Accelerator Beam Dynamics

Transcription

1 Accelerator Beam Dynamics DRubin May 3, 6 Contents Twiss Parameters Scalar invariant Computing transfer matrix with tracking 4 3 Full Turn Map 5 3 Eigenvalues 5 3 Decomposition 6 33 Propagation of Twiss Parameters 7 34 Connection to Differential Equation 8 34 Propagation of β-function 8 35 Closed Ring 35 Quadrupole error 4 Dispersion 4 Coupling of Longitudinal and Transverse Motion 4 Closed Ring 5 Symplectic Transformation 6 Coupling 4 7 Normal Mode Decomposition of N N symplectic matrices 4 7 Normal Modes 5 7 Real Basis 6 73 W-matrix 6 74 Summary 8 8 Interpretation of the Coupling Matrix 8 9 Equations of motion Solution to equation of motion Floquet transformation 3

2 Field errors 4 Linear Quadrupole Resonance 4 Matrix Method 4 3 Sexupole resonance 5 3 Hamiltonian dynamics 5 3 Point transformation 5 3 Poincare invariants 6 4 Transfer Matrices 7 5 Quadrupole 8 5 Tilted quadrupoles 9 5 Skew quadrupoles 9 6 Solenoids 9 6 Longitudinal fields 9 6 Radial fringe 3 63 Symplecticity of solenoid maps 3 7 Superimposed solenoid and quadrupole fields 3 8 Dipole 3 9 Cyclotron Equations of Motion 3 Twiss Parameters Consider propagation of a particle trajectory through a beam line The phase space -vector is x = x, p x /p) x, x ) x and x are offset and angle with respect to some reference orbit, where in the small angle approximation p x /p x The Jacobian of the mapping from x i to x f, is M = x f x i x f x i The Jacobian M for a Hamiltonian system is symplectic In a two dimensional phase space a symplectic matrix has unit determinant If the system is linear, then Scalar invariant x f x i x f x i x f = M fi x i Define the scalar s = x T Ax

3 Then define A, so that s = x x ) γ α x α β x = γx + αxx + βx A can be any 4 parameters that we like No loss of generality by setting A = A T since we only need three numbers to define the most general scalar combination of x and x α, β, γ are the twiss parameters Now let s assume linearity and propagate x b x e with the help of M Then x e = Mx b and s = x T b M T M T ) A b M Mx b = x T e M T ) A b M x e = x T e A e x e The matrix s is invariant as long as Or γe ) α e α e β e A e = M T ) A b M ) = M T ) γb α b M α b β b We propagate twiss parameters using the transfer matrix Evidently, once the twiss parameters are selected at one point in the beam line, they are defined everywhere by the mapping that propagates the phase space coordinates Another thing, from Equation we see that A e = A b The determinant of the twiss matrix is invariant We set it to unity for convenience Then γβ α = It should be clear that except for the unit determinant requirement, the twiss parameters α, β, γ) are totally unconstrained We assign them whatever values we like at one location along the beam line and they are determined everywhere else The distribution of particles can be characterized in terms of the twiss parameters The twiss parameters establish how the phase space coordinates are correlated, how x and x are related Consider the matrix of second moments The average of the first moments is zero) xx xx Σ = ) xx x x The matrix is constructed as xx T = x x x x ) and the distribution is propagated along the beam line according to x e x t e = Mx b x T b M T Then x e x t e = Mx b x T b M T = M x b x T b M T or xx xx ) xx xx xx x x = M ) xx x x M T e b which looks almost like the rule for propagating the twiss matrix A W found that A e = M T ) A b M Then A e = MA b M T 3

4 and the matrix A e transforms the same as the Σ matrix The elements of the two matrices are clearly related In particular A β α xx xx = ɛ = ) α γ xx x x where ɛ remains to be determined Since we have that A = Σ ɛ = xx x x xx σ = xx, σ ) = x x β = σ ɛ, γ = σ ) ɛ The twiss parameters are determined by the distribution of the phase space coordinates of the trajectories Computing transfer matrix with tracking Sometimes it is difficult to construct the transfer matrix from first principles The matrix conveys the focusing effect of the element but to build the matrix we essentially need to know all the gradients etc Alternatively we can determine the Jacobian directly by particle tracking Remember that the transfer matrix is the Jacobian of the map M = x f x i x f x i The strategy is essentially to compute the derivatives numerically If we know the reference trajectory uniquely defined in a circular machine, but not so straightforward in a transfer line like the entrance through the backlog iron and into the inflector), we can calculate trajectories displaced by x and x from the reference and build M In principle we need only three non degenerate trajectories to determine the X matrix for horizontal or vertical motion as well as the reference Write x f x i x f x i M i f x in x ref ) = x f x ref M i f x in M i f I)x ref = x f M i f x in x = x f ) Next construct and Equation becomes N = m Mi f x m x = m m x N xin = xf 4

