Transverse Beam Dynamicsor how to keep all particles inside beam chamber

Piotr SkowronskiIn large majority based on slides of

B.Holzer https://indico.cern.ch/event/173359/contribution/9/material/0/0.pdfF.Chautard http://cas.web.cern.ch/cas/Holland/PDF-lectures/Chautard/Chautard-final.pdfW.Herr http://zwe.home.cern.ch/zwe/O.Brüning http://bruening.web.cern.ch/bruening/Y.Papaphilippou http://yannis.web.cern.ch/yannis/teaching/

2

Betatron & Cyclotron• Particles move in magnetic field

3

• Geometric focusing– All particles are confined in

uniform magnetic field– The beam size depends on

the initial spread

• What about focusing in vertical plane ?

Betatron & Cyclotron

4

Vertical focusing in betatron & cyclotronThe weak focusing

• For vertical stability value Bx must grow in vertical direction to provide restoring force – That is deflect particles back to the center plane

• But this implies decreasing field and defocusing in horizontal plane because– This defocusing can not be stronger from geometric focusing

𝑟𝑜𝑡 𝐵=0⇒𝜕𝐵𝑥

𝜕 𝑦 =𝜕𝐵𝑦

𝜕 𝑥

5

Equation of motion in uniform magnetic field

• We choose coordinate system around the reference orbit • Restoring force

linear dependence in vicinity of ref. trajectory• For convenience define magnetic field index

– Rate that field changes in space normalized to Bz0

• ;

6

Vertical focusing in cyclotron and betatronThe weak focusing

• Both coefficients n and (1-n) must be positive so both equations are of harmonic oscillator: 0 < n < 1

7

Betatron motion• Harmonic oscillator in both planes with

– revolution frequency

• Changing the gradient increases oscillation frequency in one plane and decreases in another

• Perfect isochronism is not possible since field changes radially

8

Azimuthally Varying Filed Cyclotron• Cyclotron works only to limited energies since increasing

particle mass breaks isochronism• To restore it magnetic field must increase radially– This contradicts vertical stability condition• Solution: sectors with different fields and gradients– Orbit is not a circle, trajectory is not perpendicular to sector edge Not uniformity of the field on the edge gives vertical focusing

(we talk about it later)

9

Further optimizations• Separated sectors• Spiralled sectors increase the edge crossing angle (and foc. strength)• RF cavities instead of Dee’s

TRIUMF, Vancouver, Canada, during construction ~1972520MeV protons18m diameterStill in operation

10

Synchrocyclotron• Another solution is to modulate RF frequency– This limits big advantage of cyclotron that produces

almost continues beam: cyclotron produces one bunch per RF cycle

– But it is able to reach GeV range

11

Synchrotron• Constant radius machine– Magnetic field and RF frequency is modulated

• For the really strong guys where v/c is close to 1 also RF freq. constant

12

Separate function elements• Bends• Combined function bends

– Bend and quad together• Quadrupoles• Accelerating Cavities• Sextupoles, Skew Quadruples, Octupoles, Decapoles …• Kickers, dumpers, oscillation modulators, …• Measurement devices, collimators

• A single Hamiltonian or force can not be defined for a whole accelerator

13

Quadrupoles• Focusing with quadrupoles– Element acting as a lens– Deflection is linearly proportional to position– Magnetic field increases linearly with position

14

Focusing with quadrupoles• A magnet that fulfills requirement has 4 poles with

parabolic shape• Due to nature of magnetic field

we can not have magnet that focuses in both planes – When it focuses in one it defocuses in the other one

(𝑟𝑜𝑡 𝐵=0⇒𝜕𝐵𝑥

𝜕 𝑦 =𝜕𝐵𝑦

𝜕 𝑥 )

15

Focusing with quadrupoles• Still, focusing and defocusing lenses can be adjusted to

confine the particles within finite space in both planes•

16

FODO cell• The easiest configuration is FODO Focus (F) – Drift (O) – Defocus (D) – Drift (O)

17

Coordinate SystemCanonical Variables

• Use position along accelerator s as time like variable• x : deviation from reference trajectory

– Same for y• Canonical momentum x’ is

• Longitudinal position is z = β c dt – where dt deviation in time from reference trajectory

• Longitudinal momentum is

18

Pseudo-harmonic SolutionHill’s equation

• Equation of motion for a separate function elements can be written in form of a pseudo oscillator

– The force k(s) is different in each magnet and depends on position along the accelerator• For convince k normalized to beam rigidity

