Benasque
Accelerator Physics Transverse motion
Elena Wildner
Acknowldements to Simon Baird, Rende Steerenberg, Mats Lindroos, for course material
E.Wildner NUFACT08 School 1 Benasque Accelerator co-ordinates
Vertical Horizontal It travels on the central orbit
Longitudinal
Rotating Cartesian Co-ordinate System
E.Wildner NUFACT08 School 2 2 Benasque Two particles in a dipole field
9 What happens with two particles that travel in a dipole field with different initial angles, but with equal initial position and equal momentum ?
Particle A
Particle B
9 Assume that Bρ is the same for both particles. 9 Lets unfold these circles……
E.Wildner NUFACT08 School 3 3 Benasque The 2 trajectories unfolded
9 The horizontal displacement of particle B with respect to particle A. Particle B Particle A displacement 0 2π
9 Particle B oscillates around particle A. 9 This type of oscillation forms the basis of all transverse motion in an accelerator. 9 It is called ‘Betatron Oscillation’
E.Wildner NUFACT08 School 4 Benasque Dipole magnet
9 A dipole with a uniform dipolar field deviates a particle by an angle θ. 9 The deviation angle θ depends on the length L and the magnetic field B. 9 The angle θ can be calculated: θ ⎛θ ⎞ 1 LBL sin⎜ ⎟ == ⎝ ⎠ 22 2 ()Bρρ
9 If θ is small: L L 2 2 ⎛θ ⎞ θ sin⎜ ⎟ = ⎝ ⎠ 22 9 So we can write: LB θ θ ρ θ = 2 2 ()Bρ
E.Wildner NUFACT08 School 5 Benasque ‘Stable’ or ‘unstable’ motion ?
9 Since the horizontal trajectories close we can say that the horizontal motion in our simplified accelerator with only a horizontal dipole field is ‘stable’
9 What can we say about the vertical motion in the same simplified accelerator ? Is it ‘stable’ or ‘unstable’ and why ? 9 What can we do to make this motion stable ?
9 We need some element that ‘focuses’ the particles back to the reference trajectory.
9 This extra focusing can be done using: Quadrupole magnets
E.Wildner NUFACT08 School 6 Benasque Quadrupole Magnet
9 A Quadrupole has 4 poles, 2 north Magnetic and 2 south field
9 They are symmetrically arranged around the centre of the magnet
9 There is no magnetic field along the central axis.
Hyperbolic contour x · y = constant
E.Wildner NUFACT08 School 7 Benasque Resistive Quadrupole magnet
E.Wildner NUFACT08 School 8 Benasque Quadrupole fields
9 On the x-axis (horizontal) the field Magnetic field is vertical and given by: By ∝ x 9 On the y-axis (vertical) the field is horizontal and given by: Bx ∝ y
9 The field gradient, K is defined as: ()Byd (Tm −1 ) dx
K 9 The ‘normalised gradient’, k is defined as: m −2 )( ()Bρ
E.Wildner NUFACT08 School 9 Benasque Types of quadrupoles
Force on 9 This is a: particles Focusing Quadrupole (QF)
9 It focuses the beam horizontally and defocuses the beam vertically.
9 Rotating this magnet by 90º will give a: Defocusing Quadrupole (QD)
E.Wildner NUFACT08 School 10 Benasque Focusing and Stable motion
9 Using a combination of focusing (QF) and defocusing (QD) quadrupoles solves our problem of ‘unstable’ vertical motion. 9 It will keep the beams focused in both planes when the position in the accelerator, type and strength of the quadrupoles are well chosen. 9 By now our accelerator is composed of: 9 Dipoles, constrain the beam to some closed path (orbit). 9 Focusing and Defocusing Quadrupoles, provide horizontal and vertical focusing in order to constrain the beam in transverse directions. 9 A combination of focusing and defocusing sections that is very often used is the so called: FODO lattice. 9 This is a configuration of magnets where focusing and defocusing magnets alternate and are separated by non- focusing drift spaces.
