(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 ). These are good choices to play the role of lines on our sphere. For example, given any two non-antipodal 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 .
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:
Theorem 1.1 On a sphere of radius R, a triangle with
interior angles , , and
has area given by:
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 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.
Proof(due to Euler, 1781). From our assumptions above, Assumption (4) tells us
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 , the lune given by and the lune given by . Each of these three lunes has an antipodal lune and the intersection of these three antipodal lunes is an antipodal triangle .
We will denote the lunes by the angle that includes them, BCA, BAC, and ABC.
Starting with the triangle and the lune ABC, add to them the lune
ACB. This covers the triangle three times. Now, add in the antipodal
lune to lune CAB and the antipodal triangle . This covers the
hemisphere determined by the line , counting the triangle three times. So we
Thus, we get our desired formula:
We call the value of 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 . 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 denote the sphere of radius R centered at the origin in .
In rectangular coordinates
and in 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. . 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
on a sphere of radius R with a right angle at vertex C and sides of length a, b, and
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 and the angles determining the point
B. Here is the central angle determined by side AC, is the angle
determined by AC and is the angle determined by AB. This gives us the
following spherical coordinates for the vertices of the triangle:
Using what we know about the dot product in , we can find the cosine of the
Now, in radian measure
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:
Using this in the formula from the spherical Pythagorean theorem gives us:
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 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 , and denote the interior angles at each vertex. Then
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.
Let 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 and are right triangles with right angle at B. Additionally .
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 is a
right angle. Look at the two planes A'BC' and OAC. They intersect in the line
. Since lies in the plane OCA, and lies in the
plane OCB, the two lines must be perpendicular. Thus, is a right triangle
with right angle at C'. The classical Pythagorean theorem gives:
Combining these equations gives us
and so is a triangle with right angle at C'. Let , and 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:
We can reverse the roles of A and B and this gives
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
From the other altitude we have
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 is a right triangle on the sphere of radius
R with a right angle at C. Then