5 The goal remember is to compute M i f and x ref Choose three distinct values for x in, namely x i in, i =,, 3, track each to xi f and we get x N in x in x 3 ) in x = f x f x 3 ) f Finally x N = f x f x 3 ) f x in x in x 3 in Extract M and x ref from N as per above The strategy is readily extended to the full 6 dimensional phase space where x x x y y z δ where δ = E/E So to determine the evolution of the phase space that is the twiss parameters) through the iron and inflector into the ring we simply compute 7 trajectories We can in principle use the same 7 trajectories to determine the transfer matrix between any two points along the reference orbit 3 Full Turn Map Consider the full turn map M in a closed ring The stability of the lattice is indicated by multiturn behavior The initial phase space coordinates x in are mapped after n-turns to 3 Eigenvalues x out = M n x in It is easy to determine stability if we work in an eigen-basis The eignenvalues of the X determinant matrix M are λ ± = e ±iφ where cos φ = TrM and sin φ = TrM In view of the constraint on the trace, the full turn matrix M can be written as cos φ + x y M = z cos φ x Next define ) M = cos φ x yz = x yz = sin φ x = α sin φ, y = β sin φ, z = γ sin φ, α βγ = and the most general unit determinant matrix is cos θ + α sin θ β sin θ M = γ sin θ cos θ α sin θ 5

6 For the time being, α, β, and γ are arbitrary real numbers The relationship with the twiss parameters described above remains to be determined The normalized eigenvectors are ) β v ± = ±i α γ + β β Propagation of the phase space vector through n turns is gives v n ± = M n v ± = e ±inθ v ± The linear lattice as represented by the Jacobian of the map M) is stable if θ is real TrM < ) 3 Decomposition The similarity transformation to the eigenbasis is where U MU = Λ U = v + v = N e iθ ) e iθ β β i α β i+α β It is convenient to work in a real basis We note that cos θ sin θ = K ΛK sin θ cos θ where Then Next define K = eiπ/4 i i K U MUK = K cos θ sin θ ΛK = Rθ) sin θ cos θ G UK = β ) α β β where the constant N is chosen so that det G = In summary G MG = Rθ) In an earlier section we defined the parameters α, β, γ in terms of scalar invariant Are the two definitions consistent? The twiss parameters transform according to where M T AM = A γ α A = α β If the definitions are consistent then the β, α, γ of G are the same as the α, β, γ that are the components of A and But G T AG = I, and we have that as claimed M T AM = GRG ) T AGRG = G ) T R T G T AG ) RG A = GR T RG T = A 6 )

7 33 Propagation of Twiss Parameters Now suppose that the full turn matrix at point is written as a product of the matrices of each individual element quadrupole, drift, dipole, etc) in the ring The full turn matrix at point is and M = T T 3 M = T 3 T 34 T M = T M T The twiss parameters at are given by M In particular, cos θ = TrM, α = m m )/ sin θ, β = m / sin θ, γ = + α β and the same for point We would like to write T in terms of the twiss parameters at and We have that For an orthogonal matrix R M = T M T = G Rθ)G = T G RG Rθ) = G T R = W RW [R, W ] = W = W T If W is orthogonal we can write cos φ sin φ W = Rφ) = sin φ cos φ cos φ sin φ = sin φ cos φ T ) R G T ) G and then T = G Rφ)G 3) ) β cos φ sin φ = β β α β sin φ cos φ α β β ) cos φ α β sin φ = β sin φ ) β sin φ α cos φ β cos φ β = β α β β β cos φ α sin φ) α α ) cos φ α α +) sin φ β β β β sin φ 4) β β α sin φ + cos φ) Therefore xs) xs) x = T s) x s) 5) 7

8 xs) = x s) = ) βs) ) cos φ s α sin φ s x + βs)β sin φ s x 6) β αs) α ) cos φ s ) αs)α + ) sin φ s βs)β ) x + β βs) αs) sin φ s + cos φ s )x 7) where φ s is some phase advance from point to s How are α and β related? How is φ s defined? Since x = dx ds we have that [ x s) = dx ds = β ) ] s) βs) cos φ s α sin φ s + sin φ s α cos φ s )φ s x βs) β β [ ] β s) βs) + sin φ s + cos φ s φ s x 8) β β 34 Connection to Differential Equation Comparing Equations 7 and 8 we see that αs) = β = dβs) ds, φ s) = dφ ds = β The phase advance φ s = s ds βs) 34 Propagation of β-function As noted in an earlier section there is an invariant S = γx + αxx + βx β X 9) where γ β = α, β Consider propagation from to by matrix M x X = xx x ) x = m x + m x x = m x + m x Then and X = M X m m m m M = m m m m + m m m m ) m m m m 8

