Suppose you pick three points at random in a disk in the Euclidean plane, and draw the unique circle that contains all three. The points are picked uniformly at random, which means that the probability that each of them lies in any given region is proportional to the area of that region, and independently, which means the probability for each point being in some region is completely unaffected by the positions of the other points. It’s possible that all three points will lie on a straight line, or that two or three of them will be identical, but all of these things happen with a probability of zero.
The average radius of the circle obtained this way turns out to be infinite, in the sense that the probability distribution for the radius does not have a finite mean value. However, the average curvature, κ, which is the reciprocal of the radius, has a finite value:
Average curvature of the circle
through three random points in a
Euclidean disk of radius ρ
κEuc(ρ) = 16 (3 – 16/π2) / (15 ρ) ≈ 1.47079 / ρ
In the following sections we will derive this formula, and some related results for points on a sphere.
Our goal in this section is to find the average curvature of the circle that contains three points chosen at random from a disk of radius ρ in the Euclidean plane.
Given the coordinates of the three points, it is not hard to write down a formula for the radius R of the circle that contains them. As a geometric starting point, R is equal to the product of the three side lengths of the triangle with the chosen points as its vertices, divided by four times the area of that triangle. Then the curvature, κ, is just the reciprocal, 1/R.
However, integrating κ over the six coordinate variables for the three points lying in the disk is very unwieldy, whether we use Cartesian or polar coordinates. An easier method is to use a set of variables in which the radius of the circle in question, R, is itself one of the variables. A convenient choice for the other five variables is then:
In the figure on the right, the disk of radius ρ is shown in light blue, while the circle of radius R is shown in black. All six variables are marked here, except for θ, which determines the angular position of the centre of the circle around the disk, and has no effect on the quantities we want to measure.
To find the average value of κ = 1/R, we need to integrate it over the region(s) in our new coordinate system that correspond to configurations where all three points lie inside the disk. Since we are choosing the points uniformly with respect to areas in the Euclidean plane, we need to weight our integral in the new coordinates with the absolute value of the Jacobian determinant (also known as the “change of variables factor”) which corresponds to the product of the three infinitesimal areas swept out in the plane as we make small changes to the six new variables.
The Cartesian coordinates of the three points in terms of the new variables are:
xi = r cos θ + R cos φi
yi = r sin θ + R sin φi
The determinant of the 6×6 matrix of partial derivatives of the six Cartesian coordinates with respect to the six new coordinates is:
JEuc = R3 r (sin(φ1 – φ2) + sin(φ2 – φ3) + sin(φ3 – φ1))
Suppose the three points are known to lie on an arc of the circle of radius R such that:
–a ≤ φi ≤ a
JEuc will change sign if we swap any two of the φi coordinates, so we can integrate the absolute value of JEuc over these coordinates by integrating over a region where the coordinates are ordered in a way that makes JEuc positive, then multiplying by six, the number of ways of permuting the three coordinates.
IEuc(R, r, a) = ∫–a ≤ φi ≤ a |JEuc| dφ1 dφ2 dφ3
= 6 ∫–a ≤ φ2 ≤ φ1 ≤ φ3 ≤ a JEuc dφ1 dφ2 dφ3
= 6 ∫–a a ∫φ2 a ∫φ1 a JEuc dφ3 dφ1 dφ2
= 6 R3 r ∫–a a ∫φ2 a (sin(φ1 – φ2)(a – φ1) + cos(φ2 – a) – cos(φ2 – φ1) – cos(a – φ1) + 1) dφ1 dφ2
= 6 R3 r ∫–a a (2 a – 2 φ2 – 3 sin(a – φ2) + (a – φ2) cos(a – φ2)) dφ2
= 24 R3 r A(a)
where A(a) = a2 + a sin(a) cos(a) – 2 sin(a)2
When the circle intersects the boundary of the disk, as in the figure above, the value of a can be found by the cosine law of triangles:
cos(a) = (R2 + r2 – ρ2) / (2 R r)
When 0 ≤ r ≤ ρ – R, the circle is entirely inside the disk, and we have a = π.
Region 1
Circle smaller than disk,
entirely inside it
R < ρ
0 ≤ r ≤ ρ – R
a = π
IEuc(R, r, π) = 24 π2 R3 r
In that case we must have R < ρ for the range of r values not to be empty, but there will also be cases where R < ρ and an incomplete (but nonempty) arc of the circle lies inside the disk:
Region 2
Circle smaller than disk,
partially inside it
R < ρ
ρ – R ≤ r ≤ ρ + R
0 ≤ a ≤ π
The upper bound here on r specifies that the circle is not so far away that it never reaches the disk, while the lower bound specifies that we don’t have the previous case, where the circle lies entirely inside the disk.
