eric prebys, fnal. we will tackle accelerator physics the way we tackle most problems in classical...

Transverse Motion 1Eric Prebys, FNAL

The Journey Begins… We will tackle accelerator physics the way we

tackle most problems in classical physics – ie, with 18th and 19th century mathematics! Calculate ideal equilibrium trajectory Use linear approximations for deviations from this

trajectory Solve for motion Treat everything else as a perturbation to this

As we discussed in our last lecture, the linear term in the expansion of the magnetic field is associated with the quadrupole, so let’s start there…

USPAS, Knoxville, TN, Jan. 20-31, 2014Lecture 3 - Transverse Motion 1 2

Quadrupole Magnets*

A positive particle coming out of the page off center in the horizontal plane will experience a restoring kick

xB

y

yB

x

)()(

)(

B

lxB

B

lxBx

lB

Bf

'

)(

*or quadrupole term in a gradient magnetUSPAS, Knoxville, TN, Jan. 20-31, 2014 3Lecture 3 - Transverse Motion 1

What about the other plane?

pairs give net focusing in both planes -> “FODO cell”

xB

y

lB

Bf

'

)(

Defocusing!

Luckily, if we place equal and opposite pairs of lenses, there will be a net focusing regardless of the order.

USPAS, Knoxville, TN, Jan. 20-31, 2014 4Lecture 3 - Transverse Motion 1

Transfer matrices

The simplest magnetic lattice consists of quadrupoles and the spaces in between them (drifts). We can express each of these as a linear operation in phase space.

By combining these elements, we can represent an arbitrarily complex ring or line as the product of matrices.

)0('

)0(1

101

')0(1

)0(''

)0(

x

x

fx

x

xf

xx

xx

)0('

)0(

10

1

)('

)(

)0(')('

)0(')0()(

x

xs

sx

sx

xsx

sxxsx

Quadrupole:

s

x

Drift:

12... MMMM NUSPAS, Knoxville, TN, Jan. 20-31, 2014 5Lecture 3 - Transverse Motion 1

Example: FODO cell

At the heart of every beam line or ring is the “FODO” cell, consisting of a focusing and a defocusing element, separated by drifts:

The transfer matrix is then

f -f

L L

f

L

f

Lf

LL

f

L

f

L

f

L

f

L

1

21

11

01

10

11

101

10

1

2

22

M


Where we’re going… It might seem like we would start by looking at beam lines and

them move on to rings, but it turns out that there is no unique treatment of a standalone beam line Depends implicitly in input beam parameters

Therefore, we will initially solve for stable motion in a ring. Rings are generally periodic, made up of more or less identical cells

In addition to simplifying the design, we’ll see that periodicity is important to stability

The simplest rings are made of dipoles and FODO cells Or “combined function magnets” which couple the two


Periodic “cell”

Our goal is to de-couple the problem into two parts

The “lattice”: a mathematical description of the machine itself, based only on the magnetic fields, which is identical for each identical cell

A mathematical description for the ensemble of particles circulating in the machine

Ncellcellcellcellring MMMMM

Stability Criterion We can represent an arbitrarily complex ring as a

combination of individual matrices

We can express an arbitrary initial state as the sum of the eigenvectors of this matrix

After n turns, we have

Because the individual matrices have unit determinants, the product must as well, so


123... MMMMM nring

221 VVMVV 211 BAx

xBA

x

x

21 VVM nnn BAx

x21

21 /1

Stability Criteria (cont’d) We can therefore express the eigenvalues as

However, if a has any real component, one of the solutions will grow exponentially, so the only stable values are

