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Spherical Geometry

(These notes are based on the first chapter of Geometry from a Differential Viewpoint by John McCleary, (Cambridge University Press, 1994).)

Whereas basic plane geometry is concerned with points and lines and their interactions, most of the early geometry of the Babylonians, Arabs, and Greeks was spherical geometry--the study of the Earth, idealized as a sphere. This early science was astronomy and the need to measure time accurately by the sun.

We have to come to some agreement on what lines and line segments on a sphere are going to be.

A great circle   on a sphere is the intersection of that sphere with a plane passing through the center of the sphere.

Examples of great circles are the equator and the lines of constant longitude (such as the Greenwich Mean Time Line tex2html_wrap_inline11258). These are good choices to play the role of lines on our sphere. For example, given any two non-antipodalgif points there is a unique great circle joining those two points. This is easy to see when you remember that three non-collinear points determine a plane. Take these two points and the center of the sphere as the three non-collinear points. The intersection of the plane determined by these points and the sphere is the great circle joining the two given points.

If great circles are to be lines, then we can measure the angle between two intersecting great circles as the angle formed by the intersection of the two defining planes with the plane tangent to the sphere at the point of intersection. See the figure below (Figure 1). With this definition of angle, we can form triangles on the sphere whose interior angle sum is greater than two right angles. In fact, we will show that the interior angle sum of all triangles on the sphere is greater than two right angles. We will be working with radian measure for most of the semester (to make many of the calculations neater). Thus, we are claiming that on a sphere the interior angle sums of any triangle is greater than tex2html_wrap_inline11130.

tex2html_wrap11482
Figure 1: Angles in Spherical Geometry 

In order to discuss how to measure the angle sums, we will first discuss the concept of the area of a triangle on the sphere. We will discuss this more fully later, but for the present time we will accept the following four properties of area for polygonal regions on the sphere:

  1. The sphere of radius R has area tex2html_wrap_inline11264;
  2. the area of a union of non-overlapping regions is the sum of their areas;
  3. the areas of congruent regions are equal;
  4. the ratio of the area enclosed by two great circles to the area of the whole sphere is the same as the ratio of the angle between them to tex2html_wrap_inline11266.

Theorem 1.1  On a sphere of radius R, a triangle tex2html_wrap_inline11270 with interior angles tex2html_wrap_inline11272, tex2html_wrap_inline11274, and tex2html_wrap_inline11276 has area given by:
displaymath11278

First, some terminology: a lune  is one of the regions between two great circles and the antipodal points where they cross. Antipodal points  or figures are those that correspond to opposite ends of a diameter of the sphere. A lune is given by the angle tex2html_wrap_inline11280 at the center of the sphere between the two defining planes. Note that this is also the angle between the two great circles, cf. Figure 2.

tex2html_wrap11484
Figure 2:   A lune 

Proof(due to Euler, 1781). From our assumptions above, Assumption (4) tells us
 equation263
To make our proof easier, we may assume that the triangle lies in one hemisphere. If not you can use Assumption (2) to break the triangle up into smaller triangles that do lie in one hemisphere. You can then use the following argument on each piece and then add them together.

A triangle is the intersection of three lunes: the lune given by tex2html_wrap_inline11282, the lune given by tex2html_wrap_inline11284 and the lune given by tex2html_wrap_inline11286. Each of these three lunes has an antipodal lune and the intersection of these three antipodal lunes is an antipodal triangle tex2html_wrap_inline11288.

tex2html_wrap11486
Figure 3:   A triangle and the associated lunes

We will denote the lunes by the angle that includes them, BCA, BAC, and ABC. Starting with the triangle tex2html_wrap_inline11270 and the lune ABC, add to them the lune ACB. This covers the triangle tex2html_wrap_inline11270 three times. Now, add in the antipodal lune to lune CAB and the antipodal triangle tex2html_wrap_inline11288. This covers the hemisphere determined by the line tex2html_wrap_inline11308, counting the triangle three times. So we get:
displaymath11222
Thus, we get our desired formula:
displaymath11278
We call the value of tex2html_wrap_inline11310 the angle excess of the spherical triangle.

Note that since every triangle on the sphere has area, every triangle on the sphere has interior angle sum greater than tex2html_wrap_inline11130. If the radius of the sphere is very large and the triangle very small in area, then the angle excess will be almost 0.

Let tex2html_wrap_inline11316 denote the sphere of radius R centered at the origin in tex2html_wrap_inline11320. In rectangular coordinates
displaymath11224
and in spherical coordinates
displaymath11225

tex2html_wrap11488
Figure 4:   Spherical coordinates

The length  of a path between two points along a great circle is found by measuring the angle, in radians, made by the radii at the two endpoints and multiply it by the radius of the sphere, i.e. tex2html_wrap_inline11322. Note that the length of a segment is unchanged when we rotate the sphere around some axis or reflect it through a great circle.

