Landau Damping

22/09/11

LANDAU DAMPING

L. Palumbo

SAPIENZA UNIVERSITY ROME & LNF-INFN

L. Palumbo and M. Migliora, Landau damping in Parcle Accelerators, 2011.

• Introducon • Plasma oscillaons • Plasma waves and dispersion relaon - Vlasov-Poisson Equaon - Dispersion relaon - Landau damping in plasma waves • Mechanical system • Beams in parcle accelerators • Longitudinal instabilies in coasng beams - Dispersion relaon - monochromac beam - beam with energy spread • Longitudinal instability in single bunch beam

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Landau damping is a physical eﬀect named aer his discoverer, the Russian physicist Lev Davidovich Landau, who studied in 1946 the wave propagaon in a plasma.

According to Landau theory, an inial perturbaon of longitudinal charge density in plasma waves is prevented from developing because of a natural stabilizing mechanism.

1. Plasma oscillaon

• A cold plasma of ionized gas consists of ions and free electrons distributed over a region in space. The posive ions are very much heavier than the electrons, so that we can neglect their moon in comparison to that of electrons.

• The plasma at the equilibrium, being neutral, is 3 characterized by the same local density n0 [1/m ] for both electrons and ions.

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• If, for some reason, electrons are displaced from their equilibrium posion, the local density changes producing electrical forces that tend to restore the equilibrium.

• As in any classical harmonic oscillator, the electrons gain kinec energy, and instead of coming to rest, they start oscillang back and forth, at a frequency called ”plasma frequency”.

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! = n0e ! = "x 2 ! n0e Ex = Fx = !eEx = ! x #0 !0 2 n0e me !x!= ! x !0

2 n0e ! plasma = me"0

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2. Dispersion relaon for plasma waves

We consider now the more general case of a charge density with a disbuon funcon depending on the posion and velocity such that:

f ( x,v ,t )dx dv N ! x x =

If the charges are not in a state of equilibrium, we will observe a me evoluon of the distribuon under the eﬀect of the self electric ﬁeld.

Such a system can be studied by means of the methods developed by Boltzmann to describe the behavior of systems far from the thermodynamical equilibrium.

We have to study the moon of an ensemble of N

parcles characterized by a distribuon funcon f(x,vx,t) under the acon of self forces.

The fundamental equaon which describes the kinemacs of this ensemble is the connuity equaon for the density of the parcles in the phase space.

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It states the conservaon of the number parcles in any phase space volume during the moon.

A(t+dt)

Phase trajectory

A(t)

The phase space area enclosing a number of parcle at me t can be distorted at me t+dt but it remanins

constant. For an inﬁnitesimal area dA=dx dvx we have:

dN = f ( x,vx ,t )dxdvx = f ( x + vxdt, vx + axdt, t + dt )dxdvx

Fx where ax = me If we expand at the ﬁrst order the RHS term, simplifying the common terms, we get:

!f !f Fx !f (Boltzmann Equaon) + vx + = 0 !t !x me !vx

7 2.2.2 Vlasov-Poisson equations and the dispersion relation

In 1938 Vlasov showed[10] that Boltzmann equation is not suitable for a description of plasma dynamics due to the existence in plasma of long range 2.2.2 Vlasov-Poisson equations and the dispersion relation collective forces. Instead, a system of equations, known today as Vlasov In 1938 Vlasovequation, showed[10] was suggested that Boltzmann for the equation correct description is not suitable to take for a into account the description of plasma dynamics due to the existence in plasma of long range long range collective forces through a self-consistent field. 22/09/11 collective forces. Instead, a system of equations, known today as Vlasov Let us assume then that the collisionless system is formed by neutral equation, was suggested for the correct description to take into account the warm plasma characterized by a non-relativistic motion3, in a region with long range collective forces through a self-consistent field. Let uszero-magnetic assume then thatAn field.important the collisionlessAssuming contribuon the system case to is the of formed heavycomprehension by ions neutral (”frozen” of motion), plasma waves came first from the work of the russian warm plasmafrom characterized equation (14), by a we non-relativistic derive here the motion kinetic3, in equation a region with for the evolution of physicist Anatoly Alexandrovich Vlasov. zero-magnetica density field. perturbation Assuming the in case the of plasma, heavy ions the so-called(”frozen” Vlasov motion), equation[11]. To • In 1937, Vlasov showed that from equationsimplify (14), the we derive study,Boltzmann here we the assumeequaon kinetic theis equation suitable motion for for only the a evolutionin one dimension, of that we a density perturbation indecripon the plasma, of plasma the so-calleddynamics Vlasov only equation[11]. if To indicate with xwe, theconsider corresponding the long range velocity collecve being vx. simplify the study, we assume the motion only in one dimension, that we In this case,forces the exisng density in the plasma. function f (x, vx,t)representstheelectronsdis- indicate with x, the corresponding• Thus, a system velocity of beingequaonsvx. , known tribution functiontoday in as the Vlasov-Poisson plasma. The equaon electric, was forces generated by the charge In this case, the density function f (x, vx,t)representstheelectronsdis- distribution willsuggested act on for the the chargescorrect and descripon modify the distribution. In this case tribution function in the plasma.to take Theinto electric account forces the generatedcollecve by the charge equation (14) can be written as4 distribution will act on theforces charges through and a modifyself-consistent the distribution. field. In this case equation (14) can be written as4 ∂f ∂f e ∂f + v E =0 (15) ∂t x ∂x − m x ∂v ∂f ∂f e ∂f e x + vx Ex =0 (15) where Ex is the∂t electric∂x − fieldme satisfying∂vx where Ex is the electric fieldThe satisfyingelectric field is derived from the scalar ∂φ potenal Ex = (16) ∂φ −∂x E = (16) x −∂x φ being the electric field potential given by the Poisson equation which in turns is related to the net local density: φ being the electric field potential given by the Poisson equation We can get an approximate∂2φ ρ solutione of the Vlasov-Poisson equations by 2 ∂ φ ρ 2 e= = n0 fdvx (17) using a perturbation= =∂ technique,x −nε0 assuming−fdvε0 that− the electronic(17) distribution ∂x2 −ε −ε 0 − x We assume 0 now0 that for the system of charges there is an functionEquationsf(x, vx,t (15))isgivenbythesumoftheunperturbeddensityfunction - (17) are also known as Vlasov-Poisson equations. Equations (15) - (17)equilibrium are also known state asfo(v Vlasov-Poissonx) with a proper equations.velocity distribuon, 5 f0(vx3)andaperturbationand we considerf a 1 (smalx, v xperturbaon,t) f1(x, vx,t) around that 3 Here we have already considered non-relativistic particles. For relativistic particles we Here we have already consideredequilibrium non-relativistic: particles. For relativistic particles we had to use thehad momentum to use the instead momentum of the velocity insteadv of. the velocity v. f(x, vx,t)=f0(vx)+f1 (x, vx,t)(18) 4For our convenience4For our we convenience invert the scalar we invert products. the scalar products. Since the equilibrium state is neutral, we have that f dv = n and 7 7 0 x 0 f1dxdvx =0.Moreover,f0 does not depend on time and position, thus it results: 8 ∂f 0 =0 (19) ∂t ∂f 0 =0 (20) ∂x For what concern the electric field, we know that it vanishes in the unperturbed neutral state, and its value is related to the amplitude of the

perturbation f1 only. If we consider now the last term of the LHS of eq. (15), we can write ∂f ∂ ∂f0 Ex = Ex (f0 + f1)=Ex (21) ∂vx ∂vx ∂vx

where we have neglected Ex∂f1/∂vx, being of second order in the perturbation. By using equations (19) - (21) into eq. (15), we get

∂f1 ∂f1 e ∂f0 + vx Ex =0 (22) ∂t ∂x − me ∂vx and ∂2φ e = f dv (23) ∂x2 ε 1 x 0 A solution of equations (22) and (23) was ﬁrst derived by Vlasov who applied the double Fourier transforms from the domain (x, t)tothedomain (k, ω), getting for the perturbation and the potential

∞ ∞ i(ωt kx) f˜1(vx,k,ω)= f1(x, vx,t)e − dxdt (24) −∞ −∞ 5The perturbation is characterized by small amplitude and slope.

8 We can get an approximate solution of the Vlasov-Poisson equations by using a perturbation technique, assuming that the electronic distribution function f(x, vx,t)isgivenbythesumoftheunperturbeddensityfunction 5 f0(vx)andaperturbation f1 (x, vx,t)

f(x, vx,t)=f0(vx)+f1 (x, vx,t)(18)

Since the equilibrium state is neutral, we have that f0dvx = n0 and

f1dxdvx =0.Moreover,f0 does not depend on time and position, thus it results: ∂f 0 =0 (19) ∂t ∂f 0 =0 (20) ∂x For what concern the electric ﬁeld, we know that it vanishes in the unperturbed neutral state, and its value is related to the amplitude of the 22/09/11 perturbation f1 only. If we consider now the last term of the LHS of eq. (15), we can write ∂f ∂ ∂f0 Ex = Ex (f0 + f1)=Ex (21) ∂vx ∂vx ∂vx Since fo doesn’t depend on me and posion, neglecng the where we have neglectedsecond orderEx ∂termsf1/∂, from the vx, beingBolzmann of second equaon order in we the have perturba-: tion. By using equations (19) - (21) into eq. (15), we get

∂f1 ∂f1 e ∂f0 + vx Ex =0 (22) ∂t ∂x − me ∂vx Vlasov-Poisson Equaons and ∂2φ e = f dv (23) ∂x2 ε 1 x 0 A solution of equations (22) and (23) was ﬁrst derived by Vlasov who applied the doubleThese Fourier two transformscoupled equaons from the tell domain us that (x, t)tothedomain a density perturbaon produces an electric ﬁeld which acts back on (k, ω), getting forthe theperturbaon perturbation, both and evolve in the me. the potential

∞ ∞ i(ωt kx) f˜1(vx,k,ω)= f1(x, vx,t)e − dxdt (24) −∞ −∞ 5The perturbation is characterized by small amplitude and slope. ˜ ∞ ∞ i(ωt kx) This mechanismφ(k, ω)= can sustain plasma φ(x, t)e vaves− dxdtpropagang in (25) the medium. In order−∞ to 8−∞ﬁnd a self consistent soluon, and for the diﬀerential equation (22) Vlasov expanded the unknown funcons f1 and φ through the double Fourier transformse: ∂f i(kv ω)f˜ + i kφ˜ 0 =0 (26) x "1 − 1 me ∂vx f ( x,v ,t ) = f! ( k ,v ," )ei( kx!"t )dkd" 1 x 2! # 1 x Accordingly, eq. (23) becomes:!" 1 " #( x,v ,t ) = #!( k ,v ," )ei( kx!"t )dkd" x 2 # e x k2!φ˜!"= f˜ dv (27) − ε 1 x which applied to the Bolzmann-Poisson0 equaon produce the well known Dispersion Relaon for plasma waves: If we take f˜1 from (26) and substitute into (27), we obtain the following dispersion relation e2 ∂f /∂v 1+ 0 x dv =0 (28) ε m k ω kv x 0 e − x Integration of (28) over vx provides a relation between k and ω which de-

pends only on the slope of the unperturbed distribution function f0(vx). The dispersion relation contains a divergent integral, because of the singularity

at ω = kvx. To overcome this diﬃculty, without giving a solid explanation, 9 Vlasov calculated the principal value of the integral, getting, as result, only afrequencyshiftwithoutanykindofdamping.

2.2.3 Landau solution of the Vlasov equation

In a very original paper of 1946 Landau proposed a new method of solution of Vlasov-Poisson equations putting the basis of the theory of plasma oscillations and instabilities[1]. He demonstrated that the problem had to be considered as an initial value or Cauchy problem, with a perturbation

f1(x, vx,t)knownatt =0.TothisendheadoptedtheLaplacetransformfor the time domain and used the Fourier transform only for the space domain. Accordingly, the perturbation and the electric ﬁeld are ﬁrst transformed as

∞ ikx f˜1 (vx,k,t)= f1 (x, vx,t) e− (29) −∞

∞ ikx E˜x (k, t)= Ex (x, t) e− (30) −∞ 9 22/09/11

444 !. !. A. A. Vlasov (1937): "On Vibraon Properes of Electron Gas" (in Russian). 459

533.9 4. *)

. . (IV). , 1 . . . 2 . . 3 . * ) . 4. . 5. . 6. . 7. 4 !" " . 0 ∞ ∞ i(ωt kx) φ˜(k, ω)= φ(x, t)e − dxdt (25) 1. −∞ −∞ and for the di ﬀerential equation (22)

(?> > !) e ∂ f . i ( kv ω() f˜ + i k φ˜ 0 =0 (26) ), x − 1 m ∂ v , e x . Accordingly, eq. (23) becomes: ( ), ( ), ( 2 e). k φ˜ = f˜1dvx (27) J " − ε = m|2• (l·? ^ ' ±• ! ). 0 (37 ) (36 ) # !. r , . . , If we take f˜ from (26) and substitute into (27), we obtain the following 1 , dispersion relation ( ). , , e2 ∂f /∂v 1+ 0 x dv =0. (28) ε m k ω kv x , 0 e − x ( Integration , of (28) over v x provides a relation . between k and ω .) which de- . pends only on the slope of the unperturbed distribution function f 0 ( vx ). The • It provides a relaonship between the wave number , dispersion k=2 relationπ/λ and the contains afrequency divergent integral, ω of the becausewave of in the plasma the singularity , ω 0 =,• , : 1) , 2) at ω = kv x. To overcome this di ﬃ —culty, without giving a solid explanation, • It depends , on the 3) slope of the . equilibrium Vlasov D:, calculated the principal value of the integral, getting, as result,, only afrequencyshiftwithoutanykindofdamping. distribuon w.r.t . the velocity . 2 , D = 1/ $ , $ =— yineN /kT. . **). 2.2.3 • Landau Mathemacally solution of, the the Vlasovintegral equation shows a singular point

(zero of the denominator) % /$ = YkTIm.at ω=kvx. Vlasov overcame In a very this original difficulty paper of calculang 1946 & Landau the proposed Principal a new method Value of of solu- the , % * = % / % , * / , # * = "$ / % = " Ym/kT . = ' . tion of Vlasov-Poisson integral . equations 0 putting the basis of the theory 0 of— plasma oscillations (38) , and instabilities[1]. . He demonstrated that the problem had to . , « be considered as an initial value or Cauchy problem, with a perturbation », . . ' 2/ 2 ( ) f (x, v ,t)knownatt =0.TothisendheadoptedtheLaplacetransformfor (1 x : — % ·/«( * ») , (39 ) the time domain and used the Fourier transform only for the space domain. « », ,Accordingly, the perturbation and the electric field are first transformed as 10 , $% , & , ' , . . ˜, ∞ , ikx f 1 ( v x,k,t )= f1 ( x, vx ,t ) e − (29) , , −∞ % 0 , , ∞ . , ikx E˜x (k, t)= Ex (x, t) e− (30) , . , −∞ ., 9 , , , . , (39 ) % = 0 , * 2 = — 1 , . . , *) 8=± (3), 291 . (1938). (40) **) L. Landau, Zs. Phys. Sowjetunion 10 (2) (1936). *) « » ( . 527) . . , (36) .

