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Transcript
Space Charge Luigi Palumbo Università di Roma "La Sapienza" and LNF-INFN CAS, 25 May 2005 EQUATION OF MOTION The motion of charged particles is governed by the Lorentz force : d (mγ v ) ext = Fe.m. = e( E + v × B ) dt Where m is the rest mass, γ the relativistic factor and v the particle velocity Charged particles are accelerated, guided and confined by external electromagnetic fields. Acceleration is provided by the electric field of the RF cavity Magnetic fields are produced in the bending magnets for guiding the charges on the reference trajectory (orbit), in the quadrupoles for the transverse confinement, in the sextupoles for the chromaticity correction. CAS, 25 May 2005 SELF FIELDS AND WAKE FIELDS There is another important source of e.m. fields : the beam itself Direct self fields Space Charge Image self fields Wake fields CAS, 25 May 2005 These fields depend on the current and on the charges velocity. They are responsible of many phenomena of beam dynamics: • energy loss (wake-fields) • energy spread and emittance degradation • shift of the synchronous phase and frequency (tune) • shift of the betatron frequencies (tunes) • instabilities. CAS, 25 May 2005 Fields of a point charge with uniform motion y’ y v o’ o r E′ = q r r′ 4πε o r ′3 r B′ = 0 x,x’ q z z’ vt is the position of the point charge in the lab. frame O. • In the moving frame O’ the charge is at rest • The electric field is radial with spherical symmetry • The magnetic field is zero E ′x = q x′ 4πε o r ′3 E ′y = q y′ 4πε o r ′3 E′z = q z′ 4πε o r ′3 CAS, 25 May 2005 Relativistic transforms of the fields from O’ to O Bx = B′x E x = E ′x E y = γ ( E ′y + vB′z ) B y = γ ( B′y − vE z′ / c 2 ) E z = γ ( E ′z − vB′y ) ( ′2 ′2 Bz = γ ( B′z + vE ′y / c 2 ) r′ = x + y + z ) 1/ 2 2 ′ x′ = γ ( x − vt ) y′ = y z′ = z [ r ′ = γ 2 ( x − vt ) 2 + y 2 + z 2 ] 1/2 CAS, 25 May 2005 E x = E ′x = q 4πε o r ′3 E y = γE ′y = E z = γE ′z = x′ q y′ 4πε o r ′3 q z′ 4πε o r ′3 = = = q 4πε o q 4πε o q 4πε o [γ [γ [γ γ ( x − vt ) 2 2 2 3/ 2 2 2 3/ 2 2 2 3/ 2 ( x − vt ) + y + z γy 2 2 ( x − vt ) + y + z γz 2 2 ( x − vt ) + y + z The field pattern is moving r q with the charge and it can E = 4πε o be observed at t=0. [γ ] 2 ] ] r γr 2 2 2 x +y +z ] 2 3/ 2 The fields have lost the spherical symmetry but still keep a symmetry with respect to the x-axis. CAS, 25 May 2005 t=0 y r θ x = r cos θ z x r E= q ( γ2 4πε o 1 − β 2 sin 2 θ ) 3/ 2 r r r3 y 2 + z 2 = r 2 sin 2 θ γ=0 γ≠0 γ >>1 CAS, 25 May 2005 B is transverse to the motion direction Bx = 0 B y = −vE z / c 2 Bz = vE y / c 2 r r r v×E B⊥ = 2 c γ→∞ CAS, 25 May 2005 Two charges in the rest frame O’ q Fr′ = q 1 qq 4πε o r 2 Two charges in the laboratory frame O Relativistic transform Fr = 1 γ Fr′ = 1 qq 4πε o γr 2 Lorentz force Fr = q ( Er − vBϕ ) = q ( Er − β 2 Er ) = q γ2 Er = qq 4πε o γr 2 1 CAS, 25 May 2005 Direct Space Charge Forces CAS, 25 May 2005 What do we mean with “space charge”? It is the net effect of the Coulomb interactions in a multi-particle system. Space Charge Regime dominated by the self field produced by the particle distribution. Collective Effects CAS, 25 May 2005 Debye Length