9 We would like to determine the corresponding matrix N that propagates the 3-vector β Since β X = β X = β M M X = β M X, it follows that β = M ) T β N = M ) T ) N is constructed by replacing the elements of M with those of its inverse in Equation and transposing, yielding m m m m N = m m m m + m m m m ) m m m m Consider for example propagation of the twiss parameters through a field free region elements of M in a field free region of length s are s M = The Then N = s s s At a waist minimum β,α = ), β γ, α, β ) = /β,, β ) At a distance s from the waist, βs) = N β = /β, s/β, β + s /β ) We find that β increases quadratically with distance from the minimum Using, we write N in terms of twiss parameters β and βs) and the phase advance φ = φs) φ Suppose that β at is perturbed by a mismatch or quad error, so that β γ + γ γ γ β β α = α β + β β + β β = In view of Equations and β γ β β βs) = m γ β β + m β = βs)β sin φ)γ β β + βs) β cos φ α sin φ) β = βs) β cos φ sin φ) α cos φ sin φ) β βs) = β cos φ α sin φ) β The β error propagates as the square of the phase advance 9

10 35 Closed Ring In a closed ring, where s = C, and βc) = β, the change in β on the n th turn is where µ = πq and Q is the tune 35 Quadrupole error βn) = β cos µ α sin µ) Suppose that there is a focusing error at s, with focal length f Then where the transfer matrix for a thin lens is Q = M MQ ) /f Then M = = cos µ + α sin µ f β sin µ β sin µ γ sin µ + f cos µ α sin µ) cos µ α sin µ cos µ + α sin µ β sin µ ) γ sin µ cos µ α sin µ Then cos µ = cosµ + µ) = TrM = cos µ β f sin µ If f β, cosµ + µ) = cos µ µ sin µ = cos µ β f sin µ µ = β f 4 Dispersion The formalism developed so far describes a dimensional phase space It can be used to account for motion independently in the horizontal, vertical or longitudinal direction Now we consider systems with coupling between transverse horizontal) and longitudinal degrees of freedom The phase space vector x x p xp z p z = p p p where in the small angle limit x = p x /p and p is the reference energy mapping from x i to x f is The Jacobian of the T ij = xf j x i 3) As noted above, the Jacobian for a Hamiltonian system is symplectic A matrix is symplectic if T T ST = S

11 where and s S = s s = In a system with no coupling of horizontal and longitudinal motion Mx T = P z where M x and P z are and M z = P z = 4 Coupling of Longitudinal and Transverse Motion A bending magnet couples changes in energy p z ) with a change in horizontal angle x and position x so that m and m are in general non zero The symplectic condition then requires that p and p are also finite and the Jacobian will have the form m M x T = m p p 4) P z The dispersion is the dependence of transverse position and angle on energy offset δ, that is ) η η = The dispersion is defined at every point s along the beamline or ring so that the horizontal position can be written in terms of a combination of betatron and energy components x δ x δ xs) = x β s) + x δ s) Suppose that at some location along the beam line there is zero betatron amplitude Then x = x,δ = η δ and from 4 m x = Mη δ + δ m η = Mη + m 5) m = η Mη m Note that in a beam line without RF accelerating cavities there is no mechanism to couple longitudinal offset z) and transverse position Equation 4 is therefore a good representation of transport in such a beam line It is easy to show that if m m = m

12 then the symplectic condition implies that M z = We learned all about the formalism for a unit determinant matrix in a previous section and how to can express M z in terms of β and φ We find m = η T η 6) m where T given in Equation 4 The linear mapping from point to point can be written entirely in terms of β, η at the end points and the betatron phase advance φ We learned that the dispersion function is propagated according to Equation 5 If η and η are the closed ring dispersion at and, then dispersion errors, perhaps due to some mismatch, propagate according to η = M η = β β cos φ α sin φ) α α ) cos φ α α +) sin φ β β β β sin φ η β β α sin φ + cos φ) 4 Closed Ring In a closed ring with circumference C, ηs) = ηs + C) and full turn matrix M m T = p P Equation 4 gives the closed ring dispersion m η = Mη + m m I M)η = m η = I M) m m 5 Symplectic Transformation Consider the canonical transformation from generalized coordinates q, p to Q, P where q = q, q,, q N ) and p = p, p,, p N ) We aim to show that the Jacobian matrix J = Q, P) q, p) is symplectic We recall that the time evolution operator, namely the Hamiltonian), has the form of a generator of canonical transformations Therefore, the transformation of phase space coordinates from point η along a beam line to another point ξ is canonical and the transfer matrix is symplectic Suppose that the phase space vectors at two distinct points along the beam line are x X p x η = y and ξ p y P x = Y P y

13 and K ξ) = H η) + F t where H and K are the Hamiltonian in terms of the coordinates η and K in terms of ξ F is the generator of the transformation The generalized Hamilton s equations are H η i = S ij where S = η j Similarly We could also write K ξ i = S ij = S ij ξ j ξ j ξ i = ξ i η j + ξ i η j t Combining Equations 7 and 8 S ij H η) + F ) ξ j t H S ij + ) F ξ j t ξ j H η k S ij + ) F η k ξ j t ξ j H η) + F ) t = ξ i H S jk + ξ i η j η k t = ξ i H S jk + ξ i η j η k t = ξ i H S jk + ξ i η j η k t = ξ i H S jk + ξ i η j η k t If the generating function F is type, then F ξ j = S jk ξ k and Equation 9 becomes H η k S ij S jk ξk = ξ i H S jk + ξ i η k ξ j η j η k t H η k S ij = ξ i H S jk η k ξ j η j η k since S ij J jk = δ ik Then We recognize ξ i η j Then 7) 8) 9) ) S ij η k ξ j = ξ i η j S jk ) = M ij as the Jacobian of the mapping from η to ξ and η k ξ j SM ) T = MS S ij M kj = S ij M ) T jk = M ijs jk = M ij S jk S = MSM T = M as the inverse which is the definition of a symplectic matrix The Jacobian of a canonical transformation is a symplectic matrix 3

