nontrivial phase transition in a continuum mirror model

Journal of Theoretical Probability, Vol. 14, No. 2, 2001

Nontrivial Phase Transition in a Continuum MirrorModel

Matthew Harris1

Received November 15, 1996; revised July 14, 2000

We consider a Poisson point process on R2 with intensity *, and at eachPoisson point we place a two sided mirror of random length and orientation.The length and orientation of a mirror is taken from a fixed distribution, andis independent of the lengths and orientations of the other mirrors. We ask iflight shone from the origin will remain in a bounded region. We find that thereexists a **H with 0<**H<� for which, if *<**H , light leaving the origin in allbut a countable number of directions will travel arbitrariliy far from the originwith positive probability. Also, if *>**H , light from the origin will almost surelyremain in a bounded region.

KEY WORDS: Percolation; wind tree model; Lorenz model; phase transition.

1. INTRODUCTION

The Lorentz gas model, and the Ehrenfest wind tree model are of consider-able interest for modeling the motion of a particle in a random environ-ment.(11) In the Lorentz gas model, (12) a moving particle is elasticallyscattered by randomly placed scatterers of arbitrary shape (often taken tobe circles). The Ehrenfest wind tree model(6) is the above model where thescatterers are taken to be squares, whose diagonals are parallel to the coor-dinate axes.

A lattice version of the Ehrenfest wind tree model was introduced byRuijgrok and Cohen.(16) We describe the model in terms of light and

299

0894-9840�01�0400-0299�19.50�0 � 2001 Plenum Publishing Corporation

1 Department of Technical Mathematics and Informatics, Delft University of Technology,Mekelweg 4, 2628 CD Delft, The Netherlands. E-mail: matthew.harris�philips.com. Theauthor is currently reachable at Philips Research Labs, Weisshausstrasse 2, 52066 Aachen,Germany. +49 241 6003 521.

mirrors. Each vertex of Z2 is considered independently in turn. With prob-ability p, a double sided mirror is placed at the considered vertex of Z2.With probability : (0�:�1), the mirror is placed making angle ?�4 withthe positive x axis, and with probability (1&:) it is placed making angle3?�4. With probability 1& p no mirror is placed at the vertex. We thus endup with an arrangement of mirrors and gaps on the vertices of Z2.

Consider now shining light from the origin along one of the axes.When the light is incident on a vertex without a mirror, it continuestravelling in the same direction. If the light is incident on a mirror, it getsreflected to a perpendicular direction determined by the orientation of themirror and the direction of the incident light. One is interested in describingthe long term behaviour of such a light path.

For p=1 and :=1�2, Grimmett(8) showed that all light paths from theorigin are periodic with probability one. Bunimovich and Troubetzkoy(3)

extend this to the case p=1 and 0<:<1. Quas(14) uses ergodic theory tostudy the properties of infinite light paths if they exist.

Variations of this model exist. One can investigate different lattices (e.g.,hexagonal and triangular), and different scattering rules at the vertices.(See, for example, Refs. 3�5 and 17.) In this paper, we return to a con-tinuum model, which is closer to Ehrenfest's model.

Consider a Poisson point process on R2 with intensity *. At eachPoisson point we place a double sided mirror of random length, making aparticular angle with the positive x axis. The orientation and length of eachmirror is chosen independently at each Poisson point according to somedistribution satisfying certain conditions. We obtain a configuration ofmirrors in the plane. Light is then shone from the origin in all directions.We ask the question whether light from the origin remains in a boundedregion of R2 with probability 1.

The answer to this question depends on *. For high values of *, thereis a high intensity of mirrors, and the origin is almost surely surrounded bya closed loop of mirrors from which light from the origin cannot escape.For low values of * this does not happen, and we show that light from theorigin leaves every bounded region, in all but a countable number of initialdirections, with positive probability.

The question of recurrence is a more difficult one. Light from theorigin may travel arbitrarily far from the origin, but also may returnarbitrarily close. It is not known whether or not this happens.

This paper is divided into four sections. In the next section (Section 2)we define our notation, state our main theorem, and prove the easy partof the theorem. Section 3 is devoted to obtaining the necessary continuumpercolation theory results. In Section 4 we prove the main part of ourtheorem.

