How some metric tensors represent Black Hole?

Question

If we have a metric tensor then how do we know that it represents a Black Hole or not? Like how Schwarzschild and Kerr metric tensors represent Black Hole?

I know all metric tensors satisfies Einstein Field equations (somehow) so that is obviously not enough so how can we quickly just see the metric tensor and decide whether it is going to be a black Hole or not? What parameters do we need?

One way to do is to calculate Kretchmann scalar invariant which is not handy (one needs a software to calculate). Are there any other scalar invariants ?

Answer 1

A black hole is a region in a spacetime $M$ which, once entered, cannot be left by any physically relevant object. More rigorously, this means that there are no future-directed causal curves starting inside the region and ending outside of it. A curve is called causal if at each point along the curve, we have for the tangent $\xi^a$ $$\label{causal}\tag{1} g_{ab}\xi^a\xi^b\leq 0 $$ Where $g_{ab}$ is the metric, assumed to have signature $(-+++)$. "Future-directed" is essentially a convention. We will take it to mean "along the direction of increasing curve parameter". Now consider a submanifold $U\in M$ such that its topological closure $\bar U$ is compact, and two points $p\in U$ as well as $q\not\in U$. If there is now no curve $$ \begin{split} \gamma:[0,1]&\to M\\ s\;\;&\mapsto\gamma(s) \end{split} $$ With $\gamma(0)=p$ as well as $\gamma(1)=q$ and tangent $\xi^a$ satisfying the condition $(\ref{causal})$ for all $s\in[0,1]$, then we have no causal curves leading out of the region $U$, making it a black hole. Note that this definition makes no reference to singularities at all. The assumption that the closure $\bar U$ is compact is crucial here. Without this assumption, the region behind the cosmological event horizon, present due to the accelerated expansion of the universe, would be a black hole. This is obviously nonsensical.

Answer 2

Don't say to draw spacetime diagram

That will make things more difficult.

Either way, there are a couple things that all (reasonable) black hole metrics do:

1. They have an event horizon. If they don't, then they're either a stellar/planetary model that has mass in the middle, or they're a naked singularity, which is bad. Typically event horizons will look like coordinate singularities, which manifest in the metric tensor's components as zeroes or infinities (except at $r=0$, which is typically a "physical" singularity).
2. They are gravitating. I can come up with an exotic black hole thing that doesn't gravitate, but in most reasonable circumstances, they will. You can check for gravitational forces with the connection coefficients, which represent the apparent coordinate acceleration of geodesics near the black hole.

If it's not charged, it'll also be a vacuum solution to the Einstein field equations.

Technically, though, it can be very difficult to ascribe physical meaning to arbitrary metrics. The lack of a preferred coordinate system/reference frame means that any sufficiently-weird metric can be written as a simple one, and vice versa; “black hole”, as an astrophysical term, has come to only be applied to a set of very specific solutions that we understand very well, usually the Schwarzschild, Reissner-Nördstrom, Kerr, and Kerr-Newman solutions (and sometimes the Kerr-Newman-de Sitter solution too).

Answer 3

If the Kretschmann scalar blows up you know there is a coordinate independend singularity. That could still be a naked singularity or a big bang singularity though, but if you assume the cosmic censorship theorem that prohibits naked singularities the remaining options are a black hole, a big bang, a big crunch or a big rip.

Answer 4

Simply put, if curvature invariants diverge there is a singularity. In general, when physicists say "here is a singularity theorem", they really mean "there is a geodesic incompleteness" in this spacetime. Remember that solutions to the field equations are manifolds $(M , g)$ where the metric $g$ is found from the field equations, which should clear the "somehow" part of your question. Geodesic incompleteness is a very well understood thing and can be predicted with fairly straightforward arguments with focusing of null rays and energy conditions. And no, the Kretschmann scalar can be found without the need of a software. It is just based on the Riemann tensor, so anything like $R_{abcd}R^{abcd}$, $R_{ab}R^{ab}$, or whatever other curvature invariants you can cook up can be found fairly simply. Or use a software, your wish.

Metric tensors represent black holes because they contain terms that show a curvature invariant blow up. If your metric is of the form $ds^{2}=-dt^{2}f(r)+d\mathbf{x}^{2}g(r)$ where either $f$ or $g$ possess some sort of a pole at $r=0$ or whatever, you can guess the Riemann tensor and subsequent scalar curvatures have some sort of a corresponding pole or singularity as well. Geodesic incompleteness is just this feature where some parameter $\tau $ cannot be extended beyond a certain geodesic distance; although a Minkowski spacetime with $r=0$ deleted is still geodesically incomplete without an actual curvature singularity, but this is a mathematical singularity and not a physical one.

Tl;dr: metric terms tell you which curvature invariants blow up. It isn't that hard, take Schwarzschild metric, find the Christoffel symbols and subsequent curvature terms and see for yourself.