Synchrotron radiation

Transcription

1 Synchrotron radiation When a particle with velocity v is deflected it emits radiation : the synchrotron radiation. Relativistic particles emits in a characteristic cone 1/g The emitted power is strongly dependent from the energy (^4) and inversely proportional to the curvature radius (^2) In the accelerators particle emit in the dipoles, but also in the insertion devices. Due to the very interesting characteristics of the emitted radiation special periodic magnetic systems (wigglers, ondulators) are integrated in the storage rings (in dedicated insertion lines) to produce high brillance photon flux.

2 Electron frame Lab frame 1/g f q velocity b acceleration acceleration Transformation between frames:- tan q = g -1 sin f (1+b cos f ) -1 If f = 90 0 then q = g -1

3 SR Properties We have : X = gq, w r = bc/r (cyclotron frequency), C = e 0 c ω c (critical frequency) = 3 2 γ3 ω ρ = 3γ3 c 2ρ Frequency spectrum: dp(ω) dωdω = β2 q 2 ω 2 C 16π 3 e 2 G(ω) + e G(ω) With G(ω) = 2 1+X2 K 2X 1+X2 ω ρ γ ξ, G(ω) = K ω ρ γ ξ Respectively the components for the polarization parallel and perpendicular to the orbit plane. So : dp(ω) = 3 γ2 q 2 dωdω C 16π 3 ω ω c 2 parallel perpendicular 1 + X X K 2 3 ξ + 2 K 1+X ξ Due to the properties of the asymptotic behavior of the Bessel function the SR is negligible for ξ >1. This means that ω 2ω c 1 + X 2 3 2, θ 1 γ So the radiation has a continuous spectrum up to the critical frequency, than it decrease. The radiation for high frequency is confined in a 1/g angle. For big angles -> low frequencies. ω c ω 1 3

4 SR Properties INTEGRATING in dw => Angular distribution dp dω = 7 q2 γ 2 ω c 96 π c ε X X X 2 And INTEGRATING on the full angular range => Energy flux I ω = 2q2 γ 9ε 0 c S ω, with S(x) = 9 3 ω c 8π x K 5 3 j dj x 0 S x dx = 1 The energy flux gives the instantaneous radiated power: P γ = 1 4q 2πρ 0 I ω dω = 2 γω c or in a more convenient form: 36 π ρ ε 0 P γ = c C γ 2π E 4 ρ 2 ~E2 B 2 with C γ = m GeV 3 The critical photon energy is ħw c =0.665E 2 [GeV]B[T] The total energy radiated in one revolution is U 0 = E4 C γ 2π ds ρ 2 and for an isomagnetic ring -> U 0 = E4 C γ, average power -> P ρ γ = U 0 = c C γ T 0 2π E 4 Rρ (T 0 = βc = revolution period) 2πR

5 SR Properties

6 Quantistic point of view Emitted radiation quanta with energy u=ħw n(u)du=i(w)dw number of photons per unit time and emitted in the interval w n u = 9 3 8πu c 2 P γ u u c K 5 3 j dj The total number per second is: = 0 n u du = 15 3P γ 8 u c = αcγ 2 3ρ with a= fine structure constant. The average number per revolution is: N 2 p r/c= 5π 3 αγ The moments of the energy distribution are : u = 1 8 N 0 u n u du = u 15 3 c, u 2 = 1 N 0 u 2 n u du = 11 u 27 c 2, u 2 = 3C uc γ 4π ħc 2 E 7 mc 2 3 ρ 3 We can express the distribution of the quantum fluctuation as : N u 2 Seventh power!!!! = Cost E7 ρ 3

7 Synchrotron Cooling (heating?) Without energy losses the 6D emittance is an invariant of motion But particle loose energy by synchrotron radiation. This energy is restored in the RF cavities by a longitudinal field In longitudinal motion higher energy particles lose more energy than the lower energy ones. They recover energy with the same field -> The motion is damped -> cooling In transverse space the synchrotron radiation is emitted in a 1/g cone. Loss of transverse momentum. The energy is recovered longitudinally -> Angle damping, position unchanged -> cooling