We also have a third possibility:
Region 3
Circle larger than disk,
partially inside it
R > ρ
R – ρ ≤ r ≤ ρ + R
0 ≤ a ≤ π/2
Here the upper bound on r serves the same purpose as before, while the lower bound specifies that the disk doesn’t lie entirely inside the circle.
For each of these three regions, we will integrate three things, weighted by the change of variables factor |JEuc|:
It’s straightforward to integrate over the first region, 0 ≤ R ≤ ρ and 0 ≤ r ≤ ρ – R. We multiply by an extra factor of 2π to integrate over the polar coordinate θ for the centre of the circle.
IEuc 1(1) = 2π ∫0ρ ∫0ρ – R 24 π2 R3 r dr dR
= (2/5) (π ρ2)3
IEuc 1(R) = 2π ∫0ρ ∫0ρ – R 24 π2 R4 r dr dR
= (8ρ/35) (π ρ2)3
IEuc 1(κ) = 2π ∫0ρ ∫0ρ – R 24 π2 R2 r dr dR
= (4/(5ρ)) (π ρ2)3
For the second region, where R < ρ and ρ – R ≤ r ≤ ρ + R, the angle a decreases monotonically from π when r = ρ – R down to zero when r = ρ + R. It will be simpler to change variables from r to a rather than substituting for a as a function of R and r.
We can rearrange the cosine law into a quadratic in r:
r2 – 2 R r cos(a) + (R2 – ρ2) = 0
which is solved by:
r = R cos(a) ± √[ρ2 – R2 sin(a)2]
To change the integration variable from r to a we need:
r dr/da = d(½r2)/da = dS±(R, a)/da
where S±(R, a) = ½(r2 – ρ2) = ½R2 (cos(a)2 – sin(a)2) ± R cos(a) √[ρ2 – R2 sin(a)2]
When R < ρ we take the + sign in the solution. Rather than compute the derivative of S(R, a) we will integrate by parts with respect to a, but first we will integrate with respect to R:
S1(a) = ∫0ρ R3 S+(R, a) dR = (ρ6 / 192) (16 cos(2a) + |cos(a)| (12 (a – πa > π/2) – 3 sin(2a) – 3 sin(4a) + sin(6a)) / sin(a)5)
SR(a) = ∫0ρ R4 S+(R, a) dR = (ρ7 / 210) (15 cos(2a) + 2 cos(a) (8 + |cos(a)| (15 sin(a)6 – 3 sin(a)4 – 4 sin(a)2 – 8)) / sin(a)6)
Sκ(a) = ∫0ρ R2 S+(R, a) dR = (ρ5 / 30) (3 cos(2a) + cos(a) (|cos(a)3| (3 cos(2a) – 7) + 4)) / sin(a)4)
Integrating by parts with respect to a gives us:
IEuc 2(1) = 48π ∫π0 A(a) S1'(a) da
= 48π (A(a) S1(a) |π0 – ∫π0 A'(a) S1(a) da)
= (2/5) (9 – 80/π2) (π ρ2)3
IEuc 2(R) = 48π ∫π0 A(a) SR'(a) da
= 48π (A(a) SR(a) |π0 – ∫π0 A'(a) SR(a) da)
= (ρ/525) (1485 + (16/π2) (64 log(2) – 877)) (π ρ2)3
IEuc 2(κ) = 48π ∫π0 A(a) Sκ'(a) da
= 48π (A(a) Sκ(a) |π0 – ∫π0 A'(a) Sκ(a) da)
= ((3/5) (9 – 80/π2)/ρ) (π ρ2)3
For the third region, R > ρ and R – ρ ≤ r ≤ ρ + R, the circle is larger than the disk, and a is zero at both endpoints of the range for r, with the circle and the disk intersecting at a single point, and the interior of the disk either wholly inside or wholly outside the circle.
