Remark on a Kinetic Equation for Compton Scattering M. Escobedo - - PowerPoint PPT Presentation
Remark on a Kinetic Equation for Compton Scattering M. Escobedo UPV/EHU & BCAM Bilbao 1 Plan 1. Compton Scattering 2. Boltzmann-Compton equation: known results 3. Kompaneets approximation: known results 4. Zeldovich approximation
UPV/EHU & BCAM Bilbao
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In 1922 (!) A. H. Compton discovers the increase of wavelength of X-rays due to scattering of the incident radiation by free electrons, which implies that the scattered quanta have less energy than the quanta of the original beam. This effect, the Compton effect, illustrates the particle concept of electromagnetic radiation. In a plasma embedded in a radiation field of temperature T. The scattering
between the two components. There are regions in the Universe - like the cluster of galaxies - that contain hot, ionized gas. Cosmic microwave background radiation photons, that fills the Universe, when passes through these regions, they will be scattered by the electrons ( which are at a much high temperature) and gain energy. This will distort the cosmic microwave background radiation spectrum in the vicinity of a cluster of galaxies: Sunyaev-Zeldovich effect.
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Figure 1: The Compton experiment.
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classical equilibrium. The density of electrons is given by: e−|p|2 where, if p is the momentum of the electrons, |p|2 is their energy.
The density function is then F(t, p): represents the density of photons that at time t have momentum p. The energy of a photon of momentum p is |p|.
density function F(t, p) satisfies the following equation:
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∂F ∂t (p, t) =
(A. S. Kompaneets, JETP 1956; H. Dreicer, Phys. Fluids 1964) If we consider only radially symmetric densities (isotropic gases): k = |p|, F(t, p) = k−2f(t, k) the equation reads: ∂f ∂t (t, k) = ∞ b(k, k′)q(f, f ′)dk′, f ′ = f(t, k′) q(f, f ′) = e−kf(t, k′)(k2 + f(t, k)) − e−k′f(t, k)(k′2 + f(t, k′))
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Regular Equilibria : fµ,0(k) = k2 ek+µ − 1, µ ≥ 0 : Satisfy : q(fµ,0, f ′
µ,0) ≡ 0
The total number of particles: N(f) = ∞ f(k) dk, d dtN(f(t)) = 0 along the trajectories. Remark No regular equilibria with mass larger than N(f0,0): Nµ ≡ N(fµ,0) ≤ N(f0,0) = N0
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0.5 1.0 1.5 2.0 0.1 0.2 0.3 0.4 0.5 0.6
Figure 2: The regular equilibria fµ,0, µ = 0, 0.2, 1.
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The Boltzmann Compton equation has an entropy: H(f) = ∞ h(f, k) dk, d dtH(f(t)) > 0 h(f, k) = (k2 + f) log(k2 + f) − f log f − k2 log(k2)
If N ≤ N0 : max
N(f)=N H(f) = H(fµ,0)
for some µ ≥ 0 If N > N0 : max
N(f)=N H(f) = H(f0,α)
for some α ≥ 0 where f0,α = k2 ek − 1 + αδ(k) ≡ f0,0(k) + αδ(k) Remark . If b(k, k′) is not continuous at k′ = 0 sense of the equation for f0,α?
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What happens near k = 0 as t increases ? The Boltzmann Compton equation is simpler than the Boltzmann equation describing the boson-boson collisions in a dilute gas: ∂f ∂t (t, p) = Q(f)(t, p), t > 0, p ∈ R3. Q(f)(t, p) =
R9W (p, p2, p3, p4) q(f)dp2dp3dp4
q (f) = f3f4(1 + f)(1 + f2) − f f2(1 + f3)(1 + f4) W (p, p2, p3, p4) = δ(p + p2 − p3 − p4) δ
(Nordheim 1928)
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First results by E. Levich & V. Yakhot ’77 & ’78.... Later: Theorem 1. Under some conditions on b implying b ∈ C([0, 1) × [0, 1)) assume: fin ∈ E0 = {h ∈ L1(0, ∞), h ≥ 0, ∞ (1 + k)h(k)dk < ∞}. Then, there exists a unique global solution f ∈ C([0, ∞), E0) to Boltzmann Compton such that ∞ f(k, t)dk = ∞ fin(k)dk =: N ∀t > 0. Moreover:
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(i) if N = Nµ : lim
t→∞
∞ |f(k, t) − fµ,0(k)|dk = 0 (ii) if N > N0 : f(·, t) ⇀ f0,0 + (N0 − m)δ and: lim
t→∞
|f(k, t) − f0,0(k)|dk = 0, ∀k0 > 0. Theorem 2. Suppose that b ≡ 1 and fin(k) ∼ k as k → 0, (for example) N(fin) = N > N(f0,0). Then as t → +∞, and uniformly on 0 ≤ kt ≤ L for any L > 0 fixed:
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F(t, k) − f0,0(k) = (N − N0) t Φ(tk)
(1 + o(1)) as t → +∞ where Φ(z) = N 2 z e−N z. Theorem 1: E. & S. Mischler ’01. Theorem 2: E., S. Mischler & J.J.L.Vel´
But: Theorem 1 needs a “regular” kernel b. Theorem 2 needs b = 1.
