Buffon's Problem determines Gaussian Curvature in three Geometries
Aizelle Abelgas, Bryan Carrillo, John Palacios, David Weisbart, Adam Yassine
aa r X i v : . [ m a t h . P R ] S e p BUFFON’S PROBLEM DETERMINES GAUSSIAN CURVATURE INTHREE GEOMETRIES
AIZELLE ABELGAS ∗ , BRYAN CARRILLO † , JOHN PALACIOS ∗ , DAVID WEISBART ∗ ,AND ADAM YASSINE ‡∗ Department of Mathematics † Department of MathematicsUniversity of California, Riverside Saddleback College ‡ Department of Mathematics and Actuarial ScienceAmerican University in Cairo
Abstract.
The classical Buffon problem requires a precise presentation in order to bemeaningful. We reinterpret the classical problem in the planar setting with a needle whoselength is equal to the grating width and find analogs of this problem in the settings of thesphere and the Poincar´e disk. We show that the probability that the needle intersects thegrating in these non euclidean settings tends to the probability of the intersection in theplanar setting as the length of the needle tends to zero. Finally, we calculate the Gaussiancurvature of the spaces from probability deficits related to the generalized Buffon problem,obtaining a result similar to the Bertrand-Diguet-Puiseux Theorem.
Contents
1. Introduction 12. A General Framework 33. Probabilities for the Plane and the Sphere 74. The Needle in the Poincar´e Disk 105. Limiting Behavior and Probability Deficits 15References 211.
Introduction
In his investigation of players’ chances in the game “clean tile”, Buffon introduced thefollowing problem, its statement in his own words translated in [9] as:
E-mail address : [email protected], [email protected], [email protected],[email protected], [email protected] . I assume that in a room, the floor of which is merely divided by parallellines, a stick is thrown upwards and then one player bets the stick will notintersect any of the parallels on the floor, whereas on the contrary the otherone bets the stick will intersect some one of these lines; it is required to findthe chances of the two players. It is possible to play this game with a sewingneedle or a headless pin.”Buffon’s problem, a progenitor of geometric probability, is often simplified as it is in thepresent exposition to ask for the probability that a randomly dropped needle intersects aline on the floor. This simplification naturally leads to the solution of the original problem.Buffon’s problem has attracted considerable interest and invited many generalizations andanalogs following its introduction in the book [4] and in Buffon’s earlier lecture [3]. Alreadyin 1860, Barbier studied in [1] an analog of this problem in the planar setting but where theneedle is replaced by a rigid rectifiable plane curve, a “noodle”. Solomon gives an extensivetreatment and review of this problem and many of its generalizations in [9]. He includes areview and discussion of an application in three dimensions. Klain and Rota give a treatmentof the Buffon problem in [7]. Diaconis studies the problem in the setting of a long needle in[5], where he calculates the probability mass function for the random variable that counts thenumber of intersections and estimates its moments. He also discusses an applied motivationfor this problem in detection deployment. Peter and Tanasi study in [8] an analog of theBuffon problem on the sphere, using a grating formed by lines of longitude. Isokawa discussesan analog of Buffon’s problem on the sphere in [6], using a grating formed by lines of latitude.To dispense with the difficulty of dealing with the boundary of an actual floor of finiteextent, we reinterpret the problem as dropping a needle on an infinite floor, modeled as theplane covered with an infinite collection of evenly spaced parallel lines. The non-existenceof a uniform probability density function on all of R demands a clarification of the problemeven in this classical setting. To develop some intuition, imagine that a needle is tossed highenough over an infinitely wide floor with infinitely long parallel slats of width L so that theposition of the center of the needle approximately has a uniform distribution on conditioningit to fall on any given bounded disk D on the floor. Let A be a line that is determined by aslat. Denote by p the probability that a needle intersects A given the following conditions:(1) the needle lands uniformly randomly on a line segment of length L that is perpendicularto and bisected by A ; and (2) the angle that the needle makes with the slat is uniformlydistributed on [0 , π ]. If D is sufficiently large in comparison with the width of the slats,then the probability that the needle intersects a line formed by the slats is approximately p .With this physical interpretation in mind, we extend Buffon’s problem to dropping needleson two surfaces that are homogeneous but have non-zero Gaussian curvature, namely thesphere with an arbitrary given radius and the Poincar´e disk. While the sphere is a compactspace, making it possible to uniformly randomly select the center of the needle on this surface,the Poincar´e disk, like R , is non-compact and so choosing a point for the center of the needlerequires further interpretation. Our interpretation of this problem in the spherical setting isdifferent from the interpretation presented in [6] and in [8], but permits a natural and unified reatment of the problem in three different geometries. The solution to the problem in thesesettings gives a probabilistic way of determining the Gaussian curvature of the sphere and thePoincar´e disk through probability deficits. The relationship between probability deficits andGaussian curvature is similar to the relationship that the Bertrand-Diguet-Puiseux Theoremestablishes in [2] between area and circumference deficits and Gaussian curvature. Acknowledgements.
We thank Professor Yat Sun Poon, who as chair of the UCR Depart-ment of Mathematics has given his unwavering and enthusiastic support for undergraduateresearch and education abroad programs. Carrillo and Yassine wish to thank the UCR De-partment of Mathematics for their funding in Spring 2018 and 2019 as graduate studentmentors through the Undergraduate Research Program. We thank all those involved in theFaculty Led Education Abroad Program (FLEAP), and especially Dr. Karolyn Andrews,for the programs in London in summer 2018 and 2019 that Weisbart ran that ultimatelygave rise to the present paper. Abelgas and Palacios participated in this program in sum-mer, 2018. We thank James Alcala, Christopher Kirchgraber, Frances Lam (also a FLEAPalumn), and Alexander Wang for the time they spent as undergraduate researchers workingon this project in its various stages.2.
A General Framework
Denote respectively by R , S r , and H the plane, the sphere of radius r , and the Poincar´edisk. This section develops a uniform framework for studying Buffon’s problem in thesespaces and proves some general theorems that hold in all three settings. Denote by M either R , S r , or H .2.1. Isometries and Gratings.
The Riemannian manifold M is homogeneous, it has anisometry group G that acts transitively. Definition 2.1. An equator of M is a directed geodesic in M . Definition 2.2.
