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Accelerator Physics 1 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 ⎞ L1 LB sin⎜ ⎟= = ⎝2⎠ ρ 2 2 ()B ρ 9 If θ is small: L L 2 2 ⎛θ ⎞ θ sin⎜ ⎟ = ⎝2⎠ 2 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: d() By (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 +K( s) x = 0 ds 2 If the restoring coefficient (K) is constant ins “ is h t n e h ”t is just SHM Remember “s” s jist longitudinal displacement around u 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 +K( s) x = 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 +K( s) x = 0 ds 2 9 2nd order differential equation. 9 Guess a solution: x.= (ε β ) s cos(φ s+ (φ 0) ) 9 ε and φ0 ants, which depend on the are constinitial 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 β= ω2 9 …and let φ = (φ()s +φ 0 ) 1+α 2 Remember φ is still γ = x.= ε (s)ω cosφ a function of s β 9 In order to solve Hill’s equation we differentiate our guess, which results in: dω x'= ε φcos − ε' ωφ sin φ ds 9 ……and differentiating a second time gives: ' ' 'x '= cosεω φ − ε 'ω φ ' sinφ − εω φ 'φ − 'ε'ωφ sin ' sinφ − ε'ωφ 2 cosφ 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 substitutex.= (ε β ) s cos(φ s+ (φ 0) ) and its second derivative ' ' 'x '= cosεω φ − ε 'ω φ ' sinφ − εω φ 'φ − 'ε'ωφ sin ' sinφ − ε'ωφ 2 cosφ d 2 x +K( s) x = 0 into our initial differential equation 2 ds 9 This gives: ε ω φ' ' ε cos ω φ− ' φ ' sin ε− ω φ ' '− φ sin' ' ε sin ωφ− '2 cos φ ε ωφ φ (+K )ε s ω cos= φ 0 ine and Cosine are orthogonal and Seme timever be 0 at the sa nlilw E.Wildner NUFACT08 School 19 Benasque Solution of Hill’s equation (4) ε ω φ' ' ε cos ω φ− ' φ ' sin ε− ω φ ' '− φ sin' ' ε sin ωφ− '2 cos φ ε ωφ φ (+K )ε s ω cos= φ 0 9 Using the ‘Sin’ terms 2ω 'φ '+ωφ '= ' 02ωω 'φ '+ω 2φ '= ' 0 9 We defined β = ω 2 , which after differentiating gives 'β = 2ωω ' 9 Combining2 ωω 'φ '+ω 2φ '= 'and 0 'β = 2ωω '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 we have an exSo now pression for x and x’ x =.ε β cosφ and'αx= ε / − β cos φ− ε / β sin φ 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.
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