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?
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.