Laser-plasma particle accelerators. Zulfikar Najmudin

Transcription

1 Laserplasma particle accelerators Zulfikar Najmudin

2 Vulcan: High Power Laser

3 High intensity lasers The evolution of Power of small diameter laser systems State of art e.g. Vulcan Petawatt, 400 J in 400 fs = 1 PW (cf. UK power output 100 GW) focused to diameter spot φ =5µm, intensity I Wcm 2 (= Wm 2 ) Power Year Poynting vector: For E = E 0 cos ωt, I = E H = E B/µ 0 = E 2ˆk/cµ0 < I >= 1 2 E2 0/cµ 0,soE 0 = 2cµ 0 I NB for VulcanPW, E Vcm 1, cf Vcm 1 = E Bohr Since i.e. where m v x = ee 0 cos ωt, v x = (ee 0 /mω)sinωt, v x = a 0 c sin ωt, a 0 = ee 0 /m e cω, a 0 is normalised momentum, or normalised vector potential, numerically a (Iλ 2 ) 1 2 I [10 18 Wcm 2 ]λ[µm]

4 Some important intensities Energy associated with quiver motion: E = 1 2 mv2 x = e2 E0 2 2mω 2 = e2 Iλ 2 4π 2 0 mc 3 0.2Iλ2 ev where I is in Wcm 2 and λ in m. For 1 µm light; E 1 ev for I Wcm 2 (ionises solids) E 13.6 ev for I Wcm 2 (ionises gases) E 0.5 MeV for I Wcm 2 (electrons relativistic) Actually remember m v = e(e v B) So mv z = ev x B y cos(ωt) = a 0 ee 0 cos ωt sin(ωt) = 1 2 a 0eE 0 sin(2ωt) v z = 1 4 a2 0c[1 cos(2ωt)] = 1 2 a2 0c sin 2 (ωt)

5 Some phase space trajectories a=0.1 a=1 a=3

6 Trajectory of relativistic electrons a=0.1 a= a=0.3 a=1 5

7 High Intensity Lasers

8 Fieldionisation multiphoton ionisation tunnel ionisation barriersuppression ionisation

9 Plasma 4th State of Matter, called plasma by Langmuir because of similarity to blood plasma Makes up 99.9% of our solar system, and is therefore important for astrophysics. Important for fusion and industrial processes too (e.g. semiconductor manufacture)

10 What is a plasma? Consists of ions and electrons, and can also have neutrals (low temperature plasma), large solid bodies (dusty plasmas) etc. Can write fluid equations for a plasma: ρ u t u u = qn(e u B) P

11 Debye Length q Normal plasma Add charge q, plasma rearranges to shield charge over λd λ d = 0 k B T ne 2 1/2

12 Plasma waves A charge seperation, causes a returning force ( displacement). electrons slosh back and overshoot, resulting in an oscillation. Fe Fe ω p = ne 2 0 m 1/2

13 Plasma Oscillations Consider cold plasma of uniform density n 0. Displace a sheet of electrons by a small distance x. By Gauss s law; 0 E ds = Q So E = n0 ex/ 0. And the equation of motion of the electrons is just mẍ = ee x ẍ ω 2 px = 0, SHM with frequency ω p = (ne 2 / 0 m e ) 1/2. Numerically ω p 2π 9 n e Hz, where n e is in cm 3. (Typically microwaves for fusion plasmas) Oscillations must not be too large, or else the electrons motion will be shielded out by electrons closer to the unsheilded positive charge i.e. λ λ d.

14 Application: plasma accelerators For large amplitude plasma waves, i.e. n 1 n 0, assuming ω p = ck p i.e. a relativistic phase velocity wave; E max c me n 0 0 1/2. For n e cm 3, E max 100 GV/m, or about 1000 times the maximum accelerating field of RF linacs. Since particles traveling in such a wave at close to the speed of light stay in phase with the wave, they can be accelerated to very high energy E 2γ 2 m e c 2, where γ is the Lorentz factor associated with v φ. (For examples see the papers by A. Modena et al, [Nature 1995] and S. P. D. Mangles et al [Nature 2005]).

15 Relativistic ponderomotive force The energy of the electron is intensity dependent, U = γmc 2 = { 1a 2 0 /2}mc2 For lowenergy (a 0 1); the kinetic energy is given by T = U mc 2 (a 2 0/4)mc 2 = e 2 E 2 /4m e ω 2, (nonrelativistic form of ponderomotive potential) As a particle moves away from region of highintensity it can retain some of this energy (how much depending on the steepness of the gradient). F = T F = (e 2 /4m e ω 2 ) E 2 the ponderomotive force. Fully relativistic expression can be found by taking F = U, i.e. F = (mc 2 ) γ = (mc 2 ) (1 a 2 0/2) 1/2 = (mc 2 /2)(1 a 2 0/2) 1/2 (a 2 0/2) = (mc 2 /4γ) a 2 0 (as above but with a γ thrown in for relativity).