300 Harris

2. NOTATION AND STATEMENT OF THE THEOREM

Consider a Poisson point process of intensity * on R2 with points!0 , !1 ,... . Centred at each point, place a double sided mirror L(!i ) of ran-dom length \i in a random direction ; i . (We measure angles anticlockwisefrom the positive horizontal axis.) We take ;0 , ;1 ,... to be an i.i.d. sequenceof random variables with non-zero variance having support on [0, ?) & ?Q.Here ?Q is the set of rational multiples of ?. We also take ;0 , ;1 ,... to beindependent of the Poisson points !0 , !1 ,... . Also take \0 , \1 ,... to be ani.i.d. sequence of random variables on (0, �) with finite second momentthat are independent of the Poisson points !0 , !1 ,... and the angles;0 , ;1 ,... . We denote the (unimportant) probability space (7, A) anddenote the associated measure by P* . The expectation with respect to P*

we denote by E* .We will concentrate on proving statements about every mirror config-

uration _ in some P* measurable set. We now fix a particular configuration_ # 7, and make definitions for this _.

Let 0 :=R2_[0, 2?). We define a light element |=(x, %) # 0 to begiven by a point x # R2 called the position of the light, and an angle% # [0, 2?) called the outward angle of the light. A light element can bethought of as representing light moving from x # R2 in direction %measured anticlockwise from the positive horizontal axis. Now define

M=M(_) := .�

i=1

L(!i ) (1)

to be the set in R2 of points occupied by a mirror, and

K=K(_) := .i{ j : ;i{;j

L(!i ) & L(!j ) (2)

to be the set of corners where two mirrors intersect at a point. We use thisto define a map T=T (_): 0 � 0. We can think of T as giving the positionand direction of light just after the first reflection. A light element (x, %),not at a corner (i.e., x � K ) gets mapped to the light element ( y, %*) underT if y is the first point of M in direction % from x, and %*=(2;i&%)mod 2? if y # L(!i ). This can be interpreted as representing light beingreflected at the mirror L(!i ). A light element (x, %) at a corner (i.e., x # K )gets mapped to itself under T. We interpret this as meaning light is absorbedat corners. Repeated applications of T to the initial light element gives asequence of light elements representing the position and direction of thelight after successive reflections at mirrors in M. It will be convienient to

301Nontrivial Phase Transition in a Continuum Mirror Model

refer to the position and the outward angle of the light element T a(x1 , %1)(a�0) seperately, and we define

(T a1(x1 , %1), T a

2(x1 , %1)) :=T a(x1 , %1) (3)

Define then the orbit of (x1 , %1) to be

Orb(x1 , %1) := .n # N

T n(x1 , %1) (4)

the spatial orbit of (x1 , %1) to be

Orbs(x1 , %1) := .n # N

T n1(x1 , %1) (5)

and the angular orbit of (x1 , %1) to be

Orba(x1 , %1) := .n # N

T n2(x1 , %1) (6)

Finally, we define

Orb(O) := .% # [0, 2?)

Orb(O, %) (7)

A natural question to ask is, given *, for which values of % # [0, 2?)does light leaving O in direction % leave every bounded region? Moreprecisely, for which values of % is Orbs(O, %) a bounded set in R2? IfOrbs(O, %) is not bounded in R2, we write (O, %) � �.

This question, as we shall see, is intimately related to the percolationproperties of M, the set of points in R2 occupied by mirrors. Two pointsx and y # R2 are said to have a vacant connection (denoted x W

v y) if thereexists a continuous curve # with x and y as its endpoints, and # & M=<.The vacant cluster of the origin is defined as

C=C(O) :=[x # R2 : O Wv x] (8)

We define the critical intensity **H as

**H :=sup[* : P*( |C |=�)>0] (9)

Here, for a set A�R2, we define the diameter |A| , of A to be |A| :=sup[d(x, y) : x, y # A] where d( } , } ) is Euclidean distance. We show in Sec-tion 3 that 0<**H<�. This means that for high mirror intensity (*>**H ),the origin is encircled by a closed loop of mirrors with probability one.

302 Harris

Light starting from any point can therefore never leave every boundedregion. For low mirror intensity (*<**H ), the origin is not enclosed by anyclosed loop of mirrors with positive probability. Our theorem says, in thiscase, that with probability P*( |C |=�)>0, (O, %) � � for all but a count-able number of % # [0, 2?).

