USPAS Lecture 26 - Cornell Universitydugan/USPAS/Lect26.pdf · · 2001-12-03yy y p 1 2 0 10 0 01...

12/3/01 USPAS Lecture 26 1

LECTURE 26Collective instabilities;

Rigid beam transverse multibunch instability

The macroparticle model used in the previous lecture can beapplied to the important case of multiple bunches in a common

vacuum chamber. Long-range wakefields will couple themotion of the bunches together and can lead to tune shifts and

instabilities.

As we saw above, the wake fields generated by themacroparticle can be expressed in terms of a transverse

integrated force exerted at the location of the impedance.rF t ieI t mr r m m Zm

mm⊥

− ⊥= −( ) ( )( ) ( ) ˆ cos ˆ sin1 φ φ φ ω


For m=1, and in the vertical direction, we have

F t ieI t Zy( ) ( )= ( )⊥1 1 ω

To use the above equation, we need to know the Fourierspectrum of the dipole moment of the current. As discussed in

Lecture 25, the wake force is

F tiNe

Ty i p Q t Z p Q

y i p Q t Z p Q

y y yp

y y

( ) ˜ exp

˜ exp*

= − +( )( ) +( )( )+ − −( )( ) −( )( )

⊥

=−∞

∞

⊥

∑2

00 0 1 0

0 0 1 0

2ω ω

ω ω

in which y y i yy= + ′β . Using the symmetry property

Z Z1 1⊥ ⊥= − −( ) ( )*ω ω .


The integrated force, summed over all harmonics, can bewritten as

F tiNe

Ty i p Q t Z p Q c cy y y

p

( ) ˜ exp . .= − +( )( ) +( )( ) +⊥

=−∞

∞

∑2

00 0 1 02

ω ω

c.c represents the complex conjugate-dropped for now, addedback in equation of motion

This is the integrated force due to a single macroparticle.Suppose now that we have 2 bunches (macroparticles), of equalcharge. We’ll label the first bunch 0, and the second (trailing)

bunch 1. The wake force due to bunch 0 can be written as

F tiNe

Ty t ip t Z p Qy y

p0

2

00 0 1 02

( ) = ( ) −( ) +( )( )⊥

=−∞

∞

∑˜ exp ω ω


in which ˜ ˜ expy t y iQ ty( ) = −( )0 0ω

Suppose that bunch 1 trails bunch 0 by the time interval t t= 01.

t

0 1

t0 1(n+1)T0

0 1

t0 1

nT0

Since it arrives at the impedance at t nT t= +0 01, its current isgiven by

I tNeT

ip t tp

00

0 01( ) exp= − −( )( )=−∞

∞

∑ ω


and its betatron oscillation can be written as

˜ ˜ expy t y iQ t ty1 10 0 01( ) = − −( )( )ω

so the force created by its wake is given by

F tiNe

Ty iQ t t

ip t t Z p Q

y y

yp

1

2

010 0 01

0 01 1 0

2( ) = − −( )( ) ×

− −( )( ) +( )( )⊥

=−∞

∞

∑

˜ exp

exp

ω

ω ω

Bunch 0 arrives at the impedance at time t T T= −... , , ,...0 00 andfeels the total wake force


F F nT F nT

iNeT

y i Q n Z p Q

iNeT

y i Q ntT

ip t Z p Q

y n y y

y yp

y yp

0 0 0 1 0

2

000 1 0

2

001

01

00 01 1 0

22

22

,

˜ exp

˜ exp exp

= ( ) + ( ) =

= −( ) +( )( ) +

− −

( ) +( )( )

⊥

=−∞

∞

⊥

=

∑π ω

π ω ω−∞−∞

∞

∑

Let us defineˆ ( ) ˜ exp ˜ ( )