(« »). ( . .) !"#$%&'

()#$%&' *+,-'

Figure 2: Path of integration for Landau damping.

Accordingly, the dispersion function becomes: equation (43) can be written as 2 e ∂f0/∂vx iπ ∂f0 1+ P.V. dvx 3 =0 (43) ε m k ∂f0/∂ωvx kv kn− 0k ∂kv n0 kBT iπ ∂f0 0 e x x vx=ω/k −dvx = 2 3 4 (48) ω kvx − ω − ω me − k ∂vx The imaginary term of the− above equation produces the damping/antidamping vx=ω/k 22/09/11 effect predictedBy by using Landau, the dependingabove results, on the the slope dispersion of the distribution relation (28) func- becomes tion. With this procedure, we obtain straightforwardly the correct dispersion ω2 ω2 k T πe2 ∂f 1 p 3k2 p B i 0 =0 (49) relations via Fourier transformationω2 ofω the4 m Vlasovε equation.m k2 ∂v − − e − 0 e x vx=ω/k Example: Plasma with a Maxwellian velocity distribution As an examplenamely to clarify the use of the dispersion relation for the analysis k T πe2 ∂f Example ω2 = ω2 1+3k2 B + iω2 0 (50) of the plasma stability, we considerp a plasmaω2m with a velocityε m k2 Maxwellian∂v e 0 e x vx=ω/k distributionMaxwell functiondistribuon of a warm plasma at temperature T with ωp the plasma frequency defined by equation (10). In case of a cold n m v2 f (v )= 0 exp e x (44) plasma, with0 x T (2π0,k weT/m recover)1/2 the− same2k T results of section 2.1. → B e B where kB is theIf Boltzmann we assume constant.T =0,suchtoproduceasmallperturbationoftheplasma We can integrate by parts the principal k = is the Boltzmann constant valueB of equationfrequency (43)ω obtaining, we can write ω = ω + iδω , with ω ω ω and δω ω , p r i r p p2 i p − n0e and approximate∂f /∂v the abovef (v ) expression∞ asf (v ) ! plasma = Note thatP.V. for T 00 x dv, fo=(vx) n0 x o (coldk plasma) 0 x dv m(45)" ω kv x ω kv 2 x e 0 x x − (ω kvx) 2 − − −∞ k T − πe ∂f For a givenω2 ωwavelength2 +2iω δω , ω the 2 1+3 frequencyk2 B of + iω the 2 plasma 0 (51) r p i p ω2m p ε m k2 ∂v 12 p e 0 e x vx=ω/k wave depends on the “plasma frequency ” ωp and on the averageThe kinec real part energy of the above of the expressionelectrons is (T) k T ω2 ω2 1+3k2 B = ω2 1+3k2λ2 (52) r p ω2m p D p e where we λ = k T/m ω2 is the Debye length, having assumed kλ 1. D B e p D This result is the dispersion relation for waves [12] in warm plasma obtained by Vlasov in his paper[10]. For the imaginary part, that was not predicted by Vlasov, we have that According to Vlasov results, plasma waves can be excited πe2 ∂f and can persist forever in a interplay2 between0 pertubaon 2ωpδωi ωp 2 (53) ε0mek ∂vx and self-fields. Vlasov theory doesn’t predict anyvx= dampingω/k effect. so that

2 In a very originalπ paperωpe of 1946 Landau ∂f0 i proposedπ ωp a new 1 δωi 2 = 3 exp 2 (54) 2 ε0mek ∂vx −2 2 (kλ ) −(kλ ) method of soluon of Vlasov-Poissonvx=ω/k equaonsd pung D the basis of the theory of plasma 14 oscillaons and instabilies. He demonstrated that the problem had to be considered as an inial value or Cauchy problem, with a perturbaon

f1(x, vx, t) known at t = 0. To this end he adopted the Laplace transform for the me domain and used the Fourier transform only for the space domain.

11 and ∞ ∞ i(ωt kx) φ˜(k, ω)= φ(x, t)e − dxdt (25) −∞ −∞ and for the diﬀerential equation (22)

e ∂f0 i(kvx ω)f˜1 + i kφ˜ =0 (26) − me ∂vx Accordingly, eq. (23) becomes:

e k2φ˜ = f˜ dv (27) − ε 1 x 0

If we take f˜1 from (26) and substitute into (27), we obtain the following dispersion relation e2 ∂f /∂v 1+ 0 x dv =0 (28) ε m k ω kv x 0 e − x Integration of (28) over vx provides a relation between k and ω which de-

pends only on the slope of the unperturbed distribution function f0(vx). The dispersion relation contains a divergent integral, because of the singularity

at ω = kvx. To overcome this diﬃculty, without giving a solid explanation, Vlasov calculated the principal value of the integral, getting, as result, only afrequencyshiftwithoutanykindofdamping.

2.2.3 Landau solution of the Vlasov equation

In a very original paper of 1946 Landau proposed a new method of solution of Vlasov-Poisson equations putting the basis of the theory of plasma 22/09/11 oscillations and instabilities[1]. He demonstrated that the problem had to be considered as an initial value or Cauchy problem, with a perturbation

f1(x, vx,t)knownatt =0.TothisendheadoptedtheLaplacetransformfor the timeAccordingly domain and, the usedperturbaon the Fourier and the transformelectric only for field the are first space domain. Accordingly,Fourier- thetransformed perturbation (space and the x) electricas follows field: are first transformed as

∞ ikx so that equations (22)f˜1 ( andk, vx (23),t)= becomef1 (x, vx,t) e− dx (29) ˜ −∞ ∂f1 ˜ e ˜ ∂f0 ˜ + ikvx∞f1 Ex ikx=0 (31) Ex (∂k,t t)= E−10x m(x,e t) e∂−vx dx (30) −∞ e so that eqs. (22) and (23) becomeikE˜ = f˜ dv (32) so that equations (22) and (23) becomex −ε 1 x 0 so thatApplyingAnd equationsthen the Laplace- Laplace (22) and˜˜ transform (23)transformed become to (29) (me t): and (30), we get ∂∂f1f1 ˜˜ ee ˜˜∂f∂0f0 ++ikvikvxfx1f1 EEx x =0=0 (31) (31) ∂∂t t˜ −−mme ∂v∂vx ∂f1 ∞ ee˜ ∂xf0 pt 1(vx,k,p+ ikv)=xf˜1 f1(Ev˜x,k,t)=0e− dt (31)(33) F ∂t ˜ e0−eme ˜ ∂vx ikikE˜Ex== f˜fdv1dvx (32)(32) x −εe0 1 x ˜ −ε∞0 ˜ pt x(k,ik pE)=x = E˜x(k,f1dv t)ex− dt (32)(34) Applying the LaplaceE transform− toε0 (29) and (30), we get Applying the Laplace transform to0 (29) and (30), we get Applying the Laplace transform to (29) and (30), we get while for equations (31) and (32), reminding∞ that thept Laplace transform of (v ,k,p)= f˜ (v ,k,t)e− dt (33) F1 x ˜ ∞ ˜ 1 x pt the time derivative is1(pk, v1x,pf)=1(t =0),weobtain0 ∞f1(˜k, vx,t)e− ptdt (33) F 1F(vx,k,p− )=0 f1(vx,k,t)e− dt (33) F 0 ∞e ˜ ∂f0 pt p 1 +x(ikvk, px)=1 = Exx(k, t)+e−f˜1dt(t =0)(34) (35) E ∞∞ pt F (k, p)=F 0mE˜e E(k,∂ tv)xe− ptdt (34) x x(k, p)= Ex˜x(k, t)e− dt (34) while for equations (31)E E and (32),0 reminding that the Laplace transform of and 0 whilewhile forApplying for eqs. equations (31) the andproperes (31) (32), and reminding˜ (32), of the Laplace reminding thate the that Laplacetransforms the Laplace transform, transformVlasov- of the of the time derivative is p 1ik f1((k,t =0),weobtain p)= dv (36) F˜ − x 1 x timethe derivative timePoisson derivative isequaonp 1 is pf1 (1becomet =0),weobtainEf˜1(t =0),weobtain: −ε0 F F − F − e ∂f0 ˜ Taking 1 fromp equation1 + ikvx (35)1 = and substitutingx + f1(t it=0) in (36), we get (35) F F me E ∂∂vf0 F p + ikv =e e ∂f0 x + f˜ (t =0) (35) p 1 +1 ikvx x1 =1 x x + f˜1(1t =0) (35) F F me E ∂vx ˜ and F eF me e E ∂∂fv0x/∂vx f1(t =0) ik x (k, p)= x + dvx (37) and E −ε0 me E pe+ ikvx p + ikvx and ik x (k, p)= 1dvx (36) E −εe0 F ik x (k, p)= e 1dvx (36) Since x does not dependE on vx, we−ε findF the following expression of the Taking E 1 from equationik x (k, (35) p)= and substituting0 1dvx it in (36), we get (36) F E −ε0 F electricTaking field:1 from equation (35) and substituting it in (36), we get F ˜ ˜ Takingwhere1 from we note the eq. (35)e andpresence substitutingee/ε0∂ of the f0/∂ itfv1x in(inialt =0) (36),f1(t wecondion=0) get . Fik x (k, p)=x (k, p)= x + ˜ dvx dvx (37)(38) E E −ε0e −ikmee(Ek,p p∂)+f0/ikv∂vpx+ ikvfp1x(+t =0)ikvx Soluon of the ik x (k, p)=above coupledx equaons+˜ gives the general dvx (37) where (k, p)istheplasmadielectricfunctiondefinedasE −e ε0 eme E∂fp0+/∂ikvvx x f1p(t+=0)ikvx Since ik doesx (k, notp)= depend on v ,x we find the+ following expressiondvx of(37) the expressionx E of the −ε0 transformedmexE p + (ikvk,px ) electricp + ikvx field: E 2 Since x does not depend on vx,e we find∂ thef0/∂ followingvx expression of the electric field:E (k, p)=1+ dvx (39) electricSince field:x does not depend on vx,εe/0 wemε ek find thefip˜ (t following=0)kvx expression of the E (k, p)= 0 ˜1 − dv (38) Ex −ike/(k,ε0 p) fp1+(t ikv=0) x electricIt field: is worth notingx ( thatk, p)= equation (39), putting (xk,dv px)=0withp = (38)iω, E −ike/ε(k, p) f˜p(t+=0)ikvx − wheregives the(k, dispersionp)istheplasmadielectricfunctiondefinedas integral (28). From0 the 1 Laplace transform of the electric where (k, p)istheplasmadielectricfunctiondefinedasx (k, p)= dvx (38) E −ik(k,2 p) p + ikvx e ∂f0/∂vx field x (k, p)wecanthenobtaintheperturbationfunction2 1 as where E(k, p)istheplasmadielectricfunctiondefinedas(k, p)=1+ e ∂f0/∂vx dvx F (39) (k, p)=1+ε0mek ip kvx dvx (39) e ε0mek ∂f0ip/∂−vxkvx f˜1(t =0) 2 It is worth noting1(vx,k,p that)= equationxe(k, (39), p) putting∂f0/−∂vx+(k, p)=0withp = (40)iω, It is worthF noting(k, thatp)=1+ equationme E (39),p putting+ ikvx (dvk,px p+)=0withikvx p = (39)−iω, ε0mek ip kvx − gives the dispersion integral (28). From the Laplace− transform of the electric 12 gives the dispersion integral (28). From10 the Laplace transform of the electric field x (k, p)wecanthenobtaintheperturbationfunction 1 as fieldE (k, p)wecanthenobtaintheperturbationfunction F as Ex F1 e ∂f0/∂vx f˜1(t =0) 1(vx,k,p)= e x (k, p) ∂f0/∂vx + f˜1(t =0) (40) F 1(vx,k,p)=me E x (11k, p) p + ikvx + p + ikvx (40) F me E p + ikvx p + ikvx 10 10 so that equations (22) and (23) become

∂f˜1 e ∂f0 + ikvxf˜1 E˜x =0 (31) ∂t − me ∂vx e ikE˜ = f˜ dv (32) x −ε 1 x 0 Applying the Laplace transform to (29) and (30), we get

∞ pt (v ,k,p)= f˜ (v ,k,t)e− dt (33) F1 x 1 x 0 ∞ pt (k, p)= E˜ (k, t)e− dt (34) Ex x 0 while for equations (31) and (32), reminding that the Laplace transform of the time derivative is p f˜ (t =0),weobtain F1 − 1 e ∂f0 p 1 + ikvx 1 = x + f˜1(t =0) (35) F F me E ∂vx and e ik (k, p)= dv (36) Ex −ε F1 x 0 Taking from equation (35) and substituting it in (36), we get F1 e e ∂f /∂v f˜ (t =0) ik (k, p)= 0 x + 1 dv (37) Ex −ε m Ex p + ikv p + ikv x 0 e x x Since does not depend!"# on$ v&', we find the following expression of the 22/09/11 Ex % x electric field: e/ε f˜ (t =0) (k, p)= 0 1 dv (38) Ex −ik(k, p) p + ikv x x where (k, p)istheplasmadielectricfunctiondefinedaswhich depends on ε(k,p), the plasma dielectric funcion. e2 ∂f /∂v (k, p)=1+ 0 x dv (39) ε m k ip kv x 0 e − x ()#$%&' It is worthLandau notingshowed that that equation the asymptoc (39), putting me (behaviourk, p)=0with of the p = iω, electric field depends on the soluons of (k,p)=0. He also − gives the dispersion integral (28). From the Laplaceε transform of the electric pointed out that this condion*+ corresponds,-' to the field (k, p)wecanthenobtaintheperturbationfunction as Ex Vlasov’s dispersion relaon when p=-iω. He couldF1 also overcomeFigure the “ 2: Pathdivergence of integration” forproblem Landau damping.applying the e ∂f0/∂vx f˜1(t =0) integraon1(vx,k,p theory)= in the x (complexk, p) plane, +geng: (40) F me E p + ikvx p + ikvx Accordingly, the dispersion functionVLASOV becomes: LANDAU

2 10 e ∂f0/∂vx iπ ∂f0 1+ P.V. dvx =0 (43) ε0mek ω kvx − k ∂vx − vx=ω/k The imaginary term of the above equation produces the damping/antidamping eﬀect predicted by Landau, depending on the slope of the distribution function. With this procedure, we obtain straightforwardly the correct dispersion relations via Fourier transformation of the Vlasov equation. Example: Plasma with a Maxwellian velocity distribution If we consider the generic harmonic of the ﬁeld: As an example to clarify the use of the dispersion relation for the analysis

of the plasma stability, we consider a plasmat i( kx witht a) velocity Maxwellian E! ( k ,v ,! )e!i e !!r distribution function x x n m v2 The imaginaryf (v )= term ω produces0 exp (Landau) e x damping or (44) 0 x (2πk T/mi )1/2 −2k T andampig eﬀect, dependingB e on the signB of the slope where kof the B is thedistribuon Boltzmann constant. funcon We. can integrate by parts the principal value of equation (43) obtaining The propagaon constants k and ωr are sll derived by the real part of the ∂f0/∂vx Dispersionf0(vx) ∞ Relaon (Vlasovf0(vx) ). P.V. dvx = k 2 dvx (45) ω kvx ω kvx − (ω kv ) x − − −∞ − 12

