## Wednesday, July 27, 2016

### Spherical Geometry (Legendre 502-506)

Today is our final spherical geometry post. In today's post we will cover the last five spherical geometry propositions in Legendre, which will be numbered 502 to 506. Of these five propositions, two are corollaries to the main theorem from our last spherical geometry post (namely that the area of a triangle is equal to its "excess") and two more theorems are called "scholia." The only main theorem for today is a generalization of that previous theorem -- how is the area of any spherical polygon (not just a triangle) related to its angle sum?

Let's begin with Proposition 502:

502. Corollary I. The proposed triangle will contain as many triangles of three right angles, or eighths of the sphere (494) [Legendre actually means 495 here -- dw], as there are right angles in the measure [the excess -- dw] of this triangle. If the angles, for example, are each equal to 4/3 of a right angle, then the three angles are equal to four right angles, and the proposed triangle will be represented by [an excess of] 4 - 2 or 2; therefore it will be equal to two triangles of three right angles, or a fourth of the surface of the sphere.

Recall that Legendre uses the right angle as a unit of angle measure, and the triangle with three right angles as a unit of area (that's what Proposition 495 is for). This is convenient because a triangle with three right angles has an excess of one right angle -- recall that the "excess" is the sum of the angles of a spherical triangle minus what the sum would have been had the triangle been Euclidean (two right angles or 180 degrees). So the area of the triangle (using the equilateral right triangle as a unit) is always equal to the excess of the triangle (using the right angle as a unit).

Now that Pi Approximation Day has come and gone, we'll return to using pi radians as our circle constant, and so the excess of a triangle is its angle sum minus pi. Using radians, we can rewrite the example given by Legendre as:

If the angles, for example, are each equal to 2pi/3, then the three angles are equal to 2pi, and the proposed triangle will be represented by an excess of 2pi - pi or pi; therefore it will be equal to two triangles of three right angles, or a fourth of the surface of the sphere.

Or we can rewrite the example using degrees:

If the angles, for example, are each equal to 120 degrees, then the three angles are equal to 360 degrees, and the proposed triangle will be represented by an excess of 360 - 180 or 180; therefore it will be equal to two triangles of three right angles, or a fourth of the surface of the sphere.

OK, we now move on to Proposition 503:

503. Corollary II. The spherical triangle ABC is equivalent to a lunary surface, the angle of which is (A + B + C)/2 - 1; likewise, the spherical pyramid, the base of which is ABC, is equal to the spherical wedge, the angle of which is (A + B + C)/2 - 1.

This follows from some of Legendre's earlier propositions. Recall that a lunary surface, or lune, has its area equal to twice its angle measure, while a triangle has as its area equal to its defect. Since the defect of a triangle is A + B + C - 2 (right angles), we take half of this to find the angle of a lune with the same area. And a spherical pyramid and a spherical wedge are obtained by starting with a triangle and a lune, respectively, and digging towards the center of the sphere. So if the triangle and lune have the same area, the pyramid and wedge must also have the same volume.

Legendre's next proposition, a "scholium," continues with spherical pyramids:

504. Scholium. At the same time that we compare the spherical triangle ABC with the triangle of three right angles, the spherical pyramid, which has for its base ABC, is compared with the pyramid which has a triangle of three right angles for its base, and we obtain the same proportion in each case. The solid angle at the vertex of the pyramid is compared in like manner with the solid angle at the vertex of the pyramid having three right angles for its base. Indeed, the comparison is established by the coincidence of the parts. Now if the bases of pyramids coincide, it is evident that the pyramids themselves will coincide, as also the solid angles at the vertex. Whence we derive several consequences; [deleted -- dw]

Let's skip down to the last thing that Legendre writes about solid angles:

"The angle at the vertex of the pyramid, whose base is a triangle of three right angles, is formed by three planes perpendicular to each other; this angle, which may be called a solid right angle, is very proper to be used as the unit of measure for other solid angles. This being supposed, the same number, which gives the area of a spherical polygon, will give the measure of the corresponding solid angle. If, for example, the area of a spherical polygon is 3/4, that is, if it is 3/4 of a triangle of three right angles, the corresponding solid angle will also be 3/4 of a solid right angle."

If we write Legendre's statement using radians, we obtain the following:

The angle at the vertex of the pyramid, whose base is a triangle of three angles of measure pi/2, is formed by three planes perpendicular to each other; this angle, which may be called a solid right angle, has a measure of pi/2 steradians. This being supposed, the same number, which gives the area of a spherical polygon, will give the measure of the corresponding solid angle. If, for example, the area of a spherical polygon is 3pi/8, that is, if it is 3/4 of a triangle of three right angles, the corresponding solid angle will also be 3pi/8 steradians.

