IntroductionIntroduction to to TransverseTransverse BeamBeam OpticsOptics IIII

Bernhard Holzer, DESY-HERA

Reminder: the ideal world

0s

x xM

x x

⎛ ⎞ ⎛ ⎞= ⋅⎜ ⎟ ⎜ ⎟′ ′⎝ ⎠ ⎝ ⎠

0

1cos( ) sin(

sin( ) cos( )

⎛ ⎞⎜ ⎟

= ⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠

foc

K s K sKM

K K s K s

●

z

x

ρ

s

θ

z

●

Beam parameters of a typicalhigh energy ring: Ip = 100 mAparticles per bunch: N ≈ 10 11

The Beta Function

... question: do we really have to calculate some 1011 single particle trajectories ?

Example: HERA Bunch pattern

( ) ( ) cos( ( ) )= ⋅ +x s s sε β ψ φ

0

( )( )

= ∫s ds

ss

ψβ

2 2( )* ( ) 2 ( ) ( ) ( ) ( ) ( )′ ′= + +s x s s x s x s s x sε γ α β

Beam Emittance and Phase Space Ellipse

xεβ

εαβ

−εγ

●

●

●

●

●

●

x´

Usually we get in a quadrupole ( ) 0sα =

x●

●

● ●

●

x´

Inside foc quadrupoles β reaches maximum� largest aperture needed

●

1.) Emittance ... so sorry .constε ≠

According to Hamiltonian mechanics: q = position = xphase space diagram relates the variables q and p p = momentum = mcγβ

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 β = v/c

xp dq const mc dx mc x dxγβ γβ ′= = =∫ ∫ ∫

1x dxε

βγ′⇒ = ∝∫

the beam emittance shrinksduring acceleration ε ~ 1 / γ

… the not so ideal world

Momentum error: 0pp

Δ ≠

2 2

2( )= + − =

+ zd mv

F m x eB vxdt

ρρ

01(1 )

eBx exgx

mv mvρ ρ′′ − − = +

0 0 0

1 1 1(1 )

p

mv p p p p

Δ= ≈ −+ Δ

neglecting higher order terms …

2

1 1( )

px x k

p ρρΔ′′ + − = ⋅

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

Force acting on the particles

Question: do you remember yesterday on page 11 … sure you do:

2.) Dispersion2.) Dispersion

2

1 1( )

px x k

p ρρΔ′′ + − = ⋅

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

pρΔ′′ + ⋅ = ⋅

Normalise with respect to Δp/p:

( )( ) i

pp

x sD s Δ=

Dispersion function D(s)

* is that special orbit, an ideal particle would have for Δp/p = 1

* the orbit of any particle is the sum of the well known xβ and the dispersion

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

. ρ

xβ

Closed orbit for Δp/p > 0

( ) ( )ip

x s D sp

Δ= ⋅

Matrix formalism:

( ) ( ) ( )p

x s x s D spβ

Δ= + ⋅

0 0( ) ( ) ( ) ( )p

x s C s x S s x D sp

Δ′= ⋅ + ⋅ + ⋅ 0s

x C S x Dp

x C S x Dp

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞Δ= +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟′ ′ ′ ′ ′⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

DispersionExample: uniform dipole field

00 0 1p p

p ps

x C S D x

x C S D xΔ Δ

⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟′ ′ ′ ′ ′= ⋅⎜ ⎟ ⎜ ⎟⎜ ⎟

⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

Example HERA

3

1...2

( ) 1...2

1 10

x mm

D s m

pp

β

−

=

≈Δ ≈ ⋅

Amplitude of Orbit oscillationcontribution due to Dispersion ≈ beam size

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

= −∫ ∫% % % %

Example: Drift

1

0 1Drift

lM

⎛ ⎞= ⎜ ⎟⎝ ⎠

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

cos sin

1sin cos

Dipole

l l

Ml l

ρρ ρ

ρ ρ ρ

⎛ ⎞⎜ ⎟⎜ ⎟= →⎜ ⎟−⎜ ⎟⎝ ⎠

( ) (1 cos )

( ) sin

lD s

lD s

ρρ

ρ

= ⋅ −

′ =

3.)3.) MomentumMomentum CompactionCompaction FactorFactor::

The dispersion function relates the momentum error of a particle to the horizontal orbit coordinate.

