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I have the following metric $$ ds^2 = dt^2-dx^2 $$ and I wanted to prove to myself that the shortest path for this metric is straight. I used the following relation $x=f(t)$ and $$ S = \int_{t_1}^{t_2} \sqrt{1-[f'(t)]^2} \ dt $$ If I minimize $S$, I obtain the condition $f''(t)=0$, which implies $x=ct+x_0$, being a straight line in Euclidean space. I have difficulty understanding what it means to minimize hyperbolic path distances, because the square of the distance between two points can be zero or negative. I believe my answer is not correct.

Is the shortest path in flat hyperbolic space straight relative to Euclidean space?

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    $\begingroup$ It's important to note that $\mathbb R^2$ with the metric $ds^2$ that you wrote down is not "hyperbolic space" -- instead, it's $2$-dimensional Minkowski space. Hyperbolic space is a Riemannian manifold with constant negative sectional curvature. $\endgroup$
    – Jack Lee
    Jun 18, 2015 at 18:21

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Your answer is almost correct. You have that straight lines are the critical points of the arc-lenght functional. These lines minimize arc-lenght if they are timelike or lightlike, but they maximize arc-lenght if spacelike. This is sort of expected, since the connection of this Lorentz plane is the same as the connection from $\Bbb R^2$ and the Christoffel symbols all vanish.

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Going back to basics

That a line is the shortest path just follows from the triangle inequality.

The triangle inequality is a theorem of neutral geometry (so valid in both euclidean and hyperbolic geometry )

what else is there to say?

maybe a link https://en.wikipedia.org/wiki/Triangle_inequality

in Euclid's elements it is proposition 20 of book 1

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  • $\begingroup$ Note that the OP uses "hyperbolic space" to refer to the real plane equipped with the flat Lorentz (i.e., Minkowski) metric, not the usual (Riemannian, non-flat) hyperbolic plane. :) $\endgroup$ Jun 13, 2015 at 15:58
  • $\begingroup$ The triangle inequality is in the other direction for timelike vectors.. $\endgroup$
    – Ivo Terek
    Jun 18, 2015 at 1:38
  • $\begingroup$ @AndrewD.Hwang moved this point to a seperate question math.stackexchange.com/questions/1329906/… $\endgroup$
    – Willemien
    Jun 18, 2015 at 8:41
  • $\begingroup$ @Willemien I gave an answer to your follow-up question there, I hope it helps. $\endgroup$
    – Ivo Terek
    Jun 18, 2015 at 10:30

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