmetry point in the centerof a drift.

20 0 0( ) 2s s sβ β α γ= − +

as2

00 0

0 0

1 10,

αα γβ β

+= → = =

and we get for the β function in the neighborhood of the symmetry point

Nota bene: 1.) this is very bad !!!2.) this is a direct consequence of the

conservation of phase space density(... in our words: εεεε = const) … and there is no way out.

3.) Thank you, Mr. Liouville !!!

! ! !! ! !

Joseph Liouville,1809-1882

2

00

( )s

sβ ββ

= +

Optimisation of the beam dimension:

2

00

( )β ββ

= + ll

Find the β at the center of the drift that leads to the lowest maximum β at the end:

If we cannot fight against Liouvuille theorem ... at least we can optimise

2

20 0

ˆ1 0

d

d

ββ β

= − =l

!

0β→ = l

0ˆ 2β β→ =

If we choose ββββ0 = ℓℓℓℓ we get the smallest ββββ at the end of the drift and the maximum ββββ is just twice the distance ℓℓℓℓ

**

l l

β0

ββββββββ--------Function in a Drift:Function in a Drift:Function in a Drift:Function in a Drift:Function in a Drift:Function in a Drift:Function in a Drift:Function in a Drift:

... clearly there is another problem !!!... clearly there is another problem !!!

Example: Luminosity optics at LHC: β* = 55 cmfor smallest βmaxwe have to limit the overall length

"s” to 1.10 m

But: ... unfortunately ... in general high energy detectorsthat are installed in that drift spaces are a little bit bigger than a few centimeters ...

ZEUS detector: inelastic scattering event of e+/p

production rate of(scattering) eventsis determined bythecross section Σreactand a parameter L that is given by the design of the accelerator:… the luminosity

17.) The Mini17.) The Mini--ββββββββ Insertion:Insertion:

1 22 * *

0

*1*

4 b *x y

I IL

e fπ σ σ=

R= L * Σreact

p-Bunch

e-BunchIP 2 σσσσ

3*10 10 particles

7*10 10 particles

Luminosity

m

mrad

m

x

x

x

µσεβ

118

107

45.29

=∗=

=−

m

m

x

yx

y

µσεε

β

32

18.0

=

=

=

mAI

mAI

p

e

84

43

==

180

3.470

==

bn

kHzf

scmL 2

30 1100.34 ∗=

Example: e- p collider run at HERA

L = 1

4πe2 f 0nb

*I eI p

σ xσ y

MiniMiniMiniMiniMiniMiniMiniMini--------ββββββββ Insertions: Insertions: Insertions: Insertions: Insertions: Insertions: Insertions: Insertions: BetafunctionsBetafunctionsBetafunctionsBetafunctionsBetafunctionsBetafunctionsBetafunctionsBetafunctions

A mini-β insertion is always a kind of special symmetric drift space.�greetings from Liouville

at a symmetry point ββββ is just the ratio of beam dimension and beam divergence.

* 0α =

x´

xεβ

εα β−εγ

εα γ−

●●●●

●●●●

●●●●

●●●●

●●●●

●●●●

2*

*

1 1αγβ β

+= =

**

*

σβσ

=′

**

εσβ

′ = εβ

ββββεεεε /

8 individually powered quad magnets are needed to match the insertion ( ... at least)

Mini Beta Insertions: some guide lines

* Calculate the periodic solution in the arc

* Introduce the drift space needed for the insertion device(e.g. particle detector)

* Install a quadrupole dublet (or triplet) as close as possible

* Introduce additional quadrupol magnets to match the beam parameters To the values at the beginning of the next arc structure.

αx, βx Dx, Dy

αy, βy Qx, Qy

Are there any problems ??Are there any problems ??Are there any problems ??Are there any problems ??

sure there are...

* large ββββ values at the doublet quadrupoles ���� large contribution to chromaticity Q' … and no local correction

* aperture of mini ββββ quadrupoleslimit the luminosity

* field quality and magnet stability most critical at the high ββββ sectionseffect of a quad error:

beam envelope at the first mini β quadrupole lens in the HERA proton storage ring

���� keep distance „s“ to the first mini ββββ quadrupole as small as possible

{ }1( ) ( )

4K s s dsξ β

π= ∫oo

∫+ ∆=∆

ls

s

dsssKQ

0

0 4

)()(

ππππββββ

18.) Liouville during Acceleration18.) Liouville during Acceleration

x´

xεβ

εα β−εγ

εα γ−

●●●●

●●●●

●●●●●●●●Beam Emittance corresponds to the area covered in the x, x´ Phase Space Ellipse

Liouville: Area in phase space is constant.

