ON THE DISTRIBUTION OF THE FREE PATH LENGTH OF THE ...

arXiv:0802.1019v2 [math.DS] 8 Jul 2008

ON THE DISTRIBUTION OF THE FREE PATH LENGTH OF THE LINEAR FLOW IN A HONEYCOMB FLORIN P. BOCA AND RADU N. GOLOGAN Abstract. Let ℓ > 2 be an integer. For each ε ∈ (0, 12 ) remove from R2 the union of discs of radius ε centered at the integer lattice points (m, n), with m . n (mod ℓ). Consider a point-like particle moving linearly at unit speed, with velocity ω, along a trajectory starting at the origin, and its free path length τℓ,ε (ω) ∈ [0, ∞]. We prove the weak convergence of the probability measures associated with the random variables ετℓ,ε as ε → 0+ and explicitly compute the limiting distribution. For ℓ = 3 this leads to an asymptotic formula for the length of the trajectory of a billiard in a regular hexagon, starting at the center, with circular pockets of radius ε → 0+ removed from the corners. For ℓ = 2 this corresponds to the trajectory of a billiard in a unit square with circular pockets removed from the corners and trajectory starting at the center of the square. The limiting probability measures on [0, ∞) have a tail at infinity, which contrasts with the case of a square with pockets and trajectory starting from one of the corners, where the limiting probability measure has compact support.

1. Introduction Recent progress led to a better understanding of the statistics of the free path length of the periodic Lorentz gas in the small scatterer limit [2, 3, 5, 6, 7, 8, 9, 10]. The aim of this paper is to study situations where periodicity conditions are altered by imposing certain congruence conditions on the integer lattice points where scatterers are placed. The case of the honeycomb lattice arises as a particular example in this context. Here we only consider the situation where motion originates at the origin, extending the results from [2] and [3]. The case where the initial position is randomly chosen is more intricate and will not be treated here. Let ℓ > 2 be an integer. Consider the set Z2(ℓ) of pairs of integers (m, n) with m . n (mod ℓ). For every ε > 0 consider the “fat lattice points” (scatterers) given by small discs of radius ε centered at all points of Z2(ℓ) and the region Zℓ,ε = {x ∈ R2 : dist(x, Z2(ℓ) ) > ε}

obtained by removing all scatterers. In R2 consider a point-like particle moving at constant unit speed along a linear trajectory originating at (0, 0). The free path length (first exit time) is defined as τℓ,ε (ω) = inf{τ > 0 : τ~ ω ∈ ∂Zℓ,ε }, the distance traveled to reach the first scatterer along the direction ω ~ = eiω ∈ T, ω ∈ [0, 2π), and as +∞ when the particle escapes to infinity without reaching any scatterer. The Lebesgue measure of a measurable set A ⊆ R is denoted by |A|. This paper is concerned with the study, in the small scatterer limit (ε → 0+ ), of the asymptotic behavior of the repartition function of ετℓ,ε defined by λ 1 ω ∈ [0, 2π) : τℓ,ε (ω) > . Pℓ,ε (λ) = 2π ε

To accomplish this we first consider the situation where scatterers are obtained by translating the vertical segment Vε = {0} × [−ε, ε] by (m, n) ∈ Z2(ℓ) and estimate, for any interval I ⊆ [0, 1] of length |I| ≍ εc with fixed c ∈ (0, 1), the repartition λ Gℓ,I,ε (λ) = ω ∈ arctan I : qℓ,ε (ω) > , ε → 0+ , ε Date: June 7, 2008.

1

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F. P. BOCA AND R. N. GOLOGAN

of the horizontal free path length π ω ∈ 0, . 4 In this paper ϕ will denote Euler’s totient function. The dilogarithm is defined by Z x ∞ X ln(1 − t) xn = − Li2 (x) = dt, x ∈ [0, 1]. 2 t n 0 n=1 o n qℓ,ε (ω) = inf q : (q, q tan ω) ∈ Z2(ℓ) + Vε ,

2

Clearly Li2 (1) = ζ(2) = π6 . The main result of this paper shows that the limit of Pℓ,ε exists as ε → 0+ and this limit is explicitly computed. Theorem 1. (i) For every 0 < c1 < 1 and δ > 0, as ε → 0+ , Gℓ,I,ε (λ) = cI Gℓ (λ) + Oδ,λ,ℓ ε−δ+θ(c,c1) ,

where

λ > 0,

(1.1)

Z

1 du , θ(c, c ) = min c + c , , − 2c 1 1 1 2 2 I 1+u and the limiting repartition function Gℓ is given by  1  + A(ℓ) λ if λ ∈ 0, 12 , 1 − ζ(2)       λ Gℓ (λ) =  + A(ℓ)H2 (λ) if λ ∈ 12 , 1 , 1 − ζ(2)       2C(ℓ) H3 (λ) if λ ∈ [1, ∞), cI =

