PProceedings of the 2017 CERN–Accelerator–School course on
Vacuum for Particle Accelerators, Glumslöv, (Sweden)
Impedances and Instabilities
R. Wanzenberg
Deutsches Elektronen-Synchrotron DESY, D-22603 Hamburg, Germany
Abstract
The concepts of wake fields and impedance are introduced to describe the elec-tromagnetic interaction of a bunch of charged particles with its environment inan particle accelerator. The wake fields can act back on the beam and leadto instabilities, which may limit the achievable current per bunch, the totalcurrent, or even both. Some typical examples are used to illustrate the wakefunction and its basic properties. Then the frequency-domain view of the wakefield or impedance is explained, and basic properties of the impedance are de-rived. The impedance of a cavity mode is illustrated using an equivalent cir-cuit model. The relation of wake field effects to important beam parametersis treated in the rigid beam approximation. Several examples are employed toillustrate the impact of the geometry and the material properties of the vacuumchamber on the impedance. Finally, a basic introduction to beam instabilitiesbased on a head tail model of the beam is given.
Keywords
Wake field; impedance; instabilities. Introduction
A beam in an accelerator interacts with its vacuum chamber surroundings via electromagnetic fields. Inthis lecture the concept of wake fields is introduced to describe the electromagnetic interaction of a bunchof charged particles with its environment. The various components of the environment are the vacuumchamber, cavities, bellows, dielectric-coated pipes, and other kinds of obstacles the beam has to pass onits way through the accelerator. The wake fields can act back on the beam and lead to instabilities, whichmay limit the achievable current per bunch, the total current, or even both.This lecture builds upon a previous lecture on wake fields and impedance given by T. Weiland [1]and M. Dohlus and the author [2]. Furthermore, there are excellent textbooks available [3–6] whichcover the subject matter of this lecture.In the introduction several typical examples are presented which demonstrate the interaction of abeam with the surrounding vacuum chamber via wake fields.Then, in Section 2, the concept of wake potential is formally introduced and multipole expansionsare studied for structures with cylindrical symmetry. The Panofsky–Wenzel theorem, which links thelongitudinal and transverse wake forces, is explained.Section 3 is devoted to the analysis of wake fields due to resonant modes in a cavity. The couplingof the beam to one mode of a cavity leads to the concept of the loss parameter.In Section 4, the impedance is introduced as the Fourier transform of the wake potential. The prop-erties of the wake functions (time-domain view) are translated to properties of the impedance (frequency-domain view).
Consider a point charge q moving in free space at a velocity v close to the speed of light. The electromag-netic field is Lorentz-contracted into a thin disk perpendicular to the particle’s direction of motion [7],Available online at https://cas.web.cern.ch/previous-schools a r X i v : . [ phy s i c s . acc - ph ] J un hich we choose to be the z -axis in a cylindrical coordinate system. The opening angle of the fielddistribution is of the order of /γ , where γ = (1 − ( v/c ) ) − / . The field distribution is shown in Fig. 1.Even for an electron beam with an energy of E = 10 MeV , the opening angle φ is no greater than
50 mrad or . ◦ : φ = 1 γ = 0 .
511 MeV E = 2 . ◦ ( m c = 0 .
511 MeV is the rest mass of the electron). φ Fig. 1:
Electromagnetic field carried by a relativistic point charge q In the ultra-relativistic limit v → c (or γ → ∞ ), the disk containing the field shrinks to a δ -functiondistribution. The non-vanishing field components are E r = q π ε r δ ( z − c t ) , H ϕ = E r Z with Z = 377 Ω . Since the electric field E points strictly radially outward from the point charge, all field compo-nents are identically zero both ahead of and behind the point charge, and hence there are no forces on atest particle either preceding or following the charge q .For v slightly less than c , this is not strictly true. However, if we look at some typical bunchcharges and energies of high-energy accelerators and synchrotron light sources, as shown in Table 1, wewill notice that the space charge force V s = e/ (4 π (cid:15) d γ ) (where d is the rms distance between twoelectrons in the bunch) scales with /γ . It is then obvious that as a good approximation, any spacecharge effects can be neglected for the accelerators under consideration. Nevertheless, space chargeeffects are important in heavy ion or low-energy proton accelerators. Table 1:
Typical bunch charges and energies of high-energy accelerators and synchrotron light sources [8–10]
Machine Charge (nC) Energy (GeV) γ = (1 − ( v/c ) ) − / LHC 20 7000 7 500LEP 100 60 195 700PETRA III 20 6 11 700In the next section we will restrict ourselves to the ultra-relativistic case ( γ = ∞ , v = c ), so spacecharge effects are neglected. Consider some typical settings where electromagnetic fields occur behind a bunch with charge q movingwith velocity c through a structure. A bunch moves through a cylindrical pipe along the z -axis. All2lectric field lines terminate transversely on surface charges on the wall of the pipe, assuming a perfectlyconducting wall. There will be no wake fields behind the charge. The situation is different, however,if the cross-section of the beam pipe changes. A step-out transition is shown in Fig. 2. All fields havebeen calculated using a numerical wake field solver from MAFIA or the CST studio suite [11–14]. Here Fig. 2:
Wake fields behind a bunch generated at a step-out transition from a small to a larger beam pipe we have assumed that all pipe walls are perfect conductors. The wake field is generated because ofthe change in geometry. It should be noted that any beam pipe with finite conductivity, as well as flatbeam pipes, can generate wake fields (resistive wall wake fields) [15]. Furthermore, a dielectric-coatedpipe, which could be used as a travelling-wave acceleration section, will generate wake fields; see, forexample, [16].Another example is a cavity in a beam pipe; see Fig. 3. Again, a bunch is moving through acylindrical pipe along the z -axis. Wake fields are generated because of geometrical changes in the pipecross-section. In this respect the situation is similar to the previously considered case of a step-outtransition. The main difference is that the bunch can excite modes in the cavity and therefore long-rangewake fields, which can ring for a long time in the cavity (depending on the conductivity of the cavitywall). Fig. 3:
Wake fields in a cavity
The examples have demonstrates that wake forces are caused by geometrical changes along thebeam pipe. Space charge effects are negligible for ultra-relativistic particles. Wake fields due to the3esistive wall or dielectric coatings should always be checked in detail according to the specific situation.Furthermore, the surface roughness of the vacuum chamber can be important for special cases [17]. Wake fields
The examples above give us a qualitative understanding of wake fields and how they are generated.Before proceeding to mathematical descriptions in terms of wake potentials, let us take a closer look atthe example considered in Section 1.An ultra-relativistic point particle with charge q traverses a small cavity parallel to the z -axis,with offset ( x , y ) ; see Fig. 4. The electromagnetic force on any test charge q is given as a function of r (cid:54) z (cid:45) (cid:7)(cid:6) (cid:45) ϕ q (cid:117) v = c e z (cid:45)(cid:54) r q (cid:114) (cid:27) s Fig. 4:
