Introduction to Transverse Beam Optics
Bernhard Holzer, DESY-HERA
Part I: Lattice Elements and Equation of Motion
Lattice and Beam Optics of a typical high energy storage ring Largest storage ring: The Solar System
astronomical unit: average distance earth-sun 1AE ≈ 150 *106 km Distance Pluto-Sun ≈ 40 AE
AE Luminosity Run of a typical storage ring:
HERA storage ring: Protons accelerated and stored for 12 hours distance of particles travelling at about v ≈ c L = 1010-1011 km ... several times Sun-Pluto and back
guide the particles on a well defined orbit („design orbit“) focus the particles to keep each single particle trajectory within the vacuum chamber of the storage ring, i.e. close to the design orbit. Transverse Beam Dynamics:
0.) Introduction and basic ideas
„ ... in the end and after all it should be a kind of circular machine“ need transverse deflecting force Lorentz force FqEvB=+×*( )
8 m typical velocity in high energy machines: vc≈≈3*10 s
old greek dictum of wisdom: if you are clever, you use magnetic fields in an accelerator where ever it is possible.
But remember: magn. fields act allways perpendicular to the velocity of the particle only bending forces, no „beam acceleration“ The ideal circular orbit consider a magnetic field B is independent of the azimuthal angle θ
ˆz condition for circular orbit:
Lorentz force Fe*v*BL = ρ θ γ mv2 x 0 s centrifugal force F = Zentr ρ circular coordinate system
γmv2 ideal condition for circular movement: 0 =e* v* B ρ
p = B* ρ e
On a circular orbit, the momentum of the particle is related to the guide field B and the radius of curvature ρ. Focusing Forces: I.) the principle of weak focusing still: consider a magnetic field B is independent of the azimuthal angle θ
ˆz stability of the particle movement: small deviations of particle from ideal orbit ↔ restoring forces ● ρ z θ s x γ mv2 0 << for r ρ circular coordinate system r evB(r)z γ mv2 0 >< for r ρ r
mv2 FevB=− rest r * introduce a gradient of the magnetic field
∂∂BBxzz evB=+ ev ( B * x) = evB 1 + z 00∂∂rrBfield gradient „n“, by definition: 0z x ρ ∂Bz =−ev B n * n =− 0 1 Br∂ ρ 0
x rx()=+=ρρ1 + * develop for small x ρ mv22 mv x ≈−()1 r ρ ρ
mv2 x nx restoring force: F()evB()=−−−11 rest ρ ρρ px nx =−−−v(11 ) evB( ) ρ ρρ x =−evB * (1 − n) ρ x condition for focusing in the horizontal plane: FevB*(n)rest =−1 − ρ n < 1
Nota Bene: the condition does not exclude n = 0. there is focusing even in a homogenous field.
„Geometric focusing“ in a homogeneous field: consider three particles, starting at the same point with different angles P1 P2
xˆ
P1 P2
Problem: amplitude of betatron oscillation in this case ˆx*≈ αρ α ≈ mrad 1 for a particle beam ˆxm≈ ρ ≈m 1 several 100 weak focusing in the vertical plane:
restoring force in „z“ Fz ∝ -z we need a horizontal magnetic field component: Bx = -*const z ∂B ... or a negative horizontal field gradient x =−const ∂z Maxwells equation: ∇×B = 0 ∂B ∂∂BB xzz==<0 ∂∂∂zxr
the vertical field component has to decrease with increasing radius typical pole shape in a combined function ring ∂B z <⇔00n > ∂r Comments on weak focusing machines:
* magnetic field is independent of the azimuthal angle θ, focusing gradient is included in the dipole field
* stability of the particle movement in both planes requires
01< xnx+−=ω 2 * equation of motion (see appendix): 0 (1 ) 0 znz+=ω 2 *0 Problem: we get less than one 0 transverse oscillation per turn large oscillation amplitudes Separate the focusing gradients Example HERA: from the bending fields to obtain g==98Tm / at p 920GeV / c n >>1 n≈12420 II.) Accelerator Magnets Separate Function Machines: Split the magnets and optimise them according to their job: bending, focusing etc Dipole Magnets: homogeneous field created by two flat pole shoes calculation of the field: 3rd Maxwell equation for a static field: ∇×Hj = according to Stokes theorem: ∫ ()∇×Hnda =∫∫ Hdl = jndaNI ⋅ = ⋅ SS N*I = number of windings Hdl=+ H h H l times current per winding ∫ 0 **Fe Fe µ ≈ in matter we get with r 1000 H Hdl=+ H h0 l ≈ H h ∫ 00**Fe * µr ρ magnetic field of a dipole magnet: α µ nI B = 0 0 h ds radius of curvature ... remember p/e= B* ρ eB⋅ BT[] 1 −1 00 m.==0 2998 ρ ppGeV/c[ ] field map of a storage ring dipole magnet ds dl α =≈ bending angle of a dipole magnet: ρ ρ ∫ Bdl p for a circular machine require απ==→=22Bdl π * B*ρ ∫ q hard edge approximation: define the effective length of a magnet by +∞ Bl⋅= : Bdl 0 eff ∫ typically we get lleff≈+ iron 13 .*h −∞ Example HERA: 920 GeV Proton storage ring: N = 416 l = 8.8m, q = +1 e Nota bene: for high energy particles we can set ... Epc≈⋅ ∫ Bdl≈= N * l * B2π p / q 2π**eV 920 109 B.Tesla≈≈m 5 15 416** 3 108 *.m*e 8 8 s Quadrupole Magnets: yoke coils required: linear increasing magnetic field BgxBgzzx=− ⋅ =− ⋅ at the location of the particle trajectory: no iron, no current ∇×BBV =0 → =−∇ the magnetic field can be expressed as gradient of V(x,z)= g⋅ xz a scalalr potential ! equipotential lines (i.e. the surface of the iron contour) = hyperbolas calculation of the field: R ∫ HdsnI⋅= R 20 ∫∫ Hds⋅= H(r)dr + ∫∫ HFe ds + Hds ⋅ 012 calculation of the quadrupole field: RRB( r )dr Hds⋅= H(r)dr = = n*I ∫∫ ∫µ 000 B( r )=− g * r , r = x22 + z µ nI 1 B gradient of a g = 2 0 remember: = 0 quadrupole field: R2 normalised dipole strength: ρ p/e normalised quadrupole strength: g k = p/e focal length: 1 f:= k*l Example of a sparated function machine: heavy ion storage ring TSR „Synchrotron Magnet“: combines the homogeneous field of a dipole with a quadrupole gradient = −+⋅Bz g xz potential: V(x,z) 0 advantage: lattice with high compactness disadvantage: strong correlation of momentum (via dipole field) and beam optics. poor flexibility ∂V ==−+Bgx field index: B z0 ∂z ∂B ρρ ngk=⋅==z ρ 2 ∂xB B 00 Nota bene: Synchrotron magnet can be considered as a shifted quadrupole lens „off center quadrupole“. III.) The equation of motion Pre-requisites: * consider particles with ideal momentum or at least with only small momentum error * neglect terms of second order in x,z,and ∆p/p linear approximation * independent variable “s”, write derivative with respect to s as …´ θ 0 angle