13 equation (43) can be written as

3 ∂f0/∂vx kn0 k n0 kBT iπ ∂f0 dvx = 2 3 4 (48) ω kvx − ω − ω me − k ∂vx − vx=ω/k By using the above results, the dispersion relation (28) becomes

2 2 2 ωp 2 ωp kBT πe ∂f0 1 2 3k 4 i 2 =0 (49) − ω − ω me − ε0mek ∂vx vx=ω/k namely equation (43) can be written as 2 2 2 2 kBT 2 πe ∂f0 ω = ωp 1+3k 2 + iω 2 (50) 3ω me ε0mek ∂vx ∂f0/∂vx kn 0 k n0 kBT iπ ∂f0 vx=ω/k dvx = 2 3 4 (48) withω ωpkvthex plasma− ω frequency− ω deﬁnedme − byk equation∂vx (10). In case of a cold − vx=ω/k plasma, with T 0, we recover the same results of section 2.1. By using the above results,→ the dispersion relation (28) becomes If we assume T =0,suchtoproduceasmallperturbationoftheplasma 2 2 2 ωp ωp kBT πe ∂f0 frequency1 ωp,3k we2 can write iω = ωr + iδωi, with ωr =0ωp ωp and (49)δωi ωp, 2 4 2 − − ω − ω me − ε0mek ∂vx and approximate the above expression as vx=ω/k

namely 2 2 2 2 2 kB2T 2 πe ∂f0 ω ω +2iω δω kBTω 1+3k πe +∂fi0ω (51) ω2 = ω2r 1+3p k2 i p + iω2 ω2m p ε m k2 ∂v (50) p 2 p e2 0 e x vx=ω/k ω me ε0mek ∂vx vx=ω/k The real part of the above expression is with ωp the plasma frequency deﬁned by equation (10). In case of a cold

plasma, with T 0, we recover the same resultskBT of section 2.1. → ω2 ω2 1+3k2 = ω2 1+3k2λ2 (52) r p ω2m p D If we assume T =0,suchtoproduceasmallperturbationoftheplasma p e 22/09/11 frequency ωwherep, we we canλ write= ωk =T/mωr +ωi2δωisi, the with Debyeωr length,ωp ω havingp and δω assumedi ωpk, λ 1. D B e p − D and approximateThis result the isabove the dispersion expression relation as for waves [12] in warm plasma obtained 2 2 2 by Vlasov in his2 paper[10].2 kBT 2 πe ∂f0 ω ωr +2iωpδωi ωp 1+3k 2 + iωp 2 (51) For the Maxwell distribuon discussed beforeω m, ewe ε0mek ∂vx For the imaginary part,p that was not predicted by Vlasov,vx=ω/k we have that have: 2 The real part of the above expression is2 πe ∂f0 2ωpδωi ωp 2 (53) ε0mek ∂vx ! #!$% k T vx=ω/k " ω2 ω2 1+3k2 B = ω2 1+3k2λ2 (52) )*+,-.%-*-/0.+1)% r p ω2m p D so that p e !2)0-.%-*-/0.+1)% 2 π ωpe2 ∂f0 i π ωp 1 where we λD = kBT/meω is the Debye length, having assumed kλD 1. δωi p 2 = 3 exp 2 (54) 2 ε0mek ∂vx −2 2 (kλd) −(kλD) This result is the dispersion relation forvx= wavesω/k [12] in warm plasma obtained "% & % &% '( by Vlasov in his paper[10]. 14 Figure 14: Maxwellian velocity distribution function. ! The phase velocityFor of the plasma the imaginarywave part, is that v was= notr predicted by Vlasov, we have that References ph k [1] L. Landau, On the vibration of the electronic plasma. J. Phys. USSR 2 is the phase velocity of the wave is derivedπ e by the ∂f0 10 (1946), 25. English translation in JETP 16, 574. Reproduced in Col- 2 2ωpδωi ωp 2 (53) lectedsoluons papers of L. D. of the Landau, editeddispersion and with an introduction relaon. by D. ε m k ∂v 0 e x vx=ω/k Ter Haar, Pergamon Press, 1965, pp. 445460; and in Men of Physics: L. D. Landau, Vol. 2, Pergamon Press, D. ter Haar, ed. (1965).

[2] J. H. Malmberg, and C. B. Whartonso that Phys. Rev. Lett. 13 184 (1964).

[3] V. K. Neil, and A. M. Sessler, Rev. Sci. Instr. 6, 429 (1965). 2 π ωpe ∂f0 i π ωp 1 [4] V. G. Vaccaro, CERN Report ISR-RF/66-35 (1966). δωi 2 = 3 exp 2 (54) 2 ε0mek ∂vx −2 2 (kλ ) −(kλ ) [5] L. J. Laslett, V. K. Neil, and A. M. Sessler, Rev. Sci. Instr. 6, 46 (1965). vx=ω/k d D [6] G. Besnier, Nucl. Instr. and Meth., 164, p. 235 (1979).

[7] Y.!" Chin,#! K.$% Satoh, and K. Yokoya, Part. Acc. 13, 45 (1983). Consider a perturbaon14 in

)*+,-.%57-*-/0.+1)% the electron distribuon !2)0-.%-*-/0.+1)% such that a plasma wave propagates with a phase velocity !r &% v ph = "% &'(% k Figure 14: Maxwellian velocity distribution function. Let us imagine plasma waves References as waves in the sea, and the [1] L. Landau, On the vibration of the electronic plasma. J. Phys. USSR 10 (1946), 25. English translation in JETP 16, 574. Reproduced in Col-electrons as surfers trying to lected papers of L. D. Landau, edited and with an introduction by D. Ter Haar, Pergamon Press, 1965, pp. 445460; and in Men of Physics:catch the L. wave, all moving in D. Landau, Vol. 2, Pergamon Press, D. ter Haar, ed. (1965). the same direcon. [2] J. H. Malmberg, and C. B. Wharton Phys. Rev. Lett. 13 184 (1964).

[3] V. K. Neil, and A. M. Sessler, Rev. Sci. Instr. 6, 429 (1965).

[4] V. G. Vaccaro, CERN Report ISR-RF/66-35 (1966).

[5] L. J. Laslett, V. K. Neil, and A. M. Sessler, Rev. Sci. Instr. 6, 46 (1965).

[6] G. Besnier, Nucl. Instr. and Meth., 164, p. 235 (1979).

[7] Y. Chin, K. Satoh, and K. Yokoya, Part. Acc. 13, 45 (1983). 57 14 22/09/11

Electrons slightly faster than vph are decelerated by the wave !"#!$% )*+,-.%-*-/0.+1)% electric ﬁeld and yield energy to !2)0-.%-*-/0.+1)% the wave. Electrons slightly slower

than vph are accelerated by the wave electric ﬁeld and gain energy "% & % &% '( from the wave. Figure 14: Maxwellian velocity distribution function.

References Since vph is in the negave slope of the velocity [1] L. Landau, On the vibration of the electronic plasma. J. Phys. USSR 10 (1946),distribuon 25. English translation in JETPfuncon 16, 574. Reproduced, the in Col- number of “faster” electrons is lected papers of L. D. Landau, edited and with an introduction by D. Ter Haar,less Pergamon than Press, 1965, the pp. 445460;number and in Men of Physics: of “ L. slower” ones. Hence, there are D. Landau,more Vol. 2, Pergamonparcles Press, D. ter Haar, gaining ed. (1965). energy from the wave than losing [2] J. H. Malmberg, and C. B. Wharton Phys. Rev. Lett. 13 184 (1964).

[3] V. K.to the Neil, and A. M. Sessler,wave Rev. Sci.. The balance Instr. 6, 429 (1965). is a net energy loss which leads [4] V. G.to Vaccaro,wave CERN Report damping ISR-RF/66-35 (1966).. [5] L. J. Laslett, V. K. Neil, and A. M. Sessler, Rev. Sci. Instr. 6, 46 (1965).

[6] G. Besnier, Nucl. Instr. and Meth., 164, p. 235 (1979).

[7] Y. Chin, K. Satoh, and K. Yokoya, Part. Acc. 13, 45 (1983).

3. Mechanical System Model

The demonstraon given by Landau was purely mathemacal, an experimental behaviour was observed only 18 years later. The basic physical mechanism behind was not well understood, and sll today several papers are devoted to a beer comprehension of Landau Damping.

We wonder how is it possible that for a collisionless, lossless system there exists a physical soluon for the oscillaons characterized by an exponenal decay corresponding to a damping.

15 22/09/11 Puòessere il tipo di accoppiamento tra il singolo oscillatore ed il sistema • a produrre una instabilitàche cresce esponenzialmente nel tempo (con accoppiamento di tipo capacitivo o induttivo con il giusto segno) To this end we consider a system of N uncoupled ideal linear fissato il tipo di accoppiamento, e quindi la crescita esponenziale dell’instabilità, oscillators, with a normalized frequency distribuon G(• ω). e lo spettro, e quindi il termine di damping di Landau, si puòdeter- minare il limite di instabilità

Shaking force lj at frequency Ω 4 A simple mechanical model

The Landau damping eﬀect has been derived fromωj=√ ag/ purelj mathematical ap-

G(ωproach,) = (dN/dω)/N and there are several aspects of its physics that are still surprising. We wonder, in fact, how it is possible that a collisionless, lossless system, perturbed from the equilibrium, can show such a behavior. Ω ω In order to get a physical insight in the mechanism which is responsible of the damping, despite the free loss nature of the system, we analyze now asimplemodelconsistingofaninﬁnitesetofharmonicoscillators,with frequency distribution G (ω), such that G(ω) = (dN/dω)/N

∞ G(ω)dω =1 (55) −∞ and with an average valueω ω0. We assumeω ωο Ω that the system is driven by an external sinusoidal force of driving force central frequency frequency Ω, and that the oscillatorsωο=( doΩ+ notω)/2 interact each other. resonator frequency For a single oscillator the diﬀerential equation of motion is !x( t = 0 ) = 0 2 2 !x!+! x = Acos!t x" + ω x = A cos Ωt (56) #x!( t = 0 ) = 0 With the starting conditions x(t =0)=0andx(t =0)=0,itssolution is A x(t>0) = (cos Ωt cos ωt)(57) −Ω2 ω2 − − 16 16 22/09/11

G(ω) = (dN/dω)/N A x( t > 0 ) = (cos!t " cos!t ) !2 "! 2 δω ω Ω ωο ω

Deﬁne now |Ω – ω|=δω, and assume that δω<<ωο

) " % " % , A !" !" x( t > 0 ) ! +cos$"0 + 't ( cos$!0 ( 't. 2!0!" *+ # 2 & # 2 & -.

We also assume that all the oscillator frequencies are sufficiently close each " % other and that Ω lies within the oscillatorA frequency spectrum. Moreover,!" x( t > 0 ) ! sin "0t sin$ t' we define the difference between the resonance frequency( Ω )and a resonator 2!0!" # 2 & frequency ω as δ = Ω ω ω , such that Ω + ω 2ω . ω − 0 0 Under these assumptions, equation (57) becomes

A δ δ x(t>0) cos ω + ω t cos ω ω t = −2ω δ 0 2 − 0 − 2 0 ω A δω sin ω0t sin t (58) ω0δω 2 that can be seen as an oscillation at frequency ω0 with an amplitude mod- ulated at the lower frequency δω/2. It is convenient to wrire equation (58) as 100 At sin δω t x(t>0) = 2 sin ω t (59) δω 0 2ω0 2 t

Let’s observe now50 the motion of two oscillators, the former with δω =0, and the latter with δ = 0, as shown in Fig. 4 with blue and red curves ω respectively. Both are at rest at t =0,andtheystarttooscillateduetothe action of the same external0 force. While the amplitude of the ”on resonance” oscillator growths linearly with time, the other reaches a maximum amplitude

(beating of two close frequency) after a time t = π/δω, (when sin(δωt/2) = -50 1), after which this oscillator is ”out of resonance”, and loses the phase synchronism with the external driving force. We can reverse this argument by saying that at-100 a time t∗, only those oscillatorst=π/δω with a frequency ω, such 0 10 20 30 40 50 that δω < π/t∗ maintain a phase relation with the external force. The longer we wait, the narrower the frequencyδω bandwidth=|Ω – δωωof|=0 synchronous oscillators. Therefore, at any instant t∗ we can divide the oscillators into two groups: the oscillatorsFigure having 4: frequencies Particle oscillations suchδω that=|δΩω < due– πω/t to∗|≠which0 a sinusoidal keep their external initial force ... . phase synchronism and their amplitude grow linearly with time; the oscillators withcontributeδω > π/t∗, to which absorb are no energy longer from in resonance the force, with thebut external their frequency force. bandwidth, The ”onand resonance” thus their oscillators number, are decreases in phase with with the time. external The force net and eﬀ theyect is an absorption

of energy by the system while17 the average amplitude of oscillation remains constant. To demonstrate that, let us calculate the average of the particle displace- 17 ments, given by

∞ A (t)= G(ω) (cos Ωt cos ωt) dω (60) − Ω2 ω2 − −∞ − Since Ω ω ω , we can justify the approximation − 0 A ∞ G(ω) (t) (cos Ωt cos ωt) dω (61) −2ω0 Ω ω − −∞ − If we make a change of variable from ω to δω,weget

A ∞ sin δωt (t) sin Ωt G(Ω δω) dδω 2ω0 − δω −∞ ∞ 1 cos δωt cos Ωt G(Ω δω) − dδω (62) − − δω −∞

18 22/09/11

100 The amplitude of ”on resonance”

50 oscillator with δω=0, blue curve, growths linearly with me. The 0 oscillator with δω≠0, red curve, -50 reaches a maximum amplitude -100 aer a me t=π/δ , aer which 0 10 20 30 40 50 ω it goes ”out of resonance”, and it loses the phase

Figure 4: Particlesynchronism oscillations due to a sinusoidal with the external forceexternal ... . driving force. contribute to absorb energy from the force, but their frequency bandwidth, and thus their number, decreases with time. The net eﬀect is an absorption of energy by the system while the average amplitude of oscillation remains ∗ constant. We can say that at any me t , only those oscillators with ∗ To demonstratea frequency that, let us calculate the averageω, ofsuch the particle displace-that δ < π/t maintain a phase ments, given by ω

∞ A relaon (t)= G(ω with ) (cos the Ωt cos ωexternalt) dω (60) force, and keep absorbing − Ω2 ω2 − −∞ − Since Ω ω ω , we can justify the approximation −energy 0 from the shaking force. A ∞ G(ω) (t) (cos Ωt cos ωt) dω (61) −2ω0 Ω ω − −∞ − If we make a change of variable from ω to δω,weget

A ∞ sin δωt (t) sin Ωt G(Ω δω) dδω 2ω0 − δω −∞ ∞ 1 cos δωt cos Ωt G(Ω δω) − dδω (62) − − δω −∞

The longer we wait, the narrower the frequency

bandwidth δω of synchronous oscillators, the less the number of oscillator absorbing energy.

The center mass (CM) of the oscillator’s system, inially at rest, will start oscillang with growthing amplitude which, however, will remain bounded. The CM posion is given by the average displacement obtained weighng x(t) with the normalized distribuon G(ω):

" A x ( t ) = ! G(! ) cos!t ! cos$t d! CM # 2 2 ( ) !" " !!

18 22/09/11

Since !" = !"" << "0 and 2 2 ! !"0 ! 2"0!"