14 6 Coupling Let s begin with transverse coupling MSM T = S A 4X4 symplectic matrix The Jacobian is a 4X4 symplectic matrix M such that A B M = C D has the property that A + B = C + D = show) Diagonalizing gives us M = UEU where E is a 4X4 diagonal symplectic matrix show that the diagnonal matrix is symplectic) Then if it follows that We know that E = λ λ λ 3 λ 4 λ λ = λ 3 λ 4 = M u i = λ i u i where u i is the eigenvector with eigenvalue λ i Then the phase space evolves after n turns according to M n u i = λ n i u i If the magnitude of any of the eigenvalues λ i ) is greater or less) than, then the motion grows exponentially The eigenvalues for the full turn coupled matrix for a stable system with dimension N XN are unimodular, complex conjugate pairs 7 Normal Mode Decomposition of N N symplectic matrices Normal mode decomposition of a 4X4 symplectic matrix is a standard technique for analyzing transverse coupling in a storage ring We generalize the decomposition to any NXN symplectic matrix T and derive the transformation W from lab coordinates to normal mode coordinates U That is T = W UW ) where U is block diagonal and real and we construct the real matrix W with the form γ I C C C γ I C 3 W = C C 3 γ 3 I 3) I is the x identity, and C, C, C C = C ) etc are x If for example, n =, then γ = γ and 4

15 7 Normal Modes The block diagonal matrix where with and and G i = βi α i βi βi ) U = A B can be further decomposed as U = Y ZY 4) Rθ ) Zθ, θ,, θ n ) = Rθ ) 5) cos θ sin θ Rθ) = sin θ cos θ 6) G Y = G, 7) Since standard techniques exist for diagonalizing square matrices and identifying eigenvalues and eigenvectors, we begin by doing just that T = V DV, 8) where T is the N x N symplectic matrix, D is the diagonal matrix of eigenvalues, and V is the matrix constructed from the eigenvectors Since T is symplectic, the eigenvalues and eigenvectors, as noted above, appear as unimodular, complex conjugate pairs, λ i, λ i and v i and v i Then D can be written in the form dθ ) dθ ) e iθ D = dθ 3 ) where dθ) = e iθ 9) The n columns of the matrix V are the n eigenvectors v i The eigenvectors are not unique, but may be multiplied by an arbitrary complex number That is, v i ρ i e iφ i v i and v i ρ ie iφ i v i If V = v v v v v n v n, then V ρ, φ) = V Dρ, ρ, ρ n, φ, φ,, φ n ) ρ dφ ) ρ dφ ) = V ρ 3 dφ 3 ) Note that V ρ, φ) effects the transformation of Equation 8 for any real numbers ρ i and φ i 5

16 7 Real Basis We transform from a complex to a real basis with K where the real matrix Z Equation related to the complex matrix D Equation 9) by the similarity transformation 5) is Zθ, θ, θ 3 ) = KDθ, θ, θ 3 )K 3) where and k K = k 3) k k = i i 3) 73 W-matrix To construct W and U from V and D, we use Equations, 4 and 3 to write T = W UW = V DV = V D ρ, φ) K K ) D θ) K K ) D ρ, φ)v = V K K ) D ρ, φ) K K ) D θ) KK ) D ρ, φ) KK ) V = V K ) Z ρ, φ)z θ)z ρ, φ) K V = V ρ, φ)z θ)v ρ, φ) ) Now since the columns of V are complex conjugate pairs, V K is real similarly constructed to be real and therefore V is real So far we have W UW = V ZV W Y Z θ)y = V Z θ)v V = W Y The Z matrices are where we have used Equation 4 Next we determine the parameters ρ and φ We choose ρ so that V will be symplectic In particular, if we write V in terms of the X matrices V j i then V V V = V V = = V V ρ Rφ ) V V ρ Rφ ) ρ V Rφ ) ρ V Rφ ) ρ V Rφ ) ρ V Rφ ) 6

