Eric Prebys, FNAL. In our earlier lectures, we found the general equations of motion We initially considered only the linear fields, but now we will

Floquet Transformations and

Harmonic ResonancesEric Prebys, FNAL

Non-linear Perturbations In our earlier lectures, we found the general equations of motion

We initially considered only the linear fields, but now we will bundle all additional terms into ΔB non-linear plus linear field errors

We see that if we keep the lowestorder term in ΔB, we have

USPAS, Knoxville, TN, Jan. 20-31, 2014Lecture 11 - Floquet Transformations and Resonances 2

1

1

2

2

2

x

B

By

xx

B

Bx

x

y

),('

),(0

syByBB

sxBxBBB

xx

yy

Δ

Δ

This part gave us the Hill’s equation

),(1

)(

),(1

)(

syBB

ysKy

sxBB

xsKx

x

y

Δ

Δ

Floquet Transformation Evaluating these perturbed equations can be very

complicated, so we will seek a transformation which will simplify things

Our general equation of Motion is

This looks quite a bit like a harmonic oscillator, so not surprisingly there is a transformation which looks exactly like harmonic oscillations


)(cos)()( ssAsx

111

ds

dds

xs

Plugging back into the Equation


2/32

22

2/3

2/1

1

1

111

2

1

x

ds

d

d

dx

x

So our differential equation becomes

B

BK

sKxsKx

Δ

2/32

222

2/12/32

22

)()(

We showed a few lectures back that

So our rather messy equation simplifies


1

0)(

)()()(

)(

22

3

2

K

sw

kswsKsw

sw

k

B

B

B

BK

Δ

Δ

2/322

2/32

222

Understanding Floquet Coordinates In the absence of nonlinear terms, our equation of motion

is simply that of a harmonic oscillator

and we write down the solution

Thus, motion is a circle in the plane Using our standard formalism, we can express this as

A common mistake is to view f as the phase angle of the oscillation. νϕ the phase angle of the oscillation φ advances by 2π in one revolution, so it’s related (but NOT equal to!) the angle

around the ring.


0)()( 2

sin)(

cos)(

a

a

,

1~

where;cossin~

1sin

~cos

)(

)(

cossin)(

sincos)(

0

0

00

00

222max

2max :Note aax

unnormalized!

Perturbations In general, resonant growth will occur if the perturbation has a

component at the same frequency as the unperturbed oscillation; that is if

We will expand our magnetic errors at one point in f as

But in general, bn is a function of ϕ, as is β, so we bundle all the dependence into harmonics of ϕ

So the equation associated with the nth driving term becomes


Δ

Δ

0

2/32

2

22

2/51

2/40

2/322/32

0

33

2210

...

!

1...;)(

n

nn

n

yx

n

n

n

bB

bbbBB

B

x

B

nbxbxbxbbxB

m

imnmn

n eCbB

,

2/31

k

imnnm eC ,

22Remember!ξ,β, and bn are all functions of (only) ϕ

resonance!(...)),( Δ iaeB

Calculating Driving Terms We can calculate the coefficients in the usual way with

But we generally know things as functions of s, so we useto get

Where (for a change) we have explicitly shown the s dependent terms.

We’re going to assume small perturbations, so we can approximate β with the solution to the homogeneous equation


2

0

2/3, 2

11deb

BC im

nn

nm

dsesbs

BC im

nn

nm

)()(

2

11 2/1,

dsdsds

ddd

111

)!(!

!

j

i where;

22

1cos

)0 sopoint starting (define ;cos)(

2

n

jij

iekn

naa

a

kin

knk

nnnn

Δ

Plugging this in, we can write the nth driving term as

We see that a resonance will occur whenever

Since m and k can have either sign, we can cover all possible combinations by writing


kmi

mnm

n

knk

n

eCknna

Δ

,

2

2

22

)2( where

1

Δ

knkn

m

mk

km

k

m

1resonant

Types of Resonances


Magnet Typen k Order

|1-k|Resonant tunes

ν=m/(1-k)Fractional Tune

at Instability

Dipole 0 0 1 m 0,1

Quadrupole1 1 0 none (tune shift) -

1 -1 2 m/2 0,1/2,1

Sextupole

2 1 1 m 0,1

2 0 1 m 0,1

2 -1 3 m/3 0,1/3,2/3,1

Octupole

3 3 2 m/2 0,1/2,1

3 1 0 None -

3 -1 2 m/2 0,1/2,1

3 -3 4 m/4 0,1/4,1/2,3/4,1

Effect of Periodicity If our ring is perfectly periodic (never quite true), with a period N,

then we can express our driving term as

Where we have invoked the periodicity as Clearly, if m is any integer multiple of N, then all values are 1.

