Could someone explain how to go about proving that the meridian curves on a surface of revolution are geodesics?
Meridian curves of surfaces of revolution are geodesics
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$\begingroup$
differential-geometry
surfaces
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1I just got hold of the do Carmo book. I followed it until the top of page 256 where it says "Since the first fundamental form along the meridian u=const, v=v(s) yields $((f')^2+(g')^2)(v')^2=1$...". I don't understand how they got this from the first fundamental form? – 2012-02-08
3 Answers
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Hint: At any point along a geodesic, the normal of the geodesic is parallel to the normal of the surface.
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Alternative hint: A meridian is is the set of fixed points for a particular isometric transformation of the surface. Isometries preserve geodesics. At every point on a smooth surface, there is exactly one geodesic in each direction.
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Remembering Clairaut formula for geodesics after seeing how it was obtained is also useful. The unit normal to surface and unit normal in Frenet frame (without surface) are identical vectors for a running geodesic.
$ r_o= r*sin \psi$ ,$ \psi=0$ is the basic meridian.