In a preprint posted on the arXiv, Roy Kerr claims that there is a widespread misunderstanding related to the singularity inside the black hole that bears his name.

  • Can anyone explain his argument in "simple terms"?

  • Are there any pitfalls in his reasoning?

Here's a link to the preprint: https://arxiv.org/abs/2312.00841

Do Black Holes have Singularities?

R. P. Kerr


There is no proof that black holes contain singularities when they are generated by real physical bodies. Roger Penrose claimed sixty years ago that trapped surfaces inevitably lead to light rays of finite affine length (FALL's). Penrose and Stephen Hawking then asserted that these must end in actual singularities. When they could not prove this they decreed it to be self evident. It is shown that there are counterexamples through every point in the Kerr metric. These are asymptotic to at least one event horizon and do not end in singularities.

Here are links to talks where he discussed these ideas:

  • $\begingroup$ What experiment could test this hypothesis? $\endgroup$
    – John Doty
    Commented Dec 3, 2023 at 15:49
  • 12
    $\begingroup$ Kerr's claim, and the singularity theorem he is discussing, are mathematical claims about solutions of Einstein's equations, so they can be resolved by thinking about these solutions. $\endgroup$
    – John Baez
    Commented Dec 7, 2023 at 10:11

8 Answers 8


As far as I can tell, Kerr's point is that the singularity theorem doesn't specify whether or not a collapsing object will collapse to a "real" singularity. As a reminder, singularity theorems are usually something of the form

Given a spacetime obeying [some causality conditions] with [some appropriate set of energy conditions], the spacetime is timelike/null incomplete

There are many such theorems, some about trapped surfaces, some just about the collapse of matter in general, but Kerr's point is that the notion of singularity here is simply the curve incompleteness of a singularity, and not the more specific notion of a curvature singularity.

There are many definitions of a singularity in general relativity which are loosely connected together, the most general condition being that a spacetime is singular if given an inextendible curve with bounded acceleration, and picking some point, the generalized affine length of that curve from that point to one end of the curve is finite. "Generalized affine length" is some notion meant to capture the notion of finiteness for a curve that includes null curves, since their actual length is always zero.

This notion of singularity is called a $b$-singularity[1], which is what Penrose's theorem[2] use and what Kerr calls into question, as those are very broad and includes singularities which do not involve the divergence of the curvature at all, such as regular boundary points, which are so benign they can be removed by considering slightly larger spacetimes, or quasi-regular singularities[3][4], which have well-behaved curvature but cannot be removed. If an object travels along such a curve, it will not notice anything strange along the way before reaching the singularity and then just stop existing.

Kerr's claim is to have found two such singularities. His first example is that of Schwarzschild spacetime, where outside of the black hole, a light ray that approaches the event horizon asymptotically has such a finite length, and never touches the singularity, while for the Kerr spacetime, he gives an example of a light ray staying between the inner and outer event horizon that never dips below the inner event horizon and is always of finite length. As the curvature is bounded in both of the examples he gives (the curvature is always smaller than on the horizon in those examples), it cannot diverge and therefore they are at worse quasi-regular.

Those examples are of course by themselves not enough to disprove what he claims, ie that spacetimes of those types don't have a curvature singularity (those spacetimes also have a curvature singularity), but he shows this way that the singularity theorems by themselves are not enough to prove that this happens systematically, since it remains possible that the singularities occurring during "real" collapses are of this kind. With no constraints from the singularity theorem to have the singularity inside the horizon and just the ones on the horizon, there could be solutions which are like the Kerr metric on the outside but have no singularity on the inside, for instance.

This is by the way not an entirely new observation, since Hawking mentionned it himself in his original paper on his singularity theorem :

enter image description here

But Kerr's paper gives an example of such an occurence on pretty well-known spacetime examples.

