Pi on curved surfaces

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1. Nov 30, 2015

DaveC426913

I was lying awake the other night and thinking about Pi and flatlanders. I haven't done a lot of topology reading, so forgive my naivete.

Pi on a flat surface is a number we know well, but what happens to the ratio of a circle's diameter to its circumference on curved surfaces?

First question: it's not called Pi, is it? Pi is a constant. For the prupose of this discussion I'm going to call it rho, but pleasecorrect me if there is already a term for it.

As a surface curves positively, rho will decrease. The diameter may be constant while the circumference contracts. If the surface contracts into a sphere, then rho becomes 2 (diameter = to half the circumference). It can never become less than two because you wouldn't be able to fit that fixed sized circle on the sphere. Of course, this is only true of the surface curved evenly in all directions. Presumably, there are shapes, such as an elliptoid where you could have a high curvature in one direction but low in the other.

On a positively curved surface, I suspect that rho would have an upper limit of pi - the surface approaches a Euclidean plane.

If the surface had a negative curvature, I think you wouldn't have such problems - you should be able to fit a circle on a negatively curved surface. I think. I imagine there would be surfaces where you could not make a continuous line with all points equidistant from a central point.

What happens on a toroid? Is it possible that rho wold not exist i.e. no line could be drawn that is equidistant from a point? Ah. A toroid is an example of negatively-curved surface.

Maybe I should do a little reading.

2. Nov 30, 2015

Orodruin

Staff Emeritus
As long as your space satisfies geodesic completeness, you can make a "circle" with any radius. (It might intersect itself.) You simply take the endpoints of all geodesics with fixed length (the radius). This set will be a curve orthogonal to the radius everywhere. As you probably have figured out, the ratio between the circumference and the radius depends on the radius. In the limit of very small radius, you recover pi as long as the space is locally flat.

The curvature of the torus depends on the metric you bestow it with. It is perfectly possible to have a flat torus.

3. Nov 30, 2015

DaveC426913

Aaaaaand I'm lost.

I looked up geodesic completeness, and now I have to look up Riemannian metric and Cauchy sequences.

4. Dec 1, 2015

Orodruin

Staff Emeritus
Think of it as being able to extent a geodesic (essentially the curved space equivalent of a straight line) to an arbitrary length.

5. Dec 5, 2015

HallsofIvy

Your question makes no sense because you have not said what you mean by "$\pi$"! In Euclidean geometry, which is what, I think, you mean by "a flat surface", "$\pi$" is defined as the ratio of the circumference of a circle to its diameter. IF you are referring to the ratio of the circumference of a circle to its diameter in non-Euclidean geometry (non-flat surface) there is no such constant. It is only in Euclidean geometry that this ratio is a constant.

6. Dec 5, 2015

Hornbein

rho = the ratio of a circle's diameter to its circumference on curved surfaces.

It seems to me that the circumference of the circle goes to zero at the point which is the antipode of the center of the circle. This is where the diameter of the circle is equal to the circumference of the surface.

But it is a matter of arbitrary definitions.

7. Dec 5, 2015

DaveC426913

I presume that the diameter by definition is the shortest distance across the circumference. The moment the circumference passes the "equator", the diameter flips to the opposite hemisphere.

If we don't assume the shortest distance then what's to stop us from claiming the diameter of a circle on flat plane is infinite?

Last edited: Dec 5, 2015
8. Dec 6, 2015

HallsofIvy

The definition of diameter! And the "diameter" is NOT "the shortest distance across the circumference", it is the longest distance!

Last edited by a moderator: Dec 6, 2015
9. Dec 6, 2015

DaveC426913

Heh. Touche.

