RF EFFECTS IN PHOTO-INJECTORSRF EFFECTS IN PHOTO-INJECTORS

Massimo FerrarioMassimo Ferrario

INFN-LNFINFN-LNF

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Madison, June 28 - July 2

λ/2 λ/2(Ν−3)λ/2λ/4

1

Z Z

2

z

(5/8)λ

solenoid

B

zE

z

E

z

E

Z

C

Zb

• Trapping a plane wave in a box

• Longitudinal RF effects

• Transverse RF effects

€

dU

dt= q

r v ⋅

r E ( )

The solutions to the electromagnetic wave equation in free space are transversely polarized waves (the electric field is transverse to the propagation vector) that have phase velocity c, the speed of light. These properties are problematic from the viewpoint of charged particle acceleration, because in order for a charged particle to absorb energy from an applied electric force, the motion of the charged particle must have a component parallel to the electric field:

If the motion of the particle in an accelerating wave is rectilinear in the z-direction, the electric field must be rotated to have a longitudinal component in order for acceleration to occur. This can be accomplished by using a smooth-walled waveguide.

€

Vacc = Re Ezo

L

∫ ei ωt −kz+φ o( )dz

€

Vacc = Re Ezo

L

∫ ei ω

z

βc−kz+φ o

⎛

⎝ ⎜

⎞

⎠ ⎟dz

€

Vacc = Re Ezo

L

∫ z( )eiφ o dz

€

vz = βc

E z, t( ) = Ezei ωt−kz( )

Let us consider a charge co-propagating in the z direction with a wave having a longitudinal component Ez

the particle experiences an accelerating voltage

€

z = βct

The energy gain depends on the spatial pattern of the field and on the phase relation (phase slippage)

The wave is syncronous when

€

ωk

= βc

xy

E

H

z

€

β =0€

Ex = cBy

EEBB

Interaction with a plane wave: particle at restparticle at rest

€

˙ ̇ x =e

mo

Ex − vzBy( )cos kz − ωt( )

˙ ̇ z =e

mo

vxBy cos kz − ωt( )

x

z

Non relativistic approx:

€

kz = kβct = ωβt

€

β = 1 −1

γ 2≈ 1 −

1

2γ 2

xy

E

H

z

€

β ≈1

€

rE ' ≡(Ex 0,0)

€

Ex = cBy

EEBB

Interaction with a plane wave: particle/wave co-propagatingparticle/wave co-propagating

€

E⊥= γ E⊥' − v × B⊥

'( ) ⇒ Ex = γEx

'

B⊥= γ B⊥' +

v

c 2× E⊥

' ⎛

⎝ ⎜

⎞

⎠ ⎟⇒ By = ±γ

β

cEx

' = ±β

cEx

€

F⊥= e Ex − βcBy( )cos ωt − kz( )

= e 1 − β( )Ex cos ω 1 − β( )t( )

≅eEx

2γ 2cos

ω

2γ 2

⎛

⎝ ⎜

⎞

⎠ ⎟

β ≈ 1

E'

β ≈ 1

FE

FB=βFE

a b

using

x

z

Wave front

Interaction with a plane wave with an anglewith an angle

€

Ez = −Eo sinθ ⋅ ei ωt −k z cosθ + y sinθ( )[ ]

€

vφz =ω

kz

=ω

k cosθ=

c

cosθ> c

€

ΔφΔt

= ω 1 − β cosθ( )

Not yet suitable

€

ζ =z cosθ − x sinθ

€

ζ' = z cosθ ' + x sinθ '

€

E x,z, t( ) = E+ xo , zo , to( )eiωt−ikζ + E− xo , zo, to( )eiωt−ikζ '

z

y

θ'θ

θ

ζ

ζ'

E+ E-

H+ H -

x

z

Plane wave reflected by a perfectly conducting planereflected by a perfectly conducting plane

In the plane xz the field is given by the superposition of the incident and reflected wave

€

σ =∞

E,H

x

E⊥

H⊥

E //

H / /

n

€

∃ E⊥ and H // only

€

Ez 0,z( ) = E+ sinθ( )e−ikzcosθ − E− sinθ'( )e−ikzcosθ ' = 0

€

θ = ′ θ E+ = E−

∀z

Boundary conditions require that

€

Ez x,z, t( ) = E+ sinθ( )eiωt−ik z cosθ −x sinθ( ) − E+ sinθ( )e

iωt−ik z cosθ +x sinθ( )

