# geodesics in differential geometry

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:

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

-
..what do you mean by 'straight line'? Is $M$ embedded in $\mathbb R^n$? –  Berci Oct 8 '12 at 0:43
straight line = zero curvature –  mary Oct 8 '12 at 1:17
but that's the definition of geodesic, too, no? –  Berci Oct 8 '12 at 1:22
@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. –  Qiaochu Yuan Oct 8 '12 at 2:27
@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"). –  Jesse Madnick Oct 10 '12 at 7:00

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....
My point is that if $\kappa = 0$, then $\kappa_g = 0$. This says exactly that straight lines are geodesics. –  Jesse Madnick Oct 11 '12 at 4:34
The converse is false: geodesics in $M$ are not necessarily straight lines. –  Jesse Madnick Oct 11 '12 at 4:34