λD r C Φ (r ) = r C= e 4πεo real uniform The particle distribution around a test particle will deviate from the continuous distribution. CAS, 25 May 2005 The effective potential of a test charge can be defined as the sum of the potential of the uniform distribution and a “perturbed” term. r C − r / λD Φ p (r ) = e r λD = εo k B T e2 n kB= Boltzman constant T = Temperature kB T = average kinetic energy of the particles n = particle density (N/V) CAS, 25 May 2005 The effective interaction range of the test charge is limited to the Debye length Smooth functions for the charge and field distributions can be used as long as the Debye length remains small compared to the particle bunch size λD CAS, 25 May 2005 Longitudinal Electric field of a uniform charged cylinder <z>=1.50 [ m]Ez [ V/ m] 200000 As computed by a multiparticle tracking code 100000 1.498 1.499 1.501 z[ m] -100000 -200000 Analytical expression CAS, 25 May 2005 Space charge of a relativistic cylindrical distribution Cylindrical finite bunch, uniformly charged, With circular cross section R L Transverse electric field at r = a Er Ez 5 1.5 1 4 0.5 3 -0.002 2 0.004 0.006 -0.5 1 -0.002 0.002 -1 0.002 0.004 0.006 -1.5 Longitudinal electric field at r = 0 CAS, 25 May 2005 Relativistic Uniform Cylinder I ρ= 2 πa v Gauss’s law 2πrlε o Er = ρπr 2l Ampere’s law 2lBϕ = µ o Jlr a I J= 2 πa ∫ ε E ⋅ dS = ∫ ρdV Linear with r o Er = ρr Ir = for r ≤ a 2 2ε o 2πε o a v Bϕ = ∫ B ⋅ dl = µ ∫ J ⋅ dS o Bϕ = µ o Jr 2 Ir = µo for r ≤ a 2 2πa CAS, 25 May 2005 β c Er λo = ρπa 2 (C / m) for r < a λ ( r ) = λo ( r / a ) 2 Er ( r ) = J = β cρ ( A / m 2 ) λo r 2πε o a 2 β λo β r Bθ (r ) = Er ( r ) = 2πε o c a 2 c I = Jπa 2 = β cλ 0( A) The Lorentz Force ( ) Fr = e(Er − β cBϑ ) = e 1 − β Er = 2 eEr γ2 eλo r = 2πε oγ 2 a 2 • has only radial component • is a linear function of the transverse coordinate The attractive magnetic force, which becomes significant at high energy, tends to compensate the repulsive electric force. CAS, 25 May 2005 Longitudinal Space Charge Forces In order do derive the relationship between the longitudinal and transverse forces inside a beam, let us consider the case of cylindrical symmetry and ultra-relativistic bunches. We know that a varying magnetic field produces a rotational electric field: ∂ ∫ E ⋅ dl = − ∂t ∫ B ⋅ ndS S z z+∆z We choose as path a rectangle going through the beam pipe and the beam, parallel to the axis. CAS, 25 May 2005 b b r r E z (r , z )∆z + ∫ Er (r , z +∆z )dr − E z (b, z )∆z − ∫ Er (r , z )dr = b ∂ = −∆z ∫ Bθ (r )dr ∂t r b ∂E (r , z ) ∂Bθ (r , z ) dr − E z (r , z ) = E z (b, z ) + ∫ r ∂t ∂z r b ∂ E z (r , z ) = E z (b, z ) + ∫ [Er (r , z ) − vBθ (r , z )]dr ∂z r b E z (r , z ) = E z (b, z ) + ∂ [Er (r , z ) − vBθ (r , z )]dr ∫ ∂z r CAS, 25 May 2005 E z (r , z ) = E z (b, z ) + (1 − β 2 ) b ∂ Er (r , z )dr ∫ ∂z r where (1-β2)=1/γ2. For perfectly conducting walls Ez=0. b 1 ∂ E z (r , z ) = E (r , z )dr 2 ∂z ∫ r γ r Uniform beam in a circular p.c. pipe. b r r2 b ∂λ ( z ) Fz (r , z ) = ( 1 − + 2 ln ) 2 2 4πε 0γ a a ∂z −e CAS, 25 May 2005 Beam motion in a linear channel CAS, 25 May 2005 r.m.s. emittance εrms = γx 2 + 2αxx ′ + βx ′2 = εrms x x ′ − xx ′ 2 2 2 x’ x’rms x xrms