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Topic: geodesic question
Replies: 8   Last Post: Jul 30, 2011 2:30 PM

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Lee Rudolph

Posts: 3,143
Registered: 12/3/04
Re: geodesic question
Posted: Jul 21, 2011 10:00 AM
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mjc <martin.cohen@gmail.com> writes:

>On Jul 18, 11:43 am, Mark Steinberger <nyj...@gmail.com> wrote:
>Does anyone know an elementary proof that great circles in the sphere
>minimize arc length?
>Since arc length is defined as an integral over the path, it seems to
>me that you would inevitably be led to the calculus of variations.
>I don't know how being on a sphere (or even on a plane) would simplify

Here's how.

If a path minimizes arc length (however defined, subject
to being reasonable...) between its endpoints, it minimizes
arc length between any two of its points. If we agree to
limit the discussion to piecewise-smooth curves, therefore,
no further generality is lost by limiting it to smooth
curves parametrized by arc length, i.e., such that f'(t)
exists and has length 1 for all t in the interval on which
f is defined. Let f : [0,e] -> S^k be such a curve (say,
just to be on the safe side, with e less than 1/4 the arc
length of a great circle), let G: S^k -> R be the Riemannian
distance from f(0), and let DG be the unit gradient vector
field of G (defined at every point of S^k except f(0) and
-f(0)). By the Fundamental Theorem of Calculus, the
difference G(f(e))-G(f(0)) is the integral over [0,e]
of the inner product <DG(G(f(t))),f'(t)>, which is less
than or equal to e, and strictly less unless f'(t)=DG(G(f(t)))
for 0<t=<e. In the latter case, f *is* a parametrization
of (part of a) great circle; in the former case, the
arc of a great circle between f(e) and f(0) has length
strictly less than e, the length of f.

The same proof works for Euclidean space, _mutatis mutandis_;
but there it is easier to find a piece of your curve that is
smooth and short enough that you can apply the implicit
function theorem to it, putting you in the position of
having to show only that the arc length of the graph of
a differentiable function on [a,b] is not less than b-a,
and equals b-a if and only if the function is constant.
(Of course, in a sense that *is* the same proof; but
you don't have to utter the word "gradient", or
do any reparametrizing; you just have to notice that
the square root of 1+f'(t)^2 is at least 1, etc., etc.)

There *are* continuous parametrized curves (in any
>1-dimensional Riemannian manifold; in particular,
on the sphere or in the plane) that can reasonably
be said to have finite arc-length yet are not
differentiable on any interval (examples can easily
be constructed along the general lines of the
"snowflake" curve...I *think*). Presumably Mark
Steinberger's question, extended to them, still
has the expected answer, but I don't know how to
prove it. Maybe Gerald Edgar does? (There's
presumably a "geometric measure theory" proof
that generalizes the proof above--supposing the
generalized result is true--but then we'd be
far outside "elementary proof" territory, though
still nowhere near "calculus of variations"

Lee Rudolph

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