The equilibrium longitudinal distribution of electrons in circular accelerators is discussedfor the case of the 8\u27 wake function. Contrary to the well known fact that the solution does not exist in this case beyond a threshold, it is strongly suggested that the solution actually exists when we regularize the singularity of the 8\u27 wake function in a physical way, therefore the non-existence of the solution has no physical consequence

HIRATA Kohji

SHOBUDA Yoshihiro

光司 平田

Graduate University for Advanced Studies [SOKENDAI]  Institutional Repository

Particle Accelerators, Vol. 62, pp. 165-177Reprints available directly from the publisherPhotocopying permitted by license only© 1999 OPA (Overseas Publishers Association) N.V.Published by license underthe Gordon and Breach SciencePublishers imprint.Printed in India.THE EXISTENCE OF A STATIC SOLUTIONFOR THE HAISSINSKI EQUATION WITHPURELY INDUCTIVE WAKE FORCEYOSHIHIRO SHOBUDA* and KOHlI HIRATAKEK, High Energy Accelerator Research Organization, 1-1 Oho,Tsukuba-shi, Ibaraki-ken 305-0801, Japan(Received 19 December 1997; In final form 19 March 1998)The equilibrium longitudinal distribution of electrons in circular accelerators is dis-cussed for the case of the 8' wake function. Contrary to the well known fact that thesolution does not exist in this case beyond a threshold, it is strongly suggested that thesolution actually exists when we regularize the singularity of the 8' wake function in aphysical way, therefore the non-existence of the solution has no physical consequence.Keywords: Electron accelerators; Instabilities; Impedance; Storage rings1 INTRODUCTIONThe Haissinski equation1 describes the equilibrium longitudinal dis-tribution of electrons in circular accelerators. The analysis of the sta-bility of this solution against a small perturbation is the basis of thetheory of longitudinal instabilities.2 Nonetheless, the existence (andthe uniqueness) of the solution of this equation is not assured. It isdifficult to solve this problem in complete generality.Electrons in an accelerator interact with their environment becausethey are enclosed in metals (vacuum pipe, the RF cavities, etc.). Thiseffect is represented by the wake function. 3 The wake field acting onan electron is determined by the distribution of electrons ahead of it.*Corresponding author. Also at: Department of Physics, Tohoku University, Sendai,980-8578, Japan.165166 Y. SHOBUDA AND K. HIRATAAt the same time, the distribution of electrons is influenced by thewake field. Hence, to determine the distribution function, one shouldsolve coupled non-linear equations. The single particle equations ofmotion are as follows:d Ws-~ == --E (1)ds CO"E '~E = WS(Tf ~ _ e2LN {'XJ d~/p(()W(~' _ ~). (2)ds C ToEoc J~Here, we have used the dimensionless parameter ~ as(3)where O"E is the nominal rms relative energy spread, c is the velocity oflight, e is the electric charge of the electron, L is the total length of thepipe structure in which the wake field is generated, Eo is the referenceenergy of the beam, N is the number of electrons in a bunch, To is therevolution period of the beam, T is the time displacement between anelectron and the reference synchronous particle, E is the relativeenergy (E - Eo)/Eo with E being the electron energy, W s is the synchro-tron oscillation frequency, a is the momentum compaction factor ands is the longitudinal coordinate along the ring. The second term ofEq. (2) is the retarding force seen by a particle at ~ due to the longi-tudinal wake force which is produced by all particles infront of it; p(~)is the particle density at location ~.In the presence of radiation, two more parameters are necessary:b is the damping coefficient, and D == bO"E2 is the diffusion coefficientrepresenting the amount of quantum excitation due to photon emis-sion. The dynamics with radiation can be described by the Fokker-Planck equation for the phase space particle distribution 1/;(E, ~,s)STATIC SOLUTION FOR HAISSINSKI EQUATION 167This equation