E.Wildner NUFACT08 School 11 Benasque FODO cell
9 The ’FODO’ cell is defined as follows:
QF QD QF
L1 L2
‘FODO’ cell
Or like this…… Centre of Centre of first QF second QF
E.Wildner NUFACT08 School 12 Benasque The mechanical equivalent
9 The gutter below illustrates how the particles in our accelerator behave due to the quadrupolar fields.
9 Whenever a particle beam diverges away from the central orbit the quadrupoles focus them back towards the central orbit.
9 How can we represent the focusing gradient of a quadrupole in this mechanical equivalent ?
E.Wildner NUFACT08 School 13 Benasque The particle characterized
9 A particle during its transverse motion in our accelerator is characterized by: 9 Position or displacement from the central orbit. 9 Angle with respect to the central orbit.
x = displacement x’ x’ = angle = dx/ds
x x s ds
dx
9 This is a motion with a constant restoring coefficient, similar to the pendulum
E.Wildner NUFACT08 School 14 Benasque Hill’s equation
These transverse oscillations are called Betatron Oscillations, and they exist in both horizontal and vertical planes.
The number of such oscillations/turn is Qx or Qy. (Betatron Tune) (Hill’s Equation) describes this motion d 2 x xsK =+ 0)( ds 2 If the restoring coefficient (K) is constant in “s”then this is just SHM
Remember “s” is just longitudinal displacement around the ring
E.Wildner NUFACT08 School 15 Benasque Hill’s equation (2)
9 In a real accelerator K varies strongly with ‘s’. 9 Therefore we need to solve Hill’s equation for K varying as a function of ‘s’ d 2 x xsK =+ 0)( ds 2
9 The phase advance and the amplitude modulation of the oscillation are determined by the variation of K around (along) the machine.
9 The overall oscillation amplitude will depend on the initial conditions.
E.Wildner NUFACT08 School 16 Benasque Solution of Hill’s equation (1)
d 2 x xsK =+ 0)( ds 2 9 2nd order differential equation. 9 Guess a solution:
= ε β (φ ssx )(cos)(. +φ 0)
9 ε and φ0 are constants, which depend on the initial conditions. 9 β(s) = the amplitude modulation due to the changing focusing strength. 9 φ(s) = the phase advance, which also depends on focusing strength.
E.Wildner NUFACT08 School 17 Benasque Solution of Hill’s equation (2)
α = − β ' 9 Define some parameters 2 = ωβ2 9 …and let φ = (φ s)( +φ 0 ) 1+α 2 Remember φ is still γ = x = ε ω cos(s). φ a function of s β 9 In order to solve Hill’s equation we differentiate our guess, which results in: dω x'= ε cos − sin' φωφεφ ds 9 ……and differentiating a second time gives: x = εω φ − εω φ φ − εω φ sin''sin''cos'''' φ − εωφ sin'' φ − εωφ 2 cos' φ 9 Now we need to substitute these results in the original equation.
