dampingrings-lecture4

8/18/2019 DampingRings-Lecture4

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Insertion Devices

Ina Reichel

Berkeley Lab

June 2012

Ina Reichel (Berkeley Lab) Insertion Devices June 2012 1 / 33


2/33

Overview

1. Undulators and wigglers

2. Impact on radiation damping and equillibrium beam sizes

3. Nonlinear Effects



3/33

Undulators and Wigglers

Periodic series of dipole magnets with period λw = 2πk w with gap g

Field is periodic along the beam axis (with B̃ being the peak field)

B y = B̃ sin(k w s ) (1)



4/33

Undulators and Wigglers (cont’d)

Electromagnetic wiggler at the ATF (left) and permanent magnetundulator at the ALS (right).



5/33

Trajectory in a wiggler

Assuming y = 0 and B x = 0 the equations of motion can be written

as

ẍ = −ṡ e me γ

B z (s ) (2)

s̈ = ẋ e

me γ B z (s ) (3)

This can be approximated to (using ẋ = v x x andṡ = v s = β c =const.):

ẍ =

−

β ceB w

me γ

cos k w s (4)

Using ẋ = x β c and ẍ = x β 2c 2 this becomes

x = − eB w me β c γ

cos k w s = − eB w me β c γ

cos

2π

s

λw

(5)



6/33

Trajectory in a wiggler (cont’d)

Integration yields (β = 1):

x (s ) = λw eB w

2πme γ c sin k w s (6)

x (s ) = λ2

w eB

w 4π2me γ c cos

k w s (7)

The maximum angle of the trajectory and the wiggler axis is given by

θw = x max =

1

γ

λw eB w

2πme c (8)

If θw ≤ 1γ the device is an undulator, otherwise it’s a wiggler



7/33

Wiggler contribution to energy loss

Increased energy loss from synchrotron radiation

Ideally integrated field over the lenth is zero, therefore can be insertedinto straight section without change to overall geometry

Total energy loss per turn in a storage ring is given by(C γ = 8.846 · 10−5 mGeV3 )

U 0 = C γ

2π E 40 I 2 with I 2 =

1

ρ2 ds (9)

Need to add wiggler contribution

I 2w =

Lw 0

1

ρ2 ds =

1

(B ρ)2

Lw 0

B 2 ds = 1

(B ρ)2B 2w Lw

2 (10)

I 2w does not depend on the wiggler period!



8/33


9/33

Wiggler contribution to the momentum compaction factor

The momentum compaction factor αC has an effect on otherparameters like the synchrotron tune.

αC = 1

C 0I 1 with I 1 =

D x

ρ ds (14)

In a FODO lattice αC can be approximated using the horizontal tune:

αC ≈ 1Q 2x

with Q x ≈ 12π

C 0

β x (15)

For the ILC damping rings without wigglers (C 0 = 6.7 km andβ x ≈ 25m) one gets αC ≈ 5 × 10−4



10/33

Wiggler contribution to the momentum compaction factor(cont’d)

Need dispersion in wiggler to calculate contribution to momentumcompaction factor.

In a dipole with bending radius ρ and quadrupole gradient k 1, thedispersion is given by

d 2D x

ds 2 + K D x = 1ρ with K = 1ρ2 + k

1 (16)

Assuming k 1 = 0 we get

d 2D x

ds 2 +

B 2w (B ρ)2

D x sin2 k w s =

B w

B ρ sin k w s (17)

Try D x ≈ D 0 sin k w s For k w ρw 1, one can neglect the second term on the left:

D x ≈ −sin k w s ρw k 2w

(18)

For the ILC damping wigglers k w ρw ≈ 160Ina Reichel (Berkeley Lab) Insertion Devices June 2012 10 / 33


11/33


Have assumed all contributions to dispersion are from bending in thewiggler. Things like misaligned quadrupole components etc. wouldadd additional contributions which we neglect here.

Dispersion in generated in the wiggler is small:

|D 0| ≈ 1ρw k 2w

with ρw = B ρ

B w (19)

For the ILC damping wiggler we get |D 0| ≈ 0.39mm compared toabout 10 cm in the dipoles.



