Greg Egan on Nostr: To answer this, we can start with the fact that on a *sphere* of radius R, any two ...

To answer this, we can start with the fact that on a *sphere* of radius R, any two ants following neighbouring great circles would travel a distance of (π/2) R before their paths met. That behaviour comes from the intrinsic curvature of the sphere, which is measured by its Gaussian curvature, K = 1/R^2.

But what is the Gaussian curvature of our ellipsoid, at its equator? The equator itself has a curvature of 1, but the meridians are ellipses that are 3 times wider. What is the radius of curvature of an ellipse with semi-axis a and b, at the point where the minor semi-axis meets the ellipse?

The ellipse takes the form:

y(x) = b √[1-(x/a)^2]

For small x:

y(x) ≈ b (1-½(x/a)^2)

The second derivative is:

y''(x) = -b/a^2

If we set a = b = R this would be -1/R, so the curvature here is just the opposite of the second derivative.

The Gaussian curvature is the product of the curvatures along two perpendicular directions where it achieves it maximum and minimum values. For our ellipsoid, these curvatures are 1/b for the equator and b/a^2 for the meridian, so we have:

K = 1/a^2

In other words, the Gaussian curvature at the equator of the ellipsoid is completely independent of b (the radius of the equator) and is the same as that of a sphere of radius a (the polar radius)!

That in turn means that the ants following the red path will bump into each other after travelling a distance:

(π/2) a = 3π/2 ≈ 4.71239

So, perhaps a bit counterintuitively, the ants following the small red circle will travel *farther* before they converge than the ants following the larger blue ellipse.