An ultra-relativistic point particle with charge q traverses a small cavity parallel to the z -axis, followed bya test charge q space and time coordinates by the Lorentz equation F ( r , t ) = q (cid:0) E ( r , t ) + v × B ( r , t ) (cid:1) , where E and B are the fields generated by q ; they are solutions of the Maxwell equations ∇ × B = µ j + 1 c ∂∂t E , ∇ · B = 0 , ∇ × E = − ∂∂t B , ∇ · E = 1 (cid:15) ρ and have to satisfy several boundary conditions.In our case the charge and current distributions are ρ ( r , t ) = q δ ( x − x ) δ ( y − y ) δ ( z − ct ) , j ( r , t ) = c e z ρ ( r , t ) . After interaction of q with the cavity, there remain electromagnetic fields in the cavity. The sourceparticle has lost energy to cavity modes and excited fields that propagate in the semi-infinite beam pipes.Now consider a test charge q following q at a distance s with the same velocity v ≈ c and withoffset ( x , y ) . The Lorentz force is F ( x , y , x , y , s, t ) = q (cid:0) E ( x , y , z = ct − s, t ) + c e z × B ( x , y , z = ct − s, t ) (cid:1) . The change in momentum of the test charge can be calculated as the time-integrated Lorentz force, ∆ p ( x , y , x , y , s ) = (cid:90) F d t. E d and B d and the Lorentz Force F d of a distributed source ρ d ( r , t ) = η ( x − ¯ x , y − ¯ y ) λ ( z − ct ) can be calculated either by integration over all source points, F d (¯ x , ¯ y , x , y , s, t ) = (cid:90) F ( x , y , x , y , s, t + z /c ) η ( x − ¯ x , y − ¯ y ) λ ( z ) q d x d y d z , or by solving the electromagnetic problem for the distributed source; here λ is the line charge density, η is the transverse distribution normalized to 1, and ¯ x and ¯ y describe a transverse shift of the center ofthe distribution. A calculation of the electric fields of a distributed source is shown in Fig. 5.A fundamental difference between fields of point particles (with time dependency δ ( z − ct ) ) andfields of distributed sources (with time dependency λ ( z − ct ) ) is that the frequency spectrum of point par-ticles is not limited. In particular, long Gaussian bunches may stimulate only a few resonances (modes)in a cavity structure, or even none.We can distinguish between the long-range regime of the wake, where the interaction betweenparticles is driven by resonances, and the short-range regime, where the superposition of time-harmoniccavity fields is not sufficient to describe the effects. For instance, the fields in Fig. 2 are not determinedby oscillations, while the fields in Fig. 5 will ring harmonically on one or several frequencies after the(source) bunch has left the domain. (cid:54) (cid:45) zt (cid:27) s Fig. 5:
Wake fields generated by a Gaussian bunch traversing a cavity .2 Basic definitions Consider a point charge q traversing a structure with offset ( x , y ) parallel to the z -axis at the speed oflight (see Fig. 4). Then the wake function is defined as w ( x , y , x , y , s ) = 1 q (cid:90) ∞−∞ d z (cid:2) E ( x , y , z, t ) + c e z × B ( x , y , z, t ) (cid:3) t =( s + z ) /c . The distance s is measured from the source q in the opposite direction to v . The change in momentumof a test particle with charge q following behind at a distance s with offset ( x , y ) is given by ∆ p = 1 c q q w ( s ) . Since e z · ( e z × B ) = 0 , the longitudinal component of the wake function is simply w (cid:107) ( x , y , x , y , s ) = 1 q (cid:90) ∞−∞ d z E z ( x , y , z, ( s + z ) /c ) . Figure 6 shows the longitudinal component of the wake potential for the above example with thesmall cavity. The grey line represents the Gaussian charge distribution in the range from − σ to σ .Owing to transient wake field effects, the head of the bunch (left-hand side of the figure) is decelerated,while a test charge at a certain position behind the bunch will be accelerated. -1-0.5 0 0.5 1 0 2 4 6 8 10 12 14 W a k e / V / p C s / σ z Longitudinal WakepotentialWakeBunch
Fig. 6:
Longitudinal wake potential
The notion of wakes, as presented above, is restricted to sources and test particles that travel at thevelocity of light through a structure with semi-infinite input and output beam pipes. Therefore, for theintegrals to converge, it is necessary that there be no length-independent forces in the pure beam pipes.This is the case for v → c and perfect conductivity of the pipes. The concept of a wake per length, w (cid:48) ( x , y , x , y , s ) = 1 q (cid:2) E p ( x , y , − s,
0) + v e z × B p ( x , y , − s, (cid:3) ,
6s used to describe the effect in beam pipes ‘p’ of finite conductivity and/or velocity v ≤ c . Suppose thatthe input and output beam pipes have the same cross-section; then a generalized wake function w s ( x , y , x , y , s ) = 1 q (cid:90) ∞−∞ d z (cid:2) E s ( x , y , z, t ) + v e z × B s ( x , y , z, t ) (cid:3) t =( s + z ) /v can be defined for the scattered fields E s = E − E p and B s = B − B p . If the conditions for the wakefunction are fulfilled (i.e. convergence of the integral), then the wake function equals the generalizedwake function.The wake potential is defined similarly to the wake function, but for a distributed source: W (¯ x , ¯ y , x , y , s ) = 1 q (cid:90) ∞−∞ d z (cid:2) E d ( x , y , z, t ) + c e z × B d ( x , y , z, t ) (cid:3) t =( s + z ) /c = 1 q q (cid:90) ∞−∞ d z (cid:2) F d (¯ x , ¯ y , x , y , z, t ) (cid:3) t =( s + z ) /c . It can be calculated from the wake function by the convolution W d (¯ x , ¯ y , x , y , s ) = (cid:90) w ( x , y , x , y , s + z ) η ( x − ¯ x , y − ¯ y ) λ ( z ) q d x d y d z . Note that the s -coordinate measures in the negative z -direction while λ depends on the positive longitudi-nal coordinate. Usually numerical codes for computing wakes, such as ECHO, calculate electromagneticfields for distributed sources and therefore wake potentials. We follow the arguments of A. Chao [3, 18] to introduce the Panofsky–Wenzel theorem [19]. Thereforewe use the following different notation for the generalized wake function: w p ( x , y , x , y , s ) = w p ( x , y , r ) , with the observer vector r = x e x + y e y − s e z . We calculate curl w p with respect to the observer orthe test particle: ∇ × w p ( x , y , r ) = ∇ × vq (cid:90) ∞−∞ d t (cid:2) E s ( r + v t, t ) + v × B s ( r + v t, t ) (cid:3) . Using curl E = − ∂ B /∂t gives ∇ × w p ( x , y , r ) = vq (cid:90) ∞−∞ d t (cid:20) − ∂∂t B s ( . . . , t ) + v ( ∇ B s ( . . . , t )) − B s ( . . . , t )( ∇ v ) (cid:21) = vq (cid:90) ∞−∞ d t (cid:20) − ∂∂t − v ∂∂z (cid:21) B s ( r + v t, t )= vq (cid:90) ∞−∞ d t (cid:20) − dd t B s ( r + v t, t ) (cid:21) = − vq B s ( r + v t, t ) (cid:12)(cid:12)(cid:12) t = ∞ t = −∞ . As the scattered field is zero for negative infinite time and vanishes for positive infinite time and infi-nite distance from the scattering object, the wake is curl-free. The Panofsky–Wenzel theorem is thenreformulated in our original notation as the set of equations ∂∂s w p x = − ∂∂x w p (cid:107) , ∂s w p y = − ∂∂y w p (cid:107) ,∂∂x w p y = ∂∂y w p x . Note that the Panofsky–Wenzel theorem holds for the generalized wake function ( v ≤ c ) and for thewake function ( v = c ).Integration of the transverse gradient of the longitudinal wake function yields the transverse wakepotential w ⊥ ( x , y , x , y , s ) = − ∇ ⊥ (cid:90) s −∞ d s (cid:48) w (cid:107) ( x , y , x , y , s (cid:48) ) . The Panofsky–Wenzel theorem is applicable if the input and output beam pipes have the same cross-section.