12/33


Now lets calculate the wiggler contribution to I 1:

I 1w =

Lw 0

D x

ρ ds ≈ −

Lw 0

sin2 k w s

ρ2

w

k 2

w

ds = − Lw 2 ρ2

w

k 2

w

(20)

I 1w is negative as higher energy particles have a shorter path length inthe wiggler (which is the opposite from the path length in a storagering).

For the ILC damping wigglers (ρw k w ≈ 160 and Lw ≈ 210m) we getI 1w ≈ −0.004m which is small compared to I 1 ≈ 3.4m from thedipoles, so the contribution to the momentum compaction factor isnegligible.



13/33

Wiggler contribution to the natural energy spread

Natural energy spread (C q = 5532√

3

mc = 3.832× 10−13 m):

σ2δ = C q γ 2 I 3

J z I 2with I 3 =

1|ρ|3 (21)

I 3 does not depend on the dispersion, so the wiggler could possiblymake a large contribution to the energy spread

Bending radius in the wiggler:1

ρ =

B

B ρ =

B w

B ρ sin k w s =

1

ρw sin k w s (22)

This yields

I 3w = 1ρ3w

Lw 0

sin3 k w s ds = 4Lw 3πρ3w

(23)

For the ILC damping wigglers (Lw ≈ 210m, ρw ≈ 10.4m),I w 3 ≈ 0.079m−2 which is large compared to the dipole contribution(5.1 × 10−

4

m−2

).Ina Reichel (Berkeley Lab) Insertion Devices June 2012 13 / 33


14/33

Wiggler contribution to the natural energy spread (cont’d)

In the ILC damping rings, the damping wiggler contribution to I 2 andI 3 is large compared to the contribution of the dipoles. Therefore theenergy spread is largely determined by the wiggler:

σ

2

δ ≈ 4

3πC q

γ 2

ρw =

4

3π

e

mc C q γ B w (24)

For a damping ring, the energy spread of the extracted beam is animportant parameter: The larger it is, the more difficult thedownstream bunch compressors are to design

With a beam energy of 5 GeV and a wiggler field of 1.6 T, the naturalenergy spread is about 0.13%. This is acceptable (upper limit isaround 0.15%).



15/33

Wiggler contribution to the natural emittance

The natural emittance depends on I 2 and I 5

(C q =

55

32√ 3

mc = 3.832× 10−13m

):

ε0 = C q γ 2 I 5

J x I 2with I 2 =

1

ρ2ds and

I 5

= Hx

|ρ|3ds with

Hx = γ

x D 2

x + 2α

x D x

D px

+ β x

D 2

px (25)

The contribution of the wiggler to I 5 depends on the β -function in thewiggler. We assume αx ≈ 0. Then we get

D px ≈

dD x

ds = k w D 0 cosk w s (26)

Assuming k w 1β x we can approximate

Hx ≈ β x ρ2w k

2w

cos2 k w s (27)



16/33

Wiggler contribution to the natural emittance (cont’d)

With this we can write the wiggler contribution to I 5 as

I 5w ≈ β x ρ2w k

2w

Lw 0

cos2 k w s

|ρ|3 ds = β x ρ5w k

2w

Lw 0

| sin3 k w s | cos2 k w s ds (28)

Using| sin3 x |cos 2x = 415π we have

I 5w ≈ 415π

β x Lw ρ5w k

2w

(29)

Assuming β x ≈ 10m and the usual wiggler parameters(k w ≈ 15.7m−1), we get I 5w ≈ 5.9 × 10−6 m−1.



17/33


Lets see how this compares to the contribution from the dipoles(assuming TME lattice tuned for minimum dispersion; θ is the bendingangle in the dipoles, ρ the bending radius):

I 5 = π5√

15θ

3

ρ (30)

Assuming 120 dipoles with a field of 0.15 T and 5 GeV one getsI 5D ≈ 1.7 × 10−7 m−1, so the wiggler contribution dominates.However often a less than ideal

TME

lattice is used where the wigglercontribution can be significant. In other lattices, like FODO, the dipolecontribution can dominate.