Edit : Some notes :

  • The original type of singularities of Penrose's paper are actually $g$-singularities (singularities such that the curve in question is a geodesic), as the notion of $b$-singularity only came up later. This does not change the argument much.
  • While those singularities are not such that scalar curvature functions diverge, I did not check whether they could be non-scalar singularities (where the Riemann tensor evaluated along a parallel-transported tetrad along the curve does not converge). Those are pretty rare and as the Riemann tensor's components themselves do not diverge, I did not think it might be relevant, but the tetrads themselves can diverge as they do indeed for some tetrad fields in Schwarzschild, so it may be something to watch out for (I am slightly suspiscious about their status as quasi-regular singularities since they are typically not of that type in general and simply depend on topology and can be locally extended, which I don't think would work here, although local extendibility is a little suspiscious a concept).
  • This is slightly reminescent of some Penrose paper talking about singularities as ideal points and mentionning the existence of null-finite $\infty$-TIP, which are ideal points which are "at infinity" as measured from the perspective of timelike curve but for which the null curves are at a finite affine length, which he claims can only occur if the curvature diverges, although without much detail.
  • From the other answers, it seems that whatever Kerr may believe about the proper extension of the Kerr spacetime, the Kerr example he gives is in fact simply a regular boundary point, which can be extended, and therefore not much of a singularity.

The central point of Kerr's paper is showing examples of inextendible null geodesics with finite affine length that do not end in a singularity. To construct such an example he looks at null geodesics along the symmetry axis of the Kerr spacetime, in the region between the inner and outer horizons.

Such a curve will satisfy

$$ 0 = -dt^2 +dr^2 + \frac{2 m r}{r^2+a^2}(dr+dt)^2 ,$$

which, as Kerr notes, has two solutions

$$ \frac{dr}{dt}=-1 \quad\text{and}\quad \frac{dr}{dt}=\frac{r^2-2mr+a^2}{r^2+2mr+a^2}.$$

The first he labels as the "fast" solution, and the other as "slow". It is this slow solution that he presents as an example of a finite affine length lightlike curve that does not end in a singularity. As proof that it cannot be extended he notes that $r^2-2mr+a^2$ is zero exactly at the two horizons. Therefore, for the "slow" solution, $\frac{dr}{dt}$ goes to zero at the horizons, and the solution must asymptote to the outer horizon as $t\to-\infty$ and to the inner horizon as $t\to\infty$. Since, $r$ is an affine parameter for this curve, this (piece of) curve has finite affine length.

However, this piece of null geodesic is not in-extendible. The flaw in Kerr's logic sits in that the Kerr coordinates (which he uses) do not cover the entirety of the maximally extended spacetime. More precisely, Kerr coordinates are type of ingoing null coordinates, akin to ingoing Eddington-Finkelstein coordinates for Schwarzschild. In the ($\theta=0$ slice ) conformal diagram below, Kerr coordinates cover the regions shaded green. In the same diagram, the blue line shows an "fast" null curve, which spends all of its infinite affine length in the Kerr coordinate patch, and the red line shows the "slow" null curve. The solid piece of the red curve is the finite affine length part of this curve. As is immediately apparent this curve can be extend with the dashed piece to form an infinite affine length curve.

enter image description here

So Kerr's example of a finite affine length lightlike curve that cannot be extended and does not end on a singularity, actually isn't. The generalized version of this presented in his appendix, suffers from the same flaw.

