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Advanced laser-driven X-ray sources Stefan Karsch - PowerPoint PPT Presentation

Advanced laser-driven X-ray sources Stefan Karsch Ludwig-Maximilians-Universitt Mnchen/ MPI fr Quantenoptik Garching, Germany 1 What are x-rays ? slide courtesy A. Dpp, LOA and what do we use them for ? m mm m nm pm Wavelength


  1. Advanced laser-driven X-ray sources Stefan Karsch Ludwig-Maximilians-Universität München/ MPI für Quantenoptik Garching, Germany 1

  2. What are x-rays ? slide courtesy A. Döpp, LOA … and what do we use them for ? m mm μ m nm pm Wavelength Radio Microwave IR UV XUV X-rays γ -rays Energy meV eV keV MeV • EM-radiation in the above-keV range • X-rays were discovered in 1895 X-ray Airport security • First medical radiography 1896 NDT diffraction • Since then widely used in science, science Cargo scanners Dental industry and medicine radiography • Most applications rely on absorption X-ray CT medicine industry properties 2

  3. high harmonic sources filter „wiggly“ electron - sources a reflected t t o s e c o n p d pulse u l s e plasma target e s l Laser-driven X- u p t n e d i c n i ray sources (metal jet) bremsstrahlung and line (K- α )-sources relativistic electron beam + • undulator = undulator source, FEL • plasma fields = Betatron source • laser pulse = Thomson/Compton source 3

  4. Quality scale for x-rays ... their brilliance: photons Brilliance = mm 2 ⋅ mrad 2 ⋅ s ⋅ 0.1% bandwidth transv. emittance long. emittance (=phase space area) 1. many photons 2. small bandwidth 3. low divergence 4. small source 5. short duration That‘s where LWFA sources excel 4

  5. „Wiggly“ electron X-ray sources: Ingredients: relativistic electron beam + laser fields undulator plasma fields e - Thomson scattering undulator radiation, FEL Betatron radiation 10‘s keV - MeV 100‘s eV - keV keV – 10‘s keV λ l ≈ 1µm λ u ≈ 1cm λ b ≈ 500µm λ x − ray = λ u , b , l ⎛ ⎞ 2(4) γ 2 1 + ( K , a 0 ) 2 + γ 2 θ 2 ⎜ ⎟ 2 ⎝ ⎠ 5

  6. Laser-Plasma Accelerators slide courtesy A. Döpp, LOA Duration Accelerating field Focusing field Total size ns-ps 10-100 MV/m 500 T/m 10-1000 m Conventional : fs TV/m MT/m mm-cm Laser-plasma : • Relativistic electrons emit orders of magnitude more radiation than non-relativistic electrons need an accelerator ! Plasma cavity 6

  7. slide courtesy F. Pfeiffer, TUM brilliance [ph/ (sec mm 2 peak brilliance of laser driven sources mrad 2 0.1% BW)] 10 22 undulator deflecting magnet 10 15 average brilliance of laser driven sources 10 11 rotating anode 100 kW, 10 7 Bremsstrahlung costs (size) 1 10 100 1000 [M € (meter)] 7

  8. Larmor radiation (see e.g. Jackson) of an accelerated charged particle: d ! 2 ⎛ ⎞ e 2 p R = Radiation power: P ⎜ ⎟ ⎝ ⎠ 6 πε 0 m 0 2 c 3 dt d ! 2 ⎛ ⎞ e 2 dP p d Ω = sin 2 ϕ Angular distribution: R (Hertzian dipole) ⎜ ⎟ 16 π 2 ε 0 m 0 ⎝ ⎠ 2 c 3 dt Find relativistic invariant form of Larmor formula: d ! 2 ⎛ ⎞ 2 2 dP ⎛ ⎞ ⎛ ⎞ dt → d τ = 1 p − 1 dE µ → γ dt Transform time: and four-momentum: ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ d τ ⎝ d τ ⎠ ⎝ d τ ⎠ c 2 ⎝ ⎠ d ! d ! ⎡ ⎤ ⎡ ⎤ 2 2 2 2 ⎛ ⎞ ⎛ ⎞ e 2 c γ 2 ⎛ ⎞ ⎛ ⎞ e 2 c → lab. frame p − 1 dE p − 1 dE R = ⎢ ⎥ = ⎢ ⎥ P ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ( ) ( ) ⎝ d τ ⎠ ⎝ d τ ⎠ ⎝ ⎠ ⎝ ⎠ 2 c 2 2 dt c 2 dt ⎢ ⎥ ⎢ ⎥ 6 πε 0 m 0 c 2 6 πε 0 m 0 c 2 ⎣ ⎦ ⎣ ⎦ “general radiation formula” ! ( ) ! ( ) ! ! " β m e c , E = γ m e c 2 p = γ e 2 ⎡ ⎤ 2 2 # # ⇒ R = 6 πε 0 c γ 6 β − β × β P ⎢ ⎥ ⎣ ⎦ 8

