Lecture 5Lecture 5

RF Acceleration in RF Acceleration in LinacsLinacsPart 2Part 2

Stuart Henderson, Jeff Holmes, Stuart Henderson, Jeff Holmes, Yan ZhangYan Zhang

USPAS January 2009USPAS January 2009

OutlineOutline• Traveling-wave linear accelerators• Longitudinal beam dynamicsLongitudinal beam dynamics

M t i l f W l Ch t 3 4 6• Material from Wangler, Chapters 3, 4, 6

Guided Electromagnetic Waves in a Cylindrical Guided Electromagnetic Waves in a Cylindrical Waveguide Waveguide

• We can accomplish each of these by transporting EM waves in a waveguide

• Take a cylindrical geometry. The wave equation in cylindrical coordinates for the z field component iscoordinates for the z field component is

01112

2

22

2

22

2

=∂∂

−∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

+∂∂

tE

cE

rrEr

rrzE zzzz

φ

• Assume the EM wave propagates in the Z direction. Let’s look for a solution that has a finite electric field in that same direction:

)cos(),(),,,( 0 tzkrEtzrEE zzz ωφφ −==The azimuthal dependence must be repetitive in φ:• The azimuthal dependence must be repetitive in φ:

)cos()cos()( tzknrRE zz ωφ −=• The wave equation yields:

0)()(1)(2

22

2

2

2

2=

⎟⎟⎟⎞

⎜⎜⎜⎛

−−+∂

∂+

∂∂ rRnkrRrR

zω

22

2

2

⎟⎟⎠

⎜⎜⎝

∂∂ rcrrrck

z43421

Cylindrical WaveguidesCylindrical Waveguides

0)/1(1 222

2

=−++ RxndxdR

xdxRd

• Which results in the following differential equation for R(r) (with x=kcr)

• The solutions to this equation are Bessel functions of order n, Jn(kcr), which look like this:

Cylindrical WaveguidesCylindrical Waveguides

• The solution is:)cos()cos()( tzknrkJE cn ωφ −=

• The boundary conditions require that)cos()cos()( tzknrkJE zcnz ωφ

0)(E• Which requires that

0)( ==arEz

• Label the n-th zero of J :

n allfor 0)( =akJ cn

0)( =mnm xJLabel the n th zero of Jm: • For m=0, x01 = 2.405

)( mnm

22 ⎞⎛ 22

222

2 405.2zzc k

akk

c+⎟

⎠⎞

⎜⎝⎛=+=

ω

Cutoff Frequency and Dispersion CurveCutoff Frequency and Dispersion Curve

2220 zc kkk +=

• The cylindrically symmetric waveguide has

222 )( ckzc +=ωω0 zc zc

• A plot of ω vs. k is a hyperbola, called the Dispersion Curvep

Two cases:• ω > ωc: kz is a real number and

the wave propagates• ω < ωc: kz is an imaginary

number and the wave decays exponentially with distanceO l EM ith f• Only EM waves with frequency above cutoff are transported!

Phase Velocity and Group VelocityPhase Velocity and Group Velocity• The propagating wave solution has

tzkz ωφ −=)cos(),(0 φzrEEz =

A point of constant φ propagates ith a elocit called the phase

Th l t ti i li d i l id h h

• A point of constant φ propagates with a velocity, called the phase velocity,

zp k

v ω=

ccvc

p >−

=22 /1 ωω

• The electromagnetic wave in cylindrical waveguide has phase velocity that is faster than the speed of light:

• This won’t work to accelerate particles. We need to modify the phase velocity to something smaller than the speed of light to accelerate particlesparticles

• The group velocity is the velocity of energy flow:

UvP gRF =A d i i b• And is given by:

dkdvgω

=

Traveling Wave StructuresTraveling Wave Structures

ccvp >=

• Recall that in the cylindrical waveguide, the electromagnetic wave has phase velocity that is faster than the speed of light:

cvc

p− 22 /1 ωω

• This won’t work to accelerate particles. We need to modify the phase velocity to the speed of light (or slower) to accelerate particles in a traveling wavetraveling wave

• Imagine a situation where the EM wave phase velocity equals the particle velocity

• Then the particle “rides the wave” PARTICLEvThen the particle rides the wave

Pv

• A “disk-loaded waveguide” can be made to have a phase velocity equal to the speed of light. These structures are often used to accelerate electrons

