Geodesics in the Hyperbolic Plane

Question

In my the class the definition given for a curve $c(t)$ to be a geodesic is that $c''(t)$ is orthogonal to the surface at the point $c(t)$. The hyperbolic plane is the upper half plane with the metric: \begin{equation} ds^2 = \frac{1}{y^2}(dx^2 + dy^2) \end{equation} I have a fundamental misunderstanding somewhere, because I cannot think why the following curve is not a geodesic in $\mathbb{H}$: $c(t) = (t, t)$ for $t > 0$. This curve has second derivative as zero, but we have learnt the geodesics of $\mathbb{H}$ and this is not one of them. Where am I reasoning this incorrectly?

Answer 1

"A curve $c$ is geodesic when $c''$ is orthogonal to the surface" is for the case when your surface is isometrically embedded in $\Bbb R^n$ (usually $\Bbb R^3$) and $\Bbb R^n$ has the usual metric. Hyperbolic plane is not embedded in $\Bbb R^n$.