of ideal orbit particle trajectory θ angle of particle trajectory dl ds x´ derivative of particle amplitude with resp. to s x x ′ =−θθ design orbit 0 d(θθ− ) ρ x ′′ = 0 ds For any circular orbit path we get: ds dθ −1 dθ =− →0 = θ 0 ρ ds ρ ... quite clear, but what is d /ds ? dl Be 1 B (a) dθ = −=− dl as for any circular = ρ p orbit we know: ρ p/e as long as the angle x´ is small dl is related to s by: ρ +x (b) dl= ds ρ x dl=+ (1 )ds ρ Magnetic field: assume only dipole and quadrupole terms remember: ∂B Bgxx =− BB=+ xB =− gx definition of 00∂x field gradient Bgzz =− ppp1 (c) B=−000 kx = − kx eeeρ ρ g normalised strength k = p/e 0 putting the term (b) and (c) into the expression for the angle dθ … Be x eB ddl()dsθ =− =−1 + ⋅ pppρ +∆ 0 p 1 ekx⋅−0 x e ρ d()dsθ =−1 + ρ pp+∆ 0 pxkx1 2 dkxdsθ =−0 − + − pp+∆ ρ ρρ2 0 p ∆ p develop the momentum 0 ≈−1 pp+∆ p p for small ∆p 00 1 xkxp2 ∆∆ pxp ∆ ∆ p ddskxθ =− − + − − + kx − + kx2 ρ ρρ22ppp ρ ρ ρ p 000 0 and keep only first order terms in x, z, ∆p !! 1 xp∆ ddskxθ =− − + − ρ ρρ2 p 0 ... do you still remember the beginning ? we were looking for ... dθ dθ 11xp∆ d θ x ′′ =−0 xkx′′ =− + − + − 0 ρ ρρ2 pds ds ds 0 ∆ p x(′′ +−=11 k)x ρ 2 ρ p vertical direction: * no bending (... in general) no 1/ρ2 term * vertical gradient: ∂∂BB ∇×B =0 ⇔zx = ∂∂xz * Lorentz force gets a “-”: F=q (v x B) zkz′′ +=0 IV.) Solution of trajectory equations horizontal plane: define: 11∆ p x(′′ +−= k(s))x 2 ρ 2(s)ρ (s) p K(s)=− k(s) +1 ρ (s) ∆ p x′′ += K(s)* x 1 ρ p 2 Problems: * inhomogeneous equation set for the moment ∆p/p=0 i.e. consider particles of ideal momentum * K(s) is not constant but varies as a function of the azimuth K(s) is a “time dependent” restoring force the differential equation can only be solved numerically K(s) is prescribed by the storage ring design: given by the magnet parameters remember: hard edge model: K = const within a magnet SPS Lattice xK*x′′ += differential equation for the transverse oscillation of a 0 particle in a magnetic element of the storage ring. (... harmonic oscillator) * second order two independent solutions, * linear in x any linear combination of these „principal solutions“ will again be a solution. we choose for K > 0: C(s)== cos( Ks) , S(s)1 sin( Ks) K with the initial conditions: C(01 )== , S( 00 ) C(′′00 )== , S( 01 ) for K < 0: C(s)== cosh( Ks) , S(s)1 sinh( Ks) K Arbitrary solution of any particle: x(s)=⋅ x C(s) +⋅ x′ S(s) 00 Matrix formalism for beam transfer in a lattice: x( s)=⋅ x C( s ) +⋅ x′ S( s ) x( s) x( s) 00 ⇒=M* x′′′′ (s)=⋅ x C (s) +⋅ x S (s) x(s)′′ x(s) 00 C(s) S(s) M = where C′′ (s) S (s) HERA standard type quadrupole lens ... depending on the value of K we can establish a transfer matrix for any (linear) lattice element in the ring. 1 cos K l sin K l K horizontal focusing quadrupole: K > 0 M = − KsinKl cosKl 1 cosh K l sinh K l K vertical focusing quadrupole: K < 0 M = KsinhKl coshKl 1 l drift space: K = 0 M = 01 particle motion in the vertical plane: in general storage rings are built in the horizontal plane. no vertical bending dipoles 1 ρ = 0 define: Kk= same matrices as in x-plane. ! with the assumptions made, the motion in the horizontal and vertical planes are independent „ ... the particle motion in x & z is uncoupled“ !! dont´t forget the inhomogeneous equation transformation through a system of lattice elements combine the single element solutions by multiplication of the matrices M= M *M *M *M *M total QF D QD Be nd D*..... focusing lens xx = M( ) * ′′s2,s1 dipole magnet xx ss21 defocusing lens K. Wille x(s) 0 typical values in a strong foc. machine: s x ≈ mm, x´ ≤ mrad Dispersion: ∆ p xK(s)*x′′ +=1 inhomogeneous equation ρ p general solution = complete solution of the homogeneous equation + particular solution of inhomogeneous equation x(s)= xhi (s)+ x (s) ′′ 1 ∆p with xK(s)*xhh+=0 xK(s)*x′′+= iiρ p x(s) normalise with D( s ) = i respect to ∆p/p: ∆p/p 1 D′′ (s)+= K(s)* D(s) initial conditions: DD==′ 0 ρ 00 ∆p x(s)=⋅ x C(s) +⋅ x′ S(s) + D(s) 00 p Dispersion: xxCS D ∆p for convenience: =⋅+ xx′′ expand the matrix formalism: CS′′ p D ′ ss0 or even more convenient xCSDx xCSDx′′′′′=⋅ ∆∆pp001 pp ss 0 Determine the Dispersion from the lattice parameters: CS remember: C and S are independent solutions W = ≠ 0 of the equation of motion the Wronski determinant CS′′ dW d even more, we get: =−=−(CS′ SC ′ ) CS ′′ SC ′′ ds ds = −−=K(CS SC ) 0 Dispersion: C,C==′ 0010 →=Wconst. choose the position s = s where 0 S,S==′ 0001 W = 1 the dispersion trajectory can be calculated from the cosine and sinelike solutions: ss D(s)= S(s)11 C(s)ds− C(s) S(s)ds ∫∫ρρ(s) (s) ss00 proof: D(s) has to fulfil the equation of motion ss D′′ (s)=− S (s)11 C(s)ds C ′ (s) S(s)ds ∫∫ρρ(s) (s) ss00 ss D′′ (s)=−+− S ′′ (s)111 C(s)ds C ′′ (s) S(s)ds (CS ′ SC ′ ) ∫∫ρρρ(s) (s) ss00 = 1 D′′ =− K(s)D(s) + 1 ρ ss Example: D(s)=− S(s)11 C(s)ds C(s) S(s)ds ∫∫ρρ(s) (s) ss00 10l M = drift: Drift 010 1 ρ ==k,0 001 →=D( s )00 , D′ ( s ) = foc. Quadrupole: 1 cos K l sin K l 0 K 1 ρ ==0,K− k MKsinKlcosKl= Qfoc 0 →=D( s )00 , D′ ( s ) = 001 dipole sector magnet: design orbit angle at entrance and exit: 90° 1 ρ ==const, k dipole sector magnet: 0 K/= 1 ρ 2 ll cosρ ⋅ sin ρ ρ 2x2 matrix M = ll − 1 sin cos ρ ρρ ss D(s)=− S(s)11 C(s)ds C(s) S(s)ds ∫∫ρρ(s) (s) ss00 lslsll D(s)=⋅⋅−⋅⋅ρρ sin11 cos ds cos sin ds ρρ∫∫ ρ ρρ ρ 00 lll D( s )=⋅ρ sin2 + cos ⋅ cos −⋅1 ρ ρ ρρ l l D( s )=−ρ (1 cos ) D(s)′ = sin ) ρ ρ ll l dipole sector magnet: cosρρ sin (1 − cos ) ρ ρρ − ll l 1 MQfoc = sin cos sin ρρ ρ ρ 00 1 Example: HERA Interaction region typical values: ∆p/p ≈10-3 D ≈ 1…2m ∆ ≈ xi = D* p/p 1-2 mm DD==′ start value: 000 dispersion is generated as soon as we enter the dipole magnets where 1/ρ≠0 Remarks on Magnet Matrices: 1.) thin lens approximation: 1 cos k l sin k l k matrix of a quadrupole lens M = − ksin kl cos kl in many practical cases we have the situation: 1 fl= >> q ... focal length of the lens is much bigger than the length of the magnet klq limes:l → 0 while keeping : kl= const 10 10 M x = 1 M z = −1 1 1 f f ... usefull for fast (and in large machines still quite accurate) „back on the envelope calculations“ ... and for the guided studies ! ΨΨ 3.) edge focusing: ... dipole „box-magnets“ * horizontal plane: particle at distance x0 to the design orbit sees a „shorter magnetic field“ ∆=lxtan ⋅ ψ 0 ∆l error in the bending angle of the dipole ∆ ltanψ x ∆=α =x ⋅ 0 ρ 0 ρ corresponds to a horizontal defocusing effect ... in the approximation of ∆l = small xx= 0 10 tan ψ ⇒=M tan ψ xxx′′=+ 00 1 ρ ρ 3.) edge focusing: vertical plane Bx particle trajectory crosses the field lines at the dipole edge. Bz horizontal field component vertical focusing effect 10 fringe field effect at the edge of a dipole M z ≈ − tan ψ 1 ρ for purists only: vertical edge effect depends on the exact form of the dipole fringe field 10 M z ≈ 1 btanψ − where b = distance over which the 2 1 ρ 6 cos ψρ fringe field drops to zero Question: what will happen, if the particle performs a second turn ? ... or a third one or ... 1010 turns x 0 s Answer: ... will be discussed in the evening having a good glass of red wine ... or tomorrow in the next lecture. p V.) Résumé: beam rigidity: B ⋅=ρ q .B(T)⋅ 1 m−1 = 02998 0 bending strength of a dipole: ρ p( GeV / c ) .g⋅ km−2 = 0 2998 focusing strength of a quadrupole: p(GeV / c ) . µ nI km−2 = 02998 2 0 2 p(GeV / c ) a r 1 focal length of a quadrupole: f = kl⋅ q ∆p xKx′′ +=1 equation of motion: ρ p xMx=⋅ matrix of a foc. quadrupole: ss21 1 cos K l sin K l K M = − Ksin Kl cos Kl VI.) Appendix: Equation of motion in the case of weak focusing restoring forces are linear in the deviations x, z from the ideal orbit. Example: harmonic oscillation of a spring restoring force: Fc*r =− x xc*x +=→ x(t)x*cost = ω equation of motion: 0 0 ω = c/m x Bev Be mv in our case: F=− (11 − n)x = − ( − n)x r ρ m ρ ω 0 , the angular revolution- (or cyclotron- ) frequency is obtained from: 2 mvmρρ=→evBv = eB F(n)*x=−ω 2 − r 0 1 ωρ==v/ eB 0 m ωω==c/m *− n 0 1 As 0 < n < 1 is required for stability the frequency of the transverse oscillations ω is smaller than ω the revolution frequency 0. Maxwell‘s equations in vacuum ∇⋅E = ρ and in matter ∇⋅D = ρ δE ∇×Bj = + δD δt ∇×Hj = + δt δB ∇×E = − δB δt ∇×E = − δt ∇⋅B =0 ∇⋅B =0 Stokes integral theorem: ∫ ()∇×Anda = ∫ Adl SC ∇⋅Adx3 = Anda Gauß´ integral theorem: ∫ ∫ VS A da where Vectorfield, Surface Element of the Surface S n V Volume Normvector on the Surface S S Surface surrounding the Volume V Solution of the equation of motion: ′′ xkx+=0 k>0 foc. quadrupole in the horizontal plane Ansatz: x( t )=+ a sin(ωω t ) a cos( t ) 12 x(t)′ =− aωω cos( t) a ωω sin( t) with the derivatives: 12 x(t)′′ =− aωω22 sin(t) − a ωω cos(t) 12 =−ω 2x( t ) and we get for the differential equation: x(t)= a cos( kt)+ a sin( kt) ω = k 12with the constants a1 and a2 are determined by boundary (i.e. initial) conditions: at we require t=== x()x x()′′ x 0 000 , 0 x( )=+ a cos( ) a sin( ) →= a x 00012 10 x ′ x′ (000 )=− a k sin( ) + a k cos( ) → a = 0 12 2k x′ x(t)=+ x cos( kt)0 sin( kt) 0 k x′′ (t)=− x k sin( kt) + x cos( kt) 00 or expressed for convenience in matrix form: xx 0 = M* , xx′′ s 0 cos kt1 sin kt M = k − ksin kt cos kt