Theorem 1. For mirrors placed as above, with ;0 having supporton [0, ?) & ?Q, and non-zero variance, and with E*(\2

0)<�, we have0<**H<�, and

1. *<**H O P*((O, %) � � for Lebesgue a.e. % # [0, 2?))

= P*( |C |=�)>0

2. *>**H O P*(_%: (O, %) � �)=0

Remark 1. We take the _-algebra A on 7 to be such that this eventis measurable. It is not difficult to show that this is possible.

Remark 2. The set of % # [0, 2?) in the first part of the theorem forwhich (O, %) � � is not true is countable.

Remark 3. We need that 0<**H<� (proved in Section 3) to ensurethat neither of the statements in the theorem is void.

Proof of Theorem 1 (Part 2). If *>**H , then, by the definition of **H ,the origin is contained in a bounded vacant cluster with probability one.Thus, with probability one there is a connected circuit of mirrors aroundthe origin, and light from the origin can never escape from this region. g

3. NONTRIVIALITY OF **H

To show that 0<**H<�, we will first define some more criticalparameters for the model.

For two points x, y # R2, we say that x and y have an occupied con-nection (denote x W

o y) if x, y # M and there is a continuous path #�Mjoining x and y. Similarly, we say that x and y have a vacant connection(denoted x W

v y), if x and y are connected by a path not touching M. Let!1 be the (almost surely) unique Poisson point nearest the origin. Theoccupied cluster of the origin is defined as

Cocc=Cocc(O) :=[x # R2 : !1 Wo x] (10)


The vacant cluster of the origin is defined as

C=C(O) :=[x # R2 : O Wv x] (11)

We now define the critical densities **H and *T as follows:

**H :=sup[* : P*( |C |=�)>0] (12)

*T :=inf[* : E*( |Cocc | )=�] (13)

The following inequalities hold:

0<*T�**H (14)

The inequality 0<*T is proved in Hall.(9) The inequality *T�**H is provedin Roy.(15) (This is his inequality (4.7). His condition that \0<R is notneeded in the proof of this inequality.) We now have that **H>0.

We now show that **H<� as follows. Consider a ``thinned out'' dis-tribution of mirrors. Suppose �1 and �2 are two atoms in the distributionof ;0 with

0<p :=P*(;0=�1)�P*(;0=�2)

We can choose the axes such that (�1+�2)�2=?�2 (i.e., �1 and �2 aresymmetric about ?�2). Let P� * be a new mirror distribution with

P� *(;0=�1)=P� *(;0=�2)=1�2

and let

*� *H :=sup[* : P� *( |C |=�)>0] (15)

*� H :=inf[* : P� *( |Cocc |=�)>0] (16)

The following is now true.

2p**H�*� *H�*� H<� (17)

The inequality *� *H�*� H follows, after minor adjustments, from the proofof the analagous theorem for site percolation on Z2 in Gandolfi et al.(7) Inthis proof, it is required that the mirror distribution be symmetric underaxis reflections��a property of the P� * distribution. As the proof followsGandolfi et al.(7) almost identically, we omit it here and refer the reader toHarris.(10)

The inequality 2p**H�*� *H follows from a simple coupling argument,and the inequality *� H<� is proved in Hall.(9) We now have that0<**H<�.

304 Harris

We note that Roy(15) proved that *T=**H=*H for the case that ;0

is uniformly distributed on [0, ?) and \0 is almost surely bounded. Also,Roy(15) has a proof that **H�*H using the technical RSW lemma forbounded \0 .

4. REGIME WHERE LIGHT ESCAPES TO INFINITY

In this section we prove Theorem 1 (part 1). After defining therequired notation; we state three lemmas from which the theorem willeasily follow. The rest of this section is devoted to proving the lemmas.

