Lecture 3

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Lecture 3Lecture 3

Particle AccelerationParticle Acceleration

USPAS, January 2009USPAS, January 2009

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OutlineOutline

Electrostatic accelerators

-

RF Cavities and their properties

Material is covered in Wangler, Chapter 1(and also in Wiedemann Chapter 15)

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How do we accelerate particles?How do we accelerate particles?

We can accelerate charged particles:

electrons e- and ositrons e+

protons (p) and antiprotons (p)

1- 2+ 92+ . . , , ,

These particles are typically born at low-

e- : emission from thermionic gun at ~100 kV

~ The application usually requires that we

acce era e ese par c es o g er energy, n

order to make use of them

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Electromagnetic Forces on Charged ParticlesElectromagnetic Forces on Charged Particles

Lorentz force equation gives the force in response to electric

and magnetic fields:

The equation of motion becomes:

The kinetic energy of a charged particle increases by an

amount equal to the work done (Work-Energy Theorem)

+== ldBvqldEqldFWrr

rrrrr

)(

rrr

rr

rr

qtvvqq =+=

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Electromagnetic Forces on Charged ParticlesElectromagnetic Forces on Charged Particles

We therefore reach the important conclusion that Magnetic fields cannot be used to change the kinetic

We must rely on electric fields for particleacceleration Acceleration occurs along the direction of the electric

field nergy ga n s n epen en o e par c e ve oc y

In accelerators:

particle motion) are used for acceleration

Magnetic fields are used to bend particles for

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Acceleration by Static Fields:Acceleration by Static Fields:

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Acceleration by Static Electric FieldsAcceleration by Static Electric Fields

We can produce an electric field by establishing a potentialdifference V0 between two parallel plate electrodes,separated by a distance L:

LVEz /0=+q

E

the + electrode acquires an

increase in kinetic energy at the electrode of

V0

+ -

000

qVdzEqdzFWL

z

L

z ===

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The Simplest Electrostatic Accelerators:The Simplest Electrostatic Accelerators:

Electron GunsElectron Guns

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Electrostatic AcceleratorsElectrostatic Accelerators

Some small accelerators, such

as electron guns for TV picture

tubes, use the parallel plate

geometry just presented

Electrostatic particle

accelerators generally use a

slightly modified geometry in

which a constant electric field is

produced across anacce erat ng gap

Energy gain:Accelerating

= nVnqWelectrostatic

accelerator

= ngenerator VV

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Cascade Generators, aka CockroftCascade Generators, aka Cockroft--WaltonWalton

AcceleratorsAccelerators

Cockroft and Waltons 800 kV

accelerator, Cavendish Laboratory,

Cambridge, 1932

They accelerated protons to 800 kV

Modern Cockroft-

Waltons are still

used as proton

and observed the first artificiallyproduced nuclear reaction:

p+Li2 He

injectors for linearaccelerators

This work earned them the Nobel Prize

in 1951

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Van deVan de GraaffGraaff AcceleratorsAccelerators

Van de Graafs twin-column electrostatic accelerator (Connecticut, 1932)

Electrostatic accelerators are limited to about 25 MV terminal volta e due to

voltage breakdown

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Two Charging methods: Van de Graaff andTwo Charging methods: Van de Graaff and

Pelletron AcceleratorsPelletron Accelerators

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Highest Voltage Electrostatic Accelerator: 24Highest Voltage Electrostatic Accelerator: 24

MV (Holifield Heavy Ion Accelerator, ORNL)MV (Holifield Heavy Ion Accelerator, ORNL)

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Acceleration to Higher EnergiesAcceleration to Higher Energies

While terminal voltages of 20 MV provide sufficient beam energy for nuclear

structure research, most applications nowadays require beam energies > 1 GeV

How do we attain higher beam energies?

Analogy: How to swing a child?

Pull up to maximum height and let go: difficult and tiring (electrostatic

accelerator)

Repeatedly push in synchronism with the period of the motion

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--

Fields: RadioFields: Radio--FrequencyFrequencycce era orscce era ors

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Acceleration by Repeated Application of TimeAcceleration by Repeated Application of Time--

Varying FieldsVarying Fields wo approac es or acce era ng w me-vary ng e s

Make an electric field along the direction of particle motion with Radio-Frequency (RF) Cavities

E

E v

E

E

Circular Accelerators Linear Accelerators

cavities and make use of repeated

passage through them: This approach

leads to circular accelerators:

which the particle passes

only once:

Cyclotrons, synchrotrons and their

variants

ese are near

accelerators

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RF AcceleratorsRF Accelerators

In the earliest RF Accelerator,

Rolf Wideroe took the

electrostatic geometry we

considered earlier, but

attached alternatingconductors to a time-varying,

sinusoidal voltage source

The electric field is no longer

static but sinusoidal alternatinga per o s o acce erat on

and deceleration.

tgVtEtVtV

sin)/()(

sin)(

0

0

==

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RF AcceleratorsRF Accelerators

in Cavity

Time

This exam le oints out three ver im ortant as ects of an

RF linear accelerator

Particles must arrive bunched in time in order for efficient

Accelerating gaps must be spaced so that the particle bunches

arrive at the accelerating phase: 1===

The accelerating field is varying while the particle is in the gap;energy gain is more complicated than in the static case

2 c

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ow o e a e e s u ta e oro e a e e s u ta e or

Particle Acceleration?Particle Acceleration?

