bernhard holzer, desy-hera
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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.
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