  9. Longitudinal acceleration: rel. energy-momentum relation: p d ! v d ! E = γ m 0 c 2 , " " 2 + ! p = γ m 0 ( ) d d τ v E dE p dE p E 2 = m 0 c 2 d τ = c 2 ! d τ = ! p 2 c 2 → → d τ d τ d ! d ! d ! d ! ⎡ ⎤ 2 2 2 2 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ e 2 c e 2 c e 2 c p p p p R = ⎢ − β 2 ⎥ = = P ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ( ) ( ) ( ) ⎝ d τ ⎠ ⎝ d τ ⎠ ⎝ γ d τ ⎠ ⎝ ⎠ 2 2 2 dt ⎢ ⎥ 6 πε 0 m 0 c 2 6 πε 0 m 0 c 2 6 πε 0 m 0 c 2 ⎣ ⎦ Transverse acceleration: Deflection in magnetic field d ! d ! 2 2 ⎛ ⎞ d τ = 1/ γ dt e 2 c γ 2 ⎛ ⎞ e 2 c dE p p dt = 0 ⇒ P R = = ⎜ ⎟ ⎜ ⎟ ( ) ( ) ⎝ d τ ⎠ ⎝ ⎠ 2 2 dt 6 πε 0 m 0 c 2 6 πε 0 m 0 c 2 ⇓ For relativistic particles and the same acceleration, transverse deflection produces γ 2 -times stronger radiation 9

  10. β 2 sin 2 θ ⎡ ⎤ " dP ! e 2 e 2 β 2 sin 2 θ cos 2 φ dP d Ω = ! ⎢ ⎥ d Ω = 3 1 − ⊥ ( ) ( ) ( ) 16 π 2 ε 0 c 5 γ 2 1 − β cos θ 16 π 2 ε 0 c 1 − β cos θ ⎢ 2 ⎥ 1 − β cos θ ⎣ ⎦ Larmor radiation for longitudinal (left) and transverse (right) deflection for particle velocities of v = (0,0.3, 0.9, 0.99) c, corresponding to γ = (1, 1.4, 10, 100). Note different scale for yellow and red distributions 10

  11. Longitudinal acceleration: Injection radiation injection radiation driving laser + plasma wave laser direction density (schematic) Quasi-monochromatic electron spectrum indicates localized injection: 11

  12. How can we wiggle an electron beam: Insertion devices Slide courtesy A. Döpp, LOA � � � + � Lorentz Force � � � × � � � = � � Purely electric: Electromagnetic: Purely magnetic: Betatron Thomson/ undulator/ radiation Compton wiggler scattering 12