Now we calculate div w with respect to the observer. First, note that ∇ · w ( x , y , r ) = ∇ · cq (cid:90) ∞−∞ d t (cid:2) E ( r + c t, t ) + c × B ( r + c t, t ) (cid:3) . Using Maxwell’s equations, div E = ρ/ε and curl B = µ J + c − ∂ E /∂t , together with J = c ρ gives ∇ · w ( x , y , r ) = cq (cid:90) ∞−∞ d t (cid:2) ∇ · E + c ( ∇ × B ) (cid:3) = 1 q (cid:90) ∞−∞ d t (cid:20) − ∂∂t E z ( r + c t, t ) (cid:21) = − q (cid:90) ∞−∞ d z (cid:20) ∂∂s E z ( r + z e z , ( z + s ) /c ) (cid:21) = − ∂∂s w (cid:107) ( x , y , x , y , s ) . The term ∂w (cid:107) /∂s appears on both sides of the equation, so we can write ∂w x ∂x + ∂w y ∂y = 0 . With the Panofsky–Wenzel equations we find that the longitudinal wake is a harmonic function withrespect to the transverse coordinates of the test particle: (cid:18) ∂ ∂x + ∂ ∂y (cid:19) w (cid:107) = − ∂∂s (cid:18) ∂w x ∂x + ∂w y ∂y (cid:19) = 0 . It is required that the trajectories ( x , y , ct ) and ( x , y , ct ) do not intersect with the boundary. The longitudinal wake is also a harmonic function with respect to the transverse coordinates of thesource particle [20], i.e. L w (cid:107) = 0 with L = ∂ /∂x + ∂ /∂y . To prove this, we have to calculate ˜ E z = L E z , which is equivalent to the solution of the field problem for the source ˜ ρ = L ρ . Thesource ρ is the point particle q travelling at the speed of light along ( x , y , z = ct ) . It gives rise to theelectromagnetic fields E f = q δ ( z − ct )2 πε ( x − x ) e x + ( y − y ) e y ( x − x ) + ( y − y ) , f = c − e z × E f in free space. The fields ˜ E = L E f and ˜ B = L B f are caused by the source ˜ ρ = L ρ . These fields arezero for all points with ( x, y ) (cid:54) = ( x , y ) , as (cid:18) ∂ ∂x + ∂ ∂y (cid:19) ( x − x ) e x + ( y − y ) e y ( x − x ) + ( y − y ) = . Obviously ˜ E and ˜ B satisfy any linear boundary condition for any geometry, provided that the boundarydoes not intersect the trajectory ( x , y , z = ct ) . Therefore these fields are also solutions to the boundedwake problem, and all components of w are harmonic with respect to ( x , y ) , since (cid:18) ∂ ∂x + ∂ ∂y (cid:19) w = cq (cid:90) ∞−∞ d t (cid:2) ˜ E + c × ˜ B (cid:3) = . This information will help us to evaluate the r -dependence of the wake function in cylindricalsymmetric structures in the next subsection, and it will enable us to efficiently calculate the wake functionin fully 3D structures. r (cid:54) z (cid:45) (cid:7)(cid:6) (cid:45) ϕ (cid:54) a (cid:24)(cid:25)(cid:24)(cid:25)(cid:24)(cid:25)(cid:24)(cid:25)(cid:27)(cid:26)(cid:27)(cid:26)(cid:27)(cid:26)(cid:27)(cid:26) (cid:27)(cid:26)(cid:27)(cid:26)(cid:27)(cid:26)(cid:27)(cid:26)(cid:24)(cid:25)(cid:24)(cid:25)(cid:24)(cid:25)(cid:24)(cid:25) q (cid:117) v = c e z (cid:45)(cid:54) r q (cid:117) (cid:54) r (cid:27) s Fig. 7:
A bunch with total charge q traversing a cavity with offset r , followed by a test charge q with offset r Consider now a cylindrically symmetric acceleration cavity with side tubes of radius a (see Fig. 7).The particular shape in the region r > a is of no importance for the following investigations. Two chargespass through the structure from left to right with the speed of light: q at a radius of r and q at a radiusof r . We wish to find an expression for the net change in momentum, ∆ p ( r , ϕ , r , ϕ , s ) , experiencedby q due to the wake fields generated by q . In the following we write the wake function and potentialin polar coordinates. Let us start with the case of ϕ = 0 : ∆ p z ( r , , r , ϕ , s ) = q q w (cid:107) ( r , , r , ϕ , s ) . The wake function can be expanded in a multipole series w (cid:107) ( r , , r , ϕ , s ) = Re (cid:40) ∞ (cid:88) m = −∞ exp(i m ϕ ) G m ( r , r , s ) (cid:41) . w (cid:107) is a harmonic function in ( r , ϕ ) , we have L w (cid:107) ( r , , r , ϕ , s ) = (cid:18) r ∂∂r r ∂∂r + 1 r ∂ ∂ϕ (cid:19) w (cid:107) ( r , , r , ϕ , s )= Re (cid:40) ∞ (cid:88) m = −∞ exp(i m ϕ ) (cid:18) r ∂∂r r ∂∂r − m r (cid:19) G m ( r , r , s ) (cid:41) = 0 , where L is the transverse Laplace operator with respect to the offset of the test particle. So, for all m ,the expansion functions G m ( r , r , s ) have to satisfy the Poisson equation r ∂∂r (cid:18) r ∂∂r G m ( r , r , s ) (cid:19) − m r G m ( r , r , s ) = 0 . The solutions are G ( r , r , s ) = U ( r , s ) + V ( r , s ) ln r ,G m ( r , r , s ) = U m ( r , s ) r m + V m ( r , s ) r − m for m > . Keeping only the solutions which are regular at the origin ( r = 0 ), the longitudinal wake potentialcan be written as w (cid:107) ( r , , r , ϕ , s ) = ∞ (cid:88) m =0 r m U m ( r , s ) cos mϕ , with expansion functions U m ( r , s ) that depend on the details of the given cavity geometry.By azimuthal symmetry, the dependence on ϕ is w (cid:107) ( r , ϕ , r , ϕ , s ) = w (cid:107) ( r , , r , ϕ − ϕ , s ) ,as longitudinal fields depend only on the relative azimuthal angle of the observer with respect to thesource. Using the fact that w (cid:107) is also a harmonic function in ( r , ϕ ) , we find with the same argumentsas before that U m ( r , s ) can be factorized as r m w m ( s ) .It follows that for the general case of a charge q at ( r , ϕ ) generating fields that act on a secondcharge q at ( r , ϕ ) , the longitudinal wake function is given by w (cid:107) ( r , ϕ , r , ϕ , s ) = ∞ (cid:88) m =0 r m r m w m ( s ) cos m ( ϕ − ϕ ) . The transverse wake function is, by the Panofsky–Wenzel theorem, w ⊥ ( r , ϕ , r , ϕ , s ) = − (cid:18) e r ∂∂r + e ϕ r ∂∂ϕ (cid:19) (cid:90) s −∞ d s (cid:48) w (cid:107) ( r , ϕ , r , ϕ , s (cid:48) )= ∞ (cid:88) m =0 (cid:40) − e r m r m r m − (cid:90) s −∞ d s (cid:48) w m ( s (cid:48) ) cos m ( ϕ − ϕ )+ e ϕ m r m r m − (cid:90) s −∞ d s (cid:48) w m ( s (cid:48) ) sin m ( ϕ − ϕ ) (cid:41) . Each azimuthal order is fully characterized by a scalar function w m ( s ) . This function can becalculated by solving Maxwell’s equations for the given geometry and any choice of ( r , ϕ , r , ϕ ) ,yielding w m ( s ) = (cid:82) ∞−∞ d z E zm ( r , ϕ , z, ( z + s ) /c )) r m r m cos m ( ϕ − ϕ ) .
