macroscopic diffusion in random lorentz gases · fsu-logo. theorem fick’s law holds if and only...
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Macroscopic Diffusion in random Lorentz gases
Raphaël Lefevere1
1Université Paris Diderot (Paris 7)
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Joint work with Yann Chiffaudel
R. L. Fick’s law in a random lattice Lorentz gasArchive for Rational Mechanics and Analysis Volume 216, Issue 3(2015), 983-1008
Yann Chiffaudel and R.L. The Mirrors Model : Macroscopic DiffusionWithout Noise or Chaos
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Fick’s law
J = κ(ρL − ρR)
L
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Periodic Lorentz gas
Bunimovich-Sinai : the rescaled motion of a test particle is diffusive.
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Random Lorentz gas
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Ehrenfest random wind-tree model
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Mirrors model
Ruijgrok-Cohen(1990,1991)
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Mirrors model
FilmAvecParticule.mp4Media File (video/mp4)
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Discrete dynamics
Q = the set of midpoints of edges of an hypercube of Zd of side N and withperiodic conditions in all but the first direction : Q =
Sdi=1 Li where
Li =
z +
1
2ei : 0 ≤ z1 ≤ N − 1, (z2, . . . , zd) ∈ (Z/NZ)d−1
ff.
Possible velocities are P = {± e12, . . . ,± ed
2}
Phase space :
M = {x = (q,p) : q ∈ Q,p ∈ P s. t. if q ∈ Li then p = ±ei
2}.
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Dynamical System
Action of a mirror : for any z ∈ Zd, π(z; ·) is a bijection of P into itself thatsatisfies
π(z;−π(z; p)) = −p, ∀z ∈ Zd, ∀p ∈ P
For any (q,p) ∈M :
F (q,p) = (q + p + π(q + p; p), π(q + p; p)) .
The orbit of x ∈M is the set Ox = {y ∈M : ∃t > 0, F t(x) = y}.
F is a bijection on MF−1 = RFR, R(q,p) = (q,−p)For every x ∈M, Ox is a loop.Non-self-intersecting orbits : if y ∈ Ox and y 6= x then F (y) 6= F (x)Non-intersecting among themselves : if Ox 6= Oy , then Ox ∩ Oy = ∅.
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Law of the mirrors
Let
Π = {π(z; ·) : π(z;−π(z; p)) = −p, π(z; p) 6= −p, z ∈ Zd,p ∈ P}
Q is the uniform product measure over Π. For each z fixed, the number of πsatisfying the constraint is (2d− 1)!!.
Orbits are not Markov processes
Known facts on Zd .
Bunimovich-Troubetzkoy prove
@D > 0, such that Q[bx
εc,
t
ε2] ∼
εd
(2πDt)d2
exp−x2
Dt, ε→ 0
Kong-Cohen : numerics :
∃D > 0, such that limt→+∞
EQ[x2(t)]t
= D
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Occupation variables
Occupation variables σ(q,p; t) ∈ {0, 1}.{σ(x; 0) : x ∈M} independent Bernoulli parameter ρI ∈ (0, 1)Evolution :
σ(x; t) =
8
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Macroscopic current
Take the hyperplane Ql = {q ∈ Q : q1 = l + 12} , l ∈ {1, . . . , N − 2} as a functionof a configuration σ ∈ {0, 1}M :
J(l, t) =1
Nd+1
Xx∈M
t+N2Xs=t
σ(x, s)∆(x, l)
where ∆(x, l) = 2(p · e1)1q∈Ql ,with x = (q,p).∆(x, l) takes the value +1 (resp. −1) if x crosses the slice Ql from left to right(resp. from right to left).
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Goal
P ∼ law of original state and boundary conditionsQ ∼ law of the mirrors
Fick’s law
For any t ≥ |M|, any l ∈ {1, . . . , N − 2} and any δ > 0,
limN→∞
P×Q[|NJ(l, t)− κ(ρ− − ρ+)| > δ] = 0,
for some κ > 0.
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S± = {x ∈ B± : F 1(x) /∈ B±, . . . , F s−1(x) /∈ B±, F s(x) ∈ B∓ for some s ∈ N∗}.
N± = the numbers of crossings from B± to B∓ = |S±|
N+ = N− because every orbit is closed. Set N = N+ = N−.
Proposition
Let {σ(x; 0) : x ∈ C} be a set of independent Bernoulli random variables withE[σ(x; 0)] = ρ± ∈ (0, 1) if x ∈ B±, and E[σ(x; 0)] = ρI ∈ (0, 1) if x /∈ B− ∪B+. Forevery δ > 0, any t ≥ |M| and l ∈ {1, . . . , N − 2},
P»˛̨̨̨J(l, t)−
NNd−1
(ρ− − ρ+)˛̨̨̨≥ δ–≤ 2 exp(−4δ2Nd+1).
N is the central object
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Theorem
Fick’s law holds if and only if there exists κ > 0 such that for any � > 0
limN→∞
Q»˛̨̨̨N
Nd−2− κ˛̨̨̨> �
–= 0
If d = 2, Kozma-Sidoravicius argument shows that Fick’s law does not hold.
In d ≥ 3, if Crossing Conditions are satisfied :
1 There exists κ > 0 such that limN→∞NQ[O ∈ S] = κ.
2 limN→∞1
Nd−3P
x∈B−δ(O, x) = 0 with
δ(x, y) = Q[x ∈ S, y ∈ S]− Q[x ∈ S]Q[y ∈ S].
then Fick’s law holds in the stationary state and κ is the conductivity.
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How to prove that the conditions are verified ?
If the law of an orbit was “similar” to the law of a random walk, the firstpoint is trivial.
Recollisions
Rings model : recollision occurs only after time N , full proof is given if d islarge enough.
In the mirrors model, recollisions (“loops”) become more and more unlikely asd increases
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Crossing conditions : d ≥ 3
Figure: Q(O ∈ S) for N from 5 to 420. κ = 1.535± 0.005
Figure: δ(O, x) = Q[O ∈ S, x ∈ S]− Q[O ∈ S]Q[x ∈ S] for x = ((1/2, y, 0), e12 ), N=70with a 95% confidence interval.
Better than independent orbits !!
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Correlations : cooperation and jamming
For x ∈ B−,
δ(O, x) =XO0,Ox
(Q(O0,Ox)− Q(O0)Q(Ox))
−XO0,Ox
′Q(O0)Q(Ox). (1)
Both sums run over orbits that cross the box Q.First sum runs over compatible orbits such that O0 and Ox share a mirror.Cooperation
Second sum runs over incompatible orbits O0,Ox. Jamming
First sum is positive because Q(O0,Ox) > Q(O0)Q(Ox) when orbits share amirror.
Jamming wins !