  • $\begingroup$ Good point. The null geodesics exactly on the outer horizons are still inextendible I believe (even though they violate the "generality condition" from Hawking & Penrose). $\endgroup$
    – Void
    Commented Dec 7, 2023 at 18:05
  • 2
    $\begingroup$ In the θ=0 polar slice there is no singularity, the ring is in the equatorial plane at θ=90° so even the dashed part of the red curve in your diagram does not end in a singularity, it goes through the ring into negative space without hitting the ring itself. To hit the ring you'd have to approach it exactly equatorial since it is repulsive. The condition for the singularity is not only r=0 but also θ=90° (at r=θ=0 we have R=0 and at r=0, θ=90° it's R=a) $\endgroup$
    – Yukterez
    Commented Dec 7, 2023 at 23:04
  • $\begingroup$ @Yukterez I know, the dashed lines are only there to indicate you are entering a different region. $\endgroup$
    – TimRias
    Commented Dec 8, 2023 at 0:09
  • 6
    $\begingroup$ Sure, but you do so from t=-∞, which only exists if we're talking about eternal black holes, with realistic black holes formed by collapsing matter they can't be older than the big bang. Kerr doesn't believe the negative space and antiverse regions really exists, at least not for black holes of finite age and formed by collapsing matter. $\endgroup$
    – Yukterez
    Commented Dec 8, 2023 at 1:11
  • 2
    $\begingroup$ I was about to make the same point as Yukterez - but in a more direct way. The "maximally extended spacetime" construct has no physical relevance. Therefore, neither does the counter-argument. The paper concerns Kerr's solution, not some other solution (of questionable relevance) that contains Kerr's solution. At the very least, a global extension of the Kerr solution has to be one that is compatible with the Big Bang model. $\endgroup$
    – NinjaDarth
    Commented Dec 9, 2023 at 2:57

There are some answers here that do a good job of exhibiting the curves that Kerr has pointed out, but I don't think any have outlined his core argument in a faithful manner, and most have missed key observations at the heart of his point, so I feel compelled to chime in here a bit late. While I would say that Kerr's writing in this paper is sometimes overzealous, overly contrarian, and reads a bit like a rant, his central point has much more merit and content than some other answers would lead one to believe.

Kerr's Point

The first thing to say is that Kerr has, of course, not contradicted Penrose's theorem (or variations thereof due to Hawking, etc.), and despite his bombastic language, I don't think he purports to do so. The technical, rigorous statement of the theorem that you'll find in a GR textbook (e.g. Wald, O'Neill) is entirely correct. Moreover, Kerr has not made any earth-shattering remarks; his essential point is something that most dedicated relativists that have studied the singularity theorems have recognized themselves. Kerr has rightly observed, though, that Penrose's theorem is often strongly over-interpreted in the broader physics community.

Penrose's theorem guarantees, under certain hypotheses, that spacetime is null geodesically incomplete. In particular, it states that if a spacetime

(1) satisfies the null energy condition,

(2) is globally hyperbolic with noncompact Cauchy hypersurface, and

(3) contains a trapped surface,

then it contains at least one incomplete null geodesic, a lightlike trajectory that "ends early". This theorem is a beautiful result that made a large impact on the GR community, as it established that incompleteness is not tied to symmetries of the spacetime. Even so, Kerr has rightly pointed out that this result is actually a good deal weaker than it is sometimes given credit for.

The primary point of Kerr's paper is that this theorem has nothing to do with the central ring singularity of his namesake spacetime. The theorem does not tell us that the ring singularity is there: the singularity could, in principle, be excised from the spacetime and replaced with a self-supporting stationary matter distribution without contradicting the theorem in any way. The essential sticking point with the objections raised in some other answers (and this point is not made well by Kerr, if he intended to make it at all), is that the maximally extended version of the Kerr spacetime does not satisfy the theorem's hypotheses: condition (2) fails, as the maximal extension is not globally hyperbolic. This means that Penrose's theorem tells us nothing at all about the maximal extension-- in particular, it does not tell us that the maximal extension is null incomplete.

To illustrate, consider the following portion of the Penrose diagram of the $\theta = 0$ cross-section of the maximally extended Kerr spacetime.

$\hspace{3cm}$enter image description here

The red hypersurface $\Sigma$ cuts across the exterior asymptotically flat region, and its Cauchy development (the portion of the spacetime determined by $\Sigma$ from the vacuum Einstein equation) is shaded in gray. The inner horizon is the future Cauchy horizon $H^+(\Sigma)$ of $\Sigma$. One of the example curves that Kerr has identified is shown in blue.

