Transverse and longitudinal dynamicschiodini/didattica/phd2015/Lessons...Introduction to...

Introduction to particle accelerators and their applications - Part I: Transverse and longitudinal

dynamicsGabriele Chiodini

Istituto Nazionale di Fisica Nucleare Sezione di Lecce

PhD lessons in Physics for Università del Salento

2015-16 (20 hours, 4 CFD)

1

/59Introduction to accelerators : Transverse and Longitudinal dynamics G. Chiodini - May 2015

Transverse and longitudinal dynamics

• The longitudinal dynamics is along the direction of acceleration

• The transverse dynamics is orthogonal to the direction of acceleration

• As a first approximation we consider them separately ( synchrotron )

• We will use the synchrotron as an example but the concepts apply directly to Linac ( simpler ) and cyclotrons (more complicated )

2

G. Chiodini - Nov 2014/59Introduction to accelerators : Transverse and Longitudinal dynamics

Straight sections and arches of the synchrotron

3

L=Straight section

A=ArchΦ=arch angle

Orbit length = C = 4xL + 4xA

ρ=arch radius of curvature

Arch length = A=ρΦ[rad]

•Arches filled with magnetic dipoles for deflection and achieve closed orbit. •Straight sections used for : injection experiments, RF cavities extraction. •LHC has 8 arches and 8 straight sections

•Magnetic dipoles keep the particles along the direction of acceleration

•At each turn the particles receive an increase of energy equal to the potential V of the RF cavities.


Dipole (blu)

4

B = µ0nIh

where μ0=4π10-7H/m, n=number of turns, I coil current, h gap height

pq= ρB p(GeV / c)

z= 0.3ρ(m)B(T)magnetic rigidity

1ρ= 0.3 B(T)

p(GeV / c)


Transverse dynamics

Let’s turn off the RF: V=0

• Magnetic dipoles keep the particles along the orbit

• The particles have constant energy (constant speed)

5


Transverse dynamics problem

• The source emits a huge number of particles ( for example 1010 for bunch)

• How do you keep all these particles inside the accelerator (when particles touch the walls are lost )

• If the transverse motion is larger than the radius of the vacuum tube the particles are lost

6

Injection or source


Ideal orbit

7

Centrifugal force = Centripetal force

Fcentrifuga = ma = mv2

ρ=pvρ

Centripetal force = Magnetic force

pq= ρB p(GeV / c)

z= 0.3ρ(m)B(T)magnetic rigidity

Fcentripeta = qvB

Straight sections is a limiting case of the arcs : Radius = infinit and B = 0 .


Real orbit

8

Real orbitIdeal orbitr = ρy = 0s = vt

"

#$

%$

r = ρ + xy = ys = vt

"

#$

%$

vx = 0vy = 0

vz = v

!

"##

$##

vx = vxvy = vyvz ~ v

!

"##

$##

position

velocity

vx / v = x 'vy / v = y '

vz ~ v

!

"##

$##

velocity

orbita ideale

orbita reale

It is convenient to describe the real orbit as deviation from the ideal orbit

The ideal orbit doesn’t have transverse motion

The real orbit does have transverse motion (x,y,x’y’): betatron oscillations


Traveled distance and transverse angles

• We don’t care the time.

• We care the traveled distance (path length) s in the ring to find out where the particle is

• We care the orthogonal position to know if the particle touches the walls .

