particle accelerators: accelerating particles. · 2017. 2. 2. · • 4) cylindrical symmetry •...

Particle accelerators:Accelerating particles.

What is a beam?

• Difficult => No ‘quantitative’ definition

• What characteristics we expect?

1) Composed by particles (usually charged)

2) Propagating along a ‘direction’ -> (design)

3) Geometrically confined

4) Showing collective behavior

5) The statistical transverse thermal spread << <vpropagation>

6) If we want that there is not transition (beam also at t = inf) => in respect to2,3,4 => providing confining forces (E/B fields for charged particles)

How to accelerate a beam

BvEqF

Fdt

pd

Lorentz force 𝑭

20

22

220

2

TW

mc

cppWW

The particle total energy W is:

How to see e.m fields - cavities

dsssqEzW cos

Energy gained by a particle in a cavity of length L :

s

s z s

ds

cts

0

00

with :

with :

pTqVW cos 0

Assuming a constant velocity :

dsss

csqEzW 00

cos

dssEzV0 Cavity Voltage

dsssEz

dsssEzp

cos

sinarctan Synchronous phase

dsssEzV

T pcos0

1Transit-time factor : 0 < T < 1

Alessandro VariolaFranco-Ukranian Physics school

Electron Linac

Electrons are light fast acceleration 1 already at an energy of a few MeV

Uniform disk-loaded waveguide, travelling wave

(up to 50 GeV, f ~ 3 GHz - S-band)

Synchronism conditionphel v

kcv

)(

0),( zktieEtzE

Wave numberRF

k

2

kvph

kvg

d

d

Electric field

Phase velocity Group velocity

Particle transverse motion

• Let’s start by tracking one particle

• 1st => chose the good reference system (fundamental)

• For accelerator can be very complex (due to the design geometry) but

• We can define a design trajectory

• We obtain a reference trajectory.

• The goal is to keep all the particle ‘confined’ in respect to the reference trajectory.

Confining forces : very general considerations

Let’s Assume that the recall forces are linear in the r coordinate(parabolic gradient)

𝐹 = 𝑘 𝑟 → 𝑟 =𝑘

𝑚 𝑟 or adding a damping term 𝑟 + 𝑎 𝑟 =

𝑘

𝑚 𝑟

Solutions• ∃ 𝑢 𝑡 , 𝑣 𝑡 (𝑖𝑛𝑑𝑖𝑝𝑒𝑛𝑑𝑒𝑛𝑡) →

𝑟 𝑠 = 𝐴 𝑢 𝑠 + 𝐵 𝑣 𝑠𝑟′ 𝑠 = A 𝑢′ 𝑠 + 𝐵𝑣′ 𝑠

Where A and B are defined by the boundary conditions.

𝑟0 = 𝐴 𝑢 0 + 𝐵 𝑣 0𝑟′0 = A 𝑢′ 0 + 𝐵𝑣′ 0

So solving for A and B we finally get to

𝑟 𝑠 = 𝑎𝑟0 + 𝑏𝑟′0𝑟′ s = c𝑟0 + 𝑑𝑟′0

That can be expressed by

𝑟𝑠𝑟′𝑠

=𝑎 𝑏𝑐 𝑑

𝑟0𝑟′0

𝑜𝑟 𝑟1 = 𝑀𝑟0 𝑡ℎ𝑎𝑡 𝑖𝑛 6𝐷 𝑐𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠:

Basis for Linear Tracking (electrons, protons, photons….)

r0

r′0rs

r′s

0

0

'

0

'

0

6661

1611

'

'

....

......

......

......

......

....

E

t

y

y

x

x

aa

aa

E

t

y

y

x

x

o

o

0W(M) with )0()0(

)0()0('''

0

0

AMR

B

A

vu

vu

r

r

Back to accelerators and particles

• x,y transverse coordinates in respect the reference particle• x’, y’ can be angles defining the TRACE SPACE (x,x’) or momenta defining

the PHASE SPACE (x,Px)• The vector (x,x’) represent the dynamical state of the particle• In complex transport system, in linear approximation, we can solve the

equations for each part of the trajectory obtaining M1, M2…Mn. • The final particle state will be represented by rf= M1∙ M2 ∙ …Mn ∙ r0

with r=(x,x’)

• NOW:• 1) Let’s see the application to accelerators (coordinates transformation, eq

of motion, recall forces…)• 2) Also after this the problem is not solved : N particles (1010 in a bunch),

N turns, N elements/turn, collective effects….We must find a way to parameterize the beam and the accelerator lattice……..