has a static solution given byV(~) = 100 d~'p(Ow(( - ~),(5)(6)(7)(8)Equation (6) is the Haissinski equation (A is the normalizationconstant where Jp d~ ~ 1). Since p(~) depends only on p(~') for ~ < ~',and we know(9)Eq. (6) can be integrated from the head of the bunch to the tail for agiven value A. Let us call the result of such an integration as p(~, A)and define the "charge" Q asQ = Q(A) = I: p(~,A) df (10)If a value A exists such that Q(A) ~ 1, it gives the solution of theHaissinski equation. Usually, we find the solution of Eq. (6)numerically.Previous independent studies seem to indicate that there is always aunique solution,4 except for a single well known case, the 8' wakefunction 1,3w(~) === S8'(~). (11 )This case is called the purely inductive wake function5 and corre-sponds to a beam induced voltage V(~) === -Sp'(~) proportional to the168 Y. SHOBUDA AND K. HIRATAderivative of the particle density. This case is of practical importance:machines with many small discontinuities, bellows, masks, etc. willtend to be more inductive.3,6 For the damping ring of the StanfordLinear Collider, the observation is quite consistent with the assump-tion of the purely inductive case.6- 10 For example, S is evaluated as-7.5 for the SLAC damping ring[6] and -0.94 for KEKB. 10 Here wehave to notice that the sign of S depends on that of ex, because thesign of Ws depends on that of ex (see Eq. (8)).This paper is organized as follows: In Section 2 we review the solu-tion of the Haissinski equation with purely inductive wake force.Then in Section 3, we show that the solution actually depends on howto construct the derivative of a delta function when a parameter isover a threshold. Conclusions and discussions follow under Section 4.2 PURELY INDUCTIVE IMPEDANCEFor Eq. (11), Eq. (6) is rewritten asp' I- == -~ + Sp,P(12)where the prime denotes differentiation with respect to ~. By puttingwe get the solutionp' (13)logp - Sp == _!~2 + 10gA. (14)If S is negative, Eq. (13) has no singularity in its denominator andthere is always a continuous unique solution that is normalizable.From now on, we consider the case that S is positive. Given A, thesolution is shown in Figure 1. The manifold of the solution changessuddenly at1A == Amax == Se' e == 2.7182 ... (15)When A > Amax, the solution p(~, A) defined for -00 < ~ < 00 doesnot exist. Clearly, Q(A) increases when A increases. For p to beSTATIC SOLUTION FOR HAISSINSKI EQUATION 1692.01.5"".-...,).JJ\'-""a..1.00.50-3 -2 -1 0 1 2 3~FIGURE 1 The solution manifolds of Eq. (13) for A < Amax (- - -), A = Amax(-----) and A> Amax (-).normalizable, Q(Amax) must be greater than 1. Using Eq. (14) toexpress ~ in terms of p, this gives11 1- xS < Smax == V2 dx ~ 1.55061.- 0 Jx-Iogx-l (16)When S> Smax, no value of A can give Q== 1.3 Previous numericalstudy told Smax was 1.53 instead of 1.55061.3 This value is slightlywrong.3 A PROPERTY OF SINGULARITY POINTLet us consider more carefully why there is a threshold beyond whichno solution exists. The 8' function is an approximation. As shown inEq. (12), the 8' function induces the derivative of p in the r.h.s.In order to define the derivative, we need information around a170 Y. SHOBUDA AND K. HIRATAneighborhood. It means that we cannot determine p(~') from infor-mation for ~ >~' only. In this sense, it breaks causality.3 Further, thecase with A == Amax plays an essential role: when A < Amax, peA) isflat at ~ == 0, as expected from Eq. (13). But at A == Amax, p(o) jumps to1/S, which causes a bifurcationp' (0) == ± 1/ S. (17)This is possible only because the r.h.s. of Eq. (12) contains p'.From the above discussions, it seems reasonable to replace b' by8'(~) -; 8(~) - 8(~ - a) .a(18)Note that a must be positive in order to satisfy the causality condi-tion. Instead of Eq. (12), we getp' S-== -~+-(p(~+a)-p(~)).p a(19)The b' wake function is regained when a --+ O. With a finite a, we havethe possibility to avoid the bifurcation which occurred for the b I casewith A == Amax. Let us study this