The geometry of the surface of a sphere
REFERENCE: Hartle
Suppose you are constrained to exist on the surface of
a sphere (for all intents and purposes you are). In
the geometry of the sphere, there is no such thing as
a "straight" line - at least not in the Euclidean
sense of the word.
Every path you take is curved,
following a great circle (whose center is the center
of the sphere).
For example, in Fig. 1, the great
circle of the equator is shown as a dotted line. If
you were to measure the three angles in a triangle,
you would find that their sum is always more than .
For example in the triangle shown, angles B and C
already add up to .
In flat geometry, the relationship between the
circumference C of a circle and its radius R is given
by C = 2R. For the surface of a sphere, however, you
can see from Fig. 2 that a different relationship will
hold.
If you start at some point (for example the
pole, as illustrated) and pace off a distance R, it
will be a curved distance.
You then pace off a
circumference C such that you are always a distance R
from the pole. You have defined a general circle of
radius
R
and
circumference
C.
To
find
the
relationship between R and C, suppose our sphere has a
radius (in 3-D) of a (so we don't confuse it with our
"surface" radius of R).
Then our spherical line
element
dS2 = dr2 + r2(d2 + sin2 d2)
A
C
B
Great
circle
[Fig. 1]
R
C
Relation
between radius
and
circumference
will have a constant radius r = a, so that dr = 0.
geometry of the surface of a sphere we have
dS2 = a2(d2 + sin2 d2).
[Fig. 2]
Thus for the
[Eq. 1]
Eq. 1 is a two-dimensional line element describing the geometry of the
surface of a sphere of radius a. As far as we are concerned, we live
in a 2-D world whose geometry is governed by Eq. 1.
Thus our 2-D
surface world is described using a subset of a 3-D geometry - in this
case a sphere.
We call the geometry of Eq. 1 the geometry of a 2sphere - meaning it is a 2-dimensional subset of a 3-dimensional
sphere.
Let's find a formula for the circumference at a particular fixed value
of  = 0. Then d = 0 so the Eq. 1 becomes
dS = a sin 0 d.
[Eq. 2]
For a full circumference,  varies between 0 and 2 so that
C = 2a sin 0.
[Eq. 3]
But R can be found by doing the integral of dl2 = rd, which translates
to
[Eq. 4]
dl2 = ad.
For  varying from 0 to 0 and we have
R = a0.
[Eq. 5]
Finally, to get the relationship beween circumference C and radius R
(as we measure it) we solve Eq. 5 for 0 and substitute into Eq. 3 to
get
[Eq. 6]
C = 2a sin (R/a).
Eq. 6 predicts that
Again we get C =
described by Eq. 1
that sin x  x for
reduces to C = 2R.
C = 0 for R = 0 (as expected). But suppose R = a?
0.
Is this reasonable?
Thus in the geometry
we get unusual results.
Suppose R << a?
Recall
x << 1.
Since R << a, then R/a << 1 and Eq. 6
In principle, you could work backwards from Eq. 6 to determine Eq. 1.
Thus, if you gathered experimental data that showed that Eq. 6 held,
you could infer that the geometry of Eq. 1 held. Thus
C = 2a sin (R/a)  dS2 = a2(d2 + sin2 d2).
[Eq. 7]
If, on the other hand, you gathered experimental data that showed that
C = 2R, you could infer that you lived a world whose geometry was
determined by dS2 = dx2 + dy2. Thus
C = 2R  dS2 = dx2 + dy2.
[Eq. 8]
It is in this same way that we can determine what kind of geometries
the universe is described by.
Experiments can be devised to discover
such geometries. The plural "geometries" is used because the universe
exhibits a variety of geometries.
It turns out that the presence of
matter affects the geometry of spacetime.
Hence, the geometry near
earth is slightly different from that near the sun, which is very
different from that near a black hole, which is very different from
that in intergalactic space, where the effects of matter on the
geometry of spacetime are minimal.
Another unusual result of the geometry of Eq. 1 is that the sum of the
angles in a triangle is not  (again, see Fig. 1). Let , , and  be
the measures of three interior angles of a triangle in a geometry
described by Eq. 1. Then
( +  + ) =  + A/a2,
[Eq. 9]
where A is the area of the triangle.
Pole
We will not prove Eq. 9 in general, but it can be proved
from Eq. 1. In the problem set you will prove Eq. 9 for
the specific situation where the base of the triangle is
on the equator, and the other two sides pass through the
pole as shown in Fig. 3. This proof uses the surface area
formula for a sphere:
Asphere = 4a2.



Equator
[Fig. 3]