Practical Lattice Design

Transcription

1 Practical Lattice Design S. Alex Bogacz (JLab) and Dario Pellegrini (CERN) USPAS January, 15-19, /48 D. Pellegrini - Practical Lattice Design

2 Purpose of the Course Gain a deep understanding of the basic concepts of linear optics Develop intuition concerning optics functions and their manipulation Acquire familiarity with OptiM, an optics design program Experiment with a variety of case studies 2/48 D. Pellegrini - Practical Lattice Design

3 Content and Structure 9:00-12:00 13:30-16:30 19:00-23:00 Mon Introduction to Transverse Introduction to OptiM Homework and Optics (D) FODO Lattice (A) tutoring Tue Dispersion Suppressors Arc-to-Straight Design Homework and (A) (A) tutoring Wed Low β Optics (D) Lattice imperfections Homework and (D) tutoring Thu Radiation Damping (A) Low Emittance Lattices (A) Homework and tutoring Fri Final Exam (9:00-13:00) 3/48 D. Pellegrini - Practical Lattice Design

4 Some references 1. Mario Conte, William W. MacKay, An Introduction to the Physics of Particle Accelerators, Second Edition, World Scientific, Andrzej Wolski, Beam Dynamics in High Energy Particle Accelerators, Imperial College Press, The CERN Accelerator School (CAS) Proceedings, e.g. 1992, Jyväskylä, Finland; or 2013, Trondheim, Norway 4. Shyh-Yuan Lee, Accelerator Physics, World Scientific, Helmut Wiedemann, Particle Accelerator Physics, Springer, 4th Edition, 2015 The material collected here has been adapted from A. Latina s JUAS lectures on transverse dynamics. 4/48 D. Pellegrini - Practical Lattice Design

5 Part 1. Basics, single-particle dynamics 5/48 D. Pellegrini - Practical Lattice Design

6 Luminosity run of a typical storage ring In a storage ring: the protons are accelerated and stored for hours The distance traveled by particles running at nearly the speed of light, v c, for 12 hours is distance m this is about as going to Pluto and back! How to maintain them in a few mm pipe? 6/48 D. Pellegrini - Practical Lattice Design

7 Forces and fields Four fundamental interactions in Nature, the electromagnetic one is the most promising the Lorentz force ( ) F = q E + v B where, in high energy machines, v c m/s. Usually there is no electric field, and the transverse deflection is given by a magnetic field only. Comparison of electric and magnetic force: E = 1 MV/m B = 1 T F magnetic = evb F electric ee = βcb E β = 300 β the magnetic force is much stronger then the electric one: in an accelerator we normally have magnets although electrostatic lenses are possible at low energy. 7/48 D. Pellegrini - Practical Lattice Design

8 Stable circular motion: Lorentz force = centrifugal force Lorentz force F L = qvb Centrifugal force F centr = mv 2 ρ mv 2 ρ = q vb P = mv = m 0 γv "momentum" Bρ = "beam ridigity" P q =Bρ 8/48 D. Pellegrini - Practical Lattice Design

9 Dipole magnets: the magnetic guide Rule of thumb, in practical units: 1 ρ [m] 0.3 B [T ] P [GeV /c] Example: In the LHC, ρ = 2.53 km. The circumference 2πρ = 15.9 km 60% of the entire LHC. (R = 4.3 km, and the total circumference is C = 2πR 27 km) The field B is T The quantity 1 ρ can be seen as a normalized bending strength, i.e. the bending field normalized to the beam rigidity. 9/48 D. Pellegrini - Practical Lattice Design

10 The gutter analogy ( ) F = q E + v B Remember the 1d harmonic oscillator: F = k x 10/48 D. Pellegrini - Practical Lattice Design

11 Reminder: the 1d Harmonic oscillator Restoring force F = k x Equation of motion: which has solution: x = k m x x (t) = A cos (ωt + φ) = or a 1 cos (ωt) + a 2 sin (ωt) F, restoring force, N or MeV/m k, spring constant or focusing strength, N/m or MeV/m 2 k ω = m = 2πf, angular velocity, rad/s φ, initial phase, rad 11/48 D. Pellegrini - Practical Lattice Design f, rotation frequency, Hz A, oscillation amplitude, m m 0, particle s rest mass, MeV/c 2 m = m 0γ, particle s mass, MeV/c 2

