* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Electron Dynamics - CERN Accelerator School
Hydrogen atom wikipedia , lookup
Atomic theory wikipedia , lookup
Matter wave wikipedia , lookup
Wave–particle duality wikipedia , lookup
X-ray photoelectron spectroscopy wikipedia , lookup
Particle in a box wikipedia , lookup
Rutherford backscattering spectrometry wikipedia , lookup
X-ray fluorescence wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Radiation effects in electron storage rings Average radiated power restored by RF U ≅ 10 – 3 of E0 0 • Electron loses energy each turn • RF cavities provide voltage to accelerate electrons back to the nominal energy Electron Dynamics VRF > U0 Radiation damping • Average rate of energy loss produces DAMPING of electron oscillations in all three degrees of freedom (if properly arranged!) with radiation Quantum fluctuations • Statistical fluctuations in energy loss (from quantised emission of radiation) produce RANDOM EXCITATION of these oscillations L. Rivkin, PSI Equilibrium distributions Introduction to Accelerator Physics Course CERN Accelerator School, Baden bei Wien, September 2004 • The balance between the damping and the excitation of the electron oscillations determines the equilibrium distribution of particles in the beam BD 2 Average energy loss per turn Energy gain in the RF cavities Every turn electron radiates small amount of energy Only the longitudinal component of the momentum is increased in the RF cavity E1 = E0 – U 0 = E0 1 – U0 E0 RF Cavity Since the radiation is emitted along the tangent to the trajectory, only the amplitude of the momentum changes P P⊥ P|| U0 c eVRF = U 0 P⊥ U0 c P P|| U U P1 = P0 – c0 = P0 1 – 0 E0 The transverse momentum, or the amplitude of the betatron oscillation remains small BD 3 BD 4 Energy of betatron oscillation Exponential damping Transverse momentum corresponds to the energy of the betatron oscillation Eβ ∝ A 2 But this is just the exponential decay law! A12 = A02 1 – U0 E0 or A1 ≅ A 0 1 – ∆A = – U0 A 2E The amplitudes are exponentially U0 2E0 damped −t A= A ⋅e τ The relative change in the betatron oscillation amplitude that occurs in one turn (time T0) 0 with the damping decrement ∆A = – U0 A 2E 1 τ = U0 2 E T0 BD 5 Adiabatic damping in linear accelerators In a linear accelerator: p⊥ The electron beam “emittance”: p x′ = p⊥ decreases ∝ 1 E p p⊥ p In a storage ring beam passes many times through same RF cavity BD 6 Source area, S Angular divergence, Ω The brightness depends on the geometry of the source, i.e., on the electron beam emittance RF Emittance = S x Ω Clean loss of energy every turn (no change in x’) Every turn is re-accelerated by RF (x’ is reduced) Particle energy on average remains constant BD 7 BD 8 Damping time Emittance damping in linacs: the time it would take particle to lose all of its energy ε Ω ε 4 ε 2 Ω 2 τε = ε ∝ −1γ Ω 4 or in terms of radiated power or γ 2γ E T0 U0 remember that τε = Pγ ∝ E 4 τε ∝ γε = const. 