But so sorry ... εεεε ≠≠≠≠ const !

●●●●

Classical Mechanics:

phase space= diagram of the two canonical variables position & momentum

x px

EnergypotEnergykinVTLq

Lp

jj ..; −=−=

∂∂=&

)()()()()(2)()( 22 sxssxsxssxs ′+′+= ββββααααγγγγεεεε

According to Hamiltonian mechanics: phase space diagram relates the variables q and p

Liouvilles Theorem: p dq const=∫

for convenience (i.e. because we are lazy bones) we use in accelerator theory:

xdx dx dtx

ds dt ds

ββ

′ = = = where ββββx= vx / c

1x dxε

βγ′⇒ = ∝∫

the beam emittance shrinks during acceleration εεεε ~ 1 / γγγγ

q = position = xp = momentum = mcγβγβγβγβx

2

2

1

1

c

v−=γ

c

xx

&=β;

∫∫ = dxmcpdq xγβ

∫∫ ′= dxxmcpdq γβ

ε

Nota bene:Nota bene:

1.) A proton machine … or an electron linac … needs the highest aperture at injection energy !!!as soon as we start to accelerate the beam size shrinks as γγγγ -1/2 in both planes.

2.) At lowest energy the machine will have the major aperture problems, ���� here we have to minimise

3.) we need different beam opticsadopted to the energy: A Mini Beta concept will only be adequate at flat top.

εβσ =

β

LHC injection optics at 450 GeV

LHC mini beta optics at 7000 GeV

Example: HERA proton ring

injection energy: 40 GeV γγγγ = 43flat top energy: 920 GeV γγγγ = 980

emittance εεεε (40GeV) = 1.2 * 10 -7

εεεε (920GeV) = 5.1 * 10 -9

7 σσσσ beam envelope at E = 40 GeV

… and at E = 920 GeV

…… letlet ´s talk about acceleration s talk about acceleration

crab nebula,

burst of charged particles E = 10 20 eV

19.) The 19.) The „„ ΔΔΔΔΔΔΔΔp / p p / p ≠≠≠≠≠≠≠≠ 00““ Problem Problem

ideal accelerator:all particles will see the same accelerating voltage.���� ΔΔΔΔΔΔΔΔp / p = 0p / p = 0

„nearly ideal“ accelerator: Cockroft Walton or van de Graaf

ΔΔΔΔΔΔΔΔp / p p / p ≈≈≈≈≈≈≈≈ 10 10 --55

MP Tandem van de Graaf Accelerator at MPI for Nucl. Phys. Heidelberg

Vivitron, Straßbourg, inner structure of the acc. section

The The „„ not so ideal world not so ideal world ““

Linear Accelerator 1928, Wideroe schematic Layout:

++++ ++++ ++++ ++++---- ---- ----

* RF Acceleration: multiple application of the same acceleration voltage;brillant idea to gain higher energies

Energy Gain per „Gap“:

tUqW RFωsin** 0=

500 MHz cavities in an electron storage ring

drift tube structure at a proton linac

Problem: panta rhei !!!(Heraklit: 540-480 v. Chr.)

Z X, Y,( )

Bunch length of Electrons ≈≈≈≈ 1cmExample: HERA RF:

U0

t νλν

*

500

==

c

MHzcm60=λ

cm60=λ

994.0)84sin(

1)90sin(

==

o

o

310*6 −=∆U

U

typical momentum spread of an electron bunch: 310*1 −≈∆

p

p

Question:do you remember last session, page 11 ? … sure you do

20.) Dispersion: trajectories for 20.) Dispersion: trajectories for ΔΔΔΔΔΔΔΔp / p p / p ≠≠≠≠≠≠≠≠ 0 0

vBex

mvx

dt

dmF z−=

+−+=

ρρ

2

2

2

)( zρ

s

z

●●●● x

remember:x ≈≈≈≈ mm , ρρρρ ≈≈≈≈ m …���� develop for small x

veBxmv

dt

xdm z−=−− )1(

2

2

2

ρρ

0∂= +∂

zz

BB B x

xconsider only linear fields, and change independent variable: t→→→→ s

mv

gxe

mv

Bexx −−=−−′′ 0)1(

1

ρρ

●●●●

p=p0+ΔΔΔΔp

Forceacting on the particle

… but now take a small momentum error into account !!!