• The best and largest example of such an accelerator is the SLAC two-mile long linac

DiskDisk--loaded waveguide structureloaded waveguide structure

KEK

Energy Gain in a DiskEnergy Gain in a Disk--Loaded WaveguideLoaded Waveguide

D fiDefine• Ea: longitudinal accelerating field amplitude• U: stored energy per unit length• Pw: traveling wave power• dPw/dz: power dissipation per unit length• Shunt impedance per unit length )//(2 dzdPErL −=Shunt impedance per unit length• We have )//( dzdPUQ w−=ω

)//( dzdPEr waL

UvP gw = gw

dP ω

gwLa QvPrE /2 ω=

wwg

w PPQvdz

dP02αω

−=−=

We ha e t o choices for the accelerating str ct re consideredWe have two choices for the accelerating structure, considered now in turn

Constant Impedance Traveling Constant Impedance Traveling Wave StructureWave Structure

• Consider a disk-loaded waveguide with uniform cell geometry along the length, then Q, vg, rL, α0 are independent of z:

zezP 02)( α−=w ezP )( =

EdzdE / α−=

• Power decays exponentially along the length of the structure• The Electric field amplitude is

aa EdzdE 0/ α−=z

a eEzE 00)( α−=

• At the end of a waveguide of length Lg g02

0)( τ−= ePLPw0

0)( τ−= eELEa

QvLL

200ωατ ==

gQv2

• The energy gain isφ

τφ

τ

cos1)(cos0

00

0−−==Δ ∫

eLqEdzzEqWL

a

φτ

τ

cos120

0

0−−=Δ

eLPrqW L

Constant Impedance Structure ParametersConstant Impedance Structure Parameters

Constant Gradient Traveling Wave StructureConstant Gradient Traveling Wave Structure• A more common design keeps the gradient constant over the length

ww Pz

dzdP )(2 0α−=

• A more common design keeps the gradient constant over the length, which requires that the attenuation α0 depend on z

dz

⎥⎦⎤

⎢⎣⎡ −−= − )1(1)( 02

0τe

LzPzPw

• Which can be integrated to yield

)1)(/(11

21)(

0

0

2

2

0 τ

τ

α −

−

−−−

=eLz

eL

z• The attenuation factor is

L• The energy gain

φ

φφ

τ cos)1(

cos)(cos

020

00

−−=Δ

==Δ ∫

eLPrqW

LqEdzzEqW

L

L

a

gy gis

φcos)1(0Δ eLPrqW L

• To achieve a constant gradient, the SLAC linac structure tapers from a radius of 4 2 to 4 1 cm and the iris radiiradius of 4.2 to 4.1 cm, and the iris radii taper from 1.3 to 1.0 cm over 3 meters

Constant Gradient Traveling Wave StructureConstant Gradient Traveling Wave Structure

• The group velocity is02 )1)(/(1 τωω −eLzL

Th filli ti i

020 1

)1)(/(1)(2

)( τ

ωαω

−−−−

==e

eLzQL

zQzvg

• The filling time is

∫ ∫ =−== −−L L

FQ

eLzdze

LQ

zvdzt

0 0 022 2

)1)(/(1)1(

)( 0

0

ωτ

ω ττ

• For typical parameters the filling time is

∫ ∫ −−g eLzLzv0 0 )1)(/(1)( 0 ωω

For typical parameters, the filling time is ~1 µsec, and the beam pulse is 1-2 µsec

Constant Gradient Structure ParametersConstant Gradient Structure Parameters

SLAC LinacSLAC Linac• Largest in the world.

Reached energies of 50 GeV

Synchronicity condition in multicell RF Synchronicity condition in multicell RF structuresstructures

TM010 CavitiesDrift spaces

l1 l2 l3 l4 l5

β1 β2 β3 β4β5

• Suppose we want a particle to arrive at the center of each gap at φ=0. Then we would have to space the cavities so that the RF phase advanced by

l1 l2 l3 l4 l5

advanced by – 2π if the coupled cavity array was driven in zero-mode– Or by π if the coupled cavity array was driven in pi-mode

Synchronicity ConditionSynchronicity Condition

Zero-mode: πβλ

πλπωφ 222

====n

n

clctct

λβl λβnnl =

• RF gaps (cells) are spaced by βλ, which increases as the particle velocity increases