Waves in Free Space

E field is perpendicular to

propagation Waves confined to a

Phase velocity is greater

than speed of light

Resonant Cavity

Standing waves possible

with E-field alon direction

of particle motion

Disk-loaded Waveguide

with phase velocity equal

to speed of light

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Electromagnetic Waves in Free SpaceElectromagnetic Waves in Free Space

The wave equation is a consequence of Maxwells equations

01

2

2

2

2 =

E

E

r

r

01

2

2

2

2 =

B

B

r

r

Plane electromagnetic waves are solutions to the wave equation)cos(),( 00 txnkEtxE =

rr

rrr

rr

)cos(),( 00 tnkBtxB = rrr

Each component of E and B satisfies the wave equation provided

that ck /0 =

That is, the E and B fields are perpendicular to the direction of wave

Maxwells equations give

00=Enr

r

0=Bnr

r

)(

00),(

tzkieEtxE

= r

rr

propaga on an one ano er, an ave e same p ase. A plane wave propagating in the +z direction can be described:

)cos(),( 00 tzkEtxE = r

rr

To accelerate particles we need to i) confine the EM waves to a

specified region, and ii) generate an electric field along the directionof particle motion

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Standing WavesStanding Waves

Suppose we add two waves of equal amplitude, one moving in the+z direction, and another moving in the z direction:

tkztkzEE )cos()cos( ++=

[ ][ ] tzFtkzEEtkztkztkztkzEE

z

z

cos)(coscos2sinsincoscossinsincoscos

0

0

== ++=

The time and spatial dependence are separated in the resultingelectric-field: )()( tTzFE

z

=

s s ca e a s an ng-wave as oppose o a rave ng-wave ,since the field profile depends on position but not time

Such is the case in a radio-frequency cavity, in which the fields areconfined, and not allowed to ro a ate.

A simple cavity can be constructed by adding end walls to a

cylindrical waveguide The end-walls make reflections that add to the forward going wave

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RadioRadio--frequency (RF) Cavitiesfrequency (RF) Cavities

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Radio Frequency Cavities: The Pillbox CavityRadio Frequency Cavities: The Pillbox Cavity

Large electromagnetic (EM) fields can be built up by resonant excitation

of a radio-frequency (RF) cavity

These resonant cavities form the buildin blocks of RF article

accelerators

Many RF cavities and structures are based on the simple pillbox cavityshape

We can make one by taking a

cylindrical waveguide, and

placing conducting caps at z=0

R

and z=L

We seek solutions to the wave

equation (in cylindricall

E

coordinates), subject to theboundary conditions for perfect

conductors

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Conducting WallsConducting Walls

oun ary con t ons at t e vacuum-per ect con uctor nter ace arederived from Maxwells equations:

0 =

= BnEnrr

These boundar conditions mean:

0

0

== EnKHnrrr

Electric fields parallel to a metallic surface vanish at the surface

Magnetic fields perpendicular to a metallic surface vanish at the surface

In the pillbox-cavity case: lzzEEr ==== and0for0

For a real conductor (meaning finite conductivity) fields and currents

are not exactl zero inside the conductor but are confined to a small

RrEEz === for0

finite layer at the surface called the skin depth

.2

=

The RF surface resistance is

2/10==sR

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Wave Equation in Cylindrical CoordinatesWave Equation in Cylindrical Coordinates

-component Ezso we will start with that component.

The wave equation in cylindrical coordinates for Ez is:

0

1112

2

22

2

22

2

=

+

+

t

E

c

E

rr

E

rrrz

E zzzz

trREtzrEz cos)(),,( 0=

e w eg n w e s mp es case, assum ng anazimuthally symmetric, standing wave, trial solution

This gives the following equation for R(r) (with x=r/c)

12 dRRd2 =++ dxxdx

The solution is the Bessel function of order zero, J r/c

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Bessel FunctionsBessel Functions

= .

r/c

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Longitudinal Electric FieldLongitudinal Electric Field