Examining the (invariant) trace of the matrix

So the general stability criterion is simply


complex generalin is where;; 21 aee aa

real is where;; 21 ii ee

cos2Tr ii eeM

2Trabs M

Example

Recall our FODO cell

Our stability requirement becomes

f -f

L L

f

L

f

Lf

LL

f

L

f

L

M

1

21

2

22

fLf

L222abs

2


Twiss Parametrization We can express the transfer matrix for one period as the sum of an identity

matrix and a traceless matrix

The requirement that Det(M)=1 implies

We can already identify A=Tr(M)/2=cosμ. If we adopt the arbitrary normalization

We can identify B=sinμ and write

Note that

So we can identify it with i=sqrt(-1) and write


)()(

)()(

10

01),(

ss

ssBAsCs

M

1))()()(( 222 sssBA

1)()()( 2 sss

sincos)()(

)()(sin

10

01cos),( JI

ss

sssCsM

IJ

)()()(0

0)()()(

)()(

)()(

)()(

)()(2

22

sss

sss

ss

ss

ss

ss

)(),( sesCs JM Remember this! We’ll see it again in a few pages

Equations of Motion General equation of motion

For the moment, we will consider motion in the horizontal (x) plane, with a reference trajectory established by the dipole fields.

Solving in this coordinate system, we have


xy

s

sBvBvyBvxBvm

e

BB

vvv

syx

m

e

m

BveR

RmRmdt

d

dt

pdBveF

xyyxxsys

yx

syx ˆˆˆ

0

ˆˆˆ

xr

Reference trajectory

Particle trajectory

yysrrxrr

yyxrsrsrxryysrsrsrxrxrR

yysrxryyxrxrR

yyxrR

ˆˆ2ˆ

ˆˆˆˆ2ˆˆˆˆˆˆˆ

ˆˆˆˆˆˆ

ˆˆ

2

2

Equations of Motion (cont’d) Equating the x terms

Re-express in terms of path length s. Use

Rewrite equation


B

Bvp

Bev

mv

Bevm

Bevrr

ys

ys

s

ys

ys

2

22

2

r

s

rtvs

svrdt

dss

Note: s measured along nominal trajectory, vs measured along actual trajectory

r

v

ds

d

rv

dt

d

ds

d

rv

ds

d

dt

ds

dt

d sss

;

2

22

2

2

1 ;12

2

2

22

22

22

2

22

x

B

By

xx

B

Bx

rr

B

Bvr

B

Bv

r

vr

ds

rd

rv

xy

ys

ysss

terms)(rearrange

) (use xr

Equations of Motion Expand fields linearly about the nominal trajectory

Plug into equations of motion and keep only linear terms in x and y


x

x

sBBx

x

sBBy

y

sBx

x

sBsBsyxB

yy

sBy

y

sBx

x

sBsBsyxB

yyyyyy

x

x

xxxx

)()()()(,0,0),,(

)()()(,0,0),,(

0coupling no

dipole no

coupling no

0)(1

0)(11

112)(111

)(

2

222

2

yy

sB

By

xx

sB

Bx

xxxx

sB

B

xx

B

xx

sBB

x

x

y

y

y

Looks “kinda like” a harmonic oscillator

Piecewise solution These equations are in the form

For K (gradient) constant), these equations are quite simple. For K>0, it’s just a harmonic oscillator and we write

In terms if initial conditions, we identifyand write


sKbKsKaKsx

sKbsKasKAsx

cossin)('

sincoscos)(

K

xbxa 0

0;

0

0

cossin

sin1

cos

)(

)(

x

x

sKsKK

sKK

sK

sx

sx

0)( xsKx

For K<0, the solution becomes

For K=0 (a “drift”), the solution is simply

We can now express the transfer matrix of an arbitrarily complex beam line with

But there’s a limit to what we can do with this


0

0

coshsin

sin1

cosh

)(

)(

x

x

sKsKK

sKK

sK

sx

sx

0

0

00

10

1

)(

)(

)(

x

xs

sx

sx

sxxsx

nMMMMM ...321

Closed Form Solution Our linear equations of motion are in the form of a “Hill’s Equation”

