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Let $S\subset {\mathbb{R}}^3$ be a smooth surface. Suppose its Gaussian curvature tends to infinity in absolute value along every maximal geodesic with a finite affine-parameter endpoint. Then $S$ is an inextendible embedded surface: a smooth extension would let some geodesic reach the extension boundary in finite time while its curvature remained locally bounded.

A Riemannian manifold is geodesically incomplete when some maximal geodesic has a finite endpoint in its affine-parameter interval.

Null geodesic incompleteness means that some maximal geodesic with a null tangent has a finite endpoint of its affine parameter. This can indicate a removable missing region or a genuine obstruction; it does not alone assert curvature divergence.

A spacetime is geodesically complete if every maximal geodesic has an affine parameter ranging over all of $\mathbb{R}$. For timelike geodesics this is equivalent to unbounded proper time in both directions. An extendible geodesic segment can be prolonged in the same spacetime; an inextendible geodesic cannot. A finite coordinate endpoint need not imply finite affine parameter.

For the Kruskal spacetime, use $U,V$ withIn the right exterior $U<0<V$. A truncated ray $U=U_0<0$, $V_1<V<V_2$ is an extendible geodesic of radial null type: neither artificial endpoint is a spacetime boundary. A future ray $U=U_0>0$ in the black hole reaches $UV=1$, hence $r=0$, and is inextendible geodesic and future null-geodesically incomplete. Its Killing energy $E\ne 0$ gives $dr/d\lambda =-|E|$, so the Schwarzschild singularity occurs at finite affine parameter. The maximal continuation toward the past supplies the other half of this same null geodesic.

There is no inextendible, complete radial timelike geodesic in positive-mass Kruskal spacetime. This requested example is impossible as printed. For a radial timelike geodesic, the conserved Killing energy and normalization giveIf $E^2<1$, $r$ has at most one turning point, a maximum $r_{\max }=2M/(1-E^2)$; a maximal trajectory runs from the white hole Schwarzschild singularity to the Schwarzschild singularity. If $E^2\ge 1$, there is no finite turning point; one end can lie at infinity but the other reaches $r=0$. The exceptional $E=0$ trajectory through the bifurcation surface also reaches $r=0$ in both time directions. Near $r=0$,whose integral is finite. Constant-$r$ radial timelike curves are accelerated, not geodesics.

Two plausible repairs have different meanings. Removing “radial” permits a complete circular timelike geodesic at $r=R>6M$, with nonzero angular momentum and proper time ranging over $\mathbb{R}$. Replacing “timelike” by “null” permits a complete horizon null geodesic: $U=0$, $V\in \mathbb{R}$, with $V$ an affine parameter. The Penrose diagram shows both repairs explicitly, together with the two valid requested examples; the circular trajectory is only a radial projection and is labelled as nonradial.

The horizontal boundaries are the Schwarzschild singularities, diagonal dashed lines are the Killing horizons, and outer diagonal edges are null infinity.