= E+ sinθ( )eiωt−ikz cosθ 2ie+ikx sinθ − e−ikx sinθ

2i

= 2iE+ sinθ sin kx sinθ( )eiωt−ikz cosθ

x-SW pattern

z-TW pattern

€

d =λ

2 sinθ

y

z

β ≈ 1

d

2d

d/ 2

E+

Ez

E-

E+

Ez

E-

x

€

Ez a,z ,t( ) = 0

€

a = nd

Taking into account the boundary conditions the longitudinal component of the field becomes

Notice that

Not only on the conductor:

n=1 n=2 n=3

€

Ez a,z ,t( ) = 0

€

k sinθ =n

aπ

€

A = 2E+ sinθ

€

kz = k cosθ

€

Ez x,z, t( ) = Re iA sinnπx

a

⎛

⎝ ⎜

⎞

⎠ ⎟ei ωt−kz z( ) ⎡

⎣ ⎢

⎤

⎦ ⎥= −A sin

nπx

a

⎛

⎝ ⎜

⎞

⎠ ⎟sin ωt − kzz( )

We can confine the plane wave with a second parallel conducting plane located where Ez = 0 so that boundary condition are fulfilled

x

z

This is a guided plane wave also called TMTMnn mode mode

n indicates the number of half wavelength between the two platesn indicates the number of half wavelength between the two plates

€

sinθ =λc

2a

€

a =λc

2

€

θ =π2

€

sinθ =λ

λc

< 1

Not all angles are possible for a given distance a between the two plates

€

d =λ

2 sinθ ==>

If a would be then

Implying a normal incidence and no propagation

Cut-Off conditionCut-Off condition:Only waves with λ λ < < λλc c = 2a can propagate= 2a can propagate

ω

ωc

ω=kzc

kz€

ω 2 − ckz( )2

= ω c2€

vφz =ω

kz

=ω

k cosθ=

c

cosθ=

c

1 − sin2 θ

=c

1 − λ λ c( )2

The phase velocity is given by

> c if < c

Im if > c

The previous relation is identical to

That is a dispersion relationdispersion relation

We must slow down the wave propagation

In order to slow down the waves we have to load the cavity by introducing some periodic obstacle into it

z

d

€

Ez (z) = E0 Im an

n=−∞

∞

∑ exp i2πn +ψ( )

dz

⎡

⎣ ⎢

⎤

⎦ ⎥

With this general form of the solution, the field can be viewed as the sum of many wave components, which are termed spatial harmonics, having different longitudinal wave-numbers , and thus different phase velocities .

€

kz,n = 2πn +ψ( )/ d

€

vϕ ,n = ω / kz,n

Beam Dinamycs in Photo-Injectors:

RF effects

Micro-Bunch Production with rf photo-injectorsMicro-Bunch Production with rf photo-injectors

Luca Serafini (INFN - Milan)

Workshop on 2nd Generation Plasma Accelerators,

Kardamyli, Greece, June 1995

€

Ez = Eo cos kz( ) sin ωt +ϕ o( )€

zexit = N + 1 / 2( )λ

2

€

T = eEo cos kz( )

0

N +1 / 2( )λ / 2

∫ sin ωt +ϕ o( )dz

λ/2 λ/2(Ν−3)λ/2λ/4

1

Z Z

2

z

(5/8)λ

solenoid

B

zE

z

E

z

E

Z

C

Zb

Eo peak accelerating field

φ phase as the particle lives the cathode surface

Let us try to compute the energy gain for a particle with ββ=0=0

€

kz = kβct β = 1 ⏐ → ⏐ ⏐ ωt

Longitudinal RF effectsLongitudinal RF effects

Impulsive Approximation to Electron Capture:

assuming a constant electric field nearby the cathode surface (z~0)assuming a constant electric field nearby the cathode surface (z~0)

€

Ez ≅ Eo sinϕ

The longitudinal equation of motion becomes

€

dβ //

dt=

eEo sinϕ

mc1 − β //

2( )

β // =eEo sin ϕ( )t

mc 1 +eEo sin ϕ( )t

mc

⎛

⎝ ⎜

⎞

⎠ ⎟

2

€

γ≅ 1

1 − β //2

giving

€

z =mc 2

eEo sinϕ1 +

eEo sin ϕ( )t

mc

⎛

⎝ ⎜

⎞

⎠ ⎟

2

− 1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

€

d

dtβ //γ( ) =

eEo sinϕ

mcassuming

€

˜ z = kz

€

˜ z t →∞ ⏐ → ⏐ ⏐ kct −1

2α sinϕ

Wich has an asymptoticasymptotic behavior

€

α eEo

2mc 2kIntroducing the dimensionless vector potential amplitude

and

€

˜ z =1 + 2αkct sinϕ( )