Dispersion:Dispersion:

develop for small momentum error2000

0

11

p

p

ppppp

∆−≈∆+

⇒<<∆

ρ1− 0≈xk ∗

200

0200

02

1

p

pxeg

p

xegeB

p

p

p

Bexx

∆+−∆+−≈+−′′ρρ

ρρ1

02

∗∆=−+′′p

pkx

xx

ρρ1

)1

(0

2∗∆=−+′′

p

pkxx

Momentum spreadof the beam adds a term on the r.h.s. of the equation of motion.���� inhomogeneous differential equation.

2

1 1( )

px x k

p ρρ∆′′ + − = ⋅

general solution:

( ) ( ) ( )h ix s x s x s= +( ) ( ) ( ) 0h hx s K s x s′′ + ⋅ =

1( ) ( ) ( )i i

px s K s x s

pρ∆′′ + ⋅ = ⋅

Normalise with respect to ΔΔΔΔp/p:

( )( ) i

pp

x sD s ∆=

Dispersion function D(s)

* is that special orbit, an ideal particlewould have for ΔΔΔΔp/p = 1

* the orbit of any particleis the sumof the well known xββββ and the dispersion

* as D(s) is just another orbitit will be subject to the focusing properties of the lattice

Dispersion:Dispersion:

. ρρρρ

xββββ

Closed orbit for ΔΔΔΔp/p > 0

( ) ( )ip

x s D sp

∆= ⋅

Matrix formalism:

( ) ( ) ( )p

x s x s D spβ

∆= + ⋅

0 0( ) ( ) ( ) ( )p

x s C s x S s x D sp

∆′= ⋅ + ⋅ + ⋅ 0s

x C S x Dp

x C S x Dp

∆= + ′ ′ ′ ′ ′

DispersionDispersionDispersionDispersionDispersionDispersionDispersionDispersionExample: homogenous dipole field

00 0 1p p

p ps

x C S D x

x C S D x∆ ∆

′ ′ ′ ′ ′= ⋅

Example HERA

3

1...2

( ) 1...2

1 10

x mm

D s m

pp

β

−

=

≈∆ ≈ ⋅

Amplitude of Orbit oscillation contribution due to Dispersion ≈≈≈≈ beam size���� Dispersion must vanish at the collision point

Calculate D, D

1 1

0 0

1 1( ) ( ) ( ) ( ) ( )

s s

s s

D s S s C s ds C s S s dsρ ρ

= −∫ ∫% % % %

or expressed as 3x3 matrix

D

ββββ

(proof: see appendix)

!

Exampel: Drift

1 0

0 1 0

0 0 1Drift

l

M

=

1 1

0 0

1 1( ) ( ) ( ) ( ) ( )

s s

s s

D s S s C s ds C s S s dsρ ρ

= −∫ ∫% % % %

0= 0=

Example: Dipole

Mdrift =C S

C' S'

=

1 l

0 1

Remember: Matrix of a magnetic element

−=

)cos()sin(

)sin(1

)cos(

lKlKK

lKK

lKM foc

2

1

ρ−= kK … but in a dipole, as k = 0 …

−=

′′=

ρρρ

ρρ

ρll

ll

SC

SCM foc

cossin1

sincos

Example: Dispersion in a Sector Dipole Magnet

calculate the „D“ elements of the marix

1 1

0 0

1 1( ) ( ) ( ) ( ) ( )

s s

s s

D s S s C s ds C s S s dsρ ρ

= −∫ ∫% % % %

ρρ

ρρρρ

ρρρ

ρ ∗+−⋅∗∗−∗∗= )1cos(1

cos)sin(1

)sin()(llll

sD

)cos1()(ρ

ρ lsD −⋅=

ρl

sD sin)( =′ Dispersion elements in a sector dipole magnet

)1(coscossin)( 2 −∗+=ρρ

ρρ

ρ lllsD

x

′ x ∆p

p

s2

=

cosl

ρρsin

l

ρρ ∗ (1− cos

l

ρ)

− 1

ρsin

l

ρcos

l

ρsin

l

ρ0 0 1

∗x

′ x ∆p

p

s1

Example: Dispersion, calculated by an optics code for a real machine

pp

*)s(DxD

∆∆∆∆====

* D(s) is created by the dipole magnets

… and afterwards focused by the quadrupole fields

D(s) ≈ 1 … 2 m

sMini Beta Section,

���� no dipoles !!!