4γ E T0 E = U0 Pγ 1 E3 BD 9 Longitudinal motion: compensating radiation loss U0 Longitudinal motion: phase stability RF RF cavity provides accelerating field with frequency • h – harmonic number U0 The energy gain: BD 10 VRF U0 τ f RF = h ⋅ f 0 Particle ahead of synchronous one VRF • gets too much energy from the RF • goes on a longer orbit (not enough B) >> takes longer to go around • comes back to the RF cavity closer to synchronous part. τ U RF = eVRF (τ ) Synchronous particle: Particle behind the synchronous one • has design energy • gains from the RF on the average as much as it loses per turn U0 • gets too little energy from the RF • goes on a shorter orbit (too much B) • catches-up with the synchronous particle BD 11 BD 12 Radiation loss Longitudinal motion: damping of synchrotron oscillations Pγ ∝ E 2 B 2 During one period of synchrotron oscillation: when the particle is in the upper half-plane, it loses more energy per turn, its energy gradually reduces ε U > U0 U < U0 τ when the particle is in the lower half-plane, it loses less energy per turn, but receives U0 on the average, so its energy deviation gradually reduces Pγ∝E2B2 Displaced off the design orbit particle sees fields that are different from design values betatron oscillations: zero on average • linear term in B2 - averages to zero • quadratic term - small energy deviation Pγ ∝ E 2 • different energy: • different magnetic field particle moves on a different orbit, defined by the off-energy or dispersion function Dx The synchrotron motion is damped ⇒ both contribute to linear term in the phase space trajectory is spiraling towards the origin Pγ (ε ) BD 13 Radiation loss Pγ∝E2B2 BD 14 Energy balance Energy gain from the RF system: URF = eVRF τ = U0 + eVRF ⋅τ To first order in ε Urad = U0 + U ′ ⋅ ε electron energy changes slowly, at any instant it is moving on an orbit defined by Dx dU ′ U ≡ synchronous particle (τ = 0) will get exactly the energy loss per turn we consider only linear oscillations Each turn electron gets energy from RF and loses energy to radiation within one revolution time T0 rad dE E0 after some algebra one can write U′ = ∆ε = U0 + eVRF ⋅τ – U0 + U′ ⋅ε U0 2 +D E0 D ≠ 0 only when ρk ≠ 0 VRF = dVRF dτ τ= 0 dε = 1 eV ⋅τ – U′⋅ε dt T0 RF An electron with an energy deviation will arrive after one turn at a different time with respect to the synchronous particle dτ = –α ε E0 dt BD 15 BD 16 Synchrotron (time - energy) oscillations Synchrotron oscillations: damped harmonic oscillator d 2ε + 2α dε + Ω 2ε = 0 ε dt dt 2 αeVRF where the oscillation frequency Ω 2 ≡ T0E0 U′ typically αε <<Ω αε ≡ the damping is slow: 2T0 Combining the two equations The ratio of amplitudes at any instant τ= α ε ΩE0 Oscillations are 90 degrees out of phase θε = θτ + π 2 The motion can be viewed in the phase space of conjugate variables the solution is then: ε t =ε0e –αεtcos Ωt + θε ε, τ εˆ similarly, we can get for the time delay: τt =τ0e –αεtcos αε , Ωτ E0 ε E0 τ τˆ Ωt + θτ αε Ωτ BD 17 Orbit Length dl = 1 + ρx ds dl ds Like the tunes Qx, Qy - α depends on the whole optics ρ x Horizontal displacement has two parts: To first order xβ does not change L xε – has the same sign around the ring Length of the off-energy orbit ∆L = δ ⋅ L ε = dl = Ds ∆p ds where δ = p = ∆E E ρs A quick estimate for separated function guide field: α= x = xβ + xε Ds ds α≡ 1 L ρs Momentum compaction factor Length element depends on x BD 18 x 1 + ρε ds = L 0 + ∆L ∆L = α ⋅ δ L 1 L 0ρ0 mag D s ds = ρ = ρ0 in dipoles 1 D ⋅L mag ρ = ∞ elsewhere L 0ρ 0 But Since dispersion is approximately D ≈ R2 ⇒ α ≈ 12 typically < 1% Q Q and the orbit change for ~ 1% energy deviation L mag = 2πρ 0 α= D R ∆L = 1 ⋅ δ ≈ 10 – 4 L Q2 BD 19 BD 20 Something funny happens on the way around the ring... Revolution time changes with energy T0 = L0 cβ ∆T = ∆L – ∆β T L β dβ 1 dp Particle goes faster (not much!) β = γ 2 ⋅ p while the orbit length increases (more!) The “slip factor” η ≅ α since ∆T = α – 1 ⋅ dp = η ⋅ dp p T γ2 p Ring is above “transition energy” Not only accelerators work above transition (relativity) ∆L = α ⋅ dp p L α >> 12 γ α ≡ 12 γ tr isochronous ring: η = 0 or γ = γ tr Dante, Paradiso BD 21 Robinson theorem Damping partition numbers Transverse betatron oscillations are damped with Synchrotron oscillations are damped twice as fast Quantum nature of synchrotron radiation Damping only • If damping was the whole story, the beam emittance (size) would shrink to microscopic dimensions!