18/33


Combining I 2w and I 5w we get for the natural emittance

0 ≈ 815π

C q γ 2 β x ρ3w k

2w

(31)

Using the usual parameters this yields 0 ≈ 0.22nm. If the dipole contribution is comparable to the wiggler contribution,

the natural emittance will be larger than this by about a factor of two. The wiggler contribution can be reduced by

reducing the horizontal β -function reducing the wiggler period, i.e. increasing k w reducing the wiggler field, i.e. increasing ρw



19/33

Dynamical effects of wigglers

Aside from the effects on the natural emittance and the energyspread, the wigglers have two other effects:

1. provide linear focusing which must be included in the lattice design2. non-linear field components that can affect particles at large amplitude

and thus can limit the dynamic aperture



20/33

3D field in an ideal wiggler

If the poles are infinitely wide, the horizontal field componentvanishes:

B x = 0 (32)

B y = B w sin k z z cosh k z y (33)B z = B w cos k z z sinh k z y (34)

As B z is non-zero for a vertical offset and the particle has a horizontalvelocity thanks to the wiggler field, a particle with a horizontal offset

will experience a vertical deflecting force which leads to verticalfocusing in the wiggler.



21/33

Vertical focusing in a wiggler

For simplicity we will assume that the trajectory of a particle is

determined by the vertical field component of the wiggler. Otherforces, e.g. vertical deflections, will be trated as perturbations.

The horizontal equation of motion on the mid-plane is given by

d 2x

ds 2 = B y

B ρ = B w

B ρ sin k z s cosh k z y (35)

with the solution

x =−

B w

B ρ

1

k 2z sin k z s cosh k z y (36)

and

p x = −B w B ρ

1

k z cos k z s cosh k z y (37)



22/33

Vertical focusing in a wiggler (cont’d)

The vertical equation of motion is

dp y ds

= q p 0

p x B z = B w

B ρ p x cos k z s sinh k z y (38)

The total deflection per period is

∆p y ≈ B w B ρ sinh k z y λw

0

p x cos k z s ds (39)

Using p x as from above we find

∆p y ≈ −

B w B ρ

21

k z sinh k z y cosh k z y

λw 0

cos2 k z s ds (40)

= − π2k 2z

B w

B ρ

2sinh2k z y (41)



23/33

Vertical focusing in a wiggler (cont’d)

Series expansion in y yields

∆p y ≈ − πk z

B w

B ρ

2 y +

2

3k 2z y

3 + ...

(42)

Taking only the term linear in y into account, per wiggler period this

is equivalent to a vertically focusing quadrupole with the integratedstrength

k 1l = − πk z

B w

B ρ

2(43)

The cubic term contributes

∆p (3)y ≈ −2π

3

B w

B ρ

2k z y

3 (44)

which is often referred to as the “dynamic octupole” term.



24/33

Horizontal focusing in a wiggler

The finite width of the magnet poles leads to a decrease of the field

strength for large horizontal offsets. In a simple model the field can be written as

B x = −k x k y

B w sin k x x sinh k y y sin k z z (45)

B y = B w cos k x x cosh k y y sin k z z (46)B z =

k z

k y B w cos k x x sinh k y y cos k z z (47)

with the condition (from Maxwell’s equations)

k 2x + k 2z = k 2y (48)

A particle with a horizontal offset sees a weaker field in one set of poles and a stronger field in the other. The net effect is a horizontaldeflection that appears as a horizontal defocusing force.



25/33

Horizontal focusing in a wiggler (cont’d)

Consider a particle with the trajectory (x 0 is the inital horizontal

offset)x = x 0 + x̂ sin k z s with x̂ =

1

k 2z

B w

B ρ (49)

Assuming y = 0 the horizontal kick per period is

∆p x = − 1B ρ

λw 0

B y ds (50)

= −B w B ρ

λw

0

cos[k x (x 0 + x̂ sin k z s )] sin k z s ds (51)

= B w

B ρ λw J 1 (k x x̂ ) sin(k x x 0) (52)

≈ B w B ρ

λw k x x̂

2 k x x 0 (53)

(54)Ina Reichel (Berkeley Lab) Insertion Devices June 2012 25 / 33


26/33

Horizontal focusing in a wiggler (cont’d)

The horizontal focusing can be written as

∆p x ≈ λw 2

B w

B ρ

2k 2x k 2z

x 0 (55)

Compare to the vertical focusing

∆p y ≈ −λw 2

B w

B ρ

2 k 2y k 2z

y 0 (56)

For infinitely wide poles, k x → 0 which means there is no horizontalfocusing. In this case one also gets k y = k z . For finite horizontalpoles, the vertical focusing is enhanced due to k 2y = k

2x + k

2z .