Take *<**H . The origin is now part of an infinite vacant cluster withpositive probability. We consider individual realisations of the mirrors suchthat the origin is in an infinite vacant cluster and there are only finitelymany mirrors in any finite box. The set of considered configurations haspositive measure equal to P*( |C |=�). (Note that almost every configura-tion has only finitely many mirrors in any finite box as E*(\2

0)<�.)Take such a configuration _ # 7, and restrict our attention to light

paths inside a box Bn :=[&n, n]_[&n, n]. We wish to find the values of% # [0, 2?) for which Orbs(O, %) & Bc

n{<. To do this, it is useful to placean extra four mirrors of length 2n around the perimeter �Bn of Bn . Withthese mirrors added, it is equivalent to find the values of % for whichOrbs(O, %) & �Bn{<. (If Orbs(O, %) & Bn=Orbs(O, %), addition of mirrorsalong �Bn will not change Orbs(O, %). If Orbs(O, %) & Bc

n{<, then, additionof mirrors along �Bn will give Orbs(O, %) & �Bn{<.) In what follows wewill assume that these four mirrors are present in the configuration _ # 7.

There are finitely many mirrors in Bn and they intersect each other infinitely many places. Let V :=[v1 ,..., vn1] be the set of all points vi that iseither a point in K, the endpoint of a mirror in Bn , or the origin, O. Wecall V the vertex set, and vi # V vertices of _ in Bn . Set S to be the set ofmirror segments [vi , vj ]�M connecting two neighbouring vertices (i.e.,[vi , vj ] & V=[vi , vj ]��there are no vertices of V in the interior of anys # S ).

Given %1 # [0, 2?), and x1 # R2, the angles in Orb(x1 , %1) areenumerated in the sequence

%1 , 2;i1&%1 , 2(;i2

&;i1)+%1 , 2(; i3

&; i2+;i1

)&%1 , ...

The sequence [ij ]�j=1 gives the order of the mirrors visited. As ;1 has

support on [0, ?) & ?Q, then

Orba(x1 , %1)�[%1+#, # # ?Q]


and if %1 � ?Q, then

2(;ik&;ik&1

+ } } } +(&1)k+1 ;i1)+(&1)k %1{%1

for k odd.We define the sets 3� =3� (_, n) and 3=3(_, n) of exceptional angels.

Let

3� :=[% # [0, 2?) : T a1(v, %) # V for some a # N, v # V ] (18)

and

31 :=[% : %=(%1+#) mod 2?, %1 # 3� , # # ?Q] (19)

3 :=31 _ (?Q & [0, ?)) (20)

Light leaving a vertex v # V with angle % � 3� will never reach anothervertex. Given any point x1 # M, light leaving x1 at angle %1 � 31 will nevertake a value in 3� . Also, if %1 � ?Q & [0, ?), then then initial angle cannotbe reached after an odd number of reflections. We will need the followingproperty of 3.

Lemma 1. The set 3(_, n) is countable.

Consider (x1 , %1) # Orb(O) with x1 # s1 for s1 # S. Let E(x1 , %1)=[(s1 , %1),..., (sm ,..., %m)] with si # S, i=1,..., m, be the minimal set ofsegment angle pairs that contains �x # s1

Orb(x, %1). As ;1 has support on[0, ?) & ?Q, and there are only finitely many mirrors in Bn , the setE(x1 , %1) of segment angle pairs is finite.

If the set [x # R2 : (x, %j ) # Orb(x1 , %1)] is dense in sj for all j=1,..., m,we abuse notation slightly, and say that Orb(x1 , %1) is dense in E(x1 , %1).We will prove the following lemma.

Lemma 2. For (x1 , %1) # Orb(O), x1 # Bn and %1 � 3, we haveOrb(x1 , %1) is dense in E(x1 , %1).

We then use Lemma 2 to prove the following.

Lemma 3. If (x1 , %1) # Orb(O), x1 # Bn and %1 � 3 then Orbs(x1 , %1)& �Bn{<.

Remark 4. Aarnes(1) showed that Orbs(O, ?�4) is dense on �(C & Bn)if ?�4 � 31 in the case when ;1 # [0, ?�2] for all i. We prove Lemmas 2 and

306 Harris

3 by examining the related interval exchange transformation problem. Thisnatural step is often done in the study of billiard motion (see Boldrighiniet al.(2)).

The first part of Theorem 1 now follows from Lemmas 1 and 3.

Proof of Theorem 1. We consider configurations of mirrors _ # 7such that O is in an infinite vacant cluster, and there are only finitely manymirrors contained in finite boxes.