16 Wakefield generation plasma wave Ponderomotive force F p!"(e laser ) 2 laser field We can add the ponderomotive force to the shm equation; mẍ = ee x 1 4 mc2 a 2 ẍ ω 2 px = 1 4 c2 a 2, For laser pulse with sine pulse envelope a = a 0 sin[k(x v g t)] where we have ignored the fast oscillations, ẍω 2 px = 1 4 c2 d dx a 2 0 sin 2 [k(x v g t)] = 1 8 a2 0c 2 d dx (1 cos[2k(x v gt)]) = 1 4 kc2 a 2 0 sin[2k(x v g t)]

17 Wakefield generation 2 δx ω 2 pδx = 1 4 kc2 a 2 0 sin[2k(x v g t)] This will be resonant if 2kv g =2ω B = (the frequency of amplitude modulations) = ω p. So; δx ω 2 pδx = α[exp(iω p t)] This is a driven oscillator with solution of type δx = A(t)exp(iω p t) Insert into the above gives, Ä 2iωA = α Solving for A gives the solutions, = αt 2iω exp(iω pt)b exp(iω p t)c exp( iω p t) There is a growing term whose real part is increasing as 1 4 a2 0ct

18 Wakefield generation Note that the amplitude of the wave = 1 4 ka2 0ct/ω p We will want a relativistic wave, with kc = ω p,soa(t) = 1 4 a2 0ct If a 0 is not very large (e.g. initial experiment), then a train of modulated pulses can be used to excite the plasma wave. This is most easily done by using two laser frequencies, ω 1 and ω 2 which are chosen such that the beat between the two ω B = ω 1 ω 2 = ω p. This is called the plasma beatwave scheme. Note thought that if a 0 1, then in a plasma period t = 2π/ω p,thena(t) λ p. Hence one sufficiently short laser pulse can also generate a large amplitude plasma wave of density amplitude δn/n a2 0.

19 Energy gain in a wakefield The amplitude of a plasma accelerator is then δn = An 0 cos(k p x ω p t) In the stationary frame of the wave, δn = An 0 cos(k p x) And so from Poisson s equation 0 2 φ = e(n i n e ) = eδn 2 φ = (en 0 / 0 ) cos(k p x) φ = en 0 / 0 k 2 p cos(kp x) Let s choose ck p = w p, then φ = φ = mc 2 /e cos(k p x) n0 e 2 m e 0 1 c 2 k 2 p mc 2 e cos(k p x) NB E = φ = (mc 2 /e)(1/k p ) sin(k p x) = (mcω p /e) sin(k p x) So E max = mcω p /e So maximum energy gain [ eφ] k px=π k p x=0 = 2mc2

20 Energy Gain 2 But actually we want to have plasma wave travelling at v φ c, so that electron stays in phase with accelerating field. E is unchanged but length is Lorentz lengthened by γ (=Lorentz factor associated with wave). Hence k p k p /γ. E = (mcω p /e) sin(k p x/γ) φ = (γmc/e) cos(k p x/γ) In its rest frame, E max = [ eφ] k px=π k p x/γ=0 = 2γmc2 But actually only one half of accelerating field is focusing, usually it is accepted that E max = γmc 2 Lorentz transform back (with p = γmc), ie γ iβγ ie γ iβγ γmc 2 = cp iβγ γ cp = iβγ γ γmc 2 So E max = 2γmc 2

21 Energy gain Emax? (MeV) Maximum energy and dephasing length 1000 Emax? (MeV) Ldeph (m) 100 fs, 6 cm 500 fs, 7 m 100 fs, 3 GeV 30 fs, 300 MeV 30 fs, 1.8 mm fs, 75 GeV L deph (m) E max = 2γ 2 mc 2 In a plasma the phase velocity at which the plasma wave is excited is determined by the group velocity of the em wave inside the plasma γ (ω 0 /ω p ) (n cr /n e ) Laser energy required (J) ! p (fs) Energy requirements for LWF wavebreak E (J) break and trap E (J) 100 fs, 10 J JanUSP 30 fs, 1.0 J LOA (Jaune) ! p (fs) fs, 500 J Vulcan Petawatt NB to get higher energies need to go to lower densities. The length over which this acceleration happens is γλ p (in the dilated wave frame). Time taken to travel this distance is t γλ p /c, since that is its proper time, the time taken in the back transformed frame (lab) t γ 2 λ p /c, which travelled at the speed of light, is a dephasing length of L deph = γ 2 λ p (n cr /n e ) 3/2

22 Wakefield simulation