1st -> Recall forces

MAGNETS

• Why magnets ?:• 1) Not needed to be integrated under vacuum• 2) Efficiency given by the technological limits:

In theory the magnetic force is efficient at relativistic velocity : 𝐹𝐸 = β𝐹𝐵

But let’s take into account the technological limits. Iron saturation is 2T, Electric field breakdown threshold in vacuum ~ 10 MV/m. So also taking into account a of 0.1 we have:

𝐹𝐸𝐹𝐵

=10 𝑀𝑉/𝑚

2𝑇 ∗ 3 108= 0.1 ⇒

𝐹𝐸𝐹𝐵

=10 𝑀𝑉/𝑚

2𝑇 ∗ 0.1 ∗ 3 108=1

6

So already at non relativistic energy B field win!!!

Main type of magnets

• To bend : Dipoles

• To focalise : Quadrupoles

• Chromatic corrections :sextupoles, octupoles

Magnetic field representation : multipolar expansion

• Let’ Assume• 1) Magnetostatic fields• 2) Approximation: no fringe fields => 2Dimensions, Bz=0• 3) Maxwell equations are linear (complex fields can be described by a

sum of basic solutions• 4) Cylindrical symmetry• 5) In the vacuum chamber we have 𝑗 = 0 and

• In this approximations: the vector potential A has only the z component giving for Bx and By Chauchy-Riemann equations. So it is possible to build an analytical function that can be expressed in power series in imaginary coordinates (be careful z is x+iy, not long. coord.)

• Introducing polar coordinates (cylindrical symmetry), The general Fourier expansion is :

0 ,0 BB

CzkiyxzA n

n

n

zk, with )(~

0

Rossbach Schmuser, CAS, CERN 94/01, Vol1

yB

xB

jyxBjyxAr

nBnAnrr

B

nBnAnrr

B

nBnAr

yx

n

n

n

n

n

n

n

n

n

n

n

n

r

n

n

n

n

; and

))Im( )Re( (

: scoordinatecartesian in or

))( cos )(sin (1

))(sin )( cos (

))(sin )( cos (

nn

1

1

1

1

1

1

Multipolar expansion: cylindrical and cartesian coordinates

Coefficients meaning

• Normal lenses (not azimuthal tilt), A=0. n=1,2,3,4 -> dipole, quadrupole, sextupole, octupole…2n pole.

• The coefficient A is the rotation of the symmetry plane => it defines the skewness (skew coefficient)

• From the Cartesian coordinates expression we can deduce the equipotential lines defining the poles for a given magnet:

Dipoles : Potential linear in y => field constant

• Lorentz Force 𝐹 = 𝑞(𝐸 + 𝑣 × 𝐵)

• Dipole 𝐸 = 0, 𝑣𝐵

• The particle curvature radius is given by the equilibrium of the

Lorentz and the centrifugal force : 𝑚𝛾𝑣2

𝜌= 𝑞𝑣𝐵 →

1

𝜌=

𝑞𝐵

𝑝

• So:1

𝜌[𝑚]=

0.2998 𝐵[𝑇]

𝑝[𝐺𝑒𝑉

𝑐]

• We can so define the beam rigidity:

𝐵𝜌 =𝑝

𝑞= 3.3 β 𝐸 𝐺𝑒𝑉

• For =1 and a beam of 1 GeV a 1T field give a curvature radius of 3.3 m

Quadrupoles : Potential xy, field x linear in y

• The Field (and the Force) are linear in the transverse coordinates:𝐹𝑥 = 𝑞𝑣𝐵𝑦 = 2𝑞𝑣𝐵0𝑥 , 𝐹𝑦 = −𝑞𝑣𝐵𝑥 = −2𝑞𝑣𝐵0𝑦 ,

1) One plan is focusing, the other defocusing2) From V=-Bxy -> Shape of the poles3) Normalizing the Gradient to the particle momentum we obtain the QUADRUPOLE STRENGTH :