ℓ

with

!−1 ! ϕ(ℓ) Y 1 1 ϕ(ℓ) Y C(ℓ) = 1− 2 1− 2 , = ζ(2)ℓ p|ℓ ℓ p p p∤ℓ p prime

A(ℓ) =

2C(ℓ) 1 − , ζ(2) ℓ

p prime

! 1 − 1 − 2 Li2 (λ), H2 (λ) = 3λ − 2 + ζ(2) − (ln λ) + 2(1 − λ) ln λ ! ! 1 1 − (λ − 1) ln 1 − − 1. H3 (λ) = Li2 λ λ

(1.2)

2

(ii) For every δ > 0, as ε → 0+ ,

1 Pℓ,ε (λ) = Gℓ (λ) + Oδ,λ,ℓ ε 8 +δ ,

λ > 0.

1.0 1.0 0.8 0.8 0.6 0.6 0.4

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0.2

0.2

0

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Figure 1. The repartition function G3 and the density function g3 The continuity of Gℓ at λ = ∞ X n=1

1 2

is equivalent with the well known dilogarithm identity ! Z 1/2 1 1 1 ζ(2) − (ln 2)2 1 = = Li ln du = . 2 n 2 2 u 1−u 2 2n 0

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THE LINEAR FLOW IN A HONEYCOMB

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1 Since C(ℓ) ℓ → 0 and A(ℓ) → ζ(2) as ℓ → ∞, the compactly supported limiting repartition H(λ) from [2, Theorem 1.1] is being recovered as limℓ→∞ Gℓ (λ).

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Figure 2. The free path in a hexagonal billiard and in a honeycomb Our original motivation for considering this problem comes from the study of the exit time of the linear motion with specular cushion collisions on a hexagonal (open) billiard table with (small) circular open pockets of radius ε removed from its corners (see Figure 2). The starting remark here is that, after unfolding the hexagon to a honeycomb in R2 , one can deform the later to Z2(3) . This process converts the problem on the hexagonal billiard with pockets into one concerning the free path length of a Lorentz gas in R2 with small identical ellipses centered at the points from Z2(3) as scatterers (see Figure 8). Let τεhex (ω) denote the free path length in the hexagonal billiard with discs of radius ε removed from the corners and motion starting at the center, and let 1 λ ω ∈ [0, 2π] : τεhex (ω) > Pεhex (λ) = 2π ε

denote the repartition function of ετεhex . We prove Theorem 2. For every δ > 0, as ε → 0+ ,

! 1 2λ Pεhex (λ) = G3 √ + Oδ ε 8 −δ , 3

λ > 0.

√ Scaling ε to ε 2 one can apply Theorem 1 (ii) with ℓ = 2 to estimate the repartition function 1 |{ω ∈ [0, 2π] : τε (ω) > λε }| of the free path length τε (ω) of a billiard in the unit square with Pε (λ) = 2π pockets of radius ε at the corners and trajectory starting at the center (see Figure 3), getting Theorem 3. For every δ > 0, as ε → 0+ ,

! 1 λ = G2 √ + Oδ ε 8 −δ , λ > 0. 2 Theorem 3 should be compared with the situation where the initial position is at one of the four vertices, and where the limiting distribution has compact support [2, 3]. It is not clear whether these methods would directly extend to other concrete initial positions, such as ( 1n , 0), n ∈ N. However, it looks likely that further refinements could lead to “space-phase average” results similar to those from [5]. The main difficulty seems to arise from the increasing complexity and number of cases that need to be analyzed in detail, leading to integrals in the main terms of the asymptotic formula which are manifestly more intricate than in the case of the square. Pε (λ)

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Figure 3. The free path in a square billiard with trajectory starting at the center 2. The contribution of consecutive Farey fractions γ < γ′ with γ, γ′ ∈ FQ \ F (ℓ) The integer part of a real number x is denoted by [x]. Let ε > 0 and Q = 1ε . Denote by FQ the set of a Farey fractions γ = q in lowest terms with 0 < a 6 q 6 Q. For any interval I ⊆ [0, 1], set FI,Q = I ∩ FQ . Denote by FQ(ℓ) the set of Farey fractions γ =

(ℓ) F (ℓ) = ∪∞ Q=1 FQ . It is well known that if γ =

(ℓ) a q ∈ FQ with ℓ | (q − a), and set FI,Q ′ a a ′ q < γ = q′ are consecutive elements

= I ∩ FQ(ℓ) . Set also

in FQ if and only if

a′ q − aq′ = 1 and q + q′ > Q > max{q, q′ }. In particular we have For convenience consider

ε(q + q′ ) > ε(Q + 1) > 1 > max{εq, εq′ }.

! ˜ ℓ,I,ε (λ) := {ω ∈ arctan I : qℓ,ε (ω) > λQ} = Gℓ,I,ε λε 1 , G ε

(2.1)

λ > 0.