In between those endpoints, a reaches a maximum value when:
r = rmax a = √[R2 – ρ2]
a = amax = acos(rmax a / R) = asin(ρ/R)
Each angle a between 0 and amax is now associated with two different values of r, which are the two roots of the quadratic we described earlier. We can combine the two versions of r dr/da for the two roots into a single integral over a, taking the difference rather than the sum to account for the fact that the endpoints for a are swapped for the different ranges for r, above and below rmax a. So the integrals for the third region are:
IEuc 3(Rn) = 96π ∫ρ∞ ∫0asin(ρ/R) A(a) sin(a) Rn+4 (ρ2 + R2 (cos(a)2 – sin(a)2)) / √[ρ2 – R2 sin(a)2] da dR
We will perform a change of variable:
R = ρ / sin(b)
dR/db = –ρ cos(b) / sin(b)2
amax = asin(ρ/R) = b
The limits of ρ and ∞ for the integral over R become b = π/2 and b = 0. So we have:
IEuc 3(Rn) = 96π ∫0π/2 ∫0b A(a) sin(a) (ρ / sin(b))n+4 (ρ2 + (ρ / sin(b))2 (cos(a)2 – sin(a)2)) / √[ρ2 – (ρ / sin(b))2 sin(a)2] (ρ cos(b) / sin(b)2) da db
= 96π ρn+6 ∫0π/2 ∫0b A(a) sin(a) (cos(b) / sin(b)n+7) (sin(b)2 + cos(a)2 – sin(a)2) / √[sin(b)2 – sin(a)2]) da db
[We swap the order of integration, still covering the same region: 0 < a < b < π/2]
= 96π ρn+6 ∫0π/2 ∫aπ/2 A(a) sin(a) (cos(b) / sin(b)n+7) (sin(b)2 + cos(a)2 – sin(a)2) / √[sin(b)2 – sin(a)2]) db da
For n = 0, 1, –1:
IEuc 3(1) = 96π ρ6 ∫0π/2 ∫aπ/2 A(a) sin(a) (cos(b) / sin(b)7) (sin(b)2 + cos(a)2 – sin(a)2) / √[sin(b)2 – sin(a)2]) db da
[Substitute u = sin(b)2, du = 2 sin(b) cos(b) db]
= 48π ρ6 ∫0π/2 ∫sin(a)21 A(a) sin(a) (1 / u4) (u + cos(a)2 – sin(a)2) / √[u – sin(a)2]) du da
= 2 π ρ6 ∫0π/2 A(a) (1 / sin(a)6) (3 (4 cos(a)2 + 1) (π/2 – a) + sin(a) cos(a) (15 – 2 (sin(a)2 + 8 sin(a)6))) da
= (32/π2 – 3) (π ρ2)3
IEuc 3(R) = 96π ρ7 ∫0π/2 ∫aπ/2 A(a) sin(a) (cos(b) / sin(b)8) (sin(b)2 + cos(a)2 – sin(a)2) / √[sin(b)2 – sin(a)2]) db da
= (2π/35) ρ7 ∫0π/2 A(a) cos(a)3 (768 + 512 sin(a)2 + 480 sin(a)4 + 480 sin(a)6) / sin(a)7 da
For values of a near zero the integrand here is proportional to 1/a, so the integral diverges.
IEuc 3(κ) = 96π ρ5 ∫0π/2 ∫aπ/2 A(a) sin(a) (cos(b) / sin(b)6) (sin(b)2 + cos(a)2 – sin(a)2) / √[sin(b)2 – sin(a)2]) db da
= (64π/5) ρ5 ∫0π/2 A(a) (4 – sin(a)2 – 3 sin(a)6) cos(a) / sin(a)5 da
= ((464/(15π2) – 3)/ρ) (π ρ2)3
We can now check that the overall 6-volume is correct:
IEuc 1(1) + IEuc 2(1) + IEuc 3(1)
= (2/5 + (2/5) (9 – 80/π2) + (32/π2 – 3)) (π ρ2)3
= (π ρ2)3
This is exactly what it should be: the cube of the area of the disk, with one factor for each of the three points that lie within it.
And we can now compute the average curvature of the circle that passes through three random points within the disk:
Average curvature of the circle
through three random points in a
Euclidean disk of radius ρ
κEuc(ρ) = (IEuc 1(κ) + IEuc 2(κ) + IEuc 3(κ)) / (π ρ2)3
= (4/5 + 27/5 – 48/π2 + 464/(15π2) – 3) / ρ
= 16 (3 – 16/π2) / (15 ρ)
≈ 1.47079 / ρ
The curvature of the circle that passes through three points in a Euclidean space is sometimes known as the Menger curvature. In the context of spherical and hyperbolic geometry, the usage is a bit less clear. In that setting, we will be calculating the geodesic curvature, which measures how much the curve departs from being a geodesic, which is not the same as the Menger curvature as it is usually defined for a general metric space.
Although the average radius of the circle that passes through three random points in a Euclidean disk is not defined, the integrals we have computed allow us to find the average radius in a restricted version of the same scenario. Suppose we pick three points in the disk at random, calculate the radius of the circle, and then reject that sample if the radius of the circle is greater than the radius of the disk. We can quantify the probability of choosing three points where the radius is less than that of the disk, and give the average of the radius under those restricted conditions.