A grating G ( L ) on M with equator E and spacing L is a set of geodesicsin M that intersect E at right angles and whose set of intersections with E form an evenlyspaced set of points E L with smallest spacing equal to L .Select a directed geodesic E in M and a grating G ( L ) that is oriented by E . In the settingof the sphere S r , evenness of the spacing requires L to be equal to πrn , where n is a naturalnumber. In each of the three settings, a maximal nontrivial subgroup G ′ of G will preserveboth E and the orientation of E . There is a single element h of G ′ that generates a subgroup H of G ′ that acts transitively on E L and under which E L is invariant, and that moves pointsin E in the positive direction along E . Suppose that g is in G ′ . Denote by α g the signeddistance that g moves points along E . For each space M , the displacement function α definedby α : g α g gives an ordering on the elements of G ′ , with the additional assumption in the case of S r thatthe range of G ′ is R mod 2 πr . The function α is a homomorphism. We will verify all above ssertions as well as Propositions 2.4 and 2.3 case by case when we independently discussthe three cases, but take them now for granted throughout the remainder of this section.Further notation is useful for stating the next two propositions. Suppose that g is anelement of G and that x is an element of M . Denote henceforth by gx the result of applyingthe transformation g to the point x . If S is a subset of M , then denote by gS the set gS = { gs : s ∈ S } . Proposition 2.3. If M is any of the three spaces above, there is a unique g in G such thatif E and E ′ are two equators in M , then g E = E ′ . Proposition 2.4.
Given any two gratings G ( L ) and G ( L ) with spacing L and equator E ,there is an element g of H such that gG ( L ) = G ( L ) . Dropping the Needle.
Take X to be a random variable that is uniformly distributedon [ − ℓ, ℓ ]. The orientation of E determines a notion of signed distance between points on E .Let z be a real number. In the case when M is the sphere, z will be in R mod 2 πr . Denoteby p x ( z ) the unique point on E that lies on E a signed distance of z from x . View a needle ineach of the three geometries as a directed segment of length L of a geodesic whose midpointis p x ( X ). The tip of the needle with center p x ( z ) is a marked endpoint of the needle andits position is uniformly randomly distributed on the geodesic circle C x ( z ) of diameter L centered at p x ( z ). Denote by I x ( z ) the random variable that is 1 if the needle with midpoint p x ( z ) intersects G ( L ) and 0 otherwise. The geodesic circle C x ( z ) will intersect either onegrating line at two distinct points or two distinct grating lines at one point each. In the casewhen C x ( z ) intersects a geodesic in G ( L ) at two points, the intersection defines two arcs of C x ( z ). Denote by A x ( z ) the arc length of the smaller of the two arcs. If the geodesic circle C x ( z ) intersects two grating lines at exactly one point each, then define A x ( z ) to be 0. If C x ( z ) intersects a grating line in such a way that the intersection divides C x ( z ) into two arcsof equal length, then take A x ( z ) to be half the circumference of C x ( z ). E C x ( z ) p x ( − ℓ ) p x ( ℓ ) xp x ( z ) A x ( z )Figure 1. enote by C ( ℓ ) the circumference of a geodesic circle of radius ℓ . Take the probability P ( I x ( X ) = 1 | X = z ), the probability that the needle intersects G ( L ) given that the centerof the needle is a signed distance of z from the point x , to be uniform in the angle that theneedle makes with the equator. A needle that intersects a grating line will still intersect thesame grating line at the same point when the needle is rotated by half a circle, hence(2.1) P ( I x ( X ) = 1 | X = z ) = 2 A x ( z ) C ( ℓ ) . The Law of Total Probability implies that for each x in E , the probability P ( I x ( X ) = 1) isgiven by(2.2) P ( I x ( X ) = 1) = Z ℓ − ℓ P ( I x ( X ) = 1 | X = z ) · ℓ d z. Consequences of Homogeneity.
Suppose that L is an element of G ( L ) and that L intersects E at a point x . The present section will explicitly verify that the homogeneityof the spaces ensures that the choices of E , G ( L ), and x do not change the probability ofintersection. Lemma 2.5. If g is in H , then P ( I gx ( X ) = 1) = P ( I x ( X ) = 1) . Proof.
Suppose that g is in H and that C x ( z ) intersects at two points a grating line thatintersects x . In this case, the group element g maps the intersections of C x ( z ) with a gratingline that passes through x to the intersections of C gx ( z ) with a grating line that passesthrough gx . Since g is an isometry, it preserves the lengths of arcs and so(2.3) P ( I gx ( X ) = 1 | X = z ) = 2 A gx ( z ) C ( ℓ ) = 2 A x ( z ) C ( ℓ ) = P ( I x ( X ) = 1 | X = z ) . The Law of Total Probability implies that for each x in R , the probability P ( I x ( X ) = 1) isgiven by(2.4) P ( I x ( X ) = 1) = Z ℓ − ℓ P ( I x ( X ) = 1 | X = z ) · ℓ d z. The equalities (2.3) and (2.4) together imply that P ( I gx ( X ) = 1) = Z ℓ − ℓ P ( I gx ( X ) = 1 | X = z ) · ℓ d z = Z ℓ − ℓ P ( I x ( X ) = 1 | X = z ) · ℓ d z = P ( I x ( X ) = 1) . (2.5) (cid:3) Theorem 2.6. If g ′ is in G ′ , then P ( I g ′ x ( X ) = 1) = P ( I x ( X ) = 1) . roof. There is an element h of H and an element of g of G ′ such that α g is in [0 , L ) and hg is equal to g ′ . Given Lemma 2.5, to prove the theorem it suffices to show that P ( I gx ( X ) = 1) = P ( I x ( X ) = 1) . The Law of Total Probability implies that for each x in R , the probability P ( I gx ( X ) = 1) isgiven by P ( I gx ( X ) = 1) = Z ℓ − ℓ P ( I gx ( X ) = 1 | X = z ) · ℓ d z = 1 ℓ Z ℓ − ℓ A gx ( z ) C ( ℓ ) d z = 1 ℓ Z ℓ + α g − ℓ + α g A x ( z ) C ( ℓ ) d z (2.6)since(2.7) A gx ( z ) = A x ( z + α g ) . Split the rightmost integral in (2.6) into an integral over two regions and use the equality(2.7) but with the element h − replacing g to obtain P ( I gx ( X ) = 1) = 1 ℓ Z ℓ − ℓ + α g A x ( z ) C ( ℓ ) d z + 1 ℓ Z ℓ + α g ℓ A x ( z ) C ( ℓ ) d z = 1 ℓ Z ℓ − ℓ + α g A x ( z ) C ( ℓ ) d z + 1 ℓ Z ℓ + α g ℓ A h − x ( z ) C ( ℓ ) d z = 1 ℓ Z ℓ − ℓ + α g A x ( z ) C ( ℓ ) d z + 1 ℓ Z ℓ + α g − Lℓ − L A x ( z ) C ( ℓ ) d z = 1 ℓ Z ℓ − ℓ + α g A x ( z ) C ( ℓ ) d z + 1 ℓ Z − ℓ + α g − ℓ A x ( z ) C ( ℓ ) d z (2.8) = 1 ℓ Z ℓ − ℓ A x ( z ) C ( ℓ ) d z = P ( I x ( X ) = 1) , where Lemma 2.5 implies (2.8). (cid:3) Theorem 2.7.