8 Longitudinal damping In one revolution a particle (not the synchronous one) loses U(E) by SR and accelerate under the influence of the e.m field in the RF cavity E=qV(t) with t=arrival time θ. So the energy variation for the non synchronous ω 0 particle is: 1) d( E) dt = qv τ U(E) T 0 Now let s expand the SR power around the synchronous energy at the first order : U E = U 0 + W E, W = du de E 0 Let s consider that : α c = E C this define the path difference of the non synchronous particle in respect C E to the synchronous one. The difference in the arrival time coordinates will be: C E τ = α c = α E ce ct 0. So the time derivative of t is: E 2) dτ dt = α E c E Combining 1 and 2 we get a classical second order damped equation : d 2 τ + 2α dτ dt 2 E + ω dt s 2 τ = 0 with α E = W 2T 0 (damping term) and : ω s = α cq dv synchrotron frequency T 0 E dτ

9 Longitudinal damping Remind (longitudinal motion) d 2 τ dt 2 + 2α E dτ dt + ω s 2 τ = 0 with α E = W, W = du 2T 0 de E=E 0, U = synchrotron radiated energy, ω s = α cq V, α ET c =momentum compaction 0 Damped solution: τ t = Ae α Et cos(ω s t φ 0 )

10 Transverse Damping, Vertical motion y y In the vertical plane the particle undergoes Betatron oscillations. So the vertical displacement will be y = A βcosθ, y = A β sinθ where the amplitude A is given by the Courant Snyder invariant A 2 = γy 2 + 2αyy + βy 2 For every turn the lost energy (SR with 1/g angle) is restored by the RF cavity (zero angle). This change the longitudinal momentum. For the zero synchrotron amplitude particle : p p = y y = U E with U -> energy lost by SR and E -> nominal energy. Using the Amplitude definition it is possible to demonstrate that after many kicks, in average, (averaged on all the betatron phases) the amplitude to the first order will vary as : A A So we will have that da = U A with a damped solution Ae t αy with dt 2ET 0 α y = U -> damping decrement. 2ET 0 = U 2E

11 Transverse Damping. Horizontal motion Here the horizontal displacement is given by x = x β + x D, x = x β + x D, x D = D(s) E E x D = D s E E Emitting an energy u by SR the betatron displacement will vary by: δx β = D s u E, δx β = D s u E Always calculating the average of the Amplitude variation we get (after calculations): A A = U 0 2E D ρ 2K(s) + 1 ρ 2 ds ds ρ 2 1 = D U 0 2E

12 Transverse Damping. Horizontal motion 1) The second term is positive so the SR emission amplify the betatron motion. 2) Including the RF damping term (see vertical damping) A A = (1 D) U 0 2E so the Horizontal damping coefficient will be: α x = (1 D) U 0 2T 0 E

13 Summarizing Three dimensions : solution for the amplitude Aw for the different planes can be written as: Ae t α w Where α x = I x P γ 2E, α y = I y P γ 2E, α E = I E P γ 2E, I x = 1 D, I y = 1, I E = 2 + D -> Robinson Theorem I i = 4, or I x + I E = 3 The correspondent damping time constant will be: τ x = 2E I x P γ, τ y = 2E I y P γ, τ E = 2E I E P γ,

14 Can we go to zero emittance? NO Why? Photons are stochastically emitted in a very short time ~r/cg This is much shorter than the revolution period. So emissions are instantaneous.. In this Dt the particle makes a discontinuous energy jump. Emissions are independent so the obey the Poisson distribution. These emissions perturb the particle orbit adding a random noise spread, so increasing the average oscillations. Equilibrium is attained when the quantum fluctuation rate equals the radiation damping

15 Quantum excitation In the energy domain a particle with DE in respect to the nominal energy undergoes to synchrotron oscillations with amplitude A. If at t 1 a photon with energy u is emitted DE will be E = A 0 e jω(t t 0) ue jω t t 0 = Ae jω t t 0 with A 2 = A u 2 2A 0 u cosω t t 0 Since the emission is independent t is equally distributed in time and in average the oscillatory term will disappear giving: So we will have: δa 2 = A 2 A 0 2 t = u 2 da 2 dt = d A2 dt = Nu 2 with =emission rate