𝑘 =𝑞𝐺

𝑝𝑚−2 =

0.2998 𝐺[𝑇𝑚]

𝑝[𝐺𝑒𝑉𝑐]

=1

𝑓𝐿

With L the Qpole active thickness (magnetic length) and f the lensfocal length. kL=f-1

Multipoles

In real world

L

ff

If L<<f we have the ‘Thin Lens Model’

But how to have a Net Focusing effect in the two plans? : DOUBLETS/TRIPLETS

x

y

x

y

Focusing in both planes : doublets, triplets

2nd -> Eq of motion

• 1st of all => reference system

• Accelerator design can be very complex

• Let’s assume, for sake of simplicity, that the motion is in one plane (Horizontal -> x)

• The reference design give a reference trajectory. On the top of it we can imagine a reference particle, travelling on the reference trajectory, at the nominal momentum p0. We suppose that the reference particle has not energy losses (v=cost).This particle obviously feel only the dipoles fields.

An accelerator is designed around a reference trajectory (design orbit in circular accelerators), which is :

Reference trajectory

On this trajectory, a particle is represented by a curved abscissa : s

1. Straight line in drift and focusing element (no field on the axe)2. Arc of circle in dipole magnet, horizontal or vertical (Transfer lines)3. r is the local radius of curvature

Reference trajectory

So we will describe the particle motion as a little deviation form the reference particle, moving on the reference trajectory in s coordinate: travelling reference system! => Frenet-Serret System

syx uuu

,,

: tangent to the reference trajectory

: vertical

: in horizontal plan

su

yu

xu

Lab frame

Reference system -> riding the reference particle

Equation of motion

• To get a final motion equation we have to transform 𝑟𝑝 𝑎𝑛𝑑 𝑣𝑝 in the lab system, and d/dt ind/ds (reference)

• At the end we can plug the components in the equation given by the Lorentz force:

𝑟 =𝑞

𝑚(

𝑟 × 𝐵)

This gives the system :

deviations Little (Qpoles). mlinear ter the toup field B theof Espansionsp

q and

1

p

q5)

deviations momentum little 1p

p4)

plane) rsein transve deviations (little 11v

3)

ue)(almost tr )0,,()2

0 - vtrans)(vs negligiblemagnet in theeffect Velocity 1)

: ionsApproximatOur

11

11

0

''

2

'2

'2''

'2

'2''

22

kyBkxB

p

p

xxy

x

s

BBB

s

BxsBx

sm

qxssysy

ByBx

sm

qxssxsx

xy

yx

sx

sy

Linearization

We are looking for linear motion : we need to linearize.

HILL’s equation

• After the substitution (ref) in the equations we get to the linear equations of motion

𝑥′′ − 𝑘 −1

𝜌2𝑥 =

1

𝜌

∆𝑝

𝑝0

𝑦′′ + 𝑘𝑦 = 0

• Harmonic oscillators but k and r are k(s) and r(s)• In general the Hill’s equation will be

x′′ + 𝑘 𝑠 x = 𝑎(𝑠)

• So we defined the recall forces

• We found the Equation of motion

• We still have to see how to define the Matrices R…..

Hill’s equation. Solutions

• A solution of the general Hills equation is given by the linear combination of the Homogeneous and the Non-Homogeneous solutions

• x(s)=xH(s)+xNH(s) where :

x𝐻′′(𝑠) + 𝑘 𝑠 x𝐻(𝑠) = 0

x𝑁𝐻′′(𝑠) + 𝑘 𝑠 x𝑁𝐻(𝑠) =

1

𝜌(𝑠)

∆𝑝

𝑝0

Homogeneous solutions

• Being solution of a 2nd order differential equation the solution will be oscillatory…..we call it Sin Like and Cos like: S(s), C(s)

• They obviously satisfy: S𝐻

′′(𝑠) + 𝑘 𝑠 S𝐻(𝑠) = 0C𝐻

′′(𝑠) + 𝑘 𝑠 C𝐻(𝑠) = 0

For every transport element…so knowing k(s) we can find S and C…but we also know that from the general solution:

𝑥 𝑠 = 𝐴 𝐶 𝑠 + 𝐵 𝑆(𝑠)𝑥′ 𝑠 = 𝐴 𝐶′ 𝑠 + 𝐵 𝑆′(𝑠)

Imposing * C(0)=1, C’(0)=0, S(0)=0,S’(0)=1

𝑥 𝑠 = 𝐶 𝑠 𝑥0+ 𝑆(𝑠) 𝑥0′

𝑥′ 𝑠 = 𝐶′ 𝑠 𝑥0+ 𝑆′ 𝑠 𝑥0′

So in matrix form : 𝑥𝑥′

=𝐶 𝑆𝐶′ 𝑆′

𝑥0𝑥0

′

Like in the general case…!!!