And so we introduce a new unit of solid angle measure, the "steradian." Let's think about how ordinary radians are defined -- the radian measure of an angle equals the length of the arc of the unit circle subtended by the angle. Likewise, the steradian measure of a solid angle is equals the area of the polygon on the unit sphere subtended by the solid angle.

Notice that if we're considering circles other than the unit circle, we usually think of radian measure as the ratio of the arc length to the radius. But on the sphere, recall from the Fundamental Theorem of Similarity that the surface area of a sphere (or a spherical polygon) varies not as the radius, but as the square of the radius. Therefore, steradian measure is the ratio of the polygon area to the square of the radius -- hence another name for steradian, the "square radian." Fortunately, most of the time we will assume that the sphere is the unit sphere of radius 1.

A solid right angle measures pi/2 steradians. Therefore, each of the solid angles of a cube (of which there are eight) measures pi/2 steradians. The entire sphere measures 4pi steradians, since the surface area of the unit sphere is 4pi.

Officially, steradians are considered part of the SI system of measurement, which means that metric prefixes may be used with steradians (abbreviated as sr). As usual, we can take the earth to be our sphere for the purpose of visualizing various steradian units. For example, I found the following measures at another website:

4pi sr (= 1.2566 dasr) = entire earth's surface
1 dasr (decasteradian) = Americas plus liquid water on earth
1 sr (steradian) = Oceania plus Asia excluding Russia
1 dsr (decisteradian) = Algeria plus Libya
1 mcsr (microsteradian) = Costa Mesa, CA

Sorry, I just had to give the microsteradian example because Costa Mesa happens to be right here in Southern California. (Notice that the prefix "micro-" is often written as mc in ASCII, but the official symbol is the Greek letter mu.)

Is it possible to use degrees to convert Legendre's statement -- that is, can there possibly be such a thing as a "stedegree" or square degree? Yes, it's possible, but the conversion from steradians to square degrees is a bit awkward. Think about it -- there are three feet in a yard, yet there are nine square feet in a square yard (since 9 = 3^2). Likewise, just as there are (180/pi) degrees in a radian, there must be (180/pi)^2 square degrees in a square radian (steradian). So let's measure our solid right angle in square degrees:

solid right angle = pi/2 sr
= (16200/pi) deg^2

This works out to be approximately 5156.62 square degrees. So we can rewrite Legendre as:

The angle at the vertex of the pyramid, whose base is a triangle of three angles of measure 90 degrees, is formed by three planes perpendicular to each other; this angle, which may be called a solid right angle, has a measure of 16200/pi square degrees. This being supposed, the same number, which gives the area of a spherical polygon, will give the measure of the corresponding solid angle. If, for example, the area of a spherical polygon is 3pi/8, that is, if it is 3/4 of a triangle of three right angles, the corresponding solid angle will be 12150/pi square degrees.

Obviously, it's much easier to use steradians than square degrees. And so in practice, square degrees are almost never used.

Let's move on to Proposition 505, which is the most important theorem of the day. Before we look at the proof, let's look at the statement of the theorem:

505. The surface of a spherical polygon has for its measure the sum of its angles minus the product of two right angles by the the number of sides in the polygon minus two.

We write this using symbols as follows -- let s = angle sum, n = number of sides, and with degrees:

A = s - 180(n - 2)

And now that term 180(n - 2) should look very familiar -- it is exactly the sum of the angles of a polygon in Euclidean geometry! And indeed, if we define the "excess" of any polygon to be its angle sum minus what that sum "ought to be" in Euclidean geometry, we can write the proposition as:

505. The surface of a spherical polygon has for its measure its excess.

We've already proved this theorem for triangles -- that is, when n = 3. And in a way, we've already proved this theorem for the n = 2 case as well. Think about it -- in Euclidean geometry there is no such thing as a 2-gon, but we can plug in n - 2 into the equation s = 180(n - 2) to obtain s = 0 -- that is, the sum of the angles of a 2-gon in Euclidean geometry is 0 (which is why it doesn't exist).

Now in spherical geometry a 2-gon does exist -- the lune. The two sides of a lune are semicircles that intersect at two antipodal points -- for example, if the two sides are meridians, then the vertices are the North and South Poles. Now the excess of a lune equals its angle sum minus that of a Euclidean "2-gon" (which is zero). So the excess of a lune is its entire angle sum. A lune has two congruent angles, one at each pole, and we've already shown its area to be twice the measure of its angle. Thus the area of a lune indeed equals its excess.

So now let's give Legendre's proof of Proposition 505. As it turns out, we prove it almost the same way that we prove the angle sum of a Euclidean polygon -- by dividing it into triangles. The only difference is that the excess of a Euclidean polygon is always zero, while in spherical geometry the excess equals the area.