16 Equilibrium energy spread Let s add to the last equation the damping term as seen in the damping part: d A 2 dt = 2 A2 τ E + Nu The stationary state solution will be A 2 = 1 2 Nτ Eu 2 so for a monoenergetic emission σ E 2 = A2 2 = 1 4 Nτ Eu 2 To take into account a continuous spectrum σ E 2 = 1 4 G Eτ E Where a) G E = N u 2 = average emission on the ring in all the spectrum s b) N = 0 n u du = rate of the emitted photons in all the spectrum c) N u 2 = 0 u 2 n u du = d A2 dt = Average amplitude growth rate by photons emitted in all the spectrum Remember (slide on Q fluctuation) u 2 = 3C uc γ 4π ħc 2 E 7 mc 2 3 ρ 3 If we insert P γdesign = c C γ E 4 = orbit 2π ρ 2 P γ 1 ρ 2 ρ2 We have N u 2 design orbit = 3ħc C γ 3 2 u ρ 3 P γ 1 ρ 2 with C u = So taking (a) and τ E = 2E I E P γ

17 G E = 3ħc 2 C uγ 3 P γ 1 ρ 2 1 ρ 3 σ E 2 = 3ħγ4 mc 3 4I E C u 1 ρ 2 1 ρ 3 σ E E 2 = C q γ 2 4I E 1 ρ 2 1 ρ 3 For an isomagnetic ring : with C q = 3C uħ 4mc = m σ E E 2 γ 2 C q yields σ = E ρi E E = γ I E ρ[m]

18 Equilibrium emittance Same procedure: δx β = D s u, E δx β = D s u + u θ (θ < 1 γ) E E The second term in the angular kick is negligible Plugging in the Courant Snyder invariant to find the average amplitude oscillations (complex) 1) δ A 2 = H u E 2 with 2) H = 1 β x D 2 + β x D 1 2 β xd 2 Considering (1) the growth rate is obtained by replacing u2 by N u 2 and averaging on the accelerator circumference G x = d A2 dt = 1 2πRE 2 N u2 Hds = HN u2 s E 2 Adding the RF damping term we always get a damped equation: d A 2 dt = 2 A2 τ x + G x with solution -> A 2 = 1 2 G xτ x and σ xβx 2 = 1 2 β x A 2 Taking into account N u 2 design orbit we finally get :

19 G x = C qcr 0 γ 5 ρ 2 1 ρ 2 ε x = σ xβx 2 β x H ρ 3 = 1 G 4 xτ x = C q γ 2 I 1 x ρ 2 C q = 3C uħ 4mc = m H ρ 3 This is called natural emittance. For isomagnetic rings: ε x = C q γ 2 I x ρ H dipoles = I E I x σ E E 2 H dipoles

20 Vertical plane δy β = 0, δy β = u E θ y δ A 2 = u E 2 θ y 2 β y σ yβy 2 = 1 2 τ yβ y G y G y = N u2 θ y 2 βy s ~ N u2 2 θ y βy s ~ N u2 β y E 2 E 2 γ 2 E 2 Since N u 2 = G E yields σy 2 we get σ E 2 = τ yβ y G y τ E G E ~ I E β y I y γ 2 E 2 and using I y=1 σ y 2 ~ C q β y ρ, ε y ~ C q β y ρ

21 Application -> laser cooling (or heating?) We insert a focalized laser to impinge on the focalized electron beam. For quasi head on collisions energetic photons are emitted by Compton backscattering effect 1 st assumption : the laser acts like an undulator with magnetic field B w = 2 2 c -> free space impedance and I = laser intensity). This emits photons with: P γ = 32π r 3 e γ 2 I, r e = classical electron radius Z 0 I (Z 0 =(ce 0 ) -1 =377 2 nd assumption: the collision beam laser are located in a non dispersive region (without D the beam sizes are smaller...the minimum beam size increase the rate of production ->luminosity...next course) 3 rd assumption: laser pulse length short in respect the Rayleigh length Z R (distance of divergence). Same for the electron beam (no hourglass effect). In this way we suppose that the collision between the laser pulse and the electron beam is represented by the crossing of two cylinder with constant radius, neglecting the natural divergence in the IP. We gat back to the previous method : we have to define the damping rate and the quantum excitation. Their equilibrium will provide the equilibrium emittances and energy spread.