For any interval I ⊆ [0, 1] with |I| ≍ Q−c , 0 < c < 1, we will prove a formula of type ˜ ℓ,I,ε (λ) = cI Gℓ (λ) + Oδ ε−δ+θ(c,c1 ) , G which will immediately imply (1.1), because ε 1ε = 1 + O(ε) and for any compact set K ⊆ R+ there is C K > 0 such that |Gℓ (x′ ) − Gℓ (x′′ )| 6 C K |x′ − x′′ |, x′ , x′′ ∈ K. (2.2) ′

a+ε ε Given γ < γ′ consecutive elements in FQ , denote t0 = γ′ − qε′ = a q−ε ′ , u0 = γ + q = q . Employing (2.1) and 1 − εq ε(q + q′ ) − 1 1 − εq′ t0 − γ = , u0 − t0 = , γ ′ − u0 = , (2.3) ′ ′ qq qq qq′ we find γ 6 t0 < u0 6 γ′ . In particular if γ < γ′ < γ′′ are consecutive elements in FQ , then γ + εq 6 γ′ < γ′′ − qε′′ . Therefore the intervals [γ − qε , γ + qε ], γ = aq ∈ FQ , cover the interval [0, 1] in such a way that every element in [0, 1] belongs to at most two of these intervals. As a result any trajectory with slope tan ω ∈ (γ, γ′ ) will intersect, as in the case of the square lattice [2, 3], one of the scatterers (q, a) + Vε or (q′ , a′ ) + Vε . Since a′ q − aq′ = 1, only two situations can occur here: γ < F (ℓ) and γ′ < F (ℓ) , respectively γ ∈ F (ℓ) or γ′ ∈ F (ℓ) . The later will be discussed in Section 3. In the remainder of this section we assume that γ < F (ℓ) and γ′ < F (ℓ) , situation where the horizontal free path is given (see Figure 4) by    q if γ < tan ω < t0 ,     ′ qℓ,ε (ω) =  min{q, q } if t0 < tan ω < u0 ,     q ′ if u0 < tan ω < γ′ .


5

qℓ,ε (ω) = min{q, q′ }

qℓ,ε (ω) = q

γ

qℓ,ε (ω) = q′

γ′

u0

t0

Figure 4. The horizontal free path when γ, γ′ ∈ FI,Q \ F (ℓ) ˜ ℓ,I,ε (λ) is given by In this way the contribution of the interval [γ, γ′ ] to G    0 if max{q, q′ } 6 λQ,      ′  arctan γ − arctan γ if min{q, q′ } > λQ,    arctan γ′ − arctan u0 if q 6 λQ < q′ ,      arctan t0 − arctan γ if q′ 6 λQ < q.

This contribution is zero whenever λ > 1, so we next assume 0 < λ < 1. Using (2.3), the estimates arctan(x + h) − arctan x =

1 1 2 − 6 2(γ′ − γ) = ′ , 2 ′2 qq 1+γ 1+γ

h + O(h2 ), 1 + x2

(2.4)

and the inequality X

γ∈FQ

X 1 1 1 6 = , 2 ′2 ′ q q Qqq Q γ∈F Q

˜ ℓ,I,ε (λ) of all intervals [γ, γ′ ] ⊆ I with γ < γ′ consecutive elements in we infer that the contribution to G ′ (ℓ) FQ and γ, γ < F is given by G(1) ℓ,I,ε (λ) = AI,Q (λ) + BI,Q (λ) + C I,Q (λ) + O(ε), with AI,Q (λ) =

X

γ,γ′ ∈FI,Q \F (ℓ) min{q,q′ }>λQ

BI,Q (λ) = ′

X

γ,γ ∈FI,Q \F q6λQλQ

1 1 , · qq′ 1 + γ2

A−,2 I,Q (λ) =

X

(ℓ) γ∈FI,Q , γ′ ∈FI,Q min{q,q′ }>λQ

1 1 , · qq′ 1 + γ2

and respectively −,2 BI,Q (λ) = B+I,Q (λ) − B−,1 I,Q (λ) − BI,Q (λ),

−,1 −,2 + (λ) − C I,Q (λ) − C I,Q (λ), C I,Q (λ) = C I,Q

(2.8)

6


with B+I,Q (λ) =

X 1 − εq′ 1 · , ′ qq 1 + γ′2 γ,γ′ ∈F

+ C I,Q (λ) =

I,Q

I,Q

q′ 6λQ 0 and gcd(ℓ, m) = 1, we obtain X ϕ(ℓn) V(n) = n 16n6N

Y ϕ(pki +αi ) i

X

k1 ,...,kr >0 16i6r k p11 ···pkr r 6N

pki i

X

k 16m6N/(p11 ···pkr r )

