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*From*: John Denker <jsd@AV8N.COM>*Date*: Sun, 15 Aug 2004 15:26:20 -0400

David Bowman wrote:

.....

For a lot of extra credit points you can also find the proper formula

for the *area* of this spherical equilateral triangle in terms of the

length s of the sides of the triangle

I would argue that this "extra credit" part of the assignment

is far and away the most worthwhile part ... because if you do

this, you get the other parts practically for free.

For example, suppose we want to answer the very physically-

significant question of how much a pointer precesses if

flatlanders carry it along during a trip around the spherical

triangle. This involves "parallel transport" i.e. at each

step the new pointer orientation is aligned with (parallel to)

the old pointer orientation.

Lemma: To convince yourself that this is an interesting

question, consider the example of the spherical triangle

that has the following properties:

-- sides: length s = (pi/2) R

-- interior angles: three right angles

-- area: (pi/2) R^2

-- solid angle: one octant i.e. (pi/2)

WLoG we start out at the north pole, and depart southward

along the meridian of St. Louis (longitude 90W). Upon

reaching the equator we make a 90 degree left turn. Upon

reaching the meridian of Greenwich (longitude 0) we make

another 90 degree turn and proceed north to the pole.

Suppose the pointer is initially transverse to the path.

It will remain so all during the first leg. During the

second (eastbound) leg, the pointer will be parallel to

the path, not because the axis has changed, but because the

path is 90 degrees different. During the third (northbound)

leg, the pointer will again be transferse to the path.

Upon reaching the pole, we find that the pointer has precessed

90 degrees: it was initially transverse to the meridian of

St. Louis and now it is transverse to the meridian of Greenwich.

Note that this is a _flatland_ pointer: it necessarily

always lies along the surface; it cannot stick up into the

third dimension.

Also note that in a non-curved space, or in a situation where

the curvature is negligible because s is small compared to R,

traveling around a triangle does not cause any precession.

Now we argue that the amount of precession is proportional to

the area of the triangle. To see this, divide the triangle

into little cells. The cells won't necessarily be triangular,

but they will still be cells. Mark a direction of circulation

on each cell, all circulating the same way. This is like the

picture that you draw whenever you are explaining Stokes'

theorem, e.g. Figure 3-9 in Feynman volume II ... but use

three-sided cells instead of four-sided cells.

1) You can convice yourself that there is a serpentine path

that goes once around each of the little cells in the correct

direction.

2) The pointer remembers how much it has precessed, so each

new bit of precession just adds to whatever precession has

already occurred. Therefore the total precession is just the

sum of the contributions from the cells. This is one way of

calculating the total.

3) Another way of calculating the total is to decide that the

_interior_ lines of the diagram (where one cell touches another

cell) don't count, because you traverse each of these lines

twice, once in each direction. Whatever precession occurred

during the forward traversal is undone by the reverse traversal.

Therefore we get the same answer if we ignore all the little

cells and just travel around the exterior of the diagram, i.e.

the original triangle.

This tells us that the amount of precession is proportional

to area. The remaining work is to determine the constant of

proportionality. We can obtain this from the aforementioned

90-90-90 triangle. It has an area of (pi/2) R^2 and a

precession of (pi/2). So we conclude that the bottom-line

answer is incredibly simple:

precession = area / R^2 [1]

You can check that this has all sorts of sensible properties:

*) Precession=0 when s=0

*) If we expand precession in terms of (s/R), there is no

linear term.

*) The first nontrivial term is (s/R)^2.

Tangentially we remark that for non-spherical spaces, R^2

gets replaced by the _product_ of two radii of curvature.

As an important check, consider a cylinder of radius R.

The radius of curvature in one direction is just R, while

the radius of curvature in the other direction is infinite,

so the denominator in equation [1] is infinite, predicting

no precession ... which is correct.

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