17 Symplecticity constrains the sums of determinants of V i j so that = = = n i= n i= n i= ρ j = V i j ρ j V j i Rφ j) ρ j V j i n i= V j i In order to determine the order of the conjugate columns of V, and finally the paramters φ we expand V = W Y G) V V γ I C G V = V V = C γ I G ρ V Rφ ) ρ V Rφ ) γ G CG ρ V Rφ ) ρ V Rφ ) = C G γ G Then the diagonal blocks are required to have the form V i i = γ i G i ρ i V i i Rφ i ) = γ i βi βi α i βi ) A real solution requires that V i i > We are free to choose the order of the conjugate columns of V to ensure that this is true Note that if we reverse the order of the columns V i,j V j,i, then the sign of the determinant of the X blocks is reversed) If we reverse the order of eigenvectors in V, then we also reverse the order of eigenvalues in D θ) or equivalently θ i π θ i To find φ we proceed with our expansion of V ii and Rφ i ) and write ii ii V V ρ i ii ii V V ) ) cos φi sin φ i βi ) = γ sin φ i cos φ i βi α i i βi We choose φ i so that G i =, or tan φi = V ii V ii The ambiguity in φ i, tan φ i = tanπ φ i )) is resolved with the condition that G i = V ii cos φ i ii V sin φ > 7

18 74 Summary Find eigenvectors and eigenvalues Transform eigenvectors to a real basis 3 Construct V The columns of V are the eigenvectors The eigenvectors appear as complex conjugate pairs since T is symplectic 4 Choose the normalization for each pair of eigenvectors so that W will be symplectic In particular if c V, c V, V = c V, c V, where V i,j are X matrices, and c i = ρe iφ i then choose ρ so that ρ V, + V, + V 3, + ) = 5 Adjust the order of complex conjugate pairs so that V i,i > That is, if V i,i <, than swap the order of the columns 6 Choose the phases φ i so that has the form G i = V i,i Rθ) = β ) α β β 8 Interpretation of the Coupling Matrix Consider two dimensional coupling, that is horizontal and vertical, or horizontal and longitudinal The full turn matrix T is written in terms of normal modes where and T = V UV M m T = n N U = A B As before section on propagating twiss parameters), we can write A = G a Rµ a )G a, B = G Rµ b )G b b where G a = βa α a βa βa ), Rµ) = cos µ ) sin µ sin µ cos µ 8

19 Now where U n Rnµa = G )G Rnµ b ) Ga G = G b The real phase space vector x is related to the normalized normal mode vector u according to x = V G γi C u = C G u γi where for a 4X4 matrix, V has the form defined in Equation 7, and C is X Then after propagation through n turns, we have x n = T n x = T n G V u = V G RGV ) n V G ) u = V G Rnµ a ) u = GG V G )Rnµ a ) u = G V u n where V = GV G is the normalized coupling matrix In the limit of vanishing coupling between horizontal and vertical planes, the normal mode emittances ɛ a and ɛ b reduce to ɛ x, ɛ y as ɛ a lim a ɛ x ɛ b lim b ɛ y In electron storage rings typically ɛ b ɛ a So let s suppose that ɛ b = Then in general in the normal mode coordinates u = The normal mode coordinates propagate from one turn to the next as a rotation u a u u n Rnµa ) = u a cosnµ) + u a sinnµ) a Rnµ b ) = u a sinnµ) + u a cosnµ) u a u a It is convenient to define our starting point so that u a = Now the laboratory frame phase space coordinates on turn n are related to the normal mode coordinates according to x n = G V R n u cosnµ) γ cosnµ a ) = G V sinnµ) u a = G γ sin nµ a C cosnµ a ) C sinnµ a ) C cosnµ a ) + C sinnµ a ) u a 9

20 Now we have parameteric equations for the real space trajectory x n = β a γ cosnµ a )u a y n = β b C cosnµ a ) + C sinnµ a ))u a θ w tanθ = β b /β a ) / C - /γ w = β b /β a ) / C - /γ 9 Equations of motion For a relativistic charged particle in a static magnetic field d p dt = e v B The field is static, so energy does not change and γm is independent of time and d p dt v = e v B γm = mγ d v dt = e v B 33) Now we have to compute R = v in the curvilinear coordinate system We define ρ as the curvature The coordinates of the particle at any position along the path s are R = ρ + x)ˆx + yŷ R = ẋˆx + r ˆx + ẏŷ = ẋˆx + r θ)ŝ + ẏŷ where ˆx = θŝ The second derivative R = ẍˆx + ṙ θŝ + r θŝ + r θ ŝ + ÿŷ = ẍˆx + ṙ θŝ + r θŝ r θ ˆx + ÿŷ

21 where ŝ = ˆx θ We know that ds = ρdθ = ρ vs r dt So we can write that θ = vs r Next change dependent variable from t to s Then R = ρ v s r ) R = x ρ v s r Comparison with equation 33 gives ds = ρ v s dt r and R = ẍ v s r )ˆx + ṙ θ + r θ)ŝ + ÿŷ d = ds d dt dt ds = ρv s d r ds d dt = ρ v ) s d r ds ) v s r )ˆx + v r s r + rθ )ŝ + r θ ρ v s r x r ρ = e γm v yb z v s B y ) x r ρ e r B y γmv s ρ x e B y γmv s r ρ ) + r ρ r v s ρ x B y ) + y db y Bρ dy + xdb y dx r x ρ + xk + r ρ ρ x + ρ x ρ + xk + x + ρ ρ ρ x x ρ + xk y = e γm v sb x v x B z ) + x ) ρ y eb x γmv s y r v s ρ e γmv s B x ) + db x dx x + db x dy y) y Bρ B x) + db x dy y) y Ky and ) ŝ + y ρ v ) s ŷ r ) r + r ρ ρ + x ) ρ