Now consider the sequence of boxes B1 , B2 ,... . By Lemma 1, the set9 :=��

k=1 3(_, k) is countable. For each % # [0, 2?)�9, the light element(x1 , %1)=T (O, %) is such that %1 # [0, 2?)�9. As % � 3(_, k) for all k,Lemma 3 holds for arbitrarily large k, Orb(O, %) & K=< and the lightdoes not get absorbed at a corner. Also, we see that Orbs(x1 , %1) & �Bk{<for all k, and thus (O, %) � �. g

Proof of Lemma 1. The light path can be described in an alternativeway. When light hits a mirror L(!i ), instead of reflecting the light at themirror, we reflect Bn and all its mirrors in L(!i ), and let the light continueundiverted. The light continues in a ``reflected environment.'' Iterating thisprocedure, we obtain straight line light rays moving in a changing environ-ment. This is an equivalent description of the light path.

Consider now light emerging from vertex vi # V that reaches vertexvj # V after (a+1) iterations of T. Let s1 ,..., sa be the sequence of segments(sk # S, k=1,..., a) that the light touches en route to vj . Given that Bn isreflected around segments s1 ,..., sa , vertex vj lies in a unique position, andonly one straight line connects vi to vj . Thus, such a sequence s1 ,..., sa ofsegments connecting vi to vj in (a+1) steps corresponds to light emergingfrom vi in a unique direction.

There are therefore only finitely many directions % # [0, 2?) that con-nect two vertices in (a+1) steps, and thus countably many that connecttwo vertices in a finite number of steps. Thus 3� is countable, and clearly3 is too. g

Before proving Lemma 2, we take a little interlude, and recall the basicdefinitions about interval exchange transformations.(13)

Let Y :=[0, 1) and let l�2 be an integer. Take p=( p1 ,..., pl) to bea probability vector with pi>0 for 1�i�l. We divide Y into intervals oflength pi . Define

q0=0(21)

qk= :k

i=1

pi


and

Yk :=[qk&1 , qk) (22)

Let { be a permutation of the symbols [1,..., l]. We then rearrange theintervals according to {, and obtain a new probability vector p{=( p{&1(1) ,..., p{&1(l)), and new intervals defined by

q{0=0

(23)q{

k= :k

i=1

p{i

and

Y {k :=[q{

k&1 , q{k) (24)

The map T� : Y � Y defined by

T� y= y&qi&1+q{{(i )&1 (25)

for y # Yi and 1�i�l is an piecewise order preserving isometry on Y,and is called a ( p, {) interval exchange transformation. Note that T� iscontinuous except at the points q1 ,..., ql&1 (called separation points), and iscontinuous from the right at these points.

We say the exchange transformation T� satisfies the minimality condi-tion if

v (M1) T� is aperiodic (i.e., for each y # Y, the orbit O( y) :=[T� iy :i # Z] is infinite), and

v (M2) If F is a finite disjoint union of half open intervals (closed onthe left) with endpoints belonging to the countable set

D� := .l&1

i=0

O(q i ) _ [1]

then T� F=F implies F=Y or F=<.

We end our interlude by stating the following result from Keane.(13)

Theorem 2. T� satisfies the minimality condition if and only if O( y) isdense in Y for all y # Y.

Each (si , %i ) # E(x1 , %1) can be partitioned into ``subsegment'' anglepairs, where each point in a subsegment angle pair is mapped to the same

308 Harris

segment angle pair (sj , %j ) # E(x1 , %1). The map T then gives a rearrange-ment of the subsegment angle pairs in E(x1 , %1). We now make this moreprecise, and define an interval exchange transformation T� =T� (T, x1 , %1),where the interval [0, 1) will correspond to the disjoint union of segmentangle pairs in E(x1 , %1).

Associate with (si , %i ), i=1,..., m an interval

Ii :=_ :i&1

k=1

|sk | |sin(;(sk)&%k)|, :i

k=1

|sk | |sin(;(sk)&%k)|+ (26)

on the real line of length |si | |sin(;(si )&%i )|. Here ;(si ) is the value of ;j

corresponding to the segment si . We now define different partitions of theinterval Ii .