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More realistic kernels b(k, k′) are rather singular near the origin. Theorem 3. (M. Chane-Yook & A. Nouri ’04) For some kernel b such that b(k, k′) ∼ k−2k′−2 as k, k′ → 0 and initial data such that (1 + k−1)fin ∈ L1(R+) there exists a local (in time) weak solution of the equation such that f(t, k) k ∈ L∞((0, T), M 1(R+))
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Kompaneets in 1957: considers that the main contribution in the collision integral. comes from the region |k′ − k| << k and deduces the equation: k2∂g ∂t = ∂ ∂k
∂g ∂k + g + g2
lim
k→0 or k→+∞
∂g ∂k + g + g2
But for some solutions there is a positive time T for which this condition is not satisfied after T. (Some References: R. Caflisch & C. D. Levermore ’86; E. ,Herrero & Vel´ azquez ’98; O. Kavian ’02; N. Ben Abdallah, I. M. Gamba & G. Toscani ’11; C. D. Levermore, H.Liu & R. Pego ’11). More precisely:
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Theorem. If fin ∈ Cb(0, ∞), lim
k→∞ k4fin(k) = 0. Then there exists a unique
classical global solution such that f(k, 0) = fin(k) such that, for some C > 0: f(k, t) ≤ C k2 ∀k ∈ (0, 1), ∀t > 0; lim
k→∞ k4(∂f
∂k + f + f 2) = 0. Moreover, there are fin ∈ Cb(0, ∞) such that for some T ∗ > 0, C1 > 0, C2 > 0:
k→0 k4(∂f
∂k + f + f 2) = 0 ∀t ∈ (0, T ∗)
k→0 k4(∂f
∂k + f + f 2) > 0 ∀t ≥ T ∗
t→T ∗(T ∗ − t)f(z(T ∗ − t), t) =
C1 z2
if z > C2 if 0 < z < C2 uniformly on compact subsets of (C2, ∞) and (0, C2)
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When f >> k2 approximate: (k2 + f) → f in the equation: ∂f ∂t (t, k) = f(t, k) ∞ f(t, k′)(e−k − e−k′)b(k, k′)dk′ (Zel’dovich 1968). A more “classic” equation. This approximation is very formal. Similar “approximation” is used in other examples. For example by several authors (D. V. Semikoz et al.’74; Y. Kagan et al.’92; Y. Pomeau et al.’01) in the study
Does it really make sense? Use Theorem 2 to estimate the different terms for the case b = 1.
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As t → +∞ and for 0 < k < L t : (i) ft ∼ ktΦ′(kt) + Φ; (ii) ∞ dk′f ′fe−k ∼ m1tΦ(kt)e−k (iii) ∞ dk′f ′fe−k′ ∼ tΦ(kt) ∞ dk′f ′e−k′ as t → +∞ : − → tΦ(kt) ∞ f0e−kdk + (m1 − m0)
∞ dk′f ′k2e−k ∼ m1k2e−k; (v) f(t, k) ∞ dk′k′2e−k′ ∼ 2tΦ(kt)
converges to a positive constant as t → +∞).
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Does Zeldovich’s approximation make sense?
We are led to study the equation: ∂f ∂t (t, k) = f(t, k) ∞ f(t, k′)K(k, k′)dk′ K(k, k′) = (e−k − e−k′)b(k, k′) More generally consider kernels K(k, k′) such that: K(k, k′) = −K(k′, k) K(k, k′) < 0 if k′ < k; K(k, k′) = 0 ⇐ ⇒ k = k′ K ∈ L∞(R+ × R+) or sup
k>0
∞ |K(k, k′)|dk′ < +∞.