12 Quadrupole magnets: the focusing force Quadrupole magnets are required to keep the trajectories in vicinity of the ideal orbit They exert a linearly-increasing Lorentz force, thru a linearly-increasing magnetic field: B x = gy B y = gx Fx = qvz By = qvz g x F y = qv z B x = qv z g y Gradient of a quadrupole magnet: g = 2µ [ 0nI T raperture 2 m ] = B poles r aperture [ ] T m LHC main quadrupole magnets: g T/m the arrows show the force exerted on a particle Dividing by p/q one finds k, the normalized focusing strength k = g [ m 2 ] [ ] [ ] T P GeV g = ; q = [e] ; P/q m q = = c e Another useful rule of thumb: k [ m 2] g [T /m] 0.3 P/q [GeV /c/e]. [ GV c ] =[T m] 12/48 D. Pellegrini - Practical Lattice Design

13 Focal length of a quadrupole The focal length of a quadrupole is f = 1 length: k L [m], where L is the quadrupole 13/48 D. Pellegrini - Practical Lattice Design

14 Phase-space coordinates the ideal particle coincides with the reference orbit (perfect machine) any other particle has coordinates x, y, P x, P y 0; P P 0 with x, y ρ P x, P y P 0 The state of a particle is represented with a 6-dimensional phase-space vector: ( x, x, y, y, z, δ ) P 0 is the reference momentum and P = P 0 (1 + δ) x x = dx ds = dx dt dt ds = Vx = Px Px V z P z P 0 y y = dy ds = dy dt dt ds = Vy V z z δ = P P 0 = Py P z Py P 0 [m] [rad] [m] [rad] [m] = P P0 P 0 [#] 14/48 D. Pellegrini - Practical Lattice Design

15 The equation of motion in radial coordinates Let s consider a local segment of one particle s trajectory: ( ) and recall the radial centrifugal acceleration: a r = d2 ρ dθ 2 dt 2 ρ = d2 ρ dt dt 2 ρω2. For an ideal orbit: ρ = const dρ dt = 0 the force is F centrifugal = mρω 2 = mv 2 /ρ F Lorentz = qb y v = F centrifugal P q = By ρ For a general trajectory: ρ ρ + x : [ d 2 F centrifugal = m a r = F Lorentz m v 2 (ρ + x) dt2 ρ + x ] = qb y v 15/48 D. Pellegrini - Practical Lattice Design

16 F = m d2 (ρ + x) mv 2 = qb dt2 y v ρ + x }{{}}{{} term 1 term 2 Term 1: As ρ =const... m d2 d2 (ρ + x) = m dt2 dt 2 x Term 2: Remember: x mm whereas ρ m we develop for small x 1 ρ + x 1 ρ remember ( 1 x ρ ) Taylor expansion: f (x) = f (x 0 ) + + (x x 0 ) f (x 0 ) + (x x 0) 2 2! f (x 0 ) + We get: m d 2 x dt 2 mv 2 ( 1 x ) = qb y v ρ ρ 16/48 D. Pellegrini - Practical Lattice Design

17 The guide field in linear approximation B y = B 0 + x By x m d2 x dt 2 mv 2 ( 1 x ) = qb y v let s divide by m ρ ρ d 2 x dt 2 v 2 ρ ( 1 x ) = qby v ρ m Let s change the independent variable: t s dx dt = dx ds ds dt = x v d 2 x dt 2 = d dt = d ds x v 2 v 2 ρ x 1 ρ dx dt = d dt dx ds }{{} x ds ( x v ) = d dt ds }{{} v ( 1 x ρ ds dt }{{} v ) = qby v m ( 1 x ) = qby ρ P = d ( x v ) = dt ( x v 2) = x v 2 + x 2v dv ds let s divide by v 2 using P = mv 17/48 D. Pellegrini - Practical Lattice Design

18 The focussing field The horizontal focussing is provided by the B y field, let us Taylor expand it: B y (x) = B y0 + By x x B y 2 x 2 x B y 3! x 3 x Now we drop the suffix y and normalize to the magnetic rigidity P/q = Bρ B (x) P/q = B 0 B 0 ρ + g P/q x + 1 g 2 P/q x g 3! P/q x = 1 ρ + kx mx ! nx In the linear approximation, only the terms linear in x and y are taken into account: dipole fields, 1/ρ quadrupole fields, k Therefore we can write: x 1 ρ ( 1 x ) ( ) = By 1 ρ P/q = ρ + kx 18/48 D. Pellegrini - Practical Lattice Design