So : 𝑥 𝑠 = 𝑥0C(s)+𝑥0′S(s)+NH solution

Parenthesis : solutions matrix Wronskian

• W𝐶 𝑆𝐶′ 𝑆′

=𝐶𝑆′- 𝑆𝐶′ has an important meaning

• Let’s assume the most general equation with damping term : 𝑥′′ + 𝑣 𝑠 𝑥′ + 𝑤 𝑠 𝑥 = 0

• If C and S are solutions1)𝐶′′ + 𝑣 𝑠 𝐶′ + 𝑤 𝑠 𝐶 = 02) 𝑆′′ + 𝑣 𝑠 𝑆′ + 𝑤 𝑠 𝑆 = 0

Multiplying the first by –S and the second by C and adding the two equations:

𝐶𝑆′′ − 𝑆𝐶′′ + 𝑣 𝑠 (𝐶𝑆′−𝑆𝐶′) = 0That means :

𝑊′ 𝑠 + 𝑣 𝑠 𝑊 𝑠 = 0

With solution

𝑊 𝑠 = 𝑊0𝑒− 𝑠0

𝑠1 𝑣 𝑠 𝑑𝑠

So if the dissipating forces are zero W(s) = W0 -> The Wronskian is constant and with the conditions * is unitary.

The Wronskian is conserved if there is not energy losses.

If W ≠ 0 C and S are linear independent.

Non Homogeneous Solutions

x(s)=xH(s)+xNH(s):

x𝑁𝐻′′(𝑠) + 𝑘 𝑠 x𝑁𝐻(𝑠) =

1

𝜌(𝑠)

∆𝑝

𝑝0

In a beam (particle ~ at the same momentum) p/p0 is supposed to be a constant. So we can normalize the non homogeneous solution. The normalized solution will be the Dispersion function D(s):

𝐷 𝑠 =𝑥𝑁𝐻(𝑠)

∆𝑝𝑝0

So the general solution will be : 𝑥 𝑠 = 𝑥0C(s)+𝑥0′S(s)+𝐷 𝑠

∆𝑝

𝑝0

Green=Homogeneous / Bleu=Non Homogeneous

Physical meaning : The dispersion function give the deviation from the reference orbit due to a difference in the reference momentum. Since it is a deviation its dimensions are [m]

B

nominal trajectory

reference = design = nominal trajectory= closed orbit (circular machine)

p

P + px

p

psDsx x

s

Dx

B

p2>p

Dispersion

Off-momentum particles are not oscillating around the design orbit, but around a chromatic closed orbit, whose distance from the design orbitdepends linearly from D.

Design orbit

Design orbit

On-momentum particle trajectory Off-

momentumparticletrajectory

Chromaticclose orbit

sDsx pDp is the periodic dispersion function

Chromatic closed orbit

The matrix of one turn is :

100

sDsSsC

sDsSsC

sM

1

sD

sD

p

p

The vector giving the chromatic closedorbit at abscissa s is :

The periodic condition gives :

11

sD

sD

sMsD

sD

p

p

p

p

Leading to :

CSSCSC

DSDSsDp

1

1

CSSCSC

DCDCsDp

1

1

Periodic dispersion function

Putting all together

• If for D we take as initial conditions : D0=0, D’0=0 (particle with different energy has the same x at the start)

• General solution xH+xNH ->

𝑥 𝑠 = 𝐶 𝑠 𝑥0 + 𝑆 𝑠 𝑥′0 + 𝐷(𝑠)∆𝑝

𝑝0

• Where C,S,D satisfy:

1)𝐶′′ 𝑠 + 𝑘 𝑠 𝐶 𝑠 = 0, C 0 = 1, C′ 0 = 0

2)𝑆′′ 𝑠 + 𝑘 𝑠 𝑆 𝑠 = 0, S 0 = 0, S′ 0 = 1

3)𝐷′′ 𝑠 + 𝑘 𝑠 𝐷 𝑠 =1

𝜌(𝑠), 𝐷 0 = 0, 𝐷′ 0 = 0

The Dispersion function can be expressed by the C and S functions

𝐷 𝑠 = 𝑆(𝑠) 𝑠0

𝑠 1

𝜌𝐶 𝑡 𝑑𝑡 − 𝐶(𝑡)

𝑠0

𝑠 1

𝜌𝑆 𝑡 𝑑𝑡

(Exercise ; Prove that this solution fulfill 3)

Example: Transport Matrix of a Sector Dipole withDispersion

The Hill’s eq. For the dispersion is:

Since rho is constant the particular solution for D is:

L

The general solution for the Dispersion is:

Hence the 3x3 transport matrix for the sector dipole is:

Bend only in the horizontal plane

Summarizing

• Tracking (for sake of simplicity we took only the horizontal component, the plane of motion…in y usually D=0)

0@0

0

0

0

'

100

)(')(')('

)()()(

'

p

p

x

x

sDsSsC

sDsSsC

p

px

x

s

C, S, D defined by the Hill’s equations solutions in each element.After that we have the matrix with each element we can multiply them todefine a line or a ring for LINEAR particle tracking

Parametrization: from Floquet theorem

• Periodic boundary conditions

x"+ K s( ) × x = 0 sKSsK With :

S : focusing periodHill equation

sssx cos

Solution : Given by the Floquet theorem

sSs The beta function at position s

s

ss

dsssss

0

00/

The phase advancebetween s0 and s

The phase advance for one revolution defines the TUNE of the machine

Periodic focusing : solutions

is an invariant given by particle initial conditions

The motion is then a pseudo-harmonic oscillation with varying amplitude and frequency. This transverse motion is called the betatron oscillation.

sincos

2

UUu

cossinsin2

sin2

cos2

cos4

2

2

2

UUU

UUUu

0sincos24

2

2

2

K

Periodic Focusing: calculations

024

0

2

2

2

K

1

04

12

22

K

The phase advance variation

The envelope equation

Phase-advance, envelope equation

2

ss

s

ss

21

Let’s define :

, and are the Courant-Snyder parameters of the motion

Uuuuu 22 2 It is easy to show that :

U

U

U

U

U

u

(Erre

ur !

Sourc

e du

0-8

1

2

3

4

5

6

7

u’ This is an ellipse equation.

Particle is moving on an ellipse whose shape is given by Courant-Syder parameters.

U is the courant-Snyder invariant linked to a particle

Courant-Snyder parameters

The Beta Function

Amplitude of a particle trajectory:

Maximum size of a particle amplitude

β determines the beam size

( ... the envelope of all particle trajectories at a given position

“s” in the storage ring.

I t reflects the periodicity of the

magnet structure.

x(s) = e * b (s) * cos(y(s) + j)

Bernhard Holzer, CERN CAS Prague 2014

0000

0

sinsincos sususu

Let’s calculate the transfer matrix M(s/s0) from s0 to s.

0

0

0

0

0

0

0 sincossin1

cos sususu

is the phase advance from s0 to s.

sincossin1

cos

sinsincos

/0

0

0

0

0

00

0

0ssM

Particle transfer matrix

Motion in Phase Space

Let’s concentrate on ellipse :

rmsuttt uuuu ,

22 2

It can be also written :

IUU T 1

'u

uU is a vector given the particle position in phase-space

rmsu

tutu

tutu

uuu

uuu

,

is a beam sigma matrix

Sigma matrix : definition


Acceleration

• Up to now we treated the motion in the transverse plane when the energy is conserved. How we can conserve the energy? And how we can take a low energy particle to ultra-relativistic energies? To accelerate the main actor is the Electric Field!


