If (M,g) is a Riemannian manifold, and f(m) is a positive real-valued function on M, then f.g is another Riemannian metric on M.
If I know all the g-geodesics from x to y in M, can I find out the (f.g)-geodesics from x to y in M?
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They do not change if you use constant function for scaling. – Moishe Kohan Mar 27 '14 at 05:32
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In this answer I discuss a little how the geodesic equation changes under conformal rescaling. Some insights on how the geodesics are treated in conformal geometry can be found in this answer. – Yuri Vyatkin Mar 29 '14 at 06:30
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Almost certainly not. Think of the real plane. Geodesics from the origin are all rays from the origin. Now suppose that you multiply the metric on $[0, 1] \times [0, 1]$ by, say, $100$ (with an appropriate smooth fading off to $1$ as you approach the edge of the region). Then the new geodesic from $(0, 0)$ to, say, $(1.1, 1)$ will basically run out along the $x$ axis to $(1, 0)$, then almost vertically to the endpoint. Clearly by clever alteration of the scaling factor, (e.g., inclreasing it on the black squares of a checkerboard) I can make the geodesic be zig-zag in shape, etc. Basically, I think there's no hope for guessing it from the original geodesics.
John Hughes
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Hint: The answer is generally no you can't. This change in metric is called conformal change of the metric and there are many authors and articles discussed this change for a long time, just google it.
Semsem
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