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Why is the Kerr singularity like a ring — not a 3-D ellipsoid, or a point, or even a flattened disk? I need an intuitive answer, please.

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In Boyer-Lindquist coordinates, one of the invariant measures of curvature — the Kretschmann curvature scalar — is calculated to be:

$$ K=R_{\alpha\beta\gamma\delta}R^{\alpha\beta\gamma\delta} = \frac{48M^{2}\big{[} r^{6}-15a^{2}r^{4}\cos^{2}\theta+15a^{4}r^{2}\cos^{4}\theta -a^{6}\cos^{6}\theta \big{]}}{(r^{2}+a^{2}\cos^{2}\theta)^{6}}. $$

Wherever the Kretschmann scalar diverges, a curvature singularity is present. As you can see, in order for it to diverge in Kerr spacetime, we need both $$ r=0 \quad \text{and} \quad \theta=\frac{\pi}{2}.$$ So it is not a three-dimensional spacelike singularity (4 dimensions minus 1 constraint — $r=0$), as in the Schwarzschild metric, but a two-dimensional one that has a ring-like quality to it, related to $\theta$ being equal to $\frac{\pi}{2}$.

As for the physical interpretation, it might be difficult to try to give one. Extreme phenomena in physics, whether observed or just predicted — and singularities certainly count as such — do not always have everyday-life similarities.

Besides, in real-life scenarios, the interior of the inner horizon of the Kerr black hole is unstable under perturbations. The ring singularity might not even be present. To disturb the Kerr solution it might require as little as some interaction with weak gravitational waves. These are naturally abundant in the universe and a ring singularity might just not be feasible.

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    $\begingroup$ Schwarzschild singularity isn't timelike; it is spacelike, because the coordinate $t$ is spacelike in its vicinity. $\endgroup$ Oct 20 '21 at 16:28
  • $\begingroup$ Sorry, brain fog, of course! $\endgroup$
    – K.T.
    Oct 20 '21 at 16:30
  • $\begingroup$ Why was my question downvoted?! What's wrong with it? $\endgroup$
    – Nayeem1
    Oct 20 '21 at 17:42
  • $\begingroup$ @Nayeem1 Because s/one found it not useful/unclear/not showing a research effort? Hint: the question K.T. answered is How do you locate the singularities of the Kerr spacetime?; not the question you asked. 2nd hint: the research effort I expected from the title is how you came to the conclusion the singular locus should be a 3-D ellipsoid, or a point, or even a flattened disk. $\endgroup$ Dec 26 '21 at 14:35

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