19 The equation of motion x 1 ρ ( 1 x ) = 1 ρ ρ kx x 1 ρ + x ρ 2 = 1 ρ kx x + x ( ) 1 ρ 2 + k = 0 Equation for the vertical motion 1 ρ 2 = 0 usually there are not vertical bends k k quadrupole field changes sign y ky = 0 19/48 D. Pellegrini - Practical Lattice Design

20 Weak focussing there is a focusing force, ( ) x 1 (s) + ρ + k x (s) = 0 2 }{{} focusing effect 1 ρ 2, even without a quadrupole gradient, k = 0 x = 1 ρ 2 x even without quadrupoles there is retrieving force (focusing) in the bending plane of the dipole magnets In large machines, this effect is very weak. Mass spectrometers entirely rely on weak focusing: they have no quadrupoles; particles are separated according to their energy and focused due to the 1/ρ effect of the dipole 20/48 D. Pellegrini - Practical Lattice Design

21 Solution of the trajectory equations Definition: horizontal plane K = 1 /ρ 2 + k vertical plane K = k } x + Kx = 0 This is the differential equation of a 1d harmonic oscillator with spring constant K. We know that, for K > 0, the solution is in the form: x (s) = a 1 cos (ωs) + a 2 sin (ωs) In fact, x (s) = a 1ω sin (ωs) + a 2ω cos (ωs) x (s) = a 1ω 2 cos (ωs) + a 2ω 2 sin (ωs) = ω 2 x (s) ω = K Thus, the general solution is for K > 0. x (s) = a 1 cos ( ) ( ) Ks + a 2 sin Ks 21/48 D. Pellegrini - Practical Lattice Design

22 We determine a 1, a 2 by imposing the initial conditions: s = 0 { x (0) = x0, a 1 = x 0 x (0) = x 0, a 2 = x 0 K Horizontal focusing quadrupole, K > 0: ( ) x (s) = x 0 cos Ks + x 0 1 ( ) sin Ks K ( ) ( ) x (s) = x 0 K sin Ks + x 0 cos Ks We can use the matrix formalism: ( x x ) s 1 = M foc ( x0 For a quadrupole of length L: ( KL ) M foc = cos ( KL ) K sin x 0 ) s 0 ( 1 KL ) K sin ( KL ) cos Notice that for a drift space, i.e. when K = 0 M drift = 22/48 D. Pellegrini - Practical Lattice Design ( 1 L 0 1 ).

23 Defocusing quadrupole The equation of motion is x + Kx = 0 with K < 0 The solution is in the form: Remember: f (s) = cosh (s) f (s) = sinh (s) x (s) = a 1 cosh (ωs) + a 2 sinh (ωs) with ω = K. For a quadrupole of length L the transfer matrix reads: ( K L ) ( M defoc = cosh 1 K L ) sinh K ( K L ) ( K L ) K sinh cosh Again when K = 0 M drift = ( 1 L 0 1 ) 23/48 D. Pellegrini - Practical Lattice Design

24 Thin-lens approximation of a quadrupole magnet When the focal length f of the quadrupolar lens is much bigger than the length of the magnet itself, L Q f = 1 L Q k L Q we can derive the limit for L 0 while keeping constant f, i.e. k L Q = const. The transfer matrices are M x = ( f 1 focusing, and defocusing respectively. ) ( ) 1 0 M y = 1 f 1 This approximation is useful for fast calculations. 24/48 D. Pellegrini - Practical Lattice Design

25 Transformation through a system of lattice elements One can compute the solution of a system of elements, by multiplying the matrices of each single element: M total = M QF M D M Bend M D M QD ( ) ( ) x x x = M s1 s 2 M s0 s 1 x s 2 s 0 In each accelerator element the particle trajectory corresponds to the movement of a harmonic oscillator....typical values are: x mm x mrad 25/48 D. Pellegrini - Practical Lattice Design

26 Properties of the transfer matrix M The transfer matrix M has two important properties: Its determinant is 1 (Liouville s theorem and/or symplecticity condition for the 2D case) det (M) = 1 Provides a stable motion over N turns, with N, if and only if: (Stability condition) trace (M) 2 26/48 D. Pellegrini - Practical Lattice Design