Periodic Dispersion:

„Sawtooth Effect“ at LEP (CERN)

BPM Number

Electron course

In the arc the electron beam loses so much energyin eachoctant that the particle are runningmore and moreon a dispersion trajectory.

In the straight sections they are accelerated by therf cavitiesso much that they „overshoot“ and reach nearly theouter side of the vacuum chamber.

21.) Momentum Compaction Factor:21.) Momentum Compaction Factor:

The dispersionfunction relates the momentum errorof a particle to the horizontal orbit coordinate.

1( ) *

px K s x

pρ∆′′ + =

( ) ( ) ( )p

x s x s D spβ

∆= +

xββββ

inhomogeneous differential equation

general solution

But it does much more:it changes the length of the off - energy - orbit!!

ρρρρ

ρρρρ

dsx

dl

design orbit

particle trajectoryparticle with a displacement xto the design orbit���� path length dl...

1( )

xdl ds

sρ

→ = +

dl x

ds

ρρ+=

1( )

EE

xl dl ds

sρ∆

∆

= = +

∫ ∫

circumference of an off-energy closed orbit

remember:

( ) ( )Ep

x s D sp∆

∆=

( )

( )Ep D s

l dsp s

δρ∆

∆= ∫

* The lengthening of the orbitfor off-momentum particles is given by the dispersion function and the bending radius.

o

o

o

Momentum Compaction Factor: Momentum Compaction Factor: ααααααααpp

cpl p

L pεδ α ∆=

1 ( )

( )cpD s

dsL s

αρ

→ =

∫

For first estimates assume: 1

constρ

=

1 1 1 12cp dipolesl D D

L Lα πρ

ρ ρ= = → 2

cp

DD

L R

πα ≈ ≈

Assume: v c≈

cplT p

T L pεδδ α ∆→ = =

Definition:

ααααcp combines via the dispersion function the momentum spread with the longitudinalmotion of the particle.

o

dipoledipolesdipoles

DldssD *)()( Σ=∫

ResumeResume:: 1εβγ

∝

cpl p

L pεδ α ∆=

2cp

DD

L R

πα ≈ ≈

beam emittance

momentum compaction

20 0 0( ) 2s s sβ β α γ= − +

2

00

( )s

sβ ββ

= +

beta function in a drift

… and for αααα = 0

ρρ1

)1

(0

2∗∆=−+′′

p

pkxx

particle trajectory for ΔΔΔΔp/p ≠≠≠≠ 0inhomogenious equation

( ) ( ) ( )p

x s x s D spβ

∆= + ⋅… and ist solution

Appendix:Appendix:

Solution of the inhomogenious equation of motion

1 1

0 0

1 1( ) ( ) ( ) ( ) ( )

s s

s s

D s S s C s ds C s S s dsρ ρ

= −∫ ∫% % % %Ansatz:

SCdtSCCSdtCSsDρρρρ11

*11

*)( −′−+′=′ ∫∫

∫∫ ′−′=′ dtS

CdtC

SsDρρ

**)(

ρρρρS

CsdS

CC

SsdC

SsD ′−′′−′+′′=′′ ∫∫ ~*~*)(

( )CSSCsdS

CsdC

S ′−′+′′−′′= ∫∫ ρρρ1~*~*

= det M = 1

remember: for Cs) and S(s) to be independentsolutions the Wronski determinant has to meet the condition

0≠′′

=SC

SCW

remember: S & C are solutions of the homog. equation of motion:0*

0*

=+′′=+′′

CKC

SKS

ρρρ1~**~** ++−=′′ ∫∫ sd

SCKsd

CSKD

ρρρ1~~* +

+−=′′ ∫∫ sd

SCsd

CSKD

=D(s)