ˆ ( ) ˜ exp ˜ ( )

y n y iQ n y n

y n y iQ ntT

y nT

y

y

0 00 0

1 0101

01 0

2

2

= −( ) =

= − −

=

π

π


ˆ , ˆy n y n0 1( ) ( ) are the y variables of bunch 0,1 when bunch 0crosses the location of the impedance. These are sometimes

called the “snapshot” position of the bunch. ˆ , ˆy n y n0 1( ) ( )describe the bunch displacements and slopes, not at the same

location, but at the same time at different locations (the locationof the impedance, for bunch 0; a distance ct01 behind bunch 0,

for bunch 1).Then we have

FiNe

T

y n Z p Q

y n iptT

Z p Qy n

yp

yp

0

2

0

0 1 0

101

01 0

22

,

ˆ ( )

ˆ ( ) exp

=

+( )( ) +

+( )( )

⊥

=−∞

∞

⊥

=−∞

∞

∑

∑

ω

π ω


We now insert this into the betatron equation of motion. Theunperturbed betatron equation for the 0th bunch, written in

terms of turn number, has the form

dydn

iQ yy

ˆˆ 0

02= − π

The effect of the integrated force is to produce a change in y0

given by ∆ ∆ˆ , ,y i y iF

pvi

F

m cy yy n

yy n

0 00 0

02= ′ = =β β βγ

. Hence the

equation of motion becomes


dydn

iQ y iiNe

m c Ty A y A y B y B

A Z p Q

B iptT

Z p Q

y y

yp

yp

ˆˆ ˆ ˆ ˆ ˆ

exp

* * * *00

2

02

00 0 1 1

1 0

01

01 0

22

2

= − + − + −( )

= +( )( )

=

+( )( )

⊥

=−∞

∞

⊥

=−∞

∞

∑

∑

π βγ

ω

π ω

Now let us consider the motion of bunch 1. Since bunch 1 trailsbunch 0, it crosses the impedance at the time nT t0 01+ and feels

the force


F F nT t F nT t

iNeT

y n i QtT

ip t Z p Q

iNeT

y n Z p Q

iNe

y n y y

y yp

yp

1 0 0 01 1 0 01

2

00

01

00 01 1 0

2

01 1 0

2

22

2

,

˜ exp exp

˜

= +( ) + +( ) =

( ) −

−( ) +( )( ) +

( ) +( )( )

=

⊥

=−∞

∞

⊥

=−∞

∞

∑

∑

π ω ω

ω

222

2

0

01

00 1

01

01 0

Ti Q

tT

y B y A

B i ptT

Z p Q

y

yp

exp ˆ ˆ

exp

−

′ +( )

′ = −

+( )( )⊥

=−∞

∞

∑

π

π ω


We now insert this into the betatron equation of motion forbunch 1. The unperturbed betatron equation for bunch 1,

written in terms of turn number, has the form

dydn

iQ yy

ˆˆ 1

12= − π

The effect of the integrated force is to produce a change in y1

given by

∆ ∆ˆ exp exp,y i y iQtT

iF

m ciQ

tTy y y

y ny1 1

01

0

1

02

01

0

2 2= ′

=

β π βγ

π . Hence

the equation of motion becomes

dydn

iQ y iiNe

m c Ty A y A y B y By y

ˆˆ ˆ ˆ ˆ ˆ* * * *1

1

2

02

01 1 0 02

2= − + − + ′ − ′( )

π βγ


This, and the equation for bunch 0, are a set of coupleddifferential equations, for the 2 bunches. We can rewrite these

equations as

dydn

iQ y y y y y

dydn

iQ y y y y y

y A A B B

y A A B B

ˆˆ ˆ ˆ ˆ ˆ

ˆˆ ˆ ˆ ˆ ˆ

* * * *

* * * *

00 0 0 1 1

11 1 1 0 0

2

2

= − − + − +

= − − + − +′ ′

π

π

Γ Γ Γ Γ

Γ Γ Γ Γ

in which

Γ Γ ΓAy

By

ByANe

m c T

BNe

m c T

B Ne

m c T= = =

′′

2

02

0

2

02

0

2

02

02 2 2

βγ

βγ

βγ

We will treat the wake effects as a small perturbation: that is,we assume that


Γ2

1πQy

<< , and take the motion of the two bunches to have the

form

ˆ ˆ exp, ,y n y i n0 1 00 01( ) = −( )Ω

with Ω = + <<2 1π δ δQy ,

In this case, the complex conjugate terms in the aboveequations have the approximate forms