27/33

Nonlinear effects in wigglers

At the center of a pole (sin k z s = 1) and with y = 0 the vertical field

is given by

B y = B w cos k x x = B w

1 − 1

2k 2x x

2 + 1

24k 4x x

4 − ...

(57)

The quadratic term leads to horizontal defocusing, the sextupole field“feeds down” (when combined with the wiggling trajectory) to give alinear focusing effect.

In the same manner, the decapole component feeds down to give aoctupole component. So for finite pole width, we have a “dynamic

octupole” in both planes. Wigglers can have a significant impact on the non-linear dynamics.

They can potentially restrict the dynamic aperture. Therefore it isimportant to have a good model for analysing the nonlinear effects.



28/33

Modelling the nonlinear effects of wigglers

Modelling the nonlinear effects of wigglers is done in four steps:1. Magnetostatic codes (e.g. Tosca, Radia) are used to calculate the

magnetic field in one period.

2. An analytical model for the field (a mode decomposition) is fitted tothe field obtained in step 1.

3. The analytical model is then used to create a dynamical map whichdescribed the motion of a particle through the wiggler. This is donewith codes like MaryLie or COSY.

4. The dynamical map is then used in a tracking code to determine the

impact of the wiggler on non-linear dynamics, e.g. tune shifts,resonances or dynamic aperture.

We will have a brief look at some of the steps.


M d lli h li ff f i l 2


29/33

Modelling the nonlinear effects of wigglers, step 2

Generalise the representation used so far to inlude a series of wigglermodes:

B x = −B w m,n

c m,nmk x

k y ,mnsin mk x x sinh k y ,mny sin nk z z (58)

B y = B w m,n

c m,n cos mk x x coshk y ,mny sin nk z z (59)

B z = B w m,n

c m,nnk z

k y ,mncos mk x x sinh k y ,mny cos nk z z (60)

k 2y ,mn = m2k 2x + n

2k 2z (61)


M d lli h li ff f i l 2 ( ’d)


30/33

Modelling the nonlinear effects of wigglers, step 2 (cont’d)

From the vertical field on the mid-plane (y = 0)

B y = B w m,n

c m,n cos mk x x sin nk z z (62)

one can in principle determine the coefficients c m,n by using a 2D Fouriertransform of the field data from step 1. In practice this does not work well.The hyperbolic dependence of the field on y means that any small errorsfrom the fit increase exponentially away from the mid-plane.A better technique is to fit the field on a surface enclosing the region of

interest. The hyperbolic dependence of the field means that in this caseany small errors actually decrease exponentially towards the axis of thewiggler.


M d lli h li ff f i l 2 ( ’d)


31/33

Modelling the nonlinear effects of wigglers, step 2 (cont’d)

Using a cylindrical surface within the wiggler aperture and standard

cylindrical coordinates, we get:B ρ =

m,n

αm,nI m(nk z ρ) sin mφ sin nk z z (63)

B φ = m,n

αm,nm

nk z ρ I m(nk z ρ) cos mφ sin nk z z (64)

B z =m,n

αm,nI m(nk z ρ) sin mφ cos nk z z (65)

If we know the radial field component B ρ at a fixed radius, we can

obtain the mode coefficients αm,n by a 2D Fourier transform. Usually done as close to the poles as data quality allows. Number of modes required depends on shape of field. Once αm,n are known, we can construct the field components

everywhere. The errors are small within the cylindrical surface.


M d lli th li ff t f i l t 3


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With the mode decomposition of the field, we can now use an“algebraic” code to construct a dynamical map.

An “algebraic” code manipulates algebraic expressions instead of numbers. There are different types of algebraic codes: Differential algebra codes like COSY Lie algebra codes like MaryLie

A differential algebra code can manipulate Taylor series. Byincorporating an integrator to solve the equations of motion into adifferential algebra code, we can construct a Taylor map representingthe dynamics of a particle in the given field.


M d lli th li ff ts f i l s st 4


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The map constructed in step 3 now needs to be included in a trackingcode to look at the impact on beam dynamics

Could just track a set of particles at varying amplitudes to “measure”the dynamic aperture, however a frequency map analysis yields muchmore informtion


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