LWFA electrons: staged acceleration 1/2 Masaki Kando - - PowerPoint PPT Presentation
Lecture: LWFA electrons: staged acceleration 1/2 Masaki Kando kando.masaki@qst.go.jp Kansai Photon Science Institute QST, Japan Advanced Summer School on Laser-Driven Sources of High Energy Particles and Radiation 9-16 July 2017,
kando.masaki@qst.go.jp
Advanced Summer School on “Laser-Driven Sources of High Energy Particles and Radiation” 9-16 July 2017, CNR Conference Center, Anacapri, Capri, Italy
This work was partially funded by ImPACT Program of Council for Science, Technology and Innovation (Cabinet Office, Government of Japan).
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Advanced Summer School, 9-16 July 2017, Capri, Italy
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Advanced Summer School, 9-16 July 2017, Capri, Italy
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Coordinate with a moving particle in a horizontal plane The equation of motion in a moving particle coordinate is s x z ρ0 qev × Bz x
centrifugal force
B mγd2x dt2 = mγv2 ρ0 + x + qevBz(x) dx dt = dx ds ds dt = vdx ds d2x dt2 = v2 dx ds We rewrite this by using 1 ρ0 + x ≈ 1 ρ0
ρ0
dx x Assume that the displacement x is small: In the central orbit the forces are balanced:
(1.1)
n := − ρ0 Bz(0) dBz dx Here we define the field gradient index d2x ds2 = −1 − n ρ2 x mγv2 d2x ds2 = −mγv2 ρ2 x + qevdBz dx x
(1.3) (1.2)
(1.4)
Horizontal direction
mγv2 ρ0 + qevBz(0) = 0
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Coordinate with a moving particle in a vertical plane The equation of motion in the vertical (perpendicular to the bending plane) is s
x
z
B Similarly we use the trans. t->x and
(1.5) (1.6)
mγd2z dt2 = qevBx(z) Bx(z) ≈ Bx(0)
+ dBx(z) dz
z
In the median place, Bx is zero.
rot B = 0 dBx dz dBz dx = 0 d2z ds2 = n ρ2 z qev × Bx n := ρ0 Bz(0) dBz dx Here we use the field gradient index The equation of motion is
Vertical direction
The solutions are stable if 0<n<1. d2x ds2 = 1 n ρ2 x
Horizontal direction
(1.4)
Using Weak focusing condition
Advanced Summer School, 9-16 July 2017, Capri, Italy
The general solution of (1.7) can be expressed by a linear combination of two special solutions:
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Hill’s equation y(s) = c1w(s)eiψ(s) + c2w(s)eiψ(s) d2y ds2 = K(S )y
N.B. The solution is expressed in the form according to Floquet theory.
(1.7) (1.6)
d2z ds2 = n ρ2 z
Vertical direction
d2x ds2 = 1 n ρ2 x
Horizontal direction
(1.4)
The terms and express focusing forces. 1 n ρ2 n ρ2 Let us write (1.4) and (1.6) in the general form: Here we assume a periodicity such a ring accelerator K(s + C) = K(s) Suppose that a solution can be written as w(s) + K(s)w(s) 1 w3(s) = 0 ψ(s) = 1 w2(s) y(s) = w(s)eiψ(s)
(1.8)
Then from (1.7) the two equations are
(1.9) (1.10)
Generally it is C/w^2 but here we choose C=1.
(1.11)
Advanced Summer School, 9-16 July 2017, Capri, Italy
Let the solutions at s=s1(s2) The y2 and y1 are expressed by the following relationship.
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(1.12)
y2 y
2
w1 cos(ψ2 ψ1) w2w 1 sin(ψ2 ψ1)
w1w2 sin(ψ2 ψ1)
1w2w 2
w1w2
sin(ψ2 ψ1)
1
w2 w
2
w1
w1 w2 cos(ψ2 ψ1) + w1w 2 sin(ψ2 ψ1)
y
1
This matrix is called “Transfer matrix”.
Later, we will see transfer matrices are very useful to analyze the motion of particles in the orbit.
