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Electron beam proper.es and FEL Lecture III - PowerPoint PPT Presentation

Electron beam proper.es and FEL Lecture III Massimo.Ferrario@LNF.INFN.IT SASE FEL Electron Beam Requirements: High Brightness B n 1 + K 2 2 ( ) minimum radiation MIN wavelength r B n K


  1. ¡Electron ¡beam ¡proper.es ¡and ¡FEL ¡– ¡Lecture ¡III ¡ Massimo.Ferrario@LNF.INFN.IT ¡

  2. SASE FEL Electron Beam Requirements: High Brightness B n 1 + K 2 2 ( ) minimum radiation MIN ∝ σ δ wavelength λ r γ B n K 2 γ B n K 2 energy B n = 2I spread B n undulator 2 parameter ε n γ 3 2 gain L g ∝ length K B n n 1 + K 2 2 ( ) B n R. Saldin et al. in Conceptual Design of a 500 GeV e+e- Linear Collider with Integrated X-ray Laser Facility , DESY-1997-048

  3. Short Wavelength SASE FEL Electron Beam Requirement: High Brightness B n > 10 15 A/m 2 Bunch compressors (RF & magnetic) B n ≈ 2I 2 ε n Laser Pulse shaping Emittance compensation Cathode emittance

  4. Longitudinal ¡Manipula.on ¡

  5. The problem of relativistic bunch length Low energy electron bunch injected in a Length contraction? linac: γ = 1000 γ ≈ 1 L b ! L b = 3 mm ≈ L b " L b = γ = 3 µ m

  6. Magnetic compressor (Chicane)

  7. Magnetic compressor (Chicane) Δ Ε / Ε Δ Ε / Ε Δ Ε / Ε …or over- under- compression compression σ z 0 ‘chirp’ z z z σ z σ E / E V = V 0 sin( ωτ ) Δ z = R 56 Δ Ε / Ε RF Accelerating Path Length-Energy Courtesy Voltage Dependent Beamline P. Emma

  8. Coherent Synchrotron Radiation (CSR) • Powerful radiation generates energy spread in bends • Energy spread breaks achromatic system • Causes bend-plane emittance growth (short bunch worse) coherent radiation for λ > σ z bend-plane emittance growth σ z λ Δ Ε / Ε = 0 L 0 s R Δ x e – Δ Ε / Ε < 0 θ Δ x = R 16 (s) Δ E/E overtaking length: L0 ≈ (24 σ z R 2 ) 1/3 Courtesy P. Emma

  9. Velocity bunching concept (RF Compressor) If the beam injected in a long accelerating structure at the crossing field phase and it is slightly slower than the phase velocity of the RF wave , it will slip back to phases where the field is accelerating, but at the same time it will be chirped and compressed. The key point is that compression and acceleration take place at the same time within the same linac section, actually the first section following the gun, that typically accelerates the beam, under these conditions, from a few MeV (> 4) up to 25-35 MeV.

  10. C_Band injector FEL Single Spike Velocity Bunching THz Radiation Narrow THz Rad LASER COMB 2 pulses FEL DWFA PWFA Thomson LWFA_ext

  11. Average current vs RF compressor phase OVER- COMPRESSION 1300 1200 1100 1000 Average current (A) HIGH 900 COMPRESSION 800 700 600 MEDIUM 500 COMPRESSION 400 300 LOW COMPRESSION 200 100 -95 -90 -85 -80 -75 -70 -65 -60 RF compressor phase (deg)

  12. Transverse ¡Beam ¡Dynamics ¡ https://arxiv.org/ftp/arxiv/papers/1705/1705.10564.pdf

  13. Trace space of an ideal laminar beam " x $ p x << p z # x = dx dz = p x ! $ p z % X’ X

  14. Trace space of a laminar beam X’ X

  15. Trace space of non laminar beam X’ X

  16. In a system where all the forces acting on the particles are linear (i.e., proportional to the particle’s displacement x from the beam axis), it is useful to assume an elliptical shape for the area occupied by the beam in x-x‘ trace space. x ! x !  x + kx = 0 x

  17. ε g Geometric emittance: Ellipse equation: γ x 2 + 2 α x $ x 2 = ε g x + β $ 2 = 1 Twiss parameters: β = − 2 α ! βγ − α Ellipse area: A = πε g

  18. Trace space evolution No space charge => cross over With space charge => no cross over X’ X’ X X

  19. ε rms rms emittance + ∞ + ∞ ∫ ∫ ( ) = 0 ( ) dxd ! f x , ! x f x , ! x x = 1 ! x’ −∞ −∞ rms beam envelope: σ x' + ∞ + ∞ 2 = x 2 = ∫ ∫ x 2 ( ) dxd & f x, & x x σ x x −∞ −∞ σ x Define rms emittance: γ x 2 + 2 α x $ x 2 = ε rms x + β $ x 2 such that: βε rms σ x = = x 2 σ x' = % = γε rms

  20. x 2 = σ x = βε rms x '2 = σ x ' = γε rms α = − " β σ xx ' = x ! x = − αε rms 2 γβ − α 2 = 1 It holds also the relation: 2 " % 2 2 σ x ' σ x − σ xx ' Substituting we get α , β , γ = 1 $ ' ε rms ε rms ε rms # & We end up with the definition of rms emittance in terms of the second moments of the distribution: 2 − σ xx ' 2 = x 2 − x " ( ) 2 2 σ x ' x 2 x = p x " x ε rms = σ x " p z