The set (si , %i ) can be partitioned into subsets Pij , 1�i, j�m, asfollows:

Pij :=[(x, %i ) # (si , %i ) : T (x, %i ) # (sj , %j )] (27)

where some of the Pij 's may be empty. From Fig. 1 we see that each non-empty Pij is a union of subsegments of si , paired with angle % i . Denotethese subsegment angle pairs Pijk , k=1,..., mP(i, j ). The partition Pijk ,1� j�m, 1�k�mP(i, j ) induces a partition R ijk of the interval Ii intosubintervals in a natural way. We call this the post-partition of Ii .

The set (si , %i ) can also be partitioned into subsets Qij as follows:

Qij :=[(x, %i ) # (si , % i ) : T &1(x, %i ) # (sj , % j )] (28)

where some of the Qij 's may be empty. Again, each nonempty Qij is aunion of subsegments of si , paired with angle %i . Denote the subsegmentangle pairs Qijk , k=1,..., mQ(i, j ). Again, the partition Qijk , 1� j�m,1�k�mQ(i, j ) induces a partition S ijk of the interval Ii into subintervalsin a natural way. We call this the pre-partition of Ii .

Note that for 1�k�mP(i, j ), there exists a unique k$, 1�k$�mQ( j, i )such that TQjik$=Pijk . We associate subinterval Rijk with subinterval S jik$ .Due to the scaling of the intervals (interval Ii has length |si | |sin(;(si )&%i | ),the lengths of the subintervals associated with Pijk and Qjik$ are the same.(Associated segments have different lengths as they make different angles tothe light connecting them.)

We have defined two partitions of the interval

I :=_0, :m

k=1

|sk | |sin(;(sk)&%k)|+ (29)


File: 860J 061712 . By:XX . Date:06:03:01 . Time:13:35 LOP8M. V8.B. Page 01:01Codes: 1552 Signs: 959 . Length: 44 pic 2 pts, 186 mm

Fig. 1. A light element (x, %i ) from Pij gets mapped to a light element ( y, %j ) # (sj , %j ) under T.Due to the (finitely many) mirror obstructions, the set Pij is a union of subsegments of si ,paired with angle %i .

The sequence R1 ,..., Rl of post-partition subintervals Rijk we call the post-partition of I, and the sequence S1 ,..., Sl of the pre-partition subintervalsSijk we call the pre-partition of I. Due to the one to one association of thepre and post subintervals defined above, there is the same number of sub-intervals in the pre and post partitions of I.

The subinterval R: is associated with a subinterval S; for some; # [1,..., l]. We define

{&1(:)=; (30)

We normalise the interval I to have length 1, and define the separationpoints 0=q0 ,..., ql&1 to be the left boundary points of the post-partitionsubintervals. The permutation { defines an interval exchange transforma-tion, T� , on the interval [0, 1). We note that points in a subinterval of [0, 1)

310 Harris


Fig. 2. The orientation of a mirror segment is reversed after an odd number of reflections.

can only return to this subinterval after an even number of applicationsof T� . This is because, as [0, ?) & ?Q�3, T n

2(x1 , %1){%1 for % � 3 and nodd. The orientation of a subinterval therefore not be reversed by multipleapplications of T� (see Fig. 2). This property is required for T� to be an inter-val exchange transformation.

The correspondence between T and T� may not agree on q0 ,..., ql&1 asT� is constructed to be right continuous, and light elements at corners aremapped to themselves under T. This will not be a problem as light from theorigin that eventually enters a vertex will be excluded.

We end with a definition. Vertex separation points are separationpoints that correspond to vertices in V. The rest of the separation pointsare mapped to vertex separation points under T� . We call these non-vertexseparation points.

Proof of Lemma 2. We show that T� satisfies the minimality conditionwhen %1 � 3, and Lemma 2 then follows from Theorem 2.

To prove (M1), we note that due to our choice of (x1 , %1) (i.e., %1 � 3 ),the vertex separation points have infinite orbits. As the non-vertex separationpoints are mapped to vertices in V after one iteration of T� , these too haveinfinite orbits. It now remains to be shown that non-separation points arealso not periodic. As in Keane (1975), suppose there exists an x # [0, 1),and t>0 such that T� tx=x. We set

# :=max[T� iqj : 0�i�t&1, 0� j�l&1, T� iqj�x] (31)


If #=x, then T� tqj=qj for some 0� j�l&1, which is not the case. Wemust therefore have #<x. Each T� i with 1�i�t is an isometry on [#, x],and as T� tx=x, we have T� t#=# giving a contradiction. Non-separationpoints are thus also not periodic.