22 where we have only kept first order terms in x and y and we assume that B x ) =,and B y )/Bρ = /ρ Solution to equation of motion If Ks) is periodic with period C, such that Ks) = KC + s) then the general solution to the equations of motion is x = Aws) cos[ψs) + δ] with ws) periodic in C Substitution into the equation of motion gives x = Kx 34) = d w cosψ + δ) wψ sinψ + δ) ) + Kw cosψ + δ) ds = w cosψ + δ) w ψ sinψ + δ) wψ ) cosψ + δ) wψ sinψ + δ) + Kw cosψ + δ) If that last is true for any phase δ then = w ψ + wψ 35) Define β w and β β = β β ψ + β ψ ) = ψ ψ βψ + β ψ = d ψ ds ψ = d β ds β dψ ψ = dβ β Also = w wψ ) + Kw = β) β 3/ + Kβ = d β ds β β 3/ + Kβ ) = β β β 3/ β 3/ + Kβ β

23 Floquet transformation Let φ = ψ/ν, where ν = π ds βs) and ξ = x Then β xs) = Aβs) cosψs) + δ) x = A cosνφ + δ) β ξ = A cosνφ + δ) dξ dφ = dξ ds x ds dφ = νβ β / = ν x β / xβ ) d ξ dφ = ν β β / xβ ) β 3/ x β x β + β / x β β / xβ β / + 4 ) = ν x β 3/ xβ β / + xβ 4 β / )) = ν x β 3/ x β β / β β / ) xβ β 3/ Substitution from above gives d ξ dφ = ν d ξ dφ = ν x β / = ν ξ x β 3/ + x Kβ 3/ )) β / And if there are contributions to the equation of motion, for example from magnetic fields Bs) that do not scale linearly with displacement, then refeq:eom becomes and x + Ks)x = qcbs) Bρ d ξ dφ + ν ξ = ν β 3 Bξ, φ) Bρ 36) 37) 3

24 Field errors Linear Quadrupole Resonance Suppose there is a thin quad error field k at φ = φ Then Equation 37 becomes d ξ dφ + ν ξ = ν β 3 k β / ξδφ φ ) = ν β k ξ n= e inφ φ ) In the limit where the error is small the perturbation expansion is = Aν β k cos νφ e iφ e inφ n= = Aν β k e iφ e inφ+iνφ) + e inφ iνφ) n= The system will respond at the frequency of the driving force, which is nφ ± νφ Near resonance, that is when n ± ν) ν the solution is ξ ν β k A ν ± n ν) with resonance at ν = n/, when the tune ν is half integer or integer Matrix Method Recall the full turn map M = cos µ + α sin µ β sin µ γ sin µ cos µ α sin µ Insert a thin quadrupole with focal length f and matrix representation Q = /f Then the perturbed full turn matrix is ) M cos µ + α sin µ β sin µ = QM = f cos µ + α sin µ) γ sin µ f β sin µ + cos µ α sin µ The perturbed tune cos µ βf sin µ ) Stability requires that cos µ = cos µ = cos µ β f sin µ 4

25 3 Sexupole resonance A single sextupole with field proportional to k = B y displacment leads to the equation of motion d ξ dφ + ν ξ = ν β 5 k ξ δφ φ ) = ν β 5 k ξ n= A ν β 5 k cos νφ A ν β 5 k n= e inφ φ ) n= e iφ y Bρ e inφ φ ) scales quadratically with horizontal e inφ+φν) + e inφ φν) + e inφ)) The system responds resonantly when ν ±ν + n) = or ν ± n =, or ν = 3 n, ν = 3n, or ν = n 3 Hamiltonian dynamics Hamilton s equations ṗ = H q q = H p 3 Point transformation Consider the generating function Then So F = F 3 p, Q, t) = p ρˆx + xˆx + yŷ) q = F 3 p, P = F 3 Q P s = s F 3 = p ρ + x ) ˆx ρ s = p + x ) ŝ = p s + x ) ρ ρ P x = x = p x P y = y = p y Define canonical vector potential A s = A ŝ + x ) ρ A x = A ˆx A y = A ŷ 5

26 The new hamiltonian is just the original expressed in the new coordinates The original hamiltonian is H = p ea) c 4 + m c 4 + ev The new Hamiltonian z s) is H = c ) p s ea s ) + p x ea x ) + p y ea y ) + m c + x ρ + ev 3 Poincare invariants Consider a -dimensional region of phase space and define the area of any surface in the space J = dp i dq i i S is invariant with respect to canonical transformations of the phase space variables q, p That is J = dp i dq i = dp i dq i 38) i S i S where P i, Q i are related to p i, q i by a canonical transformation Any point on the surface can be identified by two coordinates u, v Then on the surface p i u, v), and q i u, v) The area element dp i dq i transforms to the area element dudv by the Jacobian determinant so that Then Equation 38 becomes J = i S q i, p i ) u, v) = q i u q i v p i u p i v dq i dp i = q i, p i ) u, v) dudv qi, p i ) u, v dudv = i S 39) Qi, P i ) dudv 4) u, v Since the surface is arbitrary, Equation 4 and the invariance of J is equivalent to the statement that q i q i u v p i = Q i Q i u v 4) i i u p i v where q, p) Q, P ) is a canonical transformation u, v) can be anything If we choose u q j and v p j, then Equation 4 becomes i q i q j p i q j q i p j p i p j = i Q i q j P i q j 6 P i u Q i p j P i p j P i v = i δ ij = 4)