Suppose that G ( L ) and G ( L ) are two different gratings with common equa-tor E . Denote respectively by P ( I x ( X ) = 1) and P ( I x ( X ) = 1) the probability that a needleof length L intersects G ( L ) and G ( L ) , where the center of the needle is dropped uniformlyrandomly with respect to arc length on the segment of E with midpoint equal to x and lengthequal to L . The probability P ( I x ( X ) = 1) is equal to P ( I x ( X ) = 1) .Proof. Suppose that G ( L ) and G ( L ) are gratings for E . There is an element g of G ′ suchthat G ( L ) is equal to gG ( L ). Since isometries do not change the lengths of arcs, P ( I x ( X ) = 1) = P ( I gx ( X ) = 1) = P ( I x ( X ) = 1) , where Theorem 2.6 implies the rightmost equality. The probability P ( I x ( X ) = 1) is, there-fore, independent of x and independent of the choice of grating. ix an equator E ′ that is different from E or has a different orientation. In each of thethree possibilities for M , there is an element g of G mapping E to E ′ . The probabilities aredependent only on the lengths of arcs associated to intersections of arcs with geodesics andthese lengths are invariant under isometries, implying that the probability of intersection isindependent of the chosen equator. (cid:3) Probabilities for the Plane and the Sphere
The Needle in the Plane.
Rotations about the origin, reflection across the y -axis,and the translations generate G , the isometry group of the plane. The geodesics are thestraight lines and every circle of radius ℓ will have circumference equal to 2 πℓ . Choose theequator E to be the positively oriented x -axis. Denote by G ′ the set of all translations thatfix the x -axis and let H the subgroup of G ′ generated by the vector h L, i , where vectors acton points by componentwise addition. Choose the grating G ( L ) to be the set of lines thatare the translates of the y -axis by the group H . Defining G ( L ) in this way automaticallyverifies Proposition 2.4 in the planar setting. A straightforward exercise shows that anydirected line can be mapped by an isometry to any other line, verifying Proposition 2.3 inthe planar setting. It is convenient to calculate the probability P ( I x ( X ) = 1) by taking x tobe the origin, (0 , A ( z ) the length of the arc A (0 , ( z ) and by I ( X ) the random variable I (0 , ( X ).Since reflection across the y -axis is an isometry, A ( z ) is equal to A ( − z ) and so(3.1) P ( I ( X ) = 1) = 1 ℓ Z ℓ A ( z ) πℓ d z. Take γ to be a parameterization of the arc Γ of C ( z ), where Γ has arc length equal to A ( z )and the endpoints of Γ are the points of intersection of C ( z ) with the y -axis. The curve γ isgiven by γ ( t ) = ( z + ℓ cos( t ) , ℓ sin( t )) with π − cos − (cid:0) zℓ (cid:1) ≤ t ≤ π + cos − (cid:0) zℓ (cid:1) , so that(3.2) A ( z ) = Z π +cos − (cid:0) zℓ (cid:1) π − cos − (cid:0) zℓ (cid:1) ℓ d t = 2 ℓ cos − (cid:0) zℓ (cid:1) . Combine (3.1) and (3.2) to obtain the equality P ( I x ( X ) = 1) = 1 ℓ Z ℓ A ( z ) πℓ d z = 2 πℓ Z ℓ cos − (cid:0) zℓ (cid:1) d z = 2 π Z cos − ( z ) d z = 2 π . (3.3) .2. The Needle in the Sphere.
View S r as the subset S r = { ( x, y, z ) : x + y + z = r } of R . The metric is induced by the standard euclidean metric on R . The distance betweenpoints p and q on S r is the geodesic distance. Fix a length L equal to πrn for some naturalnumber n that is greater than 1. The isometry group, G , of S r is the orthogonal group O (3).The geodesics are the great circles. Choose E , an equator for S r , to be the subset of all pointsof S r lying on the x - y plane endowed with a counterclockwise orientation when viewed fromabove. Denote by G ( L ) the set of all great circles in S r that intersect the equator at rightangles at two points in the set of points S where S = n(cid:0) r cos (cid:0) mLr (cid:1) , r sin (cid:0) mLr (cid:1) , (cid:1) : m ∈ { , . . . , n − } o . Denote by G ′ the set of rotations with a common axis A , where A is the line that passesthrough (0 , , r ) and (0 , , − r ). Denote by H the subgroup of G ′ generated by the rotation g that fixes A , where g ( r, ,
0) = (cid:16) r cos (cid:0) Lr (cid:1) , r sin (cid:0) Lr (cid:1) , (cid:17) . Defining G ( L ) as we have automatically verifies Proposition 2.4 in the S r setting. A straight-forward exercise shows that any directed geodesic can be mapped by an isometry to any othergeodesic, verifying Proposition 2.3 in the S r setting. It is convenient to calculate the proba-bility P ( I x ( X ) = 1) by taking x to be the point ( r, , w to be in [ − ℓ, ℓ ]. Denote by A ( w ) the length of the arc A ( r, , ( w ) and by I ( X ) the random variable I ( r, , ( X ). Recallthat every geodesic circle of radius ℓ on S r has circumference equal to 2 πr sin (cid:0) ℓr (cid:1) . Proposition 3.1. If P ( I ( X ) = 1) is the probability of intersection, then (3.4) P ( I ( X ) = 1) = 1 − πℓ Z ℓ sin − (cid:0) tan (cid:0) wr (cid:1) cot( ℓr ) (cid:1) d w. Proof.