  13. The radiated energy per solid angle is given by the time integral of the radiated power per solid angle: ∞ ∞ ! 2 dt dW dP ( ) ∫ ∫ d Ω = d Ω dt = c ε 0 R E t −∞ −∞ " ∞ dW ( ) 2 ∫ 2 c ε 0 R ! d Ω = E ω d ω And likewise in the frequency domain: 0 " dW 2 ( ) 2 d ω d Ω = 2 c ε 0 R 2 ! E ω Therefore, the integrand describes the angular and spectral distribution. ! Finding the fields from ⎡ ⎤ ⎡ ⎤ ! β q 1 q the Lienard-Wiechert ⎢ ⎥ ⎢ ⎥ φ = − A = − ! ! , ( ) R ( ) R 1 − ! 1 − ! 4 πε 0 4 πε 0 c Potentials (Jackson): ⎢ ⎥ ⎢ ⎥ n ⋅ β n ⋅ β ⎣ ⎦ ⎣ ⎦ ! E = −∇ φ − ∂ ret ret A ( ) ∂ t ! ! ( ) × ! ! ! " ⎡ ! ⎤ ! n × n − β β B = ∇ × A ! ! n − β e ⎢ ⎥ B = 1 ! ⎡ ⎤ ⇒ E = R 2 + n × , E ! ! ⎣ ⎦ ret ⎢ ( ) ( ) ⎥ γ 2 1 − ! c 1 − ! 4 πε 0 3 3 c n ⋅ β n ⋅ β R ⎢ ⎥ ret ⎣ ⎦ Coulomb field radiation field 13

  14. FT to frequency domain , neglecting the Coulomb field ∼ 1/R 2 γ 2 : ( ) ! ! ( ) × ! ! # n × n − β β ( ) ∞ ! ! ! ie ω ( ) d ′ B = 1 ! ( ) = ( ) ( ) / c " " " ∫ i ω t + R ′ ′ t E ω n × E ω e t , ! ( ) c 1 − ! 3 c 32 π 3 c ε 0 R n ⋅ β R −∞ This field plugged into the integrand of the angular distribution of radiated energy gives the so- called radiation integral (Jackson or Corde, Rev. Mod. Phys. 85 1 2013): ( ) ! ! ( ) × 2 ! ! " n × n − β β d 2 W e 2 i ω t − ! ( n ⋅ ! ) dt ∞ ∫ r / c d ω d Ω = e ! ( ) 1 − ! 16 π 3 ε 0 c 2 −∞ n ⋅ β ! ( ) e e 2 ω 2 2 n × ! ! i ω t − ! n ⋅ ! ( ) dt ∞ ∫ r / c = n × β 16 π 3 ε 0 c −∞ For any given trajectory β (t) and observation direction n, this integral describes the radiated energy per solid angle and energy interval. ⇒ Done! 14

  15. Electric field � ) × ˙ � � � � n − � n − � � n × (( � � ) � � + e � � E ( � x, t ) = e � 2 (1 − � (1 − � c n ) 3 R 2 n ) 3 R � · � � · � ret ret Fourier Transform Calculate Poynting vector (more or less) Change of variables t ret to t 2 � ) × ˙ � + ∞ � � n − � � d 2 I e 2 n × [( � � ] � � � · e i � ( t − � n · � r ( t ) /c ) dt d � d Ω = � � 4 � 2 c (1 − � � n ) 2 � � · � −∞ � � Radiated energy • Jackson, J.D. Classical Electrodynamics , 3 rd Edition, Chapter 14: Radiation by Moving Charges 15

  16. For the simple yet important case of highly relativistic electrons, negligible deflection ( β || = const = c) and on-axis observation, assuming without any loss of generality e.g. a sinusoidal transverse acceleration, we have: ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 0 a sin γ 2 ω u t 0 ! ! ! ⎜ ⎟ ⎜ ⎟ # ⎜ ⎟ n = β = β = 0 ⎟ , , 0 ⎜ ⎟ ⎜ 0 ⎟ ⎜ β " ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ 1 ⎝ ⎠ ⎝ ⎠ 0 ! ! ( ) × ! ! " ⎡ ⎤ n × n − β β ( ) ⎣ ⎦ 1 a sin γ 2 ω u t and becomes which is essentially just the transverse ! ( ) 1 − ! 2 1 − β ! n ⋅ β acceleration again times a constant. Conclusion: for the above conditions, the emitted spectrum is just the Fourier transform of the transverse acceleration and therefore the wiggling force. What happens if the above assumptions are violated? We have to solve the radiation integral for special cases (or numerically, if we are lazy) 16

  17. Electrons on a circular orbit: Bending magnet radiation nonrelativistic relativistic ( γ = 1) ( γ =3) y y F � C F � C v s � v s � 1 y ' 17

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