Re: spherical geometry (was Re: navigation riddle)

[John Denker <jsd@AV8N.COM>]

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.