E.Wildner NUFACT08 School 18 Benasque Solution of Hill’s equation (3)
9 So we need to substitute = ε β (φ ssx )(cos)(. +φ 0) and its second derivative x = εω φ − εω φ φ − εω φ sin''sin''cos'''' φ − εωφ sin'' φ − εωφ 2 cos' φ
d 2 x xsK =+ 0)( into our initial differential equation 2 ds 9 This gives: − − sin''sin''cos'' − sin'' − 2 cos' φωφεφωφεφφωεφφωεφωε + sK φωε= 0cos)(
Sine and Cosine are orthogonal and will never be 0 at the same time
E.Wildner NUFACT08 School 19 Benasque Solution of Hill’s equation (4)
− − sin''sin''cos'' − sin'' − 2 cos' φωφεφωφεφφωεφφωεφωε + sK φωε= 0cos)(
9 Using the ‘Sin’ terms ω φ +ωφ = 0''''2 ωω φ +ω 2φ = 0''''2
9 We defined β = ω 2 , which after differentiating gives β = ωω'2'
9 Combining ωω φ +ω 2φ = 0''''2 and β = ωω'2' gives: dβ dβ dω = β φ +βφ = ⎜⎛ βφ ⎟⎞ =0'''''' ds dω ds ⎝ ⎠ dφ 1 9 Which is the case as: βφ'= const. =1 since φ'= = ds β 9 So our guess seems to be correct
E.Wildner NUFACT08 School 20 Benasque Solution of Hill’s equation (5)
9 Since our solution was correct we have the following for x: x = ε.β cosφ
dω β' α = =− ds 2ω β
dω 9 x'= ε cosφ − εωφ'sinφ For x’ we have now: ds ω = β 9 Thus the expression for x’ finally becomes: x'= −α ε / β cosφ − ε / β sinφ
E.Wildner NUFACT08 School 21 Benasque Phase Space Ellipse
9 So now we have an expression for x and x’
x = ε β cos. φ and x −= − sin/cos/' φβεφβεα
9 If we plot x’ versus x we get an ellipse, which is called the phase space ellipse.
φ = 3π/2 x’ φ = 0 = 2π
ε.γ ε / β α ε /γ α ε / β
x ε /γ
ε.β
E.Wildner NUFACT08 School 22 Benasque Phase Space Ellipse (2)
9 As we move around the machine the shape of the ellipse will change as β changes under the influence of the quadrupoles 9 However the area of the ellipse (πε) does x’ not change x’
Area = π·r1·r2 ε / β ε / β
x ε.β
x ε.β 9 ε is called the transverse emittance and is determined by the initial beam conditions. 9 The units are meter·radians, but in practice we use more often mm·mrad.
E.Wildner NUFACT08 School 23 Benasque Phase Space Ellipse (3)
x’ x’
ε / β ε / β ε.β x
ε.β x
9 The projection of the ellipse on the x-axis gives the Physical transverse beam size. 9 The variation of β(s) around the machine will tell us how the transverse beam size will vary.
E.Wildner NUFACT08 School 24 Benasque Emittance & Acceptance
9 To be rigorous we should define the emittance slightly differently. 9 Observe all the particles at a single position on one turn and measure both their position and angle. 9 This will give a large number of points in our phase space plot, each point representing a particle with its co-ordinates x, x’. x’ emittance beam
x
acceptance 9 The emittance is the area of the ellipse, which contains a defined percentage, of the particles. 9 The acceptance is the maximum area of the ellipse, which the emittance can attain without losing particles.
E.Wildner NUFACT08 School 25 Benasque Matrix Formalism
9 Lets represent the particles transverse position and angle by a column matrix. ⎛ x ⎞ ⎜ ⎟ ⎝ x'⎠ 9 As the particle moves around the machine the values for x and x’ will vary under influence of the dipoles, quadrupoles and drift spaces. 9 These modifications due to the different types of magnets can be expressed by a Transport Matrix M
9 If we know x1 and x1’ at some point s1 then we can calculate its position and angle after the next magnet at position S2 using:
⎛ sx 2)( ⎞ ⎛ sx 1)( ⎞ ⎛ ba ⎞⎛ sx 1)( ⎞ ⎜ ⎟ = M ⎜ ⎟ = ⎜ ⎟⎜ ⎟ ⎝ sx 2)'( ⎠ ⎝ sx 1)'( ⎠ ⎝ dc ⎠⎝ sx 1)'( ⎠
