Spherical Geometry 3 -- A bit of Circular Reasoning

[wrog]

Continued from Part 2, exploring the benighted universe where "parallel" is Not a Thing.

How circles work

So, to review the weird things we've seen so far:

• When we have a circle radius of 90°, otherwise known as a straight line, and we're traversing the circumference, i.e., measuring the total length along it as we sweep out 360° from the pole in the middle, we get 360° worth of path (phrasing it this way so that if this turns out we're on a projective plane rather than a sphere and what we're really doing is traversing the same 180° path twice, I won't have been lying to you), which, being 4 times the radius, is slightly less than one might have expected (2π being roughly 6.28).
   
• If we attempt a circle of radius of 180°, we stay firmly nailed to the antipode of the center, our circumference traversal goes nowhere and thus we get a circumference of zero.

Meaning if we have to explain to the residents what "π" is, we're going to lose horribly. Best we can do is, "So: Circumference to radius? That's a ratio. It's literally all over the map. But as radius approaches zero, once you're under 90°, you'll notice the ratio is always getting bigger. If you work at it, you can prove that it's bounded and it converges to this weird transcendental number like e. And, no, don't ask us how we came up with this…"

We need to understand better how curved paths work. And rather than resort to the usual trick of chopping it into a polygon/polyline with lots of tiny segments, it's actually simpler if we cut to the chase: Walk along it, and rather than doing all of our turning at some finite number of vertices, we'll just do it continuously as we walk along. Essentially, a circular arc is a path where you have to continuously turn at a constant rate while walking along it at a constant speed in order to stay on it.

The other bit of fun is that if someone is walking alongside you but a few feet away, then they may be walking at a slightly different speed in order to be keeping pace with you. Meaning the path they're following may be longer or shorter than yours, and if they're doing the same turning that you are, that means the curvature of their path is different.

Reviewing this a bit more quantitatively, here's what we know about circular arcs from our plane geometry experience:

• The arc has a radius, and so there's a center somewhere at a distance ρ away. Put everything together and you get a wedge.
   
• The total amount you turn when traversing the curved part of the wedge is θ, the same as the angle at the center of the wedge.
   
• The area of the wedge is θρ²/2, with θ needing to be in radians in order for this to work (because in plane geometry, angles and distances annoyingly use different units)
   
• If we increase the area by increasing the radius a tiny bit (dρ), that's basically adding a uniform width strip of length θρ. In other words, the derivative of the area with respect to the radius is the length of the curved frontier.
   
• The second derivative of the area with respect to the radius is the rate of change of the frontier per unit radius added, which will just be θ.
   
• The curvature of the frontier is the amount turned per unit distance traveled, θ/θρ or 1/ρ. Or, equivalently, we could instead refer to the distance ρ we have to travel along the arc in order to turn one radian, what we call the radius of curvature, even though this doesn't actually involve measuring the radius of anything.

When we go over to our non-Euclidean world, some of our assumptions turn into lies. The angle at the center of the wedge does not have to match the angle turned. The radius of the wedge (r) need not be the same as the radius of curvature (ρ) of the frontier. E.g., recall, again, that when r = 90°, we can be sweeping some number of degrees at the center while, out on the frontier, we'll actually be on a straight line with no turning happening at all, meaning zero curvature or radius of curvature effectively infinite, so, buh.

Perhaps we should just focus on the frontier and just not worry about the center at all. This kind of thinking will come in handy when it turns out the real "center" of the curve is out in space somewhere or in the middle of a black hole where we have no fucking clue what's happening, or, worse yet, doesn't even exist (file under:  Foreshadowing for Hyperbolic Geometry).

So, imagine we have some shape with a circular frontier and the rest of it is some completely random crap:

If we can still walk that portion of the frontier, i.e., turning at a constant rate suffices to keep us on that path, and if adding a constant width (dr) strip there is the only way in which we are growing the overall shape, whatever it is, then, because we're stuck in a two-dimensional surface where the only way for the shape to grow is sideways and we've constrained what's happening at every point on that frontier, we then know that:

• The derivative of the area with respect to our "radius parameter," r, will still be the length of the frontier. Note that we don't care what the hell r actually means so long as increasing it by dr adds a strip of width dr to the frontier.
  
  
• The second derivative of the area will still be the amount we turned, θ,

because, at heart, this is still just a piece of circle we're playing with, and we know how those work.

We could now just take our curved arc by itself, with its length λ₀ and its amount of turning that we're now going to call θ₀, plant it somewhere, paint fertilizer on one side of it, turn on the grow lights, and leave it for a few hours. Or, equivalently, if you like, we can take every point along it and walk out perpendicularly a distance r and see where we get to …

a new uniformly curved arc of some length λ that turns by some angle θ, and we don't know what these are, yet, but we'll find out.