ˆ ˆ exp ˆˆ

ˆexp,

*,

*,

,*

,

y n y i Q n y ny

yi Q ny y0 1 00 01 0 1

00 01

00 01

2 4( ) ≈ ( ) ≈ ( ) ( )

π π


For Γ

21

πQy

<< , these rapidly oscillating terms may be omitted

from the equations, which then simplify to the set of coupledequations

dydn

iQ y y y

dydn

iQ y y y

y A B

y A B

ˆˆ ˆ ˆ

ˆˆ ˆ ˆ

00 0 1

11 1 0

2

2

= − − −

= − − − ′

π

π

Γ Γ

Γ Γ

or, in matrix form,

dydn

yiQ

iQy A B

B y A

ˆˆ,

rr′ = =

− − −− − −

′

M M2

2

ππ

Γ ΓΓ Γ


There will be a set of normal modes ζ m, for which theequations of motion decouple:

ˆr ry = Sζ

The normal mode eqations are

S MS S MSddn

ddn

′ = ′ = =−r

rr

r rζ ζ ζ ζ ζ 1 Λ

The matrix Λ =

λλ

0

1

0

0

in which λ0 and λ1 are the eigenvalues of the matrix M. For the

matrix given above, the eigenvalues are

λ πi y A B BiQ= − − ± ′2 Γ Γ Γ


The eigenvectors in the ( ˆ , ˆy y0 1) basis are

r rζ ζ1 2

12 1

12 1

= ′

= −

′

BB

BB,

We have, for each normal mode, the equation

ddn

ii i

′ =ζ λζ

Assuming a solution of the form

ζ ζi i in in( ) ˜ exp= −( )0 Ω . Using

ddn

iii i i i

′ = − =ζ ζ λζΩ .

The normal mode frequencies are


Ω Γ Γ Γi i y A B Bi Q i i= = − ± ′λ π2

Using the definitions of Γ and A, B from above, these become


Ω Γ Γ Γi y A B B

y

y

yp

py

Q i

i Ne

m c TA BB

i Ne

m c T

Z p Q

i p ptT

Z p Q Z

− = − ( )

= − ± ′( )

= − ×

+( )( ) ±

− ′( )

+( )( )

′

⊥

=−∞

∞

′ =−∞

∞⊥

∑

∑

2

2

2

2

2

02

0

2

02

0

1 0

01

01 0

π

βγ

βγ

ω

π ω

m

exp 11 0⊥

=−∞

∞

′ +( )( )

∑ p Qyp

ω

Consider the special case when tT

010

2= . Then


exp exp2 1 101

0π πi p p

tT

i p p p p− ′( )

= − ′( )( ) = −( ) −( ) ′

Ωi yy p

yp

Qi Ne

m c TZ p Q− = − ± −( )( ) +( )( )

⊥

=−∞

∞

∑22

1 12

02

01 0π

βγ

ω

The eigenvectors are

r rζ ζ

ζ ζ

0 1

0 0 1 1 0 1

12

1

112

1

1=

=−

⇒

= + = − +

,

ˆ ˆ , ˆ ˆ y y y y

In the sum mode, both bunches oscillate in phase; in thedifference mode, the two bunches oscillate out of phase.


The general case

Let there be M bunches in the machine, with the labelsy y yM0 1 1, ,..., − . Let the time separation between the bunches be

as shown below

mkt

0 1

t 0k

nT0

2 k m M-1

t

t0m

Following from above, the force due to the mth bunch, is givenby


F tiNe

Ty iQ t t

ip t t Z p Q

ym m y m

m yp

( ) ˜ exp

exp

= − −( )( ) ×

− −( )( ) +( )( )⊥

=−∞

∞

∑

2

00 0 0

0 0 1 0

22π ω

ω ω

The force on the kth bunch due to the mth bunch is

F n F nT t

iNeT

y iQ nt t

T

i ptT

Z p Q

yk m ym k

m ym k

mky

p

, ( ) ( )