E.Wildner NUFACT08 School 26 Benasque How to apply the formalism
9 If we want to know how a particle behaves in our machine as it moves around using the matrix formalism, we need to:
9 Split our machine into separate element as dipoles, focusing and defocusing quadrupoles, and drift spaces.
9 Find the matrices for all of these components
9 Multiply them all together
9 Calculate what happens to an individual particle as it makes one or more turns around the machine
E.Wildner NUFACT08 School 27 Benasque Matrix for a drift space
9 A drift space contains no magnetic field. 9 A drift space has length L.
x2 = x1 + L.x1’ x1’
x1 L x1’small
+= Lxxx 112 ' ⎛ x ⎞ ⎡ ⎤⎛ x ⎞ ⎜ 2 ⎟ ⎢1 L⎥⎜ 1 ⎟ ⎜ ⎟ = ⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ x2'⎠ ⎣⎢ 10 ⎦⎥⎝ x1'⎠ 2 += xx 1 '0' }
E.Wildner NUFACT08 School 28 Benasque Matrix for a quadrupole
deflection 9 A quadrupole of length L.
x1 x2
x1’ x2’ Remember By ∝ x and the deflection due to the magnetic field is: LB y LK ⋅−= x ⎜⎛B ⎟⎞ ⎜⎛Bρρ⎟⎞ ⎝ ⎠ ⎝ ⎠
= xx 12 +0 ⎛ ⎞ ⎛ x ⎞ ⎜ 01 ⎟ ⎛ x ⎞ ⎜ 2 ⎟ ⎜ ⎟ ⎜ 1 ⎟ Provided L LK ⎜ ⎟ = ⎜ LK ⎟ ⎜ ⎟ x2 ' +−= xx 11 ' − 1 ⎛ ⎞ ⎜ x '⎟ ⎜ ⎛ ⎞ ⎟ ⎜ x '⎟ is small ⎜ Bρ ⎟ ⎝ 2 ⎠ ⎜ ⎜ Bρ ⎟ ⎟ ⎝ 1 ⎠ ⎝ ⎠ } ⎝ ⎝ ⎠ ⎠
E.Wildner NUFACT08 School 29 Benasque Matrix for a quadrupole (2)
9 We found :
⎛ ⎞ ⎛ x ⎞ ⎜ 01 ⎟ ⎛ x ⎞ ⎜ 2 ⎟ ⎜ ⎟ ⎜ 1 ⎟ ⎜ ⎟ = ⎜− LK 1⎟ ⎜ ⎟ ⎜ x '⎟ ⎜ ⎛ ⎞ ⎟ ⎜ x '⎟ ⎝ 2 ⎠ ⎜ ⎜ Bρ ⎟ ⎟ ⎝ 1 ⎠ ⎝ ⎝ ⎠ ⎠
(Bρ) 9 Define the focal length of the quadrupole as f= KL
⎛ ⎞ ⎛ x ⎞ ⎜ 01 ⎟ ⎛ x ⎞ ⎜ 2 ⎟ ⎜ ⎟ ⎜ 1 ⎟ ⎜ ⎟ = 1 ⎜ ⎟ ⎜ ⎟ ⎜− 1⎟ ⎜ ⎟ ⎝ x2'⎠ ⎜ ⎟ ⎝ x1'⎠ ⎝ f ⎠
E.Wildner NUFACT08 School 30 Benasque A quick recap…….
9 We solved Hill’s equation, which led us to the definition of transverse emittance and allowed us to describe particle motion in phase space in terms of β, α, etc…
9 We constructed the Transport Matrices corresponding to drift spaces and quadrupoles.
9 Now we must combine these matrices with the solution of Hill’s equation to evaluate β, α, etc…
E.Wildner NUFACT08 School 31 Benasque Matrices & Hill’s equation
9 We can multiply the matrices of our drift spaces and quadrupoles together to form a transport matrix that describes a larger section of our accelerator. 9 These matrices will move our particle from one point (x(s1),x’(s1)) on our phase space plot to another (x(s2),x’(s2)), as shown in the matrix equation below.