2

2α sinϕ−

1

2α sinϕWe obtain

€

β → 1

€

Δϕ =ϕ −ϕ o =1

2α sinϕ€

Δϕ =ϕ −ϕ o = ωt − kz

= ω t −˜ z

kc

⎛

⎝ ⎜

⎞

⎠ ⎟

Defining the phase slippage as

It has an asymptoticasymptotic behavior given by

Which has a minimum at φ φ = = ππ/2/2 that is also the phase of maximum acceleration and it is small if α α > 1/2 > 1/2

( ex: = 2.856 GHz, Eo = 100 MV/m ==> α=1.63 )

Assuming that the photo-electrons are emitted from the cathode directly at the speed of light, at a phase equal to the asymptotic phase, the energy gain is simply given by

where the non relativistic part of the motion has been considered equivalent to a new definition of the injection phase

€

T = eEo cos ˜ z ( )

0

N +1 / 2( )π

∫ sin ˜ z +ϕ( )dz

€

γ=1 +T

mc2= 1 + 2α cos ˜ z ( )

0

N +1 / 2( )π

∫ sin ˜ z +ϕ( )d˜ z

€

γ=1 + π N + 1 / 2( )α sin ϕ( ) +α cos ϕ( )

Phase CompressionPhase Compression

Defining as the phase separation between two different photo-electrons emitted at two launching pahse, their asimptotical distance will be given by:

€

δϕ ≡ϕ + −ϕ −=Δϕ + −Δϕ −+δϕ o

≅dΔϕ

dϕδϕ o +δϕ o

€

δϕ o ≡ ϕ o+ −ϕ o

−

By defining the relative change in phase distance between the two electrons

€

Δλ =δϕδϕ o

= 1 +d Δϕ( )

dϕ

= 1 −cot ϕ

2α sinϕ

The final length of the emitted electron bunch will be given by:

€

Lbunch = 1 −cot ϕ

2α sinϕLlaser

The injection phase o which corresponds to an asymptotic phase = /2 is the treshold below which we have Phase Compression and above which Phase Expansion.

For injection phases close to o = 0 the Compression can produce bunching even down to a factor 4

€

γ=1 +α π N + 1 / 2( ) sin ϕ( ) + cos ϕ( )[ ]

Energy SpreadEnergy Spread

At the gun exit we have

€

ϕ = ϕ +Δϕsubstituting

€

γ=1 +α π N + 1 / 2( ) sin ϕ + cos ϕ[ ] 1 −1

2Δϕ 2 ⎛

⎝ ⎜

⎞

⎠ ⎟+

α π N + 1 / 2( )cos ϕ − sin ϕ[ ]Δϕ

II order expansion around < >

€

sin ϕ + Δϕ( ) = sin ϕ + Δϕ cos ϕ −Δϕ 2

2sin ϕ

cos ϕ + Δϕ( ) = cos ϕ − Δϕ sin ϕ −Δϕ 2

2cos ϕ

€

ϕ =π2

€

γ= γ +Δγ=1 +α π N + 1 / 2( )[ ] 1 −1

2Δϕ 2 ⎛

⎝ ⎜

⎞

⎠ ⎟−αΔϕ

€

γ =1 +απ N + 1 / 2( )

€

Δγ=−αΔϕ −1

2γ − 1( )Δϕ 2

€

σ Δγ =ασϕ =αkσ z

We are interested in the phase

and

giving

So that

€

σϕ = Δϕ( )2

= kσ z

€

σ Δγ = Δγ( )2Or in terms of the rms quantity

Electron phase, energy at gun exit (2nd iris): simulations vs. analytical

Exercise:Exercise: show that the Longitudinal emittance

is given by

In particular for a Gaussian distribution show that

where

Transverse RF effectsTransverse RF effects

The fields that accelerate the charged particles in a radio-frequency linac cavity also give rise to transverse components of the Lorentz force that deflect the particles.

