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<<n xnx+−=ω 2 * equation of motion (see appendix): 0 (1 ) 0 2 znz+=ω *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 ..