ϕ(m) V pk1 ···pkr (m). r 1 m

(2.9)

gcd(ℓ,m)=1

According to Lemma 1 the inner sum above can be expressed as Z N/(pk1 ···pkrr ) 1 V(pk11 · · · pkr r x) dx + Oℓ (kVk∞ + T 0N V) ln N C(ℓ) 0 Z N C(ℓ) = k V + Oℓ (kVk∞ + T 0N V) ln N . kr 1 p1 · · · pr 0

Inserting this into (2.9) and using the fact that the number of terms in the first sum in (2.9) is ≪ (ln N)ω(ℓ) and ϕ(pki i +αi ) = pki i +αi (1 − p−1 i ) we infer that the expression in (2.9) is given by ! X Z N Y 1 1 N 1+ω(ℓ) 1− . (2.10) (kVk + T V)(ln N) ℓC(ℓ) V + O ∞ ℓ 0 p k ,...,k >0 pk1 · · · pkr r 0 p|ℓ 1 1 r k

p prime

p11 ···pkr r 6N

The statement now follows from (2.10), using X 1

pk11 · · · pkr r

k1 ,...,kr >0

=

Y

p|ℓ p prime

1−

1 p

!−1

and the bound X

1

k1 ,...,kr >0 k p11 ···pkr r >N

pk11 · · · pkr r

≪ p1 ,...,pr

(ln N)r−1 . N

The following estimate [4, Proposition A4] will be employed several times. Lemma 3. Assume that q > 1 and h are two given integers, I and J are intervals of length less than q, and f : I × J → R is a C 1 function. Then for any integer T > 1 and any δ > 0 " X ϕ(q) f (x, y) dxdy + E, f (a, b) = 2 q I×J a∈I,b∈J ab=h (mod q) gcd(b,q)=1

with

1 3 1 1 k∇ f k∞ |I||J| E ≪δ T 2 k f k∞ q 2 +δ gcd(h, q) 2 + T k∇ f k∞ q 2 +δ gcd(h, q) 2 + , ∂ f ∂ f T where k f k∞ and k∇ f k∞ denote the sup-norm of f and respectively ∂x + ∂y on I × J.

Lemma 3 will be typically applied to the following situations: Let I be a subinterval of [0, 1]. For every q ∈ [1, Q] consider the intervals I = qI and J = Jλ,q = (max{λQ, Q − q}, Q], and the functions fq and gq defined on qI × Jλ,q by fq (u, v) := We clearly have k fq k∞ ≪λ

1 , Qq

1 1 · , qv 1 + u 2 q

k∇ fq k∞ ≪λ

1 , Qq2

gq (u, v) =

kgq k∞ ≪λ

1 1 − εv · 2 . qv 1 + uq

1 , Qq

k∇gq k∞ ≪λ

1 . Qq2

(2.11)

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Proposition 1. For every λ ∈ (0, 1] and c1 ∈ (0, 1),

AI,Q (λ) = cI A(ℓ)I1(λ) + Oδ,λ,ℓ (Qδ+θ1 (c1 ) ),

where  R 1−λ   1 1 ln(1 − λ) ln λ + λ ln dx =  I1 (λ) =  (ln λ)2 max{λ, 1 − x} λ x o n and θ1 (c1 ) = max 2c1 − 12 , −c1 , −1 . Z

1

1 x

1 ln 1−x dx

if λ ∈ 0, 12 , if λ ∈ 21 , 1 ,

Proof. There is at most one γ ∈ FI,Q with γ′ < I. Since |I|1 ≪ Qc and qq1 ′ 6 Q1 , the total contribution to the final asymptotic results from Theorem 1 resulting from replacing the two conditions γ, γ′ ∈ FI,Q 1 by a ∈ qI or by a′ ∈ q′ I will be ≪ Q−1 and respectively ≪ Q 8 −1 , thus negligible. As a result we shall tacitly do this in formulas (2.12), (2.15), (2.19), (2.24) and (2.27). ′ Furthermore, the summation constraints in γ = aq ∈ FI,Q and γ′ = qa′ ∈ FQ will translate, using standard properties of Farey fractions, into the following constraints on the triplet (q, a, q′):    q ∈ (λQ, Q],   a ∈ qI, q′ ∈ J , gcd(q′ , q) = 1, aq′ = −1 (mod q). λ,q

Summing first over q and then after integer pairs (a, q′) ∈ qI × Jλ,q as above (note that the number of such pairs is 6 q|I|) we infer ! X X 1 fq (a, q′ ) + O A+I,q (λ) = . (2.12) Q q∈(λQ,Q] (a,q′ )∈qI×J λ,q

gcd(q′ ,q)=1 aq′ ≡−1 (mod q)

Applying Lemma 3 with T = [Qc1 ] and (2.11), the inner sum in (2.12) can be expressed as Z Z ϕ(q) du dv c1 32 +δ −3 2 −3 −c1 2c1 21 +δ −2 Q + Q q Q + q Q Q Q q + O δ,λ q2 Jλ,q qv qI 1 + u 2 q