  21. What does rms emittance tell us about phase space distributions under linear or non-linear forces acting on the beam? x 2 − x # 2 2 = x 2 x ε rms # x’ x’ a’ a’ a x a x x = Cx n x, " x ! Assuming a generic correlation of the type: When n = 1 ==> ε rms = 0 ( ) x 2 n − x n + 1 2 2 = C 2 x 2 ε rms When n = 1 ==> ε rms = 0

  22. Normalized rms emittance: ε n,rms p x = p z ! x = m o c βγ ! x Canonical transverse momentum: p z ≈ p 2 = 1 2 − σ xp x 2 − xp x ( ) 2 2 σ p x x 2 p x ε n , rms = σ x m o c Liouville theorem: the density of particles n , or the volume V occupied by a given number of particles in phase space (x,p x ,y,p y ,z,p z ) remains invariant. It hold also in the projected phase spaces (x,p x ),(y,p y )(,z,p z ) provided that there are no couplings

  23. Limit of single particle emittance Limits are set by Quantum Mechanics on the knowledge of the two conjugate variables (x,p x ). According to Heisenberg: σ x σ p x ≥ ! 2 This limitation can be expressed by saying that the state of a particle is not exactly represented by a point, but by a small uncertainty ! 3 volume of the order of in the 6D phase space. In particular for a single electron in 2D phase space it holds: % = 0 classical limit ' ε n , rms = 1 2 ⇒ 2 − σ xp x 2 σ p x m o c = ! c ! σ x & ≥ 1 2 = 1.9 × 10 − 13 m quantum limit m o c ' 2 ( Where is the reduced Compton wavelength. ! c

  24. Normalized and un-normalized emittances p x = p z ! x = m o c βγ ! x ε n , rms = 1 2 − x βγ " 2 − xp x ( ) = βγ ε rms ( ) = 2 2 x 2 x 2 ( ) p x βγ " x x m o c Assuming small energy spread within the beam, the normalized and un-normalized emittances can be related by the above approximated relation. This approximation that is often used in conventional accelerators may be strongly misleading when adopted to describe beams with significant energy spread, as the one at present produced by plasma accelerators.

  25. When the correlations between the energy and transverse positions are negligible (as in a drift without collective effects) we can write: 2 x ! x 2 − βγ 2 2 = β 2 γ 2 x 2 x ! ε n , rms Considering now the definition of relative energy spread: β 2 γ 2 − βγ 2 2 = σ γ 2 βγ which can be inserted in the emittance definition to give: 2 x 2 = β 2 γ 2 σ γ x 2 + βγ x 2 − x ! ( ) 2 2 2 x 2 x ε n , rms ! ! Assuming relativistic electrons ( β =1) we get: 2 + ε rms ( ) 2 = γ 2 2 σ x 2 σ ! 2 ε n , rms σ γ x

  26. 2 + ε rms 2 + ε rms ( ) ( ) = γ 2 2 = γ 2 2 σ x 2 σ ! 2 2 σ o , ! 4 2 ( ) z − z o ε n , rms σ γ σ γ x x showing that beams with large energy spread an divergence undergo a significant normalized emittance growth even in a drift Simulation Formula • Energy 350 MeV • Beam divergence 1 mrad • Energy spread 1% • Beam spot-size 1 µ m < γ > ε Z

  27. Phase space, slice emittance and longitudinal correlations p x Slice Phase x Projected Phase Space Spaces FEL cooperation length

  28. x 2 = σ x = βε rms x '2 = ! σ x = γε rms σ xx ' = x ! x = − αε rms γβ − α 2 = 1 It holds also the relation: 2 " % 2 2 σ x ' σ x − σ xx ' Substituting we get α , β , γ = 1 $ ' ε rms ε rms ε rms # & We end up with the definition of rms emittance in terms of the second moments of the distribution: 2 − σ xx ' 2 = x 2 − x " ( ) 2 2 σ x ' x 2 x = p x " x ε rms = σ x " p z

  29. Envelope Equation without Acceleration Now take the derivatives: d σ x dz = d 1 d 1 x = σ x ! x 2 = dz x 2 = x 2 x ! dz 2 σ x 2 σ x σ x 2 + x !! d 2 σ x 2 2 2 x 3 = σ ! dz 2 = d = 1 d σ x ! 3 = 1 σ x ! dz − σ x ! ) − σ x ! − σ x ! x 2 + x ! ( x x x x x x ! x 3 dz σ x σ x σ x σ x σ x σ x σ x 2 − σ xx ' 2 σ x ' 2 2 + x !! x 3 + x !! x σ x = σ x = ε rms And simplify: !! 3 σ x σ x σ x σ x We obtain the rms envelope equation in which the rms emittance enters as defocusing pressure like term. 2 2 σ x − x !! x 3 ≈ T ε rms = ε rms V ≈ P !! 3 σ x σ x σ x

  30. Beam Thermodynamics Kinetic theory of gases defines temperatures in each directions and global as: 2 T = 1 ) E k = 1 2 m v 2 = 3 ( k B T x = m v x 3 T x + T y + T z 2 k B T Definition of beam temperature in analogy: 2 2 = β 2 c 2 ε rms 2 = β 2 c 2 x 2 = β 2 c 2 σ x ' 2 k B T beam , x = γ m o v x v x ! 2 σ x 2 2 = γ m o β 2 c 2 ε rms k B T beam , x = γ m o v x 2 σ x We get: 2 2 beam , x = nk B T beam , x = n γ m o β 2 c 2 ε rms 2 = N T γ m o β 2 c 2 ε rms P 2 σ x σ L σ x N 2 = N T n = πσ L σ x σ L

  31. k B T beam , x = γ m o β 2 c 2 ε rms β x ( ) S = kN log πε

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