Denote by C ℓ the circle C ℓ = n(cid:0) r cos( θ ) sin (cid:0) ℓr (cid:1) , r sin( θ ) sin (cid:0) ℓr (cid:1) , r cos (cid:0) ℓr (cid:1)(cid:1) : 0 ≤ θ ≤ π o . The circle C ℓ is a geodesic circle of radius ℓ centered at (0 , , r ). The rotation f w , given inthe standard basis by the matrix f w = − sin (cid:0) wr (cid:1) cos (cid:0) wr (cid:1) (cid:0) wr (cid:1) sin (cid:0) wr (cid:1) − , rotates C ℓ to the circle of radius ℓ centered at (cid:0) r cos (cid:0) wr (cid:1) , r sin (cid:0) wr (cid:1) , (cid:1) , the set f w ( C ℓ ) = n(cid:0) − r sin (cid:0) wr (cid:1) sin (cid:0) θ (cid:1) sin (cid:0) ℓr (cid:1) + r cos (cid:0) wr (cid:1) cos (cid:0) ℓr (cid:1) , (3.5) r cos (cid:0) wr (cid:1) sin( θ ) sin (cid:0) ℓr (cid:1) + r sin (cid:0) wr (cid:1) cos (cid:0) ℓr (cid:1) , − r cos( θ ) sin (cid:0) ℓr (cid:1)(cid:1) : 0 ≤ θ ≤ π o . he probability P ( I ( X ) = 1 | X = w ) is given by(3.6) P ( I ( X ) = 1 | X = w ) = A ( w ) πr sin (cid:0) ℓr (cid:1) . The Law of Total Probability implies that(3.7) P ( I ( X ) = 1) = Z ℓ − ℓ P ( I ( X ) = 1 | X = w ) · ℓ d w. Since the reflection across the x – z plane is an isometry, A ( w ) is equal to A ( − w ) and so theintegrand in the right hand side of (3.7) is an even function, implying that(3.8) P ( I ( X ) = 1) = Z ℓ A ( w ) πr sin (cid:0) ℓr (cid:1) · ℓ d w. Compute the arc length A ( w ) for any w in [0 , ℓ ] in order to calculate the probabilitiesgiven by (III.5). The great circle of G ( L ) that intersects ( r, ,
0) is the longitude G , where G = (cid:8) ( r sin( φ ) , , r cos( φ )) : 0 ≤ φ ≤ π (cid:9) . The circle of radius ℓ centered at the point p ( w ) is the circle f w ( C ℓ ), which intersects G when − r sin (cid:0) wr (cid:1) sin( θ ) sin (cid:0) ℓr (cid:1) + r cos (cid:0) wr (cid:1) cos (cid:0) ℓr (cid:1) = r sin( φ ) r cos (cid:0) wr (cid:1) sin( θ ) sin (cid:0) ℓr (cid:1) + r sin (cid:0) wr (cid:1) cos (cid:0) ℓr (cid:1) = 0 − r cos( θ ) sin (cid:0) ℓr (cid:1) = r cos( φ ) . The points of intersection are therefore given by,sin (cid:0) θ (cid:1) = − sin (cid:0) wr (cid:1) cos (cid:0) ℓr (cid:1) cos (cid:0) wr (cid:1) sin (cid:0) ℓr (cid:1) = − tan (cid:0) wr (cid:1) cot (cid:0) ℓr (cid:1) . Solve for θ to obtain the equality θ = − sin − (cid:0) tan (cid:0) wr (cid:1) cot (cid:0) ℓr (cid:1)(cid:1) and so(3.9) A ( w ) = πr sin (cid:0) ℓr (cid:1) − r sin (cid:0) ℓr (cid:1) sin − (cid:0) tan (cid:0) zr (cid:1) cot (cid:0) ℓr (cid:1)(cid:1) . Equations (3.8) and (3.9) together imply that P ( I ( X ) = 1) = 1 ℓ Z ℓ A ( w ) πr sin (cid:0) ℓr (cid:1) d w = 1 πrℓ sin (cid:0) ℓr (cid:1) Z ℓ A ( w ) d w = 1 πrℓ sin (cid:0) ℓr (cid:1) Z ℓ (cid:16) πr sin (cid:0) ℓr (cid:1) − r sin (cid:0) ℓr (cid:1) sin − (cid:0) tan (cid:0) wr (cid:1) cot (cid:0) ℓr (cid:1)(cid:1)(cid:17) d w and so P ( I ( X ) = 1) = 1 − πℓ Z ℓ sin − (cid:0) tan (cid:0) wr (cid:1) cot (cid:0) ℓr (cid:1)(cid:1) d w. (cid:3) . The Needle in the Poincar´e Disk
The Geometry of the Poincar´e Disk.
View H , the Poincar´e disk, as the open unitdisk in R endowed with the Riemannian metric d s , where(4.1) d s = 4(1 − ( x + y )) (d x ⊗ d x + d y ⊗ d y ) . The distance between points in H is the geodesic distance. Choose E , an equator for H , tobe the set E = { ( t,
0) : − < t < } with a left to right orientation. For any point ( x,
0) on E , if h is the signed hyperbolicdistance from (0 ,
0) to ( x, h is negative if x is negative and positive if x ispositive, then x is given by x = tanh (cid:0) h (cid:1) . The geodesic circles of H are also circles of R , but their euclidean radii and euclidean centersmay be different. In particular, a geodesic circle in H whose center is on E a signed hyperbolicdistance of h from the origin and that has a hyperbolic radius equal to λ has a euclideancenter ( x,
0) and euclidean radius r , where x = (cid:16) tanh (cid:0) h + λ (cid:1) + tanh (cid:0) h − λ (cid:1)(cid:17) and r = (cid:16) tanh (cid:0) h + λ (cid:1) − tanh (cid:0) h − λ (cid:1)(cid:17) . The signed hyperbolic distance from (0 ,
0) to ( x,
0) is h , where h = 2 tanh − ( x ) . The geodesics of H are arcs of generalized circles of the euclidean plane that intersect theunit circle at right angles.4.2. Symmetries and a Grating on the Poincar´e Disk.