– Takes the beam energy out of equation• For bending magnets k = B/(p/q) = B/(Br) = 1/r• For quadrupoles k = g/(Br), where g is the field gradient

– Useful equations to remember

�̈� (𝑠)+𝑘 (𝑠 )𝑥=0

=

𝑘𝑄=0.298⋅ 𝑔[𝑻 /𝒎]𝑝 [𝑮𝒆𝑽 /𝒄]

19

Solution of Hills Equation• We guess that solution is of the form

• Recall: s is our time-like variable s=bct, f(s)=ω(s)·s• It is a harmonic oscillator with time varying amplitude and

frequency• It is the same as

– A weight on a spring that changes it strength in time– A ball rolling in a gutter that has different radius along it

• The solution can be expressed in matrix form– Any solution can be represented of linear combination of

• pure cosine-like solution (x’i=0)• and pure sine-like solution (xi=0)

𝑥=𝑎1 (𝑠 )cos (𝜙 (𝑠 )+𝜙0 )+a2 (s )sin (𝜙(𝑠)+𝜙0 )𝑥′=𝑎3 (𝑠 )sin (𝜙 (𝑠 )+𝜙0 )+a4 (s ) cos (𝜙 (𝑠)+𝜙0 )

[ 𝑥𝑥′ ]=[ C 𝑆𝐶 ′ 𝑆′ ] [ 𝑥 𝑖

𝑥 ′𝑖❑ ]

20

Solution of Hills Equation• Insert cos like solution into Hills equation

2 �̇� (𝑠 ) �̇� (𝑠 )+2𝑎 (𝑠 ) �̈� (𝑠)=(𝑎 (𝑠 ) �̇� (𝑠 ) ) ′=0

𝜙(𝑠)=∫𝑠 1

𝑠 2 1𝑎2 (𝑠 )

𝑑𝑠

21

How to find the solution?• OK, we know how the solution looks like, but how to get

it for my particular machine• Or usually inverse: we want a given solution, how to

distribute our elements• We start from simplest cases: find solutions for each

element where k=const

22

Focusing quadrupole• k=const>0, it is harmonic oscillator with

a1, a2 and f0 depend on initial condition, i.e. coordinates of the particle entering the quadrupole, so we rewrite

Focusing quadrupole of length l transforms coordinates

[ 𝑥𝑥′ ]=[ c os (√𝑘 ∙ 𝑙) 1√𝑘sin (√𝑘 ∙ 𝑙)

√𝑘sin (√𝑘 ∙ 𝑙) c os (√𝑘 ∙𝑙) ] [ 𝑥 𝑖

𝑥 ′𝑖❑ ]

23

Transfer matrix of focusing quadrupole

[ 𝑥𝑥′ ]=[ c os (√𝑘 ∙ 𝑙) 1√𝑘sin (√𝑘 ∙ 𝑙)

√𝑘sin (√𝑘 ∙ 𝑙) c os (√𝑘 ∙𝑙) ] [ 𝑥 𝑖

𝑥 ′𝑖❑ ]

24

Defocusing quadrupole• Negative k gives exponents as solution• The same way as for focusing quadrupole

rewrite it using hyperbolic functions

[ 𝑥𝑥′ ]=[ c o sh (√𝑘 ∙ 𝑙 ) 1√𝑘si nh (√𝑘 ∙ 𝑙 )

√𝑘si nh (√𝑘 ∙ 𝑙 ) c o sh (√𝑘 ∙ 𝑙 ) ][ 𝑥𝑖𝑥 ′𝑖❑]

25

Quadrupole transfer matrix• Together for horizontal and vertical planes

Focusing:

[ 𝑥𝑖𝑥 ′𝑖𝑦 𝑖

𝑦 ′𝑖][ 𝑥𝑥 ′𝑦𝑦 ′ ]=¿

26

Quadrupole transfer matrix• Together for horizontal and vertical planes

Defocusing

[ 𝑥𝑖𝑥 ′𝑖𝑦 𝑖

𝑦 ′𝑖][ 𝑥𝑥 ′𝑦𝑦 ′ ]=¿

27

Drift• No force, a free particle motion

[ 𝑥𝑥′ ]=[1 𝑙0 1 ][ 𝑥 𝑖

𝑥′ 𝑖❑]

𝑥=𝑥𝑖+𝑙 𝑥′ 𝑖❑𝑥′=𝑥′𝑖❑

28

Accelerator Line• Accelerator is a sequence of elements• Each one has its transfer matrix• Transfer matrix of accelerator line is product of its

elements transfer matricesM=MDrift3MQuadF2MDrift2MBendMDrift1MQuadD1

This matrix describes motion of a single particle (green line)not the envelope (red line)!