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Theorem 1. For any f0 ≥ 0 in L1(R+) there exists a solution f ≥ 0 in C([0, +∞); L1(R+)). It satisfies: ∞ f(t, k) dk = ∞ f(0, k) dk supp(f(t)) = supp(f(0)) Theorem 2. The set of stationary solutions of the equation is the set of Dirac measures δ(k − a) for all a ≥ 0. Theorem 3. If f is a solution with initial data f0 with total mass N then f(t) ⇀ N δ(k − a), as t → +∞; a = inf {k; k ∈ supp(f0)}
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d dt ∞ f(t, k)ψ(k)dk = ∞ ∞ f(t, k)f(t, k′)K(k, k′)ψ(k)dxdk′. But since, ∞ ∞ f(t, k)f(t, k)K(k, k′)ψ(k)dkdk′ = − ∞ ∞ f(t, k)f(t, k′)K(k, k′)ψ(k′)dkdk′, we have, d dt ∞ f(t, k)ψ(k)dk = 1 2 ∞ ∞ f(t, k)f(t, k′)K(k, k′)(ψ(k) − ψ(k′))dkdk′.
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Taking ψ ≡ 1, we have: d dt ∞ f(t, k)dk = 0 Taking ψ = 1k>a1: d dt ∞
a
f(t, k)dk = 1 2 ∞
a
a f(t, k)f(t, k′)K(k, k′)dk′dk − 1 2 a ∞
a
f(t, k)f(t, k′)K(x, y)dk′dk. In particular if a belongs to the interior of the support of f0, d dt ∞
a
f(t, k) < 0, ∀t > 0. (If a is in the interior of the support of f0, it is in the interior of the support of f(t) for all t > 0).
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H(t) = ∞ kf(t, k)dk, dH dt = D(f)(t) D(f)(t) = 1 2 ∞ ∞ f(t, k)f(t, k′)K(k, k′)(k − k′)dkdk′ < 0
ℓ(F) = min{k ≥ 0; k ∈ supp(F)} then the following holds:
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Theorem 4. For any non negative measure F such that ∞ kdF(k) < +∞ the following are equivalent: 1.- F = m δ(k − a) for a ≥ 0. 2.- F is the solution of the minimisation problem: H(F) = min
M(F ′)=m,ℓ(F ′)=a H(F ′)
3.- D(F) = 0 and ℓ(F) = a. 4.- F ∞ K(k, k′)dF(k′) = 0 and F ≥ 0.
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Proof of 1 ⇐ ⇒ 2. 1 = ⇒ 2: Notice first that for any non negative measure F whose mass is m and ℓ(F) = a: ∞ k F(k) dk = ∞
a
k F(k) dk ≥ a ∞ F(k) dk = a m. Therefore inf
M(F ′)=m,ℓ(F ′)=a H(F) ≥ a m.
Moreover, if F = mδ(k − a), H(F) = a m, from where inf
M(F ′)=m,ℓ(F ′)=a H(F) =
min
M(F ′)=m,ℓ(F ′)=a H(F) = a m.
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2 = ⇒ 1. On the other hand, if F satisfies ℓ(F) = a, M(F) = m and ∞
a
kF(k) dk = a m then, ∞
a
(k − a)F(k) dk = 0 from where (k − a)F(k) ≡ 0, a.e. x > 0 and therefore F = m δ(k − a).
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Theorem 5. Suppose b ≡ 1. If f is a solution with initial data f0 such that ∞ f0(k)dk = N u0(k) ∼ k as k → 0 then: u(t, k) ∼ N 3t2ke−Ntk as t → +∞. Sketch of the Proof. The equation may be written: ∂f ∂t (t, k) = N e−k f(t, k) − ε(t) f(t, k) ε(t) = ∞ e−k′ f(t, k′) dk′.