Identify the open unit disk in R with the open unit disk in the complex plane. The group of M¨obius transformations arethe transformations of C of the form z az + bcz + d where a, b, c, d ∈ R and ad − bc = 0 . Denote by G the isometry group of H , the subgroup of M¨obius transformations that mapthe open unit ball to itself and consist of the transformations of the form z e iφ z + bbz + 1 with ( φ, b ) ∈ R × ( − , . The group G acts transitively on H . The subgroup G ′ of G that maps E to itself and preservesthe orientation of E is the group of transformations of the form g τ where g τ ( z ) = z + ττ z + 1 with τ ∈ ( − , . This group is the analog in the hyperbolic setting of the group of translations along the x -axis in R and of the rotations around the z -axis in the S r setting. View H once again and enceforth as a subset of R and view the elements of G as acting on R . The function g τ acts on an arbitrary point ( x, y ) in H by g τ ( x, y ) = (cid:18) τ ( x + y ) + ( τ + 1) x + τ ( τ x + 1) + τ y , (1 − τ ) y ( τ x + 1) + τ y (cid:19) , and so it acts on a point (cid:0) tanh (cid:0) h (cid:1) , (cid:1) of E by g τ (cid:0) tanh (cid:0) h (cid:1) , (cid:1) = (cid:16) tanh (cid:16) h +2 tanh − ( τ )2 (cid:17) , (cid:17) . Define for each σ in R the function F σ by F σ ( x, y ) = g tanh ( σ )( x, y )so that F σ is a function that maps E to E , preserves the orientation of E , and moves pointson E a signed hyperbolic distance of σ along E .Take G to be the bounded open line segment in R given by G = { (0 , y ) : − < y < } . The set G is an unbounded clopen geodesic in H . Define for each σ in R the geodesic G σ by G σ = F σ ( G ) . The geodesic G σ is an arc of a (generalized if σ is 0) circle in R that intersects E at a rightangle at (cid:0) − ( σ ) , (cid:1) , a point that is a signed distance of σ from (0 , σ is positive. Viewed as a subset of R with the euclidean metric, the center of G σ lies on the x -axis. A straightforward calculation reveals that the closure of G σ intersects the unit diskat right angles at the points ( a ( σ ) , ± p − ( a ( σ )) ) and that the center of G σ is ( b ( σ ) , a ( σ ) = σσ + 1 and b ( σ ) = σ + 1 σ . With a ( σ ) and b ( σ ) given above and r ( σ ) defined by r ( σ ) = b ( σ ) − σ = 1 σ , the arc G α has the parameterization G σ = n ( b ( σ )+ r ( σ ) cos( t ) , r ( σ ) sin( t )) : π − tan − (cid:16) √ − ( a ( σ )) b ( σ ) − a ( σ ) (cid:17) < t < π +tan − (cid:16) √ − ( a ( σ )) b ( σ ) − a ( σ ) (cid:17)o . If σ is negative, then G σ is the reflection of G − σ across the y -axis. Figure 2 displays ageodesic G σ in H as well as the circle in R that gives rise to the arc that is the geodesic G σ , when viewed as a subset of R . Take x ( σ ) in Figure 2 to be equal to tanh (cid:0) σ (cid:1) , so thatthe point ( x ( σ ) ,
0) is a distance of σ from (0 , , (cid:0) a ( σ ) , p − ( a ( σ ) (cid:1) ( x ( σ ) ,
0) ( b ( σ ) , G σ G − σ Figure 2.Fix a positive length L and denote by G ( L ) the grating G ( L ) = { G nL : n ∈ Z } . Defining G ( L ) as we have automatically verifies Proposition 2.4 in the H setting. A straight-forward exercise shows that any directed geodesic can be mapped by an isometry to anyother geodesic, verifying Proposition 2.3 in the H setting.4.3. The Probability of Intersection.
To simplify notation, denote for each z in R by A ( z ) the length of the arc A (0 , ( z ) and by I ( X ) the random variable I (0 , ( X ). Theorem 4.1.
For any ( x, in E , (4.2) P ( I ( x, ( X ) = 1) = 2 − πℓ sinh( ℓ ) Z ℓ (cid:0) tanh (cid:0) z + ℓ (cid:1) − tanh (cid:0) z − ℓ (cid:1)(cid:1)q(cid:0) − tanh (cid:0) z + ℓ (cid:1)(cid:1) (cid:0) − tanh (cid:0) z − ℓ (cid:1)(cid:1) · tan − s − tanh (cid:0) z − ℓ (cid:1) − tanh (cid:0) z + ℓ (cid:1) · s tanh (cid:0) ℓ + z (cid:1) tanh (cid:0) ℓ − z (cid:1) !! d z. Proof.