Pi-mode: πππωφ nlctct 22Pi mode: πβλλ

ωφ ====n

n

ctt

2/λβnnl = βnn

• RF gaps (cells) are spaced by βλ/2, which increases as the particle velocity increasesincreases as the particle velocity increases

Longitudinal DynamicsLongitudinal Dynamics• The drift space length between gaps was calculated for a particular particle with a• The drift space length between gaps was calculated for a particular particle with a

very specific energy. This is the reference particle, or the synchronous particle.• What happens to particles slightly faster or slower than the synchronous particle that

the linac was designed to accelerate?• Linacs are operated to provide longitudinal focusing to properly accelerate particle• Linacs are operated to provide longitudinal focusing to properly accelerate particle

over a range in energies or arrival time• Slower particles arrive at the next gap later than the synchronous particle

– They experience a larger accelerating fieldFaster particles arrive at the next gap earlier than the synchronous particle• Faster particles arrive at the next gap earlier than the synchronous particle

– They experience a smaller accelerating field

Equations of Motion IEquations of Motion I

• Consider an array of accelerating cells with drift tubes and accelerating gaps• The array is designed at the n-th cell for a particle with synchronous phase, kinetic

energy and velocity φsn, Wsn, βsn. Note that the synchronous phase is not zero!• We express the phase energy and velocity for an arbitrary particle in the n-th cellWe express the phase, energy and velocity for an arbitrary particle in the n th cell

as φn, Wn, βn• Assume that the particles receive a longitudinal kick at the geometric center of the

cell, and drift freely to the center of the next cell• The half-cell length isThe half cell length is

21,

1

λβ −− = ns

n

Nl

• Where N=1/2 for Pi-mode and 1 for zero-mode• The cell length (center of one drift tube to center of next) is therefore

2/)( ,1, λββ nsnsn NL += −

Equations of Motion IIEquations of Motion II• The RF phase changes as the particle advances from one gap to the

next according to

⎨⎧

++= −2 11

πωφφ ln

ππ modemode

⎩⎨++

−− 011 β

ωφφcn

nn 0 mode0 mode

• The phase change during the time an arbitrary particle travels from gap n-1 to gap n, relative to the synchronous particle is

21,

11,

1,11,, 2112)(

−

−−

−−− −≅

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=Δ−Δ=−Δ

ns

nns

nsnnsnsnns NN

βδββπ

βββπφφφφ

• Where we have used

Wδδβ

1for ,11112 <<−≅−

+=− δβ

βδβ

βδββββ ssss

• Usingssmc βγ

δβ 32=

( )1,12)( −− −Δ nsn WWNπφφ

g

• We get( )

21,1,

32,2)(

−−−=−Δ

nsnsns mc

Nβγ

πφφ

Equations of Motion IIIEquations of Motion III• Next, derive the difference in kinetic energies of the arbitrary particle

and the synchronous particle

( ) )cos(cos0TLqEWW φφ −=−Δ( ) )cos(cos ,0 nsnnns TLqEWW φφΔ

• To figure out the dynamics, we could track particles through gaps on a computer using these difference equations

• To get a feel for the dynamics “on paper”, we can convert these difference equations to differential equations by replacing the discrete action of the fields with a continuous field

• So we replace

dnd s

s)()( φφφφ −

→−Δ dnWWdWW s

s)()( −

→−Δ λβ sNsn =

πφφβγ 33 2)( WWd ss −− )cos(cos)( s TqEWWd φφ=−

• giving

λπφφβγ 2

33 2)(mcds

ssss −= )cos(cos0 sTqE

dsφφ −=

Equations of Motion IVEquations of Motion IVAssume acceleration rate is small and that E T φ and β are• Assume acceleration rate is small, and that E0T, φs and βs are constant

• We arrive at the equations of motion:

( )

20

332

qEB 2A and with

and coscos

TWWw

AwdsdB

dsdww

s

s

==−

=

−==′−==′

λβπ

φφφφ

233s

2 mcmc sλγβ

( )sABd φφφ coscos2

2

−−=

• Finally

( )sdsφφ2

Aw2

φφφφ HBAws =−+ )cos(sin

2

φφ HVAw =+2

21

φφ2• Where V is the potential energy term, and H (the Hamiltonian) is total

energy

Stable RF BucketStable RF Bucket• There is a potential well

when -π <φs< 0• There is acceleration for

/2 /2-π/2 <φs< π/2• The stable region for phase

motion is φ2 < φ < -φsTh “ t i ” d fi th• The “separatrix” defines the area within which the trajectories are stable.