QF2 QF3QF1QD2 QD3QD1

29

Twiss parameters• The one turn map, the map that is obtained for a whole

ring, is solution of the Hills equation

• If we take one particle and follow it for several turns𝑥=𝑎1 (𝑠 )cos (𝜙 (𝑠 )+𝜙0 )+a2 (s )sin (𝜙(𝑠)+𝜙0 )

𝑥

𝑝 𝑦=𝑦 ′𝑝𝑥=𝑥 ′

Phase space ellipse

30

Phase space ellipse• Every particle follows an ellipse of the same shape– “Radius” (properly called action) and starting point depend on

particle initial conditions– Of course most of the particles are in the centre and

less and less towards outside

𝑥

𝑝 𝑦=𝑦 ′𝑝𝑥=𝑥 ′

31

Beam size• For a whole bunch it gives

distribution is space– Distribution of particles

usually close to Gaussian

𝑥

𝑝𝑥=𝑥 ′

σy

σx

32

Twiss parameters and emittance• Take the ellipse that corresponds to 1 σ• Define beam sizes as

– Where Ɛ is area of the ellipse, “temperature” of the beam– What for? Because conservative forces do not change volume of the

phase space (Liouville theorem) Ɛ=const• Beta function shows how big the beam is at a given point of

accelerator– γ for momentum– α is tilt of ellipse and it is

anti-proportional to derivative of b

𝜎=√𝜖𝛽

33

Envelope• Every bunch of particles has some initial spread in

position and angle– Each starts with some different initial condition

• The function describing beam size along accelerator line is called beam envelope

Hills equation• Rewrite solution of the Hills equation (or one turn map)

for a ring using these 3 new parameters

• What is μ? (often also referred as Q)Total phase advance for one turn.• Since multiple of 2π are not important

we care only about fractional part: tuneLHC: total phase advance (at collisions)64.31 and 59.32 (hor. and vert.)LHC tunes: 0.312π and 0.322π

34

𝑥

𝑝𝑥=𝑥 ′

μ

Tune is a constant of a machine

Ring special case: Periodic solution!

35

Meaning of Twiss parameters• β(s) describes beam size at given point of accelerator• γ(s) describes spread (size) in momentum• det[M]=1 – As smaller beam size as bigger momentum spread

• Periodicity implies– Alpha is proportional to derivative of the beta function– That is divergence of the beam size (with minus sign)

• Putting above 2 relations the general solution

𝛽=1−𝛼2

𝛾

𝛽 (𝑠+𝐿 )=𝛽 (𝐿)⇒− 12𝑑𝛽𝑑𝑠

36

Equation of the phase space ellipse

37

Solution of Hills Equation• Insert cos like solution into Hills equation

2 �̇� (𝑠 ) �̇� (𝑠 )+2𝑎 (𝑠 ) �̈� (𝑠 )=(𝑎 (𝑠 ) �̇� (𝑠) ) ′=0

𝜙(𝑠)=∫𝑠 1

𝑠 2 1𝑎2 (𝑠 )

𝑑𝑠=∫𝑠1

𝑠2 1𝛽 (𝑠 )

𝑑𝑠

38

Emittance• b function describes how envelope evolves along

accelerator– The beam size , where is emittance

• Emittance definition: phase space area occupied by the beam

– You can think of it as the beam temperature

• Emittance is constant (Liouville theorem) • It is conservation of energy• In transverse there is no energy dissipation or gain,

no friction, no heating

39

Stability condition for a ring • The map obtained for a whole circular accelerator is

called One Turn Map– For each turn that a particle makes around a ring its coordinates

are modified according to the map zi+1= M zi

– We want the beam to be stable => it means that coordinates of zn = Mn z0 must be finite when n going to infinity

– Necessary and sufficient condition is that Mn is also finite– Consider eigenvectors Y and eigenvalues l of the one turn map

MY= lI, (I is identity matrix)– For arbitrary M made of a,b,c,d,

det(M- lI)=0 l2 + l(a+d) + (ad-bc)=0; (ad – bc)==det(M)=1– Solution for l exist if (a+d)/2=tr(M)/2 < 1; l1l2=1– What means that total phase advance must be real:

tr(M)=2cos(μ)

40

Hills equation solution fora line (transfer line or linac)

41

Hills equation solution fora line (transfer line or linac)

42

Twiss parameter propagation• Emittance is constant

• Coordinates transforms as

• Insert to above and compare coefficients

43

Twiss parameter propagation• Propagation can be defined via transfer matrix elements