ϕ(q) 1 2c1 − 32 +δ + Q−1−c1 . = cI 2 ln q + Oδ,λ Q q max λ, 1 − Q

Summing over q ∈ (λQ, Q] in (2.13) and employing (2.12) and [1, Lemma 2.3] we find cI A+I,Q (λ) = I1 (λ) + Oδ,λ Qδ+θ1 (c1 ) . ζ(2)

(2.13)

(2.14)

To estimate A−,1 I,Q (λ), note first that gcd(ℓ, q) = 1 because ℓ | q − a and gcd(q, a) = 1. As a result, putting w = q − a = ℓu, v = q′ and using a′ q − aq′ = 1 and q′ > Q − q, we find (with ℓ¯ denoting the multiplicative inverse of ℓ (mod q)) ! X X 1 1 1 −,1 · +O AI,Q (λ) = qv 1 + q−w 2 Q λQ Q and γ∈FI,Q qq1 ′ 6 1, it follows from (3.2) that the total contribution to G ′ ′ (ℓ) [γ, γ ] ⊆ I with γ < γ consecutive in FQ and γ ∈ F is given by ! ! ∞ X X 1 1 1 1 − εq (2) +O · = S I,Q (λ) + O . Gℓ,I,ε (λ) = q(kq + q′ ) 1 + γ2 Q Q (ℓ) k=0 γ∈FI,Q qk 6λQ 0, uniformly in λ on compacts of [2, ∞), Z cI C(ℓ) 1 1 − u λ G(2) (λ) = ln du + Oδ,ℓ ε−δ+θ(c,c1 ) . ℓ,I,ε ℓ u λ−u 0

An identical formula holds for G(3) ℓ,I,ε (λ).

(0, 1)

•

λ−1 ,1 3 •

λ−1 ,1 2 •

(λ − 1, 1) •

• λ−1 3−λ , 2 2

(λ − 1, 2 − λ)

•

(1, 1)

•

(1, λ − 1)

•

(1, 0)

•

Figure 7. The set ∪∞ k=1 Ωk ∩ T when 1 < λ < 2

(3.7)


15

When 1 < λ < 2 the contribution to S I,Q (λ) of γ with q + q′ ≤ λQ is ∞ X X fk,q (q′ , a) AI,Q (λ) = k=1

=

∞ X k=2

+

(ℓ) γ∈FI,Q (q,q′ )∈Q(Ωk ∩T )

X

X

q∈QIk gcd(ℓ,q)=1

(λ−1)Q6q6Q gcd(ℓ,q)=1

X

fk−1,q (q′ , a)

(0) (a,q′ )∈qI×QJk,q aq′ ≡−1 (mod q), ℓ|(q−a)

(1) (a,q′ )∈qI×QJk,q aq′ ≡−1 (mod q), ℓ|(q−a)

X

X

fk,q (q′ , a) +

!

f1,q (q′ , a),

(1) (a,q′ )∈qI×QJ1,q aq′ ≡−1 (mod q), ℓ|(q−a)

and the contribution of γ with q + q′ > λQ is X BI,Q (λ) =

′

X

f0,q (q′ , a).

(λ−1)Q 1, k f0,q k∞ 6

1 (λ−1)qQ

and

× qI, and summing as in the case λ > 2 we also obtain

Proposition 4. For every δ > 0, uniformly in λ on compacts of (1, 2], Z λ cI C(ℓ) 1 1 − u ln du + Oδ,ℓ ε−δ+θ(c,c1 ) . G(2) (λ) = ℓ,I,ε ℓ u λ−u 0

(3.8)

An identical asymptotic formula holds for G(3) ℓ,I,ε (λ).

As a result formula (3.7) will also hold when 1 < λ 6 2. Consider finally the case 0 < λ 6 1. When q′ > λQ we have qk > λQ for all k > 0. So the (ℓ) contribution of each γ ∈ FI,Q with q′ > λQ is in this case ! 1 1 1 arctan γ′ − arctan γ = ′ · , + O qq 1 + γ2 q2 q′2 summing up to ! X X 1 1 1 . (3.9) C I,Q (λ) = · + O qq′ 1 + a 2 Q 16q6Q (a,q′ )∈qI×J λ,q

gcd(ℓ,q)=1 gcd(q′ ,q)=1, ℓ|(q−a) aq′ ≡−1 (mod q)

q

Using the same estimates as in case λ > 2, (3.9) leads to Z cI C(ℓ) 1 1 1 C I,Q (λ) = ln du + Oδ,ℓ Qδ−θ(c,c1 ) ℓ max{1 − u, λ} 0 u ! Z 1−λ 1 1 cI C(ℓ) ln du + ln(1 − λ) ln λ + Oδ,ℓ Qδ−θ(c,c1 ) . = ℓ u 1−u 0

(3.10)