1( ) *

px K s x

pρΔ′′ + =

( ) ( ) ( )p

x s x s D spβ

Δ= +

xβ

inhomogeneous differential equation

general solution

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

ρ

ρ

dsx

dl

design orbit

particle trajectory

particle with a displacement x to the design orbit� path length dl ...

1( )

xdl ds

sρ⎛ ⎞

→ = +⎜ ⎟⎝ ⎠

dl x

ds

ρρ+=

1( )

EE

xl dl ds

sρΔ

Δ⎛ ⎞

= = +⎜ ⎟⎝ ⎠

∫ ∫

circumference of an off-energy closed orbit

remember:

( ) ( )Ep

x s D spΔ

Δ=

( )

( )Ep D s

l dsp s

δρΔ

⎛ ⎞Δ= ⎜ ⎟⎝ ⎠∫

* The lengthening of the orbit for off-momentumparticles is given by the dispersion functionand the bending radius.

o

o

o

cpl p

L pεδ α Δ=

1 ( )

( )cpD s

dsL s

αρ

⎛ ⎞→ = ⎜ ⎟

⎝ ⎠∫

For first estimates assume: 1

constρ

=

( ) dipoles dipoledipoles

D s ds l D= ⋅∫

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 Δ→ = =

Definition:

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

o

Question: what will happen, if you do not make toomany mistakes and your particle performsone complete turn ?

Tune and Quadrupoles

( )

( )

0 00

0 0 0

0

cos sin sin

( )cos (1 )sincos sin

⎛ ⎞+⎜ ⎟

⎜ ⎟= ⎜ ⎟− − +⎜ ⎟−⎜ ⎟⎝ ⎠

ss s s s

s s s ss s s

s

M

s

β ψ α ψ β β ψβ

α α ψ α α ψ β ψ α ψββ β

Transfer Matrix from point „0“ in the lattice to point „s“:

cos sin sin

sin cos sinturn turn turn

turnturn turn turn

C SM

C S

ψ α ψ β ψγ ψ ψ α ψ

+⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟′ ′ − −⎝ ⎠ ⎝ ⎠

Definition: phase advanceof the particle oscillationper revolution in units of 2πis called tune

2 2turnQ

ψ μπ π

Δ= =

x(s)

s

0

Matrix for one complete turn

the Twiss parameters are periodic in L:( ) ( )

( ) ( )

( ) ( )

s L s

s L s

s L s

β βα αγ γ

+ =+ =+ =

Quadrupole Error in the Lattice

optic perturbation described by thin lens quadrupole

00

cos sin sin1 0

sin cos sin1turn turn turn

dist kturn turn turn

M M Mkds

ψ α ψ β ψγ ψ ψ α ψΔ

+⎛ ⎞⎛ ⎞= ⋅ = ⋅⎜ ⎟⎜ ⎟ − −−Δ⎝ ⎠ ⎝ ⎠

cos sin sin

(cos sin ) sin sin cos sindistMkds kds

ψ α ψ β ψψ α ψ γ ψ β ψ ψ α ψ

+⎛ ⎞=⎜ ⎟Δ ⋅ + − −Δ ⋅ + −⎝ ⎠

0 0( ) 2cos 2cos sinTrace M kdsψ ψ β ψ= = +Δ

0ψ ψ ψ= +Δ

rule for getting the tune

Quadrupole error� Tune Shift

ideal storage ringquad error

00 0

sincos( ) cos

2

kdsβ ψψ ψ ψ Δ+Δ = +

00 0 0

sincos cos sin sin cos

2

kdsβ ψψ ψ ψ ψ ψ Δ⋅ Δ − ⋅ Δ = +

remember the old fashioned trigonometric stuff and assume that the error is small !!!

1≈ ψ≈Δ

2 Qψ π=

2

kdsβψ ΔΔ =

and referring to Q instead of ψ:

0

0

( ) ( )

4

s l

s

k s s dsQ

βπ

+ ΔΔ = ∫

0

0

( )

4 4

s lquad

s

klk sQ ds

ββπ π

+ ΔΔΔ = ≈∫

a quadrupol error leads to a shift of the tune:

! the tune shift is proportional to the β-function at the quadrupole!! field quality, power supply tolerances etc are much tighter at places where β is large!!! mini beta quads: β ≈ 1900

arc quads: β ≈ 80

!!!! β is a measure for the sensitivity of the beam

Example: measurement of β in a storage ring:

GI06 NR

y = -6.7863x + 0.3883

y = -3E-12x + 0.28140.2800

0.2850

0.2900

0.2950

0.3000

0.3050

0.01250 0.01300 0.01350 0.01400 0.01450

k*L

Qx,

Qy

tune spectrum ... tune shift as a function of a gadient change

ChromaticityChromaticity: : ξξ

Influence of external fields on the beam: prop. to magn. field & prop. zu 1/p

*B dl

pe

α = ∫dipole magnet

focusing lensg

kp

e

=

particle having ... to high energyto low energyideal energy

( )*D

px D s

p

Δ=

definition of chromaticity:

gk

pe

=

ChromaticityChromaticity: : ξξ

in case of a momentum spread:

00 0 0

(1 )e g e p

k g k kp p p p

⋅ Δ= ≈ − ⋅ = − Δ+ Δ

00

pk k

p

ΔΔ = ⋅

… which acts like a quadrupole error in the machine and leads to a tune spread:

00

1( )

4

pdQ k s ds

pβ

πΔ= ⋅ ⋅

0

pQ

pξ ΔΔ = ⋅

0p p p= + Δ

Problem: chromaticity is generated by the lattice itself !!

ξ is a number indicating the size of the tune spot in the working diagram, ξ is always created if the beam is focussed� it is determined by the focusing strength k of all quadrupoles

1: * ( ) ( )

4k s s dsξ β

π= ∫

k = quadrupole strengthβ = betafunction indicates the beam size … and even more the sensitivity of

the beam to external fields

Example: HERA

HERA-p: ξ = -70 … -80Δ p/p = 0.5 *10-3

Q = 0.257 … 0.337

�Some particles get very close to resonances and are lost

o

N

Sextupole Magnets:

CorrectionCorrection of of ξξ::

1.) sort the particles acording to their momentum ( ) ( )Dp

x s D sp

Δ=

2.) apply a magnetic field that rises quadratically with x (sextupole field)

xB gxz= %

2 21( )

2zB g x z= −%

linear rising„gradient“:

x zB Bgx

z x

∂ ∂= =∂ ∂

%

S

SN

corrected chromaticity:

{ }1( ) ( ) ( )

4k s mD s s dsξ β

π−= −∫

normalised quadrupole strength:

./sext sextgx

k m xp e

= =%

.sext sextp

k m Dp

Δ=

o

ˆ1*

4 Q

Nf

β βξπ

∨

−≈ −

ChromaticityChromaticity in in thethe FoDoFoDo LatticeLattice

1( )* ( )

4s k s dsξ β

π= − ∫

(1 sin )2ˆ

sin

Lμ

βμ

+=

(1 sin )2

sin

Lμ

βμ

∨ −=β-Function in a FoDo

(1 sin ) (1 sin )1 1 2 2* *4 sinQ

L LN

f

μ μ

ξπ μ

⎧ ⎫+ − −⎪ ⎪= − ⎨ ⎬

⎪ ⎪⎩ ⎭

x

β

yβ

2 s in1 1 24 s inQ

LN

f

μ

ξπ μ

= −

remember ...

sin 2sin cos2 2

x xx =

tan1 2*4 sin

2

CellQ

L

f

μ

ξ μπ= − putting ...

sin2 4 Q

L

f

μ =

1* tan

2Cell

μξπ

= −

contribution of one FoDo Cell to the chromaticity of the ring:

using some TLC transformations ... ξ can be expressed in a very simple form:

sin1 1 24 sin cos

2 2Q

LN

f

μ

ξ μ μπ= −

1( ) ( )

4k s s dsξ β

π−= ∫

question: main contribution to ξ in a lattice … ?

Chromaticity

interaction region

o

ResumeResume´:´:

1εβγ

∝

( ) ( ) ( )p

x s x s D spβ

Δ= +

cpl p

L pεδ α Δ=

2cp

DD

L R

πα ≈ ≈

0

0

( ) ( )

4

s l

s

k s s dsQ

βπ

+ ΔΔ = ∫

1: * ( ) ( )

4k s s dsξ β

π= ∫

beam emittance

dispersion orbit

momentum compaction

quadrupole error

chromaticity o

QuestionQuestion::

… after all the very exciting question:

why do our particles not obey gravity and just fall down in the storage ring ????

Answer will be discussed in the evening, having a good glass of red wine at the bar.