10. Dec 7, 2015

mathwonk

this question of the ratio between a diameter and a circumference is the basic one to measure curvature. when it equals π the surface has zero curvature, when it exceeds π the surface has positive curvature, and when it fall short of π the surface has negative curvature. (π here could be defined analytically by a series if desired.) of course i am simplifying and assuming the surfaces considered have constant curvature. in more general surfaces one can avoid this concern by considering very small circles about a given center, and taking limits to obtain a local measure of curvature. (one must also incorporate the area of the circle here to compensate for the fact that on smaller circles the surface is more nearly flat.) but to me the basic principle is that if you want to explain to someone what it means for a surface to have positive or zero or negative curvature at a point, just show the difference between a flat disc and its border circle, versus a rounded cap - shaped "disc" cut from a sphere, and finally a scalloped disc with wavy boundary "circle". i.e. a flat tray, a skull cap, and a saddle.

on page 3B45 of Spivak's diff geom vol 2, Prop 10, one finds the explicit formula of Bertrand and Puiseux (1848) for the curvature k(p) of a surface at a point p in terms of the discrepancy between the circumference C(r) of a circle of radius r centered at p, and the expected value 2πr, namely k(p) is the limit as r-->0, of (3/πr^3).(2πr-C(r)). Notice we are dividing once by r to get the ratio between the circumference and the radius, and again by another r^2 to bring in the area of the circle, as mentioned above.

On page 3B46 Spivak gives also the formula of Diquet (1848) expressing curvature in terms of area A(r) of a circle of radius r, namely k(p) is the limit as r-->0 of (12/πr^4)(πr^2 - A(r)).

Last edited: Dec 8, 2015
11. Dec 9, 2015

meldraft

Assuming that the radius vector should always be normal to the tangent of the circumference, am I right to assume that if the surface curvature is such that rho=constant, the ratio will be exactly the same (Pi)? Otherwise, it's obviously no longer a constant.

12. Dec 9, 2015

DaveC426913

No. On a consistently curved surface (such as a sphere) the radius will still be perpendicular to the circumference, but rho ( C / 2r ) will be smaller than pi (2.0 in fact).

Last edited: Dec 9, 2015
13. Dec 9, 2015

mathwonk

I think he means assume rho is constant for all circles. in the case of a sphere it seems indeed always less than π, but not always equal to 2, and apparently approaches π as the radius approaches 0, doesn't it? i.e. as the portion of the sphere we are looking at becomes smaller, hence closer to planar. and it seems to approach zero as the radius approaches its maximal value of half the circumference of the sphere. does this scan?

14. Dec 9, 2015

DaveC426913

Yes.

15. Dec 10, 2015

mathwonk

I just noticed that actually on a sphere a line segment can be extended to a line segment of arbitrary length, except after it gets longer than the circumference of the sphere, it loops back on itself and covers the same points again and again. So that means that circles exist on the sphere of arbitrary positive radius, and their circumferences have lengths that range from 0 to the circumference of the sphere and then back to zero and then grow again, and continue to oscillate like that. So as r ranges from 0 to infinity, the number rho also varies from π to zero and then becomes positive again, but smaller, then zero, then again positive,.....

16. Dec 10, 2015

DaveC426913

I'm not sure that a radius that loops around the circle multiple times really counts as a radius. If the distance from a point on C to the centre/pole point is r, then it seems to me, that is the distance. I don't think it is sensible to say that the distance can also be 5r, 9r or nr.

Circles of diameter greater than C simply cannot exist on a sphere of diameter greater than 2r.

Last edited: Dec 10, 2015
17. Dec 10, 2015

Orodruin

Staff Emeritus
But this is just a special case of what I was mentioning in post #2 ...