If K is a constant >0, then so try a solution of the form

If we plug this into the equation, we get

Coefficients must independently vanish, so the sin term gives

If we re-express our general solution


)()( ;0)( sKCsKxsKx Consider only periodic systems at the moment

)(cos)()( ssAwsx

0sin)2(cos)( 2 wwAKwwwAKxx

2

22by mutliply

'0'202w

kwwwwww

w

sincos

sincos

sincos

1221

1221

21

w

kAwA

w

kAwA

wAwAwAwAx

AAwx

sKAsx cos)(

)()(

BUT ),()( assume

sCs

swCsw

03

2

Kww

kw

We’ll see this much later

Solving for periodic motion Plug in initial condition (s=0Ψ=0) Define phase advances over one period

and we have

But wait! We’ve seen this before…


k

wxwxA

w

xA

002

01

)(cos);(sin CCCS

0

0

2

2

2

0

2

2

20

200

0

0

2

0

02

00

/

1)(

)(

'

)('

)(

x

x

Sk

wwCS

kwkww

Sk

wS

k

wwC

Cx

Cx

xSk

wwCSx

k

w

w

k

Sw

kx

k

wx

k

wwxCxCx

xSk

wxS

k

wwC

Sk

wwxwxCxCx

This form will make sense in a minute

Recall the Twiss representation of a Period We showed a flew slides ago, that we could write

We quickly identify

We also showed some time ago that a requirement of the Hill’s Equation was that


sin)(cossin)(

sin)(sin)(cos)()(

)()(sin

10

01cos),(

ss

ssss

sssCsM

)(

)(1)(

)(

2

1

2

1)(

)(

2

2

2

s

ss

ds

sd

k

w

ds

d

k

wws

k

ws

dss

Csw

kCs

s

0

0)(

1)(

)(

12

)(C

Phase advance over one period

Super important!Remember forever!

Closing the Loop We’ve got a general equation of motion in terms of initial

conditions and a single “betatron function” β(s)

β(s) is a parameter of the machine, but we still don’t know its form! Important note!

(s) (and therefore (s) and Υ(s)) are defined to have the periodicity of the machine!

In general Ψ(s) (and therefore x(s)) DO NOT! Indeed, we’ll see it’s very bad if they do

So far, we have used the lattice functions at a point s to propagate the particle to the same point in the next period of the machine. We now generalize this to transport the beam from one point to another, knowing only initial conditions and the lattice functions at both points


)(

)( ;)(cos)()(0s

s

dssssAsx

)(sin)(cos)()(

1

)(sin)(

)()(cos)(

)(

1

2

1)(

)(cos)()(

ssss

A

sds

sdsAs

ds

sd

sAsx

ssAsx

cellcellNs

ds

2

1

)(2

1 Define “tune” as the number of

pseudo-oscillations around the ring

We use this to define the trigonometric terms at the initial point as

We can then use the sum angle formulas to define the trigonometric terms at any point Ψ(s1) as


0

00

0000

0

000

)(sin

1)(cos

Ax

AxsS

AxsC

sincossin1

sincos)(sin)(sin

sinsincos1

sincos)(cos)(cos

00

0

0

0

0

00011

00

0

0

0

0

00011

Ax

AAx

CSsSs

Ax

AAx

SCsCs

General Transfer Matrix We plug the previous angular identities for C1 and S1 into the

general transport equations

And (after a little tedious algebra) we find

This is a mess, but we’ll oftenrestrict ourselves to the extremaof b, where


111

1

1

11111

1

cos

SCAx

CAAx

02

1

ds

d

0

0

1

0

10

100

1

1

1

cossin1

sincos

x

x

x

x

About the Tune The tune is the number of oscillations that a particle makes

around equilibrium in one orbit. For a round machine, we can approximate

Note also that in general


RRds

2

2

1

2

1

yx

eric prebys, fnal. we will tackle accelerator physics the way we tackle most problems in classical...

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