Accelerator Physics. Elena Wildner. Transverse motion. Benasque. Acknowldements to Simon Baird, Rende Steerenberg, Mats Lindroos, for course material

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1 Accelerator Physics Transverse motion Elena Wildner Acknowldements to Simon Baird, Rende Steerenberg, Mats Lindroos, for course material E.Wildner NUFACT08 School

2 Accelerator co-ordinates Horizontal Longitudinal Vertical It travels on the central orbit Rotating Cartesian Co-ordinate System E.Wildner NUFACT08 School

3 Two particles in a dipole field 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 Assume that Bρ is the same for both particles. Lets unfold these circles E.Wildner NUFACT08 School 3 3

4 The trajectories unfolded The horizontal displacement of particle B with respect to particle A. Particle B Particle A displacement 0 π Particle B oscillates around particle A. This type of oscillation forms the basis of all transverse motion in an accelerator. It is called Betatron Oscillation E.Wildner NUFACT08 School 4

5 Dipole magnet A dipole with a uniform dipolar field deviates a particle by an angle θ. The deviation angle θ depends on the length L and the magnetic field B. The angle θ can be calculated: sin θ L = ρ = If θ is small: θ θ sin = So we can write: LB θ = ( Bρ ) LB ( Bρ ) L θ θ θ L ρ E.Wildner NUFACT08 School 5

6 Stable or unstable motion? Since the horizontal trajectories close we can say that the horizontal motion in our simplified accelerator with only a horizontal dipole field is stable What can we say about the vertical motion in the same simplified accelerator? Is it stable or unstable and why? What can we do to make this motion stable? We need some element that focuses the particles back to the reference trajectory. This etra focusing can be done using: Quadrupole magnets E.Wildner NUFACT08 School 6

7 Quadrupole Magnet A Quadrupole has 4 poles, north and south Magnetic field They are symmetrically arranged around the centre of the magnet There is no magnetic field along the central ais. Hyperbolic contour y = constant E.Wildner NUFACT08 School 7

8 Resistive Quadrupole magnet E.Wildner NUFACT08 School 8

9 Quadrupole fields Magnetic field On the -ais (horizontal) the field is vertical and given by: B y On the y-ais (vertical) the field is horizontal and given by: B y The field gradient, K is defined as: Tm ( By) d d ( ) The normalised gradient, k is defined as: K B ( ρ ) ( m ) E.Wildner NUFACT08 School 9

10 Types of quadrupoles Force on particles This is a: Focusing Quadrupole (QF) It focuses the beam horizontally and defocuses the beam vertically. Rotating this magnet by 90º will give a: Defocusing Quadrupole (QD) E.Wildner NUFACT08 School 0

11 Focusing and Stable motion Using a combination of focusing (QF) and defocusing (QD) quadrupoles solves our problem of unstable vertical motion. It will keep the beams focused in both planes when the position in the accelerator, type and strength of the quadrupoles are well chosen. By now our accelerator is composed of: Dipoles, constrain the beam to some closed path (orbit). Focusing and Defocusing Quadrupoles, provide horizontal and vertical focusing in order to constrain the beam in transverse directions. A combination of focusing and defocusing sections that is very often used is the so called: FODO lattice. This is a configuration of magnets where focusing and defocusing magnets alternate and are separated by nonfocusing drift spaces. E.Wildner NUFACT08 School

12 FODO cell The FODO cell is defined as follows: QF QD QF L L Centre of first QF FODO cell Or like this Centre of second QF E.Wildner NUFACT08 School

13 The mechanical equivalent The gutter below illustrates how the particles in our accelerator behave due to the quadrupolar fields. Whenever a particle beam diverges away from the central orbit the quadrupoles focus them back towards the central orbit. How can we represent the focusing gradient of a quadrupole in this mechanical equivalent? E.Wildner NUFACT08 School 3

