# Understanding the distance metric for a 3D space of uniform positive curvature.

I am reading a cosmology textbook, and the distance metrics for three dimensional spaces exhibiting various curvatures are being presented. My question is about their treatment of a three dimensional space under unifom positive curvature:

In polar coordinates, on the two dimensional surface of a sphere, we can express the distance $d\ell$ between two points as a function in their separation in the radial coordinate $r$, and

$d\ell^2 = dr^2 + R^2\sin^2(r/R)d\theta^2$

In three dimensions, this extends to

$d\ell^2 = dr^2 + R^2\sin^2(r/R)[d\theta^2 + \sin^2\theta d\phi^2]$.

Now, my texbook asserts that when the two points whose separation we are measuring are at antipolar locations, we have $r = \pi R \rightarrow r/R=\pi$, which gives

$sin^2(r/R) = 0\rightarrow d\ell^2 = dr^2$.

But this makes no sense to me. This isn't how spherical coordinates work at all, right...? If I have two point at antipolar points on a sphere, and I measure each of their $r$ coordinates (the length to the point along a line forming angles $\theta$ and $\phi$ from the $z$ and $x$ axes, respectively) as $r_1$ and $r_2$, then shoudln't their separation $dr = |r_1-r_2|$ simply be $2R$? This would mean that $r/R = 2 \neq \pi$

Claiming that $r/R = \pi$ implies that $r$ refers to the actual path length between the points along the surface of the sphere, which is not how spherical coordinates work.

Is my error in applying these spherical coordinates to a two dimensional sphere surface, when I am supposed to be thinking about the three dimensional surface under uniform positive curvature? I.e. the image I have in my head of how spherical coordinates work in this context is all wrong, or at least my placement of the points?

I am picturing a "three dimensional space under uniform positive curvature" as a sphere, like the Earth. But that isn't right, is it? That is just me imagining a 2D surface being curved into a third dimension, when the more accurate analog is to somehow imagine a 3D space being "curved" into a fourth dimension?

• What are polar coordinates on the surface of a sphere? I only know of polar coordinates in the plane. Does the book give an explanation? – Rahul May 11 '17 at 19:20
• @Rahul. Yes, polar coordinates for the sphere. – Rafa Budría May 11 '17 at 20:21
• @Rafa: Thanks for explaining that polar coordinates on a sphere are polar coordinates on a sphere, that certainly clears it up. – Rahul May 11 '17 at 22:38

Indeed, these two examples are for $S^2 \subset \mathbb R^3$ and $S^3 \subset \mathbb R^4$, and all distances are being measured on the (hyper)surface of the sphere.

I think the main point of confusion is the usage of the label $r$. The coordinate $r$ here is not the same $r$ from spherical coordinates on $\mathbb R^3$ - it is instead the distance from the north pole, measured within the surface. From the point of view of a ($2$-dimensional) observer sitting at the north pole, these really are the closest thing available to polar coordinates on $S^2$. In general they're known as geodesic polar coordinates.

To make the relationship more obvious, let's take the usual metric for $\mathbb R^3$ in spherical coordinates and rename a variable: I'm going to call the radial coordinate $\rho$. The metric is then

$$d\ell^2 = d\rho^2 + \rho^2\left( d \phi^2 + \sin^2(\phi) d\theta^2 \right).$$

Restricting this to the sphere $\rho = R$ we get $$d \ell^2 = R^2 \left(d \phi^2 + \sin^2(\phi) d\theta^2\right).$$

Now we just need to switch the polar angle $\phi$ to the north-pole distance $r=R \phi$, which yields $$d \ell^2 = dr^2 + R^2 \sin^2(r/R) d\theta^2.$$

If we expand the sphere (i.e. decrease the curvature) by taking $R \to \infty$ we get $R^2 \sin^2(r/R) \to r^2$ and thus this converges in some sense to the usual metric for polar coordinates $(r,\theta)$ on $\mathbb R^2$.

If you do the same thing with hyperspherical coordinates on $\mathbb R^4$ you should get the three-dimensional version.

To complete the excellent answer Anthony has offered us, I'd like to remark what the $S^3$ example means to illustrate.

It's true that the parametrisation of $S^3$ can be obtained from $\mathbb R^4$ with the euclidean metric in spherical coordinates, but it has too a good interpretation without any appeal to $\mathbb R^4$.

We can imagine ourselves as living in a space (of three dimensions) with intrinsic constant curvature $1/R$. To describe this space with spherical coordinates seems as natural as do it for a flat space: we send lines from some point and two special lines we chose and measure two angles wrt this special lines to determine each of the other lines. The position of any point can be established with this two angles and the distance along the line corresponding to this angles. Remark again that this is what we do for spherical coordinates in a flat space.

The curvature thing it's shown when, obviously, we need to calculate (or measure!) angles, relative distances, displacements, relative displacements... We do all these with the aid of the metric that somebody discovered and that it is described by:

$$d\ell^2 = dr^2 + R^2\sin^2(r/R)[d\theta^2 + \sin^2\theta d\phi^2]$$

With $R$ some very large parameter. If we don't intend to travel very far away, we have that $\sin(r/R)\approx r/R$, an we can do well our calculations with:

$$d\ell^2 = dr^2 + r[d\theta^2 + \sin^2\theta d\phi^2]$$

The old line element for euclidean space expresed in spherical coordinates.

But considering large distances we have to deal with the $\sin$ thing in the line element.

We send two ships, one along the ray with $\theta=0,\phi=0$ the other along the ray with $\theta=0,\phi=\pi/4$, all the way in a straight path and with the same speed. We observe initially they, as expected, are getting apart... until some point from where they start to get closer. That point, they measure, is at a distance of $\pi R/4$ from the departure. How do we understand $R$ now? simply as some parameter that, in this universe, explains the weird thing that at some well determined distance two ships traveling in straight and initialy diverging lines start to get closer. Further, eventually they meet! and, it must be repeated, traveling in straight lines.

From the point of view of a physicist, it's not necessary to explain that metric saying that our universe is some subset of some larger space. At all, we have not any evidence of that larger space. We say that the space we live is curved in nature, simply, being the fact of that we can understand $S^3$ as part of $\mathbb R^4$ no more than a good thing to make calculations easier. So is, the number $R$ is no more than a parameter to express what we can call size of the universe and $\pi$ a constant to deal with the sinus part of the metric.