• The stable area is called theThe stable area is called the “bucket”

• Stable motion means that particles follow a trajectory p j yabout the stable phase, with constant amplitude given by Hφ

Hamiltonian and Hamiltonian and SeparatrixSeparatrix ParametersParameters

• We can calculate the Hamiltonian to complete the discussion• At the potential maximum where, φ = -φs , φ’=0 and w=0

))()( i (BH φφφ ))cos()(sin( sssBH φφφφ −−−=

• The points on the separatrix must therefore satisfy2Aw

• We can calculate the “size” of the separatrix. We will do the energy width. Th i idth d t φ φ

)cos(sin)cos(sin2 ssss BBAw φφφφφφ −−=−+

The maximum energy width corresponds to φ = φs

)cos(sin)cos(sin2

2max

ssssss BBAw φφφφφφ −−=−+2

33

• Giving for the energy half-width of the separatrix. The energy acceptanceis twice this value:

)sincos(22

330

2max

max sssss

mcTqE

mcWw φφφ

πλγβ

−=Δ

=

Phase WidthPhase Width

• The maximum phase width is determined from the two solutions for w=0. One solution is φ1= -φs. The other solution φ2 is given byg y

• The total phase width isssss φφφφφφ sincoscossin 22 −=−

φφΨ• The total phase width is• The phase width is zero at φs=0 and

i t φ /2 i i 2 (

2φφ −−=Ψ s

maximum at φs=-π/2, giving ψ=2π (see Wangler figure 6.4)

Small Amplitude OscillationsSmall Amplitude Oscillations• Look at small amplitude oscillations. Letting φ-φs be

small, 0))(sin( =−−+′′ ssAB φφφφ

• This is an equation for simple harmonic motion with an angular frequency given by

))(( ss φφφφ

ss

sl mc

TqEβγπφλωω 32

02

2

2)sin(−

=

• Note that as the beam becomes relativistic the frequency goes to• Note that as the beam becomes relativistic, the frequency goes to zero

• From the equation of motion we can calculate the trajectory of a particle:particle:

1)()(2

0

2

20

2

=Δ−

+φφφ s

ww 22

033

020 2/)sin( mcTqEmc

Ww sss πφφλγβ Δ−=Δ

=

• This is the equation of an ellipse in w, φ-φs phase space• Particles on a particular ellipse circulate indefinitely on that

trajectory

Longitudinal Phase Space MotionLongitudinal Phase Space Motion• We studied the approximation of small acceleration rate, and constant

velocity, synchronous phase, etc.• In a real linac, the velocity increases, and the phase space motion and

β = const β ≠ const

separatrix becomes more complicated.• The “acceptance” takes a shape called the “golf-club”

βγ = const βγ ≠ const

Longitudinal Dynamics: Real data from SNS Longitudinal Dynamics: Real data from SNS Drift Tube LinacDrift Tube Linac

• Longitudinal “Acceptance Scan”Simulated DTL1 Acceptance

DataData

FWHM=24 deg

Measurement of SNS SC Linac Acceptance (Y. Measurement of SNS SC Linac Acceptance (Y. Zhang)Zhang)

SimulationMeasurement

Adiabatic Phase DampingAdiabatic Phase Damping• Louiville’s theorem:

The density in phase space of non-interacting particles in a conservative or Hamiltonian system measured along the trajectory of a particle is y g j y pinvariant.

• Or, if you prefer: phase space area is conserved• Area of ellipse: WA ΔΔφArea of ellipse:

00 WArea ΔΔ= φπ

4/30 )( ss

constγβ

φ =Δ 4/30 )( ssconstW γβ×=Δ

• Which gives)( ssγβ

• Since area is conserved an initial distribution with phase width (Δφ)iacquired a smaller phase widthacquired a smaller phase width after acceleration:

4/30 )()( if βγφΔ

4/30

0

)()(

)()(

f

i

i

f

βγβγ

φφ

=Δ

The EndThe End

• That concludes our whirlwind tour of Linear Accelerators

• Now, on to Rings….Now, on to Rings….