possibility and show that Eq. (19)has a solution for arbitrary a and S.When a is sufficiently large, it is easily found that Eq. (19) reducesto the case of the b-wake where the existence of the solution is wellknown: 3,!! the term p(~+a) of Eq. (19) can be neglected comparedwith the p(~) termp' S- ~ -~ - -p(~).p a(20)It is obvious that a continuous and normalizable p exists for arbi-trary S. A solution that satisfies the normalization condition can bewritten as3,!!e-(e/2)p(~) = J(7i72)(Sja) (coth(Sj2a) + erf(~jv'2)) . (21)Numerically we confirm that the solution of Eq. (19) can be approxi-mated by that of Eq. (20) (see Figure 2). When a is small and S is(a) 0.50.40.3~uP"-"0-0.20.1a-4 -2 0 2 4~-( \I \I \!I \~ '\>" ..o0.40.30.20.1(b) 0.5-4 -2 o~2 4FIGURE 2 (a) The solution of Eq. (19) at a=3 and 8=2; (b) Its approximatesolution (Eq. (21)).172 Y. SHOBUDA AND K. HIRATAsmaller than Smax, the solution of Eq. (19) can be well approximatedby that ofEq. (13).Let us numerically study the case where a is small and S is largerthan Smax. Numerical study shows that Eq. (19) has a solution forarbitrary small a. Let us see the feature of this solution and show thereason why Eq. (19) can have solution. In Figure 3, we show the solu-tion manifold for the case with a ~ 0.1. The most remarkable changeis that the solution is not symmetric with respect to ~. It is obvious,because Eq. (19) is not symmetric when a is finite. Further, p is notsymmetric even when a ---+ O. It is no surprise that a solution is notsymmetric even when its original equation is symmetric. This kind ofsymmetry breaking really occurs. Actually, ~o that satisfies p'(~o) ~ 0is leaving away from 0 as a ---+ 0 (see Figure 4). We expect that remainstrue for much smaller values of a.0.81\I I0.6~uP"-'"c.0.40.2 1-- ; + ;#./ + 11'1 ; ; --1-4 -2 o~2 4FIGURE 3 The solution of Eq. (19) at a=O.l for A=0.17785 (-), A=0.19119(- - -), A = 0.19208 (_00 '-00 .-) and A = 0.19230 (-.-.-).STATIC SOLUTION FOR HAISSINSKI EQUATION 173-0. ,-0.20-0.3uP-0.4 ........................._ ~ n...... a'. .. . 1 ao · ·.· ·..· .-0.5 - 1'- .:, ·························i···························· --\101o. ,-0.6 -==---...L..-.L...-...L...--l-..L....I....1.-l.-------I-..-I~I.-L--J,......1-.L.--____I...._._..L~0.01aFIGURE 4 The ~o-a relation.This symmetry breaking can also be seen by Q(A) point of view.We saw in the a == 0 case that Q(A) had an upper bound because pee)must be continuous and symmetric. However, if symmetry breakingoccurs for A == Amax, the upper bound of Q(A) due to its bifurcation isremoved. It strongly suggests that, with finite a, a continuous solutionexists for all S and that the restriction of Eq. (16) does not apply.Actually, Q(A) is a smooth function and monotonically increasing.There exists always one and only one value of A satisfying Q(A) == 1(see Figure 5).In the a == 0 case, the bifurcation occurs and there is a thresholdbeyond which no solution exists. In the a i- 0 case, the bifurcationdoes not occur and a solution always exists. Further, the solutionobtained by a~ +0 is not smoothly connected to the solutionobtained by a~ -0 when S is larger than Smax. Thus, a == 0 ismathematically a branch point and the solution manifolds are174 Y. SHOBUDA AND K. HIRATA0.1 9230.1922A1J/I/././'~~/.-----.../L-"""'-...-..........'1--"""'"1.1 51.001.101.200.950.1920.....-..«~Ci1.05FIGURE 5 The function Q(A) for a = 0.1.topologically different from those of the a i- 0 case. In the a-S plane,the line a == 0, S> Smax forms a "cut" and only the positive side of a isphysical as we have mentioned before.In the following way, we can also make the approximate formulawhere a is so small and S is larger than Smax' First, we chooseEq. (14) in which A == Amax (see Figure 6). Then, we find ~o: ~o can bedecided by integrating this function in order to satisfy the normali-zation condition.In the above discussions, we used Eq. (18) in order to regularize the8'(~). There can be other regularization of 8'(~). It is interesting thatour discussion is valid also for other choice of the regularization. Letus see briefly the case that we use a resonator wake3 as a regulariza-tion function because it is sometimes used to parametrize the wakefunction. For simplicity, we consider the case with Q== 1/2. In thiscase a resonator wake is written by the following form: 3W(~) == Lw~e-WR~(1 - WR~)W(~) == 0for ~ ~ 0,for ~ :::; o.