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In O'Neill's "The Geometry of Kerr Black Holes" he explicitly states that the $z$-axis has a ± isometry; a.k.a. Z2 symmetry of reflection in z=0 plane that exchanges south and north hemispheres. Is this necessary? Could it be replaced by "analytic continuation" or some such. This would imply the singular ring acquires a twist rotation, but that doesn't (superficially) appear to be unreasonable; either mathematically or physically.
Prior work/References would be fine :). Although, a succinct answer would save me some more "education". I have a lot of reference papers, and some books, but they seem to skirt justifying this symmetry. Plowing through this has really been educational, but it is getting out of hand.
Expanding/Explaining my interest: I find the unbalanced geometry immediately around the ring and the "time machine", arising from a seemingly reasonable differential manifold equations, disturbing.

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    $\begingroup$ during the formation or collision of the black hole(s) that will be the case since the process won't be perfectly symmetric, but after the ringdown the no hair theorem comes into play since all quadru- and higher multipoles get flattened out $\endgroup$
    – Yukterez
    Jun 18, 2022 at 22:40
  • $\begingroup$ As I mentioned below, I am thinking about after ringdown processes. In particular, perturbations around the ring that might not get out into "the wild"; universe. There are a series of papers attacking the physical situation from the viewpoint of the field of scalars. They did mention that the scalars might be observable effects; presumably by inferences, not by causal "forces". Unfortunately, I never studied (yet) the multipole radiation theory. Which might blow away any long term changes. $\endgroup$
    – rrogers
    Jun 19, 2022 at 2:42
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    $\begingroup$ What do you mean by “+- isometry of $z$-axis”? $\mathbb{Z}_2$ symmetry of reflection in $z=0$ plane that exchanges south and north hemispheres? $\endgroup$
    – A.V.S.
    Jun 19, 2022 at 4:23
  • $\begingroup$ @A.V.S. Yes, I will add that to the question. $\endgroup$
    – rrogers
    Jun 19, 2022 at 12:08

2 Answers 2

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Why is polar ($z$-axis) symmetry assumed in Kerr Black Holes?

Maybe I'm being too pedantic, by I would like to note that Kerr black hole is not a theorem to have assumptions. It is a family of solutions of Einstein field equations that all have a property, namely the $\mathbb{Z}_2$ symmetry of reflection in the equatorial plane. So in a sense this is a question of definition: if a black hole does not have this equatorial reflection symmetry it would not be a Kerr black hole.

On the other hand specific derivations of the Kerr metric may impose reflection symmetry as one of the simplifying assumptions. Alternatively, Kerr metric could be obtained by “spinning up” Schwarzschild solution, in this case equatorial reflection symmetry would already be present in the “seed” metric. No justification for imposing or relaxing this symmetry is needed except for the ability to arrive at solution of EFEs, if one could do this without equatorial reflection symmetry all the better.

There are indeed families of black hole solutions that generically do not have this reflection symmetry, for example Plebański–Demiański family of solutions (see here for a review). Kerr black holes could be obtained within that family by setting to zero all but two (mass and spin) of its parameters.

The prevalence of reflection symmetry in black hole metrics considered in the literature at least in part could be explained by the special role that Kerr (and Kerr–Newman, if we consider electrovacuum spacetimes) families of solutions play. And this role is guaranteed by various “no-hair theorems” that establish Kerr/Kerr–Newman families as the only “physically reasonable” stationary asymptotically flat black hole solutions. (There are a lot of technical details in multiple possible definitions of “reasonable”, see this review).

So a black hole without reflection symmetry must somehow evade these no-hair theorems.

First alternative is to consider matter that would allow nontrivial “hair” (matter fields outside the horizon) that would break reflection symmetry. For example this paper considers non-minimally coupled scalar field that would produce an egg-shaped event horizon:

Euclidean emdedding of asymmetric horizon geometry

Second alternative is to consider black holes that are not asymptotically flat. In the mentioned above Plebański–Demiański family solutions with non-zero NUT and/or acceleration parameters are not asymptotically flat and do not have equatorial reflection symmetry.