Average radius of the circle
through three random points in a
Euclidean disk of radius ρ,
if we only accept samples where the circle’s
radius R < ρ
Prob(R < ρ) = (IEuc 1(1) + IEuc 2(1)) / (π ρ2)3
= (4 – 32/π2)
≈ 0.757722
Average {R | R < ρ} = (IEuc 1(R) + IEuc 2(R)) / (IEuc 1(1) + IEuc 2(1))
= (1605 π2 + 16 (64 log(2) – 877)) ρ / (2100 (π2 – 8))
≈ 0.641464 ρ
|
|
We can compute the probability density function (PDF) for R by omitting the integration over R in the computations for the three 6-volumes that we carried out earlier, then dividing by the total 6-volume. For regions 1 and 2 we can obtain a closed-form result as a function of R/ρ, but for region 3 the integral over a for each value of R needs to be performed numerically.
The figures on the right show the PDFs for the radius R and the curvature κ. We have set ρ=1, so we are choosing the three points from a unit disk, but for other values of ρ the curves are just rescaled horizontally, and in fact they will look exactly like these figures if we choose to plot R/ρ and κ ρ along the horizontal axes.
The red portions of the curves correspond to region 3, where R > ρ. The green curves give the asymptotic behaviour for large values of R. That the PDF for R is asymptotic to a constant times 1/R2 makes it clear that R cannot have a finite mean, since the integral of 1/R from any point out to infinity does not converge.
The PDF for R < ρ, with contributions from region 1 and 2, is:
PDFR(R, ρ) = (6 μ/π2)
[μ (μ (π2 (μ2+4)–15μ)–9) + 5 (3μ4–4μ2+1) atan(μ) + (4–24μ2) Li2(μ) + (6μ2–1) Li2(μ2)]
where μ = R/ρ
and Li2 is the polylogarithm function of order 2
The PDF for the curvature κ > 1/ρ is then:
PDFκ(κ, ρ)
= PDFR(1/κ, ρ) | dR/dκ |
= PDFR(1/κ, ρ) / κ2
For region 3, the contribution to the 6-volume is:
IEuc 3(1) = 96π ∫ρ∞ ∫0asin(ρ/R) A(a) sin(a) R4 (ρ2 + R2 (cos(a)2 – sin(a)2)) /
√[ρ2 – R2 sin(a)2] da dR
If we drop the integration over R, divide by the total 6-volume (π ρ2)3, set R = 1/κ and change variables from a to α = a/(ρ κ), then to the lowest order in κ:
PDFR(R, ρ) ≈ (64 ρ / (15 π2)) κ2 ∫01 α7 / √[1 – α2] dα
= (64 ρ / (15 π2)) κ2 (16/35)
= 1024 ρ / (525 π2 R2)
This gives us the green curves that show the asymptotic behaviour for large values of R, or equivalently, low values of κ.
On a sphere, every circle has two centres. For example, every circle of latitude on the Earth has a centre at both the north and south poles: both points are equidistant from every point on the circle. By the radius of a circle on the sphere, we will mean the angle between the points of the circle and the nearest of the two centres, which can never exceed π/2.
The radius R in this context can be thought of either as an angle, or as the distance from the circle’s centre, on a unit sphere, as measured along the surface. It is not the radius of the same circle when considered as a circle in three-dimensional Euclidean space, where the centre will lie in the same plane as the circle, rather than on the surface of the sphere.
The geodesic curvature of a circle with radius R on a unit sphere is not the reciprocal of the radius, but the reciprocal of tan(R):
κg = 1/tan(R) = cot(R)
Using this definition, any great circle, which has R = π/2, has zero geodesic curvature, which is what we would expect.
There are two easy calculations we can make for the case where we pick three points from the entire surface of the sphere. Because the situation is so symmetrical and there are no constraints on the location of the points, we can just weight each configuration by the probability of all three points lying in an infinitesimally thick annulus containing the circle in question. For each point this is proportional to sin(R), and so for three points it is proportional to sin(R)3.