14 The particle characterized A particle during its transverse motion in our accelerator is characterized by: Position or displacement from the central orbit. Angle with respect to the central orbit. = displacement = angle = d/ds ds s d This is a motion with a constant restoring coefficient, similar to the pendulum E.Wildner NUFACT08 School 4

15 Hill s equation These transverse oscillations are called Betatron Oscillations, and they eist in both horizontal and vertical planes. The number of such oscillations/turn is Q or Q y. (Betatron Tune) (Hill s Equation) describes this motion d + K( s) = ds If the restoring coefficient (K) is constant in s then this is just SHM Remember s is just longitudinal displacement around the ring 0 E.Wildner NUFACT08 School 5

16 Hill s equation () In a real accelerator K varies strongly with s. Therefore we need to solve Hill s equation for K varying as a function of s d + K( s) = ds The phase advance and the amplitude modulation of the oscillation are determined by the variation of K around (along) the machine. The overall oscillation amplitude will depend on the initial conditions. 0 E.Wildner NUFACT08 School 6

17 Solution of Hill s equation () d + K( s) = ds nd order differential equation. Guess a solution: ε and φ 0 are constants, which depend on the initial conditions. β(s) = the amplitude modulation due to the changing focusing strength. φ(s) = the phase advance, which also depends on focusing strength. E.Wildner NUFACT08 School 7 0 ( φ( ) 0) = ε. β ( s) cos s + φ

18 Solution of Hill s equation () Define some parameters and let φ = ( φ( s) + φ 0) = ε. ω(s) cosφ Remember φ is still a function of s α = β ' β = ω + α γ = β In order to solve Hill s equation we differentiate our guess, which results in: ' = dω ε cosφ ds εωφ' sinφ and differentiating a second time gives: '' = εω''cosφ εω' φ'sinφ εω' φ'sinφ εωφ''sinφ εωφ' cosφ Now we need to substitute these results in the original equation. E.Wildner NUFACT08 School 8

19 Solution of Hill s equation (3) So we need to substitute = ε. β ( s) cos( φ( s) + φ 0) and its second derivative '' = εω''cosφ εω' φ'sinφ εω' φ'sinφ εωφ''sinφ εωφ' cosφ d ds into our initial differential equation + K( s) = 0 This gives: εω''cosφ εω' φ'sinφ εω' φ'sinφ εωφ''sinφ + K( s) εω cosφ = 0 Sine and Cosine are orthogonal and will never be 0 at the same time εωφ' cosφ E.Wildner NUFACT08 School 9

20 Solution of Hill s equation (4) εω''cosφ εω' φ'sinφ εω' φ'sinφ εωφ''sinφ εωφ' cosφ + K( s) εω cosφ = 0 Using the Sin terms ω' φ' +ωφ'' = 0 ωω' φ' + ω φ'' = 0 We defined β = ω, which after differentiating gives β ' = ωω' Combining ωω' φ' + ω φ'' = 0 and β ' = ωω' β ' φ' + βφ'' = βφ' ' = 0 gives: dβ = ds dβ dω dω ds dφ Which is the case as: βφ' = const. = since φ' = = ds So our guess seems to be correct β E.Wildner NUFACT08 School 0

21 Solution of Hill s equation (5) Since our solution was correct we have the following for : = ε. β cosφ dω β' = = ds ω α β For we have now: ' = dω ε cosφ ds εωφ' sinφ ω = β Thus the epression for finally becomes: ' = α ε / β cosφ ε / β sinφ E.Wildner NUFACT08 School

22 Phase Space Ellipse So now we have an epression for and = ε. β cosφ and ' = α ε / β cosφ ε / β sinφ If we plot versus we get an ellipse, which is called the phase space ellipse. φ = 3π/ φ = 0 = π ε.γ ε / β α ε / γ α ε / β ε /γ ε.β E.Wildner NUFACT08 School

23 Phase Space Ellipse () As we move around the machine the shape of the ellipse will change as β changes under the influence of the quadrupoles However the area of the ellipse (πε) does not change ε / β Area = π r r ε / β ε.β ε is called the transverse emittance and is determined by the initial beam conditions. The units are meter radians, but in practice we use more often mm mrad. ε.β E.Wildner NUFACT08 School 3

