Mechanisms for beam loss LECTURE 18 1. Scattering from atoms of the residual gas in the beam vacuum Beam loss and beam emittance growth chamber Mechanisms for emittance growth and beam loss Large angle Coulomb scattering-can cause beam loss if scattered particle hits an aperture Beam lifetime: Bremsstrahlung: (electrons only) Large radiative energy losses from residual gas interactions; Touschek effect; quantum lifetimes in electron machines; Beam lifetime due to beam-beam collisions Inelastic nuclear scattering (protons only): Beam loss through nuclear reactions Emittance growth: 2. Scattering of one particle by another particle in the bunch: from residual gas interactions; intrabeam scattering; random noise sources Coulomb scattering of one particle by another particle in the bunch is called intrabeam scattering. Both the angle and the energy of both particles can change in this process. If the energy change is 11/26/01 USPAS Lecture 18 1 11/26/01 USPAS Lecture 18 2 large enough, the particles may find themselves outside the energy Mechanisms for emittance growth aperture and be lost. This type of loss is called the Touschek effect. 1. Scattering from atoms of the residual gas in the beam vacuum 3. (Electron machines only): The quantum fluctuations due to chamber photon radiation can cause a particle to exceed the energy aperture Elastic Coulomb scattering: Random, small angle scattering or physical aperture of the machine. The resulting lifetime is called (multiple Coulomb scattering) causes transverse emittance growth the “quantum lifetime”. 2. Small angle intrabeam scattering of one particle by another 4. Beam loss at the interaction point in colliders particle in the bunch: Electron-positron colliders: beam loss occurs through radiative Small angle scattering equalizes the beam temperature in all Bhabha scattering (e+ + e- -> e+ + e- + γ), in which the energy of one dimensions: it causes a transfer of emittance from one dimension of the electrons falls outside the energy aperture. to another. Above transition, the emittance in all three degrees of Hadron colliders: beam loss occurs through inelastic reactions freedom can grow. p+−> p() p hadrons 3. Emittance growth from random noise sources: 11/26/01 USPAS Lecture 18 3 11/26/01 USPAS Lecture 18 4 Random power supply noise, and ground motion, can cause ρ transverse emittance growth Beam loss from residual gas interactions N We start by reviewing the concept of “cross section”. The cross A section for a particular reaction between two particles is the ∆s effective area which one particle presents to the other. Consider the volume element shown below, of infinitesimal length ∆s, area A, The probability of a reaction is σ containing a gas with atomic number density n. In this volume, ∆∆==σ ∆ ∆∆= σ PN0 ns there are NnAs0 atoms, which have a reaction cross section A with the incident beam particles. so the change in the number of beam particles is dN ∆∆NNPNns=− =−σσ ∆ ⇒ =−Nn ds 11/26/01 USPAS Lecture 18 5 11/26/01 USPAS Lecture 18 6 If the beam is relativistic, then ds= cdt and the change in N with Coulomb scattering time is The differential cross section for Coulomb scattering of a dN =−σ =−N relativistic charge e, from a material whose nuclei have charge Ze, Nnc 2 dt τ dσσd 4Zr2 2 mc for small angles, is ==0 e 1 Ω πθ θ 2 This gives the equation for the beam lifetime τ = associated d 2 d (θθ2 + 2 ) p ncσ 1 with beam loss when passing through the gas. The physics of the in which θ is the polar angle. (“Handbook”, p. 213) This is just the interaction which causes the loss of beam is contained in the cross mc section σ. We now consider the cross sections for the important Rutherford scattering formula. The angle θα= Z13 e , 1 p beam loss mechanisms. The equations for these cross sections, and α for many of the other formulae quoted in this lecture, have been =1/137, accounts for electron screening at small angles. taken from “Handbook of Accelerator Physics and Engineering”, If a particle is scattered into a polar angle θ, at a point where the A.Chao and M. Tigner, eds, World Scientific (1999). The course β, web page has a link to the Handbook web page. lattice function is then the maximum excursion of the resulting 11/26/01 USPAS Lecture 18 7 11/26/01 USPAS Lecture 18 8 = θββ betatron oscillation is zmax max 2 (for either plane). The Example: Z=7 (nitrogen gas-not