10 particular choice of r can be used to avoid the infinite integration range: since E z vanishes at themetallic tube boundary, only the cavity gap contributes to the integral. The integration range is reducedto the cavity gap by setting r to the radius of the beam tube. This trick is possible if no obstacle intersectswith the infinite cylindrical beam pipe.This type of wake integration is utilized by computer codes such as ECHO [21, 22] for bunches offinite length. Wake potentials can be calculated by such programs in the time domain, but wake functions(of point sources) need asymptotic considerations; see [23].It should be mentioned that in many practical cases, due to the ( r/a ) m dependence, the longitudi-nal wake is dominated by the monopole term and the transverse wakes by the dipole term: w (cid:107) ( r , ϕ , r , ϕ , s ) = w ( s ) , w ⊥ ( r , ϕ , r , ϕ , s ) = r (cid:90) s −∞ d s (cid:48) w ( s (cid:48) ) (cid:2) − e r cos( ϕ − ϕ ) + e ϕ sin( ϕ − ϕ ) (cid:3) . While for cylindrical symmetric structures the dependence of the wake on transverse coordinates isexplicitly known and can be used to reduce the integration range and domain of the field calculation,more general structures require us to use the harmonic property of the wake for a beam tube of arbitraryshape. The simple 3D cavity in Fig. 8, with a beam tube of square cross-section, is used to demonstratethis. We suppress the dependence of the wake function (or potential) on the offset of the source and writesimply ˜ W (cid:107) ( x, y, s ) = W (cid:107) ( x , y , x, y, s ) . This function is harmonic in the observer offset, ∇ ⊥ ˜ W (cid:107) ( x, y, s ) = 0 . Fig. 8:
A 3D cavity structure with two symmetry planes (top) and a quarter of the structure (bottom)
For points x and y on the surface of the beam tube, we can calculate the wake by a finite-rangeintegration through the cavity gap, as shown in Fig. 9. If we know ˜ W for all surface points, we cancalculate the wake for any point inside the tube by numerical solution of the boundary value problem.Therefore a 2D Poisson problem has to be solved. In our example, with two transverse symmetries, onlya quarter of the structure needs to be considered to calculate the wake of a source in the center.The transverse wake potential can be calculated from the longitudinal one using the Panofsky–Wenzel theorem. The transverse gradient of the longitudinal wake potential in a beam tube is alsoindicated in Fig. 9. 11 estbeamBeam Fig. 9:
Illustration of the indirect test beam method. The upper pictures show lines of constant longitudinal wakepotential and the gradient of the longitudinal wake potential; an integration gives the transverse wake potentialaccording to the Panofsky–Wenzel theorem. The lower diagram depicts the paths of the beam and the test beam. Cavities, resonant structures and eigenmodes
Many structures in an accelerator environment can be considered as a hollow space with semi-infinitebeam pipes on both sides. Usually this vacuum volume is bounded by metal surfaces with high con-ductivity. As a good approximation, the cavity walls can be treated as perfect electric conducting (PEC)boundaries, and sometimes the beam pipes are even neglected so that the volume is closed.Electromagnetic fields with frequencies below the lowest cutoff frequency of the beam pipes aretrapped in the volume, and the fields oscillate at discrete frequencies: E ( r , t ) = (cid:88) ν ˆ A ν ˆ E ν ( r ) cos(ˆ ω ν t + ˆ ϕ ν ) , B ( r , t ) = (cid:88) ν ˆ A ν ˆ B ν ( r ) sin(ˆ ω ν t + ˆ ϕ ν ) . These oscillations are called eigenmodes or cavity modes. They are characterized by their field patterns ˆ E ν ( r ) and ˆ B ν ( r ) and their eigenfrequencies ˆ ω ν . The modes may ring with any amplitude ˆ A ν and phase ˆ ϕ ν , and the amplitude normalization of the eigenfields is arbitrary. Such modes are called standing-wavemodes, as the electric and magnetic fields ring at all spatial points with the same phase, but the electricfield is phase-shifted by ◦ relative to the magnetic field. For simplicity, in the following we omit themode index ν but indicate all indexable (mode-specific) quantities with a hat. We will introduce furthermode-specific quantities, such as the quality ˆ Q , the modal longitudinal loss parameter ˆ k , and the modeenergy ˆ W = 12 (cid:90) ε ˆ E d V = 12 (cid:90) µ − ˆ B d V, which depends on the arbitrary amplitude normalization. The total electromagnetic field energy of allthe modes is W = 12 (cid:90) εE ( r, t ) d V + 12 (cid:90) µ − B ( r, t ) d V = (cid:88) | ˆ A | ˆ W . The field energy of a particular mode does not depend on the stimulation of other modes, as the mode fields are orthogonalto each other; see Appendix A.
Fig. 10:
Electric field of a mode in a rotationally symmetric cavity with beam pipes
In structures with symmetries (e.g. rotational symmetry), eigenmodes and beam-pipe modes of thesame symmetry condition are coupled. Therefore the lowest cutoff frequency for a particular symmetrydefines the highest possible eigenfrequency for the corresponding eigenmodes. For instance, monopolemodes may have resonance frequencies that are above the lowest dipole mode cutoff frequency, whichis lower than the lowest monopole mode cutoff frequency. Beyond that, there can exist quasi-trappedmodes above the lowest cutoff frequency that have very weak coupling to the pipes. The energy flow(per period) of such fields into the beam pipes may be comparable to the energy loss (per period) ofnon-trapped modes to non-perfectly conducting metallic boundaries.
We consider a cavity of length L and a bunch with charge q , offset ( x , y ) and velocity c , which entersthe cavity at time t = 0 . The electromagnetic fields after the charge has left the cavity, namely E ( r , t > L/c ) = (cid:88) Re (cid:8) ˆ A ˆ E ( r ) exp(i ˆ ωt ) (cid:9) + E r ( r , t ) , B ( r , t > L/c ) = (cid:88) Im (cid:8) ˆ A ˆ B ( r ) exp(i ˆ ωt ) (cid:9) + B r ( r , t ) , can be split into eigenfields and a residual part, E r or B r . The long-range interaction between bunchesor particles is essentially driven by the modal part, as the residual fields decay or are not stimulatedresonantly. The complex mode amplitudes are proportional to the source charge and depend on thesource offset. Hence they can be expressed as ˆ A = q ˆ f ( x , y ) .Suppose that a small test charge δq follows the source particle on the same trajectory at a distanceof s > . It induces the additional amplitude δ ˆ A = δq exp( − i ˆ ωs/c ) ˆ f ( x , y ) . Therefore the energy ofthe modes is increased by δ W modes = (cid:88)(cid:0) | ˆ A + δ ˆ A | − | ˆ A | (cid:1) ˆ W≈ (cid:88) (cid:8) ˆ Aδ ˆ A ∗ (cid:9) ˆ W≈ q δq (cid:88) | ˆ f ( x , y ) | Re (cid:8) exp(i ˆ ωs/c ) (cid:9) ˆ W . The relevant length is not exactly the length of the cavity, but rather the length with non-zero field of the modes. For openstructures, with beam pipes, this length is in principle infinite, but for practical considerations the field has decayed sufficientlyafter a pipe length of a few times the widest dimension of the cross-section.
