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

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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 PN0 nsσ 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

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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 dσσd 4Zr2 2  mc τ for small angles, is ==0  e  τ = 1 d Ω 2 d 2 + 2 2  p  This gives the equation for the beam lifetime associated πθ θ ( 1 ) nc σ θθ 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 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 particleββ 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 d 4 Zr2 2  mc  mc field of a charge Z is σπ==2 ∫ d 0  ee = max 2 Zr2 2  Coulomb Ω 2   2 0   2 2 θ dσ p b p dσα16 r  184 1−+uu 0. 75 min min ≈+0 ZZ( 1)ln θθ π θθ<<  13 for the typical case of 1 min. du 3 Z u θ ββ π

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∆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 2   are larger than the machine angular acceptance. Consequently, we σ =≈d 16 r0 ( +)  184 1− 5 Brem ∫ σαdu 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 = > 20 hrs. This implies a residual gas pressure For protons, we use Coulomb Nuclear in ncσ . 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[][]nTorr barn σ

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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 nc eff 00 photodesorption. The synchrotron radiation photons produced by eff σ 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. E particles will be lost. The expression for the Touschek lifetime is quite complex. It depends on the details of the lattice, the machine energy acceptance, the particle energy, and the emittance of the beam. The

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formula for the lifetime is given in “Handbook”, p. 125-126. The (Electron machines only): quantum lifetime basic structure is The quantum fluctuations due to photon radiation may cause a 1 rcN2 particle to exceed the energy aperture or physical aperture of the ∝ 0 b fEE( ,,,,,,∆ ) τ 4 xyxy x a machine. The resulting lifetime is called the “quantum lifetime”. Touschek xyL γεεε ββεεδη Let’s look at the loss process in the horizontal transverse plane The following is a plot of the Touschek lifetime for the CESR first. When we discussed synchrotron radiation damping, we saw operating parameters, as a function of the relative energy that the balance between damping and quantum excitation (due to ∆ acceptance Ea/E: photon emission in dispersive regions) led to an equilibrium horizontal emittance. The process which leads to this equilibrium Lifetime hrs 50 involves the random emission of large numbers of photons; such a 40 H L random process lead to a Gaussian distribution, in both x and x′. 30 20 In phase-amplitude variables (r,φ), the Gaussian phase space 10

E E % distribution can be written as 0.1 0.2 0.3 0.4 0.5 D a

ê H L 11/26/01 USPAS Lecture 18 19 11/26/01 USPAS Lecture 18 20 2 dN Nr  dN/dr2 =−exp , rdrdφπε22 ε

dN/dt (fluctuations) in which ε is the rms equilibrium horizontal emittance.

Integrating over φ, the number of electrons in an interval of dN/dt (damping)  2  dN 2 dN Nr amplitude squared dr 2 is dr , where =−exp  r 2 2 2 εε 2 dr dr 22  rw dN Consider the following plot of : The number of particles per unit time (the particle flux) crossing a dr2 22= line at rrw due to damping is given by 2 dN = dN dr dt 2 2 dt damping dr rw damping

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222 dr rr2 Now imagine that we have an aperture limit at rw. Since there are in which =− =− w . So no particles at r>r , there will no longer be an inward damping dt ττ w damping 2 2 x r rw flux, only the outward flux due to fluctuations, which constitutes 22 beam loss. The distribution function must go to zero at this point, dN Nr r  2 =−w exp − w so it will not longer be Gaussian. However, if r >>ε, the ετ ε   w dt damping x  2  distribution will not change very much, and we can use the above But since the distribution is in equilibrium, the total expression to estimate the rate of beam loss: dN Nr22 r N dN =+=dN dN ≈−w exp − w =− 0. Thus, we know that τετε   dt dt dt dt x  2  q fluctuations damping which gives for the quantum lifetime τ dN Nr22 r  q =−w exp w . dt ετ ε  2   2  fluctuations x ττ= ε rw qx2 exp  rw 2  ε 11/26/01 USPAS Lecture 18 23 11/26/01 USPAS Lecture 18 24 ≥ σ This can be written in terms of the limiting value of the aperture, Usually one designs for xa 10 to provide adequate safety 22= β σβε2 = xraw, and the mean square horizontal beam size , as margin. 2  2  ττ= σ xa qx2 exp 2  xa 2  The equivalent set of considerations with regard to energy σ fluctuations leads to the following result for the quantum lifetime Note the extremely rapid dependence of the lifetime on the ratio due to energy fluctuations: x2 a . To obtain quantum lifetimes of 10 hours, for typical damping 2  2  2 E Ea σ ττ= ε σ exp∆  q ∆ 2 2 times of order 10 ms, we need to have Ea 2 E  τ 2  2  2 σ q xx in which τε is the energy damping time, σΕ is the rms energy = σ exp aa ≥×4106 ⇒ ≈ 24 ⇒x ≈ 5. τ 2 2 2 a ∆ x xa 2  spread, and Ea is the energy aperture (due either to the bucket σσ Going below this limit reduces the lifetime extremely rapidly. In height, or to aperture limits at a dispersive point). Again, one typically designs for ∆E ≥ 10σ . this regime, the lifetime will be very sensitive to the aperture.σ aE

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4. Beam loss at the interaction point in colliders From the definition of the luminosity, the loss rate for one species Electron-positron colliders: beam loss occurs through radiative of particle will be + - + - + γ dN Bhabha scattering (e + e -> e + e ), in which the final energy = σ L RBS of one of the electrons falls outside the energy aperture. dt = 2 The differential cross section for radiative Bhabha scattering is Since L kN , where k is a constant if the cross sectional areas of given in “Handbook”, p. 220. When integrated to give the cross the beams don’t change, we have section corresponding to an energy loss sufficient to leave the dL dN L ==22kN kNLσ =− machine, the result is a slowly varying function of the relative dt dt RBS energy aperture and the beam energy. For a wide range of energies τL τ and apertures, the cross section is in the range of which gives for the initial luminosity lifetime L, σ ≈−× −25 2 RBS 2310 cm . τ ==1 N0 L 22kN L The lifetime for beam loss from this process depends on the 0 σσRBS o RBS luminosity, since the loss occurs due to beam-beam collisions.