(ℓ) Finally the contribution of each γ ∈ FI,Q with q′ 6 λQ is

arctan t0 − arctan γ =

! 1 1 1 − εq , · + O qq′ 1 + γ2 q2 q′2

X

1 − εq 1 1 · +O ′ 2 qq Q 1 + aq

summing up to DI,Q (λ) =

X

(1−λ)Q 0, uniformly in λ on compacts of (0, 1], cI C(ℓ)(ζ(2) − λ) + Oδ,ℓ ε−δ+θ(c,c1 ) . G(2) ℓ,I,ε (λ) = ℓ An identical formula holds for G(3) ℓ,I,ε (λ). 4. End of the proof of Theorem 1 From Corollary 1 and Propositions 3, 4, 5 we gather ˜ ℓ,I,ε (λ) = cI Gℓ (λ) + Oδ,ℓ ε−δ+θ(c,c1) , G

with repartition function Gℓ given by  λ  + A(ℓ)H1 (λ) if λ ∈ 0, 12 , 1 − ζ(2)       λ Gℓ (λ) =  1 − ζ(2) + A(ℓ)H2 (λ) if λ ∈ 12 , 1 ,       2C(ℓ) H3 (λ) if λ ∈ [1, ∞), ℓ where

Z 1 1−u 1 λ 1 ln du + 2 ln du, H1 (λ) = λ − ζ(2) + (ln λ) ln(1 − λ) + 2 u 1−u u 1−u 1−λ λ Z 1 Z λ 1 1 1 1 = −λ − (ln λ) ln(1 − λ) + ln du − ln du 1−u 1−u 1−λ u 0 u = −λ, Z 1 1−u λ 2 H2 (λ) = λ − ζ(2) + (ln λ) + 2 ln du u 1−u λ Z 1 1 1 1 = 3λ − 2 − ζ(2) − (ln λ)2 + 2(1 − λ) ln −1 +2 ln du, λ 1−u λ u Z 1 λ 1−u ln du. H3 (λ) = u λ−u 0 Z

1−λ

This establishes part (i) in Theorem 1. Using also ! ! ! Z 1 Z 1−u 1 1 1 1−u 1 1 1 1 H3′ (λ) = du = − ln 1 − , − du = − + 1 − u λ λ−u λ 0 λ−u λ λ λ 0 it follows that the density of the repartition function Gℓ is  1    ζ(2) + A(ℓ)    1 ′ 2 1 gℓ (λ) = −Gℓ (λ) =  ζ(2) + A(ℓ) − 3 + λ − 2 λ − 1 ln      2C(ℓ) 1 − 1 − 1 ln 1 − 1 ℓ λ λ λ

1 λ

−1

λ > 1,

if λ ∈ 0, 21 , if λ ∈ 21 , 1 ,

if λ ∈ (1, ∞).

Part (ii) of Theorem 1 can now be deduced from part (i) by a standard approximation argument qℓ,ε (ω) c [1, 2, 3, 6] based on τℓ,ε (ω) ≈ cos ω for tan ω ∈ I with I ⊆ [0, 1] interval of length |I| ≍ ε . We skip the detailed proof, which can be easily reconstructed from some of the arguments in the next section.


17

5. The conversion from a honeycomb to a square lattice with congruence constraints In the situation of the honeycomb it suffices to consider ω ∈ [0, π6 ]. The linear transformation ! y 2y T R2 ∋ (x, y) −→ (x′ , y′ ) = x − √ , √ ∈ R2 3 3 √

maps the first sextant Γ+ := {(q + a2 , a 2 3 ) : q, a ∈ Z, 0 6 a 6 q} of the grid of equilateral triangles of side 1 onto the first quadrant of the square lattice Z2 = {(q, a) : a, q ∈ Z}. Elements of the subset Γhex + ⊆ Γ+ of vertices from the honeycomb grid map to integer lattice points (q, a) with q . a (mod 3) (see Figure 8).

H0,0L

H0,0L

Figure 8. The free path length in the honeycomb and in the deformed honeycomb

12

Hx0 ,y0 L+H12,3

2LΕ+

T Hx¢0 ,y¢0 L

Hx0 ,y0 L

Hx0 ,y0 L-H12,312 2LΕH0,0L

Hx¢0 ,y0¢ +Ε+ L

T-1

Hx¢0 ,y0¢ -Ε- L H0,0L

Hx¢0 ,0L Hx¢0 +y0¢ 2L

Figure 9. Change of scatterers under the linear transformation T √

√

T also maps circular scatterers (q + a2 , a 2 3 ) + ε(cos θ, sin θ) centered at (x0 , y0 ) = (q + a2 , a 2 3 ) to y √ θ , 2 sin √ θ ) centered at (x′ , y′ ) = (q, a). Denote ω0 = arctan 0 and ellipsoidal scatterers (q, a) + ε(cos θ − sin 0 0 x0 3 3 ω′0 = arctan

y′0 x′0 .

Denote also S ε = {δ(cos π3 , sin π3 ) : |δ| 6 ε}.

