The A, B, C and D are determined by these 4 BCs to obtain

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1 Solution:. Floquet transformation: (a) Defining a new coordinate η = y/ β and φ = (/ν) s 0 ds/β, we find ds/dφ = νβ, and dη dφ = ds dη dφ d 2 η dφ 2 = ν2 β ( β y ) ( 2 β 3/2 β y = ν β /2 y ) 2 β /2 β y, ds = νβ ( β /2 y 2 β /2 β y 4 β 3/2 β 2 y Thus the equation of motion becomes d 2 η dφ + 2 ν2 η = ν 2 3/2 B β Bρ, where we have used the identity of Eq. (2.64). (b) The Green function is ). { A cos νφ + B sin νφ 0 φ < φ G(φ, φ ) = C cos νφ + D sin νφ φ φ 2π with the boundary conditions: G(0, φ ) = G(2π, φ ), G (0, φ ) = G (2π, φ ), G(φ +, φ ) = G(φ, φ ), G (φ +, φ ) G (φ, φ ) =. The A, B, C and D are determined by these 4 BCs to obtain G(φ φ ) = 2ν sin πν cos ν[π (φ φ ) ]. Taking deivatives of the Green function with respect to φ, we can also verify the Green function solution. Using the Green function, we obtain easily the closed orbit solution given by Eq. (2.57). The solution of part (a) can be attained by expanding the right-hand side in Fourier series as 3/2 B β Bρ = k= f k e jkφ with f k = β B 2πν Bρ e jkφ ds, where f k are integer stopband integrals. Expressing the closed orbit as η co (s) = Fk e ikφ, we find ν 2 f k y co (s) = β(s) ν 2 k 2 ejkφ. k= 7

2 (c) Using a single stopband approximation and limiting the closed orbit deviation to less than 20% of the beam size, we find βν 2 f [ν] y co 2 ν 2 [ν] 0.20 βɛ 2 rms. Approximating ν 2 [ν] 2 2νΓ [ν], where Γ [ν] is the stopband width, we find Γ [ν] 5ν f [ν] / ɛ rms. 2. When a quadrupole is misaligned, it introduces dipole error. (a) In the thin lens approximation, the dipole kick angle in the horizontal plane is θ x = ( x)/f. The misalignment does not change the vertical closed orbit. The horizontal closed orbit is x co (s) = G(s, s 0 )θ x = x β F f β x (s) 2 sin πν x cos(πν ψ x (s) ψ F ). Here β F is the horizontal betatron amplitude function given by Eq. (2.74) at the misaligned quadrupole location. Using the AGS Booster as an example, we find that the phase advance of 85 produces ν x = 5.67 with β x = 4.4 m at the focusing quadrupole location. The stopband width is f k = (b) For the vertical misalignment errors, the the dipole kick angle in the vertical plane is θ z = ( z)/f. The misalignment does not change the horizontal closed orbit. The vertical closed orbit is z co (s) = G(s, s 0 )θ z = z β D f β z (s) 2 sin πν z cos(πν ψ z (s) ψ D ). Here β D is the vertical betatron amplitude function given by Eq. (2.75) at the misaligned quadrupole location. Since β D < β F, the resulting closed orbit error is slightly smaller. Using the AGS Booster as an example, we find β D = 2.74 m. The stopband width due to the mis-alignment is f k = 0.005, which is smaller than that of the horizontal mis-alignment. 3. When quadrupole strength is increased by 5%, then it produces tune shift and stopband at the half integer stopbands. We use the AGS Booster as an example, where C = m with 24 superperiods. For the phase advance of 85, we find that the tune shift are ν x = 0.08 and ν z = 0.004, and the stopband widths are J p,x = and J p,z = See Eq.(2.84) in the Lecture Note. 5. Transfer matrix for the dispersion function (a) The solution of the homogeneous equation, D + KD = 0 with K > 0 is D = a cos Ks + b sin Ks, 8

3 and the special solution for D + KD = /ρ is D = /Kρ. Thus the general solution is D = a cos Ks + b sin Ks + /Kρ. Using the initial conditions: D = D 0 and D = D 0 at s = 0, we obtain a = D 0 Kρ b = D 0. K The transfer matrix the 3 3 matrix becomes D cos Ks K sin Ks ( cos Ks) D Kρ = K sin Ks cos Ks ρ sin D 0 Ks. D K 0. (b) The general solution for K < 0 is D = a cosh K s + b sinh With initial conditions, the solutions becomes D = D 0 cosh K s + D 0 D = D 0 K sinh K s + D 0 cosh K s K ρ. K sinh K s + K ρ ( + cosh K s), K s + ρ K sin K s. (c) For a sector magnet, we substitute K = /ρ 2 and Ks = θ into part (a) and obtain cos θ ρ sin θ ρ( cos θ) M x = sin θ cos θ sin θ ρ (d) In a rectangle magnet, edge angle δ = θ/2. The transfer matrix becomes 0 0 cos θ ρ sin θ ρ( cos θ) 0 0 M x = tan θ 0 ρ 2 sin θ cos θ sin θ ρ tan θ 0 ρ 2 ρ sin θ ρ( cos θ) = 0 2 tan θ 2 (e) In thin lens approximation, we have l 0 and Kl /f for a quadrupole; and θ 0, ρ sin θ l and cos θ for a dipole, the transfer matrix becomes l lθ/2 M x = 0 θ 9