Theorem 2.6 implies that for any x in ( − , P ( I ( x, ( X ) = 1) isindependent of x . It is convenient to calculate this probability by taking x to be zero. Recallthat every geodesic of radius ℓ on H has circumference equal to 2 πr sinh( ℓ ). The probability P ( I ( X ) = 1 | X = z ) that the needle with hyperbolic center p ( z ) on E intersects G is givenby(4.3) P ( I ( X ) = 1 | X = z ) = A ( z ) π sinh( ℓ ) , and so the Law of Total Probability implies P ( I ( X ) = 1) = Z ℓ − ℓ P ( I ( X ) = 1 | X = z ) · ℓ d z Z ℓ − ℓ A ( z ) π sinh ( ℓ ) · ℓ d z. (4.4)Since reflection across the y -axis is an isometry, A ( z ) is equal to A ( − z ) and so the integrandin (4.4) is an even function, implying that(4.5) P ( I ( X ) = 1) = Z ℓ A ( z ) π sinh( ℓ ) · ℓ d z. Denote by x ( z ) and r ( z ) the real values with the property that the euclidean center of C ( z ) is ( x ( z ) ,
0) and the euclidean radius of C ( z ) is r ( z ). The circle C ( z ) intersects G atthe points (0 , ± y ), where y = p r ( z ) − x ( z ) . Let R ( z ) be the ray from ( x ( z ) ,
0) to the point of intersection of C ( z ) and G in the upperhalf real plane. Define by θ ( z ) the angle formed by R ( z ) and the ray lying along the x -axiswith its tail (or origin) at ( x ( z ) , x -axis is an isometry of H together imply that the hyperbolic arc length A ( z ) is given by A ( z ) = Z A (0 , ( z ) d s = 2 Z πθ ( z ) kh− r ( z ) sin( t ) , r ( z ) cos( t ) ik (1 − ( x ( z ) + r ( z ) cos( t ))) − ( r ( z ) sin( t )) d t = 4 r ( z ) Z πθ ( z ) d t (1 − ( x ( z ) + r ( z ) cos( t ))) − ( r ( z ) sin( t )) . On taking α ( z ) = 1 − x ( z ) − r ( z ) and β ( z ) = 2 r ( z ) x ( z ) , the equality Z d tα ( z ) − β ( z ) cos( t ) = 2 p α ( z ) − β ( z ) tan − (cid:18)q α ( z )+ β ( z ) α ( z ) − β ( z ) · tan (cid:0) t (cid:1)(cid:19) + C implies that A ( z ) = 8 r ( z ) p α ( z ) − β ( z ) tan − (cid:18)q α ( z )+ β ( z ) α ( z ) − β ( z ) · tan (cid:0) t (cid:1)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) πθ ( z ) . Since the integral describing the arc length gives the arc length of the top half of a geodesiccircle when the lower bound is zero, A ( z ) = 2 π sinh( ℓ ) − r ( z ) p α ( z ) − β ( z ) tan − (cid:18)q α ( z )+ β ( z ) α ( z ) − β ( z ) · tan (cid:0) t (cid:1)(cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) πθ ( z ) π sinh( ℓ ) − r ( z ) p α ( z ) − β ( z ) tan − (cid:18)q α ( z )+ β ( z ) α ( z ) − β ( z ) · tan (cid:16) θ ( z )2 (cid:17)(cid:19)! . The equalities α ( z ) − β ( z ) = 1 − ( x ( z ) + r ( z )) , α ( z ) + β ( z ) = 1 − ( x ( z ) − r ( z )) and α ( z ) − β ( z ) = ( α ( z ) + β ( z ))( α ( z ) − β ( z )) = (cid:0) − ( x ( z ) + r ( z )) (cid:1) (cid:0) − ( x ( z ) − r ( z )) (cid:1) together with the equalities x ( z ) − r ( z ) = tanh (cid:0) z − ℓ (cid:1) and x ( z ) + r ( z ) = tanh (cid:0) z + ℓ (cid:1) imply that(4.6) A ( z ) = 2 π sinh ( ℓ ) − (cid:0) tanh (cid:0) z + ℓ (cid:1) − tanh (cid:0) z − ℓ (cid:1)(cid:1)q(cid:0) − tanh (cid:0) z + ℓ (cid:1)(cid:1) (cid:0) − tanh (cid:0) z − ℓ (cid:1)(cid:1) · tan − s − tanh ( z − ℓ ) − tanh (cid:18) z + ℓ (cid:19) · tan (cid:16) θ ( z )2 (cid:17) . Use the equality tan (cid:16) θ ( z )2 (cid:17) = − cos( θ ( z ))sin( θ ( z )) to show after some simplification that A ( z ) = 2 π sinh ( ℓ ) − (cid:0) tanh (cid:0) z + ℓ (cid:1) − tanh (cid:0) z − ℓ (cid:1)(cid:1)q(cid:0) − tanh (cid:0) z + ℓ (cid:1)(cid:1) (cid:0) − tanh (cid:0) z − ℓ (cid:1)(cid:1) · tan − s − tanh (cid:0) z − ℓ (cid:1) − tanh (cid:0) z + ℓ (cid:1) · s tanh (cid:0) ℓ + z (cid:1) tanh (cid:0) ℓ − z (cid:1) !! . Since P ( I ( X ) = 1 | X = z ) = A ( z ) π sinh( ℓ )= 2 − π sinh ( ℓ ) (cid:0) tanh (cid:0) z + ℓ (cid:1) − tanh (cid:0) z − ℓ (cid:1)(cid:1)q(cid:0) − tanh (cid:0) z + ℓ (cid:1)(cid:1) (cid:0) − tanh (cid:0) z − ℓ (cid:1)(cid:1) · tan − s − tanh (cid:0) z − ℓ (cid:1) − tanh (cid:0) z + ℓ (cid:1) · s tanh (cid:0) ℓ + z (cid:1) tanh (cid:0) ℓ − z (cid:1) !!! , he equalities P ( I ( X ) = 1) = Z ℓ − ℓ P ( I ( X ) = 1 | X = z ) · ℓ d z = Z ℓ P ( I ( X ) = 1 | X = z ) · ℓ d z. imply the equality given in the statement of the theorem. (cid:3) Limiting Behavior and Probability Deficits
This section computes series expansions for the probabilities given by (3.4) and (4.2) thatdescribe the probabilities up to zeroth, first, and second order in the parameter ℓ . Zerothorder terms give the probability in the planar case, first order terms vanish, and second orderterms depend on the Gaussian curvature of the space. Probability deficits determine theGaussian curvature of the space in a way that is very similar to the way in which the lengthand area deficits of the Bertrand-Diguet-Puiseux Theorem determine Gaussian curvature.The following subsections utilize the standard “little-o” notation to simplify expressions. Let M be either R , S r , or H and in each case denote by I ( X ) the probability of intersectionfirst defined by (2.2) in Section 2. Note that (3.3) already implies that if M is R , then I ( X ) = 2 π and there is no dependency on ℓ .5.1. Order Estimates for Spheres.Proposition 5.1. If M is S r , then P ( I ( X ) = 1) = 2 π + 49 πr ℓ + o ( ℓ ) . Proof.