22 Damping Consider a laser with E L = I dx dy dz For an electron travelling in z the emitted power will be: c dz P γ = 32π c 3 r e γ 2 E L Z R L with L = laser wavelength (from lasers Z R L 4π transverse sizes ). = σ w 2. The waist is the location with minimum This is the electron energy loss DE!!! So we can provide the number of collisions (turns if 1 collision per turn) necessary to damp all the beam energy The average damping rate will be : n d = E E = L μm Z R [mm] E L J E[MeV] τ L = 1 n d T 0 = 1 ε n,x,y dy dx = E E T 0

23 Quantum excitation Zero dispersion region Emission in a straight session of un undulator Compton is a kinematical effect. The most efficient in boost the photon energy. The photon are emitted in a Spectrum with an energy cut off w c. dn γ = 1 de γ = 3 E 1 2Υ + dω ħω dω ħω 2 2Υ2, c with ω c = 4γ 2 ω L, Υ = ω ω c and w L = laser frequency Kinematic -> energy angle emitted photon relationship with q = emission angle ω γ = ω C 1+γ 2 θ 2

24 Emittance The diffusion recoil of the particle after the photon emission is δφ = ħω θ E Integrating on the full photon spectrum and projecting on the two planes we obtain the emittance diffusion: ε n,x\y = γβ ω c IPx,y δφ 2 dn γ dω dω = 3 10 With c = compton wavelength = h/mc = [m] c E L E β IPx,y So the average quantum excitation rate will be dε dt = 3 10 c E L E β IPx,y T 0 The balance between the Q.Excitation and the damping rate will give : n,x\y equil. = dε dt = 3 c β 10 IPx,y L

25 Energy spread In this case the quantum excitation rate is d σ E 2 dt = 1 ω c ħω 2 dn γ T 0 0 dω = 7 ħω dω 10 c Balancing with the damping rate: E T 0 σ δ = σ E E equil = 7 10 c L γ

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27 a E To get the parameter as a function of the accelerators parameters we have to calculate W: U(radiated energy) = P γ dt = P γ dt ds ds = 1 c P γ(1 + x ρ )ds = 1 c P γ 1 + D E ρe 0 ds So: 1)W = du de = 1 dp γ c de + DP γ ds, ρe E0 Taking P γ ~E 2 B 2 dp γ = 2 P γ + 2 P γ db de E0 E 0 B de = 2 P γ + 2 P γ dx db E 0 B de dx = 2 P γ + 2 P γd db E 0 BE 0 dx Substituting in (1) du de = 1 c 2 P γ + 2 P γd db E 0 BE 0 dx + DP γ ds = U db DP ρe E0 E 0 cu γ 0 B dx + 1 ds ρ E0 So : α E = W 2T 0 = U 0 T 0 E 2 + D where D is the partition number D = 1 2 db DP cu γ + 1 ds = D 2K(s) B dx ρ E0 ρ ρ ds 2 ds ρ 2 1, with K s = 1 db Bρ dx (Qpole gradient) For an isomagnetic ring: D = 1 2π D 2K(s) + 1 ds and if the isomagnetic ring has separated functions magnets : ρ 2 Dipole D = 1 2πρ D(s) ρ ds = α cr ρ So for separate functions machines α E = P γ E = U 0 ET 0

28 Coupling Linear coupling. Coupling coefficient κ Errors, solenoids, skew Qpoles Quantum excitation and damping is shared between the two planes If we take into account that, since also in the vertical plane we feel the dispersion, ε x + ε y =ε natural -> ε x = 1 1+κ ε natural, ε y = κ 1+κ ε natural