Demonstration. From the same vertex A, let there be drawn to the other vertices the diagonals AC, AD; the polygon ABCDE will be divided into as many triangles minus two as it has sides. But the surface has for its measure the sum of its angles minus two right angles, and it is evident that the sum of all the angles of the triangles is equal to the sum of the angles of the polygon; therefore the surface of the polygon is equal to the sum of its angles diminished by as many times two right angles as there are sides minus two. QED

We may compare Legendre's proof to the proof of the Polygon-Sum Theorem (Euclidean geometry, of course) given in Lesson 5-7 of the U of Chicago text. If we wish, we may even combine the Euclidean and spherical proofs, as follows:

Sum(ABCDE...) = Sum(ABC) + Sum(ACD) + Sum(ADE) + ...
= pi + Excess(ABC) + pi + Excess(ACD) + pi + Excess(ADE) + ...
= pi(n - 2) + Excess(ABCDE...)

and then we replace all the "Excess" terms with 0 for Euclidean and "Area" for spherical geometry.

The final theorem to cover is a "scholium" -- basically the same theorem except given as multiples of the Legendre's favorite unit, the right angle:

506. Scholium. Let s be the sum of the angles of a spherical polygon, n the number of sides; the right angle being supposed unity, the surface of the polygon will have for its measure s - 2n(-2) = s - 2n +4.

We have finished our study (which took more than year!) of Legendre's spherical geometry. But there are a few more issues that I want to discuss before I leave spherical geometry altogether.

Last month, I mentioned that there's another way to look at spherical geometry. From the perspective of Legendre, the geometry of the sphere is basically a part of Euclidean geometry, as the sphere itself lies in Euclidean space. The other perspective is that of Riemann -- as in Bernhard Riemann, the German mathematician who lived about a generation after Legendre.

Riemann is probably best known for the Riemann hypothesis, an unsolved conjecture. In fact, mathematicians have sought a solution for so long that it is the first of the Millennium Problems, with a million-dollar prize to be awarded to the one who gives a solution.

But today we will discuss the Riemann sphere, a model of non-Euclidean spherical geometry. For Riemann, we don't consider a sphere as sitting in a Euclidean universe -- instead the Riemann sphere is the entire non-Euclidean universe. Last month, I wrote:

And so another way to approach spherical geometry -- without referring to the sphere itself or any of its Euclidean properties -- is to start with some postulates that describe spherical geometry and use these to prove the theorems of spherical geometry. We aren't allowed to use phrases like "great circle" which betray knowledge of the underlying sphere. Instead, we must refer to great circles by the role that they serve in spherical geometry -- as lines.

Here are some new postulates that I suggested might work to generate a spherical geometry:

-- Any two lines intersect.
-- Every line has exactly two poles. (Here "pole" is an undefined term.)
-- Every point is the pole of exactly one line.
-- The pole of a line never lies on the line.
-- If l and m are lines and a pole of l lies on m, then the other pole of l lies on m, and both poles of m lie on l. (Two lines l and m satisfying this property are defined to be perpendicular.)

Using these postulates, we can prove the following theorem:

Theorem: Any two lines intersect in at least two antipodal points.

Notice that here we need to define "antipodal points." Well, that's easy -- every line has two poles and every point is the pole of exactly one line. So the other pole of the line of which the point is a pole is its antipodal point. To make it easier, let's use the word "equator" to refer to the line of which a point is the pole. So the antipodes of a point is just its equator's other pole.

Proof:
Let a and b be the two lines. Our postulates already give us one point at which they intersect, say P, so P lies on both a and b. Let p be the equator of P. Then p has another pole, P', the antipodes of P.

Now another postulate states that if l and m are lines and a pole of l lies on m, then the other pole of l lies on m. Thus since p and a are lines and a pole of p (namely P) lies on a, the other pole of p (namely P') lies on a. Likewise P' lies on b. So a and b intersect at both P and P'.

But as good as these postulates sound, I was unable to prove that the two lines must intersect in at most two antipodal points. Of course, I could just start adding postulates to guarantee that two lines intersect in exactly two points, but continuing to add axioms is a bit ad hoc.

Instead, it's more elegant if we just start with Euclid's five postulates, and keep only the postulates that hold in Riemann's spherical geometry. Here's a link to Euclid's postulates:

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/bookI.html

Postulate 1.
To draw a straight line from any point to any point.
Postulate 2.
To produce a finite straight line continuously in a straight line.
Postulate 3.
To describe a circle with any center and radius.
Postulate 4.
That all right angles equal one another.
That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.

Now I've mentioned before that in the other form of non-Euclidean geometry, hyperbolic geometry, the first four postulates of Euclid hold, but the famous Fifth Postulate fails -- and indeed, we usually think of Euclidean geometry as one that satisfies the Fifth Postulate and non-Euclidean geometry as one that violates it.

