Geodesic Incompleteness and the Kretschmann Scalar

Question

If my understanding is correct, singularities (whether time-like or space-like) are defined by geodesic incompleteness. Since it is easier, we use the Kretschmann scalar $R^{\alpha\beta\gamma\delta}R_{\alpha\beta\gamma\delta}$ to identify whether the given metric has a true singularity or not.

My question is,

1. Is there an example where it is geodesic complete, but the Kretschmann scalar diverges.

2. Or the other way around, where it is geodesic incomplete, but the Kretschmann scalar is finite.

3. Is there a criterion for what kind of metric, one can use the Kretschmann scalar to identify singularities.

I am trying to wrap my head around this relation between geodesic incompleteness and the Kretschmann scalar (and other quantities I heard that are used). So it would be great for some clarification between these two.

Answer 1

Singularities are defined properly by the following condition : For a spacetime $M$, such that every standard measurable quantity on it is defined (ie we won't allow the Riemann tensor to be divergent at a point and so forth), that spacetime is singular if there exists inextendible curves of bounded acceleration for which the curve's half-length (measured by some parallel transport of a tetrad) is finite.

These are a lot of specific conditions, but there are a lot of types of singularities, or non-singular spacetimes that may appear to be singular otherwise. Here's a few things :

If a spacetime is geodesically complete, it is still possible that it is singular. There are a few explicit examples of this [1][2]. They are fairly artificial examples, but they do exist, and although not explicitely said, the Riemann tensor, and therefore the Kretschmann scalar, diverges along non-geodesic curves (you can see it because part of the construction involves sections of spacetime conformally equivalent to AdS space, with factor $2^{-n}$, $n \to \infty$).

In the opposite direction, there are many examples where the spacetime has curvature scalars well-defined, but is singular. The big example are quasi-regular singularities, for instance. The example of a cone minus its apex is a standard example : it is locally flat, singular at the apex, but no extension of it is possible. As far as spacetimes go, there are many cases, such as the spacetimes for cosmic strings with angular deficits, or the Israel-Khan spacetime. Another case is when all curvature scalars are well-behaved, but the Riemann tensor evaluated with some local tetrad is divergent, such as the conformal Misner singularities, or tilted homogeneous cosmologies.

The singularities for which the Kretschmann scalar is divergent are a type of scalar singularities (singularities where polynomial scalar quantities built from the derivatives of the Riemann tensor diverge). I don't know if there is a specific test for metrics for this, outside of the obvious.

Answer 2

There is a lot more than one curvature scalar out there. When considering scalars strictly linear in the curvature tensor, the only independent scalar is the Ricci scalar $R$. When considering scalars quadratic in the curvature tensor, one can choose the set of three independent scalars to be the Kretschmann scalar $K_1 = R^{\mu\nu\kappa\lambda}R_{\mu\nu\kappa\lambda}$, the Chern-Pontryagin scalar $K_2 = R_{\mu\nu \kappa\lambda}\epsilon^{\mu\nu\gamma\delta}R^{\kappa\lambda}_{\;\;\;\gamma\delta}$, and the Euler scalar $K_3 = R_{\mu\nu\kappa\lambda}R_{\gamma\delta\phi\omega} \epsilon^{\mu\nu \gamma \delta} \epsilon^{\kappa\lambda\phi \omega}$. (In vacuum, $K_2$ and $K_3$ are not independent).

However, even this list is not exhaustive, since you can still take covariant derivatives of the curvature tensor and keep contracting. The tensor itself has 20 independent components and there is physical information in it that has to be captured in other ways. Probably the most robust approach is to choose a tetrad and make really sure it is well normalized and defined everywhere. If and only if you did everything right and the components of the curvature tensor blow up, you are dealing with a curvature singularity. When one sees the curvature "gradually" blowing up in any frame, it is very easy to show that this implies geodesic incompleteness.

Nevertheless, there a few sneaky curvature singularities, in particular those that are of a distributional character, which diverge "only at a single point", i.e., they correspond to the Dirac delta function or its derivatives. These will also provide a counter-example to the statement that a curvature singularity means geodesic incompleteness.

Consider a razor-thin disk of matter in your space-time located at a coordinate $z=0$. This corresponds to a stress-energy tensor $\sim \delta(z)$, and thus a diverging Ricci scalar precisely at $z=0$. However, it is possible to uniquely extend geodesics through the disk if the geodesic equation is solved in an integral form across the jump. I.e., a localized (distributional) curvature singularity does not need to mean geodesic incompleteness. Specifically, it is not a big deal as long as it is sufficiently extended. For example, a matter source located on an infinitely thin string will cause geodesic incompleteness for geodesics trying to pass through the string.