Change variables to rewrite (3.4) as(5.1) P ( I ( X ) = 1) = 1 − π Z sin − (cid:0) tan (cid:0) ℓzr (cid:1) cot (cid:0) ℓr (cid:1)(cid:1) d z. The series expansionstan( x ) = x + x o ( x ) and cot( x ) = 1 x − x − x
45 + o ( x )imply that(5.2) P ( I ( X ) = 1) = 1 − π Z sin − (cid:0) z + ℓ (cid:0) z r − z r (cid:1) + o ( ℓ ) (cid:1) d z. Define F by(5.3) F ( ℓ ) = Z sin − (cid:0) z + ℓ (cid:0) z r − z r (cid:1) + o ( ℓ ) (cid:1) d z o that(5.4) F ′ ( ℓ ) = Z ℓ (cid:0) z r − z r (cid:1) + o ( ℓ ) q − (cid:0) z + ℓ (cid:0) z r − z r (cid:1) + o ( ℓ ) (cid:1) d z and(5.5) F ′′ ( ℓ ) = Z (cid:0) z r − z r (cid:1)q − (cid:0) z + ℓ (cid:0) z r − z r (cid:1) + o ( ℓ ) (cid:1) d z + o ( ℓ ) . Equations (5.3), (5.4), and (5.5) respectively imply that(5.6) F (0) = Z sin − ( z ) d z = π − , (5.7) F ′ (0) = 0 , and(5.8) F ′′ (0) = − r Z z − z √ − z d z = − r Z z √ − z d z = − r . Equalities (5.6), (5.7), and (5.8) together imply that(5.9) F ( ℓ ) = π − ℓ r + o ( ℓ ) . Combine (5.2), (5.6), and (5.9) to obtain the equality(5.10) P ( I ( X ) = 1) = 2 π + 49 πr ℓ + o ( ℓ ) . (cid:3) Order Estimates for the Poincar´e Disk.Proposition 5.2. If M is H , then P ( I ( X ) = 1) = 2 (cid:18) − ℓπ sinh( ℓ ) (cid:18) ( π −
1) + ℓ (cid:18) π + 13 (cid:19) + o ( ℓ ) (cid:19)(cid:19) . Proof.
Change variables to rewrite (4.2) as(5.11) P ( I ( X ) = 1) = 2 − π sinh( ℓ ) Z (cid:0) tanh (cid:0) ℓ · z +12 (cid:1) − tanh (cid:0) ℓ · z − (cid:1)(cid:1)q(cid:0) − tanh (cid:0) ℓ · z +12 (cid:1)(cid:1) (cid:0) − tanh (cid:0) ℓ · z − (cid:1)(cid:1) · tan − s − tanh (cid:0) ℓ · z − (cid:1) − tanh (cid:0) ℓ · z +12 (cid:1) · s tanh (cid:0) ℓ · z (cid:1) tanh (cid:0) ℓ · − z (cid:1) !! d z. se the series expansions for tan( x ), √ − x , and − x to obtain the equalities(5.12) tanh (cid:18) ℓ · z + 12 (cid:19) − tanh (cid:18) ℓ · z − (cid:19) = ℓ (cid:18) − ℓ
12 (3 z + 1) (cid:19) + o ( ℓ ) , (5.13) 1 q(cid:0) − tanh (cid:0) ℓ · z +12 (cid:1)(cid:1) (cid:0) − tanh (cid:0) ℓ · z − (cid:1)(cid:1) = 1 + ℓ z + 1) + o ( ℓ ) , (5.14) s − tanh (cid:0) ℓ · z − (cid:1) − tanh (cid:0) ℓ · z +12 (cid:1) = 1 + ℓ z o ( ℓ ) , and(5.15) tanh (cid:0) ℓ · z (cid:1) tanh (cid:0) ℓ · − z (cid:1) = z + 11 − z (cid:18) − ℓ z o ( ℓ ) (cid:19) . Equations (5.12) and (5.13) together imply that(5.16) tanh (cid:0) ℓ · z +12 (cid:1) − tanh (cid:0) ℓ · z − (cid:1)q(cid:0) − tanh (cid:0) ℓ · z +12 (cid:1)(cid:1) (cid:0) − tanh (cid:0) ℓ · z − (cid:1)(cid:1) = ℓ (cid:18) ℓ (cid:19) + o ( ℓ ) . Equations (5.14) and (5.15) together imply that(5.17) s − tanh (cid:0) ℓ · z − (cid:1) − tanh (cid:0) ℓ · z +12 (cid:1) · s tanh (cid:0) ℓ · z (cid:1) tanh (cid:0) ℓ · − z (cid:1) = (cid:18) ℓ z o ( ℓ ) (cid:19) s z + 11 − z (cid:18) − ℓ z o ( ℓ ) (cid:19) . Define the function H by(5.18) H ( ℓ ) = 1 ℓ Z tanh (cid:0) ℓ · z +12 (cid:1) − tanh (cid:0) ℓ · z − (cid:1)q(cid:0) − tanh (cid:0) ℓ · z +12 (cid:1)(cid:1) (cid:0) − tanh (cid:0) ℓ · z − (cid:1)(cid:1) · tan − s − tanh (cid:0) ℓ · z − (cid:1) − tanh (cid:0) ℓ · z +12 (cid:1) · s tanh (cid:0) ℓ · z (cid:1) tanh (cid:0) ℓ · − z (cid:1) ! d z. Combine (5.16) and (5.17) to obtain the equality(5.19) H ( ℓ ) = Z (cid:18) ℓ o ( ℓ ) (cid:19) · tan − (cid:18) ℓ z o ( ℓ ) (cid:19) s z + 11 − z (cid:18) − ℓ z o ( ℓ ) (cid:19)! d z. implify (5.11) using the expression for H ( ℓ ) to obtain the equality(5.20) P ( I ( X ) = 1) = 2 (cid:18) − ℓπ sinh( ℓ ) H ( ℓ ) (cid:19) . Define by F the function(5.21) F ( ℓ ) = Z tan − (cid:18) ℓ z o ( ℓ ) (cid:19) s z + 11 − z (cid:18) − ℓ z o ( ℓ ) (cid:19)! d z. Since s z + 11 − z (cid:18) − ℓ z o ( ℓ ) (cid:19) = (cid:18) − ℓ z o ( ℓ ) (cid:19) r z + 11 − z , (5.22) F ( ℓ ) = Z tan − (cid:18) ℓ z o ( ℓ ) (cid:19) r z + 11 − z ! d z. Take derivatives with respect to ℓ to see that(5.23) F ′ ( ℓ ) = 2 ℓ Z z q z +11 − z ℓ z ) (cid:0) z +11 − z (cid:1) d z + o ( ℓ )and(5.24) F ′′ ( ℓ ) = 23 Z z q z +11 − z ℓ z ) (cid:0) z +11 − z (cid:1) d z + 2 ℓ Z dd ℓ z q z +11 − z ℓ z ) (cid:0) z +11 − z (cid:1) d z + o ( ℓ ) . Equations (5.22), (5.23), and (5.24) respectively imply that F (0) = Z tan − r z + 11 − z ! d z = lim c → − √ − x + x tan − r z + 11 − z !!(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) c = −
12 + π , (5.25)(5.26) F ′ (0) = 0 , and F ′′ (0) = 23 Z z q z +11 − z (cid:0) z +11 − z (cid:1) d z = 13 Z z √ − z d z = 19 . (5.27)Equalities (5.25), (5.26), and (5.27) together imply that(5.28) F ( ℓ ) = −
12 + π ℓ + o ( ℓ ) , nd so H ( ℓ ) = (cid:18) ℓ o ( ℓ ) (cid:19) F ( ℓ )= (cid:18) ℓ o ( ℓ ) (cid:19) (cid:18) −
12 + π ℓ + o ( ℓ ) (cid:19) = (cid:18) −
12 + π (cid:19) + ℓ (cid:18) π
12 + 136 (cid:19) + o ( ℓ ) . (5.29)Equalities (5.20) and (5.29) together imply that(5.30) P ( I ( X ) = 1) = 2 (cid:18) − ℓπ sinh( ℓ ) (cid:18)(cid:18) −
12 + π (cid:19) + ℓ (cid:18) π
12 + 136 (cid:19) + o ( ℓ ) (cid:19)(cid:19) , verifying the proposition. (cid:3) Probability Deficits and Gaussian Curvature.
