Re: spherical geometry (was Re: navigation riddle)

[John Denker <jsd@AV8N.COM>]

David Bowman wrote:

> John's comments and argument noting and arguing for the
> proportionality of the enclosed area and the precession (or
> excess turning) angle are *extremely* useful in helping to do the
> problem and its extra credit part in a relatively painless manner.
> However, I seem to disagree with the first quote above assessing
> what part of the problem is "free" and what part is the key to what
> other part.

Fair enough.  The area <==> angles formula is easier to
derive in the area ==> angles direction, but once you've
got the formula it makes sense to turn it around and use
it in the angles ==> area direction.

For those who want a hint on how to find the angles, let me
point out that we have here a veritable poster child for
geometric algebra.  You can do the problem on a half sheet
of paper, without cramping the writing.

Here's the outline:

 1) The legs of the triangle are by definition geodesically
  straight.
 2) That means each leg is a piece of a great circle, since
  we are dealing with a spherical world.
 3) There is a one-to-one correspondence between a great circle
  and the set of rotation operators that rotate things in the
  plane containing that great circle.
 4) There is also a one-to-one correspondence between a great circle
  and a unit bivector.  The bivector is the generator of rotations
  along that great circle.  To specify a _particular_ rotation
  you need not only the generator (a bivector telling you the
  orientation) but also an angle (a scalar telling you how far
  to rotate).  To represent a great circle, find two orthonormal
  vectors in the plane of that circle and multiply them together.
  (FWIW these unit bivectors are isomorphic to quaternions, but
  today I'm just gonna call them unit bivectors.)
 5) WLoG assume that the base of our triangle lies along the
  gamma1/\gamma2 great circle.  WLoG let gamma1 be the vector
  from the center of the sphere to the middle of the base.
 6) To construct another leg of the triangle, make a copy of
  the gamma1/\gamma2 circle and rotate it in the gamma2/\gamma3
  direction by an angle A.  We don't know the value of A yet;
  we will solve for A later.  Then rotate this thing in the
  gamma1/\gamma2 plane by an angle -sigma/2, where sigma := s/R,
  where we were given s (the length of a side of the triangle)
  and R (the radius of the sphere).
 7) To construct the third leg, repeat the previous step, but
  reverse the sign of A and the sign of sigma.
 8) We have now constructed an isoceles triangle with base angles
  equal to A.  We need to solve for the apex angle.  If we then
  require that the apex angle is also equal to A, we have one
  equation in one unknown, telling us the value of A that
  produces an equilateral triangle.
 9) To find the angle at which two bivectors meet, just multiply
  them together.  The product has a scalar piece equal to cos(theta),
  and a grade=2 piece involving sin(theta).  For present purposes
  the scalar piece suffices to tell us what we need to know.
  Actually, to find the angle between B and C, it is convenient
  to multiply B by C~, where C~ is the reverse of C.  This gets
  rid of some ugly minus signs.  If you don't believe me,
  construct two bivectors having a known angle between them,
  multiply them, and see what you get.  The beautiful thing is
  that if it works for your example, it must work for all
  bivectors, because the operations involved (multiplication
  and projecting out the scalar part) are manifestly rotation-
  invariant.
 10) Here is the only place where plug-and-chug doesn't work; you
  need to actually draw the picture and/or think a little bit.
  When two planes (or bivectors, or great circles) meet, there
  are actually four angles:  theta, pi-theta, theta, pi-theta.
  The way I set up the problem, the formalism was giving me
  theta when I wanted pi-theta, so I had to intervene to fix it.

I get the following equation:
  (1+cos(sigma)) cos^2(A) + cos(A) - cos(sigma) = 0
It checks out in the obvious limiting cases i.e. sigma->0 and
sigma=pi/2.  Don't think of it as a scary transcendental
equation;  just solve the quadratic to find cos(A) in terms
of the given cos(sigma).

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

There must be other ways of solving this problem, but this
one seems elegant and convenient.  Also IMHO it's a nice
illustration of why people study physics:  a little bit of
formalism makes the problem a whole lot easier.  The effort
of learning the formalism pays for itself the second or
third time you use it.  And there are lots of applications
for this formalism, e.g. when you need to compound rotations
or compare rotations in three (or more!) dimensions.

For hints on how to use bivectors for rotating things (including
rotating other bivectors), see
  http://www.av8n.com/physics/rotations.htm