But we've seen before on the blog that spherical geometry doesn't fit this idea perfectly. Indeed, we see that there are propositions of Euclid that fail in spherical geometry, even though the proof avoids the fifth postulate. The first theorem such is Proposition 16:

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI16.html

This theorem is often known as the Triangle Exterior Angle Inequality or TEAI:

There are geometries besides Euclidean geometry. Two of the more important geometries are elliptic geometry and hyperbolic geometry, which were developed in the nineteenth century. The first 15 propositions in Book I hold in elliptic geometry, but not this one.

Geometries satisfying the first four postulates are called "neutral geometry." Both Euclidean and hyperbolic geometry satisfy Postulates 1-4, so they both count as neutral geometry. But Proposition 16 fails in spherical geometry despite requiring only Postulates 1-4, so spherical geometry is not a neutral geometry.

A tricky question is, which of the first four postulates does spherical geometry violate? According to the link above, we see:

Elliptic geometry satisfies some of the postulates of Euclidean geometry, but not all of them under all interpretations. Usually, I.Post.1, to draw a straight line from any point to any point, is interpreted to include the uniqueness of that line. But in elliptic geometry a completed “straight line” is topologically a circle so that any pair of points on it divide it into two arcs. Therefore, in elliptic geometry exactly two “straight lines” join any two given “points.”

Also, I.Post.2, to produce a finite straight line continuously in a straight line, is sometimes interpreted to include the condition that its ends don’t meet when extended. Under that interpretation, elliptic geometry fails Postulate 2.

In spherical geometry (which is slightly different from "elliptic geometry," by the way), there are infinitely many lines through two antipodal points. So under my own interpretation, spherical geometry absolutely violates the First Postulate.

It appears that spherical geometry must violate either the First or Second Postulates, since Proposition 16, which fails in spherical geometry, depends on both of them. But we see that Proposition 5, the Isosceles Triangle Theorem, also depends on Postulates 1-2 -- and yet ITT is valid in both Euclidean and spherical geometry! Of course, the reason is obvious -- the First Postulate is invoked in Proposition 5 only to show the existence of a line joining two points, not its uniqueness.

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI5.html

The best way to see why Proposition 16, the TEAI, fails in spherical geometry is to take a triangle for which it fails and see whether we end up trying to join two antipodal points, which isn't unique.

Let ABC be a triangle, and let one side of it BC be produced to D.
I say that the exterior angle ACD is greater than either of the interior and opposite angles CBA and BAC.

Let ABC be a triangle for which we know TEAI fails -- for example, let's take Legendre's favorite triangle of three right (interior) angles. All the exterior angles are also right angles, and so no exterior angle is greater than any interior angle. We'll place B at the North Pole, and so both A and C end up on the Equator.

Bisect AC at E. Join BE, and
produce it in a straight line to F.

Since E is the midpoint of AC, E must also be on the Equator.

Make EF equal to BE

And that does it. Since BE is a quadrant, so is EF, which places F at the South Pole. So B and F are now antipodal points, which renders the use of the First Postulate with B and F invalid.

I will consider the Second Postulate to be valid in spherical geometry (so as far as we're concerned, any line segment can be extended into a full "line," or great circle). Meanwhile, as I've pointed out before, I consider the Fifth Postulate -- believe it or not -- to be valid in spherical geometry! This postulate is vacuously true in spherical geometry, especially if we rewrite it as Playfair's Postulate (given a point and a line, there is at most one line parallel to the line through the point).

So to me, Postulates 2-5 hold in both Euclidean and spherical geometry, just as Postulates 1-4 hold in both Euclidean and hyperbolic geometry. Just as we have the name "neutral geometry" to refer to one in which Postulates 1-4 hold, I propose a new name, "normal geometry," to refer to one in which Postulates 2-5 hold. The name "normal" sounds a bit like "neutral" -- and besides, there are only two geometries that people "normally" think about, namely Euclidean geometry on a plane and spherical geometry to describe the earth (for example, when flying on an airplane). On the other hand, we don't "normally" think about hyperbolic geometry at all.

One theorem of normal geometry is "the sum of the angles of a triangle is at least 180 degrees," just as one theorem of neutral geometry is "the sum of the angles of a triangle is at most 180 degrees."

I've fascinated by what some normal proofs might look like. I'm curious, for example, whether the Fifth Postulate can be used to prove something nontrivial in normal geometry (that is, other than something like "if two parallel lines are cut by a transversal, then..." which is only vacuously true in spherical geometry).

But of course, this blog focus is no longer on geometry. It's about time that I spend my exclusive focus on my new middle school classroom.