9

t→ s = vtvx → x ' = vx / vax → ax ' = ax / v

2

• Time multiplied by tangential velocity is the traveled distance

• The transverse velocities divided by v becomes the angles

• The transverse acceleration divided by

the square of the tangential velocity v2 is the curvature

vvx x ' = vx / v = x / s

v

ax ' = ax / v2


Equation of motion

It is the harmonic oscillator equation with angular frequency ω=2πf

10

ax ' = −xρ2

ax (t) = −ω2t

(proof in appendix)


x(t = 32T) = 0

Harmonic oscillator

11

ax (t) = −ω2t

x(t) = Acos(ω t +ϕ ) = Acos(2πft +ϕ )

Harmonic oscillator equation

Solution with amplitude A and phase φ which depend on the initial conditions

ϕ = 0

x(t = 14T) = 0

x(t = 12T) = −A

x(t = 0) = A x(t = T) = A

T = 1f

cos(0) = 1

cos(π2) = 0

cos(π ) = −1cos(2π ) = 0


Spring motion

12

Felastica = max(t) = −kx

C is the elastic constant of the spring which exerts a force proportional to the displacement x and of opposite sign (recoiling force).

x(t) = Acos(ω t +ϕ )vx (t) = −Aω sin(ω t +ϕ )ax (t) = −Aω

2 cos(ω t +ϕ )

$

%&

'&

ax (t) = −kmx→ω =

km

ax = −ω2x

Verify by substitution

C C-Cx


Equation of motion along x (bending direction)

In the bending direction there is e recoiling force F=m/ρ2

13

ax ' = −xρ2

kx =1ρ→ωx =

vρ

λx =2πkx

→ Tx =2πωx

We formulate the equation in term of space variables (wavenumber k and wavelength λ) and not in term of time variables (angular frequency ω and period T).


Betatron oscillations• The real particle around the ideal orbit makes radial

oscillations called betatron oscillations ( horizontal )

• The horizontal betatron oscillations are just due to the centripetal force of the dipoles and for a circular accelerator they have a wave length equal to the accelerator circumference (only one oscillation)

• For larger machines the horizontal recoil force is small and the particle accomplishes very large betatron oscillations ( "weak focusing" )

• Without a "strong focusing" the opening of the magnets must be enormous and the magnets must be arranged along all the accelerator.

14

λx = 2πρ (solo dipolo)

The oscillation in along x but the wave length is along the orbit

λx(Only dipole)


Vertical motion

• In the vertical direction there is neither the "weak focusing" and the motion is a parabola.

• Sooner or later the particle touches the walls because there the force of gravity in the vertical direction is “defocusing”

• How the accelerators works without "strong focusing”?

15

Equation of motion in the vertical direction taking into account the gravity and its solution (uniformly accelerated motion ) .

y(t) = y0 + v0yt −12gt2

ay = −g = −9.8ms2

hcaduta =12gtcaduta

2 → tcaduta = 2 hcadutag

hcaduta=10cm → tcaduta=0.14s


Weak vertical focusing in the cyclotron

16

If the magnetic field B decreases away from the centre arises a return force in the vertical direction ( vertical betatron oscillations ) .

Poles are shimmed to reduceB radially


Need of “strong focalisation”

17

• In the circular accelerators exists a weak natural focus in both directions thanks to the bending dipoles:

• curvature in the horizontal direction ( weak for large machines ) • edge effect in the vertical direction ( weak )

•Unfortunately the focusing effect is weak : • significant betatron fluctuations in the horizontal and vertical planes • large aperture magnets in large aperture ( low fields and expensive)


Quadrupole (red)

18

g = 2µ0nIr2

•g[25-220T/m] quadrupole gradient •r=quadrupole aperture •If r is large then g is very small

Bx = −gyBy = −gx

"#$

%$

k = gp / e

= 0.3 g(T / m)p(GeV / c) normalised gradient

•A particle displaced along x experiences an increase of vertical component of B then a recoil force along x (horizontal focusing) •A particle displaced along y experiences an increase of the horizontal component of B then a repulsive force along y (vertical defocusing)

Particle enters in the slide

Fx = −qvgxFy = qvgy

"#$

%$

Fy = qvgy

Fx = −qvgx

ξ


Equation of motion with quadrupole

19

ax ' = − k + 1ρ2

#

$%&

'(x

ay ' = kx

)

*++

,++

k = 0.3 g(T / m)p(GeV / c)