18. Dec 10, 2015

mathwonk

Dave, It depends on your notion of a line segment. Euclid has a postulate that says any segment can be extended arbitrarily far. You can take that to mean that given a positive real number d, one can travel in the direction of an oriented line segment until one has traveled through a distance d, even if one has covered the same ground multiple times. It is true the points one reaches will not have distance d from the original point, if distance is defined as length along shortest path. This is in the context of the concept mentioned in post #2 of "geodesic completeness", since as I understand it, a geodesic is a path which is only locally length minimizing, but may not be globally so. So one could consider all paths to be parametrized, and measure their "length" by the length of the parameter. In Riemannian terminology, one looks at the image of a Euclidean line segment in the tangent space, under the exponential map to the sphere, and considers this a geodesic. In this context the set of points of a circle on the sphere do not determine uniquely the "radius" of the circle, i.e. the length of the geodesics chosen as the "radii". Rather, to form the circle, you just choose any positive number r as your radius, then you move away from the north pole along each great circle until you have traveled a distance r along each geodesic. the set of points you reach would make up the circle whose radius equals one of those geodesics of length r reaching from the north pole to the circle (and of course having also crossed that circle multiple times). You may not care for this theory, but since it satisfies many of the usual axioms, suitably interpreted, it has interest, at least for me. For one thing it helps point out gaps in reasoning in proving the usual theorems in geometry since Euclid's axiom for extending line segments nowhere excludes this posssibility that they may loop back on themselves. He in particular does not say anywhere that the line segment joining two distinct points is unique, which has often been pointed out. So it is of interest to ask what geometries one gets from allowing these more liberal interpretations of the axioms. Even if you want to exclude them, this search reveals what further restrictions must be made to the usual axioms to rule them out.

In particular, your observation that the equatorial circle has rho = 2, leads further to noticing that in the multiple radius setting suggested here, the values of rho vary downwards from π to 2 at th hemisphere, then down to zero at the south pole, then if we continue they increase again to 2/3 at the hemisphere, then decrease again to zero at the second encounter with the north pole, then increase again to 2/5 at the equator, etc,.....oscillating btween zero and 2/(2n-1), on the nth encounter with the equator.

Last edited: Dec 18, 2015
19. Dec 18, 2015

lavinia

- When asking questions about the relationship of curvature to geometry, there is an important general theorem. It says that the integral of minus Gauss curvature over a disk shaped region on any surface - e.g. a polar ice cap on a sphere - is equal to the total geodesic curvature of the bounding curve - e.g. a circle parallel to the equator - minus 2π. This theorem always works even when the bounding curve is far from being a circle and the disk shaped region itself is wobbly and warped.

To understand this theorem one needs to know what geodesic curvature is. In Physics one says that a particle moving at a constant speed is moving in a straight line if no forces are acting on it to change its direction of motion. This means that the particle is not accelerating in any direction. If on the other hand, the particle is accelerating then it follows a curved path and the amount that the path is curved is measured by its rate of acceleration. Its acceleration is its geodesic curvature in the flat plane. For instance, a particle moving at constant speed along a circle has a rate of acceleration equal to $1/r$, the reciprocal of its radius. This is the geodesic curvature of a circle in the flat plane.

So what does the theorem say in the case of a circle in the plane? Since the Gauss curvature is zero and the geodesic curvature of a circle is 1/r,

$∫_{C}1/r ds - 2π = 0$ or since $r$ is constant, $1/r∫_{C}ds = 2π$ i.e the ratio of the circumference of a circle to its radius is $2π$

While this theorem seems to just verify what was already known since the time of Euclid, there is an important difference. In Euclidean geometry $1/r$ is reciprocal of the length of a radius line from the center of the disk bounded by the circle. But in Physics and Differential Geometry it is the acceleration of a particle moving at constant unit speed and does not require any reference to objects outside of the curve itself. It is an intrinsic measure of the radius of the circle and could be measured by a instrument attached to the particle without any measurements of lengths in the surrounding space.

On a surface, the acceleration of a unit speed curve is also its geodesic curvature but it is not the usual acceleration but rather, the covariant derivative of the tangent vector field to the curve with respect to itself. If the surface is embedded in 3 space then the geodesic curvature is the length(up to a sign) of the orthogonal projection of the ordinary acceleration vector onto the tangent space of the surface. One can think of geodesic curvature as the intrinsic infinitesimal radius of a curve and one can think of the analogue of a circle on a curved surface as a closed curve of constant geodesic curvature.

For such a curve,$C$ ,with constant geodesic curvature,$k_C$, the theorem says

$k_C∫_Cds = 2π - ∫_DK$ where $K$ is the Gauss curvature and $D$ is a disk shaped domain bounded by $C$. If one interprets the geodesic curvature as the reciprocal of the intrinsic radius of the curve,$C$, then one has an analogous formula to the formula for flat Euclidean space ,except now there is a correction term involving the Gauss curvature.