Third alternative is to consider black holes that are not stationary. For example, accretion disks or other forms of matter around black hole may break reflection symmetry and backreaction would ensure that this symmetry breaking carries over to the metric itself. Another possibility is that binary black hole coalescence can occur without reflection symmetry.

In standard general relativity deviations from equatorial reflection symmetry in realistic astrophysical scenarios are expected to be either very small or very transient, however they may become more pronounced for theories differing substantially from GR in the strong gravity regions.

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    $\begingroup$ But why does the interior of.a Kerr black hole have symmetry? An asymmetric interior and a symmetric exterior wouldn't violate the no-hair theorem. $\endgroup$ Jun 20, 2022 at 20:43
  • $\begingroup$ I marked this as a good answer. But, I am thinking more about "late arrivals" to the Black hole; that is, deliberately (?) inserted masses. I am going over the paper I mentioned; and tracking the history of uniqueness. That paper seemed pretty adamant. I know of the causality constraints of the differential equations Cauchy initial conditions, but am not familiar with the actual theorems and proofs. So I have to tread carefully if I want to "believe"; one way or the other. Thanks! I will read your references and use Sagemanifolds to verify the deductions. $\endgroup$
    – rrogers
    Jun 21, 2022 at 2:14
  • $\begingroup$ @PeterShor: Analyticity would prevent this. But if we relax the smoothness to $C^k$ for some small $k$, then yes, I believe it is possible to cut and paste into the interior the asymmetric region. $\endgroup$
    – A.V.S.
    Jun 21, 2022 at 13:03
  • $\begingroup$ @A.V.S.: if you form a Kerr black hole by smashing two Schwarzchild black holes together, it seems very unlikely to me that the two original singularities will ever merge. And in the limit of infinite time, won't you get a perfectly symmetrical Kerr black hole on the outside and some asymmetric interior? I guess analyticity might not preserved when you interchange limits, but it seems that assuming a symmetric solution for the interior is probably unphysical. $\endgroup$ Jun 21, 2022 at 14:16
  • $\begingroup$ @PeterShor Both your suggestion and my "dropping" something in later violate the "stationary" condition of the uniqueness proofs I have looked at so far. But that doesn't mean a lot; since I am just starting on them. I am still imagining the internal ring; ringing :) And the idea of the Time Machine gives new meaning to "stable" oscillation :) And you can't arbitrarily purport "friction" since we have superconductors; which, if left in an isolated part of space, would happily conduct current forever (it is said). $\endgroup$
    – rrogers
    Jun 23, 2022 at 12:18
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This is sort of a chicken-and-egg situation.

You get this symmetry in the math because that's the math the model produces.

The reason the Kerr model produces symmetry about the z-axis is because that's exactly what it's designed to do.

The Kerr model does not claim to model all rotating black holes. It models black holes that have that symmetry and are eternal. It's used because it's the simplest model that does describe a rotating black hole. Just like the basic Schwarzchild model it is eternal and unchanging so it does not model real black holes which can't be eternal and unchanging.

It's a model that does "enough" to be useful without getting too complex to work with.

Extending the model really hits a basic problem : the Einstein Field Equations are very hard to find solutions for. It's not that people don't try, but the resulting metrics, if they can be found in a closed form at all, tend to have limited applications. For many purposes the Kerr or Kerr-Newman models are more than adequate.

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  • $\begingroup$ So it is an axiom; which may or not be assumed? So an infalling mass, say into the polar region, could have physical constants that might not be portioned into mass and Kerr angular momentum; and usually these are assumed to be radiated out; through the horizons? Without proof? $\endgroup$
    – rrogers
    Jun 19, 2022 at 12:42
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    $\begingroup$ Actually you can prove that all stationary asymptotically flat solutions belong to the Kerr family of solutions. $\endgroup$
    – TimRias
    Jun 19, 2022 at 15:07
  • $\begingroup$ @TimRias Would you consider: "Uniqueness properties of the Kerr metric M Mars" arxiv.org/pdf/gr-qc/0004018.pdf an adequate read/proof? $\endgroup$
    – rrogers
    Jun 20, 2022 at 11:28

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