1ρ= 0.3 B(T)

p(GeV / c)

x(s) = Ax cos(2πsλx

+ϕx )Strong focalisation in the horizontal direction

y(s) = Ay cosh(2πsλy

+ϕy )Strong defocalisation in the vertical direction

cosh(z) = ez + e−z

2

Coseno iperbolico

1ρ2

→ k + 1ρ2

Magnetic rigidity (dipole)

Normalized gradient (quadrupole)

(Demonstration in appendix)

The quadrupole introduce a recoiling force in one orthogonal direction and a repulsive force in the other one

ξ

ξ

ξx

ξ


Strong focalisation with alternating gradient

!• The quadrupoles focus in a transverse direction but defocus in the orthogonal transverse direction •Two quadrupoles in series one focussing and one de focu s i n g a re s t rong focusing in both orthogonal directions

20

ax ' = − k + 1ρ2

#

$%&

'(x

ay ' = kx

)

*++

,++

ax ' = − −k + 1ρ2

#

$%&

'(x

ay ' = −kx

)

*++

,++

D=Defocusing along x F=Focusing along x

DF

General principle: magnetic fields with alternating gradient are strong focusing

ξ

ξx ξx

ξ


Analogy with lenses

21

f1s0+1s1=1f

1s0−1s1= −

1f

s0 s1

fs0

s1

F=Convergent

D=Divergent

Parallel rays from left are focused in the lens focus f in the right. Rays from s0 on the left are focused in s1 on the right.

Parallel rays from left are defocused in the lens focus f in the left. Rays from s0 on the left are defocused in s1 on the left.


Composite lens D+F=F

22

1f + d

+1

fcomposto=1f→

1fcomposto

=1f−

1f + d

=d

f(f + d)> 0

fcompostof d

For the lens F the parallel ray from left is like coming from the focus of lens D placed at a distance d+f from F then is focalised in fcomposto according to the lens law.

D F


Composite lens F+D=F

23

1−(f − d)

+1

fcomposto= −

1f→

1fcomposto

= −1f+

1f − d

=d

f(f − d)> 0

fcomposto fd

For the lens D the parallel ray from left is like coming from the focus of lens F placed at a distance f-d from D then is defocalised in fcomposto according to the lens law.

F D

- -+

-<


Beta function

24

x(s) = Ax cos( K xs +ϕx )

Kx =1βx2

Ax = εxβx

#

$%%

&%%

βx is called beta function. It has dimension of a length [L] and depends on the position along the ring.

The beta function is the local spatial periodicity of the betatron oscillation.

λx = 2πβxβx =1Kx

=1

k(Quadrupolo) + 1ρ2(Dipolo)

The beta function is a property of the ring optics (straight section, dipoles, quadrupoles, …)

Kx = k +1ρ2

x(s) = εxβx cos(sβx

+ϕx )

ξ

ξ(Quadrupole)

βx=

εxβxπεxβx

π


Single particle motion and beta function

25

!•The single particle makes betatron oscillations (one horizontally and one vertically) along the ring with local spatial period given by 2πβ !

•Easy case is β = constant along the ring

x

s

λx=2πβx

Ax = εxβx


Tune Q

26

Qx=number of horizontal betatron oscillations along the ring = C/βx

Qy=number of vertical betatron oscillations along the ring= C/βy

4<Q<5 C=2πρ


Stability

27

Q=3/2=1.5 orbit closes after 2 turns

Q=5/3=1.66 orbit closes after 3 turns

A perturbation in phase with the revolution motion every p turns continuously growth like in a child swing with the right push.

It is necessary to keep the TUNE far away from the resonances Qx≠qx/px, Qy≠qy/py (with q and p small integer numbers)


Trajectory in phase space

28


+ϕx )

x '(s) = − εxβxsin( s

βx+ϕx ) x

x’

Area Elisse = πab = π εxβxεx

βx

= εx

The particle in the phase space describes an ellipse of area εx

Maximum amplitude

Max

imum

an

gle

εxβxπ

εxβxπ

εxβxπ

εxπβx

εxπβx

εxπβx


Particle after many turns

29

xt=0

xt=T

xt=2T

x t=nT

• As the tune Q must be far from a rational number, the orbit is not closed on itself but at every turn the phase is different.