˜ exp

exp

= + =

− − −

×

+( )( )⊥

=−∞

∞

∑

0 0

2

00

0 0

0

01 0

22

2

π

π ω

in which


t t tmk m k= −0 0

Using ˆ ( ) ˜ expy n y iQ ntTm m y

m= − −

2 0

0

π , the force on the kth

bunch is

F niNe

Ty n iQ

tT

i ptT

Z p Q

yk m m yk

mky

p

, ( ) ˜ exp

exp

= ( ) −

×

+( )( )⊥

=−∞

∞

∑

2

0

0

0

01 0

22

2

π

π ω

The total force on the kth bunch is


F niNe

TiQ

tT

y n

i ptT

Z p Q

yk yk

mm

M

mky

p

( ) exp ˜

exp

= −

× ( )

+( )( )=

−

⊥

=−∞

∞

∑

∑

2

0

0

0 0

1

01 0

22

2

π

π ω

We now insert this into the betatron equation of motion forbunch k. The unperturbed betatron equation for bunch k, written

in terms of turn number, has the form

dydn

iQ yky k

ˆˆ = −2π

The effect of the integrated force is to produce a change in yk

given by


∆ ∆ˆ exp expy i y iQtT

iF n

m ciQ

tTk y k y

ky

yky

k= ′

=( )

β π βγ

π2 20

0 02

0

0

. Hence

the equation of motion becomes

dydn

iQ y

N e

m c Ty n

i ptT

Z p Q

ky k

ym

m

M

mky

p

ˆˆ

˜

exp

= − −× ( )

+( )( )

=

−

⊥

=−∞

∞

∑

∑2

2

2

2

02

0 0

1

01 0

π

βγ

π ω

This is a set of M coupled differential equations for the Mbunches. In matrix form, it can be written as


dydn

y

iQN e

m c Ti p

tT

Z p Qkm y kmy mk

yp

ˆˆ,

exp

rr

=

= − −

+( )( )⊥

=−∞

∞

∑

M

M 22

22

02

0 01 0π δ

βγ

π ω

There will be a set of M normal modes ζ m, for which theequations of motion decouple:

r ry = Sζ

S MS S MSddn

ddn

′ = ′ = =−r

rr

r rζ ζ ζ ζ ζ 1 Λ

The matrix Λ ij ij i= δ λ contains the eigenvalues λi of the matrix

M.


As in the two-bunch case, the normal mode frequencies aregiven by the eigenvalues of M:

Ωi ii= λ

For any bunch spacing and impedance, the matrix given abovemay be diagonalized numerically and the normal mode

frequencies obtained. However, a general analytical solution forthe normal mode frequencies for M bunches is only possible in

special cases.

For example, suppose that the M bunches are uniformlydistributed around the ring.


Then, we can write

tm

MTm0 0

1= −( )

and

Mkm y kmy

yp

iQN e

m c Ti p

m kM

Z p Q= − − −

+( )( )⊥

=−∞

∞

∑22

22

02

01 0π δ

βγ

π ωexp

By analogy with the 2-bunch case, the matrix which gives thenormal modes has the form