To prove (M2) we take F as in condition (M2) with T� F=F. Takex # �F, x{1. Then x=T� kqj for some k # Z and some j # [0, 1,..., l&1].Consider first the case when k>0. Now, T� x is a boundary point of F,or x # D :=[q0 ,..., ql&1]. Suppose that [x, T� x, T� 2x,...] consists entirely ofboundary points of F not in D. As �F is finite, this would imply that x iseventually periodic. As T� is one to one, this would imply that x is periodic,and then qj also, which is not the case. There must therefore be an integert�0 such that T� tx=qi # D for some i. Thus qi=T� t+kqj . As non-vertexseparation points map to vertex separation points under T� , and as theorbits of vertex separation points are infinite and distinct, k+t may onlytake on the value 1. The only possibility is k=1 and t=0, qi is a non-vertex separation point, and x is a vertex separation point.

Now consider the case when k�0. Then x=T� &lqj with l�0. Thepoint T� &1x is a boundary point of F, or T� &1x # D. Again, as the q j 's arenot periodic, T� is one to one, and there are only finitely many boundarypoints of F, there must be an integer t>0 such that T� &tx=qi # D forsome i. Then qj=T� l+tqi , and the same argument as above gives that xmust be a vertex separation point.

We have that F must be the union of whole intervals Ii , 1�i�m. IfF is non-empty, it must contain at least one interval Ik , 1�k�m, corre-sponding to the segment angle pair (sk , %k) # E(x1 , %1). As (sk , %k) #E(x1 , %1), there exists an x # Ik & O(x1 ) where x1 is the point in I corre-sponding to (x1 , %1). As T� F=F, we see x1 # F, and thus I1�F. This meansthat there are points from every interval Ii , 1�i�m in F, and as F is theunion of whole intervals Ii , then F=I=[0, 1), and the minimality conditionis satisfied. g

In the proof of Lemma 3 we will argue that if light from the originreaches a particular segment travelling in a particular direction, then lightfrom the origin will also reach neighbouring segments travelling in thesame direction. This will then imply that light from the origin travelling inthis direction will reach the boundary of the box Bn .

We will need to specify the side of the mirror at which light is reflected.To this end, we define an orientated segment to be a segment angle pair (s, ,s)with s # S and ,s # [0, 2?) perpendicular to s (i.e., |(,s&;(s)) mod ?|=?�2.)We say that the light element (x, %) is entering orientated segment (s, ,s) if?�2<|%&,s | and that light element (x, %) is leaving the orientated segment(s, ,s) if |%&,s |<?�2. Note that Orb(x, %) consists only of light elements

312 Harris


Fig. 3. Light element (x, (2;&%) mod 2?) entering the oriented segment (s, ,s) and beingreflected to light element (x, %) leaving the same oriented segment.

leaving oriented segments. The light element (x, (2;&%) mod 2?) enteringthe oriented segment at x is reflected to the light element (x, %) leaving theoriented segment. See Fig. 3. Note that the light is reflected on the side ofthe mirror given by ,s .

Proof of Lemma 3. In the sequel we will adjust the definition ofOrb(x, %) to include light elements entering oriented segments. This newdefinition will make the presentation of Lemma 3 clearer.

Two oriented segments (r, ,r) and (s, ,s) are said to be adjacent if

Case 1: They have one vertex in common and |,r&,s |<?, or

Case 2: They are the same mirror segment, one of both of the verticesis not an intersection point of two mirrors, and |,r&,s |=?.

See Fig. 4.We now study the orbit of a light element (x1 , %1) # Orb(O), and

prove the following claim:

Claim. If there exists a light element in (x, %) # Orb(x1 , %1) thatleaves (enters) an oriented segment (r, ,r) at angle %, and (s, ,s) is an adja-cent oriented segment with

Case 1: |,r&,s |<?, or

Case 2: |,r&,s |=?,



Fig. 4. Two neighbouring oriented segments. In Case 1 the two oriented segments are ondifferent mirrors, and in Case 2 the two oriented segments are opposite sides of the samemirror.

then

Case 1: there exists another light element in Orb(x1 , %1) that leaves(enters) the adjacent oriented segment (s, ,s) at angle %, or

Case 2: there exists another light element in Orb(x1 , %1) that enters(leaves) (s, ,s)=(r, |?&,r | mod 2?) at angle %.