The significance of this curve is that, since the full maximal extension is not globally hyperbolic, curves like this are all one can get from Penrose's theorem. Indeed, if one restricts their attention to the gray region, this subset is globally hyperbolic, and the theorem does apply here: it tells us the gray region is null geodesically incomplete as a spacetime in its own right. Of course, that this region is incomplete is nothing new since we already know we can extend the metric beyond it, and the incompleteness of the gray region has nothing to do with any singular behavior that may or may not arise beyond the Cauchy horizon under such an extension. Since they are incomplete in the gray region, Kerr's curves provide explicit examples of the null geodesics guaranteed by the theorem. These curves, however, do not demonstrate any kind of singular behavior-- that we know they can be extended in a larger spacetime means there's nothing singular about them. All curvature scalars are finite, etc. All Penrose's theorem tells us, then, is that null curves like Kerr's (which are decidedly nonsingular) exist in the gray region, and they are incomplete as geodesics within this globally hyperbolic set. As soon as you extend the spacetime beyond this subset by any amount whatsoever (perhaps to try to include the ring singularity), then Penrose's theorem is simply inapplicable, telling us nothing about the extension.

The upshot, then, is that Penrose's theorem does not offer any definitive proof that the interior of a stationary, axially symmetric black hole must be singular-- the incomplete null geodesics it can give us are only incomplete as geodesics restricted to the globally hyperbolic region, and this incompleteness has nothing to do with the ring singularity that may exist beyond the Cauchy horizon. While Kerr has not explicitly provided an alternative extension beyond the Cauchy horizon that is nonsingular, there is no solid reason to believe one doesn't exist. Note that the possibility of replacing the singularity here does not at all necessitate one's going so far as to model the dynamical collapse of a rotating star; it may well be the case that the ring singularity can be replaced by a stationary matter distribution beyond the Cauchy horizon while leaving the gray region entirely unchanged.

The SCC Caveat

A significant caveat to the above discussion is that Cauchy horizons are generally thought to be unstable; this is physically expected due to the infinite blueshift of signals from matter in the exterior that an observer would experience upon crossing $H^+(\Sigma)$. This hypothesis is formalized in the (unproven) Strong Cosmic Censorship Conjecture, which states that the Cauchy development of generic physical initial data (e.g. data specified on $\Sigma$) will be inextendible to a larger (physically meaningful) spacetime. If one accepts this as true, in our diagram it would mean that arbitrarily small perturbations to Kerr initial data on $\Sigma$ would essentially result in the full maximal extension collapsing to just the gray region. In this case, any incomplete curves like Kerr's examples actually would terminate in an inextendible manner, a scenario many relativists would readily call singular.

Even if one accepts Strong Cosmic Censorship, however, Penrose's theorem is still quite limited. It does not ensure that its incomplete null curves experience curvature divergence, and it still doesn't ensure that any massive matter, which follows timelike curves, collapses along singular trajectories.

  • $\begingroup$ Please correct me if I'm mistaken but for asymptotically flat spacetimes, the weaker version of the Strong Cosmic Censorship conjecture (due to Christodoulou) is true, right? Moreover, Cauchy horizon is a weak null singularity (as demonstrated by Ori). We can definitely extend the metric beyond the inner horizon but such extensions will not have locally square integrable Christoffel symbols, afaik. $\endgroup$
    – noir1993
    Commented Jan 9 at 20:06
  • $\begingroup$ @noir1993 what you say is exactly true - for asymptotically flat spacetimes the Christodoulou formulation of SCC holds (as obtained when introducing classical matter perturbations), and indeed the spacetime is extendible (at least classically). In the asymptotically deSitter case SCC fails classically for certain black hole parameters (as demonstrated by Cardoso et al). $\endgroup$
    – Whyka
    Commented Feb 21 at 19:01
  • $\begingroup$ @jawheele thank you for this answer, I found it the most satisfying. $\endgroup$
    – Whyka
    Commented Feb 21 at 19:15
  • $\begingroup$ If I understood this correctly, has Kerr shown that the Penrose theorem should really be understood as an incompleteness theorem rather than necessarily a singularity theorem? $\endgroup$ Commented Apr 23 at 15:17
  • $\begingroup$ @AwkwardWhale That Penrose's result is best described as an incompleteness theorem is a fair statement. Kerr's paper indeed emphasizes this, but I wouldn't say he's "shown" it in a way that implies novelty: he observed it, like many before him. Mathematical relativists have recognized this, and many have insisted on the language "incompleteness theorem" over "singularity theorem", for quite a while. $\endgroup$
    – jawheele
    Commented Apr 23 at 18:15