24 Phase Space Ellipse (3) ε / β ε.β ε / β ε.β The projection of the ellipse on the -ais gives the Physical transverse beam size. The variation of β(s) around the machine will tell us how the transverse beam size will vary. E.Wildner NUFACT08 School 4

25 Emittance & Acceptance To be rigorous we should define the emittance slightly differently. Observe all the particles at a single position on one turn and measure both their position and angle. This will give a large number of points in our phase space plot, each point representing a particle with its co-ordinates,. emittance beam The emittance is the area of the ellipse, which contains a defined percentage, of the particles. acceptance The acceptance is the maimum area of the ellipse, which the emittance can attain without losing particles. E.Wildner NUFACT08 School 5

26 Matri Formalism Lets represent the particles transverse position and angle by a column matri. ' As the particle moves around the machine the values for and will vary under influence of the dipoles, quadrupoles and drift spaces. These modifications due to the different types of magnets can be epressed by a Transport Matri M If we know and at some point s then we can calculate its position and angle after the net magnet at position S using: ( s ( s ) )' = M ( s) ( s)' = a c b d ( s) ( s)' E.Wildner NUFACT08 School 6

27 How to apply the formalism If we want to know how a particle behaves in our machine as it moves around using the matri formalism, we need to: Split our machine into separate element as dipoles, focusing and defocusing quadrupoles, and drift spaces. Find the matrices for all of these components Multiply them all together Calculate what happens to an individual particle as it makes one or more turns around the machine E.Wildner NUFACT08 School 7

28 E.Wildner NUFACT08 School 8 Benasque = + L. L A drift space Matri for a drift space contains no magnetic field. A drift space has length L. ' 0 ' ' L + = + = small = ' 0 ' L }

29 E.Wildner NUFACT08 School 9 Benasque Matri for a quadrupole A quadrupole of length L. deflection Remember B y and the deflection due to the magnetic field is: B LK B LB y = ρ ρ ' ' 0 B LK + = + = ρ Provided L is small = ' 0 ' B LK ρ }

30 E.Wildner NUFACT08 School 30 Benasque Matri for a quadrupole () Define the focal length of the quadrupole as f= ( ) KL Bρ = ' 0 ' f = ' 0 ' B LK ρ We found :

31 A quick recap. 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 We constructed the Transport Matrices corresponding to drift spaces and quadrupoles. Now we must combine these matrices with the solution of Hill s equation to evaluate β, α, etc E.Wildner NUFACT08 School 3

32 Matrices & Hill s equation We can multiply the matrices of our drift spaces and quadrupoles together to form a transport matri that describes a larger section of our accelerator. These matrices will move our particle from one point ((s ), (s )) on our phase space plot to another ((s ), (s )), as shown in the matri equation below. ( s '( s ) ) a = c b d ( s) '( s) The elements of this matri are fied by the elements through which the particles pass from point s to point s. However, we can also epress (, ) as solutions of Hill s equation. = ε. β cosφ and ' = α ε / β cosφ ε / β sinφ E.Wildner NUFACT08 School 3

33 Matrices & Hill s equation () = ε. β cos( μ + φ) = ε. β cosφ ( s '( s ) ) a = c b d ( s) '( s) ' = α ε / β cos( μ + φ) ε / β sin( μ + φ) ' = α ε / β cosφ ε / β sinφ Assume that our transport matri describes a complete turn around the machine. Therefore : β(s ) = β(s ) Let μ be the change in betatron phase over one complete turn. Then we get for (s ): ( s) = ε. β cos( μ + φ) = a ε. β cosφ bα ε / β cosφ b ε / β sinφ E.Wildner NUFACT08 School 33