We get: A & $ G(! ) ) x ( t ) G( )sin t cos t P.V. d CM ! (! " " # " % !+ 2!0 ' #$ ! #" *

The average oscillaon amplitude of the system does not increase with me, il remains limited as me goes to inﬁnity.

The masses oscillate inchoerently, the center of mass moon will be bounded.

Shaking force Shaking force at frequency Ω at frequency Ω

G(ω) = (dN/dω)/N

Ω ω

19 22/09/11

A & $ G(! ) ) x ( t ) G( )sin t cos t P.V. d CM ! (! " " + " % !+ 2!0 ' #$ ! #" *

Dispersion relaon? Let us assume that the shaking force is proporonal to the displacement of the center of mass.

Acos!t = " x e#i!t ( CM ) Asin!t = $ x e#i!t ( CM ) #i!t ( & + i t x e G(! ) x e# ! CM P.V. d i G( ) CM % * ' ! # " ! - 2!0 ) #& ! #! ,

Example - Uniform distribuon

G(ω) ! 1 # , !1 < ! < !2 G(! ) = " !2 !!1 # $ 0; elsewhere ω1 ω2

A * $" "#' - x ( t ) ! ,! sin#t + ln& 2 )cos#t / CM ( ) & ) 2!0 !2 "!1 +, % #"!1 ( ./

20 that represents an oscillation at frequency Ω with time independent amplitude. It is worth noting that the ”sine” term is responsible of the power absorption, because, by doing the time derivative of equation (67), we get the velocity of the average particle displacement, whose ”cosine” term is in phase with the external force producing an absorption of energy by the that represents an oscillation at frequency Ω with time independent ampli- system. that represents an oscillation at frequency Ω with time independent amplitude. ItThis istude. statement worth It noting is can worth be that demonstrated noting the that ”sine” by the reminding term ”sine” is term responsible that is the responsible energy ofU theof of an power the power absorption,harmonicabsorption, because, oscillator byis because, proportional doing bythe doing to time the the derivativesquare time of derivative the of oscillation equation of equation amplitude. (67), we(67), get we get From equationthe velocity (58) we of get the average particle displacement, whose ”cosine” term is the velocity of the average particle displacement, whose ”cosine” term is 22/09/11 in phase within phase the external with the force external producingA force2 producingδ an absorption an absorption of energy of energy by the by the U sin2 ω t (68) system. ω2δ2 2 system. ∝ 0 ω This statement can be demonstrated by reminding that the energy U of an Thiswhich statement leads to the can total be demonstrated energy of the system by remindingUtot that the energy U of an harmonic oscillator is proportional to the square of the oscillation amplitude. 2 2 δω A ∞ sin t harmonic oscillator isU proportionalN toG the(Ω squareδ ) of2 thedδ oscillation amplitude.(69) FromLooking equationtot at (58) the we2 getenergy absorbedω 2 ω by the system of ∝ ω0 − δω From equation (58) we get −∞ oscillators: 2 where N is the total number of particles of theA system.2 δω As time increases, the 2 U 2 2 sin t (68) 2 2 A ∝ ωδδ 2 function sin (xt/2)/x becomesU peakedsin around2 0ωωtx =0andtendstoaDirac(68) ∝ ω2δ2 2 delta functionwhich leads to the total energy0 ω of the system Utot sin2 xt/2 πt which leads to the total energylim of the system2 = δ(Ux)(70)tot 2 δω t x2 A ∞2 sin 2 t →∞U N G(Ω δ ) dδ (69) tot ω2 ω δ2 ω that, substituted in the integral2 ∝ of equation0 (69) gives−2 δω ω A ∞ −∞ sin 2 t Utot N G(Ω δω) dδω (69) where N is the total2 number of2 particles of2 the system. As time increases, the ∝ ω0 A π− δω U −∞N G(Ω)t (71) function sin2(xt/2)/xtot2 becomes∝ ω2 2 peaked around x =0andtendstoaDirac where N is the total number of particles0 of the system. As time increases, the Equationdelta (71) function shows that the energy of the system increases linearly with function sin2(xt/We2)/x ﬁnd2 becomes that it peakedgrowthssin around with me !!!! 2 xt/2 x =0andtendstoaDiracπt lim = δ(x)(70) time. This energy cannot be regardedt as thermalx2 energy2 of the system be- delta function →∞ cause is notthat, distributed substituted over in al thesin particles, integral2 xt/2 but ofπ equation itt is stored (69) in gives a time narrowing lim = δ(x)(70) range of frequencies aroundt the drivingx2 frequency.2 Only the particles with →∞ A2 π frequency such that x Ω < 1/t contributeUtot N to2 theG ”sine”(Ω)t response and are (71) that, substituted in the| integral− | of equation∝ (69)ω0 2 gives in resonance with the external force, but since their number decreases with Equation (71) shows thatA the2 π energy of the system increases linearly with time, the net contribution toU the averageN displacementG(Ω)t remains constant. (71) tot 2 time.4. Beams in parcle accelerators This energy cannot∝ beω0 regarded2 as thermal energy of the system because is not distributed over al particles, but it is stored in a time narrowing Equation (71) shows that the energy20 of the system increases linearly with range of frequencies around the driving frequency. Only the particles with time. This energyWe cannot consider be regarded a beam as thermal circulang energy of inside the system an be- frequencyaccelerator, such that x and Ω < assume 1/t contribute that to for the this ”sine” system response and are cause is not distributed over al| particles,− | but it is stored in a time narrowing in resonancethere exists an equilibrium state. with the external force, but since their number decreases with range of frequenciestime, the net around contribution the driving to the frequency. average displacement Only the remains particles constant. with frequency such that x Ω < 1/t contribute to the ”sine” response and are We | − wander | whether a small perturbaon in resonance witharound the external the force, equilibrium but since20 their state number will decreases grow with time, the net contribution(instability) or decay (stability). to the average displacement remains constant.

21 22/09/11

BEAM perturbaon

Beam-Fields Beam-Wall interacon interacon

e.m. ﬁelds

as a first example,as we a first consider example, we consider the the longitudinal longitudinal beam dynamics of beam a coast- dynamics of a coasting beam subjected to the space charge and smooth wall interaction forces ing beam subjectedonly. to Additionally, the space we assume that charge the beam current and is given smooth by a stationary wall interaction forces Longitudinal Instabilies in coasng Beamsconstant current I0 plus a sinusoidal perturbation ∆I of the kind i(ks ωt) only. Additionally, we assume thatI(s, t)=I0 the+ ∆Ie beam− current(74) is given by a stationary As shown in figure 4.1, the perturbation behaves like a wave traveling constant current I0alongplus the ring a moving sinusoidal with the same velocity perturbation of the charges. According∆ toI of the kind the notation adopted in particle accelerators, the longitudinal coordinate s represents the azimuthal position of the charge along itsi(ks orbit ofω radiust) R0. The wavelengthI of( thes, perturbation t)=I is a+ submultiple∆Ie of the machine− length (74) L =2πR 0 0 0 I(s,t=0) As shown in figure 4.1, the perturbation behaves like a wave traveling R0

Fsc Fsc along the ring moving with the same velocity ofΔI the charges. According to the notation adopted in particle accelerators, the longitudinal coordinate s

represents the azimuthal position of the charge along its orbit of radius R0. The wavelength of the perturbation is a submultiples of the machine length

The wavelength of the perturbaonFigure is 6: a Longitudinalsubmulple beam distribution of the machine for a coasting beam. length L such that k=2π/λ=2πn/L = n/R 0 L0, such that:0 0 I(s,t=0) 2π 2πn n k = = = (75) In the LHS picture the number of perturbaonλ wavelengthsL0 R0 in the ring Following from equation (74), also the electromagnetic ﬁelds produced is n=4, therefore k=4/R0 by the beam can be seen as a sum of those of the stationary distribution,

Fsc Fsc ΔI

Figure 6: Longitudinal beam distribution for a coasting beam.

L0, such that: 2π 2πn n k = = = (75) λ L0 R0 Following from equation (74), also the electromagnetic ﬁelds produced by the beam can be seen as a sum of those of the stationary distribution,

22 that we do not examine here because they modify the stationary distribution but do not lead to instabilities, plus those of the perturbation on which we focus our attention. These fields, acting back to the beam, cause an energy variation which can be positive in case of energy gain or negative in case22/09/11 of energythat we loss. do not Due examine to the here phase because relationship they modify the between stationary perturbation distribution current and perturbationbut do not lead electromagnetic to instabilities, plus forces, those of we the may perturbation have also on which a pure we ”reactive” focus our attention. These fields, acting back to the beam, cause an energy interaction, in which the net exchange of energy is zero. variation whichCoherent can be instabilies positive in are casecaused of energy by the gain orelectromagnec negative in case Inof order energy to loss.interacon calculate Due to the theof phase the ratebeam relationship of energyperturbaon between variation perturbationcurrent of a with single current the particle in one turnand perturbation due towalls the of the electromagnetic beam-wallvacuum electromagnetic forces, chamber we. may have interaction, also a pure ”reactive” we introduce the longitudinalinteraction, wake in which function the netW exchange(∆z), of defined energy is as: zero. In orderThe to calculatefield generates the rate|| by the of energybeam variation perturbaon of a single is modified particle in by the walls and causes e.m. forces, proporonal to the one turn due to the beam-wall electromagneticF (∆ interaction,z) L0 we introduce the current, thatW acts(∆ back on the z)= || beam. They can lead to a (76) longitudinal wake function W|| (∆z), defined− as:qq1 coherent instability|| .

F (∆z) L0 where F (∆z) is the electromagnetic|| force averaged along the accelerator || The averageW e.m(∆. force over z)= one turn is: (76) || − qq1 L0 where F (∆z) is the electromagnetic force1 averaged along the accelerator that we do not examine|| here becauseF they(∆z modify) = the stationaryF ds distribution (77) If|| we indicateL with0 E0 the|| beam nominal energy, and with ε = ∆E/E0 1 L00 but do not lead to instabilities,its plusF relative( those∆z) of= variation, the perturbationF andds if we on take which into we account(77) the effects of the fields || L || acting on the charge q1, q is the charge0 0 producing the electromagnetic fields, focus our attention. These fields,generated acting back by all to the particles beam, cause belonging an energy to the longitudinal perturbation, the acting on the charge q1, q is the charge producing the electromagnetic fields, ∆variationz is the which distance can be positivebetweenrate in caseq ofand relative of energyq1. energy gain variation or negativeε can in casebe written in terms of wake function of energy∆z loss.is the Due distance to the between phase relationshipq and q1. between perturbation current Equation (76) representsas the energy lost (in J/C) in one turn by q1 due Equation (76) represents the energy lost (in J/C) in onet turn by q1 due and perturbation electromagnetic forces, we may∂ε have∆ε also a puree ”reactive” i(ks ωt ) In order to calculate the rate = of energy variaonW (ct of ct) ∆ a Ie − dt (78) tointeraction, theto electromagnetic the in electromagnetic which the net exchange fields fields produced produced of energy∂t by isq by∆ zero.leadingt q −leadingE the0T0 motion the|| andmotion− passing and passing single parcle in one turn due to −∞ the beam-wall through a machine devicethat at an is, byearlier changing time t the such integration that ∆z variable,= c(t t). throughIn order a to machine calculate devicethe rate ofat energy an earlier variation time of at single such particle that in∆−z = c(t t). interacon, we introduce the longitudinali(ks wakeωt) 0 funcon − If we indicate with E the beam nominal∂ε energy,e∆Ie and− with ε = ∆E/E ω one turn due to the beam-wall electromagnetic0 interaction, we introduce the 0 i c y If we indicatedefined with asE the 0 theaverage beam energy nominal gain/= energy,loss per unit and charge withW (y). εe−= dy∆E/E0 (79) ∂t − cE0T0 || longitudinalits relative wake function variation,W and(∆z), if defined we take as: into account the effects of−∞ the fields || q q1 its relative variation, and ifHere weT0 takerepresents into the account beam revolution the effects time given of the by T fields0 = L0/c. generated by all the particles belonging to the longitudinal perturbation, the F (∆z) L0 W (∆z)=For ultra-relativistic|| charges, due(76) to the causality principle, the wake generatedrate of by relative all the energy particles|| variation belongingε can be written to the in longitudinal terms of wake perturbation, function the − qq1 function is zero ahead of the test particle so that the integral becomes c ratewhere ofasF relative(∆z) is the energy electromagnetic variation forceε can averaged be alongwritten the accelerator in terms of wake function || t ∂ε ∆ε e i(ks ωt ) = times the FourierW (ct transform ct) ∆Ie of the− wakedt field which,(78) by definition, is the L0 as ∂t ∆t −E0T01 || − F (∆zlongitudinal) = −∞ t couplingF ds impedance[15]; we(77) then get that is, by∂ε changing∆ε|| the integrationeL0 0 variable,|| i(i(ksks ωωtt)) = z=ctW (ct ∂εct) ∆e∆IeIe − dt (78) acting on the charge q , q is the chargei producing(ks ωt) 0 the electromagnetic fields, ∂t 1 ∆t −E0T0 || − =ω Z (ω)(80) ∂ε e∆Ie − i y || = −∞ W (y)∂et− c −dy E0T0 (79) ∆z is the distance between∂tq and− q1.cE0T0 || that is, by changing the integrationBefore concluding variable,−∞ this section we show some common longitudinal cou- Equation (76) represents theZ energyZ’ lost (in J/C) in one turn by q1 due pling impedancesi(ks 23ωt) that0 are generally found in a particle accelerator[10, 15]. to the electromagnetic fields∂ε producede∆Ie by q −leading the motion andi passingω y = W (y) e− c dy (79) For a perfectly conducting|| smooth and circular vacuum chamber of radius b, through a machine device∂t at an− earliercE time0T0t such that ∆z = c(t t). also the space charge−∞ effect due to− the non relativistic velocity of the charges If we indicate with E0 the beam nominal energy, and with ε = ∆E/E0 its relative variation, and if wecan take be into written account23 in terms the e offfects coupling of the impedance, fields and it gives 23 generated by all the particles belonging to the longitudinal perturbation, theR0ω b Z (ω)=iZ0 ln (81) || c(βγ)2 r rate of relative energy variation ε can be written in terms of wake function with Z the impedance of the free space, and r the transverse position of the as 0 t ∂ε ∆ε e test charge q1. i(ks ωt ) = W (ct ct) ∆Ie − dt (78) ∂t ∆t −E0T0 || − −∞In case of the resistive wall of the circular pipe with beam at center and that is, by changing the integration variable, 2 2 high conductivity σc, such that c /(ω b)andb are much bigger that the skin i(ks ωt) 0 ∂ε e∆Ie − i ω y = depth, we haveW (y) e− c dy (79) ∂t − cE0T0 || −∞ R Z ω Z (ω)= 0 0| | [1 i sign(ω)] (82) 23 || b 2cσc −