Average radius and curvature of the circle
passing through three random points on a
unit sphere
RSphere = ∫0π/2 R sin(R)3 dR / ∫0π/2 sin(R)3 dR
= (7/9) / (2/3)
= 7/6
κSphere = ∫0π/2 cot(R) sin(R)3 dR / ∫0π/2 sin(R)3 dR
= (1/3) / (2/3)
= 1/2
 |
 |
The first of these calculations is enough to tell us that, for the unit sphere, the probability density function for R is:
PDFR(R) = (3/2) sin(R)3
We can then derive the PDF for κ = cot(R):
PDFκ(κ)
= PDFR(acot(κ)) | d(acot(κ))/dκ |
= 3 / [2 (κ2 + 1)5/2]
The corresponding cumulative density functions are the integrals of the PDFs:
CDFR(R) = ∫0R (3/2) sin(R)3 dR
= ½ (cos(R)3 – 3 cos(R) + 2)
CDFκ(κ) = ∫0κ 3 / [2 (κ2 + 1)5/2] dκ
= κ (2κ2 + 3) / [2 (κ2 + 1)3/2]
These CDFs allow us to write the medians of the distributions, i.e. the values where the CDF equals ½, as simple functions of a root of a cubic polynomial:
Median R = acos(y) ≈ 1.21611
where y3 – 3y + 1 = 0 and |y| < 1
give y ≈ 0.347296
Median κ = cot(Median R)
= cot(acos(y))
= y / √[1 – y2]
≈ 0.370349
Just as an ordinary sphere (the surface of a 3-dimensional ball) is a 2-dimensional space with constant positive Gaussian curvature, a hyperbolic plane is a 2-dimensional space with constant negative Gaussian curvature.
Unlike a sphere, though, a hyperbolic plane is infinite, and though it is possible to embed portions of it in 3-dimensional Euclidean space (as surfaces known as pseudospheres), embedding the whole thing in a flat 3-dimensional space is only possible with two “spacelike” and one “timelike” dimensions.
This means that if we have 3 coordinates on the space, say (t, x, y), we make use of a metric g that gives us a “dot product” between vectors v = (vt, vx, vy) and w = (wt, wx, wy):
g(v, w) = – vt wt + vx wx + vy wy
The minus sign on the product of the t components of the vectors is what makes the t dimension “timelike”. Readers familiar with special relativity will recognise this as a version of the flat spacetime that acts as the backdrop for relativistic physics in the absence of gravitational curvature, albeit with only two dimensions of space in place of the usual three.
We will embed the hyperbolic plane with Gaussian curvature of –1 in this space as the surface consisting of all points (t, x, y) that satisfy the equation:
g((t, x, y), (t, x, y)) = –t2 + x2 + y2 = –1
where t ≥ 1
This surface is one sheet of a two-sheeted hyperboloid, and it is known as the hyperboloid model of the hyperbolic plane. A crucial difference from the same algebraic surface in ordinary Euclidean 3-dimensional space is that we use the spacetime metric, g, to measure distances. The length of any vector v tangent to the surface is given by:
|v| = √g(v, v) = – vt2 + vx2 + vy2
The length of any curve on the hyperboloid can then be found by integrating the length of the tangent vector to that curve (the derivative of the coordinates along the curve with respect to the curve’s parameter), from the start to the end of the curve.
We can also describe the surface in terms of pairs of coordinates that we map to the 3-dimensional space. For example, we can use polar coordinates r and φ:
H(r, φ) = (cosh(r), sinh(r) cos(φ), sinh(r) sin(φ))
In both the flat Euclidean plane and the sphere, there are only two qualitatively different kinds of curve with constant geodesic curvature: geodesics, with curvature zero (i.e. straight lines on the plane, or great circles on the sphere), and ordinary circles, which can have any positive value for their curvature. But in the hyperbolic plane with Gaussian curvature of –1, the curvature of any circle will always be greater than 1! To encompass all possible values for the curvature, κ, we need four distinct cases:
All four of these types of curves can be found in the hyperboloid model as the intersections between the hyperboloid and suitably chosen planes.
Plane through the origin, with spacelike normal κ = 0 |
Plane not through the origin, with spacelike normal 0 < κ < 1 |
Plane not through the origin, with lightlike normal κ = 1 |
Plane not through the origin, with timelike normal κ > 1 |
For each type of curve, we will give either a single example or a 1-parameter family of examples, but then all the other curves of that type can be found by rotating and/or boosting the example(s).
An example of a geodesic is:
G(s) = (cosh(s), sinh(s), 0)
This is the intersection of the hyperboloid with the plane y = 0, which contains the origin and has a spacelike normal vector, (0,0,1). The unit-length tangent to this curve with respect to the arc-length parameter s, and its derivative, the curvature vector, are:
T(s) = G '(s) = (sinh(s), cosh(s), 0)
K(s) = T '(s) = (cosh(s), sinh(s), 0)
Because the whole hyperboloid lies at a constant spacetime interval from the origin, as defined by the metric g, a unit-length vector normal to the surface — orthogonal to it using the dot product defined by the same metric — is just the vector from the origin to the surface itself:
N(s) = G(s) = (cosh(s), sinh(s), 0)
The geodesic curvature of a curve embedded in a surface is the length of the projection of the curvature vector into the tangent plane, i.e. the length of the component of the curvature vector K(s) orthogonal to the normal to the surface, N(s). But K(s) = N(s), so the orthogonal component is zero, and the geodesic curvature is κ = 0.