34 Matrices & Hill s equation (3) So, for the position at s we have ε. β cos( μ + φ) = a ε. β cosφ bα ε / β cosφ b ε / β sinφ cosφ cos μ sinφ sin μ Equating the sin terms gives: ε. β sin μ sinφ = b ε / β sinφ Which leads to: b = β sin μ Equating the cos terms gives: ε. β cos μ cosφ = a ε. β cosφ α ε. β sin μ cosφ Which leads to: a = cos u +α sin μ We can repeat this for c and d. E.Wildner NUFACT08 School 34

35 Matrices & Twiss parameters Remember previously we defined: These are called TWISS parameters α = β' β = ω + α γ = β = ωω' Remember also that m is the total betatron phase advance over one complete turn is. Our transport matri becomes now: cosμ + α sin μ γ sin μ Q = μ π β sin μ cosμ α sin μ Number of betatron oscillations per turn E.Wildner NUFACT08 School 35

36 Lattice parameters cosμ + α sin μ γ sin μ β sin μ cosμ α sin μ This matri describes one complete turn around our machine and will vary depending on the starting point (s). 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 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

37 Lattice calculations and codes 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 net. 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. 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. The insertion section have to be calculated separately. E.Wildner NUFACT08 School 37

38 The (s) and Q relation. μ Q = π, where μ = φ over a complete turn But we also know: ( s) dφ = ds β () s Over one complete turn This leads to: Q = π s ds o β () s Increasing the focusing strength decreases the size of the beam envelope (β) and increases Q and vice versa. E.Wildner NUFACT08 School 38

39 Tune corrections What happens if we change the focusing strength slightly? The Twiss matri for our FODO cell is given by: cos μ + α sin μ γ sin μ β sin μ cos μ α sin μ Add a small QF quadrupole, with strength dk and length ds. This will modify the FODO lattice, and add a horizontal focusing term: 0 dk ( Bρ ) dk = f = dkds ( Bρ ) dkds The new Twiss matri representing the modified lattice is: 0cos μ + α sin μ dkds γ sin μ β sin μ cos μ α sin μ E.Wildner NUFACT08 School 39

40 Tune corrections () This gives cos μ+ α sin μ dkds( cosμ+ sinμ ) γ sin μ β sin μ dkdsβsinμ+ cos μ α sin μ This etra quadrupole will modify the phase advance μ for the FODO cell. μ = μ + dμ New phase advance If dμ is small then we can ignore changes in β So the new Twiss matri is just: Change in phase advance cos μ + α sin μ γ sin μ β sin μ cos μ α sin μ E.Wildner NUFACT08 School 40

41 Tune corrections (3) These two matrices represent the same FODO cell therefore: cos μ+ α sin μ dkds( cosμ+ sinμ ) γ sin μ β sin μ dkdsβsinμ+ cos μ α sin μ Which equals: cos μ + α sin μ γ sin μ β sin μ cos μ α sin μ Combining and compare the first and the fourth terms of these two matrices gives: Only valid for change in << cos μ = cos μ dk dsβsin μ E.Wildner NUFACT08 School 4

42 Tune corrections (4) cos μ = cos μ dk dsβsin μ Remember μ = μ + dμ and dμ is small cosμ sinμdμ In the horizontal plane this is a QF sinμd μ=dkdsβsinμ dμ= dkdsβ,but: dq = dμ/π dqh = + dk. ds. βh 4π If we follow the same reasoning for both transverse planes for both QF and QD quadrupoles QD dqv = + βv. dk 4π dqh = βh. dk 4π E.Wildner NUFACT08 School 4 D D. ds. ds D D βv. dk 4π + βh. dk 4π F F. ds. ds F F QF

43 Tune corrections (5) Let dk F = dk for QF and dk D = dk for QD β hf, β vf = β at QF and β hd, β vd = β at QD Then: dqv dqh β = 4π β 4π vd hd β 4π β 4π vf hf dk dk D F ds ds This matri relates the change in the tune to the change in strength of the quadrupoles. We can invert this matri to calculate change in quadrupole field needed for a given change in tune E.Wildner NUFACT08 School 43