• Synchronicity implies that the (time) distance between cavities is a multiple integer of the RF wavelength = kv. The correspondantsynchronous RF phase will be Ys=t-kz

• This is one of the reasons for which the longitudinal phase space in Acc physics is referred to the coordinates (, E)


Convention

1. For circular accelerators, the origin of time is taken at the zero crossing of the RF voltage with positive slope

Time t= 0 chosen such that:

1

f1

tRF

tEtE RFsin)(2

2E 2

2

tRF

tEtE RFcos)(2

2E

2. For linear accelerators, the origin of time is taken at the positive crest of the RF voltage


f2

f2 - The particle is decelerated

- The particle arrives later – tends toward f0

f1 - The particle is accelerated

- The particle arrives earlier – tends toward f0

f1

f0

RFV

tRF

Synchrotron oscillations


1

0

RFV

t2

Synchrotron oscillations

pp

Phase space picture

Ring - Synchrotron oscillation4-6.

One gets:

sincos1cossin

2

0

23

ss

ssRF

TqEds

Ed

mc

E

ds

d

sincos1cossin

223

0

2

2

ss

ssRF mc

TqE

ds

d

This is a non-linear oscillator equation describing the synchrotron oscillation

21 s

Phase-space trajectory

When js = 0° or 180°, the synchronous particle is not acceleratedBucket

Separatrix

0 30 60 90 120

150

180

-180

-150

-120

-90 -60 -30


• So also for the longitudinal coordinates (phase space) the particle follow an oscillatory motion characterized by a frequency (a wavelength for ) and a tune.

• The area of the longitudinal phase space that represent a stable motion is called “bucket”

• Before we neglect the t. But this is important (cooling). This is possible if the energy loss is dependent from the particle energy itself. t=-(1/2T0)(dU/dE)E0

Synchroton and Betatron motion are damped by synchrotron radiation emission!!!!!

Summary

• Linear equation -> harmonic oscillators

- Transport matrix for tracking

- Still too difficult -> parametrization

- Twiss parameters, CS invariant, phase advance and tunes

- Also longitudinal, but non linearities by RF

Spares slides: Matrix examples

0 L

x’

x

x’L

s

0xLx

LxxLx 00

10

10/

LLM

Drift length : L

Drift

Focal length : f

0

x’

x

f

io xx

f

xxx i

io

11

01

/

f

oiM

Thin lens ( f>>L )

Length (m) : L Gradient (T/m) : G

zp

GqsK

Strength :

sKuKsKusu

sKKusKusu

sin0cos0

sin0cos0

x’

0 L

x

s

xGB

yGB

y

xB

Quadrupole

LKLKK

LKK

LK

cossin

sin1

cos

LKLKK

LKK

LK

coshsinh

sinh1

cosh

K > 0(focusing quadrupole)

K < 0(defocusing quadrupole)

kLcoshkLsinhk00

kLsinhk

1kLcosh00

00kLcoskLsink

00kLsink

1kLcos Quadrupole

L=lengthk2=(B/a)(1/B)a=radius (aperture)

In 2 D

Quadrupole 2

Sector dipole

B

LBL

2

1

22sin

Sector dipole 2

coscos

sin2000

010000

00cos

000

00)cos

1ln(cos

100

0000cos

0

0000)cos

1ln(cos

1

EE

E

EE

E

EE

EE

E

E

EL

EE

EE

E

E

EL

CavityL=lengthE= energy at the startE=energy gain=phase

<1 : adiabatic damping !!!!

Accelerating structure

Focusing thin Lens : 1 0

−1

𝑓1 ; Defocusing thin Lens:

1 01

𝑓1

DOUBLET:

Exercise : Triplet !!!

Doublet

Periodic systems -> the most simple transport line : FODO channel (doublet extension)

Derivation next course

Stability

Stability

sincossin

sinsincos/

00

00

00 sSsM

If M is the matrix of one period (or lattice) from s0 to s0+S :

sin2

sin

sin

2

1cos

22110

210

120

2211

MM

M

M

MM

Particle Transfer matrix over 1 period

One defines the tune per lattice as :

In synchrotron, the tune is the number of betatron oscillations over one turn.

Resonance : nQnQn yyxx

Resonance’s order : yx nn

Qx

Qy

Int(Qx) Int(Qx)+1

Int(Qy)

Int(Qy)+1

Avoid resonances :

find the best working point in tune

diagram

Betatron tune

particle accelerators: accelerating particles. · 2017. 2. 2. · • 4) cylindrical symmetry •...

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