25 generated by all the particles belonging to the longitudinal perturbation, the rate of relative energy variation ε can be written in terms of wake function as ∂ε ∆ε e ∞ i(ks ωt ) generated by all the particles= belonging toW the( longitudinalct ct) ∆Ie perturbation,− dt the(78) ∂t ∆t −E0T0 || − rate of relative energy variation ε can−∞ be written in terms of wake function that is, by changing the integration variable, as ∂ε ∆ε e ∞i(ks ωt) ∂ε e∆Ie − ∞ i(ksi ωωyt) =If we= indicate withW (Ect0 thect beamW) ∆(Iey nominal) e−−c dy energy,dt and(78) with(79)ε = ∆E/E0 || ∂t ∆t ∂−t E0−T0 cE0T0 − || its relative variation,−∞ and if−∞ we take into account the effects of the fields that is, by changing the integration variable, Here T0 representsgenerated the by beamall the revolution particles belonging time given to the by longitudinalT0 = L0/(β perturbation,c). the i(ks ωt) For ultra-relativistic∂ε e∆Ie charge,− with∞β =1,duetothecausalityprinciple,thei ω y rate= of relative energy variationW (y)εecan− c bedy written in terms(79) of wake function ∂t − cE0T0 || wake functionas is zero ahead of the test−∞ particle so that the integral becomes ∂ε ∆ε e t i(ks ωt) Herec timesT0 represents the Fourier the transform beam revolution of the= wake time field givenW which, by(ctT 0 by=ct)L definition,∆0/Ie(βc).− isdt the (78) ∂t ∆t −E0T0 || − −∞ Forlongitudinal ultra-relativistic couplingthat charge, is, by impedance[15]; changing with β the=1,duetothecausalityprinciple,the integration we then get variable, wake function is zero ahead of the test particle soi(ks thatωt) the0 integral becomes ∂ε ∂εe∆Ieie(∆ksIeωt) − i ω y = = − Z (ω)(80)W (y) e− c dy (79) ∂t − cE0T0 || c times the Fourier transform∂ oft the− wakeE0 fieldT0 which,|| −∞ by definition, is the Here T0 represents the beam revolution time given by T0 = L0/c. longitudinalBefore coupling concluding impedance[15]; this section we we then show get some common longitudinal cou- 22/09/11 For ultra-relativistic charges, due to the causality principle, the wake pling impedances that∂ε are generallye∆Iei(ks foundωt) in a particle accelerator[10, 15]. function is= zero ahead− of theZ test(ω)(80) particle so that the integral becomes c For a perfectly conducting∂t − smoothE0T0 and circular|| vacuum chamber of radius times the Fourier transform of the wake field which, by definition, is the Beforeb, the concluding space chargelongitudinal this eff sectionect couplingdue we to show the impedance[15]; non some relativistic common we then velocity longitudinal get of the cou- charges

gives i(ks ωt) pling impedances that are generally found∂ε in a particlee∆Ie − accelerator[10, 15]. Impedance R=0ω b Z (ω)(80) Z (ω)=iZ0∂t − lnE0T0 || (81) For a perfectly conducting smooth|| and circularc(βγ)2 vacuumr chamber of radius b, the space charge eﬀectBefore due concluding to the non this relativistic section we velocity show some of common the charges longitudinal cou- with Z0 the impedance1) Perfectly of the free conducng space, and circularr the transverse beal pipe of positionradius of the b pling impedances that are generally found in a particle accelerator[10, 15]. gives test charge q1. For a perfectly conductingR0ω smoothb and circular vacuum chamber of radius b, In case of the resistiveZ (ω)= walliZ of0 the circular2 ln pipe with beam at center(81) and also the space|| chargec e(ﬀβγect) due tor the non relativistic velocity of the charges 2 2 high conductivity σc, such that c /(ω b)andb are much bigger that the skin with Z0 the impedancecan ofbe the written free in space, terms and of couplingr the transverse impedance, position and it gives of the depth, we have 2) Resisve wall circular beal pipe of radius b test charge q1. R ω b Z (ω)=iZ 0 ln (81) R Z ω 0 2 In case of the resistiveZ wall(ω)= of theFor0 circular a resonating0| ||| [1 pipei withsign( modec(βγ beamω)] of) a atr cavity, center the and longitudinal (82) coupling impedance || b 2cσc − with Z0 the impedance2 2 of the free space, and r the transverse position of the high conductivityFor a resonatingσc, such mode that ofcan ac / cavity,( beω b written)and the longitudinalb asare much bigger coupling that impedance the skin test charge q1. Rs depth,can be we written have as 3) RF Resonator mode Z (ω)= (83) In case of the resistive wall of the circular|| pipe1+ withiQ beamωr at centerω and R ω ωr R0 Z0 ω s − Z (ω)=Z (ω)= | | [1 i sign(2 ω2)] (82)(83) high conductivity|| σc, such25 thatωr c ω/(ω b)andb are much bigger that the skin || bwith1+2Rcσs iQc the− shunt resistance, Q the quality factor, and ωr the resonant ω ωr depth, we have − frequency. with Rs the shunt resistance, Q the quality factor, and ωr the resonant R0 Z0 ω NoteZ that(ω)= the sign| of| [1 thei imaginarysign(ω)] part of the coupling (82) impedance de- frequency. || b 2cσc − pends25 on the notations that has been used (j or i to indicate the imaginary Note that the sign of the imaginary part of the25 coupling impedance depends on the notations that hasterm). been used (j or i to indicate the imaginary

term). Consider now a parcle with nominal energy E0 which moves in 4.2 the circular Longitudinal machine beam with velocity dynamics βc on of a coasting beams closed orbit, called the reference orbit, of length 4.2 Longitudinal beam dynamics of coasting beams L0 = 2πR0. A particle with nominal energy E0 moves in the circular machine with ve-

A particle with nominal energylocityE0 βmovesc on a in closed the circular orbit, called machine the with reference ve- orbit, of length L0 =2πR0. A parcle with a small energy deviaon ∆E, with locity βc on a closed orbit, calledA particle the reference with a orbit, small of energy length deviationL =2πR∆E. , with ∆E = βc∆p, travels ∆E = βc∆p, travels along a different path with a 0 different0 A particle with aspeed small. The energyalongchange deviation a di ff∆erentω∆ of E path,its with revoluon with∆E a= diffβ erentfrequencyc∆p, speed. travels is The due change ∆ω of its revolu- along a different pathto a withcombinaon ation different frequency of speed.two is The due effects changeto a combination: ∆ the ω ofspeed its revolu- of two and eff the ects[16]: the speed and the tion frequency is duedispersion to a combinationdispersion, in the magnet of so two that efieldffects[16]:. the speed and the dispersion, so that: ω0 ω¯0 ∆ω 1 ∆p − = = αc 2 = ω¯0 ω¯0 − − γ p0 ω0 ω¯0 ∆ω 1 ∆p − = = αc 2 = 1 1 ∆E η ∆E η ω¯0 ω¯0 − − γ p0 = αc 2 2 = 2 = 2 ε (84) 1 1 ∆E η ∆−E −ηγ β E0 −β E0 −β = α = = ε(84) − c − γ2 β2 E −β2 E −β2 with ω ¯0 the revolution0 frequency0 of a particle with nominal energy E0, αc the

withω ¯0 the revolution frequencymomentum of a particle compaction with nominal (a property energy ofE0 the, αc guidethe fields) and γ the relativistic momentum compaction (a propertyfactor. of When the guideη > fields)0themachineworksabovethetransitionenergyanda and γ the relativistic 24 factor. When η > 0themachineworksabovethetransitionenergyandapositive deviation ε causes a longer trajectory which produces a reduction in positive deviation ε causes a longerthe revolution trajectory frequency. which produces a reduction in the revolution frequency. The change in the revolution frequency influences the longitudinal posi- The change in the revolutiontion frequency of a particle. influences If we the use longitudinal the quantity posi-z to define the longitudinal coor- tion of a particle. If we use thedinate quantity of a particlez to define with the respect longitudinal to the reference coor- one, which has a nominal dinate of a particle with respect to the reference one, which has a nominal 26 26 For a resonating mode of a cavity, the longitudinal coupling impedance can be written as R Z (ω)= s (83) || 1+iQ ωr ω For aω resonating− ωr mode of a cavity, the longitudinal coupling impedance with Rs the shunt resistance, Qcanthe be quality written factor, as and ωr the resonant R frequency. Z (ω)= s (83) || 1+iQ ωr ω Note that the sign of the imaginary part of the coupling impedance de-ω − ωr pends on the notations that has beenwith usedRs the (j or shunti to resistance, indicate theQ imaginarythe quality factor, and ωr the resonant frequency. term). Note that the sign of the imaginary part of the coupling impedance depends on the notations that has been used (j or i to indicate the imaginary 4.2 Longitudinal beam dynamics of coasting beams term).

A particle with nominal energy E0 moves in the circular machine with velocity βc on a closed orbit, called4.2 the reference Longitudinal orbit, of length beamL dynamics0 =2πR0. of coasting beams A particle with a small energy deviation ∆E, with ∆E = βc∆p, travels A particle with nominal energy E0 moves in the circular machine with ve- along a different path with a differentlocity speed.βc on a The closed change orbit,∆ calledω of theits revolu- reference orbit, of length L =2πR . 22/09/11 0 0 tion frequency is due to a combinationA particle of two with effects[16]: a small energy the speed deviation and the∆E, with ∆E = βc∆p, travels dispersion, so that along a different path with a different speed. The change ∆ω of its revolution frequency is due to a combination of two effects[16]: the speed and the ω0 ω¯0 ∆ω 1 ∆p − = = α dispersion,= so that ω¯ ω¯ − c − γ2 p 0 0 0 ω0 1 ω¯0 1 ∆∆Eω η ∆E 1 ∆ηp = αc − = == α = ε= (84) − − γω¯2 β2 Eω¯ −β2 cE− γ2 −pβ2 0 00 0 0 1 1 ∆E η ∆E η withω ¯ the revolution frequency of a particle with nominal= energyαc E , α the = = ε (84) 0 When η > 0 the machine works −above − the 0γ2 transionc β2 E −β2 E −β2 0 0 momentum compaction (aenergy property, a posive of the guidedeviaon fields) ε andcausesγ the a longer relativistic trajectory which produceswithω ¯0 a thereducon revolution in the frequencyrevoluon of a particle frequency with. nominal energy E0, αc the factor. When η > 0themachineworksabovethetransitionenergyanda momentum compaction (a property of the guide fields) and γ the relativistic positive deviation ε causesThe a longerchange trajectory in the revoluon which produces frequency a reduction influences in the longitudinalfactor. posion of a When η >parcle0themachineworksabovethetransitionenergyanda. We use the quanty z the revolution frequency. to definepositive the longitudinal deviation ε coordinate causes a longer of a trajectoryparcle with which produces a reduction in respect to the reference one, which has a nominal energy The change in the revolution frequencythe revolution influences frequency. the longitudinal posi- E0. tion of a particle. If we use the quantityThe changez to define in the the revolution longitudinal frequency coor- influences the longitudinal posi- dinate of a particle with respect totion the of reference a particle. one, If we which use the has quantity a nominalz to define the longitudinal coordinate of a particle with respect to the reference one, which has a nominal energy E0, we observe that a revolution frequency different fromω ¯0 produces 26 achangeinthelongitudinalpositionz in one turn given by the relation We observe that a revoluon frequency different26 from energy E0, we observe that a revolution frequency different∆z from∆ω ω ¯0 produces ω0 produces a change in the longitudinal= posion z in (85) as a first example, we consider the longitudinal beamL dynamics0 of aω¯ coast-0 achangeinthelongitudinalpositionone turn inggiven beam subjectedz by the relaon: in to theone space turn charge and givensmooth wall interaction by the forces relation from whichonly. Additionally, we assume that the beam current is given by a stationary constant∆z current I∆0 plusω a sinusoidal perturbation∆∆I zof the kind i(ks ωt) = I(s, t)=I0 + ∆Ie − = ∆ωR(74)0 (85) (86) LAs0 shown in figureω¯0 4.1, the perturbation behavesT0 like a wave traveling In thealong above the ring moving relations with the same velocity we of have the charges. assumed According to z>0 ahead of the reference from which For examplethe notation, the adopted inlongitudinal particle accelerators, the longitudinaleffect coordinate of the s space charge ∆representsz the azimuthal position of the charge along its orbit of radius R0. particle.in a perfectlyThe wavelength conducng of the perturbation is pipe a submultipleis of a force the machine lengthproporonal to = ∆ωR0 (86) T I(s,t=0) −∂I/∂s. Even though0 we start with a monochromatic beam, all the particles hav- In the above relations we have assumed z>0 ahead of the reference ing the same energy E , space charge and beam-wall interaction will produce 0 Fsc Fsc particle. ΔI electromagnetic forces that, interacting back on the beam, modify the parti- Even though we startcle with energy. a monochromatic beam, all the particles having the same energy E0, space charge and beam-wall interaction will produce For example, the longitudinal effects of the space charge in a perfectly electromagnetic forces that,conducting interacting pipe is back a force on the proportional beam, modify to ∂ theI/∂ parti-s[17]. As a consequence, Figure 6: Longitudinal beam distribution for a coasting beam. − cle energy. the particlesL0, such that that: are on the front slope of the sinusoidal perturbation will 2π 2πn n k = = = (75) λ L0 R0 For example, the longitudinalexperienceFollowing a e positiveffect from equation of force, the (74), also space the and, electromagnetic in charge one fields produced turn, in a their perfectly energy will increase. The25 by the beam can be seen as a sum of those of the stationary distribution, conducting pipe is a forcecontrary proportional will happen to to∂ theI/22 ∂ rears[17]. slope As of a the consequence, perturbation. If we are above − the particles that are ontransition, the front from slope equation of the (84),sinusoidal an increase perturbation of energy will implies a decrease of experience a positive force,the revolution and, in one frequency. turn, their Therefore energy the will particle increase. in the The front slope will delay contrary will happen toand the those rear in slope the back of the crest perturbation. will anticipate, If we giving, are asabove a net result, an increase transition, from equationof the (84), height an increase of the crest. of energy The initial implies perturbation a decrease is thus of increased leading to instability, known as negative mass instability. On the contrary, below the revolution frequency. Therefore the particle in the front slope will delay transition, the longitudinal space charge forces stabilize the beam. and those in the back crest will anticipate, giving, as a net result, an increase of the height of the crest. The initial perturbation is thus increased leading 4.3 Dispersion relation of longitudinal coasting beam to instability, known as negative mass instability. On the contrary, below transition, the longitudinalThis space dynamics charge of the forces coasting stabilize beam the can beam. be formalized and generalized by treating the motion of the particles by means of the Vlasov equation. The 4.3 Dispersion relationformalism is of very longitudinal similar to that we coasting have used beamfor the waves in a perturbed plasma. Here we use f(z,ε; t)forthebeamdistributionfunctionsuchthat This dynamics of the coasting beam can be formalized and generalized by 27 treating the motion of the particles by means of the Vlasov equation. The formalism is very similar to that we have used for the waves in a perturbed plasma. Here we use f(z,ε; t)forthebeamdistributionfunctionsuchthat

27 22/09/11

as a ﬁrst example, we consider the longitudinal beam dynamics of a coasting beam subjected to the space charge and smooth wall interaction forces only. Additionally, we assume that the beam current is given by a stationary

constant current I0 plus a sinusoidal perturbation ∆I of the kind

i(ks ωt) Parcles that on the I(s, t)= front I0 + ∆Ieslope− experience(74) a posive As shown in ﬁgure 4.1, the perturbation behaves like a wave traveling force, and, in along the ring movingone turn, with the sametheir velocity energy of the charges. will According increase to . The thecontrary notation adopted will in particlehappen accelerators, to the the longitudinalrear coordinate slopes of the perturbaonrepresents the. azimuthal position of the charge along its orbit of radius R0. The wavelength of the perturbation is a submultiple of the machine length

I(s,t=0)

Fsc Fsc ΔI

Figure 6: Longitudinal beam distribution for a coasting beam.