The solution for the longitudinal electric field is

tcrJEEz cos)/(00= To satisfy the boundary conditions, Ezmust vanish at the

cavity radius:0==RrE

Which is only possible if the Bessel function equals zero

00 rc

Using the first zero, J0(2.405)=0, gives

Rcc /405.2= , ,

which satisfies the boundary conditions

The cavity is resonant at that frequency

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Magnetic Field ComponentMagnetic Field Component

The electric field is

trkJEE rz cos)(00=

(Amperes law)

=

CSd

EldB

rr

rr

00S

rdrtrkJErB

r

= 2sin)(20

0000

Using

We find

= )()( 10 xxJdxxxJ

trkJcEB r sin)()/( 10=

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The Pillbox Cavity FieldsThe Pillbox Cavity Fields

trkJEE cos=

-

trkJcEB r sin)()/( 10= Rkr 405.2=

boundaryconditions are

satisfied!

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The Pillbox Cavity FieldsThe Pillbox Cavity Fields

We have found the solution for one particular

normal mode of the pillbox cavity

This is a Transverse Magnetic (TM) mode,because the axial magnetic field is zero (Bz=0)

For reasons explained in a moment, this particular

mode is called the TM010 mode It is the most frequently used mode in RF cavities

for accelerating a beam

e s ou no e surpr se a e p ox cav yhas an infinite number of normal modes of

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Normal Modes of OscillationNormal Modes of Oscillation

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Mechanical Normal ModesMechanical Normal Modes

Drumhead modes

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Transverse Magnetic ModesTransverse Magnetic Modes

But, we selected one solution out of an infinite number

of solutions to the wave equation with cylindrical

Our trial solution had no azimuthal dependence, and noz-dependence

whereas the general solution for Ez is

trz cos0=

tzkmrREE zz cos)cos()cos()(0=

e wave equa on y e s

)(1)( 2222

mrRrR

22

2

2

rcrrrck

z

43421

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Which results in the following differential equation

for R(r) (with x=kcr)

0)/1(1 222

2

=++ RxmdxdRxdxRd

With solutions Jm(kcr), Bessel functions of order m

The solution is:

The boundar conditions re uire that

tzkmrkJEE zcmz cos)cos()cos()(0=

0)( == RrE

Which requires that

= cm

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- m

0)( =mnm xJ

A mode labeled TMmnp haslpkz /=

-

n zeros of the axial field component in the radial direction

p half-period variations in z

ox cav y as a scre e spec rum o requenc es, w c

depends on the mode. The dispersion relation is222

x

There also exist Transverse Electric modes (Ez= 0) with

2 +=+= lRc zmn

22

2

2

zmn

kkc

+=

Rxk mnmn /= lpkz /=

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Mode Frequencies of a Pillbox CavityMode Frequencies of a Pillbox Cavity

ac mo e as s resonan requency e ne y egeometry of the pillbox cavity

Di i CDi i C

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Dispersion CurveDispersion Curve

A plot of frequency versus wavenumber, (k), is called the dispersion curve

One finds that there is a minimum frequency, the cutoff frequency, below

which no modes exist

The dispersion relation is the same as for a cylindrical waveguide, except

that the longitudinal wavenumber is restricted to discrete values, as requiredby the boundary conditions

C it P tC it P t

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Cavity ParametersCavity Parameters

Stored energy:+= dVBEU )/(

10

22

0

The electric and magnetic stored energy oscillate in time 90degrees out of phase. In practice, we can use either the electric or

magnetic energy using the peak value.

Power dissi ation:

.2

;1

;2 === ss RdsHR

P

where Rs is the surface resistance, is the dc conductivity and is

Power dissipation always requires external cooling to remove heat;

Superconducting cavities have very small power dissipation

C it P t tdC it P t td

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Cavity Parameters, contdCavity Parameters, contd

Quality factor:The quality factor is defined as 2 times the stored energy divided by the

energy dissipated per cycle

P

Q =

The ualit factor is related to the dam in of the electroma netic

Q

UP

dt

dU ==

oscillation:

= -

tQeUtU

)/(

00)( =

Since U is proportional to the square of the electric field:

)cos()( 0)2/(

00

+= teEtE tQ

, ,

time

0

/2 Q=

R t B h i f El t i l O ill tR t B h i f El t i l O ill t

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Resonant Behavior of Electrical OscillatorsResonant Behavior of Electrical Oscillators

The frequency dependence of the electric field can be obtained by Fourier

Transform:

22

2

02 )2/()( Q

E

1.2

00

This has a full-width at half maximum of the power, , equal to Q

0

=

0.8

1

Q=1000

Q=2000

Q=10000

0.6

squared

0.2

.E

-0.2

497.00 498.00 499.00 500.00 501.00 502.00 503.00

Frequency (MHz)

Th Pillb C it P tTh Pillb C it P t

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The Pillbox Cavity ParametersThe Pillbox Cavity Parameters