• Ellipse circumference is filled uniformly after many turns.

x’

x’

x’

x’


Many particles

30

ϕx = ϕx1 ϕx = ϕx1,ϕx2 ϕx = ϕx1,ϕx2 ,ϕx3 ϕx = ϕx1,ϕx2 ,ϕx3,ϕx4

vx

x

vx

x

vx

x

vx

x

vx

x

•1010 particles of different phases •Ellipse circumference is filled uniformly by many particles.


Motion of many particles = beam

31

x

•1010 particles of different phases, positions, velocity and ampiezze (different amplitudes) •Full ellipse

vx

This is the beam in one transverse direction


Beam emittance

32

x

x’

εxβxπ

εxπβx

εx is called emittance. It is a property of the beam (it depends on the source) and it is constant along the ring.

Many particles in the injection have different position and slope filling the ellipse of the emittance


+ϕx )

x '(s) = − εxβxsin( s

βx+ϕx )

Area Elisse = πab = π εxβxεx

βx

= εx

εxβxπ

εxβxπ

εxπβx

εxπβx


Beam size and divergence

33

This is the beam in one transverse direction

Beam size = εxβxπ

εxπβx

Beam divergence =

If the beta function increase the beam increase and the divergence decrease.

xεxβxπ

εxπβx

x’


Beam size and acceptance

34

sezione del fascio = π × raggio orizzontale × raggio verticale = εxβx εyβy

The beam size depends on the horizontal and vertical emittance (depends on the source) and from the ring optics which propagates

the beam ( straight sections , dipoles , quadrupoles, ... )

Dense beams are obtained with low emittance and low beta

The acceptance is the vacuum tube transverse area where the beam circulates and must be larger than beam size otherwise the beam will be cut along the exceeding dimensions.

Beam size horizontal radius x vertical radius


Liouville’s theorem

35

In a conservative physical system the particles density in phase space is preserved: qv = constant. •εx is the area of the beam phase space ellipse and is constant along the entire ring •The shape of the ellipse follows the periodicity of the ring beta function.

βx (s) is a local properties of the ring optics and has the same spatial periodicity s

xvx

Focusing: Large βx: Large amplitude and small slope (large beam - small divergence) Defocusing: Small βx: Small amplitude and large slope (narrow beam - large divergence)

F FD


Transport along the ring

36

In a strong focusing machine typical values are x~mm e x’~mrad

After one turn the particle makes about Q~1.66 oscillation also if β along the ring has periodicity 8 like the ring lattice.

s

xx ' = dx

ds

The single particle motion (green line) has periodicity Q and β determine the envelope (√εβ/π) of all beam particles (green area) then the beam has periodicity β

FODO Cell


Longitudinal dynamicsLet’s turn the RF on V>0

• The particles at every RF cavity crossing receive an increase of energy equal to the potential V of the RF cavity.

• The magnetic dipoles keep the particles along the orbit increasing the intensity of the magnetic field B field synchronously with the increase of the particle energy .

37


Increase of energy

38

Orbit length = C Particle velocity = v

Revolution period = T = C/v

V(t) = V0 sin(2πfRFt)

T = harmonicoTRFIdeal particle synchronous condition. !Example:

C=3000m=3E3m v=c/10=3E7m/s T=1E-4s=100us f=10kHz fRF=200MHz hharmonic=fRF/f=20,000

At each turn the particle gain an energy equal to V0 (1-100MV)→E=Einitial+(turns)xV0

Acceleration voltage of the RF cavity

- +


Harmonic number

39

T = harmonicoTRF

4 BUNCHES in the ring then harmonico-4 BUCKETS are empty

The minimum distance between bunches is dmin=v*TRF

Example: v=c/10=3E7m/s fRF=200MHz=2E8Hz TRF=1/fRF =5E-9s=5ns dmin=15E-2m=15 cm

The particle groups synchronous with the radio-frequency are said BUNCHES and are at most equal to the harmonic number, which is equal to the number of BUCKETS, this is the possible synchronous particle groups.