Otherwise, the sum describes a closed path in the complex plane, which adds to zero, so

That is, we are only sensitive to terms where m is a multiple of the periodicity. This reduces the effect of the periodic non-linearities


1

0

22

0

2

0

22

2

0

22

0

2

0

,

)(

...)2

2()2

()()(

N

l

NlimN

im

NN

imNN

imNimim

mn

edef

deN

fdeN

fdefdefC

fN

f

2

jNm

jNmdefNCN

immn

if 0

if )(

2

0

,

Resonant Behavior Remember that our unperturbed motion is just

A resonance will modify the shape and size ofthis trajectory, so we replace a with a variable rand we can now express the postion in the rξ plane

We express r2 in terms of our variables and we have


sin)(

cos)(

a

a

1tan

sinsin)(

coscos)(

rr

rr

m

imnm

nn

m

imnnm eCreC

rd

d

r

,1

,

222

2

2

22

sincos22

222

Plug in nth driving term for this

Example: Third Order Resonance Sextupole terms (n=2) can drive a third order resonance

We will consider one value of |m| at a time

We’ll redefine things in terms of all real components by combining the positive and negative m values


m

imm

m

imm

eCrd

d

eCrd

dr

2,3

2,23

2

cos1

sincos2

dse

B

beC imim

m

22/3

2, 2

1

mBmA

dsmB

bmdsm

B

bm

dsmB

beCeC

mm

imm

imm

sincos1

sinsin1

coscos1

cos1

2,2,

22/322/3

22/32,2,

Evolution of Angular Variable We have

These are our general equations to evaluate the effects of particular types of field errors. Remember that our sensitivity to these errors is actually contained in the Cm,n coefficients


m

imnm

nn

m

imnm

n

m

imnm

n

eCr

r

eCeC

d

d

,11

22

,12

222

,12

22222

2

2

1

cos1

11

1

1

tan

2,

2

use

m

imnnm eC

So we have define real driving terms

So we plug this into the formulas

For unperturbed motion and we’re interested in behavior near the third order resonance, where n ~ m/3, so

All other terms will oscillate rapidly and not lead to resonant behavior


dsmB

bB

dsmB

bA

m

m

sin

cos

22/32,

22/32,

mmmmB

mmmmAr

mBmArd

dr

m

m

mm

3coscos3coscos

3sinsin3sinsin4

1

sincossincos2

2,

2,3

2,2,23

2

3133m

m

So we’re left with

The angular coordinate is given by

We perform yet another transformation to the (rotating) coordinate system

We then divide the two differentials to get the behavior of r2 in this plane


about caret don' weterms3sin3coscos8

1

sincoscos1

2,2,3

2,2,3

mBmAr

mBmArd

d

mm

mm

mBmArd

drmm 3cos 3sin

4

12,2,

32

3

~3

~

m

d

d

d

d

m

3

~ mNote: in an unperturbed system, this would just be

~3sin

~3coscos

81

3

~3cos

~3sin

41

~~2,2,

3

2,2,3

2

2

mm

mm

BArm

BAr

dd

ddr

d

dr

This equation can be integrated to yield

A and B are related to the angular distribution of the driving elements around the ring. We can always define our starting point so B=0, so let’s look at

a is an integration constant which is equal to the emittance in the absence of the resonance.

This is ugly, but let’s examine some general features


12

~3sin

~3cos

312

~3sin

~3cos 2,2,322,322 mmm BA

rrm

BArra

12

~3cos

2,322

mArra

A largefor solution no3

5,

2,

3

~3

4,

3

2,0

~6

11,

2

3,

6

7,

6

5,

2,

6

~

2max

2

2min

2

22

rr

rr

ar

The separatrix is defined by a triangle. We’d like to solve for the maximum a as a function of the driving term A. When a corresponds to the maximum bounded by the separatrix, we have that at


sepd

r

~

sepsepsepsep aadar2

3

6cos

6

~

Plug this in when the angle =0, and we have

max

2,

2,

2,3

3

2

2

2

3

824

31

3

4

12

~3cos

2

3

2

3

m

sep

sepm

msepsepsep

Aa

aA

Aaaa

In general

8

3

3

64

22,

22,

22,

22,

22

max

mm

mm

BA

BA

dsB

BB

dsB

BA

m

m

3sin2

3cos2

2/32,

2/32,

Behavior in Phase Space We convert back to our normal Floquet angle

So as we move around the ring, f advances andthe shape will rotate by an amount (m/3)f

Since m/3~ν is a non-integer, particles must alwaysmake three circuits (Δϕ=6π) before the shape completelyrotates at the origin.


123

3sin3

3cos 2,2,322

mB

mA

rramm

3

m

Application of Resonance If we increase the driving term (or move the tune closer to m/3),

then the area of the triangle will shrink, and particles which were inside the separatrix will now find themselves outside

These will stream out along the asymtotesat the corners.

These particles can be interceptedby an extraction channel Slow extraction Very common technique


Extraction Field

Septum

Unstable beam motion in N(order)

turns

Lost beam

Extracted beam

Hamiltonian Approach to Resonances


We now want to define a new coordinate which represents the “flutter” with respect to the average phase advance.

We define a new coordinate θ, such that

We want to transform to new variables θ and I. Try

unperturbed Hamiltonian

Third Order Resonances Revisited


In the x plane + a sextupole

sextupole moment

We have

We expand this in a Fourier series

The rest proceeds as before


Expand the cos3 terms and just keep the cos terms.

Looking at Hamilton’s Equations, we have

Examine near

Define a new variable


The part of the Hamiltonian which drives the resonance is

We now have the equations of motion

the fixed points are when the two are zero so

The rest proceeds in a similar fashion as before…

Documents

Eric Prebys, FNAL. In our earlier lectures, we found the general equations of motion We initially considered only the linear fields, but now we will