27 Stability condition Question: Given a periodic lattice with generic transport map M, ( a b M = c d under which condition the matrix M provides stable motion after N turns (with N )? ) x N = M... M M M x }{{} 0 = M N x 0 N turns, with N The answer is simple: the motion is stable when all elements of M N are finite, with N. The difficult question is... how do we compute M N with N? Remember: det (M) = ad bc = 1 trace (M) = a + d If we diagonalize M, we can rewrite it as: ( ) λ1 0 M = U U T 0 λ 2 where U is some unitary matrix, λ 1 and λ 2 are the eigenvalues. 27/48 D. Pellegrini - Practical Lattice Design

28 Stability condition (cont.) What happens if we consider N turns? ( M N λ N = U λ N 2 ) U T Notice that λ 1 and λ 2 can be complex numbers. Given that det (M) = 1, then to have a stable motion, x must be real: x R. Now let us write down the characteristic equation: From which derives the stability condition: λ 1 λ 2 = 1 λ 1 = 1 λ 2 λ 1,2 = e ±i x ( a λ b det (M λi ) = det c d λ λ 2 (a + d) λ + (ad bc) = 0 λ 2 trace (M) λ + 1 = 0 trace (M) = λ + 1/λ = = e ix + e ix = 2 cos x since x R trace (M) 2 ) = 0 28/48 D. Pellegrini - Practical Lattice Design

29 Orbit and tune Tune: the number of oscillations per turn. Example for the current LHC: Q x = Q y = The non-integer part is crucial! 29/48 D. Pellegrini - Practical Lattice Design

30 Summary beam rigidity: Bρ = P q bending strength of a dipole: focusing strength of a quadruple: 1 ρ [m 1 ] = B 0 [T] P [GeV/c] k [m 2 ] = g P [GeV/c] focal length of a quadrupole: f = 1 k L Q ( ) equation of motion: x k x = 0 ρ 2 solution of the eq. of motion: x s2 = M x s1... with M ( C S C S ) ( KL ) e.g.: M QF = cos ( KL ) K sin ( K L ) M QD = cosh ( K L ) K sinh ( 1 KL ) K sin ( KL ) Thin Lense cos 1 sinh K ( K L ) ( K L ) cosh, M D = ( f 1 ( 1 L 0 1 ) ), 30/48 D. Pellegrini - Practical Lattice Design

31 Part 2. Optics functions and Twiss parameters 31/48 D. Pellegrini - Practical Lattice Design

32 Envelope So far we have studied the motion of a particle. Question: what will happen, if the particle performs a second turn?... or a third one or turns... 32/48 D. Pellegrini - Practical Lattice Design

33 The Hill s equation In 19th century George William Hill, one of the greatest masters of celestial mechanics of his time, studied the differential equation for motions with periodic focusing properties : the Hill s equation x (s) + K (s) x (s) = 0 with: a restoring force const K (s) depends on the position s K (s + L) = K (s) periodic function, where L is the lattice period We expect a solution in the form of a quasi harmonic oscillation: amplitude and phase will depend on the position s in the ring. 33/48 D. Pellegrini - Practical Lattice Design

34 The beta function General solution of Hill s equation: x (s) = β x (s) J x cos (µ x (s) + µ x,0 ) (1) J x, µ 0 =integration constants determined by initial conditions β x (s) is a periodic function given by the focusing properties of the lattice quadrupoles β x (s + L) = β x (s) Inserting Eq. (1) in the equation of motion, we get (Floquet s theorem) the following result ˆ s ds µ x (s) = 0 β x (s) where µ x (s) is the phase advance between the points 0 and s, in the phase space. For one complete revolution, µ x (s) is the number of oscillations per turn, or tune when normalized to 2π Q x = 1 ds 2π β x (s) J x is a constant of motion, called the Courant-Snyder invariant or action. 34/48 D. Pellegrini - Practical Lattice Design

35 The orbit in the phase space is an ellipse General solution of the Hill s equation x (s) = β x (s) J x cos (µ x (s) + µ x,0 ) (1) x Jx (s) = βx (s) {αx (s) cos (µx (s) + µ x,0) + sin (µ x (s) + µ x,0 )} (2) From Eq. (1) we get cos (µ (s) + µ 0 ) = x (s) Jx βx (s) α x (s) = 1 2 β x (s) γ x (s) = 1 + αx (s)2 β x (s) Insert into Eq. (2) and solve for J J x = γ x (s) x (s) 2 + 2α x (s) x (s) x (s) + β x (s) x (s) 2 J x is a constant of the motion, i.e. the Courant-Snyder invariant or Action it is a parametric representation of an ellipse in the xx space the shape and the orientation of the ellipse are given by α x, β x, and γ x these are the Twiss parameters 35/48 D. Pellegrini - Practical Lattice Design

