# Spherical Geometry

** Spherical geometry** is the study of geometric objects located on the surface of a sphere. Spherical geometry works similarly to

**in that there still exist points, lines, and angles. For instance, a "line" between two points on a sphere is actually a great circle of the sphere, which is also the projection of a line in three-dimensional space onto the sphere.**

*Euclidean geometry*However, *it differs from typical Euclidean geometry* in several substantial ways:

① There are no parallel lines in spherical geometry. In fact, all great circles intersect in two antipodal points.

② Angles in a triangle (each side of which is an arc of a great circle) add up to more than \(180\) degrees.

③ Line segments (arcs of great circles) have bounded length, and regions on the surface of the sphere have bounded area.

Spherical geometry is useful for accurate calculations of angle measure, area, and distance on Earth; the study of astronomy, cosmology, and navigation; and applications of stereographic projection throughout complex analysis, linear algebra, and arithmetic geometry. Whether or not they know it, nearly everybody relies on spherical geometry some time. Just consider the best way to chart a plane's course.

#### Contents

## Distance

The minimal distance between two points is known as a geodesic and is the spherical analog of a line segment: an arc of a great circle. The distance between two points is therefore \(R \varphi\), where \(R\) is the radius of the sphere and \(\varphi\) is the measure (in radians) of the central angle subtended by the radii to the two points.

What is the minimal distance

on the sphere, centered at the origin and of radius \(2\), between points \((1, \, 1, \, \sqrt{2})\) and \((-1, \, 1, \, \sqrt{2})\)?

The dot product of the vectors to the two points is \(R^2 \cos \varphi = 1 \cdot (-1) + 1 \cdot 1 + \sqrt{2} \cdot \sqrt{2} = 2\), implying \(\cos \varphi = \tfrac{1}{2}\). Therefore, \(\varphi = \tfrac{\pi}{3}\), and the distance along the sphere is \(R \varphi = \boxed{\frac{2\pi}{3}.}\)

The equator has a latitude of \(0^\text{o}.\) Find its azimuthal position as measured from the North Pole.

In order to get from the North Pole to the equator, one must travel \(\frac14\) of the way around the circumference of the earth, so the azimuthal position of the equator is \(\frac14\) of \(360^\text{o}\) or

\(90^\text{o}.\)So, since latitude is measured off the equator, the convention is actually \(90^\text{o}\) off of a true azimuthal measurement.

## Area

The angle between two geodesics is taken to be the angle between the planes of their great circles. The measure of that angle is obtained normally, yielding a value between \(0^\circ\) and \(180^\circ\).

A spherical triangle is a "triangle" whose vertices are on the sphere and whose edges are geodesics of the sphere.

If a spherical triangle has angles of measure (in radians) \(\alpha\), \(\beta\), and \(\gamma\), then it has area \((\alpha + \beta + \gamma - \pi) \cdot R^2\). All polygons may be created from triangles sharing edges.

A circle on a sphere has the same boundary as a sphere in three-dimensional space: namely, the intersection of a plane with the sphere. The interior of the circle is the portion of the sphere on one side of the boundary. A circle with radius \(r\) in three-dimensional space has area on the sphere \(2\pi R \cdot (R \pm \sqrt{R^2 - r^2})\), where the sign is determined by whether or not a great circle is contained within the interior of the circle.

Pretend the Earth is a perfectly spherical object with a radius of 4,000 miles. (In reality, the Earth is not perfectly spherical, and its average radius is about 1% less than that.)

What is the area of the region within \(\dfrac{4000\pi}{3}\) miles from the North Pole?

## Properties

Just as on Earth a straight line eventually becomes a great circle and a triangle is actually a spherical triangle, line refers to great circle and triangle refers to spherical triangle in this section.

All similar triangles are congruent.

This means that the angles *completely* characterize a triangle! So there is no such thing as similar triangles of different sizes.

Given a line \(l\), there exist two antipodal points through which any line perpendicular to \(l\) must pass.

Upon remembering a line really is a great circle, this makes a bit more sense; the antipodal points also determine the diameter perpendicular to the interior of the great circle.

Identifying each point with the point antipodal to it gives rise to the real projective plane.

The projective plane helps provide a stronger understanding of geometric concepts, both in projective and Euclidean geometry, and the connection between spherical geometry and projective geometry yields a variety of results in complex and real mathematics.

The projection \((x, \, y, \, z) \mapsto (\frac{x}{R - z}, \, \frac{y}{R - z})\) is a bijective, smooth map from points on the sphere to the real plane and the point at infinity. This projection preserves angle measure.

Stereographic projection has a variety of other nice properties that help make spherical geometry even more powerful.

## Spherical Coordinates

Three-dimensional Cartesian space could be interpreted with spherical geometry by selecting various values of the sphere radius \(R = \rho\). The point \((x, \, y, \, z)\) may be converted into spherical coordinates with the following formulas.

\[ \begin{align*} x = \rho \sin \varphi \cos \theta, \\ y = \rho \sin \varphi \sin \theta, \\ z = \rho \cos \varphi. \end{align*} \]

Then, \(x^2 + y^2 + z^2 = \rho^2\). Here, \(\rho\) represents the radius of the sphere, \(\varphi\) represents the angle off of the \(z\)-axis, and \(\theta\) represents the angle off of the \(x\)-axis. Note that \(\theta\) serves the same purpose as in polar coordinates. By convention, the variables take values satisfying \(\rho \ge 0\), \(0 \le \theta < 2\pi\), and \(0 \le \varphi \le \pi\). With the exception of the origin, there is a one-to-one correspondence between \((x, \, y, \, z)\) and \((\rho, \, \theta, \, \varphi)\).

Spherical coordinates are useful because they allow triple integrals to be rewritten. For instance, given a multivariable function \(F\),

\[ \int \int \int_E F(x, \, y, \, z) \, dV = \int \int \int_E \rho^2 \sin \varphi F(\rho \sin \varphi \cos \theta, \, \rho \sin \varphi \sin \theta, \, \rho \cos \varphi) \, d\rho \, d\theta \, d\varphi.\]

This allows for more types of triple integrals to be evaluated and provides another method for certain cases of volume of revolution.

## See Also

**Cite as:**Spherical Geometry.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/spherical-geometry/