Let us introduce parameters as following. Twiss parameters µ := ψ2 ψ1 1 + α2 β = γ Twiss parameters are NOT independent. β := w2 α := ww γ := 1 + (ww)2 w2
(1.13) (1.14)
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Beam ellipse y2 y
2
cos µ + α sin µ β sin µ γ sin µ cos µ α sin µ y1 y
1
w(s) = 1 2β1/2β = α β, ψ(s) = 1 β y(s) = c √β {−α cos(ψ(s) + δ) − sin(ψ(s) + δ)} φ = ψ(s) + δ cos φ = y cw sin φ = α cos φ y c γy2 + 2αyy + βy2 = c2 ε := (phase space area)/π = c2
Courant-Snyder Invariant
γy2 + 2αyy + βy2 = ε Using (1.13), Eq. (1.12) is rewritten as
(1.15)
Recalling y(s) is a linear combination of two solutions: y(s) = c1w(s)eiψ(s) + c2w(s)eiψ(s) and y(s) is a real number, then we get
(1.11) (1.16)
c is a constant
Differentiating (1.16) we obtain
(1.17)
Here using the following relations and express cos and sin function using y and y’: Thus we obtain This equation exhibits an ellipse and its area equals to . πc2 y(s) = cw(s) cos(ψ(s) + δ) cw(s)ψ(s) sin(ψ(s) + δ) we obtain Here we define an unnormalized emittance as
(1.18)
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max = γε
Diverging beam (alpha<0)
The beta function gives the amplitude of betatron oscillation as xmax =
(1.19)
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The beta function gives the amplitude of betatron oscillation as xmax =
From Eq. (1.10) ψ0(s) = 1 w(s)2 = 1 β(s) ψ(s2) − ψ(s1) = Z s2
s1
1 β(s)ds
definition of beta
Integrating this we obatin
Phase advance
Recalling Eq. (1.16) and the definition of emittance: y(s) = cw(s) cos(ψ(s) + δ) = p β(s)ε cos(ψ(s) + δ) The integration of the inverse of beta function gives the phase advance of betatron
(1.20)
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However, 1/e2 values are used in laser, while rms values are used in electron
Practical laser beams are characterized by a M2 factor.
Hence,
This must be “un-normalize emittance” because it is related to beam size from these.
λ=0.8 µm, εUN=0.064 mm-mrad ~ Ee=80 MeV, εN=10 mm-mrad
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express as follows the value itself is same.
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Let us consider a particle exerted by a sinusoidal electric field. s stands for the synchronous particle that moves at the same velocity of the electric field. d(W − Ws) ds = qeE0(cos φ − cos φs) d(φ − φs) ds = ω c 1 β − 1 βs
λ W − Ws γ3
sβ3 smc2
In the final form, we expanded to the first order of and . 1 β γ
In electron LINACs, is often used.
φs = 0
Differentiating (1.22) and using Multiplying and integrate by s, we
we obtain
C: constant
ˆ s = ks, ∆γ = (W − Ws)/(mc2), ∆φ = φ − φs
a0 = qeE0 kmc2 , b0 = 1 γ3
sβ3 s
d∆φ dˆ s
∆γ dˆ s = a0(cos φ − cos φs) ∆φ dˆ s = −b0(∆γ + γs)
1 2(∆γ + γs)2 = −a0 b0 [sin(∆φ + φs) − ∆φ cos φs] + C
This equation gives the longitudinal phase space plots.
We can rewrite this formula in Hamilton formalism.
H = − 1 2b0 (∆γ + γs)2 − a0[sin(∆φ + φs) − ∆φ cos φs]
d∆φ dˆ s = ∂H ∂∆γ , d∆γ dˆ s = − ∂H ∂∆φ
(1.21) (1.22) (1.23) (1.24) (1.25) (1.26)
W = γmc2
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∆γ
unstable stationary points separatrix
φs = π/2 φs = 0 ∆γ ∆φ We can estimate the required acceptance (phase space area to achieve some beam performance).