* • Lots of problems! (e.g. coherent radiation) U0 1 1 τx = τ z = 2ET0 U0 1 τε = ET0 Quantum fluctuations • Because the radiation is emitted in quanta, radiation itself takes care of the problem! • It is sufficient to use quasi-classical picture: The total amount of damping (Robinson theorem) depends only on energy and loss per turn » Emission time is very short » Emission times are statistically independent U0 1 1 1 2U0 τx + τ y + τ ε = ET0 = 2ET0 J x + J y + J ε the sum of the partition numbers BD 22 (each emission - only a small change in electron energy) Purely stochastic (Poisson) process Jx + Jz + Jε = 4 BD 23 BD 24 Quantum nature of synchrotron radiation Visible quantum effects Damping only I have always been somewhat amazed that a purely quantum effect can have gross macroscopic effects in large machines; • If damping was the whole story, the beam emittance (size) would shrink to microscopic dimensions!* • Lots of problems! (e.g. coherent radiation) and, even more, that Planck’s constant has just the right magnitude needed to make practical the construction of large electron storage rings. * How small? On the order of electron wavelength h = λC E =γmc 2 = hν = hc ⇒ λ e = 1γ mc γ λe h A significantly larger or smaller value of λ C = 2.4⋅ 10 –12m – Compton wavelength Diffraction limited electron emittance ε≥ λC 1 ×N 3 4πγ – would have posed serious -- perhaps insurmountable -problems for the realization of large rings. fermions Mathew Sands BD 25 Quantum excitation of energy oscillations Photons are emitted with typical energy u ph ≈ h ω typ at the rate (photons/second) N γ3 = hc ρ P = γ u ph Fluctuations in this rate excite oscillations During a small interval ∆t electron emits photons losing energy of Actually, because of fluctuations, the number is resulting in spread in energy loss N = N ⋅ ∆t N ⋅ u ph N ± N ± N ⋅ u ph For large time intervals RF compensates the energy loss, providing damping towards the design energy E0 Steady state: typical deviations from E0 ≈ typical fluctuations in energy during a damping time τε BD 26 Equilibrium energy spread: rough estimate We then expect the rms energy spread to be τε ≈ and since σε ≈ E0 ⋅ u ph E0 Pγ and σ ε ≈ N ⋅τ ε ⋅ u ph Pγ = N ⋅ u ph geometric mean of the electron and photon energies! Relative energy spread can be written then as: σε ≈γ E0 λ– e ρ λ– e = mh c ∼ 4 ⋅ 10 – 13m it is roughly constant for all rings • typically E ∝ ρ2 e σε ~ const ~ 10 – 3 E0 BD 28 Equilibrium bunch length Equilibrium energy spread Bunch length is related to the energy spread ε More detailed calculations give ρ • for the case of an ‘isomagnetic’ lattice ρ s = ∞0 σε E with 2 = Cq E in dipoles elsewhere Energy deviation and time of arrival (or position along the bunch) are conjugate variables (synchrotron oscillations) σ στ = α ε Ωs E 2 Jερ0 recall that Ωs ∝ VRF τ τ= α ε Ωs E Two ways to obtain short bunches: hc = 1.468 ⋅ 10 – 6 m Cq = 55 32 3 m ec 2 3 GeV 2 It is difficult to obtain energy spread < 0.1% στ ∝ 1 