- +


Longitudinal phase

40

V(t) = V0 sin(2πfRFt)

Varrivo = V0 sin(2πfRFtaarrivo ) = V0 sin(ϕarrivo )

t = tarrivo

Varrivo

At each turn the particle gain and energy given by Varrivo< V0

More realistically an ideal particle has a phase with respect to the radio-frequency.

- +


No acceleration

41

If the phase is zero the ideal particle is not accelerated and go head at constant velocity

- +

ϕarrivo = 0


A late particle at the 1st turn

42

- +

A late particle (black) gains more energy with respect to the ideal particle (red).


Late particle 2nd turn

43

- +

The ideal particle (red) gains the correct energy to arrive to the same phase after a turn. The late particle (black) gained more energy with respect to the ideal particle (white and higher position on the radio-frequency potential in the 1st turn) reducing its delay in the 2nd turn.


Late particle 3rd turn

44

- +

The ideal particle (red) gains the correct energy to arrive to the same phase also after two turns. The late particle (black) gained more energy with respect to the ideal particle (white and higher position on the radio-frequency potential in the 2nd turn) zeroing the delay in the 3rd turn and higher velocity with respect to the ideal particle.


Late particle 4th turn

45

- +

The ideal particle (red) gains the correct energy to arrive to the same phase also after three turns. The late particle (black) gained the same energy like the ideal particle (white and same position on the radio-frequency potential in the 3rd turn) but with higher velocity and advance the ideal particle.



46

- +

The ideal particle (red) gains the correct energy to arrive to the same phase also after four turns. The late particle (black) gained less energy with respect to the ideal particle (white and lower position on the radio-frequency potential in the 4th turn) and decreases the advance with respect to the ideal particle.



47

- +

The ideal particle (red) gains the correct energy to arrive to the same phase also after five turns. The late particle (black) gained much less energy with respect to the ideal particle (white and lower position on the radio-frequency potential in the 5th turn) and decreases the advance even more with respect to the ideal particle.



48

- +

The ideal particle (red) gains the correct energy to arrive to the same phase also after six turns. The late particle (black) gained much less energy with respect to the ideal particle (white and lower position on the radio-frequency potential in the 5th turn) zeroing the advance but with less velocity with respect to the ideal particle.


Phase stability principle

49

The velocity (or energy or phase) of the synchronous particles oscillate between a minimum and a maximum ( BUCKET ) determined by the resonant cavity radio-frequency ( synchrotron oscillations ).

The particles with a phase corresponding to the increase of the RF acceleration potential are in a stable region ( PRINCIPLE OF PHASE STABILITY ).

Phase stability

Phase instability Δφ

Δp

Longitudinal emittance


Transition energy

50

So far we have neglected the increase of trajectory due to the increase of the momentum that for the same magnetic field causes an increase of the radius of curvature in the arcs .

At relativistic energies the speed of light is reached and it can not rise more but the momentum continues to increase . The particle will perform a larger trajectories and the revolution is increased ( particle delayed).

At low energies the acceleration causes an increase of velocity overcompensating the increase of momentum and the time of revolution is reduced ( particle advance ) .

Stability region -π/2<φ<π/2

Stability region π/2<φ<2π

- +

+ -

In the first synchrotron the transition energy crossing turn-out safe for the beam!!!