In the setting of the sphere, theprobability of intersection given by Proposition 5.1 implies thatlim ℓ → + P ( I ( X ) = 1) = lim ℓ → + π + 49 πr ℓ + o ( ℓ ) = 2 π . (5.31)Similarly, in the setting of the Poincar´e disk, the probability of intersection given by Propo-sition 5.2 together with the fact that lim ℓ → + sinh( ℓ ) ℓ = 1implies that(5.32) lim ℓ → + P ( I ( X ) = 1)= lim ℓ → + (cid:18) − ℓπ sinh( ℓ ) (cid:18)(cid:18) −
12 + π (cid:19) + ℓ (cid:18) π
12 + 136 (cid:19) + o ( ℓ ) (cid:19)(cid:19) = 2 π . The probability of intersection in the planar case is π . Theorem 5.3 summarize these calcu-lations. Theorem 5.3. If M is R , S r , or H , then lim ℓ → + P ( I ( X ) = 1) = 2 π . Definition 5.4.
The probability deficit of M is the difference P ( I ( X ) = 1) − π and is a function of the parameter ℓ .Denote by κ ( M ) the Gaussian curvature of M . Similar to the way in which circumferenceand area deficits determine through a limiting procedure the Gaussian curvature of a surface,the probability deficits determine the Gaussian curvature of M . heorem 5.5. If M is R , S r , or H , then lim ℓ → + π P ( I ( X ) = 1) − π ℓ = κ ( M ) . Proof.
In the planar setting, the result is immediate because the Gaussian curvature of theplane is 0. To prove the theorem in the setting of spheres, recall first that the Gaussiancurvature of the sphere is the square reciprocal of the sphere’s radius. Use Proposition 5.1to obtain the equalitieslim ℓ → + π P ( I ( X ) = 1) − π ℓ = lim ℓ → + π π + πr ℓ + o ( ℓ ) − π ℓ = lim ℓ → + π πr ℓ + o ( ℓ ) ℓ = 1 r . To prove the theorem in the Poincar´e disk setting, recall first that the Gaussian curvatureof the Poincar´e disk is −
1. Use Proposition 5.2 to obtain the equalitieslim ℓ → + π P ( I ( X ) = 1) − π ℓ = lim ℓ → + π (cid:16) − ℓπ sinh( ℓ ) (cid:0)(cid:0) − + π (cid:1) + ℓ (cid:0) π + (cid:1) + o ( ℓ ) (cid:1)(cid:17) − π ℓ = lim ℓ → + π − ℓπ sinh( ℓ ) (cid:18) π
12 + 136 (cid:19) + 2 (cid:16) − ℓπ sinh( ℓ ) (cid:0) − + π (cid:1)(cid:17) − π ℓ = 9 π − lim ℓ → + ℓπ sinh( ℓ ) (cid:18) π
12 + 136 (cid:19) + lim ℓ → + (cid:16) − ℓπ sinh( ℓ ) (cid:0) − + π (cid:1)(cid:17) − π ℓ = 9 π − − π + lim ℓ → + (cid:16) − ℓπ sinh( ℓ ) (cid:0) − + π (cid:1)(cid:17) − π ℓ . (5.33)Use Bernoulli’s rule for indeterminate forms to obtainlim ℓ → + (cid:16) − ℓπ sinh( ℓ ) (cid:0) − + π (cid:1)(cid:17) − π ℓ = lim ℓ → + − dd ℓ (cid:16) ℓπ sinh( ℓ ) (cid:0) − + π (cid:1)(cid:17) ℓ = 13 (cid:18) − π (cid:19) . (5.34)Substitute (5.34) into the limit in (5.33) to obtain the limitlim ℓ → + π P ( I ( X ) = 1) − π ℓ = 9 π (cid:26) − − π + 13 (cid:18) − π (cid:19)(cid:27) = − . (cid:3) o obtain the Gaussian curvature from circumference and area deficits requires a limit ofthe quotient of the deficit by resepctively the cube and the fourth power of the radius ofthe circle and these are the quotients that appear in the Bertrand-Diguet-Puiseux Theorem.The determination of the Gaussian curvature in the settings we have considered involves aquotient of the probability deficit by the square of the needle length. This indicates that thecurrent probabilistic method for determining Gaussian curvature is different from the priormethods, although arc length determines the probability of intersection on conditioning theneedle to fall at a particular location. It should be possible to extend this method to moregeneral surfaces, although the lack of homogeneity of a more general surface complicates thecalculations involved in determining the probabilities. References [1] Barbier, E.:
Note sur le problme de l’aiguille et le jeu du joint couvert . Journal de Math´ematiques Pureset Appliqu´ees, 2e s´erie, (1860) 273–286. (Referred to on page 2.)[2] Bertrand, J., Diguet, C. F., Puiseux, V.: D´emonstration d’un th´eor`eme de Gauss . Journal deMath´ematiques, 13: 80–90, (1848). (Referred to on page 3.)[3] Buffon, G.: Editor’s note concerning a lecture given 1733 by Mr. Le Clerc de Buffon to the Royal Academyof Sciences in Paris. Histoire de l’Acad. Roy. des Sci., (1733) 43–45. (Referred to on page 2.)[4] Buffon, G.:
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