Proof of Claim. There are four cases, shown in Figs. 5 and 6. Thefirst two, labelled 1a and 1b, covered in statement 1 above, and the lasttwo, labelled 2a and 2b, covered in statement 2 above can be provedsimilarly. We only give the proof of the two cases 1a and 2a.

Case 1a. A light element in Orb(x1 , %1) leaves (r, ,r) at angle %. AsOrb(x1 , %1) is dense on E(x1 , %1), there are light elements in Orb(x1 , %1)with positions in r arbitrarily close to v, leaving at angle %. These aremapped under T to light elements entering an oriented segment (r$, ,r$) atangle %. By Lemma 2, Orb(x1 , %1) is dense on (r$, ,r$) with this angle, andthere are light elements in Orb(x1 , %1) entering (r$, ,r$) whose preimageswere light elements leaving (s, ,s) at angle %. (Here we say that Orb(x1 , %1)is dense on (r$, ,r$) with angle % if Orb(x1 , %1) is dense on (r$, %).)

Case 2a. Now s=r and we denote ,r=, (1)r , ,s=, (2)

r . Furthermorewe let v be an endpoint of r. A light element in Orb(x1 , %1) leaves (r, , (1)

r )at angle %. As Orb(x1 , %1) is dense on E(x1 , %1), there are light elements inOrb(x1 , %1) with positions on r, arbitrarily close to v, leaving at angle %.

314 Harris


Fig. 5. Mirror segment arrangement as described in Case 1.

Fig. 6. Mirror segment arrangement as described in Case 2.



These are mapped under T to light elements entering an oriented segment(r$, ,r$) at angle %. By Lemma 2, Orb(x1 , %1) is dense on (r$, ,r$) with thisangle, and there are then light elements in Orb(x1 , %1) that are mappedfrom light elements leaving an oriented segment (r", ,r") (the first orientedsegment in direction % from v) with this angle. There are light elementsleaving (r", ,r") at angle % that are mapped under T to light elements entering(r, , (2)

r ) at angle %.This completes the proof of the claim.By repeated applications of the claim; we see that light elements in

Orb(x1 , %1) either enter or leave (or both) every mirror segment inCocc(x1) & �C. (Recall that Cocc(x1) :=[x # R2 : x1 W

o x] is the occupiedcluster containing x1 # R2, and that �C is the boundary of the vacantcluster containing the origin. The set �C is a function of n, and containspart of the boundary �Bn of the box Bn . To emphasize this fact, we willwrite �Cn :=�C.)

Without loss of generality, suppose that %1 # [0, ?�2). (%1 points upand to the right.) Now, either Cocc(x1) & �Cn & �Bn{<, and the lemmafollows, or Cocc(x1) & �Cn & �Bn=<. Then there exist light elements inOrb(x1 , %1) that leave Cocc(x1) at angle %1 and enter a mirror at angle%1 at a point y � Cocc(x1) which is higher than any mirror in Cocc(x1).(See Fig. 7). We begin the argument again, obtaining that light elements in

Fig. 7. Light from the origin at angle % reaches the boundary box.

316 Harris

Orb(x1 , %1) enter or leave every mirror segment in Cocc( y) & �Cn . IfCocc( y) & �Cn & �Bn{< then we are finished, otherwise we repeat theargument with z � Cocc(x1) _ Cocc( y) that is higher than every point inCocc(x1) _ Cocc( y).

These repetitions must stop as there are only finitely many mirrors in Bn ,and the lemma follows. g

ACKNOWLEDGMENTS

I would like to thank Michael Keane, Alberto Gandolfi, GeofferyGrimmett, and Ronald Meester for their input, and the stimulating discus-sions we had on this problem. Also I would like to thank the anonymousreferee for his helpful comments, and in particular his suggestions forsimplifying the proof of Lemma 3.

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nontrivial phase transition in a continuum mirror model

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