(22)AI \i\I \-------V \ ~1.0(a)0.80.6---.uP"-'"Q.0.40.20-4 2 o 2 4~1.0(b)0.80.6---.uP"-'"Q.0.40.20-4 -2 2 4~o ~FIGURE 6 (a) The solution of Eq. (19) at a=O.l and 8=2; (b) Its approximatesolution (Eq. (14) where A = Amax and 8 = 2).176 Y. SHOBUDA AND K. HIRATAHere Land WR are parameters. Impedance of this wake is written as. wiZ(w) == -lwL . 2'(WR - lW) (23)Since Z(w) -+ -iwL for WR -+ 00, WR can be seen as a regularizationparameter. Given A, the solution is shown in Figure 7. For small A,the solution is almost symmetric. As A becomes larger, p becomesasymmetric. Further, there is no upper bound for A. These propertiesare identical with what we saw in the case of Eq. (18). Thus, weexpect that this regularization is equivalent to the previous one whenWR -+ 00. It may imply that our conclusion does not depend on thechoice of regularization function.2.5 ..-----.-----,-,\r-------r------r:-I-~--__,\'/ "\2·0 ~ i ,. ; ..~!..\ \ /J \ \ I\ i\ III3j \ 1~·········· ········ ·..·..···i···..·········· j.J~..; / /- \, \ \'1.5 / / \ \ \ Ij / -.l.. \ ..\!f ; ,. ···;····· · ··..··..· 1~··..·····························i·············· ····"'j"" iJ. •••••••••/ ~"\~•••,.<•••••••-v \/ / / \ Y:.. .1,\ , , ,1.0 ..-·····..··· ..···· ·..·····,..·······..···..;··-1-·/..···)[·1·····;10~~~~.\~,~'\\ i , ..,j /7 if / "'2\\~0.5 ::::::::::·.:::::::::::::::::::::.·::::/::.·~::~/:~I/·f....J/r...; ..f. ../..:'·~~~~~:~"\·:-,\·0"..·~..\,......···:::::....:::::....:::::··::::::: .... :::::....:::::... :::::::::::11··· ..···.. ····'// r---. 'L,,'\~~·····..········..·..·· ·..···:/·.;/!Ih/v ~,."'~~.:~~.~..:\".~~~ , : I-~~......................... ~ ..O~·-3 -2 -1 0 1 2FIGURE 7 The solution manifolds for several values of A for Eq. (22) at WR = 2 (-)and the solution manifold of Eq. (13) for A = Amax (- - -).STATIC SOLUTION FOR HAISSINSKI EQUATION 1774 CONCLUSIONS AND DISCUSSIONSThe purely inductive wake function (S8') was the only known exam-ple where the solution of the Haissinski equation may not exist.With physical and appropriate regularization of this wake function,we have shown that the solution still exists. It is of great interest tosee whether the Haissinski equation has at least one solution for anyphysical wake function. Although it seems true intuitively, we are farfrom its proof. The present work is a first step towards a solution ofthis problem. More extensive work will be published elsewhere.AcknowledgementsThe authors thank S. Matsumoto, E. Forest and K. Bane for discus-sions. One of the authors (Y.S.) thanks the members of the KEKAccelerator Theory Group for their hospitality and help.References[1] J. Haissinski, Nuovo Cimento 18B, 72, 1973.[2] K. Oide and K. Yokoya, KEK-Preprint-90-10.[3] A.W. Chao, Physics of Collective Beam Instabilities in High Energy Accelerators,Wiley-Interscience, 1993.[4] K. Oide, private communication.[5] Even if we add the resistive part, w(~) = So8(~) + S8'(~), still there is a thresholdfor S beyond which the solution does not exist.[6] K. Bane, SLAC-PUB-5177, 1990.[7] K. Bane, Proc. of the first European Particle Ace. Con!, Rome, p. 637, 1988.[8] K. Bane and R. Ruth, Proc. of the 1989 IEEE Particle Ace. Con!, Chicago,p. 789, 1989.[9] K. Bane et a/., Proc. of the first European Particle Ace. Con!, Rome, p. 878, 1988.[10] E.S. Kim and K. Hirata, KEK Report No. 95-5,1995.[11] A. Ruggiero et al., IEEE Trans. Nucl. Sci. NS-24(3), 1205, 1977.

The Existence of a Static Solution for the Haissinski Equation with

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The Existence of a Static Solution for the Haissinski Equation with

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