€

Er ≅ −1

r

∂Ez

∂z r=0

˜ r d˜ r

0

r

∫ = −r

2

∂Ez

∂z r=0

Assuming cylindrical symmetry, transverse electromagnetic fields in an accelerating structure can be obtained by a linear expansion near the axis of the Maxwell equations:

€

Bϕ ≅1

rc 2

∂Ez

∂t r=0

˜ r d˜ r

0

r

∫ =r

2c 2

∂Ez

∂t r=0

€

r∇ ⋅

rD = ρ e

€

r∇ ×

rH =

∂r

D

∂t+

r J e

The radial component of the Lorentz force becomes

€

Fr = q Er − v0Bϕ( ) = −qr

2

∂Ez

∂z r=0

+β0

c

∂Ez

∂t r=0

⎡

⎣ ⎢

⎤

⎦ ⎥

≅ −qr

2

∂Ez

∂z r=0

+1

c

∂Ez

∂t r=0

⎡

⎣ ⎢

⎤

⎦ ⎥

= −qr

2

dEz

dz r=0

(v0 → c ).

Thus in the ultra-relativistic limit , the net radial Lorentz force is simply proportional to the total z-derivative of the accelerating field.

€

Er r,z, t( ) = −r

2

dEz z( )

dzsin ωt +ϕ o( )

Bϑ r,z, t( ) =r

2

k

cEz z( )cos ωt +ϕ o( )

€

Ez r,z, t( ) = Ez z( ) sin ωt +ϕ o( ) = Eo cos kz( ) sin ωt +ϕ o( )

Considering as a longitudinal component

€

Δpr =−er

2mc 2

dEz

dtTc

Tnc

∫ dt =−er

2mc 2Ez Tnc( ) − Ez Tc( )[ ]

=−er

2mc 2Eo sin ϕ + 2nπ( ) − Eo sin ϕ( )[ ] = 0

Symmetric structure

Tc Tnc

If we integrate the radial forces through an isolated electromagnetic structure with constant velocity and radial offset we obtain for the full momentum transfer

This result explicitly shows that the sum of all inward and outward impulses applied within the structure cancel for an ultra-relativistic particle traversing a cylindrically symmetric structure.

€

pr =−er

2mc 2

dEz

dtTc

Texit

∫ dt =−er

2mc2Eo sin ϕ( ) = αkr sinϕ

z

Anti-symmetric structure

Tc Texit

But in the last half cell

This represents a defocussing kick when 0 < <

Δ ′ r =Δpr

γ2≅10 mrad @ S−band, 100 MV /m

Thus it usually will be necessary to focus the beam immediately after leaving the cavity

Transverse RF emittanceTransverse RF emittance

Rewriting pr in Cartesian coordinates:

It gives the phase space distribution: a collection of lines with different slopes corresponding to different

The normalized transverse emittance is:

By inserting px we obtain

€

sin ϕ + Δϕ( ) = sin ϕ + Δϕ cos ϕ −Δϕ 2

2sin ϕ

sin2 ϕ + Δϕ( ) = sin2 ϕ + Δϕ 2cos ϕ sin ϕ −Δϕ 2

2cos2 ϕ − sin2 ϕ( )

writing and assuming that is small

one obtains from:

so that

It has a minimum for <>= /2

€

ϕ =π2

€

ϕ ≠π2

Away from the minimum we have

For a Gaussian beam distriution

€

ρ x, y,Δz( ) =1

2π( )3 / 2

σ x2σ z

e−

1

2

x 2 +y 2

σ x2

+Δz 2

σ z2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

€

Δϕ 2 = k 2σ z2

Δϕ 4 = k 4σ z4

€

εx−minrf =

αk 3 x 2 σ z2

2

The rms quantitie are

And the emittance becomes:

Exercise:Exercise: show that for a uniform distribution in a cylinder of radius a and length L, the relevant moments for the distribution are:

And the transverse emittance is:

Second order transverse focusing in electromagnetic Second order transverse focusing in electromagnetic accelerating fieldsaccelerating fields

if one relaxes the assumptions of constant velocity and offset from the design orbit and examines the effects of the alternating gradient forces, one finds that they give rise to a second order secular focusing force

Symmetric structure

Tc Tnc

The approximation we will employ here assumes that the motion can be broken down into two components, one which contains the small amplitude fast oscillatory motion (the perturbed part of the motion), and the other that contains the slowly varying or secular, large amplitude variations in the trajectory.