L0, such that: 2π 2πn n k = = = (75) λ L0 R0 Following from equation (74), also the electromagnetic ﬁelds produced by the beam can be seen as a sum of those of the stationary distribution, Above transion, an increase of energy implies a 22 decrease of the revoluon frequency. Therefore the as a ﬁrst example, we consider the longitudinal beam dynamics of a coast- parcle in the ing beam front subjected to theslope space charge will and smooth delay wall interaction and forces those in the only. Additionally, we assume that the beam current is given by a stationary back crest willconstant current ancipate, I0 plus a sinusoidal perturbationgiving∆I of, theas kind a net result, an

i(ks ωt) increase of the height of the I(s, t)=I0 + ∆Iecrest− . (74) As shown in ﬁgure 4.1, the perturbation behaves like a wave traveling along the ring moving with the same velocity of the charges. According to The inial perturbaonthe notation adopted in particle accelerators,is thus the longitudinal increased coordinate s leading to instability, knownrepresents the azimuthalas negave mass position of the charge along its orbitinstability of radius R0. . The wavelength of the perturbation is a submultiple of the machine length

I(s,t=0)

Fsc Fsc ΔI

Figure 6: Longitudinal beam distribution for a coasting beam.

L0, such that: 2π 2πn n k = = = (75) λ L0 R0 Following from equation (74), also the electromagnetic ﬁelds produced by the beam can be seen as a sum of those of the stationary distribution,

22 26 electromagnetic forces that, interacting back on the beam, modify the particle energy. For example, the longitudinal eﬀect of the space charge in a perfectly conducting pipe is a force proportional to ∂I/∂s[17]. As a consequence, − the particles that are on the front slope of the sinusoidal perturbation will experience a positive force, and, in one turn, their energy will increase. The contrary will happen to the rear slope of the perturbation. If we are above transition, fromelectromagnetic equation (84), forces an increase that, interacting of energy back implies on the a decrease beam, modify of the parti- the revolutioncle frequency. energy. Therefore the particle in the front slope will delay and those in theFor back example, crest will the anticipate, longitudinal giving, eﬀ asect a net of theresult, space an increase charge in a perfectly 22/09/11 conducting pipe is a force proportional to ∂I/∂s[17]. As a consequence, of the height of the crest. The initial perturbation is thus− increased leading to instability,the known particles as negative that are mass on the instability. front slope On of the the contrary, sinusoidal below perturbation will transition, theexperience longitudinal a positive space charge force, and,forces in stabilize one turn, the their beam. energy will increase. The By usingcontrary equation will (75) happen into (74), to the we can rear write slope of the perturbation. If we are above

transition, fromDispersion Relaon for coasng beams equation (84),i[kz an(ω increasenω¯0)t] of energy implies a decrease of 4.3 Dispersion relationI(z,t)=I0 of+ ∆ longitudinalIe − − coasting beam(89) the revolution frequency. Therefore the particle in the front slope will delay This dynamics of the coasting beam can be formalized and generalized by and also theand beam those distribution inThe thedynamics back function crest will of can a anticipate, thencoasng be written ( giving,unbunched as as a net) result,beam an can increase be treating the motion offormalized the particles by by meansmeans of of the the Vlasov equation.Vlasov equaon The . The i[kz (ω nω¯0)t] of the heightf(z,ε; oft)= thef crest.(ε)+f The(ε)e initial− − perturbation is thus(90) increased leading formalism is very similarformalism to that0 we is have very1 used similar for the to wavesthat inwe a perturbedhave used for the to instability, known as negative mass instability. On the contrary, below plasma.In writing Here we the use abovef(wavesz, equationsε; t)forthebeamdistributionfunctionsuchthat in a perturbed we have considered plasma. Here that thewe stationary use f(z,ε;t) for the transition,beam the longitudinal distribuon space funcon charge such forces that stabilize: the beam. itsdistribution integration does over not longitudinal depend either space on andtime energy or on z givesbeing the the total circular number machine of particlesazimuthallyN in symmetric. the beam BEAM PLASMA 4.3 Dispersion relation of longitudinal coasting beam The beam distribution function satisfies the Vlasov equation[3]f ( x,v that,t )d wex dv N f(z,ε; t)dzdε = N ! x (87) x = This dynamics of the coasting beam can be formalized and generalized by write here in the form F The beamtreating current theI, defined motion∂f ∂ by off ∂ equation thez particles∂f ∂ε (74), by can means be obtained! off the from! Vlasovf thex equation.!f The + + =0+ vx + (91) = 0 beam distributionformalism function is very∂ ast similar∂z ∂t to∂ε that∂t we have used! fort the! wavesx me in!v ax perturbed The termsplasma.∂z/∂t Hereand ∂ε we/∂ uset representf(z,ε; t the)forthebeamdistributionfunctionsuchthat rate of change of the longitudi- I(z; t)=ec f(z,ε; t)dε (88) nal positionits and integration energy of overthe particle. longitudinal The characteristicspace and energy time gives in which the totalall number of theHere involved we haveparticles variables performedN havein thea significant change beam of changes,variable from also ins to presencez = s ct ofassuming, instability, − fromis generally now on, much ultra-relativistic longer than velocities the revolution with β time=1.T0; therefore we may as- f(z,ε; t)dzdε = N (87) sume T0 to be the minimumThe beam time deviation.current can Under be thisobtained assumption from we can the beam 26 write The beamBydistribuon using current equationI funcon, defined (75) intoas by: equation (74), we (74),can write can be obtained from the ∂z ∆z beam distribution function= as∆ωR0 (92) ∂t T0 i[kz (ω nω¯0)t] I(z,t)=I0 + ∆Ie − − (89) where the last identity has been obtained by using equation (86). I(z; t)=ec f(z,ε; t)dε (88) The relative energyand also variation the beam in distribution one turn arises function from can the then longitudinal be written as

forces producedHere by the we have interaction performed of the a change beam withof variable the surroundings fromi[kz s(ωtonzω¯0=) andt] s ct assuming, f(z,ε; t)=f0(ε)+f1(ε)e − − − (90) by the spacefrom charge.By now using on,These equation ultra-relativistic forces vanish (75) into if velocities the (74), longitudinal we with canβ write=1. distribution is uniform along the acceleratorIn writing and, the as above we have equations shown, we they have can considered be expressed that the stationary i[kz (ω nω¯0)t] I(z,t)=I0 +26∆Ie − − (89) by means of the longitudinaldistribution wakedoes not function[15]. depend either on time or on z being the circular machine As a matterand of alsoazimuthally facts, the by beam applying symmetric. distribution equation (88), function the instantaneous can then be current written as can also be written asThe beam distribution function satisfies the Vlasov equation[3] that we i[kz (ω nω¯0)t] write here in thef( formz,ε; t)=f0(ε)+f1(ε)e − − (90) ecN i[kz (ω nω¯0)t] I(z; t)= + ece − − f1(ε)dε (93) L0 ∂f ∂f ∂z ∂f ∂ε In writing the above equations+ we have+ considered=0 that the stationary (91) ∂t ∂z ∂t ∂ε ∂t distribution does not depend27 either on time or on z being the circular machine The terms ∂z/∂t and ∂ε/∂t represent the rate of change of the longitudi- azimuthally symmetric. nal position and energy of the particle. The characteristic time in which all 27 The beam distribution function satisfies the Vlasov equation[3] that we the involved variables have significant changes, also in presence of instability, write here in the form is generally much longer than the revolution time T0; therefore we may as- ∂f ∂f ∂z ∂f ∂ε sume T0 to be the minimum+ time deviation.+ =0 Under this assumption we (91) can ∂t ∂z ∂t ∂ε ∂t write The terms ∂z/∂t and ∂ε/∂t represent∂z ∆z the rate of change of the longitudi- = ∆ωR0 (92) ∂t T0 nal position and energy of the particle. The characteristic time in which all where the last identity has been obtained by using equation (86). the involvedThe variables relative energy have significant variation in changes, one turn also arises in presence from the of longitudinal instability, is generallyforces produced much longer by the than interaction the revolution of the beam time withT0; therefore the surroundings we may and as- sumebyT0 theto be space the charge. minimum These time forces deviation. vanish if Under the longitudinal this assumption distribution we can is writeuniform along the accelerator and, as we have shown, they can be expressed ∂z ∆z by means of the longitudinal wake function[15].= ∆ωR0 (92) ∂t T0 As a matter of facts, by applying equation (88), the instantaneous current where the last identity has been obtained by using equation (86). can also be written as The relative energy variation in one turn arises from the longitudinal

ecN i[kz (ω nω¯0)t] forces produced by theI(z interaction; t)= + ofece the− beam− withf1 the(ε)d surroundingsε (93) and L 0 by the space charge. These forces vanish if the longitudinal distribution is 27 uniform along the accelerator and, as we have shown, they can be expressed by means of the longitudinal wake function[15]. As a matter of facts, by applying equation (88), the instantaneous current can also be written as

ecN i[kz (ω nω¯0)t] I(z; t)= + ece − − f (ε)dε (93) L 1 0 27 By using equation (75) into (74), we can write

i[kz (ω nω¯0)t] I(z,t)=I0 + ∆Ie − − (89)

and also the beam distribution function can then be written as

i[kz (ω nω¯0)t] f(z,ε; t)=f0(ε)+f1(ε)e − − (90)

By using equationIn writing (75) the into above (74), equations we can wewrite have considered that the stationary distribution does not depend either on time or on z being the circular machine i[kz (ω nω¯0)t] I(z,t)=I + ∆Ie − − (89) azimuthally symmetric.0 andBy also using the beam equationThe distribution beam (75)distribution into function (74),function can we thensatisfies can be write the written Vlasov as equation[3] that we write here in the form i[kz i[(kzω n(ω¯ω0)t]nω¯0)t] f(z,ε; It)=(z,tf0)=(ε)+∂fI0f1+(∂εf)∆e∂zIe−∂f−−∂ε − (90) (89) + + =0 (91) ∂t ∂z ∂t ∂ε ∂t In writing the above equations we have considered that the stationary and also the beamThe terms distribution∂z/∂t and ∂ε function/∂t represent can the then rate of be change written of the as longitudi- distribution does not depend either on time or on z being the circular machine nal position and energy of the particle. The characteristic time in which all azimuthally symmetric. i[kz (ω nω¯0)t] the involvedf( variablesz,ε; t)= havef significant0(ε)+f1 changes,(ε)e also− in− presence of instability, (90) The beam distribution function satisfies the Vlasov equation[3] that we is generally much longer than the revolution time T0; therefore we may as- writeIn here writing in the the form above equations we have considered that the stationary sume T0 to be the minimum time deviation. Under this assumption we can distributionwrite does not depend∂f ∂ eitherf ∂z on∂f ∂ε time or on z being the circular machine 22/09/11 + +∂z ∆z =0 (91) ∂t ∂z ∂t ∂ε ∂t = ∆ωR0 (92) azimuthally symmetric. ∂t T0 The terms ∂z/∂t and ∂ε/∂t represent the rate of change of the longitudi- The beamwhere distribution the last identity function has been satisfiesobtained by the using Vlasov equation equation[3] (86). that we nal position andThe energy relative of the energy particle. variation The in characteristic one turn arises time from in the which longitudinal all write here in the form the involvedforces variables produced have significantby the interaction changes, of also the beam in presence with the of surroundings instability, and is generallyby much the longer space charge. than∂f the These revolution∂f forces∂z vanish time∂f if∂εT the; therefore longitudinal we maydistribution as- is + + 0 =0 (91) sume T0 to beuniform the minimum along the accelerator time∂t deviation.∂z and,∂t as Under we∂ε have∂ thist shown, assumption they can we be expressed can writeThe termsby means∂z/∂ oft and the longitudinal∂ε/∂t represent wake function[15]. the rate of change of the longitudi- As a matter of facts,∂z by∆ applyingz equation (88), the instantaneous current = ∆ωR0 (92) nal position and energy of∂t the T particle.0 The characteristic time in which all can also be writtenAs we have as shown in section (4.1), the rate of energy variation is then where the last identity has been obtained by using equation (86). the involved variables have significanti(ecNks ωt) changes,2 i also[kz (ω innω¯0) presencet] of instability, As we∂ε have showne∆Ie in section− (4.1),i[kz ( theω cenω¯ ratee0)t] − of− energy variation is then =I(z; t)= +Zece(ω)=− − f1(ε)dεZ (ω) f1(ε)d(93)ε (94) The relative energy variation inL one turn|| arises from the longitudinal|| ∂t −i(ks Eω0t)T00 2−i[kz (ω En0ω¯T0)0t] is generally much∂ε longerAse∆ thanIe we have− the shown revolution in sectionce e − (4.1), time− theT0 rate; therefore of energy variation we may is then as- forces produced by the=We interaction now linearize ofZ the(ω)= Vlasov beam equation with the by surroundings substitutingZ (ω) f1 equation(ε) anddε (90)(94) into ∂t − E0T0 || −27 E0T0 || i(ks ωt) 2 i[kz (ω nω¯0)t] sume T to be the minimum∂ε timee∆Ie deviation.− Underce e − this− assumption we can by the0 space charge.We(91) Thesenow and linearizeAs ignoringforces we= have the vanish second shown Vlasov if order in equation the sectionZ ( termsω longitudinal)= (4.1), by in substituting the the perturbation, rate distribution of energy equationZ that( variationω) (90)is isf the into1( isε) term thendε (94) ∂t − E0T0 || − E0T0 || that contains (∂f /∂εi)(ks fωt)(ε)dε, with the2 i[ usekz (ω ofnω¯ the0)t] equations (75), (92), writeuniform along the(91) accelerator and ignoring∂ε and, seconde as∆1 Ie we order have− 1 terms shown, in thece they perturbation,e can− − be expressed that is the term We now= linearize theZ Vlasov(ω)= equation by substitutingZ (ω) f1 equation(ε)dε (90)(94) into and (94),∂ andt obtain− E0T0 || − E0T0 || by means of thethat longitudinal contains (∂ wakef1/∂∂εz function[15].) f1(∆ε)dzε, with the use of the equations (75), (92), (91)We and now ignoring linearize second the= Vlasov order∆ω equationR terms0 in by the substituting perturbation, equation that (90) is the into(92) term and (94), and obtain∂t T0 i[kz (ω nω¯0)t] As a matter of facts, byi (ω applyingnω¯0 n equation∆ω) f1e (88),− − the= instantaneous current −that(91) contains− and ignoring− (∂f1 second/∂ε) orderf1(ε)d termsε, with in the the perturbation, use of the equations that is the (75), term (92), 2 i[kz (ω nω¯0)t] ∂f e cZ (nω¯ ) where the last identityi (ω n hasω¯ n been∆ω) f obtainede − − by= 0 using equation0 i[kz (ω nω¯ (86).0)t] can also be written as andthat (94),0 contains andobtain (∂1f1/∂ε) f1(=ε)dε, with|| the usee of− the− equationsf d (75),ε (95) (92), − − − ∂ε E T 1 ∂f e2cZ (nω¯ 0) 0 and (94), and obtain 0 0 i[kz (ω nω¯0)t] The relative energyecN variation in one= i[kz turn(ω ||nω¯ arises0)t] e − from− thef d longitudinalε (95) To firsti (ω ordernω¯ of0i[ perturbationkzñ∆(ωω)nωf¯10)et] the∞− coupling− ∞ = impedancei(ωt kx) has been1 evaluated I(z; t)=− +− ece −φ−(k,−ω)=∂ε fE10(Tε0)φdε(x, t)e − dxdt(93) (25) L i[kz (ω nω¯0)t] i0(ω nω¯ n∆ω) f e − − = 2 forces produced byat the the unperturbed interaction0 frequency of the1 nω¯0−∞ beam. −∞∂f0 withe cZ ( thenω¯0) surroundings and To first order− of− perturbation− the coupling impedance|| hasi been[kz (ω evaluatednω¯0)t] = 2 e − − f1dε (95) and for the differential equation (22)∂f∂ε0 e cZ E(n0ω¯T00) at the unperturbedBy using now frequency equation27 nω¯ (84). the above equation|| cani[kz be(ω writtennω¯0)t] as by the space charge. These forces vanish0 if= the longitudinale − − distributionf1dε (95) is To first order of perturbation the∂ε2 coupling2eE0T0∂f impedance has been evaluated By using now equation (84)i( thekv∂f0/ above∂εω)f˜ equation+e ic Z k( cannφ˜ω¯0) be0 =0 written as (26) To first orderf1 = ofi perturbationx the1 coupling|| impedancef1dε has been evaluated(96) uniform along the acceleratorat the unperturbed and,ω as frequencyn weω¯0 +− havenω¯n0ω¯ηε. shown,mEe0L0 ∂ theyvx can be expressed ∂f /∂ε e2c20Z (nω¯ ) at the unperturbed frequency−0 nω¯ . 0 IfAccordingly, we integratef1 = i eq. both (23) the becomes: members0 by|| ε, andf use1dε the definition(96) of the By using nowω n equationω¯0 + nω¯0 (84)ηε theE0 aboveL0 equation can be written as by means of the longitudinalBy using wake now− equation function[15]. (84) the above equation can be written as average current of equation (93) then we obtain2 2 the dispersion integral If we integrate both the members∂f by0/∂εε,e ande usec Z the(nω¯ definition0) of the 2 ˜ 2 2˜ || f1 = i ∂fk0/φ∂ε= e cfZ1dv(nxω¯0) f1dε (27)(96) As a matter of facts, by applyingf = equationi(ωZ /n−n)ω¯I00L+0 (88),nω¯ε0ηε∂f the0/∂ε||E instantaneous0L0 f dε current(96) average current of equation1=1 (93)i then|| we obtain0 the dispersiondε integral1 (97) ω −nω¯0 + nω¯0ηε(ω nω¯E0)0L0 2πN(E0/e)η − If we integrate both− the membersnω¯ byη ε+,ε and use the definition of the can also be written asIf we take f˜ from(Z (26)/n)I0 andL0 substitute∂f0/∂ε0 into (27), we obtain the following If we integrate1=1 i both|| the members by ε,d andε use the definition(97) of the (ω nω¯0) averageObserve currentthat the of above2π equationN(E dispersion0/e)η (93) integral, then− we+ derived obtainε from the dispersion the Vlasov integral equa- dispersionaverage relation current of equation (93) thennω¯0 weη obtain the dispersion integral ecN i[kz (ω2 nω¯ )t] tion, has a very close similarity( toZe the/n) equationI00L∂f0 0/∂v (28)x ∂f0 obtained/∂ε for the plasma ObserveI(z; t)= that the above+ ece dispersion−(Z integral,/n− )I0L0 derivedf∂1f(0 from/ε∂ε)d theε Vlasov equa- (93) 1=1+i |||| dvx =0dε (28)(97) 1=i (ω nω¯0) dε (97) oscillations.L As0 in that case, weε20πm nowNek(E know0/eω)η thatkv(ω xn weω−¯0) must execute the above tion, has a very close similarity to the2πN equation(E0/e)η (28)− obtainednω¯ η+ ε+ forε the plasma − nω¯0η 0 Integration of (28) over v provides a relation between k and ω which de- oscillations.integralObserve AsObserve in inthe that complex that that case, theε weabove-plane above now dispersionx dispersionby know deforming that integral, integral, we the must contourderived derived execute from of fromthe the the integration, theVlasovabove Vlasov equa- equa- pends only on the slope27 of the8 unperturbed distribution function f (v ). The integralin orderintion,tion, the hasto complex has avoid a a very very theε-plane close singularity similarity similarity by deforming. toIf to we the the the do equation equation that, contour by (28) (28)usingof obtained the obtained the integration, same for the for rule0 plasma the ofx plasma in orderequationdispersionoscillations. tooscillations. avoid (43), relation the we singularityAs As get in contains that that8 case,. case, If a we we divergent we donow now that, know know integral, by that using that we because we themust must same execute of execute rule the the of singularity above the above