Kerr writes:

The original Kerr-Schild coordinates were deliberately chosen to be a generalisation of Eddington’s, avoiding any coordinate singularities on either horizon. It will be shown in Section 5 that through every point of these spaces there are light rays that are asymptotically tangential to one or other horizon, do not have endpoints and yet their affine lengths are finite.


The simplest example of a FALL [finite-affine-length null geodesic] [...] lies on the rotation axis between the two event horizons and is asymptotic to each of these. It is what one gets when a torch is shone ”backwards” while falling into a black hole down the axis. It does not cross either horizon.

He's evidently talking about geodesics like the one shown in green on this diagram:

enter image description here

This is a Penrose diagram of the maximally extended Kerr spacetime with the portion covered by Kerr-Schild coordinates highlighted. The green geodesic "ends" at the boundary of the coordinate patch only in the sense that the rest of it can't be described using those coordinates. This isn't a counterexample to the singularity theorems unless you interpret them in such a way that they're trivially wrong. The green geodesic isn't "asymptotic to" "the two event horizons" (by which he means A and D); it just looks like it is in those particular coordinates.

He argues that the horizons other than A and D aren't real. There actually are good arguments that they aren't real, but they don't support his position.

On the past side, to describe a realistic black hole that forms from a collapsing star, you need to graft a portion of the vacuum Kerr solution to some sort of spacetime describing the star. The portion that you use doesn't include the past boundary of the Kerr-Schild patch (C and E) or anything below it in the diagram. But the geodesic doesn't end there; it continues into the rest of the spacetime, which just happens to not be Kerr-shaped.

On the future side, there's an argument that if the Kerr solution is even slightly perturbed (e.g. by the presence of any matter at all) then there is a curvature singularity at A and B due to the mass-inflation instability.

Putting these together, it's plausible that the only physically relevant part of the Kerr spacetime is (part of) the middle and lower shaded squares. But Kerr's argument is that the green geodesic ends in a finite affine distance without a singularity, and that's just not true. In the future direction, it ends at a singularity. In the past direction, it extends an infinite affine distance into the pre-black-hole universe (or at least he hasn't shown that it doesn't).

The preprint contains some cranky passages:

Why do so many believe that the star inside must become singular at this moment? Faith, not science! Sixty years without a proof, but they believe!.

(italics his). If it's accepted by a reputable journal then I guess I'm wrong, but I think it won't be.

  • 2
    $\begingroup$ Since you're using a θ=0 slice just like TimRias my argument under his post also applies to yours: in the polar plane there is no singularity, if extended the green path in your diagram goes through the ring without hitting it. $\endgroup$
    – Yukterez
    Commented Dec 7, 2023 at 22:56
  • 2
    $\begingroup$ @Yukterez True, but then it continues for an infinite affine distance in the $r<0$ region. I don't think that's what Kerr was talking about in the paper. He really seems to think the green segment is the entire geodesic. $\endgroup$
    – benrg
    Commented Dec 7, 2023 at 23:49
  • 3
    $\begingroup$ That's because he doesn't believe the negative space behind the ring to where the geodesics extend is physical for black holes formed by collapsing matter, that is only valid for eternal vacuum black holes with infinite past. Also the region crossings of your green worldline through the inner and outer horizons would be at t=-∞, but that does not exist for a realistic black hole in a big bang universe. $\endgroup$
    – Yukterez
    Commented Dec 8, 2023 at 1:23
  • 2
    $\begingroup$ @Yukterez In the text I quoted he says the geodesic lies between the horizons and is asymptotic to them, so if you're saying he thinks it extends farther than my green line, I don't see how you can be correct. "the region crossings [...] would be at $t=-∞, but that does not exist for a realistic black hole" - I covered that in the answer. Realistically, the geodesic ends at a singularity in the positive time direction, which doesn't help his argument, and it begins long before the black hole forms, e.g. at past null infinity or at a big bang singularity, which also doesn't help his argument. $\endgroup$
    – benrg
    Commented Dec 8, 2023 at 2:02
  • 3
    $\begingroup$ @Yukterez It hits the mass-inflation singularity, not the ring singularity, as I said in the answer. $\endgroup$
    – benrg
    Commented Dec 8, 2023 at 2:23