Adiabatic shrinking of the beam emittance

51

ε1p1 = ε0p0 → ε1 = ε0p0p1→

ε1β1π

=ε0β0π

p0p1

xy

εx,0βxπ

spazio fasi classico = xv→ ε = cos tan tespazio fasi relativistico = xp→ εp = cos tan te

The Liouville theorem at relativistic energies must be generalized taking into account the new dynamic variables

εy,0βyπ

xy

εx,1βxπ

εy,1βyπ

At relativistic energies the beam radius is reduced with the acceleration as the square root of the moment

Relativistic acceleration

NB: For electrons Liouville 's theorem does not apply because they emit a lot of energy in the form of X-rays when they follow orbits curves ( synchrotron radiation )


Syncrotron light

52

• The electrons in curved orbits emit a lot of energy in the form of X-rays ( synchrotron radiation )

• The emission of synchrotron light is a dissipative phenomenon that naturally dumps the betatron and synchrotron oscillations.

• The transverse dynamic( betatron oscillations ) and longitudinal ( synchrotron oscillations ) are coupled and can not be treated independently.

• The Liouville theorem for electrons is not applicable because they emit energy to the outside. The reduction of the emittance is not due to the adiabatic shrinking like protons but to the emission of X-rays that make them lose energy.

• The energy emitted by radiation is so high that the RF cavities are many and powerful

Collimated X ray cones

Circulating electrons

The synchrotron light has many applications in physics, chemistry,

biology , materials , ...


Momentum dispersion

53

p→ p + Δp

What happens if the energy of the particle is different from that of the ideal particle?

• The dipoles cause the horizontal displacement of the orbit called DISPERSION

• The quadrupole cause the change of the TUNE called that CHROMATICITY

Reasons : source and acceleration involve a finite longitudinal emittance ( synchrotron oscillation )

ΔφΔp


Dispersion function

54

ρ

xp+Δp (s) = Dx(s)Δpp

xp(s) Typical values

Δpp~ 10−3;Dx ~ 1m;xS ~ 1mm

Dx(s) =xp+Δp (s)Δp / p

The dispersion function is the ideal orbit for Δp/p=1

The dispersion function Dx( s) is the proportionality between the ideal orbit displacement and

the percentage change of the moment of the ideal particle.

p→ p + Δp


Dispersion and beam section

55

raggio verticale = εyβy

π

raggio orizzontale = εxβx

π +Dx

Δpp

Vertical radius

Horizontal radius


Dipoles are sources of dispersions

56

Dx~(Ldipolo)2/ρ !

Dy= 0


Quadrupoles are sources of chromaticity

57

Ideal energy

Energy < Ideal energy

Energy > Ideal energy

ΔQ = Q' Δpp→ Q' = ΔQ / Δp

p#

$%&

'(

The chromaticity Q is the change of the number of revolutions (TUNE) for Δp/p=1 or equivalently the proportionality factor between the TUNE change and the relative change of the moment p .

Higher the particle momentum less focussing the quadrupole because harder to bend.

How the TUNE changes with momentum

p→ p + Δp


Sextupole (green)

58

Un modo per correggere la cromaticita’ della macchina e’ mettere dei sestupoli dopo ogni quadruple The TUNE determines the stability of the beam and should always be kept away from rational numbers q/p with q and p small for all momentum values between p-Δp and p+Δp.. The chromaticity is very important because it determines the stability of the accelerator. One way to correct the chromaticity is to put a correction sextupole after each quadrupole

Quadrupole Sextupole

Bx = −gyBy = −gx

"#$

%$

Bx = %gyxBy = %g x2 − y2( ) / 2

"#$

%$Linear in the position Quadratic in the position


Summary

59

• Transverse dynamics

• Betatron oscillations

• Weak focusing: dipoles

• Strong focusing: quadrupoles F+D

• Beta function and tune

• Emittance and beam

• Longitudinal dynamics

• Phase stability

• Synchrotron oscillations

• Transition energy

• Momentum dispersion

• Dispersion in dipoles

• Chromaticity in quadrupoles: correction sextupoles


Appendix

60


Equzione del moto (1)• Radialmente (orizzontalmente) agisce la forza centrifuga e la

forza magnetica

• Verticalmente agisce la forza di gravita’ che pero’ trascuriamo

• Longitudinalmente agisce l’accelerazione delle cavita’ risonanti che in questa lezione non consideriamo (vedi lezione sulla dinamica longitudinale)