A family of hypercycles is given by:
Hδ(s) = (cosh(δ) cosh(s/cosh(δ)), cosh(δ) sinh(s/cosh(δ)), sinh(δ))
where s is an arc-length parameter along the curve, and the parameter δ > 0 gives the different curves in the family. Each of these curves is the intersection of the hyperboloid with the plane y = sinh(δ), which does not contain the origin (since δ > 0) and has a spacelike normal vector, (0,0,1).
We then have:
Tδ(s) = Hδ'(s) = (sinh(s/cosh(δ)), cosh(s/cosh(δ)), 0)
Kδ(s) = Tδ'(s) = (cosh(s/cosh(δ))/cosh(δ), sinh(s/cosh(δ))/cosh(δ), 0)
To find the component of Kδ(s) orthogonal to the normal to the surface, note that:
g(Hδ(s), Hδ(s)) = –1
g(Hδ(s), Kδ(s)) = –1
g(Hδ(s), Kδ(s) – Hδ(s)) = 0
This tells us that the projection of Kδ(s) into the tangent plane is Kδ(s) – Hδ(s). The squared length of this is:
g(Kδ(s) – Hδ(s), Kδ(s) – Hδ(s))
= g(Kδ(s), Kδ(s)) – 2g(Hδ(s), Kδ(s)) + g(Hδ(s), Hδ(s))
= –1/cosh(δ)2 + 2 – 1
= tanh(δ)2
So the geodesic curvature of these hypercycles is:
κδ = tanh(δ)
Since δ > 0, we have 0 < κδ < 1.
The parameter δ measures how far the hypercycle Hδ(s) is from the geodesic H0(s). This is an intrinsic property of these curves within the hyperbolic plane itself, independent of the hyperboloid model: the curve that lies a constant distance δ from some chosen geodesic will have a constant geodesic curvature of tan(δ).
An example of a horocycle is:
Hor(s) = (½s2+1, ½s2, s)
This is the intersection of the hyperboloid with the plane t – x = 1, which has a lightlike normal vector, (1,1,0). We have:
THor(s) = Hor'(s) = (s, s, 1)
KHor(s) = THor'(s) = (1, 1, 0)
KHor(s) – Hor(s) = (–½s2, 1 – ½s2, –s)
g(KHor(s) – Hor(s), Hor(s)) = 0
g(KHor(s) – Hor(s), KHor(s) – Hor(s)) = 1
So the geodesic curvature of this horocycle (and in fact, every horocycle) is κ = 1.
A family of circles is given by:
CR(s) = (cosh(R), sinh(R) cos(s/sinh(R)), sinh(R) sin(s/sinh(R)))
where s is an arc-length parameter along the curve, and the parameter R > 0 gives the radius of each circle in the family. Each of these circles is the intersection of the hyperboloid with the plane t = cosh(R), which does not contain the origin, and has a timelike normal vector, (1,0,0).
The unit-length tangent and curvature vectors are:
TR(s) = CR'(s) = (0, –sin(s/sinh(R)), cos(s/sinh(R)))
KR(s) = TR'(s) = (0, –cos(s/sinh(R))/sinh(R), –sin(s/sinh(R))/sinh(R))
Then the calculations are very similar to those for the hypercycles:
g(CR(s), CR(s)) = –1
g(CR(s), KR(s)) = –1
g(CR(s), KR(s) – CR(s)) = 0
g(KR(s) – CR(s), KR(s) – CR(s))
= g(KR(s), KR(s)) – 2g(CR(s), KR(s)) + g(CR(s), CR(s))
= 1/sinh(R)2 + 2 – 1
= coth(R)2
This means the geodesic curvature of a circle of radius R in the hyperbolic plane is:
κR = coth(R)
If we choose three points in a disk of radius ρ in the hyperbolic plane, in order to cover all the possibilities we need to consider cases where the three points lie on a circle, and also cases where they lie on a hypercycle.