4 6. From (2.226), the emittance growth due to kick angles is ɛ = βθ 2 β θ 2 (a) The particle numbers per volume N v is N v = n V N a = N a Pg RT, The target thickness in [gm/cm 2 ] per second is then x = N v m g βc = P g RT A gβc = βp g [ntorr] A g [gm/cm 2 ]. Use the above thickness to get the scattering angle θ 2 per second, and further the emittance growth per second, or the emittance growth rate as shown in the problem. (b) With the scattering angle shown in the problem, the emittance growth of H beam per passage is then β [m] t[µg/cm 2 ] ɛ = 7.8 β 2 (pc[mev]) 2 X 0 [g/cm 2 ] [πmm mrad] For 7 MeV H passing through carbon foil, we have β = , pc = MeV, X 0 = g/cm 2, θ = mrad, and the emittance growth per passage is ɛ = πmm-mrad. 7. This exercise derives the equation of motion for the orbit distortion function. It is an alternative method to discuss the half integer stopband and betabeat in accelerators. (a) Let A = α β 0 /β α 0 β /β 0, B = β /β 0 β 0 /β, we obtain ( db ds = A + ), β 0 β da ds = β ( ) β 0 β0 + β β0 β β 0 β (b) In a region without gradient error, K = 0, A 2 + B 2 is constant. ( + (K K 0 ) β β 0 = B + ) + K β β 0. β 0 β A da ds + B db ds = 0 or d ds (A2 + B 2 ) = 0 (c) In the presence of small quadrupole field error, we integrate the equations in part (a) to obtain A = β β 0 Kds β 0 g, B 0, where g = Kds is the integrated gradient error, and we have used thin lens approximation with β β 0 to obtain the above results. 0

5 (d) In a sextupole with a non-zero closed orbit x co, the effective quadrupole field is K = B 2 Bρ x co. B Thus A β 2 s 0 x Bρ co. (e) Using the definition, d φ ds = ( + ), 2 ν β 0 β db db = d φ ds ds = 2 νa d φ we find or d 2 B d φ 2 da = 2 ν ds ds ( d φ = 4 ν2 B 2 4 ν 2 + ) β β 0 K β 0 β d 2 B d φ ν2 B 2 = 4 ν 2 (β 3 0β ) 2 K 2 ν 2 β 2 K. β 0 + β 8. There are many ways to derive this result. In the presence of quadrupole error, Hill s equation becomes y + (K(s) + k(s))y, where k(s) = K(s). The Hill s equation can be deerived from the Hamiltonian: H = 2 y K(s)y2 + 2 k(s)y2. Now, we carried out Floquet transformation and change the independent coordinate from s to the orbiting angle θ. This can be accomplished by substituting y = into the Hamiltonian (see Sec. II.4), i.e. 2βJ cos(ψ + χ(s) νθ) H = νj + Jk(s)β(s) [ + cos 2(ψ + χ(s) νθ)]. 4 Here χ(s) = s 0 (ds/β), and θ = s/r. Now, we expand the perturbation in Fourier Harmonics: Note that the zeroth harmonic of the first term gives the tune shift ν = 4π k(s)β(s)ds. The resonance can be derived from the second term of the perturbation, i.e. 4 Jk(s)β(s) cos 2(ψ + χ(s) νθ) = 2 J p G p cos(2ψ pθ + ξ p ),

6 where G p = 2π = 2π k(s)β(s)e j[pθ+2νθ 2χ(s)] ds k(s)β(s)e jpθ ds and ξ p is the phase of G p. Only one perturbation term is important if ν is near a half-integer, i.e. ν = p/2. In the one term approximation, the Hamiltonian becomes H = νj + 2 G p J cos(2ψ pθ + ξ p ). Now, we carry out canonical transformation with the generating function: to obtain a new Hamiltonian: F 2 = (ψ p 2 θ + ξ p 2 )I H = δi + 2 G p I cos 2φ, where I and φ = ψ p θ + ξp are conjugate variables and δ = (ν p ) is the resonance proximity parameter. Hamilton s equation of motion becomes I = G p I sin 2φ, ψ = δ + 2 G p cos 2φ Carry out the second derivative on the action I, you find Ï = G p 2 I + 2δ G p I cos 2ψ. Now, if ν = p/2, we have δ = 0, and the second term is zero. The solution of the action is given by G = A exp{ G p θ} + B exp{ G p θ} A exp{ G p θ}. 2

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