13n the other hand, the test particle gains kinetic energy δ W k = (cid:90) ∞−∞ δq E z ( x , y , z − s, z/c ) d z = q δq (cid:88) Re (cid:26) ˆ f ( x , y ) (cid:90) ∞−∞ ˆ E z ( x , y , z ) exp(i ˆ ω ( z + s ) /c ) d z (cid:27) + · · · . The sum of the field energy and the kinetic energy is conserved, if terms with the same oscillationfrequency exp(i ˆ ωs/c ) cancel: | ˆ f ( x , y ) | ˆ W + ˆ f ( x , y ) (cid:90) ∞−∞ ˆ E z ( x , y , z ) exp(i ˆ ω ( z ) /c ) d z = 0 . This is satisfied with ˆ f ( x , y ) = − ˆ v ∗ ( x , y ) / (cid:112) ˆ W and the normalized mode voltages ˆ v ( x, y ) = 12 (cid:112) ˆ W (cid:90) ∞−∞ ˆ E z ( x, y, z ) exp(i ˆ ωz/c ) d z, which do not depend on the arbitrary normalization mode fields.The amplitude excited by the charge q is ˆ A = q ˆ f ( x , y ) = − q ˆ v ∗ ( x , y ) (cid:14)(cid:112) ˆ W , and the energy of all modes is W EM , modes = (cid:88) | ˆ A | ˆ W = q (cid:88) ˆ k, with the per-mode loss parameter ˆ k = | ˆ v ( x , y ) | = 14 ˆ W (cid:12)(cid:12)(cid:12)(cid:12)(cid:90) ∞−∞ ˆ E z ( x , y , z ) exp(i ˆ ω ( z ) /c ) d z (cid:12)(cid:12)(cid:12)(cid:12) . The excitation of mode amplitudes depends linearly on the source distribution: another particlewith charge q and offset ( x , y ) at a distance s gives rise to the additional amplitude ˆ A = − q ˆ v ∗ ( x , y ) exp( − i ˆ ωs/c ) (cid:14)(cid:112) ˆ W , with phase shift − ˆ ωs/c due to the time shift s/c . Therefore it is possible to calculate the mode excitationfor arbitrary charge distributions; for example, for a one-dimensional bunch with offset ( x , y ) and linecharge density λ ( z, t ) = λ ( z − ct ) , ˆ A = ˆ v ∗ ( x , y ) (cid:90) λ ( − s ) exp( − i ˆ ωs/c ) d s. In particular, a Gaussian bunch with charge q and rms length σ excites the amplitudes ˆ A = q ˆ v ∗ ( x , y ) exp( − (ˆ ωσ/c ) / .We introduce the shape-dependent per-mode loss parameter ˆ k σ = ˆ k exp( − (ˆ ωσ/c ) ) . The electromagnetic field energy of all modes, after such a bunch has traversed the cavity, is W EM , modes ,σ = (cid:88) | ˆ A | ˆ W = q (cid:88) ˆ k σ . .3 Contribution of eigenmodes to the wake function After the source particle has traversed the cavity, the electromagnetic fields are E ( r , t > L/c ) = q (cid:88) Re (cid:110) − ˆ v ∗ ( x , y ) ˆ W − / ˆ E ( r ) exp(i ˆ ωt ) (cid:111) + E r ( r , t ) , B ( r , t > L/c ) = q (cid:88) Im (cid:110) − ˆ v ∗ ( x , y ) ˆ W − / ˆ B ( r ) exp(i ˆ ωt ) (cid:111) + B r ( r , t ) , Therefore the momentum of a test charge q at a distance s > L behind the source, with offset ( x , y ) and velocity c , is changed by ∆ p = q q c (cid:88) Re (cid:26) − ˆ v ∗ ( x , y ) ˆ W − / (cid:90) ∞−∞ d z (cid:104)(cid:0) ˆ E − i c × ˆ B (cid:1) exp(i ˆ ω ( z + s ) /c ) (cid:105)(cid:27) + ∆ p r , where the term ∆ p r stands for the contribution of the residual fields. Likewise, we can split the wakeinto modal and residual parts: w ( x , y , x , y , s > L ) = c ∆ p q q = (cid:88) ˆ w ( x , y , x , y , s ) + w r ( x , y , x , y , s ) , where ˆ w ( x , y , x , y , s > L ) = − (cid:8) ˆ v ∗ ( x , y )ˆ v ( x , y ) exp(i ˆ ωs/c ) (cid:9) , with the normalized vectorial voltages ˆ v ( x, y ) = 12 (cid:112) ˆ W (cid:90) ∞−∞ d z (cid:104)(cid:0) ˆ E ( x, y, z ) − i c × ˆ B ( x, y, z ) (cid:1) exp(i ˆ ωz/c ) (cid:105) . Loss parameters describe the loss of energy of a source particle or source distribution to electromagneticfield energy.We have seen that the total energy loss of a point particle is given by the wake function for x = x , y = y and s = 0 , so that k tot = − w ( x , y , x , y ,
0) = W EM , total /q , and we know that the loss to eigenmodes is given by the per-mode loss parameters ˆ k = | v ( x , y ) | . The sum of all the per-mode loss parameters converges for cavities with beam pipes to a value belowthe total loss parameter k tot , as not only are there modes excited, but also field energy is scattered andpropagates along the beam pipes. (The wake of a closed cavity is completely determined by oscillatingmodes, but the sum is divergent.)The wake potential (of distributed sources) and the shape-dependent loss parameter are usuallycalculated directly using electromagnetic time-domain solvers. The shape-dependent total loss parameteris the convolution of the longitudinal wake potential with the charge density function; for instance, forbunches with longitudinal profile λ ( z, t ) = λ ( z − ct ) and negligible transverse dimensions, k tot ,σ = − (cid:90) ∞−∞ W ( x , y , x , y , z ) λ ( z ) d z. ˆ k σ = ˆ k exp (cid:0) − (ˆ ωσ/c ) (cid:1) , and for a general longitudinal profile λ it is ˆ k σ = ˆ k (cid:12)(cid:12)(cid:12)(cid:12)(cid:90) ∞−∞ λ ( z ) cos(ˆ ωz/c ) d z (cid:12)(cid:12)(cid:12)(cid:12) . For closed cavities, the sum of the per-mode loss parameters ˆ k σ converges to the total loss param-eter k tot ,σ . This is also true for long bunches with σ (cid:29) c/ω cutoff , which cannot excite frequencies abovethe lowest cutoff frequency ω cutoff ∝ πc/a of the beam pipes, where a is the characteristic transversedimension of the pipes.For the extreme case of ultra-short bunches it is difficult to calculate the wake potential, as a veryhigh spatial resolution is required. In this case, only a small fraction of energy is lost to resonant modesand only a small part of the wakes is caused by resonances. Impedances
The Fourier transform of the negative wake function is called the impedance or coupling-impedance: Z (cid:107) ( x , y , x , y , ω ) = − c (cid:90) ∞−∞ d s w (cid:107) ( x , y , x , y , s ) exp( − i ωs/c ) . The wake function and impedance are two descriptions of the same thing, namely the couplingbetween the beam and its environment. The wake function is the time-domain description, while theimpedance is the frequency-domain description: w (cid:107) ( x , y , x , y , s ) = − π (cid:90) ∞−∞ d ω Z (cid:107) ( x , y , x , y , ω ) exp(i ωs/c ) . The reason for the usefulness of the impedance is that it often contains a number of sharply definedfrequencies corresponding to the modes of the cavity or the long-range part of the wake. Figure 11shows the real part of the impedance for a cavity. Below the cutoff frequency of the beam pipe, there isa sharp peak for each cavity mode. The spectrum above the cutoff frequency is continuous, caused byresidual fields (not related to eigenmodes) and by the ‘turn-on’ of the harmonic eigen-oscillations. Thecontinuous part of the spectrum is important for short-range wakes, especially for very short bunches.For the transverse impedance, it is often convenient to use a definition containing an extra factor i : Z ⊥ ( x , y , x , y , ω ) = i c (cid:90) ∞−∞ d s w ⊥ ( x , y , x , y , s ) exp( − i ωs/c ) . The reason is that the transverse–longitudinal relations due to the Panofsky–Wenzel theorem then readas follows in the frequency domain: ωc Z ⊥ ( x , y , x , y , ω ) = (cid:18) e x ∂∂x + e y ∂∂y (cid:19) Z (cid:107) ( x , y , x , y , ω ) . The sign is chosen so as to obtain a non-negative real part for x = x and y = y . ig. 11: Real part of the impedance for a cavity with side pipes; the peaks correspond to cavity modes. The resultswere obtained with the CST wakefield solver.