27 Now let s suppose that q i, p i are the phase space coordinates of a particle at time t, and Q i, P i the coordinates at a later time t The two sets of coordinates are related by a canonical transformation Therefore, the elements of the Jacobian matrix relating q, p) to Q, P ) satisfy the relationship of Equation 4 There is one other invariance that we will need to demonstrate symplecticity of the Jacobian It turns out that Equation 38 can be generalized for a 4-dimensional surface, or a 6-dimensional surface, etc In particular the integral J = dp i dq i dp k dq k 43) i S is invariant with respect to canonical transformations Then as before we conclude that the Jacobian determinant q i, p i, q k, p k ) 44) u, v, w, z) is also invariant and that 4 Transfer Matrices Q i, P i, Q k, P k ) q i, p i, q k, p k ) = 45) Consider a quadrupole magnet The magnet is consists of 4 poles with alternating pole tip field If the magnet is very long, then far from the ends the field is purely transverse Expand the vertical component of the field in the horizontal plane B y = B y + x B y x + 46) By symmetry B x y = ) =, Consider a magnet with 4 poles with alternating pole tip field Place the center of each pole on the diagonals By symmetry In the two-dimensional limit, B z = Then B y θ) = B x θ + π/) = B y θ + π) = B x θ + 3π/) B x θ) = B y θ + π/) = B x θ + π) = B y θ + 3π/) B = B y y B = B x y = B x x = B y x B y = x B y x + B x = y B x y + 7

28 5 Quadrupole The linear equations of motion for a magnet with quadrupole symmetry are p t p s s t p s v z = e γm p zb = e v B = e γm p B = e γm p B The total momentum is constant and p z = Substituting into the above p p x p y p = p x y px p py p x γm p x y ) = e γm p x y ) B For small angles x p p s v z = e γmŝ B x = e pŝ B For the quadrupole x = e B y p x x = Kx y = e B x p y y = Ky where K = ep By x and K has dimensions of L The solutions to the equations of motion are x = A cos Ks + B sin Ks y = C cos Ks + D sin Ks If xs = ) = x and x s = ) = x then A = x and B = x / K and similary C = y and D = y / K If K > the quadrupole is horizontally focusing and vertically defocusing and The matrix for the transverse phase space becomes cos Kl K sin Kl K sin Kl cos Kl M = cosh Kl K sinh 47) Kl K sinh Kl cosh Kl 8

29 and x xl) M x y = x l) yl) y y l) 5 Tilted quadrupoles The 4 4 transfer matrix for a horizontally focusing quad with strength k and length l can be written as?? [] Kf M quad = 48) K d where K f,d are appropriate matrices The matrix for a quad rotated by an angle θ about the z-axis is Q rot = R θ)m quad Rθ) 49) where 5 Skew quadrupoles [] I cos θ I sin θ Rθ) = I sin θ I cos θ 5) A skew quad is rotated by θ = 45, Q skew = [] Kf + K d K f + K d K f + K d For a thin skew quad, l and k sin kl) f, [] I Kt Q thin =, K I t = K t K f + K d [] f 5) 5) 6 Solenoids 6 Longitudinal fields In the longitudinal field of a solenoid p p s v z = e p γm p ŝ B z 53) x = y e p B z = k s y The coupled equations are solved by substituting one into the other y = x e p B z = k s x 54) x = k sx y = k sy 9

30 The solutions are x = A cos k s l + B sin k s l x = A sin k s l + B cos k s l) + a k s y = C cos k s l + D sin k s l y = C sin k s l + D cos k s l) + b k s Substitution into 54 gives The boundary conditions fix the constants k s A sin k s l B cos k s l) = k s C cos k s l + D sin k s l) x = A, x = B k s + a y = B, y = A k s + b A = D, B = C A = x, B = y, a = x y k s, b = y + x k s The transfer matrix for the motion through a longitudinal field is [ M s M long = M s )] M s M s 55) where k s = e pc B z and 6 Radial fringe [ M s )] = k s sin k s z cos k s z [ M s )] = k s cos k s z ) sin k s z 56) 57) The fringe field of a solenoid is radial, of equal magnitude and opposite direction at each end The elements of the transfer matrix for the radial fringe are x x ) = cos χ cosh χ x x ) = kr sin χ cosh χ + sinh χ cos χ) x y ) = sin χ sinh χ x y ) = kr sin χ cosh χ sinh χ cos χ) y x ) = x y ) y x ) = x y ) y y ) = cos χ cosh χ y y ) = kr sin χ cosh χ + cos χ sinh χ) 58) 3

