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The Essence of Spherical Geometry, Part 1 
10th-Mar-2019 01:04 am
toyz

So, as part of my possibly-continuing "Geometry on Drugs" series, here is a prequel to my post on spherical geometry, which was more of a "hey, this is useful" post in which much there's a whole lot you're expected to take on faith. It was really more intended for the hardcore engineering type who needs to see that use case up front.

This version is going back to first principles, where we do the axiom wanking and you (hopefully) get a sense of why things turn out the way they do.

Also, this is the practice run before I launch into the Essence of Hyperbolic Geometry, so, … Onward …

The Geometry Axiom Everybody Hates

Start with this diagram and the inevitable question that comes up:

Start with a line ℓ and a point A not on it. How do you put a line through A that doesn't intersect ℓ?

(In other news, I am now convinced that the Unicode committee contained at least one disgruntled geometry teacher. How else to explain why there's this isolated script ℓ code point?)

We can drop a perpendicular from A meeting ℓ at some point X, and then it's obvious that the line you want (dotted) is the one perpendicular to XA. If you tilt it even slightly away from 90°, then it simply must intersect ℓ somewhere.

Proof by diagram. We're allowed to do that, right?

Thus, we come to the boundary between Mathematics and Handwaving. Here is something that Euclid was convinced was true but he could not prove it from what he had. So he did the next best thing and added it to his list of axioms, the list of facts you need to bootstrap your geometry: two points define a unique line, all right angles are congruent, and so on. Four axioms, all simple and elegant, and then this annoyingly complicated monster, otherwise known as the Parallel Postulate, shows up and has to be #5. He was not a happy camper.

And so we have 2000 years of Greeks, Romans, Arabs, and basically everybody, beating their heads against this particular wall trying to find a proof, or some simpler, more obvious idea that it can follow from that would work better as an axiom, until finally it occurs to someone (really, several someones within the space of a few decades; well okay, I'm guessing Gauss had it first, but the guy was a super-anal perfectionist who hated Actually Publishing anything, and so you'd go up to him with some brilliant discovery of yours and he'd be all, "Oh, yeah; I did that already. Here are my notes from 20 years ago when I just sort of dashed it off one Saturday afternoon and forgot to tell anyone. Sorry." … Just a complete asshole and also the PhD candidate's worst nightmare; he randomly shows up in the audience on the day of your defense, and you may as well just kiss it all goodbye right there. … Yeah, okay, I'm digressing way too much)

… that maybe the reason they couldn't find a proof was because (drumroll…) there exist ways to have this Not Be True.

Once you start thinking along these lines, it's not actually that hard to come up with alternatives — everything is, of course, easy once somebody has thought to do it and figured it out. So let's try out an alternative and see where it gets us. (… cue Twilight Zone music …)

Taking Red Pill #1

So, you want a line that goes through A but is entirely disjoint from ℓ? What odd notions you have!

You do realize that this is simply not possible. All pairs of lines intersect. No exceptions.

And yes, that means there is no such thing as "parallel". Parallelism is a lie perpetrated by those Deep State military-industrial-complex warmongers who believe in vaccines and evolution and stuff.

If we have something perpendicular to AX coming out of A, then it must intersect ℓ somewhere, … perhaps Really, Really Far Away, but still somewhere, which is a big difference from nowhere, because somewhere provides this handy intersection point P that we can then use to play nasty geometry pranks.

For example, we can drop a perpendicular bisector from P to AX.

(… and yes, all of those lines are supposed to be straight, our problem being that we're stuck having to depict things on a flat screen, so you're going to be seeing lots of bendy shit from now on.)

We're already having to face up to having APX be this isoceles right triangle — and you thought that was never supposed to happen in real life; hahahaha — meaning AP and XP, the sides opposite the two (equal) right angles must be the same length. And now this bisector BP turns out to also be the same length as well.

But we don't have to stop there; we can drop a whole lot of bisectors to get as close as we want to any desired point along AX. Meaning every point along AX must be the same distance from P.

We can also reflect everything we're doing through XP to put yet more triangles below ℓ, or through AP to get triangles above, to get this mess:

… and now the conclusion is inescapable that every point along the line containing AX is the same magic distance from P. If we were to draw a circle of this radius centered at P, it would contain AX and all possible extensions of it in either direction.

We also see that the distance between X and A, or, rather the distance between any two points on that line, depends solely on the angle those points form at P, and it's a linear relationship; doubling the angle doubles the distance and so on. Which then gives us a completely natural way to measure distances in this world. E.g., if ∠APX is 15°, then we may as well say that the distance from A to X is 15°.

If we so happen to pick two points along the line AX such that the angle at P is 90° (which is clearly possible to do), we then have this equilateral triangle with three 90° angles and thus three 90° sides. And so the radius of the circle at P has to be 90°.

Finally, since the line AX is contained within the circle at P, and there's only 360° worth of angle available at P, it should be clear that if you start at X, or anywhere on that line, and travel a distance of 360° in one of the directions, you will come back to where you started. We will have traversed every point on the line through X and A and therefore, i.e., the line also contains the circle centered at P.

The line and the circle are one and the same. And since we made no assumptions about the points we started with, we then see that all lines are circles of radius 90° and each line has at least one point that I'll call a "pole" for that line, which can serve as its center.

Hopefully, you will forgive me if I occasionally slip up and for no particular reason refer to lines as "equators", or if I refer to a bunch of lines passing through a particular point (pole) as "meridians". But just remember that we're not wedded to any coordinate system yet; North, South, and Greenwich can be anywhere we want them to be. (And also this might not necessarily be a sphere.)

Now you might be thinking, "But wait, there are parallel lines here!" If we think of "parallel" being not so much the "never intersects" definition but rather the "all of the points a particular distance away from the line on one side of it" definition, what then? In other words:

Stare at this for a while and you'll realize X₂A₂ is also part of a circle, since all of its points are the same distance from P, but the radius of this circle is less than 90°. Having a smaller radius suggests it'll have more curvature than a line, but our intuitions on curvature may not be entirely reliable here. We'll get back to this.

But we at least know that the circle containing X₂A₂ — which I may occasionally refer to as a "latitude line", — while it may indeed be "parallel" to XA in some sense, cannot be a line. Because if it were, it would have to intersect XA somewhere.

Some further thought will show that all circles in this world have radius ≤ 90°, or, rather, can be described that way as long as you pick the right center point and look at what the actual shortest distances are. Even if you might have originally thought you were creating a larger circle, you won't be.

Consider, e.g., the particularly weird case of "circles of radius 180°" which turn out to consist of single points (i.e., circles of radius zero), because if you set out from a point P, which will be a pole for something, because we can draw circle of radius 90° around any point, you will cross that equator after traveling 90° out from P in any direction, and, after continuing for another 90° you will necessarily end up at the pole for the other side of that line …

(… which might actually also be P, the place where you started out; nothing rules this out, but I'll save the Projective Plane for another day.)

At the very least, it's clear that the relationship between a circle's radius and circumference is a bit more complicated than you might have thought, what with "π" somehow ranging between 0 and 4 depending on how big our radius is. How can we get a handle on this?

(… to be continued in Part 2) This entry was originally posted at https://wrog.dreamwidth.org/62392.html. Please comment there using OpenID.

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