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(1.28) (1.1)
Taking only the 1st-order terms and remembering the forces are balanced at x=0, i.e. Let’s consider particles whose energy is different from the central energy: mγd2x dt2 = mγv2 ρ0 + x + qevBz(x) mγd2x dt2 = m(γ+∆γ)(v+∆v)2 ρ0 + x + qe(v+∆v)Bz(x) γ → γ + ∆γ, v → v + ∆v mγv2 ρ0 + qevBz(0) = 0 we obtain d2x ds2 + 1 − n ρ2 x = 1 ρ0 ∆p p This is HIll’s equation with the right-hand side term. If is constant in an element, K = 1 − n ρ2 x = 1 ρ0K ∆p p is a special solution of (1.29). Thus a general solution of (1.29) can be written as (K>0) x = C1 cos( √ Ks) + C2 sin( √ Ks) + 1 ρ0K ∆p p Now we consider the new transfer matrix x x0 ! x x0 ∆p/p instead of Setting x0 and x0’ as the initial values at s=0
B B B B B B B B @ x x0 ∆p/p 1 C C C C C C C C A = B B B B B B B B B B @ cos( p Ks)
1 p K sin(
p Ks)
1 ρ0K
n 1 cos( p Ks)
K sin( p Ks) cos( p Ks)
1 ρ0K sin(
p Ks) 1 1 C C C C C C C C C C A B B B B B B B B @ x0 x0 ∆p/p 1 C C C C C C C C A
(1.29) (1.30) (1.31)
for K>0
(1.27)
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(1.32)
Similarly, we can obtain the case K<0 and let us summarize the results.
(1.31)
B B B B B B B B @ x x0 ∆p/p 1 C C C C C C C C A = B B B B B B B B B B @ cosh( p|K|s)
1 p|K| sinh( p|K|s) 1 ρ0|K|
n 1 cosh( p|K|s)
cosh( p|K|s)
1 ρ0 p|K| sin( p|K|s)
1 1 C C C C C C C C C C A B B B B B B B B @ x0 x0 ∆p/p 1 C C C C C C C C A B B B B B B B B @ x x0 ∆p/p 1 C C C C C C C C A = B B B B B B B B B B @ cos( p Ks)
1 p K sin(
p Ks)
1 ρ0K
n 1 cos( p Ks)
K sin( p Ks) cos( p Ks)
1 ρ0 p K sin(
p Ks) 1 1 C C C C C C C C C C A B B B B B B B B @ x0 x0 ∆p/p 1 C C C C C C C C A for K<0 for K>0
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To design matching sections, focusing is a
components. R = Rxx | Rxy | Rxz + + Ryx | Ryy | Ryz + + Rzx | Rzy | Rzz
X = x x y y z
∆p p
Sometimes we do not need calculate all six dimensions. You can specify what is needed in your applications. If the beamline is composed of several optics we can calculate the beam parameters after the optics using matrix calculations XN = RNRN−1 · · · R1X0 s s=s0 X0 s=s XN 1 2 3 4 5 6 7
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Twiss parameters calculation Let be a transfer matrix from s1 to s2 . Twiss parameters at s2 can be calculated by those at S1 by the equation: M = m11 m12 m21 m22 !