⎛ sx 2)( ⎞ ⎛ ba ⎞ ⎛ sx 1)( ⎞ ⎜ ⎟ = ⎜ ⎟⋅⎜ ⎟ ⎝ sx 2)(' ⎠ ⎝ dc ⎠ ⎝ sx 1)(' ⎠
9 The elements of this matrix are fixed by the elements through which the particles pass from point s1 to point s2. 9 However, we can also express (x,x’) as solutions of Hill’s equation. x = ε β cos. φ and x −= − sin/cos/' φβεφβεα
E.Wildner NUFACT08 School 32 Benasque Matrices & Hill’s equation (2)
x = ε β μ +φ)cos(. x = ε β cos. φ
⎛ sx 2)( ⎞ ⎛ ba ⎞ ⎛ sx 1)( ⎞ ⎜ ⎟ = ⎜ ⎟⋅⎜ ⎟ ⎝ sx 2)(' ⎠ ⎝ dc ⎠ ⎝ sx 1)(' ⎠
x −= −+ +φμβεφμβεα)sin(/)cos(/' x −= − sin/cos/' φβεφβεα
9 Assume that our transport matrix describes a complete turn around the machine.
9 Therefore : β(s2) = β(s1) 9 Let μ be the change in betatron phase over one complete turn.
9 Then we get for x(s2):
sx 2 = =+ a cos.)cos(.)( − b cos/ − b sin/ φ βε φ βεα φ βε φμ βε
E.Wildner NUFACT08 School 33 Benasque Matrices & Hill’s equation (3)
9 So, for the position x at s2 we have… =+ a cos.)cos(. − b − b sin/cos/ φ βε φ βεα φ βε φμ βε
φ μ − φ sinsincoscos μ
9 Equating the ‘sin’ terms gives: − ε β μ sinsin. φ = −b ε β sin/ φ 9 Which leads to: b = β sin μ
9 Equating the ‘cos’ terms gives: = a − cossin.cos.coscos. φ μ βεαφ βε φ μ βε
9 Which leads to: = ua +α sincos μ
9 We can repeat this for c and d.
E.Wildner NUFACT08 School 34 Benasque Matrices & Twiss parameters
9 α = − β' = −ωω' Remember previously we defined: 2 β = ω 2
9 2 These are called TWISS parameters 1+α γ = β
9 Remember also that m is the total betatron phase advance over one complete turn is. Number of betatron μ oscillations per turn Q = 2π 9 Our transport matrix becomes now:
⎛ ⎞ ⎜ μ +α sincos μ β sinμ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ − sin − sincos μαμμγ⎠
E.Wildner NUFACT08 School 35 Benasque Lattice parameters
⎛ ⎞ ⎜ μ +α sincos μ β sinμ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ − sin − sincos μαμμγ⎠
9 This matrix describes one complete turn around our machine and will vary depending on the starting point (s). 9 If we start at any point and multiply all of the matrices representing each element all around the machine we can calculate α, β, γ and μ for that specific point, which then will give us β(s) and Q 9 If we repeat this many times for many different initial positions (s) we can calculate our Lattice Parameters for all points around the machine.
E.Wildner NUFACT08 School 36 Benasque Lattice calculations and codes
9 Obviously m (or Q) is not dependent on the initial position ‘s’, but we can calculate the change in betatron phase, dm, from one element to the next.
9 Computer codes like “MAD” or “Transport” vary lengths, positions and strengths of the individual elements to obtain the desired beam dimensions or envelope ‘β(s)’ and the desired ‘Q’.
9 Often a machine is made of many individual and identical sections (FODO cells). In that case we only calculate a single cell and not the whole machine, as the the functions β (s) and dμ will repeat themselves for each identical section.
9 The insertion section have to be calculated separately.
E.Wildner NUFACT08 School 37 Benasque The I(s) and Q relation.