a typical gas in an accelerator > particle will be lost if zbmax , where b is the radius of the vacuum system, but we’ll use it in this example). Take a vacuum 2b chamber with radius b=30 mm, and a machine with β =30 m, vacuum chamber at β . So for all angles θθ>= , the max max min ββ β = σ = max 15 m, and p=5 GeV/c. Then we find Coulomb 013. barn, particle will be lost. The cross section for loss of a particle due to a where 1 barn=10-24 cm2. large angle Coulomb scatter, averaged around the ring, is Bremsstrahlung (electrons only) The differential cross section for bremsstrahlung in the nuclear ∞ 2 2 dσ 4πZr2 2 mc ββ mc field of a charge Z is σπ==2 ∫ θθd 0 ee = max 2πZr2 2 Coulomb Ω 2 2 0 2 2 θ d θ p b p dσα16 r 184 1−+uu 0. 75 min min ≈+0 ZZ( 1)ln θθ<< 13 for the typical case of 1 min. du 3 Z u 11/26/01 USPAS Lecture 18 9 11/26/01 USPAS Lecture 18 10 ∆E in whichu = is the relative energy lost to radiation. Nuclear scattering (protons only). E Protons can be lost through nuclear absorption reactions with (“Handbook”, p 213). If the energy aperture of the machine is ∆E , a nuclei of the residual gas. Even most nuclear elastic scattering then the cross section for particle loss due to bremsstrahlung is reactions will cause losses, as the typical elastic scattering angles 1 σα2 are larger than the machine angular acceptance. Consequently, we σ =≈d 16 r0 ( +) 184 1− 5 Brem ∫ du ZZ 1 ln ln σ 13 simply take the total proton-nucleus cross section nuclear as the u du 3 Z ua 8 a cross section for proton loss. Curves are given in the “Handbook”, ∆ = Ea p. 216. Typical values of σ for protons on nitrogen are in the for ua <<1. nuclear E range of 0.4 barn, roughly independent of energy from 3 GeV to σ = Example: Z=7 and ua=0.003. We find Brem 4 barn more than a TeV. 11/26/01 USPAS Lecture 18 11 11/26/01 USPAS Lecture 18 12 Beam lifetime from residual gas interactions (protons) For proton storage rings, a typical requirement on the lifetime is τ σσ=+ σ 1 = σ > 20 hrs. This implies a residual gas pressure For protons, we use Coulomb Nuclear in nc . The τ o p 0. 474T[]K molecular density of the (ideal) residual gas is given by n = , p[]nTorr ≤ mol kT 20σ[]barn with p=pressure, T=absolute temperature, and k=Boltzmann’s Examples: Fermilab antiproton accumulator p=8 GeV/c. Then constant. Numerically, this is σ = σ = o -8 Coulomb 008. barn, Nuclear 04. barn, T=293 K=> p<1.7 x10 − p[]Torr []324=× Torr nmol m 966. 10 o T[]K σ Fermilab Tevatron p=1000 GeV/c. Then Coulomb is negligible, σ = o -10 which gives for the lifetime, for a diatomic gas, Nuclear 04. barn, T=4 K=> p<2.3x10 Torr. 0. 474T[]o K τ[]hr = p[][]nTorrσ barn 11/26/01 USPAS Lecture 18 13 11/26/01 USPAS Lecture 18 14 Beam lifetime from residual gas interactions (electrons) dN =−σσ =− + Nnc Nc( n0 GN) σσ=+ σ dt For electrons, we use Coulomb Brem. The equation for the 1 lifetime is more complicated than for protons, because of The lifetime at t=0, τ ==+, nnGN, depends on the σ eff 00 photodesorption. The synchrotron radiation photons produced by nceff the electrons, striking the walls of the vacuum chamber, desorb initial beam intensity N0. For electron storage rings, a typical substantial quantities of gas. This amounts to a contribution to the requirement on the lifetime is τ > 10 hrs. This implies a residual residual gas density that is proportional to the beam intensity: gas pressure nn=+ GN 0 0. 474T[]o K in which n is the atomic gas density due to beam-unrelated gas p []nTorr ≤ 0 eff 10σ[]barn sources, such as thermal outgassing, and G is the density produced = by photodesorption by one electron. The equation for beam loss in which peff kTn eff 2 includes the effects of the photodesorbed then becomes gas (assumed to be diatomic). 11/26/01 USPAS Lecture 18 15 11/26/01 USPAS Lecture 18 16 σ = σ = Example: CESR p=5 GeV/c. Then Coulomb 013. barn, Brem 4 o -9 -p x barn, T=293 K=> peff<3.4 x10 Torr -p x s p s p x s 2. Loss due to scattering of one particle by another particle in the The above figure is in the rest frame of the bunch. In the laboratory bunch (Touschek effect): frame, the momenta in the s direction get Lorentz boosted to γ ≈≈γγ′ In an elastic Coulomb scattering event between two particles in the ppxpsx . same bunch, there may be an exchange of transverse momenta for ∆E If γx′ > a , the relative energy acceptance of the machine, the longitudinal momenta.