In the honeycomb with scatterers {(x0 , y0 )+S ε : (x0 , y0 ) ∈ Γ+hex } of same length, the ”local” repartition function π ˜ hex (λ) = ω ∈ arctan I : q˜ εhex (ω) > λ , interval, (5.1) I ⊆ 0, G I,ε ε 6 of the ”horizontal” free path length   √       hex q + a/2 a 3/2 hex q ∈ N : ∃a ∈ N, ∃(x , y ) ∈ Γ , q˜ ε (ω) = inf  ∈ (x , y ) + S ,  0 0 0 0 ε +  ,   cos ω sin ω

turns out to be closely related with G3,I,ε (λ), being estimated through the same approximation procedure as for the later when I ⊆ [0, 1] is a short interval of length |I| ≍ εc , 0 < ε < 1. Indeed, the equality √ ! y′0 + δ (y′0 + δ) 3 ′ −1 ′ ′ , , T (x0 , y0 + δ) = x0 + 2 2

18


shows that T maps the oblique scatterer (x0 , y0 ) + S ε from the honeycomb onto the vertical scatterer y′ +δ (x′0 , y′0 ) + Vε from the square lattice with x′0 ≡ y′0 (mod 3), and the line through the origin with slope 0x′ 0 y′ +δ onto the line through the origin with slope Φ 0x′ , where Φ is the bijection 0

" √ # 3 , Φ : [0, 1] → 0, 3

Φ(µ) =

√ µ 3 , 2+µ

2x Φ−1 (x) = √ . 3−x

In the process the condition q˜ εhex (ω) > λε above is being replaced by q3,ε (ω′ ) > λε in the square lattice. The only difference arises from replacing expressions (with x′0 = q, y′0 = a and 0 6 δ 6 q1 ) ! ! y′0 + δ y′0 δq δ 1 δ2 δ2 arctan − arctan ′ = ′ · +O 2 + O ′2 = 2 x′0 x0 x0 1 + y′20 q + a2 q x0 x′2 0

that collect the contribution of angles ω for G3,I,ε (λ), by √ √ ! ! y′0 3 y′0 + δ (y′0 + δ) 3 y′0 arctan Φ − arctan ′ − arctan Φ ′ = arctan ′ x′0 x0 2x0 + y′0 + δ 2x0 + y′0 √ √ ! y′0 3 (y′0 + δ) 3 1 − = y ′ √3 2 2x′0 + y′0 + δ 2x′0 + y′0 1 + 2x0′ +y′ 0 0 √ ! ′ 2x0 δ 3 1 δ = 1+O ′ y ′ √3 2 2x0 + y′0 (2x′0 + y′0 )2 1 + 2x0′ +y′ 0 0 √ ! δq 3 δ2 = +O 2 . 2(q2 + aq + a2 ) q R dx := The effect will only be on the main term, where c J = J 1+x2 , J ⊆ [0, 1], will be replaced by chex I √ R 1 3 dx, I ⊆ 0, √3 . In this way we obtain, uniformly in λ on compacts in R+ \ {1}, Φ−1 (I) 2(1+x+x2 ) ˜ hex (λ) = chexG3 (λ) + Oδ ε−δ+θ(c,c1 ) . (5.2) G I,ε I

The change of variable x = Φ(µ) gives √ Z Φ−1 (tan ω ) Z tan ω1 1 dµ dx 3 ∆ c[tan ω0 ,tan ω1 ] = = = ω1 − ω0 . 2+1 2 Φ−1 (tan ω0 ) x2 + x + 1 µ tan ω0

(5.3)

In this case q˜ εhex (ω) and the free path length τ˜ εhex (ω) are related (by the rule of Sines) by √ sin 2π q˜ εhex (ω) 3 3 hex hex τ˜ ε (ω) = q ˜ (ω) = · , 2 cos π6 + ω sin π3 − ω ε

so that

2λ cos π6 + ω λ hex hex hex ˜PI,ε . (5.4) (λ) := ω ∈ arctan I : τ˜ ε (ω) > = ω ∈ arctan I : q˜ ε (ω) > √ ε ε 3 Fix ωI ∈ arctan I. Using cos π6 + ω − cos π6 + ωI 6 |ω − ωI | ≪ εc , (2.2), and chex ≍ εc , equalities I (5.2) and (5.4) yield ! 2λ cos π6 + ωI ) hex hex ˜PI,ε (5.5) + Oδ ε2c + ε−δ+θ(c,c1) . (λ) = cI G3 √ 3 Let ε± = ε± (ω0 , ε) as in Figure 9. Consider the case of scatterers (x0 , y0 ) + S ω0 ,ε , (x0 , y0 ) ∈ Γ+hex , where S ω0 ,ε denotes the segment {δ(cos π3 , sin π3 ) : −ε− 6 δ 6 ε+ }. Let τ˜ εhex (ω) denote the free path