μ 9 Q = φ 2π , where μ = ∆ over a complete turn
dφ(s) 1 9 But we also know: = β ()sds
Over one complete turn 1 s ds 9 This leads to: Q = ∫ 2 o βπ()s
9 Increasing the focusing strength decreases the size of the beam envelope (β) and increases Q and vice versa.
E.Wildner NUFACT08 School 38 Benasque Tune corrections
9 What happens if we change the focusing strength slightly? 9 The Twiss matrix for our ‘FODO’ cell is given by: ⎛ μ +α sincos μ β sin μ ⎞ ⎜ ⎟ ⎝ − sin − sincos μαμμγ⎠
9 Add a small QF quadrupole, with strength dK and length ds. 9 This will modify the ‘FODO’ lattice, and add a horizontal focusing term: ⎛ 01 ⎞ dK (Bρ) ⎜ ⎟ dk = f = ⎝− dsdk 1⎠ ()Bρ dKds
9 The new Twiss matrix representing the modified lattice is: ⎛ 01 ⎞⎛ μ +α sincos μ β sin μ ⎞ ⎜ ⎟⎜ ⎟ ⎝− dsdk 1⎠⎝ − sin − sincos μαμμγ⎠
E.Wildner NUFACT08 School 39 Benasque Tune corrections (2)
⎛ μ+α sincos μ β sin μ ⎞ 9 This gives ⎜ ⎟ ⎝− dsdk ()sinsincos −−+ dsdk −+ sincossin μαμμβμγμμ⎠
9 This extra quadrupole will modify the phase advance μ for the FODO cell. New phase advance μ1 = μ + dμ Change in phase advance
9 If dμ is small then we can ignore changes in β
9 So the new Twiss matrix is just:
⎛cos μ 1 +α sin μ 1 β sin μ 1 ⎞ ⎜ ⎟ ⎝ −γ sin μ 1 cos μ 1 −α sin μ 1 ⎠
E.Wildner NUFACT08 School 40 Benasque Tune corrections (3)
9 These two matrices represent the same FODO cell therefore: ⎛ μ+α sincos μ β sin μ ⎞ ⎜ ⎟ ⎝− dsdk ()sinsincos −−+ dsdk −+ sincossin μαμμβμγμμ⎠
9 Which equals:
⎛cos μ 1 +α sin μ 1 β sin μ 1 ⎞ ⎜ ⎟ ⎝ − sin μγ1 cos 1 − sin μαμ1 ⎠
9 Combining and compare the first and the fourth terms of these two matrices gives:
22cosμ 1 = cosμ −dk ds β sinμ
Only valid for change in I <<
E.Wildner NUFACT08 School 41 Benasque Tune corrections (4)
Remember μ = μ + dμ 22cosμ 1 = cosμ −dk ds β sinμ 1 and dμ is small 2cosμ−2sinμdμ 2sinμdμ=dkdsβsinμ 1 dμ= dkdsβ ,but: dQ = dμ/2π In the horizontal 2 plane this is a QF 1 dQh = + dk.ds.βh 4π If we follow the same reasoning for both transverse planes for both QF and QD quadrupoles 1 1 dQv = + βv.dk .ds − βv.dk .ds 4π D D 4π F F QD 1 1 QF dQh = − βh.dk .ds + βh.dk .ds 4π D D 4π F F
E.Wildner NUFACT08 School 42 Benasque Tune corrections (5)
Let dkF = dk for QF and dkD = dk for QD
βhF, βvF = β at QF and βhD, βvD = β at QD
Then: ⎛ 1 −1 ⎞ β β ⎛ dQv⎞ ⎜ vD vF ⎟⎛ dsdk ⎞ ⎜ ⎟ = ⎜ 4π 4π ⎟⎜ D ⎟ dQh −1 1 ⎜ dsdk ⎟ ⎝ ⎠ ⎜ β β ⎟⎝ F ⎠ ⎝ 4π hD 4π hF ⎠
This matrix relates the change in the tune to the change in strength of the quadrupoles. We can invert this matrix to calculate change in quadrupole field needed for a given change in tune
E.Wildner NUFACT08 School 43