In the spatial s -domain, the relationship between the wake potential of a line charge density λ ( z − ct ) and the wake functions of a point particle is described by the convolution W ( x , y , x , y , s ) = (cid:90) ∞−∞ d z w ( x , y , x , y , s + z ) λ ( z ) . The corresponding equation in the frequency domain for the Fourier transform of the negative wakepotential is V ( x , y , x , y , ω ) = Z (cid:107) ( x , y , x , y , ω ) I ( ω ) , where I ( ω ) = (cid:90) ∞−∞ i ( t ) exp( − i ωt ) d t = (cid:90) ∞−∞ cλ ( − ct ) exp( − i ωt ) d t is the beam current in the frequency domain. The energy loss of the bunch to electromagnetic fields, W loss = (cid:90) ∞−∞ W ( x , y , x , y , s ) λ ( − s ) d s = 12 π (cid:90) ∞−∞ V ( x , y , x , y , ω ) I ( ω ) ∗ d ω = 1 π (cid:90) ∞ Re { Z ( x , y , x , y , ω ) }| I ( ω ) | d ω, has to be non-negative for any bunch shape λ . Therefore the real part of the longitudinal impedancemust be non-negative for all offsets with x = x and y = y . The real part can be negative for, say, x = − x and y = − y , for a structure with azimuthal symmetry and frequency close to a dipoleresonance.As the wake potential is a real function, the real part of the impedance is an even function of thefrequency while the imaginary part is an odd function of it: Re { Z (cid:107) ( . . . , ω ) } = Re { Z (cid:107) ( . . . , − ω ) } , Im { Z (cid:107) ( . . . , ω ) } = − Im { Z (cid:107) ( . . . , − ω ) } . Hence, the wake function is given in terms of the impedance as w (cid:107) ( . . . , s ) = − π (cid:90) ∞−∞ d ω Z (cid:107) ( . . . , ω ) exp(i ωs/c ) − π (cid:90) ∞−∞ d ω (cid:0) Re { Z (cid:107) ( . . . , ω ) } cos( ωs/c ) − Im { Z (cid:107) ( . . . , ω ) } sin( ωs/c ) (cid:1) . Furthermore, the electromagnetic field ahead of the source particle is zero for v = c , as electromagneticwaves cannot overtake the source. Therefore, the wake function is causal , and the real and imaginaryparts of the impedance are dependent on each other. From w (cid:107) ( . . . , s <
0) = 0 it follows that for u = − s > , (cid:90) ∞−∞ d ω Re { Z (cid:107) ( . . . , ω ) } cos( ωu/c ) = − (cid:90) ∞−∞ d ω Im { Z (cid:107) ( . . . , ω ) } sin( ωu/c ) , so only the real (or imaginary) part of the impedance is really needed: w (cid:107) ( . . . , s >
0) = 1 π (cid:90) ∞−∞ d ω Re { Z (cid:107) ( . . . , ω ) } cos( ωs/c ) . The modal part of the wake function is ˆ w ( x , y , x , y , s ) = − (cid:8) ˆ v ∗ ( x , y )ˆ v ( x , y ) exp(i ˆ ωs/c ) (cid:9) for s < , for s = 0 , otherwise . We are interested in the longitudinal component on the axis ( x = y = x = y = 0 ), and for simplicitywe omit the transverse coordinates, so ˆ w (cid:107) ( s ) = − k cos(ˆ ωs/c ) for s < , for s = 0 , otherwise . Then the impedance per mode, ˆ Z (cid:107) ( ω ) = − c (cid:90) ∞−∞ ˆ w (cid:107) ( s ) exp( − i ωs/c ) d s, is calculated as ˆ Z (cid:107) ( ω ) = 2ˆ k (cid:26) πδ ( ω + ˆ ω ) + πδ ( ω − ˆ ω ) + i ω ˆ ω − ω (cid:27) . This is equivalent to the impedance of a parallel resonant circuit (see Fig. 12), ˆ Z (cid:107) ( ω ) = lim ˆ R →∞ (cid:18) i ω ˆ C + 1i ω ˆ L + 1ˆ R (cid:19) − with ˆ C = 1 / (2ˆ k ) , ˆ L = 2ˆ k/ ˆ ω and ˆ R → ∞ .Although the resistor ˆ R was introduced for obvious formal reasons, it is helpful to consider weaklosses of a resonator with a high quality factor ˆ Q = ˆ R/ (ˆ ω ˆ L ) . The impedance per mode of a resonatorwith weak losses is ˆ Z (cid:107) ( ω ) = 2ˆ k i ω ˆ ω − ω + i ω ˆ ω/ ˆ Q .
The resistor R = ˆ Z (cid:107) (ˆ ω ) = 2ˆ k ˆ Q/ ˆ ω is called the shunt impedance . The quality factor ˆ Q describes thedecay time ˆ τ = 2 ˆ Q/ ˆ ω, C R
Fig. 12:
Equivalent circuit model of the impedance of one mode the resonance bandwidth ∆ˆ ω = ω/ ˆ Q, with | ˆ Z (cid:107) (ˆ ω ) / ˆ Z (cid:107) (ˆ ω ± ∆ˆ ω/ | ≈ , and the energy loss per unit time ˆ P = ˆ ω ˆ W / ˆ Q. The last relation is used to determine the quality factor by perturbation theory: the energy loss (withoutbeam) is caused by wall losses; as a good approximation these can be calculated from the fields obtainedfor the mode with infinite conductivity. If ˆ H t is the magnetic field tangential to the surface, the totalpower dissipated into the wall is given by a surface integral ˆ P = 12 (cid:90) ∂V Re (cid:8) ˆ E s × ˆ H ∗ (cid:9) · d A = 12 (cid:90) ∂V (cid:114) ˆ ωµ κ | ˆ H | d A where ˆ E s = Z s n × ˆ H is the tangential component of the electric field on the surface, with surfaceimpedance Z s = (cid:112) i ˆ ωµ/κ for conductivity κ . Instabilities
In the previous section the properties of the wake potential and the related impedance has been discussedin detail. Now the effect on the beam motion will be studied. The kick on a test charge due to a transversedipole wake is (cid:126)θ ( s ) = eE q W (1) ⊥ ( s ) (cid:126)r, where E is the beam energy, q the charge of the bunch with transverse offset (cid:126)r .In the rigid bunch approximation the transverse equation of a bunch in a storage can be written as: d ds y ( s ) + (cid:16) ω β c (cid:17) y ( s ) = 0 , were s is the longitudinal position in the storage ring, y is a transverse coordinate of the bunch and ω β the betatron frequency.In order to include wakefield effects one has to modify the above equation [3]: d ds y ( s ) + (cid:16) ω β c (cid:17) y ( s ) = e qmc γ ∞ (cid:88) n =0 C y ( s − n C ) W (1) ⊥ ( n C ) , where C is the circumference of the storage ring and γ the relativistic γ -factor. The right hand side ofthe equation is a sum of all the wakefield kicks during each turn in the storage ring.19n ansatz for the solution of the above equation is y ( s ) ∼ exp( − i Ω s/c ) , with the complex frequency Ω : Ω = ω β + i τ , including the betatron frequency ω β and the growth or damping rate /τ .The ansatz leads to a relation for the complex frequency Ω : Ω − ω β = − i e qmc γ cT ∞ (cid:88) p = −∞ Z (1) ⊥ (Ω + pω ) . The transformation from the time domain picture to the frequency picture is based on the relation: ∞ (cid:88) n =0 exp( i n Ω T ) W (1) ⊥ ( n C ) = iT ∞ (cid:88) p = −∞ Z (1) ⊥ (Ω + pω ) , which allows to replace the wake potential with the impedance Z (1) ⊥ . T is the revolution time in thestorage ring.If one further assumes that Ω does not deviate much from ω β ( Ω + ω β ≈ ω β ) one obtains anrelation for the betatron tune shift and the growth rate with the transverse impedance: Ω − ω β = − i ω β e qmc γ cT ∞ (cid:88) p = −∞ Z (1) ⊥ ( ω β + pω ) . The imaginary part is contributing to a mode frequency shift ∆Ω = Re(Ω − ω β , while the real part ofthe impedance corresponds to a growth rate /τ = Im(Ω − ω β ) : τ − = − ω β e qmc γ cT ∞ (cid:88) p = −∞ Re( Z (1) ⊥ )( ω β + pω ) . This demonstrates how the impedance can be use to calculate instability growth rates for a simplifiedmodel of transverse rigid bunch motion. In the next section a two macro particle is used to get a basicunderstanding of a head tail instability in an storage ring.
In the previous section the bunch was modelled just as a rigid bunch without an internal structure. Nowit is assumed that the bunch consists of two parts, a head and a tail as show in Fig. 13. This two particlemodel can be used to included synchrotron oscillations.
Storage ring
Circumference: C
Bunch
Fig. 13:
Headtail model of a bunch in a storage ring. θ tail = eE q W ⊥ y head where E is the beam energy, q the charge of the bunch with transverse offset y head of the head. W ⊥ isthe effective wake of the head of the bunch acting on the tail of the bunch.The equation of motion for the two particles are: y (cid:48)(cid:48) + (cid:16) ω β c (cid:17) y = 0 for the head particle and y (cid:48)(cid:48) + (cid:16) ω β c (cid:17) y = N r γ C W ⊥ y for the tail particle, where r = 14 π (cid:15) e m c = 2 . · − m is the classical radius of the electron and N is the bunch population. N/ electrons are in the head. Thepositions of the head and the tail are exchanged after one synchrotron oscillation period as illustrated inFig. 14. Therefor the above equation of motion are only valid for one half of a period of synchrotronoscillations. For the second half of the synchrotron oscillation period the indices’s have to interchanged. Fig. 14:
Synchrotron oscillation of the head and tail macro particles in the bunch.
