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Variational minimization is one approach. The details of the proof would come down to what sort of space you're working in, though. Euclidian spaces are a lot s
by wyager 10y ago
Variational minimization is one approach. The details of the proof would come down to what sort of space you're working in, though. Euclidian spaces are a lot simpler to prove this in than more complicated spaces with different notions of intrinsic and extrinsic curvature. For example, the shortest path on a sphere is also a line, but it's extrinsically curved and also might not be unique. On the other hand, the proof in a one-dimensional Euclidean space sort of comes from the definition of the metric over the space.
- wfunction 10y agoWe're just talking plain Euclidean space here. It's still not a basic geometry problem as the parent was suggesting.
- m00n 10y agoSince the notion of curved path (maybe aside from circular arcs) is not discussed in basic geometry, the problem would be impossible to state. If you assume given the notion of arc-length as the least-upper-bound on lengths of piecewise-linear paths between the two points, then the triangle inequality of basic geometry suffices. This seems not more hand-wavy a proof as something about "light-travelling-at-constant-speed" analogies.
- wfunction 10y ago> Since the notion of curved path (maybe aside from circular arcs) is not discussed in basic geometry, the problem would be impossible to state. >> It's still not a basic geometry problem seems we're in agreement?