Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.



Re: geodesic question
Posted:
Jul 21, 2011 10:00 AM


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 >things.
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 piecewisesmooth 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 ba, and equals ba 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 >1dimensional Riemannian manifold; in particular, on the sphere or in the plane) that can reasonably be said to have finite arclength 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 abovesupposing the generalized result is truebut then we'd be far outside "elementary proof" territory, though still nowhere near "calculus of variations" territory.)
Lee Rudolph