The equation of motions can also be rewritten in a complex phasor notion for both particles withthe index 1 and 2 [3]: ˜ y , = y , + i cω β y (cid:48) , For the first half of the synchrotron oscillation period the following relations holds: (cid:18) ˜ y ˜ y (cid:19) s = c T s / = exp( − i ω β T s / (cid:18) i Υ 1 (cid:19) (cid:18) ˜ y ˜ y (cid:19) s =0 , where the wakefield effect is now presented with the parameter Υ , defined as: Υ = π N r c γ Cω β ω s W ⊥ . For one complete synchrotron period one obtains now the relation (cid:18) ˜ y ˜ y (cid:19) s = c T s = exp( − i ω β T s ) (cid:18) i Υ0 1 (cid:19) (cid:18) i Υ 1 (cid:19) (cid:18) ˜ y ˜ y (cid:19) s =0 . Υ : Υ < . For a known effective wakefield W ⊥ this can be directly translated into an limit for the bunch populationwhich will be stable with respect to head tail instabilities: N < γ Cω β ω s π N r c W ⊥ . Finally, the effect of ion trapping will be discussed, which can result in an increased beam emittance,betatron tune shifts and reduced beam lifetime. The effect of the ion cloud on the beam can be modelledas a broad band resonator wake field [25]. One can apply the classical theory of instabilities to obtainstability criteria with respect to ion effects. Nevertheless, it is interesting to have a simple criteria athand to know whether one can expect trapped ions in a beam. This was already analyzed by Kohaupt in1971 [26] for the storage ring DORIS at DESY.The residual gas density d gas at room temperature (300 K) can be calculated from vacuum pres-sure: d gas = p gas N Avo R gas
300 K = 24 . · cm − , where the numbers have been calculated for p gas = 1 · − mbar . R gas = 8 . / (K mol) is thegeneral gas constant and N A = 6 . · is the Avogadro number.Assuming now a typical cross section of 2 Mbarn ( · − cm ) for the ionization process, oneobtains an ion density of λ ion = d gas σ ion N ≈ d gas N , where N is the bunch population. After the passage of one bunch with a bunch population of · theion density is already . ions/cm or ions/cm after the passage of 960 bunches, which is one of thefilling modes of the synchrotron radiation facility PETRA III at DESY.The bunch train in a storage ring is acting as sequence of quadrupole lenses on the ion beam. In alinear approximation the interaction of the bunch with the ion can be can be presented by a matrix: M = (cid:18) L b (cid:19) (cid:18) − a (cid:19) , where L b = c ∆ t is the bunch spacing, and a = N b r p σ y ( σ x + σ y ) 1 A is the linear force of the beam on the ion. N b is the bunch population, r p = 1 . · − m, σ x , σ y are the transverse rms beam dimensions and A is the mass number of the ion. For CO and N the massnumber is A = 28 , while for water A = 18 . The ion motion will be only stable if the trace of the matrix M is smaller than 2, or T r ( M ) = 2 − a L b < . The ion will be trapped, when the mass number A islarger than a critical mass number A c : A > A c = N b L b r p σ y ( σ x + σ y ) . ppendix A: Eigenmodes of a closed cavity We consider a (simply connected) cavity volume V c , with perfectly conducting walls (boundary ∂V c ) andwithout current density. We search for time-harmonic eigensolutions, which can be written as E ( r , t ) = ˆ E ( r ) cos(ˆ ωt ) , B ( r , t ) = ˆ B ( r ) sin(ˆ ωt ) , where ˆ E and ˆ B are the eigenfields and ˆ ω the (angular) eigenfrequencies. Substituting these into Maxwell’sequations gives ∇ ε ˆ E = ˆ ρ, ∇ × ˆ E = − ˆ ω ˆ B , ∇ ˆ B = 0 , ∇ × µ − ˆ B = − ˆ ωε ˆ E . We apply the operator ∇ × µ − to the first curl equation and use the second curl equation to eliminatethe magnetic flux density, thus obtaining the eigenproblem ε − ∇ × µ − ∇ × ˆ E = ˆ λ ˆ E , with the eigenvalues ˆ λ = ˆ ω and the boundary condition n × ˆ E = . The operator ε − ∇× µ − ∇× isself-adjoint with scalar product (cid:104) A , B (cid:105) = 12 (cid:90) V c ε A · B d V. Therefore the problem has an infinite number of discrete real eigenvalues and a complete orthogonalsystem of eigenvectors, (cid:104) ˆ E ξ , ˆ E τ (cid:105) = ˆ W ξ δ ξτ , where ˆ W ξ is the electromagnetic field energy of mode ξ . The eigenvalues ˆ λ are non-negative so that alleigenfrequencies ˆ ω are real. There are obviously two types of eigensolutions: ˆ ω = 0 , ˆ ω (cid:54) = 0 , ∇ ε ˆ E (cid:54)≡ , ∇ ε ˆ E = 0 , ∇ × ˆ E = 0 , ∇ × ˆ E = − ˆ ω ˆ B , ˆ B ≡ , ˆ B (cid:54)≡ . Eigenfields for ˆ ω = 0 are curl-free and are just solutions to the electrostatic problem for any sourcedistribution ˆ ρ and the boundary condition n × ˆ E = 0 . Oscillating eigenfields are free of divergence; thisis a consequence of Maxwell’s second curl equation.In Appendix B we use the property that any linear combination of eigensolutions with ˆ ω = 0 isorthogonal to any linear combination of oscillating eigenfields. The property (cid:104) ε − ∇ × µ − ∇ × A , B (cid:105) = (cid:104) A , ε − ∇ × µ − ∇ × B (cid:105) can be shown with help of the identity ∇ [ A × µ − ∇ × B − B × µ − ∇ × A ] = B × ∇ × µ − ∇ × A − A × ∇ × µ − ∇ × B and the divergence theorem. The left-hand side gives asurface integral that is zero because of the boundary conditions. The right-hand side corresponds to the assertion. This property can be shown by using the identity ∇ [ ˆ E × µ − ∇ × ˆ E ] = µ − ( ∇ × ˆ E ) − ˆ E ∇ × µ − ∇ × ˆ E and thedivergence theorem. The left-hand side gives a surface integral that is zero because of the boundary conditions. The volumeintegral of the first term on the right-hand side is non-negative; the integral of the second term gives − λ ˆ W . As ˆ W is positive, ˆ λ cannot be negative. ppendix B: Wake of a closed cavity We consider a (simply connected) cavity volume V c of arbitrary shape, with perfectly conducting walls,that is located between the planes z = 0 and z = L . It is traversed by a point particle with charge q ,offset ( x , y ) and velocity v = c . The stimulating charge and current density are ρ ( r , t ) = q δ ( x − x ) δ ( y − y ) δ ( z − ct ) , j ( r , t ) = c e z ρ ( r , t ) . We use the complete orthogonal system of