€

x = xosc + xsec

-1.50

-1.00

-0.50

0.00

0.50

1.00

1.50

0.0 1.0 2.0 3.0 4.0 5.0 6.0

x

k

sec

z

€

′ ′ x + κ 02 sin kpz( )x = 0

The oscillatory component is analyzed by making the approximation that the offset x=xx=xsecsec is constant over an oscillation so that:

€

′ ′ x osc + κ 02 sin kpz( )xsec = 0

€

xosc = sin kpz( )κ 0

2

kp2

xsecWhich has a simple solution

The original equation becomes

€

′ ′ x = −κ 02 sin(kpz)x ≅ −κ 0

2 sin(kpz) 1 + sin kpz( )κ 0

2

kp2

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥xsec

The last step is to convert it into an averaged expression over a period, that gives the behavior of the secular component of the

motion

€

xsec ≡ x

€

xsec'' ≅ −κ 0

2 sin(kpz) 1 + sin kpz( )κ 0

2

kp2

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥xsec

€

′ ′ x sec +κ 0

4

2kp2

xsec = 0

z

d

€

Ez (z) = E0 Im an

n=−∞

∞

∑ exp i2πn +ψ( )

dz

⎡

⎣ ⎢

⎤

⎦ ⎥

Simple Fourier decomposition of the on-axis solution then gives the useful form

With this general form of the solution, the field can be viewed as the sum of many wave components, which are termed spatial harmonics, having different longitudinal wave-numbers , and thus different phase velocities .

€

kz,n = 2πn +ψ( )/ d

€

vϕ ,n = ω / kz,n

we assume the n=0 spatial harmonic is synchronous with the relativistic particle motion, or

For a particle located at a phase with respect to maximum acceleration, we have

€

ω / kz,0 = c

€

Ez = E0 Im ane i (2k0nz+ϕ n )

n=−∞

∞

∑ ⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

€

Fρ ≅ −qρ

2

dEz

dz ρ =0

= −qE0ρ Im ik0nane i (2k0nz+ϕ n )

n=−∞

∞

∑ ⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

If the energy variation is ignorably small over a period, then one may proceed to find the secular radial equation of motion by averaging the lowest order oscillatory motion given by the previous equation over a structure period.

€

′ ′ ρ osc =Fρ ρ = ρ sec

β 2γm0c2≅ −

qE0ρ sec

γm0c2Im ik0nane i (2k0nz+ϕ n )

n=−∞

∞

∑ ⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

witch has a steady-state solution

€

ρosc ≅ ρ sec 1 −qE0

4γm0c2Im

an

ik0ne i (2k0nz+ϕ n )

n=−∞

∞

∑ ⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

Substitution of the value of and averaging, we have

€

ρosc

€

′ ′ ρ sec = −F ρ

β 2γm0c2

≅ −1

8

qE0

γm0c2

⎡

⎣ ⎢

⎤

⎦ ⎥

2

ρ sec an2 + a−n

2 + 2ana−n sin(2ϕ )( )

n=1

∞

∑

This radial focusing force, like that derived from the solenoid provides equal focusing in both x and y, which is second order in applied field strength . In the case of the solenoid, the net radial force is second order due to the accompanying rotation, while in the present case it is of second order because of the fast radial oscillatory motion due to alternating gradient focusing

€

η(ϕ ) ≡ an2 + a−n

2 + 2ana−n sin(2ϕ )( )

n=1

∞

∑represents a sum over all spatial harmonics that contribute to the represents a sum over all spatial harmonics that contribute to the alternating gradient force, and is equal to 1 for a pure harmonic alternating gradient force, and is equal to 1 for a pure harmonic standing wave.standing wave. Note that the synchronous harmonic (n=0) does Note that the synchronous harmonic (n=0) does not contribute to the first or second order force in the ultra-not contribute to the first or second order force in the ultra-relativistic limit, so that relativistic limit, so that ηη=0 for a pure forward traveling wave=0 for a pure forward traveling wave (ao=1, all other an vanishing).

the alternating gradient focusing effect arises from the existence of non-synchronous spatial harmonics

€

′ ′ x +′ γ

γ

⎛

⎝ ⎜

⎞

⎠ ⎟ ′ x +

η ϕ( )

8 sin2 ϕ( )

′ γ

γ

⎛

⎝ ⎜

⎞

⎠ ⎟2

x = 0

Because we have kept the energy constant, we have not kept the effects of adiabatic damping in the equation of motion. Acceleration can be taken into account by use of the damping term.

Because the focusing is symmetric, the resulting equations of motion in x, y and are all equivalent, and are of the form