ρ1

* +−=′′ DKD … or ρ1

* =+′′ DKD

qed

ρρρ1~*~* +′′−′′=′′ ∫∫ sd

SCsd

CSD

and as it is independent of the variable „s“

0)()( =−−=′′−′′=′−′= SCCSKCSSCCSSCds

d

ds

dW

we get for the initital conditions that we had chosen … 1,0

0,1

00

00

=′==′=

SS

CC1=

′′=

SC

SCW

Dispersion in a FoDo Cell:Dispersion in a FoDo Cell:

in analogy to the derivations ofˆ,β β∨

* thin lens approximation:1

QQ

fk

= >> ll

10,

2Q D L≈ → =l l

1 1

2 ff=

%

* * * * length of quad negligible

* * * * start at half quadrupole

Calculate the matrix of the FoDo half cell in thin lens approximation:

!! we have now introduced dipole magnets in the FoDo:���� westill neglectthe weak focusingcontribution 1/ρρρρ2

���� but take into account 1/ρρρρ for the dispersion effectassume: length of the dipole = lD

2 2

* *HalfCell QD B QFM M M M=

Matrix of the half cell

1 0 1 01

* *1 11 10 1Half CellM

f f

= −

l

% %

2

1

' '1

Half Cell

C S fM

C S

f f

− = = − +

ll

%

l l

% %

∫∫ −= s~d)s~(S)s~(

*)s(Cs~d)s~(C)s~(

*)s(S)s(Dρρρρρρρρ

11

calculate the dispersion terms D, D´ from the matrix elements

0 0

1 1( ) * * 1 1 * *

sD ds s ds

f fρ ρ

= − − − ∫ ∫l l

ll l

% %

2 2 2 3 2 31( ) 1 * *

2 22 2 2D

f f f fρ ρ ρ ρρ ρ

= − − − = − − +

l l l l l l l ll l

% % % %

2

( )2

Dρ

= ll

in full analogy on derives for D:

'( ) 12

D sfρ

= +

l l

%

S(s) C(s) C(s) S(s)S(s) C(s) C(s) S(s)S(s) C(s) C(s) S(s)S(s) C(s) C(s) S(s)S(s) C(s) C(s) S(s)S(s) C(s) C(s) S(s)S(s) C(s) C(s) S(s)S(s) C(s) C(s) S(s)

2

2

12

' ' ' 1 (1 )2

0 0 10 0 1

halfCell

fC S D

M C S Df f f

ρ

ρ

−

− = = + +

l ll

%

l l l l

% % %

and we get the complete matrix includingthe dispersion terms D, D

boundary conditionsfor the transfer fromthe center of the foc. to the center of thedefoc. quadrupole

1/ 2

ˆ

0 * 0

1 1

D D

M

∨ =

D

ββββ

Dispersion in a FoDo CellDispersion in a FoDo Cell2

ˆ (1 )2

D Df ρ

∨

→ = − +l l

%

2ˆ0 * (1 )

2D

f fρ→ = − + +l l l

% %

2

221

1

2

2

µµµµ

µµµµ

ρρρρ sin

)sin(*D

++++==== l

2

221

1

2

2

µµµµ

µµµµ

ρρρρ sin

)sin(*D

−−−−====

∨∨∨∨ lwhere μμμμ denotes the phase advance of the full celland l/f = sin(μμμμ/2)

0 30 60 90 120 150 1800

2

4

6

8

1010

0.5

D max µ( )

D min µ( )

1801 µ

∨

D

D

Nota bene:

! small dispersionneedsstrong focusing→→→→ large phase advance

!! ↔↔↔↔ there is an optimum phase for smallββββ

!!! ...do you remember thestability criterion?½ trace = cosμμμμ ↔↔↔↔ μμμμ < 180°°°°

!!!! … life is not easy

Measuring the Dispersion

Idea: apply a well definedmomentum shiftΔΔΔΔp /pof the beam

without changing the magnetic fieldsαααα cp

revolution time

v

LT 0

0 =v

dv

L

dL

T

dT −=→0

0

0

0

v

dv

p

dp

T

dTcp −= α

0

0p

dp

L

dLcp /=α

p

dp

p

dp

f

dfcp 2

1

γα −=− p

dp

v

dv*

12γ

=

)1

(2γ

α −=− cpp

dp

f

df

Measuring the Dispersion

)1

(2 cpp

dpf

df ααααγγγγ

−=

≈≈≈≈ 0 0 0 0 for high energy beams.

Example:

310*2.1

208

360

−===

cp

MHzf

Hzdf

α

310*2.1*208

360/ −=−=

MHz

Hz

f

df

p

dpcpα

310*4.1 −=p

dp

Dispersion Function in the arc:

mD 2ˆ =

mmxD 3≈→

HERA Dispersion Orbit