equationat ω (43),integral= kv we(Zx. inget/n To the)I0 overcomeL complex0 ε-plane this∂ dif byﬃ0/culty,∂ε deforming without the∂ contourf giving0 of a thesolid integration, explanation, integral1=i in|| the complexP.V.ε-plane by deformingdε iπ the contour of the integration,(98) 28 (ω nω¯80) (nω¯ ω) 2πN(E0/e)η − + ε − ∂ε 0− Vlasovin order calculated to avoid the the principal singularitynω¯ valueη.8 If weof the do that, integral, by usingε getting,= nω¯ theη same as result, rule of only in(Z order/n)I to0L0 avoid the singularity∂f0/∂ε0 . If we do∂f that,0 by using0 the same rule of 1=i || P.V. dε iπ (98) 8 equation (43), we get (ω nω¯0) afrequencyshiftwithoutanykindofdamping.Observe that the singularity exists only if the frequency shift (ωnω¯0 nωω¯) due to the cou- equation2πN(E0/e (43),)η we get − + ε − ∂ε ε= − 0 nω¯0η n−ω¯0η pling impedance lies within the frequency spread due to the energy distribution. Outside 8 (Z /n)I0L0 ∂f0/∂ε ∂f0 Observe that1= thei singularity(Z ||/n)I existsL P.V. only if the frequency∂f /∂εd shiftε iωπ nω¯0 ∂duef to the cou- (98) 0 0 (ω nω¯0)0 0 this frequency rangeπN|| thereE /e isη no Landau damping. − ∂ε (nω¯0 ω) 2.2.31= Landaui 2 ( solution0 ) P.V. of the− Vlasov+ ε equationd−ε iπ ε= − (98) pling impedance lies within the frequency spread (nω dueω¯0ηnω¯ to0) the energy distribution. nω¯ Outside0(ηnω¯ ω) 2πN(E0/e)η − + ε − ∂ε ε= 0− nω¯0η nω¯ η 8 0 this frequency rangeObserve there that is the no Landausingularity damping. exists28 only if the frequency shift ω nω¯0 due to the cou- In a8 veryObserve original that the paper singularity of 1946 exists Landau only if the proposed frequency ashift new− ω methodnω¯ due to of the solu- cou- pling impedance lies within the frequency spread due to the energy distribution.− 0 Outside tionplingthis of frequencyimpedance Vlasov-Poisson range lies therewithin equations is the no Landau28 frequency putting damping. spread the due basis to the of energy the theory distribution. of plasma Outside oscillationsthis frequency and range instabilities[1]. there is no Landau He damping.demonstrated that the problem had to 28 be considered as an initial value or Cauchy28 problem, with a perturbation

∞ ikx f˜1 (vx,k,t)= f1 (x, vx,t) e− (29) −∞

∞ ikx E˜x (k, t)= Ex (x, t) e− (30) −∞ 9 As we have shown in section (4.1), the rate of energy variation is then

i(ks ωt) 2 i[kz (ω nω¯0)t] ∂ε e∆Ie − ce e − − = Z (ω)= Z (ω) f1(ε)dε (94) ∂t − E T || − E T || 0 0 0 0 We now linearize the Vlasov equation by substituting equation (90) into (91) and ignoring second order terms in the perturbation, that is the term

that contains (∂f1/∂ε) f1(ε)dε, with the use of the equations (75), (92), and (94), and obtain

i[kz (ω nω¯0)t] i (ω nω¯ n∆ω) f e − − = − − 0 − 1 e2cZ (nω¯ ) ∂f0 0 i[kz (ω nω¯0)t] = || e − − f dε (95) ∂ε E T 1 0 0 To ﬁrst order of perturbation the coupling impedance has been evaluated

at the unperturbed frequency nω¯0. By using now equation (84) the above equation can be written as 2 2 ∂f0/∂ε e c Z (nω¯0) f = i || f dε (96) 1 ω nω¯ + nω¯ ηε E L 1 − 0 0 0 0 If we integrate both the members by ε, and use the definition of the average current of equation!"# (93)$%&' then we obtain the dispersion integral (Z /n)I0L0 ∂f0/∂ε 1=i || dε (97) (ω nω¯0) 22/09/11 2πN(E0/e)η − + ε nω¯0η Observe that the above dispersion integral, derived from the Vlasov equation, has a very close similarity to the equation (28) obtained for the plasma oscillations. As in that case, we now know that we must execute the above integral in the complex ε-plane by deforming the contour of the integration, in orderCoasng to avoid Beams the singularity 8. If we do that, by using()# the$ same%&' rule of equation (43), we get *+,-' (Z /n)I0L0 ∂f0/∂ε ∂f0 1=i || P.V. dε iπ (98) (ω nω¯0) (nω¯ ω) 2πN(E0/e)η − + ε − ∂ε ε= 0− nω¯0η nω¯0η Figure 2: Path of integration for Landau damping. 8Observe that the singularity exists only if the frequency shift ω nω¯ due to the cou- − 0 pling impedancePlasma lieswaves within the frequency spread due to the energy distribution. Outside Accordingly, the dispersion function becomes: this frequency range there is no Landau damping. 2 e ∂f0/∂vx iπ ∂f0 1+ P.V. 28dvx =0 (43) ε0mek ω kvx − k ∂vx − vx=ω/k The imaginary term of the above equation produces the damping/antidamping 4.4effect predicted Monocromatic by Landau, beam depending on the slope of the distribution func- 4.4 Monocromatic beam Lettion. us With use the this dispersion procedure, integral we obtain to discuss straightforwardly the stability the of a correct monochromatic dispersion 4.4 MonocromaticLet us use the dispersion beam integral to discuss the stability of a monochromatic beam,relations namely via Fourier a beam transformation without energy of spread.the Vlasov In equation. this case the stationary beam, namely a beam without energy spread. In this case the stationary Let usdistribution use the dispersion can be written integral as to discuss the stability of a monochromatic distributionExample: Plasma can be written with a as Maxwellian velocity distribution beam, namely a beam withoutMonochromac beam energy spread.δ(ε) In this case the stationary As an example to clarify thef ( useε)= ofN theδ( dispersionε) relation for the analysis(99) 0f (ε)=N (99) distribution can be written as 0 L0 of the plasma stability, we consider a plasmaL0 with a velocity Maxwellian with δ the Dirac delta function. Since in this case there is no energy spread, with δ the Dirac delta function. Since inδ( thisε) case there is no energy spread, thedistribution dispersion function integral doesf not0(ε diverge,)=N and we can use directly equation(99) the dispersion integral does not diverge,L and0 we can use directly equation 2 (97).(97). With With the the above above relation, relation, we we getn get0 mevx with δ the Dirac deltaf0 function.(vx)= Since in this1/2 exp case there is no energy spread,(44) (2πkBT/me) −2kBT (Z(/nZ /n)I0)I0 δδ(ε()ε) ((ZZ /n/n)I0 ∂∂ 11 the dispersion1=1=i i || integral|| does not diverge,ddεε== ii and|||| we can use directly equation where kB is the Boltzmann(ω(ωnω¯n0ω¯) constant.0) We can integrate by((ωω partsnnω¯ω0¯)0) the principal 2π(2Eπ0(E/e0)/eη)η − − ++εε −− 22ππ((EE00/e)η ∂ε∂ε −− ++εε (97). With the above relation, nω¯n0ω¯η0η we get nnω¯ω¯0η0η ε=0ε=0 value of equation (43) obtaining (100)(100) thatthat( isZ is/n)I0 δ(ε) (Z /n)I0 ∂ 1 ∂f0/∂vx f0(vx) ∞ 2 f0(vx) 1=i || ηd(Zε =/n)I0i ||nω¯0 2 P.V. (ω nω¯01=) dvηxi(=Z /n|| )I0 nω¯0 k (ω nω¯dv0) x (101)(45) 2π(E0/e)η − 1=+i ε || − 2π(E0/e)η ∂ε − 2 + ε(101) ω nω¯0kvη x 2π(ωE0/ekv) x ω n−ω¯0 (ω kvnxω¯)0η ε=0 2π(E0/e) ω nω¯0 − − −∞− − (100) so that the frequency is − so that the frequency is that is 12 η(Z /n)I0 2 ωη=(Znω¯/n0 )In0ω¯0 iη(nZω¯||0/n)I0 (102) 1=ω i= nω¯||0 ±nω¯0 i 2π(||E0/e) (102)(101) 2π(E±0/e) ω2π(nEω¯0/e0 ) When ω has an imaginary part ωi, we obtain− a perturbation with a time so thatWhen the frequencyω has an isimaginary part ω , we obtain a perturbation with a time exponential growing amplitude thati leads to instability (actually in the above exponential growing amplitude that leads to instability (actually in the above equation there is a second solution that producesη(Z /n an)I exponential0 decay) . The || equation there is a secondω = solutionnω¯0 n thatω¯0 producesi an exponential decay) . The(102) real part of ω, ωr, gives the frequency± of2 theπ(E perturbed0/e) current term. If we 29 real part of ω, ω , gives the frequency of the perturbed current term. If we ignore the machiner coupling impedance Z this frequency is nω¯0. When ω has an imaginary part ωi, we obtain|| a perturbation with a time ignoreIf the the machine machine coupling coupling impedance impedance ZZ hasthis a frequency real part, that is nω¯ is0. a resistive exponential growing amplitude that leads to|||| instability (actually in the above component,If the machineω will coupling always impedance have an imaginaryZ has a part real and part, therefore that is the a resistive beam equation there is a second solution that produces|| an exponential decay) . The component,will be unstable.ω will always For a pure have imaginary an imaginary impedance part andZ = thereforeiZ ,i, stability the beam or || || real part of ω, ωr, gives the frequency of the perturbed current term. If we willinstability be unstable. will depend For a pure on the imaginary sign of η impedanceand Z ,i. AboveZ = transitioniZ ,i, stability energy or || || || ignore the(η > machine0), if Z coupling> 0(acapacitiveimpedanceduetothespacechargeasinthe impedance Z this frequency is nω¯0. instability will,i depend on the sign of η and|| Z ,i. Above transition energy || || previous example) we find the negative mass instability. The overall behavior If(η the> 0), machine if Z ,i > coupling0(acapacitiveimpedanceduetothespacechargeasinthe impedance Z has a real part, that is a resistive || || can be summarized by saying that when ηZ ,i < 0, the beam is stable. component,previousω example)will always we find have the negative an imaginary mass|| instability. part and The therefore overall behavior the beam

will becan unstable. be summarized For aby pure saying imaginary that when29 impedanceηZ ,i < 0, theZ beam= iZ is,i stable., stability or || || || instability will depend on the sign of η and Z ,i. Above transition energy 29 || (η > 0), if Z ,i > 0(acapacitiveimpedanceduetothespacechargeasinthe || previous example) we ﬁnd the negative mass instability. The overall behavior

can be summarized by saying that when ηZ ,i < 0, the beam is stable. || 29 4.4 Monocromatic beam

Let us use the dispersion integral to discuss the stability of a monochromatic beam, namely a beam without energy spread. In this case the stationary distribution can be written as δ(ε) f0(ε)=N (99) L0 with δ the Dirac delta function. Since in this case there is no energy spread, the dispersion integral does not diverge, and we can use directly equation (97). With the above relation, we get

(Z /n)I0 δ(ε) (Z /n)I0 ∂ 1 1=i || dε = i || (ω nω¯0) (ω nω¯0) 22/09/11 2π(E0/e)η − + ε − 2π(E0/e)η ∂ε − + ε nω¯0η nω¯0η ε=0 The equation (102) can also be used to determine the instability(100) growth rate once we know the machine coupling impedance. For example, if we plot that is sign(η)Z ,i as a function of sign(η2)Z ,r at constant values of ωi,weobtain − η(Z /n|| )I0 nω¯0 || 1=thei ﬁgure|| (7) that represents the stability diagram for zero energy(101) spread. 2π(E0/e) ω nω¯0 Observe that ωi is related− to the instability rise time by the relation: so that the frequency is 1 ω = (103) i τ η(Z /n)I0 ω = nPositiveω¯0 n valueω¯0 ofiτ produce|| instability and the curves allow(102) to evaluate ± 2π(E0/e) the rise time once the coupling impedance is known.