SM

iabMab =

1 2exp

π

and the eigenvalue matrix is then


Λab akm

M

k

M

km mb

y aby

yp

k

M

m

M

S M S

iQN e

m c TZ p Q

MiamM

i pm k

Mikb

M

=

= − − +( )( ) ×

−

−

−

=

−

=

−

⊥

=−∞

∞

=

−

=

−

∑∑

∑

∑∑

1

0

1

0

1

2

02

01 0

0

1

0

1

22

1 22

2

π δβ

γω

π π πexp exp exp

Using the identity

exp , 20

1

π δimi jM

Mm

M

i j rM

−

=

=

−

−∑where r is any integer, we find


1 22

2

12 2

2

0

1

0

1

0

1

0

1

MiamM

i pm k

Mikb

M

Mim

p aM

ikb p

M

imb a

M

k

M

m

M

k

M

m

M

p b rM

exp exp exp

exp exp

exp,

−

−

= −

−

= −

=

−

=

−

=

−

=

−

+

∑∑

∑∑

π π π

π π

δ πmm

M

p b rM b aM=

−

+∑ =0

1

δ δ, ,

so the eigenvalues are

λ πβ

γωm y

yy

r

iQNM e

m c TZ rM m Q= − − + +( )( )⊥

=−∞

∞

∑22

2

02

01 0


and the normal mode frequencies are

Ωm yy

yr

QiNM e

m c TZ rM m Q− = − + +( )( )⊥

=−∞

∞

∑22

2

02

01 0π

βγ

ω

The tune shift and damping rate for mode m are related toΩm yQ− 2π by

Ω ∆m y m mQ Q i− = −2 2π π α

so the tune shift is

∆QNMe

m c TZ rM m Qm

yy

r

= + +( )( )

⊥

=−∞

∞

∑βπ γ

ω2

02

01 04

Im

and the damping rate is


αβ

γωm

yy

r

NMe

m c TZ rM m Q= + +( )( )

⊥

=−∞

∞

∑2

02

01 02

Re

The eigenmodes are

ζ πb ba

a

M

a aa

M

S yM

iabM

y= = −

−

=

−

=

−

∑ ∑1

0

1

0

11 2ˆ exp ˆ

The damping rate (or instability growth rate, if it is negative)for the multibunch instability is proportional to the total numberof bunches, that is, the total current. The impedance is sampledat frequencies spaced by Mω0, rather than ω0, as in the single

bunch case. If the frequency structure of the impedance is muchbroader than Mω0, then the sparse sampling roughly cancelsthe factor of M in front, and the damping or growth rates are

roughly the same for multiple bunches as for one bunch.


(This is because the wakefields for a broadband impedance areshort range, and do not couple the bunches together).

But if the impedance is narrow-band compared to Mω0 (long-range wakefield), then the bunches are strongly coupled and themultibunch growth rates can be M times larger than for a single

bunch.

Example: the transverse resistive wall instability. Theimpedance is (Lecture 19, p 23)

Z Ci

b

c1 3

212

⊥ = − ( )( )

sgnω ωωπ

ω µσ

The impedance enters the damping rate in the form


Re Z pM m QCb

c pM m Q

pM m Qyp

y

yp1 0 3

2

02⊥

=−∞

∞

=−∞

∞

+ +( )( )[ ] =+ +

+ +∑ ∑ωπ

µω σ

The multibunch mode which is most strongly driven will be theone for which the denominator is the smallest. The denominator

is pM m n+ + + ∆β, in which n is the integral part of the tune.Consider, for example, the Tevatron Collider, with M=36

bunches, and an integral tune of n=19. The denominator will be36 19p m+ + + ∆β, which is just ∆β for p=-1 if the mode number

is m=17. Thus, the mode m=17 will be the dominantmultibunch mode. The snapshot mode pattern for m=17,

ˆ expyia

aa

M

=

=

−

∑16

17180

1 π


is shown below:

5 10 15 20 25 30 35

-0.15

-0.1

-0.05

0.05

0.1

0.15

This is a low frequency oscillation, which can be easily dampedwith a narrow band feedback system.


The damping rate per turn is

αβ

π γµ

ω σ β= ( )M Ne c

b m cfy

2 2

30

202

∆

in which f ∆β( ) is the function defined in Lecture 25. Taking

the fractional tune to be ∆β=-0.4, and with other parameters for

the Tevatron as follows:

βy=100 m, N=1011, b=2.5 cm, γ=103, T0=21 µs,

σ=3.5x107 Ω-1m-1 (aluminum), we find a damping time ofT0 3 2α

= − . s. (a weak instability). This is a gross overestimate,

in fact, since most of the Tevatron vacuum chamber is cold, andthe resistance is therefore much less than assumed above.


USPAS Lecture 26 - Cornell Universitydugan/USPAS/Lect26.pdf · · 2001-12-03yy y p 1 2 0 10 0 01...

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