α2 β2 γ2 = m11m22 + m12m21 −m11m21 −m12m22 −2m11m12 m2
11
m2
12
−2m21m22 m2
21
m2
22
α1 β1 γ1
[proof: exercise. Hint: Definition of transfer matrix Eq.(1.12)] The phase advance is also obtained by using the transfer matrix. ∆µ tan ∆µ = m12 β1m11 − α1m12
B B B B B B B B @ x x0 ∆p/p 1 C C C C C C C C A = B B B B B B B B B B @ cos( p Ks)
1 p K sin(
p Ks)
1 ρ0K
n 1 cos( p Ks)
K sin( p Ks) cos( p Ks)
1 ρ0K sin(
p Ks) 1 1 C C C C C C C C C C A B B B B B B B B @ x0 x0 ∆p/p 1 C C C C C C C C A
Drift space In this case we input K = 1 − n ρ2 → 0, ρ0 → ∞ into Eq. (1.31) we obtain sin √ Ks √ K = ssin √ Ks √ Ks →
√ Ks→0
s Here we used
(1.31) (1.33) (1.34) (1.35)
Mdrift = 1 s 1 1
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Bending magnet (n=0) In this case we input into Eq. (1.31) we obtain Quadrupole In this case we input into Eq. (1.31) MH = cos θ ρ0 sin θ ρ0(1 − cos θ) − sin θ
ρ0 cos θ
cos θ sin θ 1 MV = 1 ρ0θ 1 1 MQF = cos( √ Ks)
1 √ K sin(
√ Ks) − √ K sin( √ Ks) cos( √ Ks) 1 MQD = cosh( √−|K|s)
1 √|K| sinh( √|K|s)
− √|K| sinh( √ Ks) cosh( √|K|s) 1 KHoriz = 1 − n ρ2 → 1 ρ2 , KVert = 0, L = ρ0θ K = G Bρ, ρ0 → ∞ Bρ = pc/(qe) Magnetic rigidity
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Focus The drift space with the length=L is at s=L we want to know here ✓ x x0 ◆ = ✓ 1 L 1 ◆ ✓ x⇤ x0⇤ ◆ β = −2Lα∗ + β∗ + L2 ∗ γ∗ α∗ = 0, β∗γ∗ − α∗2 = 1 γ∗ = 1/β∗ β(L) = β∗ + L2 β∗ x(s) = p βε x(L) = x∗ s 1 + ✓ L β∗ ◆2 w(L) = w∗ s 1 + ✓ L ZR ◆2 At focus alpha=0 and using the relationship (1.14) Using the formula (1.33) and we obtain It is same as the Gaussian beam propagation. Beta-function at focus shows the Rayleigh length.
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LWFA (Linear, a0<<1)
0.0 0.5 1.0 2.0 1.5 1.0 0.5 0.0
Phase (π)
Accelerating Focusing
Useful Phase
Longitudinal field Focusing field eEz(r, ζ) = mc2 √π 2 k2
pσza2
× exp − r2 2σ2
r
− k2
pσ2 z
4 cos kpζ eEr(r, ζ) = −mc2 √πkpσza2 r σ2
r
× exp − r2 2σ2
r
− k2
pσ2 z
4 sin kpζ Linac (TM01): Disk loaded type Focusing field Longitudinal field The focusing force depends on the phase! This causes emittance growth of the whole beam!
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Linac (TM01) Focusing field Longitudinal field Es = AJ0(kr) exp i(ωtβ
0s)
Er = iAβ0 k J1(kr) exp i(ωt − β0s) Bθ = iA ω kc2 exp i(ωt − β0s) β0 = 2π/λg wavelength in the cavity Fs = qA cos(ωt − β0s) Fr = −qAω 2 ω c 1 − βeβw βw sin(ωt − β0s) In RF accelerators, the focusing force decreases when the beam energy becomes relativistic. The focusing force dependence of the longitudinal coordinate is
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Liouville’s theorem The phase-space density is UNCHANGED provided that Hamiltonian forces are acting the particles.
18-36 (CERN-1989-005), DOI10.5170/CERN-1989-005.18
Without scattering, nonlinear forces (space charge), coupling, and non- Hamiltonian friction forces, an emittance is conserved. We see that this situation happens in the limited case (transverse motion).
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18-36 (CERN-1989-005), DOI10.5170/CERN-1989-005.18
Even if Lioville’s theorem works the effective emittance may increase! The evolution of particles in a Hamiltonian system. What Liouville’s theorem tells is thatt the hatched area is conserved. But the beam quality is indeed degraded.
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18-36 (CERN-1989-005), DOI10.5170/CERN-1989-005.18
So decreasing the emittance is really tough! It does worsen naturally. In linear accelerators thus the initial emittance is very important. This is what “stochastic cooling” does in phase space within Hamiltonian forces. By the way, is it really impossible to reduce emittances? The answer is NO.
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Advanced Summer School, 9-16 July 2017, Capri, Italy
Acceleration”
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