Here is how this looks on the sphere.

Parametrize the sphere by $X = (sin(α)cos(θ),sin(α)sin(θ),cos(α))$

Here, the north pole, $(0,0,1)$, is the point with $α = 0$ For each $α$ between $0$ and $π/2$, the parameter, $θ$, traces out a circle of radius $sin(α)$. So the ordinary curvature in 3 space of this circle is $1/sin(α)$. But its geodesic curvature, the length of its projection onto the tangent space of the sphere, is $cos(α)/sin(α)$. It is less that its curvature in 3 space. For a person living on the sphere, the circle seems less curved, i.e. it is straighter than the same circle in Euclidean 3 space. For instance, for the equator, the ordinary curvature vector is orthogonal to the sphere so its projection onto the tangent space is zero. This means that the equator is a straight line, i.e. it is a geodesic on the sphere. On the other hand, as $α$ approaches $0$, the geodesic curvature diverges and from the Taylor series for the cotangent one sees that it is close to $1/α$.

For one of these circles with $α$ equal to a constant,$c$, the theorem says

$cos(c)/sin(c)∫_{C}ds =2π - ∫_{0≤α≤c}KdA$ where $K$ is the Gauss curvature. On the unit sphere, $K = 1$ everywhere so the integral of the curvature is the area of the polar cap bounded by the circle. For small $c$, this is close to $1/c∫_{C}ds$ so the formula is close to the Euclidean formula.

When $c = π/2$ one is left with $0 = 2π - ∫_{0≤α≤c}KdA$ which says that the area of the hemisphere is equal to $2π$. Beyond the equator the integral of the geodesic curvature changes sign and in the limit of the south pole, one has $-2π = 2π - ∫_{sphere}KdA$. In other words the integral of the Gauss curvature over the entire unit sphere is $4π$ , the area of the sphere.

Since this theorem works for any disk shaped region, it works for the sphere(minus a point) with a different metric. That is: it works for any surface that is topologically a sphere. For instance, it works for an ellipsoid or for a dented sphere. For all possible shapes, even those in which the Gauss curvature is not constant and can even be negative in some places, the integral of the Gauss curvature is always $4π$. This means that this integral is more than just an area calculation. It is an invariant of the topological sphere. It is independent of its geometry.

Notice that this theorem works because the entire sphere except for one point, e.g. the south pole, can be covered by an open disk shaped region. On an arbitrary surface, e.g. a torus, this is not true but still the integral of the Gauss curvature is a topological invariant of the torus. The general theorem says: For any compact surface without boundary,

$∫_{surface}KdA = 2πχ_{surface}$ where $χ_{surface}$ is called the Euler characteristic of the surface. For a sphere, the Euler characteristic is 2. For other surfaces, it is always an integer and is independent of the metric. For instance, since a torus can be given a flat metric, its Euler characteristic is zero.This shows that $π$ is intimately linked to the topology of all surfaces, and is not only a number that has significance for flat Euclidean space. It is a universal constant.

Last edited: Dec 19, 2015
20. Dec 18, 2015

lavinia

- Around any point on a surface, there is as analogue of polar coordinates called geodesic normal coordinates. In these coordinates, radial lines from the center of the coordinate system are geodesics, the equivalent on a curved surface of straight lines, and the curves orthogonal to these radial lines are the analogue of polar circles. They are like latitude and longitude lines around the north pole. Geodesic polar coordinates generally exist only in a neighborhood of a point and cannot be extended indefinitely across the surface.

Nevertheless, if the manifold is geodesically complete, then individual geodesics can be extended indefinitely and may, as has been noted. circle back on themselves. (There are even examples of surfaces where a geodesic turns around and crosses over itself on it way back to its starting point. )However, the geodesic normal coordinates will not follow the geodesic beyond a certain point so beyond that point it no longer makes sense to talk about the ratio of the circumference of a circle to its radius.

Last edited: Dec 19, 2015