Theorem 1.2:[Spherical Pythagorean Theorem]   For a right triangle tex2html_wrap_inline11270 on a sphere of radius R with a right angle at vertex C and sides of length a, b, and c, then
displaymath11226

tex2html_wrap11490
Figure 5:   Spherical right triangle

Proof: Rotate the sphere so that the point A has coordinates (R,0,0) and the point C lies in the xy-plane. This will make tex2html_wrap_inline11274 and tex2html_wrap_inline11346 the angles determining the point B. Here tex2html_wrap_inline11274 is the central angle determined by side AC, tex2html_wrap_inline11272 is the angle determined by AC and tex2html_wrap_inline11276 is the angle determined by AB. This gives us the following spherical coordinates for the vertices of the triangle:
eqnarray310

Using what we know about the dot product in tex2html_wrap_inline11320, we can find the cosine of the angle tex2html_wrap_inline11276.
displaymath11227
Now, in radian measure
displaymath11228
and we are done.

Why is this called the spherical Pythagorean theorem? First, note that it does give a relationship of the hypotenuse of a right triangle with the two sides of this triangle. To see how it is related to the classical Pythagorean theorem, recall the Maclaurin series (Taylor series at a=0) for the cosine function:
displaymath11229
Using this in the formula from the spherical Pythagorean theorem gives us:
eqnarray329
Now, as R gets arbitrarily large while a, b, and c stay small, we get the classical Pythagorean theorem as a limiting case.

Theorem 1.3:[Spherical Sine Theorem]  Let tex2html_wrap_inline11270 be a spherical triangle on a sphere of radius R. Let a, b, and c denote the lengths of the sides in radians, and let tex2html_wrap_inline11392, tex2html_wrap_inline11394 and tex2html_wrap_inline11396 denote the interior angles at each vertex. Then
displaymath11230

Proof: We will first prove this for a right triangle and then use that result to prove the general case. In addition we can restrict our attention to triangles that lie entirely within a quarter of a hemisphere.

tex2html_wrap11492

Let tex2html_wrap_inline11270 be a right triangle with right angle at C. Extend the radius OA to OA', where A' is the on the intersection of the tangent plane to the sphere at B with the line OA. Likewise, extend OC to OC'. This means that the triangles tex2html_wrap_inline11416 and tex2html_wrap_inline11418 are right triangles with right angle at B. Additionally tex2html_wrap_inline11422.

Now, we have other right angles as well. First, planes A'BC' and OBC are perpendicular, from above. Now, plane OCB is perpendiculare to plane OCA since tex2html_wrap_inline11432 is a right angle. Look at the two planes A'BC' and OAC. They intersect in the line tex2html_wrap_inline11438. Since tex2html_wrap_inline11438 lies in the plane OCA, and tex2html_wrap_inline11444 lies in the plane OCB, the two lines must be perpendicular. Thus, tex2html_wrap_inline11448 is a right triangle with right angle at C'. The classical Pythagorean theorem gives:
eqnarray375
Combining these equations gives us
displaymath11231
and so tex2html_wrap_inline11452 is a triangle with right angle at C'. Let tex2html_wrap_inline11272, tex2html_wrap_inline11274 and tex2html_wrap_inline11276 denote the angles at the center of the sphere determined by the arcs BC, AC, and AB, respectively. The definition of the sine function gives us:
displaymath11232
and so
displaymath11233
We can reverse the roles of A and B and this gives
displaymath11234
and so
displaymath11235

tex2html_wrap11494
Altitudes

Now, if we have an arbitrary triangle, we can construct the spherical analogue of the altitude to reduce the relation for two sides to the case of a right triangle, which we just finished. For example in the following figure, inserting the altitudes from A and from B, we get
displaymath11236
From the other altitude we have
displaymath11237
and the theorem follows.

These theorems show that there are exactly six related pieces of information determining a spherical triangle -- the three sides and the three angles. Given any three of these pieces you can determine the other three pieces. This means that similar triangles must be congruent! This is not immediate. We need one other piece which we are able to prove, but will not do so now.

Theorem 1.4: Suppose that tex2html_wrap_inline11270 is a right triangle on the sphere of radius R with a right angle at C. Then
displaymath11238

Homework

  1. Prove that two great circles intersect.
  2. Prove that two great circles bisect each other.
  3. The sphere of radius one can be considered as
    displaymath11239
    In these two coordinate systems determine the distance along great circles between two arbitrary points on the sphere as a function of the coordinates.
  4. A pole of a great circle is one of the endpoints of a diameter of the sphere perpendicular to the plane of the great circle. For example, the North and South poles are both poles for the equator. Prove that through a given point, not on a given great circle and not the pole of the great circle, there is a unique great circle ghrough the given point perpendicular to the given great circle.

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Next: Logic and the Axiomatic Up: Neutral and Non-Euclidean Geometries Previous: The Origins of Geometry

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