# How is a non-rotating black hole a special case of a rotating black hole?

The Schwarzschild metric for an eternal non-rotating black hole is a special case of the Kerr metric for an eternal rotating black hole. But the Penrose diagrams for the maximal analytic extensions (MAEs) of these two spacetimes look qualitatively different from each other, as shown in sections 2.4 and 4.2 in https://arxiv.org/abs/gr-qc/9707012. Is MAE-Schwarzschild somehow still a special (or limiting) case of MAE-Kerr, despite this qualitative difference? Are there any simpler examples to illustrate this?

I understand that we don't expect these eternal black hole solutions to have direct relevance to astrophysics, but I would still like to understand this mathematical aspect of GR.

Since Penrose diagrams are a bit abstract, I'll add some spice to the question by relating it to the fates of infalling test-objects. When a test-object falls into a Schwarzschild black hole, it hits the singularity in a finite proper time. A test-object falling into a Kerr black hole along the axis of symmetry has a different fate. If I'm not mistaken (please correct me if I am), the infalling object never even reaches the plane of the ring singularity. Instead, it turns around and falls back out (by crossing a white hole horizon in the MAE), then falls back in (by crossing a different BH horizon in the MAE), then falls back out, then back in, and so on forever. This is qualitatively different than the fate of the test-object in the Schwarzschild case. Can the behavior of the test-object in the the Schwarzschild case be understood as a special (or limiting) case of the behavior of the axially-falling test-object in the Kerr case?

• There are several Penrose diagrams in section 4.2 of the Townsend paper. Is there a particular one you think is most relevant? – Ben Crowell Oct 25 '18 at 1:15
• @BenCrowell The one on page 84, immediately before section 4.2.1. There is no caption, but I was interpreting it as the Penrose diagram of a non-maximally-rotating Kerr black hole. I'm not very experienced with these extended solutions or their Penrose diagrams, so please correct me if I'm wrong. – Chiral Anomaly Oct 25 '18 at 1:33
• I don't know the answer, but I wouldn't wonder if Schwarzschild cannot be obtained as a limit case of Kerr. There are several examples in physics of this behaviour. A well-known case is the famous paradox of the two capacitors, one charged, the other uncharged. If you connect them, charge is shared, and energy appears not being conserved. But if the connecting wire has a non-null resistance $R$, you can see that the lacking energy is dissipated as Joule heat in $R$. The paradox is that the energy dissipated does not depend on $R$, so that the limit $R\to0$ would still give the same value. – Elio Fabri Oct 26 '18 at 6:27
• – Ben Crowell Oct 28 '18 at 17:17
• @ElioFabri, the two-capacitor missing energy paradox is only a paradox in the context of ideal circuit theory which essentially ignores loss of energy through electromagnetic radiation. Unless you're speculating that one is blocked from getting to Schwarzschild from Kerr due to some idealization that 'throws away' some relevant physics, I don't see the connection with the capacitor paradox. – Alfred Centauri Oct 29 '18 at 14:38

As one can easily check the metric functions of the Kerr spacetime written in any common coordinate systems will approach the Schwarzschild metric as the spin parameter $$a=J/M$$ approaches zero (mass $$M$$ is assumed constant). For example, Boyer-Lindquist coordinates would reduce to ordinary Schwarzschild coordinates. So for the patches of spacetimes covered by this coordinates we can say that Schwarzschild spacetime is a limit of the Kerr spacetime at every point away from singularities.
For the global structure of spacetime one should remember that while Penrose diagrams are useful tools, they do leave out a lot of information. In case of Kerr geometry, they do not give idea about how the geometry varies with varying spin parameter $$a$$. As $$a$$ approaches zero, the radial sizes of inner features of the Kerr metric: the inner horizon, ring singularity, CTC region and inner ergosphere all go to zero, while the values of curvature tensor at these features diverge. So if we ascribe to our test observer some realistic features such as a finite size and a maximal value of spacetime curvature she might endure before being destroyed, then no matter what values we choose, for a given mass of a black hole there would be a minimal value of $$a$$ for which there are trajectories that enter the black hole and leave the white hole in a parallel universe such that observer could survive the travel. For all smaller values of the spin parameter the observer would be destroyed by encountering a region of high curvature.
Thus the full Kerr MAE manifold in the limit $$a\to 0$$ becomes a countable sequence of Schwarzschild MAE manifolds disconnected from each other, as the regions inside the inner horizons become a sequence of Schwarzschild black and white hole singularities with finer features of Kerr solution becoming inaccessible to any physical observer.