*Continued from Part 1, in which we discover at least one consequence to doing away with the concept of "parallel lines". (and yeah, I did a bit of a George Lucas thing here; sorry…)*

### Let's talk about Area

Having noticed that isoceles right triangles give us a natural way to define/measure distances, we see that we can do area this way as well. That is, the area of ΔAPX is clearly the angle at P times some constant, which we may as well just take to be 1 if we haven't defined a unit of area yet.

There is one small caveat in that, if we measure angles in degrees, we need to be clear that the corresponding unit of area, what I'll call 1 *spherical* degree, is going to differ from the "square degree" you might have expected us to use (i.e., if you can imagine a square that's 1° on a side, never mind that shape will turn out to be *impossible to build* in this world, as we'll see in our next episode). With radians we get lucky and that proportionality constant turns out to be 1 but people still tend to sleep easier if you use a different name ("steradian" rather than radian) for area units.

We next notice that the area of our entire world is *finite*. If there are only 360° available at any given pole, then the entire "hemisphere" on that side of the corresponding line is likewise bounded to be no more than 360° spherical degrees or 2π steradians (can be less if points are repeating themselves (*cough*-projective-plane-*cough*, admittedly the PP also screws up what "on that side of" might mean)).

Knowing how to get the area of isoceles right triangles, how do we do general triangles? The usual argument goes like this:

Here, we have cleverly divided up our entire world into 3 "lunes" — or, if you prefer, "biangles with both sides being 180°, — (the 3rd one being the white background because we're having to do a wacky projection to be able to see the entire world here) whose areas are known, plus two copies of the triangle we care about, one of them mirror-reversed on the far side of the world. Adding everything up and dividing by 2, we see that (π−A)+(π−B)+(π−C) + area(ΔABC) = some constant probably having to do with the Area of the Entire World. Considering the special case where, say, both B and C are right angles (π/2) and our prior stipulation for that case that the area of the triangle is just A, this constant is evidently 2π.

Why? We could stand to pick at this oddly magical scab a bit more.

### Let's talk about ~~Area ~~Deficits

What exactly *is* (π−A)+(π−B)+(π−C)?

It's the sum of the exterior angles of our triangle, but it's *also* the total amount of turning you have to do when traversing the exterior in order that when you go around it once and get back to your starting point you'll be facing the same direction as when you started out. For any simple closed curve, you would expect this to be 360° (2π). After all, it always is in the plane. But here, for some reason, it's *not*.

What we have here is an *exterior angle deficit*. It's particularly egregious in the case of that Congo-Ecuador-NorthPole equilateral 90° triangle, where we only get 270° worth of turns out of it before we're done. But this turns out to be true for *all* triangles with non-zero area.

This is what tells us that we're not actually living in a plane, that we have *curvature* to contend with.

"What's curvature?" you ask, if you've lived your entire life inside of a 2-dimensional surface with no concept of how it might be bending around or stretched or bubbly or whatever.

Ok, fine, maybe we'll just use the deficits to *define* curvature for you and the other 2D folks who will have no other way of understanding it. (Note that this particular outlook will come in handy when it comes time to consider what happens in three and four dimensional spaces that are curved.)

One key observation about angle deficits is that they are *strictly additive*: If I take two polygons and glue them together at a vertex or along a side, the deficit of the combined shape will be the sum of the deficits of the individual shapes (I'll leave that as an exercise).

Once we have that, we can subdivide to our heart's content, i.e., slice a given shape up into billions of little triangles, each with its eentsy piece of the overall deficit. Take the limit as triangle area goes to zero, we get a value for this ratio at every point, and we can give that a name:

* curvature* =

*exterior angle deficit per unit area enclosed*,

and the only reason Euclid missed out on this is that he stumbled onto the

*one and only*geometry where this number happens to be

**zero everywhere**and hence

**useless**.

But, there's more. Because we're in a Euclidean-ish world where we can, due to the other four axioms, take any triangle with particular measurements (and hence a particular deficit) and build it *anywhere else we please*, it follows that in all such worlds **the curvature must be the same everywhere**.

Which means we can turn this around: if we *have* shapes with non-zero deficits at all, we can use deficits to define area. Which is essentially what we've been doing here all along anyway. Recall that we started out defining area using polar angles once we noticed that we can build an isoceles right triangle — which is just an in-your-face, screaming deficit — and now we've generalized that statement.

*(and now we can make sense out of circles in Part 3)*This entry was originally posted at https://wrog.dreamwidth.org/62716.html. Please comment there using OpenID.