19

hex length and P˜ I,ε (λ) = |{ω ∈ arctan I : τ˜ εhex (ω) > λε }|. Since | cos ω± − cos ωI | ≪ εc and cos( 6π + ωI ) > 21 there exists C1 > 0 such that ! ! 1 1 ′′ c c 6 ε 6 ε 6 ε := ε . (5.6) − C ε + C ε ε′ := ε − + 1 1 cos( 6π + ωI ) cos( 6π + ωI )

The inequalities τ˜ ∆ε′′ (ω) 6 τ˜ ∆ε (ω) 6 τ˜ ∆ε′ (ω) and (5.6) combined with formula (5.5) and (2.2) lead to λ ˜P hex (λ) 6 ω ∈ arctan I : τ˜ hex ε′ (ω) > I,ε ε ( ) λ hex = ω ∈ arctan I : τ˜ ε′ (ω) > ′ ε cos( π6 + ωI ) − C1 εc ! λ hex ˜ = Pε cos( 6π + ωI ) − C1 εc ! cos( π6 + ωI ) 2λ G + Oδ ε2c + ε−δ+θ(c,c1) = chex · √ 3 I π c 3 cos( 6 + ωI ) − C1 ε ! 2λ + Oδ ε2c + ε−δ+θ(c,c1 ) , G = chex 3 √ I 3 ˜ hex and to a similar lower bound for PI,ε (λ), and so we get ! ˜P hex (λ) = chexG 2λ (5.7) + Oδ ε2c + ε−δ+θ(c,c1) . 3 √ I I,ε 3 The trivial inequality |τ∆ε (ω) − τ˜ ∆ε (ω)| 6 2ε and (5.7) now provide the formula ! 2λ hex G (λ) = chex PI,ε (5.8) + Oδ ε2c + ε−δ+θ(c,c1 ) √ 3 I 3 hex for the repartition function PI,ε (λ) = |{ω ∈ arctan I : τεhex (ω) > λε }| of ετεhex . Finally we choose a partition 0, √13 = ∪Nj=1 I j with intervals I j of equal size N1 ≍ εc . Applying (5.8) to each individual interval I j with c = c1 = hex √ (λ) = P[0,1/ 3],ε

N X j=1

cIhex G3 j

1 8

and summing over j we find ! ! 1 π 1 2λ 2λ √ + Oδ ε 8 −δ = G3 √ + Oδ ε 8 −δ . 6 3 3

(5.9)

Theorem 2 now follows immediately from (5.9) and obvious symmetry properties of the honeycomb. Acknowledgments We would like to thank the referee for careful reading and useful comments. The second author thanks Department of Mathematics, UIUC, for hospitality during his Fall 2007 visit, when most of this research has been completed. References [1] F. P. Boca, C. Cobeli, A. Zaharescu, Distribution of lattice points visible from the origin, Comm. Math. Phys., 213 (2000), 433–470. [2] F. P. Boca, R. N. Gologan, A. Zaharescu, The statistics of the trajectory of a certain billiard in a flat two-torus, Comm. Math. Phys. 240 (2003), 53–73. [3] F. P. Boca, R. N. Gologan, A. Zaharescu, The average length of a trajectory in a certain billiard in a flat two-torus, New York J. Math. (electronic) 9 (2003), 303–330. [4] F. P. Boca, A. Zaharescu, On the correlations of directions in the Euclidean plane, Trans. Amer. Math. Soc. 358 (2006), 1797–1825. [5] F. P. Boca, A. Zaharescu, The distribution of the free path lengths in the periodic two-dimensional Lorentz gas in the smallscatterer limit, Comm. Math. Phys. 269 (2007), 425–471. [6] E. Caglioti, F. Golse, On the distribution of free path lengths for the periodic Lorentz gas III., Comm. Math. Phys. 236 (2003), 199–221. [7] E. Caglioti, F. Golse, The Boltzmann-Grad limit of the periodic Lorentz gas in two space dimensions, C. R. Math. Acad. Sci. Paris 346 (2008), no.7-8, 477–482.

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[8] P. Dahlqvist, The Lyapunov exponent in the Sinai billiard in the small scatterer limit, Nonlinearity 10 (1997), 159–173. [9] F. Golse, The periodic Lorentz gas in the Boltzamann-Grad limit, Proc. ICM 2006 Madrid, Spain, Vol.III (invited lectures), pp.183–201. [10] J. Marklof, A. Str¨ombergsson, The distribution of free path lengths in the periodic Lorentz gas and related lattice point problems, preprint math.DS 0706.4395, to appear in Ann. of Math. Department of Mathematics, University of Illinois at Urbana-Champaign, 1409 W. Green St., Urbana, IL 61801, USA Institute of Mathematics of the Romanian Academy, P.O.Box 1-764, Bucharest RO-014700, Romania E-mail address: [email protected] Institute of Mathematics of the Romanian Academy, P.O. Box 1-764, Bucharest RO-014700, Romania E-mail address: [email protected]