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# Re: spherical geometry

• From: John Denker <jsd@AV8N.COM>
• Date: Wed, 18 Aug 2004 17:45:04 -0400

Carl E. Mungan wrote:
I just took a look at google and here is a splendid derivation of the
relation between area and angles of a spherical triangle:

http://math.rice.edu/~pcmi/sphere/gos4.html

I still prefer my way of looking at it.

For starters, I derive a more general result, concerning
the precession of a pointer when it is transported around
a path of _any_ shape, not restricted to triangles.

(I have no idea whether my derivation is original or not.
Most likely not, but I don't know the literature well
enough to know what attribution to cite. The result is
sufficiently intuitive that it must have been reinvented
on many occasions. The only halfway tricky step goes back
to Stokes if not earlier.)

1) We start with the observation that if you parallel-transport
a pointer along a given path from point A to point B, and
then transport it back to point A along the *same* path,
there is no net precession. This seems self-evident to
me, an immediate corollary of the definition of parallel
transport. You don't even need the equation that describes
parallel transport; just ask yourself how you would perform
parallel transport operationally ... Schild's ladder, or
masking tape, or whatever. It's *got* to be reversible.

If this is not 100% obvious to you, you might want to
spend some time playing with masking-tape geodesics, as
described at
http://www.av8n.com/physics/geodesics.htm

Then, if you are still mystified by this step, get back
to me and I'll make another attempt to explain it.

2) Please refer to the diagram I created just now:
http://www.av8n.com/physics/img48/stokes.png

This is similar to the figure that you draw whenever
you try to explain Stokes' Theorem ... although we
are putting the figure to a slightly different use.

3) We start in the lower-left corner; call this "figure 1a".
There are four three-sided shapes. (They do not need to be
triangular, or even three-sided; they just need to "fit
together" with their neighbors, the way jigsaw puzzle pieces
do.) Imagine four separate experiments, transporting a pointer
around each of the four shapes, in the indicated direction.
Imagine how much precession there would be.

4) We shift attention to the upper center; call this "figure 1b".
It is almost a three-sided figure, but each side is indented with
a narrow dart. Imagine parallel-transporting a pointer around
this figure in the indicated direction.

5) In the lower-left corner ("figure 1c") is the same thing,
without the darts.

6) Note that each of the sub-figures can be made only
infinitesimally different from the others. We can pass from
1b to 1a by making the darts longer until the big shape falls
apart into four smaller shapes. We can pass from 1b to 1c by
making the darts narrower until they vanish entirely. We
can pass from 1a to 1c by pushing the four shapes together
until they touch.

7) Putting together all the ideas: The precession around
figure 1c is the same as the precession around figure 1b,
because the darts make no net contribution. The precession
in figure 1b is the same as the total of the four precessions
in figure 1a.

8) We conclude that we can find the precession in the
large shape (1c) by integrating the precession in the
sub-shapes (1a) that make up its area.

9) This tells us there is a certain amount of precession
associated with each patch of area. We then invoke the
symmetry of the sphere to argue that the precession per
unit area is the same everywhere on the sphere. So the
total precession is proportional to area.

10) To determine the proportionality constant, one
reference point suffices. The 90-90-90 spherical
triangle has area (pi/2) R^2 and precession (pi/2),
so in general the rule must be:
precession = area / R^2
where R is the radius of the sphere.

11) It should go without saying that you can apply the "dart
decomposition" recursively, to chop an area into arbitrarily
small pieces.

=================

For me the critical step was visualizing the "dart decomposition"
shown in figure 1b. A millisecond after seeing that figure, I
knew the precession was going to be expressible as an integral
over area.

I guess another ingredient is some familiarity with Stokes'
Theorem. It's a good thing to know ... not just the statement
of the theorem, but the logic behind it.

Also: They say "education is the process of cultivating
your intuition". I have spent enough time educating myself
(using masking-tape geodesics etc.) that it seemed intuitively
clear that parallel transport was reversible, so going out-and-back
along a dart would make no net contribution.