In a nutshell, the Penrose Singularity Theorem states that light inside the horizon cannot escape to infinity and therefore must end up in a singularity. This means a singularity must exist inside the horizon.

However, the Kerr spacetime allows light to escape to infinity in a different region, often referred to as "another" or "parallel" universe. Well, if light inside the horizon can escape to infinity (no matter in what region of spacetime), then it does not have to end up in a singularity. This means that in general a singularity does not have to exist inside the horizon.

This argument does not disprove the Singularity Theorem, but only indicates that the conditions required by this theorem are not satisfied by all possible types of spacetime.

Apparently this is the logic in the abstract. I have not read the whole paper, but perhaps by "physical bodies" Roy means that all black holes rotate, so Schwarzschild-like static spacetimes with no escape cannot exist in reality. This is just my guess, I am sure his actual arguments are much smarter.

enter image description here

  • 2
    $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – ACuriousMind
    Commented Dec 8, 2023 at 17:58
  • $\begingroup$ Nice answer. I have a new question on black holes:physics.stackexchange.com/questions/808139/… $\endgroup$ Commented Mar 28 at 23:28
  • $\begingroup$ To clarify: Kerr has always maintained (despite very rarely being listened to) that the ringlike singularity in his solution is a placeholder for a physical object, one that creates centrifugal force. He does not think an object like a dead star or collapsed gas cloud would open the way to a parallel universe. $\endgroup$
    – Wookie
    Commented Apr 20 at 20:46

The author’s opinion is that gravitational clumping leads inevitably to black holes in our universe, confirming what is observed, but this does not lead to singularities.

Kerr's arguments are stringent and worth to be considered. The idea of gravitational collapse ending in so-called singularity originates from Oppenheimer-Synder model which however takes advantage of nonphysical assumption of pressureless matter (dust). Infinite energy density at star center is an unavoidable consequence of such an approach. Dust particles follow geodesics but for general perfect fluid with pressure the proper time along the trajectory does not coincide with the proper time along the corresponding geodesics - see Faraoni’s Quasi-geodesics in relativistic gravity.

In Schwarzschild interior solution the central energy density is always finite. It is the pressure that for some critical compactness parameter $r_S/R$ diverges giving rise to the universal maximal force $2c^4/G$, see https://physics.stackexchange.com/a/707944/281096.

In the interview with Prada Penrose says (page 19):

What the singularities tell us is that the laws of classical general relativity are limited. I’ve always regarded this as strength in general relativity. It tells you where its own limitations are. Some people thought it was a weakness of the theory because it has these blemishes, but the fact that it really tells you where you need to bring in other physics is a powerful ingredient in the theory.

Roy Kerr, and Einstein I presume, apparently does not agree with him. He states

When theory predicts singularities, the theory is wrong!

Regarding his argument about equation of state I would like to quote Dennis Lehmkuhl General Relativity as a Hybrid theory: The Genesis of Einstein’s work on the problem of motion

I believe that Einstein was very aware of the fact that the energy-momentum tensor is not a representation of matter itself, but only of its energetic properties. And this was fine in the context of GR: the theory was supposed to be a theory of gravity first and foremost, and in this context all one needs to know about matter is its mass-energy distribution. Einstein’s 1936 claim that the energy-momentum tensor is only a phenomenological representation of matter means exactly that: the tensor does not tell us all there is to know about matter, but only what we need to know in the context of a pure theory of gravity.