x 2 = βεrms and x ′2 = γεrms α =− β′ 2 =− 1 2εrms xx ′ d 2 x =− dz εrms γβ − α 2 = 1 CAS, 25 May 2005 Emittance degradation Longitudinal correlation along the bunch induced by e.m non linear e.m. fields x’ x’ x x RF fields, solenoidal fields, space charge, wake fields CAS, 25 May 2005 Equation of motion in a drift space: d 2 r eE r eI γm 2 = 2 = r 2 2 dt γ 2πγ εo a v 2 d2r d r 2 2 = β c dt 2 dz 2 K= d2r eI K = r= 2 r 2 3 2 3 dz 2πmγ εo a v a eI 2I = 2πmγ 3εov 3 Ioβ 3 γ 3 4πεo mc 3 Io = e Generalized perveance Alfven current CAS, 25 May 2005 γ=1 γ=5 γ = 10 L(t) Rs(t) ∆t CAS, 25 May 2005 Transport in a Long Solenoid ks = qB 2mcβγ K (ζ ) = 2Ig(ζ ) Io (βγ ) 3 0.9 0.8 g(ζ) g 0.7 0.6 0.5 0 0.0005 0.001 0.0015 metri 0.002 0.0025 ζ K (ζ ) R′′ + k R = R 2 s R′′ = 0 ==> Equilibrium solution ==> Req (ζ ) = K (ζ ) ks CAS, 25 May 2005 Small perturbations around the equilibrium solution CAS, 25 May 2005 Emittance Oscillations are driven by space charge differential defocusing in core and tails of the beam px Projected Phase Space x Slice Phase Space CAS, 25 May 2005 Envelope oscillations drive Emittance oscillations 2 1.5 R(ζ) envelopes 1 0.5 0 δγ =0 γ -0.5 0 1 2 3 4 5 metri 80 60 ε(z)emi 40 qB ks = 2mcβγ 20 0 0 1 2 3 4 5 metri ε(z) = R 2 R′2 − RR′ 2 ÷ sin ( 2k z) s CAS, 25 May 2005 BEAM DYNAMICS MODELING Space charge On Axis ∆t RF Field + Solenoids Off Axis ∆t CAS, 25 May 2005 On axis Envelope equation (linear fields) Space charge Solenoid field RF field 4ε e &x& + βγ 2 β&x + (k rf + k sol ) 2 x = 3 E xsc (ξ s , Axs , x) + γ m γ th n 4ε e &y& + βγ 2 β&y + (k rf + k sol ) 2 y = 3 E xsc (ξ s , Axs , y ) + γ m γ 2 1 3 x th n 2 1 3 y CAS, 25 May 2005 CODES used for simulations of Space Charge Effects PARMELA, ASTRA Multi-particle tracking code, includes space charge but not wake fields HOMDYN Relies on a multi-envelope model based on the time dependent evolution of a uniform bunch CAS, 25 May 2005 CAS, 25 May 2005 CAS, 25 May 2005 CAS, 25 May 2005 CAS, 25 May 2005 Slice analysis through the bunch CAS, 25 May 2005 Space charge with image currents CAS, 25 May 2005 Effects of conducting or magnetic screens Let us consider a point charge q close to a conducting screen. The electrostatic field can be derived through the "image method". Since the metallic screen is an equi-potential plane, it can be removed provided that a "virtual" charge is introduced such that the potential is constant on the screen q q -q CAS, 25 May 2005 A constant current in the free space produces circular magnetic field. If µr≈1, the material, even in the case of a good conductor, does not affect the field lines. I CAS, 25 May 2005 For ferromagnetic type, with µr>>1, the very high magnetic permeability makes the tangential magnetic field zero at the boundary so that the magnetic field is perpendicular to the surface, just like the electric field lines close to a conductor. I I I In analogy with the image method we get the magnetic field, in the region outside the material, as superposition of the fields due to two symmetric equal currents flowing in the same direction. CAS, 25 May 2005 CAS, 25 May 2005 Time-varying fields Static electric fields vanish inside a conductor for any finite conductivity, while