When ω has an imaginary part ωi, we obtain a perturbation with a time -sign(η)Z ||,i ω =0 increasing ωi exponential growing amplitude that leads to instabilityi (actually in the above equation there is a second solution that producesstability an exponential decay) . The

real part of ω, ωr, gives the frequency of the perturbed current term. If we

ignore the machine coupling impedance Z this frequency is nω¯0. || If the machine coupling impedance Z has a real part, that is a resistive || component, ω will always have an imaginary part and therefore thesign( beamη)Z ||,r

will be unstable. For a pure imaginary(arbitrary units) impedance Z = iZ ,i, stability or || || instability will depend on the sign of η and Z ,i. Above transition energy || (η > 0), if Z ,i > 0(acapacitiveimpedanceduetothespacechargeasintheFigure 7: Stability diagram relating growth rate and impedance for zero || previous example) we ﬁnd theenergy negative spread (in mass arbitrary instability. units). The overall behavior

can be summarized by saying that when ηZ ,i < 0, the beam is stable. || 4.5 Beam with energy spread 29 Till now we have seen that a monochromatic beam is stable only when the machine impedance is purely imaginary with a proper sign. This, however,

If the machine coupling impedance30 has a real part, that is a resisve component, ω will always have an imaginary part and therefore the beam will be unstable.

For a pure imaginary impedance stability or instability depends on the sign of η and Zi.

Above transion energy (η>0), the beam is unstable if Zi>0 (capacive impedance) and stable if Zi<0 (inducve impedance).

30 -sign(η)Z ||,i

ωi=0 increasing ωi stability

sign(η)Z ||,r we obtain the dispersion relation 22/09/11 (arbitrary units) 2 1 2 1/2 4(Z /n)I0β x (1 x ) 1/2 || 2 1= i 2 2 P.V. − dx + iπy 1 y (110) − π (E0/e)ηεm 1 y + x-sign(η)Z − − ||,i Figure 7: Stability diagram relatingwith y growthgiven rate by andequation impedance (106). for The zero P.V. of the integral can be easily done, energy spread (in arbitrary units).and weParabolic get energy distribuon increasing ωi 2 The relave energy deviaon ranges4(Z /n between)I0β −ε and ε 1/2 1 astationaryparabolicenergydistributionofthekind[18]1= || y m1 y2 m i y2 (111) π(E /e)ηε2 − − 2 − 3N ε 2 0 m f0(ε)= 1 (104) 4L εThe− realε and imaginary part of the impedance are represented in ﬁgure 0 m m

-sign(η)Z ||,i such that the relative energy deviation9asafunctionof ranges betweenω withεm and constantεm.Duetoω ω> i =00. The curves are similar to those of − i the fact that the frequency ω must lie within the frequency spread produced ﬁgure 8 except that in this case the stable area is a circle thesign( radiusη)Z ||,r of which increasing ωi by the energy distribution, if the real part of frequency(arbitraryωr lies units) within the can be found by the condition that y beSTABILITY REGION real, from which we get frequency spread given by the above energy distribution, by using eq. (98), π(E /e)ηε2 we can write 0 m 2 1/2 Z ,r/n = 2 4y 1 y (112) 1 ω i =0|| − 4I0β − 3(Z /n)I0 x 1= i || P.V. Figuredx 8:+ Stabilityiπy diagram(105)sign( relatingη)Z ||,r growth rate and impedance for a − 4π(E /e)ηε2 y + x 0 andm (arbitrary1 units) − parabolic energy distribution (in arbitrary2 units). with π(E0/e)ηεm 2 ω nω¯ Z ,i/n = 4y 2 (113) 0 || − 4I β2 − A.G. Ruggero, V.G. Vaccaro, CERN Report ISR-TH/68-33 (1968) y = − If the coupling impedance(106)Z is0 inside the stable area, then the coherent Figurenω¯ 8:0ηε Stabilitym diagram relating growth rate and impedance for a thatparabolic is energy distribution (in arbitrary units). We remind that the imaginary term in eq.oscillation (105) exists energy only of if the beam is transferred to2 the incoherent kinetic energy Z π(E0/e) η εm || If the couplingof impedance a smallerZ is insideand thesmaller stable area, number= then the of coherent particles|2 | inside the beam, thus stabilizing(114) 1

5 Longitudinal dynamics and Landau damping in bunched beams 31 The longitudinal beam dynamics of bunched beams, as for the coasting beam case, is described by the Vlasov equation (91), but with diﬀerent equations of motion. Actually the rate of change of the longitudinal coordinate z is still given by equation (92), but due to the presence of the longitudinal focusing force of RF cavities, which is responsible of the synchrotron oscillations, the rate of change of energy is given by the contribution of two terms, one due to the RF and the other due to the wake ﬁeld, such that, for small synchrotron

34 22/09/11

Bunched Beam (Longitudinal)

The effect of Landau damping on bunched beam dynamics is a complex problem. However, a simplified and approximated expression, similar to the Keil-Schnell stability criterion for the coasng beam, has been proposed by D. Boussard in case of short range wake fields (acng on the sigle bunch) at high frequencies. The idea is that for high frequency fields generated by the beam, a bunched beam can be considered as a coasng one, provided we use the bunched beam peak current in Thethe effectthreshold of Landau criterion damping. on bunched beam dynamics is a complex problem. However, a simplified and approximated expression, similar to the Keil-Schnell stability criterion for the coasting beam, has been proposed by D. Boussard[29] in case of short range wake field and broad band impedance[15]. The idea is that at the high frequency of the signals emitted by the bunch in the instability regime, a bunched beam can be considered as a coasting beam with a current equal to the bunched beam peak current. As a consequence, ˆ 2 2 we can useThus eq., (116)considering by substituting a GaussianI0 with I ,energyε1/2 with distribuon 2 ln 2σε (we consider, we aGaussianenergydistribution),and,since0end up with the Boussard criterion.68 2ln2=0.94 1, then we × end up with the Boussard criterion

2 Z 2π(E0/e) η σε || =< | | (119) n ˆ I The Boussard criterion can be used to give a first evaluation of the threshold singleThe bunchBoussard current incriterion a storage ringcan before be used the microwave to get instability a first occurs[30].evaluaon of the threshold current of a ring for a given Let usimpedance now evaluate Z/n. the eItff ectsdepends of the on the Landau dampingparcle for energy a more, the simple energy spread, and on the factor η. case, by considering an instability of Nb equally spaced bunches in a storage

ring, produced by a single high order resonant mode at the frequency pω¯0 (long range wake ﬁeld), and by supposing that this resonant mode drives the instability of a single azimuthal beam oscillation mode m. We omit here the details of the calculations, but it is possible to obtain a dispersion integral similar to eq. (97), which assumes now the form[31] 32 mcItot Z (pω¯0) ∞ ∂f0 pω¯0zˆ 1 1= i || J 2 dzˆ (120) − (E /e) p ∂zˆ m c ω mω (ˆz) 0 0 − s

with Itot = ceNNb/L0 the total beam current, Jm the Bessel function of

the ﬁrst kind and m-th order, ωs(ˆz)theamplitudedependentsynchrotron

frequency, and f0(ˆz)theunperturbeddistributionfunctionexpressedinterms of the synchrotron oscillation amplitudez ˆ. If the beam particles have all

the same synchrotron frequency, that means ωs(ˆz)isconstant,thereisno

37 22/09/11

The effect of Landau damping on bunched beam dynamics is a complex problem. However, a simplified and approximated expression, similar to the The threshold corersponds to the maximum single Keil-Schnell stability criterion for the coasting beam, has been proposed by D. bunch current one can store in a storage ring, keeping Boussard[29]the beam in case stable of short. Above range the wakethreshold field and broad current band, impedance[15].we enter The ideain the is that regime at the high of the frequency “microwave of the signals instability emitted”, byhowever the bunch in the instabilitythe beam regime, is not a bunched lost. beam can be considered as a coasting beam with a current equal to the bunched beam peak current. As a consequence, The microwave instabilies willˆ 2 heat the 2 beam, we can use eq. (116) by substituting I0 with I, ε1/2 with 2 ln 2σε (we consider aGaussianenergydistribution),and,since0increasing the energy spread .68such2ln2=0 to restore.94 1, the then we threshold condion. × end up with the Boussard criterion

2 Z 2π(E0/e) η σε || = | | (119) n ˆ I The Boussard criterion can be used to give a ﬁrst evaluation of the threshold single bunch current in a storage ring before the microwave instability occurs[30]. Let us now evaluate the eﬀects of the Landau damping for a more simple

case, by considering an instability of Nb equally spaced bunches in a storage ring,References produced by a single high order resonant mode at the frequency pω¯ 0 (long[1] L. Landau, On the range wake ﬁeld),vibraon and of the by supposingelectronic plasma. that thisJ. Phys resonant. USSR 10 (1946), 25. English mode drives the translaon in JETP 16, 574. Reproduced in Collected papers of L. D. Landau, edited and with instabilityan introducon of a single by D. Ter azimuthalHaar, Pergamon beam Press, 1965, pp. 445460; and in Men of oscillation mode m. We omitPhysics here: L. the D. Landau, Vol. 2, Pergamon Press, D. ter Haar, ed. (1965). details[2] J. H. of theMalmberg calculations,, and C. B. Wharton Phys. Rev. but it is possibleLe. 13 184 (1964). to obtain a dispersion integral similar[3] V. K. Neil, and A. M. to eq. (97), whichSessler assumes, Rev. Sci. Instr. 6, 429 (1965). now the form[31] [4] V. G. Vaccaro, CERN Report ISR-RF/66-35 (1966). [5] L. J. Lasle, V. K. Neil, and A. M. Sessler, Rev. Sci. Instr. 6, 46 (1965). mcItot Z (pω¯0) ∞ ∂f0 pω¯0zˆ 1 [6] G. Besnier, Nucl. Instr. and Meth., 164, p. 235 (1979). || 2 1= i Jm dzˆ (120) [7] Y. Chin, K. Satoh, and K. − (E0/e) pYokoya, Part. Acc. 13, 45 (1983). 0 ∂zˆ c ω mωs(ˆz) [8] see, e.g., Y. Elskens, In Topics in Kinec Theory (2005), T. Passot− , C. Sulem, and P. L. withSulemItot , Eds., vol. 46, 2003. = ceNNb/L0 the total beam current, Jm the Bessel function of [9] I. Langmuir, Phys. Rev., 33: 954 (1929). the ﬁrst[10] A. W. kindChao and, Physics m-th of order,collecveω beams(ˆz)theamplitudedependentsynchrotron instabilies in high energy accelerators, John Wiley & Sons, p. 274, (1993). frequency,[11] A. Vlasov, J. Exp. Theor. Phys. 8 (3), 291 (1938). [12] A. Vlasov, J. Phys. 9, 25 (1945). and f0(ˆz)theunperturbeddistributionfunctionexpressedinterms of the[13] I. Langmuir and L. synchrotron oscillationTonks Phys. Rev. 33, 195 (1929). amplitudez ˆ. If the beam particles have all [14] A. W. Chao, ibidem, p. 221. the same[15] L. Palumbo, V. G. Vaccaro, M. synchrotron frequency,Zobov that, CERN 95-06, pp. 331-390, means ωs(ˆz)isconstant,thereisnoNovember 1995.

33 22/09/11

[16] A. Hofmann, CERN 77-13, pp. 139-174, November 1976. [17] A. W. Chao, ibidem, p. 21. [18] J. L. Laclare, CERN 94-01, pp. 349-384, January 1994. [19] H. G. Hereward, CERN report 65-20 (1965). [20] A. W. Chao, ibidem, p. 257. [21] A.G. Ruggero, V.G. Vaccaro, CERN Report ISR-TH/68-33 (1968), see also K. Hubner, V. G. Vaccaro, CERN Report ISR-TH/70-44 (1970). [22] E. Keil, W. Schnell, CERN Report ISR-TH-RF/69-48 (1969) [23] F. Sacherer, CERN Report SI-BR/72-5 (1972), F. Sacherer, IEEE Trans. on Nucl. Sci., NS-20, p. 825 (1973), F. Sacherer, CERN Reports PS- BR/77-5 and 77-6, (1977). [24] A. W. Chao, ibidem. [25] J. L. Laclare, CERN 87-03, pp. 264-305, (1987). [26] B. Zoer, CERN Reports SPS/81-18, SPS/81-19, and SPS/81-20 (1981). [27] C. Pellegrini, BNL 51538, (1982). [28] see, e. g., T. Suzuki and K. Yokoya, Nucl. Instr. and Meth., 203, p. 45 (1982). [29] D. Boussard, CERN Reports LABII/RF/INT/75-2 (1975). 60 [30] M. Migliora, Il Nuovo Cimento, Vol. 112 A, N. 5 (1999), pp. 485-490. [31] M. Migliora, L. Palumbo, M. Zobov, DAΦNE Technical Note G-21, Frasca (1993) [32] See, e. g., A. Hofmann, S. Myers, LEP Note 158 (1979). [33] M. Migliora, L. Palumbo, M. Zobov, Nucl. Instr. and Meth., A 354, p. 235 (1995) [34] A. W. Chao, ibidem, p. 254. [35] E. Wilson in CERN 85-19, p. 111 (1985). [36] D. F. Escande, Waveparcle interacon in plasmas: a qualitave approach. Oxford Univ. Press, 2009