It is a hybrid theory, fundamental in its treatment of purely gravitational fields in regions of spacetime in which nothing else is present, and effective/phenomenological in regions of spacetime in which matter is present.


There is an interesting story behind the Roy Kerr's claim:

My solution (Kerr) assumed that there was no actual matter present (empty Einstein space) and so there had to be some singularity inside. Since we are talking about a real collapsed object, we do not need this singularity. What happens is that as the star collapses it spins faster and faster until the centrifugal force and can counterbalance the attractive gravitational forces. Why do some people think there must be a place inside where the density is infinite? Soon after I constructed the Kerr metric many tried to prove “singularity theorems”. The first two that I saw where presented at seminars at UT Austin. I gave counterexamples to these during the seminars so they disappeared. Then in the late 1960’s Stephen Hawking came to Austin for a long weekend. He told us that he had proved that either singularities or closed time-like loops would form in the future if they didn’t already exist! He said he used Raychadhuri’s theorem but did not give details. I spent the next two days trying to prove this myself. During a party for Stephen on Monday night I told him and George Ellis that I couldn’t quite prove what he said. I could prove almost closed loops but couldn’t quite close them. I presumed he could (I was very naive then) but he said no more. A few months later he published exactly what I had said, with no attribution, of course. Then Roger Penrose published his famous “theorem” that said that if there is a trapped surface then there is a singularity. Since there is a trapped surface inside the outer event horizon of Kerr this “proved” that there must be a singularity inside! He assumed Raychadhuri’s theorem and whatever else he needed to “prove” this. I never believed these theorems, thought of them as typical physics proofs - if you need but cannot prove something, just assume it. Last year when a Nobel prize was given for this theorem I thought I better have a closer look at. Clearly it needed Raychadhuri’s theorem to be true so I looked for a counter example to that. I knew from a small calculation in 1963 that there were two light rays along the axis of rotation between the two horizons. The incoming one goes into the interior, the “outgoing” one also falls in but never crosses the inner horizon. I suspected this had a bounded affine parameter, contradicting Raychadhuri. At this point I had the help of a brilliant undergraduate student, Alex Goodenbour. He confirmed my conjecture. Could we do even better? The Kerr metric was discovered from the properties of it curvature tensor. It has a special set of light rays that are repeated eigenvectors of the Curvature tensor and these were used in its construction. It also has a second set with the same properties but “outgoing” outside the outer horizon. These try to go outwards between the horizons but cannot do so and fall towards the inner horizon. These curves are asymptotic to each horizon at their ends. Alex and I showed that their affine parameters are bounded, contradicting Raychadhuri’s theorem. This shows that ALL singularity theorems are valueless and so there is no proof that singularities form inside a rotating black hole. The body inside cannot be touching the inner horizon but light rays from it can. There are no trapped surfaces inside this horizon, unlike the outer one.

I have found it accidentally as Roy's answer on quora.


Kerr is talking about realistic black holes...

Roy Kerr wrote: "...generated by real physical bodies"

so in the Penrose diagrams in the other answers where the worldlines are extended behind t=-∞ you can't do that for real.

Also there is no singularity in those diagrams since the ring is in the equatorial plane, while those diagrams are polar slices, so they do indeed show that the worldlines depicted in them do not hit the singularity, but either pass through the ring if we're talking about eternal black holes, or hit the nonsingular remains of the collapsed star if we're talking about realistic ones.

Even if there was a rotating ring singularity it would be gravitationally repulsive, so you couldn't hit it since it would deflect your trajectory away from it. Only a perfectly equatorial ray could hit the singular ring, everything else gets either deflected through it or away from it. Therefore it would be hard for the singularity to form in the first place due to the centrifugal repulsion.

In an eternal black hole you could travel into the infinite past and cross into other regions while getting fried by the infinitely blueshifted light that fell into the black hole during all that time, but in reality that t=-∞ surfaces don't exist since the black hole must be younger than the big bang, so the worldlines he is talking about are indeed of finite affine length and don't hit the singularity.