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Define: h(t, k) = f(t, k) exp t ε(s) ds
we have ∂h ∂t (t, k) = N e−k h(t, k) = ⇒ h(t, k) = f0(k) eNe−kt f(t, k) = f0(k)eNe−kt exp
t ε(s) ds
λ(t) = ε(t) exp t ε(s) ds
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By the definition of ε one has: λ(t) = d dt exp t ε(s) ds
∞ e−yh(t, y) dy = ∞ e−yf0(y) eNe−yt dy (1) We just need the behavior of the right hand side of (1) as t → +∞. That only depends on the behavior of f0(k) as k → 0. If, for example f0(k) = k + o(k) as k → 0 then ∞ e−k′f0(k′) eNe−k′t dk′ = eNt
N 2t2 + O 1 t3
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Integration of this estimate between 0 and t gives: exp t ε(s) ds
N 3t2 + O 1 t3
Therefore, for all k > 0 and t → +∞: f(t, k) = h(t, k) exp
t ε(s) ds
f0(k) eNe−kt eNt
1 N3t2 + O
1
t3
= N 3 t2 k e−Nkt (1 + · · · ) = N tΦ(tk) (1 + · · · ) From the two results on the long time behavior near k = 0 we deduce:
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Corollary Let f and F be such that: ∂f ∂t (t, k) = ∞
∂F ∂t (t, k) = F(t, k) ∞ F(t, k′)(e−k − e−k′)dk′ f(0, k) = F0(k) = fin(k), N = ∞ fin(x)dx > N0 Then, ∀L > 0 : lim
t→+∞
L/t |(f(t, k) − f0,0(k)) − (N − N0)F(t, k)|dk = 0 Proof We have shown: f(t, x) − f0,0(x) = (N − N0)tΦ(tk)(1 + o(1)) for k ∈ (0, L/t) as t → +∞ F(t, x) = tΦ(tk)(1 + o(1)) as t → +∞.
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Remark 6. Same scale as in Compton-Boltzmann. Not the same mass. The reduced equation may be writen: ∂F ∂t (t, k) = F(t, k) ∞ F(t, k′)
dk′ = N e−k F(t, k) − F(t, k) ∞ e−k′ F(t, k′) dk′. The Boltzmann-Compton equation may be writen: f(t, k) = f0,0 + v(t, k), ∞ v(t, k)dk = N − N0 ∂v ∂t(t, k) = (N − N0)e−kv(t, k) − v(t, k) ∞ v(t, k′)e−k′dk′ − −N0(1 − e−k)v(t, k) − f0,0(k) ∞ v(t, k′)e−k′dk′ + (N − N0)
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∂v ∂t(t, k) = (N − N0)e−kv(t, k) − v(t, k) ∞ v(t, k′)e−k′dk′ − −N0(1 − e−k)v(t, k) − f0,0(k) ∞ v(t, k′)e−k′dk′ + (N − N0)
(N − N0)3 t2 k e−(N−N0)kt (1 + · · · ) , as t → +∞ But we must instead write the complete Boltzmann Compton equation: ∂v ∂t(t, k) = Nv(t, x)(e−k − 1) + (v + f0,0) ∞ (1 − e−k′)v(t, k′)dk whose solution behaves like: (N − N0)N 2 t2 k e−Nkt (1 + · · · ) , as t → +∞
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A reasonable estimate of a more physical kernel gives: b(k, k′) = eβ k+k′
2
k2k′2 (k+k′)2
(k−k′)2
e−β |k−k′|2
2
−βs
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√s ds for some β > 0. As k → 0 and k′ → 0 we have: b(k, k′) ∼ Cmin(k, k′) k2k′2 For the sake of simplicity we consider kernels of the form b(k, k′) = (kk′)−α. We can prove: existence of a local classical solution. Problem: global or not?
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A regular type of kernel may be as follows (Zel’dovich, 68): Ga(k) = (4aπ)−1/2e−|k|2
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Ka(k, k′) = ∂Ga ∂k (k − k′) = (4aπ)−1/2(k′ − k) 2a e−|k−k′|2
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All what was said for good kernels applies to Ka. For every a > 0 we can then solve the problem: ∂f ∂t (t, k) = f(t, k) ∞ Ka(k, k′) f(t, k′) dk′ = f(t, k) ∂ ∂k (Ga ∗ f) (t, k). It is formally clear that as a → 0 we obtain in the limit the conservation law: ∂f ∂t (t, k) = f(t, k)∂f ∂k(t, k) but · · ·
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Theorem 7. Suppose that K satisfies: κ = sup
k>0
∞ |K(k, k′)|dk′ < +∞ Then, for any initial data f0 ∈ L∞(R+), f0 ≥ 0 there exists a T ∗ > 0 such that there is a unique non negative solution f ∈ C([0, T ∗); L∞(R+)) of the equation. Moreover, for every t ∈ [0, T ∗): ||f(t)||L∞(R+) ≤ ||f0||∞eκ
t
0 ||f(s)||∞ds.
Finally, either T ∗ = +∞ or limt→T ∗ ||f(t)||∞ = +∞.
Question: global or blow up in finite time?
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