magnetic fields pass through unless of an high permeability. This is no longer true for time changing fields, which can penetrate inside the material only in a region δw called skin depth. Inside the conducting material we write the following Maxwell equations: ∂B B = µH ∇ × = − E ∂t D = εE ∂D ∇×H = J + J = σE ∂t Copper σ = 5.8 107 (Ωm)-1 Aluminium σ = 3.5 107 (Ωm)-1 Stainless steel σ = 1.4 106 (Ωm)-1 . CAS, 25 May 2005 Consider a plane wave (Hz,Ey) propagating in the material ∂2Ey ∂x 2 − εµ ∂2Ey ∂t 2 − σµ ∂E y ∂t =0 ( the same equation holds for Hz). Assuming that fields propagate in the x-direction with the law: Ey i ω t − γ x H z = H˜ oe Ey = E˜ oe iωt − γx 2 2 iωt− γx (γ + εµω − i ωµσ )E˜o e =0 x Hz ∆w We say that the material behaves like a conductor if σ>>ωε thus: γ ≅ (1 + i ) σµε 2 CAS, 25 May 2005 Fields propagating along “x” are attenuated. The attenuation constant measured in meters is called skin depth δw: δw δw ≅ 1 2 = ℜ(γ ) ωσµ ∆w The skin depth depends on the material properties and the frequency. Fields pass through the conductor wall if the skin depth is larger than the wall thickness ∆w. This happens at relatively low frequency. At higher frequency, for a good conductor δw<< ∆w and both electric and magnetic fields vanish inside the wall. For the copper δw ≅ 6.66 (cm) f For a pipe 2mm thick, the fields pass through the wall up to 1 kHz. (Skin depth of Aluminium is larger by a factor 1.27) CAS, 25 May 2005 • compare the wall thickness and the skin depth (region of penetration of the e.m. fields) in the conductor. • If the fields penetrate and pass through the material, they can interact with bodies in the outer region. • If the skin depth is very small, fields do not penetrate, the electric filed lines are perpendicular to the wall, as in the static case, while the magnetic field line are tangent to the surface. I I -I -I CAS, 25 May 2005 Circular Perfectly Conducting Pipe (Beam at Center) In the case of charge distribution, and γ→∞, the electric field lines are perpendicular to the direction of motion. The transverse fields intensity can be computed like in the static case, applying the Gauss and Ampere laws. 2 r λ(r) = λo ; a ∫ S E r (2πr)∆z = λ (r)∆z εo λ(r)∆z β ; Bθ = E r 2πεo r c λ r λ oβ r E r (r) = o ; B (r ) = θ 2π εo a 2 2πεoc a 2 Er = F⊥ (r) = e( E r − βc Bθ ) = e γ 2 Er CAS, 25 May 2005 • Due to the symmetry, the transverse fields produced by an ultrarelativistic charge inside the pipe are the same as in the free space. • For a distribution with cylindrical symmetry, in the ultrarelativistic regime, there is a cancellation of the electric and magnetic forces. • The uniform beam produces exactly the same forces as in the free space. • This result does not depend on the longitudinal distribution of the beam. In general one has to consider the local charge density λ(z) CAS, 25 May 2005 Parallel Plates (Beam at Center) q -q q -q q 2h In some cases, the beam pipe cross section is such that we can consider only the surfaces closer to the beam, which behave like two parallel plates. In this case, we use the image method to a charge distribution of radius a between two conducting plates 2h apart. By applying the superposition