  • 1
    $\begingroup$ If we are talking about realistic black holes then the inner horizon is singular, and Kerr’s example of a finite affine length curve does in fact end in a singularity. $\endgroup$
    – TimRias
    Commented Dec 8, 2023 at 6:58
  • $\begingroup$ Towards the past the lines would extend for an infinite affine length into the region occupied by the astrophysical matter that formed the black hole, possibly through the origin of the coordinate system,back out into the vacuum region outside the matter, and all the way back to past null infinity. $\endgroup$
    – TimRias
    Commented Dec 8, 2023 at 17:50
  • 1
    $\begingroup$ @Tim Rias: but is the mass inflation singularity at the inner horizon a proven thing or just a hypothesis? I'm not so sure about that, if that's not proven Kerr's still right in claiming there's no proof. One thing's for sure at least, these worldlines don't hit the ring singularity. $\endgroup$
    – Yukterez
    Commented Dec 9, 2023 at 19:24
  • 1
    $\begingroup$ Whether these lines hit the ring singularity (which indeed they don’t) is completely beside the point. The question is whether they can be extended. In pure vacuum symmetric Kerr the answer is unambiguously, yes. In perturbations away from Kerr, the answer is generically: only as a C^0 curve (which you might say is a no). $\endgroup$
    – TimRias
    Commented Dec 9, 2023 at 20:03
  • $\begingroup$ Sabine Hossenfelder just released a video on the subject: youtube.com/watch?v=nz55jONtFAU $\endgroup$
    – Yukterez
    Commented Dec 13, 2023 at 21:50

Kerr here opposes a research that was awarded the Nobel Prize (Penrose 2020), but I believe he is fundamentally right. Among other things, this is a classical scenario. In the quantum scenario, things change further. In 2014 Vaz demonstrated that, in the quantum scenario, the non-homogeneous gravitational collapse of dust ends with the matter condensing naturally on the Schwarzschild surface, which is not therefore a real horizon but an apparent one. This therefore confirms what Einstein predicted in 1939. I showed in two recent papers published in mainstream journals that Vaz's result was correct and can be obtained, again in the quantum scenario, also from the historical Oppenimer-Snyder homogeneous gravitational collapse, see https://doi.org/10.1088/1572-9494/ace4b2 and https://doi.org/10.1002/prop.202300028. Contrary to common belief, quantum corrections become important at the Schwarzschild scale rather than the Planck scale, and such corrections predict that neither real horizons nor singularities are formed. Clearly, the various firewalls, fuzzballs, complementarity and other postulated weird things are no longer necessary, although this is unlikely to be accepted by most black hole researchers.

  • 2
    $\begingroup$ Your post is exactly correct and is a relief that not everyone yet has gone to the la-la-land of non-falsifiable nonsense +1 $\endgroup$
    – safesphere
    Commented Dec 9, 2023 at 19:40
  • $\begingroup$ @safesphere Huh, confusing comment to say the least! Your answer is itself 100% la-la-land as far as I can see, since nothing falsifiable happens inside the horizon. $\endgroup$
    – m4r35n357
    Commented Dec 13, 2023 at 9:51
  • $\begingroup$ @m4r35n357 Correct, because the question is what someone else may think, not me. $\endgroup$
    – safesphere
    Commented Dec 14, 2023 at 10:21
  • $\begingroup$ I don't grok the relevance of the Nobel; the significance of the agreement with a purely classical result in (Einstein 1939), who had been unaware that the horizon singularity was coordinate only (albeit physically significant indeed); and the wording "real horizon". I'm intrigued although. I doubt gravity being quantisable sensu stricto; semi-classical theories seem more promising. I had time only to quickly scan heads and tails of your two papers. You're perhaps aware of firewall resolution in (Maldacena, Susskind 2013), and (t'Hooft 2015, 18, 19) on BH quantum information, ain't you? $\endgroup$ Commented Feb 23 at 12:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.