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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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    ..what do you mean by 'straight line'? Is $M$ embedded in $\mathbb R^n$?2012-10-08
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    straight line = zero curvature2012-10-08
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    but that's the definition of geodesic, too, no?2012-10-08
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    Sorry, if gamma is a straight line, then we can say there exists a point xo that lies on every tangent line to gamma. Does this do the job?2012-10-08
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    @mary: no. Tangent lines don't lie in $M$, they lie in some tangent space. I don't see any way of answering this question that isn't tautologous. The correct definition of "straight line" _is_ geodesic.2012-10-08
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    There's an excellent pdf explaining geodesics on Riemannian manifolds here- http://page.mi.fu-berlin.de/atariah/Files/WhatIs_AGeodesic.pdf They also have links to some nice animations to build intuition.2012-10-08
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    @QiaochuYuan: I think the OP is considering a surface (isometrically immersed) in $\mathbb{R}^3$. By a "straight line in $M$," I think the OP means a geodesic in $\mathbb{R}^3$ that happens to lie in $M$. The question is then why such a geodesic in $\mathbb{R}^3$ is then a geodesic in $M$.2012-10-10
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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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    so then if kappa = 0 and kappag = 0, then the result is that kappan = 0 so everything is zero. But how can you incorporate this argument in an iff statement2012-10-11
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    My point is that if $\kappa = 0$, then $\kappa_g = 0$. This says exactly that straight lines are geodesics.2012-10-11
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    The converse is false: geodesics in $M$ are not necessarily straight lines.2012-10-11
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    how do we know that kappag = 0?2012-10-11
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    Why is it if kappa=0 then Kappasubg =0?2012-10-11
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    I'll let you think about that one on your own.2012-10-11