• The Deficit Thing gives us the area of the shape we just grew. Sum the exterior angles angles plus the net turning we do along the two curves (the θ₀ bit being negative because it's bimping the wrong way) to get:
  
     90° + 90° + θ + 90° + 90° − θ₀ = 360° − (θ₀ − θ).
  
  Meaning area = θ₀ − θ.
   

• First derivative gets us the frontier, λ = −θ′
  
  
• Second derivative is −θ″ but is also θ,

and, wow, it looks like we have a differential equation, θ″ = −θ.

Oh c'mon, please don't all run away. You learned this one the third week of calculus. θ as a function of r has to be one of our old friends, the sine or cosine functions, or some combination of the two. And the only combination that gives the right θ and λ for the case r = 0 is θ = θ₀cos r − λ₀sin r

and I am inordinately amused that in "θ cos r", we are using the letters r and θ backwards from the way you usually see them. Yeah okay, I'm weird.

Now, at this point, you might want to recall the addition formula for cosines:

• cos(r+r₀) = cos r₀ cos r − sin r₀ sin r

meaning if there's a way to make θ₀ look like cos r₀ and λ₀ look like sin r₀, that would be a way to combine the stuff on the right into one term.

Which we can just do by finding an r₀ such that tan r₀ = λ₀/θ₀, and since the range of tan() is all real numbers including infinity, this will always be possible even when θ₀ = 0 (pedants, shut up), and so we have the following clever shuffle/rewrite:

θ =	θ₀ cos(r + r₀)/cos r₀
λ = −θ′ =	λ₀ sin(r + r₀)/sin r₀

the first observation being that we should have grown things inward rather than outward, because, clearly, something weird is going to happen when we get to r = −r₀: the frontier length λ decreases all the way to 0, meaning those top and bottom sides perpendicular to the original curve have now converged on a "center" point.

Always. Even if we'd started out with a straight line (λ₀ > 0, θ₀ = 0 implies r₀ = 90°). Meaning if we attempt to build a rectangle with base λ₀ and height r, we'll get

area = −θ =	λ₀sin r	(instead of λ₀r)
frontier λ =	λ₀cos r	(instead of λ₀)

the problem being that, with there being three straight sides and four right angles, all of the deficit has to come from the frontier, which is caved in to get us our −θ, which is why we end up with this trapezoid thing that's been beaten up a bit.

Yes, actual rectangles are just Not Allowed here. They come really close to behaving like rectangles when r is small, but that's the best we can do.

Notice that when we get to r = 90°, λ hits zero, meaning we've reached the pole and thus have an actual isoceles right triangle, whose area λ₀, as it happens, also matches the length λ₀ of the equator piece that we started with…

… most of which we already knew. But now we've derived it solely from The Deficit Thing, so we really could have started with that as our 5th axiom (well okay, it's wouldn't actually be 5th because of the stuff that Euclid overlooked, like the definition of "between" and various other axioms David Hilbert figured out were needed to make it all actually rigorous, and I'm probably also cheating by using calculus, bleah …).

The second particular case we care about is if we start from a center with a zero-length segment and try to construct a wedge. With θ₀ > 0, setting λ₀ = 0 implies r₀ = 0, and turning the crank the rest of the way gives us

θ =	θ₀cos r
area =	θ₀(1 − cos r)
frontier λ =	θ₀ sin r
radius of curvature = λ/θ =	tan r

A few things to notice here

• If r is really small, these numbers come out roughly as you would expect for a planar wedge (once you recall that, for small r, sin r and tan r are really close to r and (1 − cos r) is really close to r²/2).
  
  
• If you plug in θ₀=2π, we see that the full circumference of a circle radius r is 2π sin r, just like the latitude line on a sphere that's (angular) distance r from the pole.
  
  Not only that, but if you were to construct a planar curve with the same length and curvature, you'd have part of a circle of radius of tan r but there'd only be enough of it to go 2π cos r of the way around, i.e., you're making a circular wedge with that as your central angle … and then you grab the scissors, cut out the wedge, and glue the two radial edges together to make a cone that will turn out to be exactly the right size and shape so that you can perch it on the sphere and have it be tangent to that latitude line. (File under: How to do Conic Projections.) The curvature being the same means if you're walking along the latitude line, you won't be able to tell whether you're walking on the sphere or walking on the cone; your turn rate will be the same either way.
  
  … the moral of this story being that if you try to construct a model for this geometry in Euclidean 3-space, one that properly preserves all of the distances and angles the way you'd want — this may or may not be possible, but if it is (spoiler alert: it is) — then (drumroll …) a sphere (… or at least a piece of it) is pretty much the only way to do it.
  
  … hence, why this is Spherical Geometry, just in case you were wondering.

(Continued in Part 4 in which we tackle trigonometry.)