61

Fcentrifuga =mv2

r=mv2

ρ + xFdipolo = qB0vmax = Fcentrifuga + Fdipolo

max =mv2

ρ + x+ qB0v

r

- qB0v

- qB0v


Equzione del moto (2)

62

max =mv2

ρ + x+ qB0v = mv

2 1ρ1− x

ρ

#

$%&

'(+ qB0v

<<xρ) →)) max = mv2 1ρ1− x

ρ

#

$%&

'(+ qB0v

1ρ + x

=1ρ

11+ x

ρ

"

#$

%

&'

x<<ρ( →((1ρ(1− x

ρ)

max = mv2 1ρ(1− x

ρ) + qB0v

11+ x

ρ

x<<ρ" →"" 1− xρ

1ρ + x

=1ρ

11+ x

ρ

"

#$

%

&'

- qB0v

- qB0v - qB0v


Equzione del moto (3)• Il tempo t non ci interessa.

• Interessa sapere la distanza s percorsa nell’anello per sapere dove sta la particella e se tocca le pareti.

• Il tempo e la distanza percorsa che differiscono solo per la velocita’ tangenziale: s=vt.

63

t→ s = vt

max(s)v2 = mv2 1

ρ(1− x

ρ) + qB0v

vx → vx / v

ax → ax (s) / v2 → ax (s) = axv

2

• Velocita’ trasversa diventa angolo

• Acce leraz ione t rasversa s i moltiplica per v2

- qB0v


Equzione del moto (4)

E’ l’equazione dell’oscillatore armonico con frequenza angolare ω=2πf

64

ax (s) =1ρ(1− x

ρ) + qB0

mv

ax (s) = −xρ2

ax (t) = −ω2x

pq= ρB-


Dispersione

65

qB0mv

=qB0p '

=qB0p + Δp

=qB0p

1− Δpp

#

$%&

'(=qB0p

−qB0Δpp2

=1ρ−Δpρp

ax (s) + xy =1ρΔpp x(s) = xomogenea (s) + xspeciale (s)

La dispersione del momento introduce un termine omogeneo nella equazione del moto quindi la soluzione generale e’ data dalla somma delle soluzioni dell’equazione omogena (vedi slide precedenti) ed una soluzione speciale. La soluzione speciale normalizzata rispetto alla variazione percentuale del momento e’ detta funzione di dispersione Dx(s).

p→ p + Δp

Dx(s) =xspeciale (s)Δp / p La funzione di dispersione e’ l’orbita ideale per Δp/p=1


Equazione del moto con quadrupolo

66

ax (s) = − k + 1ρ2

#

$%&

'(x

ay(s) = kx

)

*++

,++

k = 0.3 g(T / m)p(GeV / c)

1ρ= 0.3 B(T)

p(GeV / c)

x(s) = Ax cos(2πsλx

+ϕx )Soluzione focalizzata forte in direzione orizzontale

y(s) = Ay cosh(2πsλy

+ϕy )Soluzione defocalizzata in direzione verticale

cosh(z) = ez + e−z

2

Coseno iperbolico

max =mv2

ρ + x+ qB0v − qvgx =

mv2

ρ + x+ qB0v −

mv2gxmv / q

=

mv2

ρ + x+ qB0v −mv

2kx ~ mv2

ρ(1− x

ρ) + qB0v −mv

2kx =

1ρ2

→ k + 1ρ2

Rigidita’ magnetica (dipolo)

Gradiente normalizzato (quadrupolo)

- qB0v - qB0v

- qB0v - qB0v

Transverse and longitudinal dynamicschiodini/didattica/phd2015/Lessons...Introduction to...

Documents

Transcript of Transverse and longitudinal dynamicschiodini/didattica/phd2015/Lessons...Introduction to...