11/26/01 USPAS Lecture 18 27 11/26/01 USPAS Lecture 18 28 N in which L 0 is the initial luminosity, and N0 is the initial number of τ = 0 particles. L 2L oppσ Example: Consider a high luminosity electron-positron collider, for which =3x1034 cm-2 s-1, and for which N = 5x1013. The L 0 0 34 -2 -1 Example: The LHC, for which L0=10 cm s , and for which N0 = luminosity lifetime due to radiative Bhabha scattering will be 14 about an hour. 3.6x10 . The luminosity lifetime due to pp collisions will be about 50 hours. Hadron colliders: beam loss occurs through inelastic reactions p+−> p() p hadrons At very high energies, this cross section is slowly varying with σ ≈ -25 2. energy, and is about pp 100 mbarn = 10 cm The luminosity lifetime will be Mechanisms for emittance growth 1. Scattering from atoms of the residual gas in the beam vacuum chamber

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 sin2 Q Elastic Coulomb scattering: Random, small angle scattering ξ cos2 Q   (multiple Coulomb scattering) causes transverse emittance growth   =   n   úξ π Q  úú+ ∆  n+1 −QQsin22 cos π Q n To discuss this, we will start by deriving a general relation for ξ amplitude growth from any source of random angular kicks Then we look at the ππchange in the amplitude ξξ delivered to the beam. 2 dr ==−=−∆βθφβθ2 22 + 2 θ rrnn+1 r2 r nnnnsin Let n be an angular kick delivered to a particle at a particular dn location in the ring, on the nth turn. Let’s look at the change in the In general, this will depend on the value of θ .. But if we average phase-amplitude variables at that point in the ring. n θ over many turns, and the kicks n are truly random from turn to Let the initial values of the phase-amplitude variables on turn n be θ φ ξ ξú ′ turn, then the piece linear in n averages to zero, and we will have rn and n, with corresponding Floquet variables n and n. When x changes by θ , the change in Floquet coordinate is ∆ξβθú = Q . dr2 n n = βθ2 We add this to the original Floquet coordinate and propagate this dn around the ring once, to find

11/26/01 USPAS Lecture 18 31 11/26/01 USPAS Lecture 18 32 that is, the amplitude (and emittance) will grow proportional to the Using these expressions in the formula for the amplitude growth average of the square of the kick angle. Let us now apply this to above due to Coulomb scattering in the residual gas, and the case of multiple Coulomb scattering. integrating around the ring, gives 2 2 dr 13. 6 MeV C = β()s     The average scattering angle squared for Coulomb scattering, due dn pc X0 to passage through a distance s in a medium, is given in the β “Handbook”, p 213, as in which C= machine circumference. 2 This emittance growth is generally consequential only in proton 13. 6 MeV s θ 2 ≈   machines, for which there is no natural damping mechanism.  pc  X β 0 2. Coulomb scattering of one particle by other particles in the The quantity X0 is called the radiation length; a formula is also bunch: given on the same page in the “Handbook” This is called intrabeam scattering.

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Small angle Coulomb scattering equalizes the beam temperature in dr2 all dimensions: it causes a transfer of emittance from one = βθ2 dn dimension to another. Above transition, the emittance in all three degrees of freedom can grow. The relations for the growth time are is applicable. The source of the random kicks could be dipole quite complex: they are given in p 125-127 of the “Handbook”. power supply noise, producing random field noise with a mean The growth times have a form very similar to that of the Touschek square value (∆B)2 : lifetime: (∆BL)2 2 1 rcN2 θ 2 = ∝ 0 b f ( ,,,,,) ( )2 T γεεε4 xy,, p xyxy x B0 xy,,∆ p xyL ββεεδη ρ ∆ (∆ )2 3. Emittance growth from random noise sources: Random ground motion, with a mean square amplitude x , will produce angular kicks (by misaligning quadrupoles) given by Random power supply noise, and ground motion, can cause transverse emittance growth (∆x)2 θ 2 = The general relation f 2

11/26/01 USPAS Lecture 18 35 11/26/01 USPAS Lecture 18 36 in which f is the focal length of the quadrupole in question. Plugging in the numbers, such a vibration would produce an emittance growth time of about 5x10-14 m/s. Thus, the emittance Example: If a machine has a revolution period T, then the rms would increase by 2x10-9 in about 11 hours. Since this is a typical emittance growth rate per unit time due to random quadrupole store length, this would be a serious problem. motion will be εβ(∆x)2 d = 1 dt2 T f 2 Consider the Tevatron collider, in which the rms emittance is approximately 2x10-9 m-rad, and for which T=21 µs. Let a quadrupole with a focal length of 5 m, at a β of 50 m, suffer random noise vibrations. Let the rms amplitude be 1 nm, and assume that all this vibration is directly reflected in the field. How long will it take for the emittance to double?

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