principle we get the total image field at a position y inside the beam. CAS, 25 May 2005 λ(z ) E (z ,y) = 2π ε o im y ∞ ∑ λ (z ) π 2 (−1) ≅ y 2 2 2 (2nh) − y 4 π ε o h 12 n=1 n −2y ∞ 1 λ(z ) 1 im n Ey (z ,y) = (−1) − 2π ε o 2nh + y 2nh − y n=1 ∑ y Where we have assumed h>>a>y. For d.c. or slowly varying currents, the boundary condition imposed by the conducting plates does not affect the magnetic field. CAS, 25 May 2005 From the divergence equation we derive also the other transverse component: ∂ im ∂ im − λ(z ) π 2 im Ex = − Ey ⇒ Ex (z,x) = x 2 ∂x ∂y 4 π ε o h 12 Including also the direct space charge force, we get: eλ(z )x 1 π2 Fx (z ,x) = 2 2− 2 π ε o 2a γ 48h eλ(z )y 1 π2 Fy (z ,x) = 2 2+ 2 π ε o 2a γ 48h There is no cancellation of the electric and magnetic forces due to the "image" charges. CAS, 25 May 2005 Parallel Plates (Beam at Center) a.c. currents Usually, the frequency beam spectrum is quite rich of harmonics, especially for bunched beams. It is convenient to decompose the current into a d.c. component, I, for which δw>>∆w, and an a.c. component, Î, for which δw<< ∆w. While the d.c. component of the magnetic does not perceives the presence of the material, its a.c. component is obliged to be tangent at the wall. For a charge density λ we have I=λv. We can see that this current produces a magnetic field able to cancel the effect of the electrostatic force. CAS, 25 May 2005 ˜ (z)y π 2 λ ; E˜ y (z, x) = π ε o 48h2 I -I I -I I β B˜ x (z,x) = E˜ y (z,x) c 2 ˜ ˜Fy (z,x) = e λ (z)y π π ε oγ 2 48h2 2 ˜ ( z) x 1 π λ e F˜x (z, x ) = 2 2 2 − 24 h 2π ε oγ a 2 ˜ ( z)y 1 π λ e F˜y ( z, x ) = 2 2 2 + 24h 2π ε oγ a There is cancellation of the electric and magnetic forces !! CAS, 25 May 2005 Parallel Plates - General expression of the force Taking into account all the boundary conditions for d.c. and a.c. currents, we can write the expression of the force as: 2 2 1 1 π2 π π 2 λ m β λ u Fu = + 2 2 m 2 2 2 2π ε o γ a 24h 24h 12 g u = x, y e where λ is the total current, and λ its d.c. part. We take the sign (+) if u=y, and the sign (–) if u=x. CAS, 25 May 2005 Space charge effects in storage rings CAS, 25 May 2005 Self Fields and betatron motion Consider a perfectly circular accelerator with radius ρx. The beam circulates inside the beam pipe. The transverse single particle motion in the linear regime, is derived from the equation of motion. Including the self field forces in the motion equation, we have O y ρ x z x d (mγ v ) r r ext self = F (r ) + F (r ) dt dv = dt r r self F (r ) + F (r ) ext mγ CAS, 25 May 2005 For the single particle "transverse dynamics" we write: r r = (ρ x+ x )eˆx + yeˆ y r v = x&eˆx + y& eˆ y + ωo (ρ x + x )eˆz r a = &x& − ωo2 (ρ x+ x ) eˆx + &y&eˆ y + [ω& o (ρ x + x ) + 2ωo x& ]eˆz [ ] For the motion along x: &x& − ωo2 (ρ x + x ) = ( 1 Fxext + Fxself mγ ) Which, with respect to the azimuthal position s=vzt becomes: &x& = v z2 x′′ = ωo2 (ρ x + x )2 x′′ x′′ − ( 1 1 Fxext + Fxself = ρ x + x mv z2γ ) CAS, 25 May 2005 We assume small transverse displacements x with respect to the closed orbit, and only dipoles for bending and quadrupole to keep the beam around the closed orbit: ext x F ≈F ext o ∂Fxext + ∂x x x =0 x <<ρx Around