Eric Prebys, FNAL. Our previous discussion implicitly assumed that all particles were at the same...

Off Momentum Particles

Eric Prebys, FNAL

Off-Momentum Particles Our previous discussion implicitly assumed that all particles

were at the same momentum Each quad has a constant focal length There is a single nominal trajectory

In practice, this is never true. Particles will have a distribution about the nominal momentum

We will characterize the behavior of off-momentum particles in the following ways “Dispersion” (D): the dependence of position on deviations from the

nominal momentum

D has units of length “Chromaticity” (η) : the change in the tune caused by the different focal

lengths for off-momentum particles

Path length changes (momentum compaction)

USPAS, Knoxville, TN, Jan. 20-31, 2014Lecture 5 - Off Momentum Particles 2

0

)()(p

psDsx x

00

sometimes p

p

p

px

x

xxx

x

p

p

L

L

Treating off-momentum Particle Motion We have our original equations of motion

Note that ρ is still the nominal orbit. The momentum is the (Bρ) term. If we leave (Bρ) in the equation as the nominal value, then we must replace (Bρ) with (Bρ)p/p0, so

Where we have kept only the lowest term in p/p. If we also keep only the lowest terms in B, then this term vanishes except in bend sectors, where


2

2

2

1

1

x

B

By

xx

B

Bx

x

y

0

2

0

2

0

02

2

02

2

0

...111

...111

p

p

B

Bx

p

p

B

Bx

p

p

B

By

p

p

B

Bxx

p

p

B

Bxx

p

p

B

Bx

xxx

yyy

1

)(10 xsKx

B

B

B

By

This is a second order differential inhomogeneous differential equation, so the solution is

Where d(s) is the solution particular solution of the differential equation

We solve this piecewise, for K constant and find

Recall


)()()()(

)()()()(

00

00

sdsSxsCxsx

sdsSxsCxsx

1

Kdd

sK

Ksd

sKK

sdK

sKK

sd

sKK

sdK

sinh1

)(

cosh11

)(:0

sin1

)(

cos11

)(:0

yy BKxBx22

11

Example: FODO Cell We look at our symmetric FODO cell, but assume that the drifts are

bend magnets

For a thin lens d~d’~0. For a pure bend magnet

Leading to the transfer Matrix


2f -f

L L

2fs

10084

122

142

412

212

21

100

012

1001

100

102

1

100

011

001

100

102

1

100

012

1001

2

2

2

2

3

2

2

2

2

f

L

f

L

f

L

f

L

f

L

f

LL

f

LL

f

L

f

LL

f

LL

fM

Solving for Lattice Functions Solve for

As you’ll solve in the homework


02

sin

2sin

21

1

sin2

sin12

22sin

,,

2,

,

DFDF

DF

DF

D

LD

L

f

L

11

D

D

D

D

M

Momentum Compaction and Slip Factor In general, particles with a high momentum will travel a longer

path length. We have

The slip factor is defined as the fractional change in the orbital period

Note that


Dx l

D

lll 1

D

ds

dsD

C

C

dsp

pDppC

dspC

00

0

1)(

)(

p

p

p

p

p

p

p

p

C

C

v

v

C

C

T

Tv

CT

T

22

2

11

1

"transition"0:

around go to time takeparticlesenergy higher 0:

around go to time takeparticlesenergy higher 0:

T

T

T

more

less

Transition γ In homework, you showed that for a simple FODO CELL

If we assume they vary ~linearly between maxima, then for small μ

It follows

This approximation generally works better than it shouldUSPAS, Knoxville, TN, Jan. 20-31, 2014Lecture 5 - Off Momentum Particles 8

2sin

2sin

21

1 and ;

sin2

sin12

2minmax,minmax,

LDL

2

D

R

C

t

C DdsD

C

1

1112

Chromaticity In general, momentum changes will lead to

a tune shift by changing the effective focal lengths of the magnets

As we are passing trough a magnet, we can find the focal length as

But remember that our general equation of motion is

Clearly, a change in momentum will have the same effect on the entire focusing term, so we can write in general


00

00

00

0

0

1

4

1

11

111

p

p

p

p

f

p

p

ff

p

p

fp

p

B

lB

B

lB

f

i ii

dsB

B

f

L

00

1

xsKxxB

sBx )(0

)(12

dssKs )()(4

10

Chromaticity in Terms of Lattice Functions A long time ago, we derived two relationships when solving our

Hill’s equation

(We’re going to use that in a few lectures), but for now, divide by β to get

So our general expression for chromaticity becomes


1

2

1

1

2

1

01

0)(

)()()(

)(

22

2/3

2

2/3

2/33

2

K

Ksw

kswsKsw

sw

k

21 K

Multiply by β3/2

dsss )()(4

1

Chromaticity and Sextupoles we can write the field of a sextupole magnet as

If we put a sextupole in a dispersive regionthen off momentum particles will see a gradient

which is effectively like a positiondependent quadrupole, with a focallength given by

So we write down the tune-shift as

Note, this is only valid when the motion due to mometum is large compared to the particle spread (homework problem)


22

2 expressedoften 2

1)( xbxBxB

x

yB

Nominal momentum

p=p0+Dp

p

pDx x

0

'p

pDBDxB

Eric Prebys, FNAL. Our previous discussion implicitly assumed that all particles were at the same...

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