When the three points lie on a circle, we can use six coordinates that describe the configuration in a manner that is directly analogous to the case of the Euclidean disk, with one coordinate, R, giving the radius of the circle on which they lie, while r and θ specify the centre of the circle relative to the centre of the disk, and φ1, φ2, φ3 are angular coordinates for the three points on the circle. The Jacobian determinant is:
JHC = 4 sinh(R)3 sinh(r) sin(½(φ1–φ2)) sin(½(φ2–φ3)) sin(½(φ3–φ1))
As in the Euclidean case, we can integrate the absolute value of the Jacobian over –a ≤ φi ≤ a, giving us:
IHC(R, r, a) = 24 sinh(R)3 sinh(r) A(a)
where A(a) = a2 + a sin(a) cos(a) – 2 sin(a)2
The variable a is now related to the radius of the disk, ρ, and the R and r coordinates by a hyperbolic cosine law of triangles in place of the Euclidean one:
cos(a) = (cosh(R) cosh(r) – cosh(ρ)) / (sinh(R) sinh(r))
This gives us what we need to compute integrals for the hyperbolic versions of the three kinds of configuration we called regions 1, 2 and 3, covering all possibilities for the relative sizes and positions of the disk and the circle.
But we also need to include a fourth kind of configuration, for cases where the three points lie on a hypercycle rather than a circle. For region 4, we replace R, the radius of the circle, with δ, the distance of the hypercycle from its associated geodesic, and we replace the angular coordinates φ1, φ2, φ3 with coordinates μ1, μ2, μ3 that correspond to s/cosh(δ) in the formula we gave earlier for a family of hypercycles. The coordinate r is now the distance from the centre of the disk to the closest point on the associated geodesic, and θ gives the angular position of that point around the disk. In terms of these coordinates, the Jacobian determinant is:
JHH = 4 cosh(δ)3 cosh(r) sinh(½(μ1–μ2)) sinh(½(μ2–μ3)) sinh(½(μ3–μ1))
and the integral of |JHH| over –a ≤ μi ≤ a is:
IHH(δ, r, a) = 24 cosh(δ)3 cosh(r) AHH(a)
where AHH(a) = a2 + a sinh(a) cosh(a) – 2 sinh(a)2
The value of a that gives the portion of the hypercycle lying within the disk of radius ρ can be found from:
cosh(a) = (sinh(δ) sinh(r) + cosh(ρ)) / (cosh(δ) cosh(r))
The area of a hyperbolic disk of radius ρ is:
DiskH(ρ) = 2π (cosh(ρ) – 1)
= 4π sinh(ρ/2)2
For region 1, where the three points lie on a circle that sits entirely within the disk of radius ρ, the 6-volume of this region is given by:
IHC 1(1) = 2π ∫0ρ ∫0ρ–R 24 π2 sinh(R)3 sinh(r) dr dR
= 2π3 (cosh(ρ)3 + 15 cosh(ρ) – 9 ρ sinh(ρ) – 16)
If we divide this by the total 6-volume, DiskH(ρ)3, we get the probability of the selection of points landing in region 1, which has limits:
limρ→0 IHC 1(1) / DiskH(ρ)3 = 2/5
limρ→∞ IHC 1(1) / DiskH(ρ)3 = 1/4
The region 1 integral weighted by the geodesic curvature, coth(R), is:
IHC 1(coth(R)) = 2π ∫0ρ ∫0ρ–R 24 π2 cosh(R) sinh(R)2 sinh(r) dr dR
= 2π3 (sinh(ρ)3 + 3 sinh(ρ) – 3 ρ cosh(ρ))
As a fraction of the total 6-volume this gives the contribution from region 1 to the average geodesic curvature. The limits are:
limρ→0 IHC 1(coth(R)) / DiskH(ρ)3 = ∞
limρ→∞ IHC 1(coth(R)) / DiskH(ρ)3 = 1/4
For region 4, where the three points lie on a hypercycle, the 6-volume is:
IHH 4(1) = 2π ∫–∞∞ ∫rρ+r 24 AHH(a) cosh(δ)3 cosh(r) dδ dr
where a = acosh((sinh(δ) sinh(r) + cosh(ρ)) / (cosh(δ) cosh(r)))
This cannot be evaluated analytically, but the numerical result divided by the total 6-volume, which gives the probability that the three points will lie on a hypercycle, is plotted in the figure on the right.
The asymptotic value of this probability as ρ goes to infinity isn’t easy to evaluate from the integral above, but it can be found using a different approach. Given two points in the hyperbolic plane, there will always be two horocycles that pass through both points, and the area that lies between those horocycles will depend solely on the distance between the points. This is illustrated in the figure on the left, using the Poincaré disk model of the hyperbolic plane.