the closed orbit, putting vz= βc, we get 1 ∂Fxext 1 − x′′ + 2 2 ρ x β Eo ∂x x = 1 Fxself ( x) 2 Eo β x =0 where Eo is the particle energy. This equation expressed as function of “s” reads: 1 1 x ′′(s)+ 2 + K x (s) x(s) = 2 Fxself (x,s) β Eo ρ x (s) CAS, 25 May 2005 •In the analysis of the motion of the particles in presence of the self field, we will adopt a simplified model where particles execute simple harmonic oscillations around the reference orbit. •This is the case where the focussing term is constant. Although this condition in never fulfilled in a real accelerator, it provides a reliable model for the description of the beam instabilities 1 x ′′(s)+ K x x(s) = 2 Fxself (x) β Eo Q x 2 1 self x ′′( s) + x( s) = F ( x , s) x 2 β Eo ρx x(s) = Ax cos [ K s− ϕ ] x x ax βx = Ax ⇒ βx const. K x (s)βx2 = 1 1 1 βx = ' = µx Kx µ x (s) = K x s ω 1 Qx = x = ω o 2π L ∫ o Q 2 ds′ = ρx K x ⇒ K x = x β( s′) ρx CAS, 25 May 2005 Transverse Incoherent Effects We take the linear term of the transverse force in the betatron equation: ∂Fxs.c. s.c. Fx ( x, z ) ≅ ∂x x = 0 x 2 Qx 1 ∂Fxs.c. x = x′′ + x 2 β Eo ∂x x = 0 ρx (Qx + ∆Qx ) 2 ≅ Qx2 ∂ Fxs.c. + 2Qx ∆Qx ⇒ ∆Qx = − 2 β 2 EoQx ∂x ρ x2 The betatron shift is negative since the space charge forces are defocusing on both planes. Notice that the tune shift is in general function of “z”, therefore there is a tune spread inside the beam. CAS, 25 May 2005 Consequences of the space charge tune shifts In circular accelerators the values of the betatron tunes should not be close to rational numbers in order to avoid the crossing of linear and non-linear resonances where the beam becomes unstable. The tune spread induced by the space charge force can make hard to satisfy this basic requirement. Typically, in order to avoid major resonances the stability requires ∆Qu < 0.3 CAS, 25 May 2005 Example: Incoherent betatron tune shift for an uniform electron beam of radius a, length lo, inside circular perfectly conducting Pipe ∂ Fxs.c. ∂ eλo x eλo = = 2 2 2 2 ∂ x ∂ x 2 πε o γ a 2 πε o γ a re , p = e2 4πεomoc 2 ρ x2 Ne 2 ∆Qx = − 4πεoa 2 β 2γ 2 EoQxolo (electrons : 2.82 10 −15 m, protons :1.53 10 −18 m) ρ x2 Nre, p ∆Qx = − 2 2 3 a β γ Qxolo For a real bunched beams the space charge forces, and the tune shift depend on the longitudinal and radial position of the charge. CAS, 25 May 2005 PS Booster, accelerate proton bunches From 50 to 800 MeV in about 0.6 s. The tunes occupied by the particle are indicated in the diagram by the shaded area. As time goes on, the energy increase and the space charge tune spread gets smaller covering at t=100 ms the tune area shown by the darker area. The point of highest tune correspond to the particles which are least affected by the space charge. This point moves in the Q diagram since the external focusing is adjusted such that the reduced tune spread lies in a region free of harmful resonances. Finally, the small dark area shows the situation at t=600 ms when the beam has Reached the energy of 800 MeV. The tune spread reduction is lower than expected with the energy increase (1/γ3) dependence since the bunch dimensions also decrease during the acceleration. CAS, 25 May 2005 END CAS, 25 May 2005