eigensolutions to describe the time-dependent electricfield: E ( r , t ) = (cid:88) ν ∈ C ˆ a ν ( t ) ˆ E ν ( r ) , where ν is the mode index, C is the set of all indexes and ˆ a ν ( t ) are the time-dependent coefficients. Asin the main text, we shall write all mode-specific quantities with a hat and omit the index ν . We solveMaxwell’s equations ∇ ε E = ρ, ∇ × E = − ∂∂t B , ∇ B = 0 , ∇ × µ − B = J + ε ∂∂t E by applying the operator ε − ∇ × µ − to the first curl equation and eliminating the magnetic flux densitywith the help of the second curl equation: ε − ∇ × µ − ∇ × E = − ε − ∂∂t J − ∂ ∂t E . By using the modal expansion and the eigenmode equation, we obtain (cid:88) ν ∈ C ˆ a ( t )ˆ ω ˆ E = − ε − ∂∂t J − ∂ ∂t (cid:88) ν ∈ C ˆ a ( t ) ˆ E . This set of scalar equations can be decoupled by applying the operator (cid:104) ˆ E ξ , · · · (cid:105) to both sides and usingthe orthogonality condition: ˆ a ξ ( t )ˆ ω ξ ˆ W ξ = − ε − ∂∂t (cid:104) ˆ E ξ , J (cid:105) − ∂ ∂t ˆ a ξ ( t ) ˆ W ξ . Finally, we substitute the Dirac current density and suppress the index, to arrive at (cid:18) ˆ ω + ∂ ∂t (cid:19) ˆ a ( t ) = − W ε ∂∂t (cid:104) ˆ E , J (cid:105) = − cq W ∂∂t ˆ E ( x , y , ct ) . This ordinary differential equation can be solved to give ˆ a ( t ) = − q (cid:112) ˆ W Re (cid:8) ˆ v ∗ ( x , y , ct ) exp(i ˆ ωt ) (cid:9) with v ( x, y, z ) = 12 (cid:112) ˆ W (cid:90) z −∞ ˆ E z ( x, y, s ) exp(i ˆ ωs/c ) d s The causal solution of ¨ a + ω a = ˙ b is a ( t ) = Re { (cid:82) t −∞ b ( τ ) exp(i ω ( t − τ )) d τ } . ∂∂z v ( x, y, z ) = 12 (cid:112) ˆ W ˆ E z ( x, y, z ) exp(i ˆ ωz/c ) . The longitudinal wake function is the sum over all modes, w (cid:107) ( x , y , x , y , s ) = (cid:88) ν ∈ C ˆ w (cid:107) ( x , y , x , y , s ) , with the ‘per-mode’ contributions ˆ w (cid:107) ( x , y , x , y , s ) = 1 q (cid:90) ∞−∞ ˆ a (cid:18) z + sc (cid:19) ˆ E z ( x , y , z ) d z = − (cid:112) ˆ W (cid:90) ∞−∞ Re (cid:26) ˆ v ∗ ( x , y , z + s ) exp (cid:18) i ˆ ω z + sc (cid:19)(cid:27) ˆ E z ( x , y , z ) d z = − (cid:26) exp(i ˆ ωs/c ) (cid:90) ∞−∞ ˆ v ∗ ( x , y , z + s ) ∂∂z v ( x , y , z ) d z (cid:27) . None of these terms is causal, i.e. ˆ w (cid:107) ( x , y , x , y , s < (cid:54)≡ , but the sum has to be! In the followingwe use causality to find the simplified representation of the longitudinal wake function w (cid:107) ( x , y , x , y , s >
0) = − (cid:88) ˆ ω (cid:54) =0 ˆ k ( x , y ) cos(ˆ ωs/c ) for x = x and y = y , where ˆ k ( x , y ) is the longitudinal per-mode loss parameter, as defined in themain text. We therefore split the summation over all modes into the components w (cid:107) d ( x , y , x , y , s ) = (cid:88) ˆ ω =0 ˆ w (cid:107) ( x , y , x , y , s ) ,w (cid:107) c ( x , y , x , y , s ) = (cid:88) ˆ ω (cid:54) =0 ˆ w (cid:107) ( x , y , x , y , s ) and use the causality relation w (cid:107) d ( x , y , x , y , s <
0) + w (cid:107) c ( x , y , x , y , s <
0) = 0 together with the anti-symmetry of the non-resonant part, w (cid:107) d ( x , y , x , y , s ) = − w (cid:107) d ( x , y , x , y , − s ) proved in (A) below, to eliminate w (cid:107) d , yielding w (cid:107) ( x , y , x , y , s >
0) = w (cid:107) c ( x , y , x , y , s ) + w (cid:107) c ( x , y , x , y , − s ) . To get the simplified representation for x = x and y = y , we have to show that the condition(B) ˆ w (cid:107) ( x , y , x , y , s ) + ˆ w (cid:107) ( x , y , x , y , − s ) = − k ( x , y ) cos(ˆ ωs/c ) is fulfilled for eigenmodes with ˆ ω (cid:54) = 0 .We will now prove (A) and (B).(A) For non-oscillating modes ( ˆ ω = 0 ), the normalized voltage integrals ˆ v ( x, y, z ) are real and thecontribution per mode is ˆ w (cid:107) ( x , y , x , y , s ) = − (cid:90) ∞−∞ ˆ v ( x , y , z + s ) ∂∂z ˆ v ( x , y , z ) d z. ˆ w (cid:107) ( x , y , x , y , s ) = − (cid:90) ∞−∞ ˆ v ( x , y , z + s ) ∂∂z ˆ v ( x , y , z ) d z = 2 (cid:90) ∞−∞ ˆ v ( x , y , z ) ∂∂z ˆ v ( x , y , z + s ) d z = 2 (cid:90) ∞−∞ ˆ v ( x , y , z − s ) ∂∂z ˆ v ( x , y , z ) d z = − ˆ w (cid:107) ( x , y , x , y , − s ) . The physical meaning of this symmetry is that the energy transfer from particle 1 to particle 2 (by ˆ w (cid:107) ( x , y , x , y , s ) ) plus the reverse energy transfer (by ˆ w (cid:107) ( x , y , x , y , − s ) ) is zero. This is obviousas no energy is left to the non-resonant mode after both particles have departed the volume. The voltage ˆ v ( x, y, z ) is zero for z < before the source q entered the cavity and, as the eigensolution is curl-free,it is zero for z ≥ L . Therefore ˆ w (cid:107) ( · · · , s ) = 0 for | s | > L . Two particles can interact only throughnon-oscillating modes if they are simultaneously in the cavity at any time.(B) The normalized voltage integral for oscillating modes ( ω (cid:54) = 0 ) does not depend on z after q has left the cavity: v ( x, y, z > L ) = 12 (cid:112) ˆ W (cid:90) L −∞ ˆ E z ( x, y, s ) exp(i ˆ ωs/c ) d s = v ( x, y ) . Therefore the following integral relation can be derived: ˆ v ∗ ( x , y )ˆ v ( x , y ) = (cid:90) ∞−∞ ∂∂z (cid:8) ˆ v ∗ ( x , y , z + s )ˆ v ( x , y , z ) (cid:9) d z = (cid:90) ∞−∞ ˆ v ( · · · , z ) ∂∂z ˆ v ∗ ( · · · , z + s ) d z + (cid:90) ∞−∞ ˆ v ∗ ( · · · , z + s ) ∂∂z ˆ v ( · · · , z ) d z = (cid:90) ∞−∞ ˆ v ( · · · , z − s ) ∂∂z ˆ v ∗ ( · · · , z ) d z + (cid:90) ∞−∞ ˆ v ∗ ( · · · , z + s ) ∂∂z ˆ v ( · · · , z ) d z. This relation is needed to prove the symmetry: ˆ w ( x , y , x , y , s )= − (cid:26) exp(i ˆ ωs/c ) (cid:90) ∞−∞ ˆ v ∗ ( x , y , z + s ) ∂∂z ˆ v ( x , y , z ) d z (cid:27) = − (cid:26) exp(i ˆ ωs/c ) (cid:20) ˆ v ∗ ( x , y )ˆ v ( x , y ) − (cid:90) ∞−∞ ˆ v ∗ ( · · · , z + s ) ∂∂z ˆ v ( · · · , z ) d z (cid:21)(cid:27) = − (cid:8) exp(i ˆ ωs/c )ˆ v ∗ ( x , y )ˆ v ( x , y ) (cid:9) − ˆ w ( x , y , x , y , − s ) . With x = x and y = y , we find that ˆ w ( x , y , x , y , s ) + ˆ w ( x , y , x , y , − s ) = − k ( x , y ) cos(ˆ ωs/c ) , where ˆ k ( x , y ) = | ˆ v ( x , y ) | ; in particular, for the origin, ˆ w ( x , y , x , y ,
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