36 The phase-space ellipse J x = γ x (s) x (s) 2 + 2α x (s) x (s) x (s) + β x (s) x (s) 2 The area of ellipse, π J x, is an intrinsic beam parameter and cannot be changed by the focal properties. Important remarks: A large β-function corresponds to a large beam size and a small beam divergence wherever β reaches a maximum or a minimum, α = 0. 36/48 D. Pellegrini - Practical Lattice Design

37 Particles distribution and beam ellipse For each turn x, x at a given position s 1 and plot in the phase-space diagram Plane: x x 37/48 D. Pellegrini - Practical Lattice Design

38 Particles distribution and beam matrix In the phase space a realistic particles distribution matches the shape of an ellipse, and can be described using a beam matrix Σ Where Σ is defined as ( ) ( σ11 σ Σ = 12 x 2 xx = ) σ 21 σ 22 x x x 2 the covariance matrix of the particles distribution The determinant of the covariance matrix of a distribution, can be used to define the geometric emittance, corresponding to the area of the distribution ɛ = det Σ = det ( cov ( x, x )) = σ 11 σ 22 σ 12 σ 21 = area of the beam ellipse Remember: when accelerating the beam the preserved quantity is the normalised emittance: 38/48 D. Pellegrini - Practical Lattice Design ɛ normalized def = β rel γ rel ɛ geometric

39 The transfer matrix in terms of Twiss parameters As we have already seen, a general solution of the Hill s equation is: x (s) = β x (s) J x cos (µ x (s) + µ x,0 ) x J x (s) = β x (s) [αx (s) cos (µx (s) + µ x,0) + sin (µ x (s) + µ x,0 )] Let s remember some trigonometric formulæ: then, sin (a ± b) = sin a cos b ± cos a sin b, cos (a ± b) = cos a cos b sin a sin b,... x (s)= β x (s) J x (cos µ x (s) cos µ x,0 sin µ x (s) sin µ x,0 ) x J x (s)= β x (s) [αx (s) (cos µx (s) cos µ x,0 sin µ x (s) sin µ x,0 ) + + sin µ x (s) cos µ x,0 + cos µ x (s) sin µ x,0 ] 39/48 D. Pellegrini - Practical Lattice Design

40 At the starting point, s (0) = s 0, we put µ (0) = 0. Therefore we have cos µ 0 = x0 β0j sin µ 0 = 1 (x ) 0 β0 + α0x0 J β0 If we replace this in the formulæ, we obtain: x (s)= x (s)= β s { } {cos µ s + α 0 sin µ s } x 0 + βs β 0 sin µ s x 0 β 0 1 βs β 0 {(α 0 α s ) cos µ s (1 + α 0α s ) sin µ s } x 0 + β 0 β s {cos µ s α s sin µ s } x 0 The linear map follows easily, ( x x ) s ( ) x = M x 0 M = βs β (cos µ s + α 0 sin µ s ) 0 (α 0 αs ) cos µs (1+α 0 αs ) sin µs βs β0 βs β 0 sin µ s β0 (cos µs αs sin µs ) βs We can compute the single particle trajectories between two locations in the ring, if we know the α, β, and γ at these positions! Exercise: prove that det(m) = 1 40/48 D. Pellegrini - Practical Lattice Design

41 Periodic lattices, 1-turn map The transfer matrix for a particle trajectory βs (cos µ β s + α 0 sin µ M 0 s = 0 s) (α 0 α s ) cos µ s (1+α 0 α s ) sin µ s βs β 0 βsβ 0 sin µ s β0 β s (cos µ s α s sin µ s) simplifies considerably if we consider one complete turn: ( ) cos µl + α M = s sin µ L β s sin µ L γ s sin µ L cos µ L α s sin µ L where µ L is the phase advance per period ˆ s+l ds µ L = s β (s) Remember: the tune is the phase advance in units of 2π: Q = 1 ds 2π β (s) = µ L 2π 41/48 D. Pellegrini - Practical Lattice Design

42 Evolution of α, β, and γ Consider two positions in the storage ring: s 0, s ( x x ) s ( x = M x ) s 0 with M = M QF M D M Bend M D M QD ( ) ( C S M = C S M 1 S S = C C ) Since the Liouville theorem holds, J = const: 42/48 D. Pellegrini - Practical Lattice Design J = βx 2 + 2αxx + γx 2 J = β 0x α 0x 0x 0 + γ 0x 2 0