If a third point lies anywhere between these horocycles then the curve passing through all three points will lie on a hypercycle. So the probability that three points lie on a hypercycle can be found by integrating this area for all choices of the two points within the disk of radius ρ, multiplying by a factor of 3 to account for each of the points being the middle one along the hypercycle, then dividing by the total 6-volume, which is the cube of the area of the disk. Note that because the boundary of the disk has a curvature of κ = coth(ρ) > 1, the arcs of the horocycles joining two points within the disk can never lie outside the disk itself, since the curvature of every horocycle is 1. So the region whose area we are calculating will always lie entirely within the disk.
The area between the horocycles passing through a pair of points separated by a distance d can be found fairly easily by applying the Gauss-Bonnet formula, which expresses the integral of the Gaussian curvature over a region in terms of the integral of the geodesic curvature along its boundary, and the angles at any corners.
Since the Gaussian curvature of the hyperbolic plane is a constant, K = –1, and the geodesic curvature along the horocycles is also a constant, κ = 1, we can express the area simply in terms of the length of the arcs of the horocycles between the two points, and the angles between the tangents to those arcs.
In the hyperboloid model, two horocycles that pass through two points, (cosh(d/2), ±sinh(d/2), 0) that are a distance of d apart, are described by arclength-parameterised curves:
Hord±(s) = (½((1+s2)/cosh(d/2) + cosh(d/2)), s, ±½((1+s2)/cosh(d/2) + cosh(d/2)))
Both horocyles pass through the points when s = ±sinh(d/2), so the distance between the points along the horocycle is 2 sinh(d/2). The unit-length tangents to the horocycles are:
Hord±'(s) = (s/cosh(d/2), 1, ±s/cosh(d/2))
and the spacetime dot product between these tangents at either of the points is:
cos(φ) = g(Hord+'(sinh(d/2)), Hord–'(sinh(d/2))) = 1 – 2 tanh(d)2
where φ is the angle between the tangents. This is equivalent to:
tan(φ/2) = sinh(d/2)
The Gauss-Bonnet formula gives us, when κ and K are constant:
κ × (length of perimeter of region) + K × (area of region) = 2π – Σ (exterior angles at corners)
Given κ = 1 and K = –1, and two corners with exterior angles that are π minus the angle between the tangents, we have for the area:
A(d) = 4 sinh(d/2) – 4 atan(sinh(d/2))
In the asymptotic case where ρ goes to infinity and the integral is dominated by configurations where the distance between the two points is large, we can drop the inverse tan term from the area, since it is bounded by 2π, and we can replace sinh by half an exponential. This gives us:
limρ→∞ PHypercycle(ρ)
= limρ→∞ 6π ∫01 ∫01 ∫–ππ 2 exp(½d) ½exp(r1 ρ) ½exp(r2 ρ) ρ2 dφ dr1 dr2 / DiskH(ρ)3
Here d is the distance between the points with polar coordinates (r1 ρ, 0) and (r2 ρ, φ), which satisfies:
cosh(d) = cosh(r1 ρ) cosh(r2 ρ) – cos(φ) sinh(r1 ρ) sinh(r2 ρ)
exp(d) ≈ ½ exp((r1+r2) ρ) (1 – cos(φ))
exp(½d) ≈ exp(½(r1+r2) ρ) |sin(½φ)|
Substituting this into the integral, and also approximating the disk area with an exponential, we end up with:
limρ→∞ PHypercycle(ρ)
= limρ→∞ 6π ∫01 ∫01 ∫–ππ 2 exp(½(r1+r2) ρ) |sin(½φ)| ½exp(r1 ρ) ½exp(r2 ρ) ρ2 dφ dr1 dr2 / (π exp(ρ))3
= limρ→∞ 6/π2 exp(–3ρ) ∫01 ∫01 ∫0π exp((3/2)r1 ρ) exp((3/2)r2 ρ) sin(½φ) ρ2 dφ dr1 dr2
= limρ→∞ 6/π2 exp(–3ρ) ρ2 [∫01 exp((3/2)r ρ) dr]2 [∫0π sin(½φ) dφ]
= limρ→∞ 12/π2 exp(–3ρ) ρ2 [(exp((3/2) ρ) – 1) (2/(3ρ))]2
= 16/(3π2)
The figure on the right shows the average geodesic curvature for the curve through three points chosen at random in a hyperbolic or Euclidean disk of radius ρ. For the Euclidean disk, the average curvature is just inversely proportional to ρ, so it decreases monotonically towards an asymptotic value of zero.
The result for the hyperbolic disk is more complicated. Its behaviour is similar to the Euclidean function for small ρ, but it reaches a minimum of around 0.828 at ρ ≈ 2.994, then it increases monotonically towards an asymptotic value of 1.
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