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Let gamma be a straight line in a surface M. How can we prove that gamma is a geodesic?

ALl I note is that a geodesic on a surface M is a unit speed curve on M with geodesic curvature = 0 everywhere.

Update: To making it not look like the question is a tautology, check out this:

http://www.physicsforums.com/showthread.php?t=407105

I'm trying to fill in the gaps and understand the argument for proving this theorem. Thanks

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    @Berci: Yes and no. The definition of a geodesic _in_ $\mathbb{R}^3$ is that $\kappa = 0$ (what the OP refers to as "zero curvature"). The definition of a geodesic _in_ $M$ is $\kappa_g = 0$ ("zero geodesic curvature").2012-10-10

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Use the formula $\kappa^2 = \kappa_g^2 + \kappa_n^2.$

Here, $\kappa_g$ is the geodesic curvature, $\kappa_n$ is the normal curvature, and $\kappa$ is (unfortunately) just called the "curvature" (it is the $\kappa$ that appears in the Frenet-Serret Formulas).

A straight line has $\kappa = 0$, so....

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    I'll let you think about that one on your own.2012-10-11