43 We express x 0 and x 0 as a function of x and x : ( x x ) s 0 = M 1 ( x x ) s x0 = S x Sx x 0 = C x + Cx Substituting x 0 and x 0 into the expression of J, we obtain: J = βx 2 + 2αxx + γx 2 J = β 0 ( C x + Cx )2 + 2α 0 ( S x Sx ) ( C x + Cx ) + γ 0 ( S x Sx ) 2 We need to sort by x and x : β (s) = C 2 β 0 2SCα 0 + S 2 γ 0 α (s) = CC β 0 + ( SC + S C ) α 0 SS γ 0 γ (s) = C 2 β 0 2S C α 0 + S 2 γ 0 43/48 D. Pellegrini - Practical Lattice Design

44 Evolution of α, β, and γ in matrix form The beam ellipse transformation in matrix notation: T 0 s = C 2 2SC S 2 CC SC + S C SS C 2 2S C S 2 β α γ s = T 0 s β α γ 0 This expression is important, and useful: 1. given the twiss parameters α, β, γ at any point in the lattice we can transform them and compute their values at any other point in the ring 2. the transfer matrix is given by the focusing properties of the lattice elements, the elements of M are just those that we used to compute single particle trajectories 44/48 D. Pellegrini - Practical Lattice Design

45 Exercise: Twiss transport matrix, T Compute the Twiss transport matrix, T, C 2 2SC S 2 T = CC SC + S C SS C 2 2S C S 2 for: 1. the identity matrix: M = ±I 2. a drift of length L β α γ s = T 3. a thin quadrupole with focal length ±f β α γ 0 45/48 D. Pellegrini - Practical Lattice Design

46 Beam ellipse evolution (another approach) Let s write the ellipse equation: ( ) J = γx 2 + 2αxx (s) + βx 2 x in matrix form, for X = x : ( X T Ω 1 β α X = J with: Ω = α γ ) = Σ ε 2 At a later point if the lattice the coordinates of an individual particle are given using the transfer matrix M from s 0 to s 1 : X 1 = M X 0 Solving for X 0, i.e. X 0 = M 1 X 1, and inserting in X0 T Ω 1 0 X 0 = J, one obtains: ( ) M 1 T X 1 Ω 1 ( ) 0 M 1 X 1 = J ( ( ) ) X1 T M T 1 Ω 1 ( ) 0 M 1 X 1 = J ( ) X1 T M T 1 Ω 1 0 M 1 X 1 = J }{{} Ω 1 1 Which gives Ω 1 = M Ω 0 M T 46/48 D. Pellegrini - Practical Lattice Design

47 Summary Hill s equation: x (s) + K (s) x (s) = 0, K (s) = K (s + L) general solution of the Hill s equation: x (s) = Jβ (s) cos (µ (s) + µ 0) phase advance & tune: µ 12 = s 2 s 1 ds β(s), Q = 1 2π ds β(s) beam ellipse: J = γ (s) x (s) 2 + 2α (s) x (s) x (s) + β (s) x (s) 2 beam emittance: ɛ = Area of the beam ellipse = det ( cov ( x, x )) transfer matrix s 1 s 2: M = βs β (cos µ s + α 0 sin µ s ) 0 (α 0 αs ) cos µs (1+α 0 αs ) sin µs βs β0 βs β 0 sin µ s β0 (cos µs αs sin µs ) βs stability criterion: trace (M) 2 47/48 D. Pellegrini - Practical Lattice Design

48 Summary: beam matrix, emittance, and Twiss parameters The beam matrix is the covariance matrix of the particle distribution ( ) ( σ11 σ Σ = 12 x 2 xx ) = σ 21 σ 22 x x x 2 this matrix can be also expressed in terms of Twiss parameters α, β, γ and of the emittance ɛ: ( x 2 xx ) ( ) Σ = x x x 2 = ɛ 2 β α α γ ( ) C S Given M = C S, we can transport the beam matrix, or the twiss parameters, from 0 to 0 s s in two equivalent ways: 1. Twiss 3 3 transport matrix: β α γ s = CC SC + S C SS C 2 2SC S 2 C 2 2S C S 2 β α γ 0 2. Recalling that Σ s = M Σ 0 M T : ( β α α γ 48/48 D. Pellegrini - Practical Lattice Design ) s ( β α = M α γ ) M T 0