Stored energy:

)405.2(2

1

2

0

2

0 JElRU

=

Power dissipation:

])[405.2(2

1

2

0

0

0 RlJERRP s +=

Quality factor:

== cU

Q405.20

+ ls 1

Pictures of Pillbox RF CavitiesPictures of Pillbox RF Cavities

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Pictures of Pillbox RF CavitiesPictures of Pillbox RF Cavities

Superconducting CavitiesSuperconducting Cavities

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Superconducting CavitiesSuperconducting Cavities

conductor: copper has 1/=1.7x10-8 -m

At 500 MHz, R =5.8m

20=sR

RF Surface resistance for

superconducting niobium, withTc=9.2K, Rres=10

-9-10-8 [ ]

res

TT

s ReGHzf

R c += /92.12

5109)(

At 500 MHz, with Rres= 10-8 , T=4.2K,

Rs= 9x10-8

Superconducting RF structures

have RF surface resistance ~5orders of magnitude smaller thanfor copper

Removal of heat from a high- duty-factor normal-conducting cavity is amajor engineering challenge Gradients are limited to a few MeV/m as

a result

RF power systems are asubstantial fraction of the cost of alinac

RecapRecap

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RecapRecap

We found a solution to the wave equation with cylindrical boundary

conditions appropriate for a pillbox-cavity

This solution has two non-zero field com onents

Longitudinal Electric field (yeah! We can accelerate particles with

this) that depends on radius

- .

radius

This cavity has a resonant frequency that depends on the

Because of finite conductivity, the cavity has a finite quality factor,

and therefore the cavity resonates over a narrow range of

, .

An infinite number of modes can be excited in a pillbox cavity; theirfrequencies are determined by their mode numbers

010

ExampleExample

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ExampleExample

Design a copper (1/ = 1.7x10-8 m) pillbox cavity with

TM010 resonant frequency of 1 GHz, field of 1.5 MV/m

a) What are the RF surface resistance and skin depth?b What is the cavit radius?

c) What is the power dissipation?

d) What is the quality factor?e) If instead of copper, the cavity was made with

superconducting niobium at 4K (assume Rres = 10-8),

f) Calculate the frequencies of the TM01p modes for p=

0, 1, 2

Additional MaterialAdditional Material

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Additional MaterialAdditional Material

Guided Electromagnetic Waves in a CylindricalGuided Electromagnetic Waves in a Cylindrical

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WaveguideWaveguide

We can accomplish each of these by transporting EM waves in awaveguide

Take a cylindrical geometry. The wave equation in cylindrical

01112

2

22

2

22

2

=+

+ t

Ec

Err

Errrz

E zzzz

Assume the EM wave propagates in the Z direction. Lets look fora solution that has a finite electric field in that same direction:

)cos(),(),,,( 0 tzkrEtzrEE zzz == e az mu a epen ence mus e repe ve n :

)cos()cos()( tzknrRE zz = The wave equation yields:

0)()(1)( 22

22

=

+

+

rRn

krRrR

2

rcrrr

ck

43421

Cylindrical WaveguidesCylindrical Waveguides

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0)/1(1 22

2

2

=++ Rxndx

dR

xdx

Rd

Which results in the following differential equation for R(r) (withx=kcr)

The solutions to this equation are Bessel functions of order n, Jn(kcr),

which look like this:


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The solution is:

coscos tknrkJE =

The boundary conditions require that

zcnz

Which requires that

==arz

-

nallfor0)( =akJ cn

0=xJ m

For m=0, x01 = 2.405

222

2

405.2zzc k

akk

c+

=+=

Cutoff Frequency and Dispersion CurveCutoff Frequency and Dispersion Curve

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Cutoff Frequency and Dispersion CurveCutoff Frequency and Dispersion Curve

222kkk +=

The cylindrically symmetric waveguide has

222 )( ckc +=

A plot of vs. k is a hyperbola,called the Dis ersion Curve

Two cases:

> c: kz is a real number and

the wave propagates < c: kz is an imaginary

number and the wave decaysexponentially with distance

n y waves w requencyabove cutoff are transported!

Phase Velocity and Group VelocityPhase Velocity and Group Velocity

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Phase Velocity and Group VelocityPhase Velocity and Group Velocity

The propagating wave solution has

tzkz =)cos(),(0 zrEEz=

po n o cons an propaga es w a ve oc y, ca e e p asevelocity,

z

pk

v =

cc

v

c

p >

=22

/1

e e ec romagne c wave n cy n r ca wavegu e as p asevelocity that is faster than the speed of light:

This wont work to accelerate particles. We need to modify the phasevelocity to something smaller than the speed of light to accelerate

The group velocityis the velocity of energy flow:

UvP gRF= And is given by:

dk

dvg

=

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