RF voltage (power!) Momentum compaction factor in the limit of α = 0 isochronous ring: particle position along the bunch is frozen VRF στ ∝ α • limit on undulator brightness! BD 29 Horizontal oscillations: equilibrium After an electron emits a photon u ph its energy decreases: = E0 1 + δ E = E0 1 – E0 E = E0 - uph Neither its position nor angle change after emission its reference orbit has smaller radius (Dispersion) xref = D ⋅ δ It will start a betatron oscillation around this new reference orbit xβ = D ⋅ δ Horizontal oscillations excitation Emission of photons is a random process Again we have random walk, now in x. How far particle will wander away is limited by the radiation damping The balance is achieved on the time scale of the damping time τx = 2 τε σ xβ ≈ In N ⋅ τx ⋅ D ⋅ δ = 2 ⋅ D ⋅ smooth approximation for D or, typically 10-3 of R, reduced further by Q2 focusing! In large rings Q2 ~ R, so D ~ 1m σ ⋅ E σ xβ ≈ 2 R 2 E Q Typical horizontal beam size ~ 1 mm Quantum effect visible to the naked eye! Vertical size - determined by coupling σε E Equilibrium horizontal emittance Beam emittance Detailed calculations for isomagnetic lattice Betatron oscillations εx0 ≡ 2 σxβ CqE 2 H mag = ⋅ ρ Jx β • Particles in the beam execute betatron oscillations with x’ different amplitudes. and = 1 D 2 + βD′ + αD β H mag • Gaussian (electrons) • “Typical” particle: 1 - σ ellipse (in a place where α = β’ = 0) 2 is average value in the bending magnets Emittance ≡ 2 H ~ Dβ ~ R3 Q For simple lattices (smooth approximation) σx’ Transverse beam distribution H = γD 2 + 2αDD′ + βD′ 2 where Area = π ⋅ ε Units of ε m ⋅ rad σ x2 β σx = ε β σ x′ = ε / β 2 εx 0 ≈ x σx CqE R 1 ⋅ρ⋅ 3 Jx Q BD 33 1 e –x 2 / 2σ 2dx 2π σ µ x σx′ Area = π εx Probability to be inside n-σ ellipse P1 = 1 – e – 1 2 = 0.39 Pn = 1 – e – n 2 2 2 E ε∝ θ 3FFODO µ Jx Emittance x’ Probability to be inside 1-σ ellipse x′ BD 34 FODO Lattice emittance σx Electron rings emittance definition 1 - σ ellipse σ β = σx 100 2-D Gaussian distribution n x dx = ε = σ x ⋅ σ x′ ε ∝ 13 Q 10 1 0 BD 35 20 40 60 80 100 120 140 160 180 BD 36 Phase advance per cell [degrees] Ionization cooling Summary of radiation integrals E p⊥ p|| absorber acceleration σ0 σ′0 σ′ = σ′ 0 >> σ′ MS 2 σ′ 02 + σ′ MS similar to radiation damping, but there is multiple scattering in the absorber that blows up the emittance to minimize the blow up due to multiple scattering in the absorber we can focus the beam Momentum compaction factor α= I1 2πR I1 = D ds ρ I2 = ρ2 ds ds Energy loss per turn I3 = U 0 = 1 Cγ E 4 ⋅ I 2 2π I4 = D 2k + 1 ds 2 ρ I5 = H ds 3 re Cγ = 4π = 8.858 ⋅ 10 – 5 m 3 3 m ec 2 3 GeV ρ3 ρ ρ BD 37 Summary of radiation integrals (2) Damping parameter D = I4 I2 I1 = Damping times, partition numbers I2 = J ε = 2 + D , Jx = 1 – D , J y = 1 I3 = τ0 2ET0 τ0 = Ji U0 Equilibrium energy spread τi= σε E 2 Cq E 2 I 3 = ⋅ Jε I2 Equilibrium emittance σ2 C E2 I εx 0 = xβ = q ⋅ 5 β Jx I2 BD 38 Smooth approximation ds ρ2 x ≈ a β n cos s – ϕ0 βn ds ρ x s = a β s cos ϕ s – ϕ0 Betatron oscillation approximated by harmonic oscillation D ds ρ 3 I4 = D 2k + 1 ds 2 ρ I5 = H ds 3 ρ ρ hc = 1.468 ⋅ 10 – 6 m Cq = 55 32 3 m ec 2 3 GeV 2 H = γD 2 + 2αDD ′ + βD′ 2 BD 39 ϕs = s 0 ⇐ x′′ + k eff ⋅ x = 0, k eff = 12 βn Phase advance around the ring Dispersion obeys the equation Momentum compaction factor α 2π Q = ds βs β s = β n = const ds = 1 ⋅2 π R βn βn D′′ + k eff D = 1 R α= ⇒ βn = R Q ⇒ Dn = D β2 = n2 R R ⇒ β n2 R = R Q2 α ≈ 12 Qx BD 40