31 k where χ = r a, k r = e a pc B z, and a is the length along z) of the pole tips, that is, the effective length of the radial field [x x ), x x ), etc are obtained by differentiating x x ), x x ), etc with respect to z] In the limit of a thin radial fringe a z and k r z = ksz a ks, the transfer matrix becomes [] [] I Ks M fringe =, where K K s I s = 59) 63 Symplecticity of solenoid maps k s Eqs55), 58) are not symplectic, but the solenoid matrix M sol = M fringe M long M See Eq??), Sec?? fringe is symplectic Solenoid lens M sol can also be written as a combination of rotations of an angle θ = k s l/4 and a thick lens that focuses in both planes with focusing strength k = k s /), where k s = e pc B z and l is the solenoid length, M sol = R k s 4 l ) F k s 4, l R k s 4 l ) 6) [] Kf k, l) F = K f k, l) K f is as in Eq48) 7 Superimposed solenoid and quadrupole fields The matrices M QL and M QR are given for superimposed quadrupole and longitudinal fields and for superimposed quadrupole and radial fields The elements of M QL are x x ) = fk q g + θ cos θ + z g θ + cosh θ z) x x ) = fk q g + θ sin θ + z + g θ + sinh θ z) x y ) = k qk s f k q θ sin θ + z + θ + sinh θ z) x y ) = k s f cos θ + z cosh θ z) y x ) = x y ) y x ) = x y ) y y ) = f g cos θ + z g + cosh θ z) y y ) = fk q g θ + sin θ + z + g + θ sinh θ z) 6) The focusing strength of the quadrupole is k q, k s = e pc B z, f = ks 4 + 4kq, θ ± = k s ± f), and g ± = k q k s ± f) Elements x x ), x x ), etc are obtained by differentiating x x ), x x ), etc with respect to z 3

32 If k q > k r where k r = k s /a)), elements of M QR are If k r > k q, then x x ) = x x ) = x y ) = x y ) = h g+ cos φz g cosh φz) φ h g + sin φz g sinh φz) k r h cosh φz cos φz) sinh φz sin φz) k rφ h y x ) = x y ) y x ) = x y ) y y ) = y y ) = h g+ cosh φz g cos φz) φ h g + sinh φz g sin φz) 6) x x ) = h φ cos ζ cosh ζ + k q sin ζ sinh ζ) x x ) = α hφ g sin ζ cosh ζ + g + sinh ζ cos ζ) k x y ) = r h sin ζ sinh ζ) x y ) k = rα hφ cos ζ sinh ζ sin ζ cosh ζ) y x ) = x y ) y x ) = x y ) y y ) = h φ cos ζ cosh ζ k q sin ζ sinh ζ) y y ) = α hφ g + cosh ζ sin ζ g sinh ζ cos ζ) 63) where φ = kq kr 4, h = φ, g ± = ±φ k q, α = φ/ and ζ = αz Again x x ), x x ), etc are obtained by differentiating x x ), x x ), etc with respect to z Matrices M QL and M QR are not symplectic, but the combination M thin QR M QLM thin QR ) is symplectic in the limit of zero length fringe M thin QR is M QR in the limit of zero length) See Eq59) 8 Dipole Sextupoles skew quad with strength k = eb pcr When there is a vertical closed orbit y at a sextupole, the sextupole behaves as a, where B and r are the field and the radius at the pole tip The sensitivity of the luminosity in e + e colliders to the details of the vertical orbit is related to this property 9 Cyclotron Equations of Motion In cylindrical coordinates F = m d r dt dr dt = d dt rˆr = ṙˆr + r ˆr = ṙˆr + rˆθ θ d r dt = rˆr + ṙ ˆr + ṙˆθ θ + r ˆθ θ + rˆθ θ = rˆr + ṙˆθ θ + ṙˆθ θ rˆr θ + rˆθ θ = rˆr + ṙˆθ θ rˆr θ + rˆθ θ 3

33 where we used ˆr = θˆθ and ˆθ = ˆr θ The Lorentz force is F = q v B + E) In a cyclotron with uniform vertical magnetic field and electrostatic focusing F r = qv s B + q E x r r r ) where r = qb mv s, and v s is the velocity in the azimuthal direction The radial force Let r = r + x and Equation 64 becomes F r = qv s B + q E r r r r ) = m r r θ ) 64) where ω = qvsb mr and The vertical force ẍ r θ = m qvb + q E r r x) ẍ = m qv sb + q E r r x) + r vs r v s ẍ = m qv sb + q E r r x) + x + r ẍ v s + q E r r m r x + v s x ) r r ẍ v s r ωx = ω n) q m Q x = ω x ω = n n = r v s E r r )x ) q E r m r = r E r v s B r ) F z = q E z z z = m z z q E z m z z = ωz = q E z m z z Since E =, Ez z = Er r and ωz = q E r m r vs = n = ω n r Q z = ω z ω = n 33