• And in a matrix notation

– Having matrix to propagate a single particle we obtain matrix to propagate the ensemble

44

Adiabatic damping• Phase space area is conserved only for a given energy– Is Louiville wrong? NO!– It is coordinate system that we use:

we defined canonical momentum as• When particle accelerates – p0 increases– x’ decreases– emittance shrinks as 1/γrel

• And the beam size also shrinks as• Normalized emittance that is conserved εN=ε*γrel

• To avoid confusion ε is referred as geometric emittance

𝑥′=𝑝 𝑥

𝑝0

45

Example optics• So how we design optics?– Every accelerator is different – Its optics is optimized for • a given job• to circumvent given problems

• Example: LHC– In the arcs there is regular smooth optics: FODO cell• FODO is one of the easiest and permissive solutions, use it whenever the beam needs to

be simply transported through and there are no other constraints– Prepare the beam for collisions: focus it as much as possible to maximize collision

cross section (luminosity)– ATLAS(1) and CMS (5) need smallest beams, ALICE (2) and LHCb (8) less demanding– Other straight sections: accelerating cavities (4), ejection to dump (6),

halo cleaning (7), off momentum cleaning (3)

3 4 7 85 61 2

46

FODO cell• The easiest configuration Focus (F) – Drift (O) – Defocus (D) – Drift (O) - …

47

FODO cell matrix

48

FODO cell matrix

49

FODO CELL parameters

max,min

Stability diagram

Min β at μ=~76°

50

Reaching small beam sizes for collisions

• Beta function along drift is parabolic

• If we start from minimum point α0 = 0 and γ0=1/β0

– It is called “waist”

𝛽 (𝑠 )=𝛽0−2 ⋅𝛼0⋅𝑙+𝛾0 ⋅ 𝑠2

𝛽 (𝑠 )=𝛽0−𝑠2𝛽0

51

Reaching small beam sizes for collisions• As smaller β0 as faster β grows with s– β at collision point is referred as β*

– Distance from last quad to collision point is L*

s

β ~10m

52

• The experiment asks for highest lumi– That is smallest beam size at Interaction Point (IP)

• But they don’t let you put any quadrupole close• Make β as big as possible at the last (Final Focus) quads– Technological challenge: highest aperture and strength– In LHC to squeeze beam to 60cm it goes to 4.5km in FF quads

Reaching small beam sizes for collisions

53

Chromatic effects• All above was for an ideal case:

particles with the same energy– Monochromatic beam

• It is impossible to produce all particles with the same momentum– Even if it was possible via some “laser like” process then

particles in a bunch interact with each other and also with electromagnetic fields exchanging photons and changing its energy. They also emit photons when changing direction (synchrotron light)

• Every beam has some energy (momentum) spread

54

Hills equation with off-momentum• Deflection angle of a bend magnet and orbit change

with particle momentum

55

Hills equation with off-momentum• Deflection angle of a bend magnet and orbit change

with particle momentum

56

Dispersion• Deflection angle of a bend magnet and orbit change

with particle momentum

Dispersion:

Dispersive orbit

57

Transfer matrices with dispersion

58

Transfer matrices with dispersion

59

Dispersion equation• Use definition of dispersion and plug it to Hills equation

• Can be solved for each element type to obtain dispersion transfer map

• General solution

60

Qauds affect dispersion

• But you can not null dispersion with quads only

61

Transfer maps with dispersion• Bend

• Focusing quad

• Defocusing quad

62

Chromaticity• As particle has higher energy and it is more rigid,

quadrupoles focus them less– Strength of quadruple is reduced by

• It makes less oscillations along accelerator line:phase advance depends of particle energy

Δ k=− Δpp0⋅𝑘

𝜙=𝑄 ′ Δpp0Q’ is chromaticityNatural chroma (purely linear optics) is always negative

63

Chromaticity correction• We need an element to correct chromaticity– It needs to revert the effect of quadrupole– Focusing linearly increases with position offset • Its strength should be quadratic with position offset

α≈x2

64

Chromaticity correction• Sextupole

• In a bunch particles with different energies are mixed, we have to sort them first: dispersion• Sextupole in a dispersive region corrects chroma– As bigger dispersion and beta as weaker sextupole

is sufficient– Equation developed next time• Sextupoles can also harm– Coupling– Nonlinear force can lead to

chaotic motion

65

END• Next in transvers beam dynamics to follow– Errors from different kinds of elements– Resonances– Tune diagram– Amplitude detuning– Coupling– Dynamic aperture– Filamentation– Decoherence– Orbit correction