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Comment by dhosek | Hacker News Reader

I figure at least some of it comes from the idea that mathematically, a singularity is a point (e.g., in the graph of z=1/w, there is a singularity at the point w=0, and in the graph of z=(1-w)²/(1-w) there is a removable singularity at w=1 (that is, the function is undefined at w=1, but if you put a point at (1,0), the graph will be continuous and no longer have any holes in it). The fact that both have the same name and the similar behavior of a black hole singularity to a mathematical singularity¹ can lead people to make an incorrect assumption.

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1. I must admit to a lack of sufficient GR education to feel confident in this, but I think that one of the issues that made physicists unwilling to accept the idea of black holes when they were first postulated was that there ended up being a division by zero in the mathematics.

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>The fact that both have the same name

They don't just have the same name, they are the same thing.

A Schwarzschild black hole has both: a removable singularity at the event horizon that is just an artefact of a particular choice of coordinates and a true non-removable mathematical singularity at r=0 where curvature really does go to infinity. It also wouldn't be much of an issue in classical physics, because this singularity is always hidden from outside observers, so the mathematical weirdness there can't screw with your normal predictions in space outside the black hole. The problems start once you consider quantum mechanics, because any such singularity will break unitarity (a fancy way of saying that probabilities must add up to 1), which means your theory as a whole can no longer make predictions. This has opened a whole can of worms with a bunch of solution attempts, which are all sadly untestable for the foreseeable future.

OSM - slight generalization of Schwarzshild BH, where you take evolving spherically-symmetric mass distribution instead of point mass - shows that point singularity in the middle can be naked (aka observable), so it's not just QM that causes worms...

https://en.wikipedia.org/wiki/Oppenheimer–Snyder_model

It is not difficult to construct geometries with naked singularities. Reissner, Nordström, Weyl and others individually came up with one long before Oppenheimer. You can also construct geometries where faster than light travel is possible. But all of these suffer from fundamentally unphysical energy conditions. Quantum mechanics may change the picture because it allows weirder energy states than normal physics because everything fluctuates. But it is unknown if and how this actually affects gravity.

Thanks, I enjoyed your answers !

> The problems start once you consider quantum mechanics, because any such singularity will break unitarity (a fancy way of saying that probabilities must add up to 1), which means your theory as a whole can no longer make predictions.

How is this any different than classical? Isn't it still just an ~impossibility hidden behind an event horizon in either model?

With our current understanding, baryon and lepton number are not conserved as a black hole radiates. I think this is a better demonstration of the incompatibility with classical and quantum mechanics.

Baryon and lepton number conservation are what's called "accidental symmetries" in the standard models, meaning there is no real underlying symmetry that would conserve them. In fact many extensions of the Standard Model don't, while still retaining unitarity. The problems already start once you try to calculate any time evolution of anything, because the Hilbert operator not being unitary means that everything breaks. Even the total energy in a closed system might vanish or blow up to infinity. You can't calculate anything under these conditions.

Even without quantum mechanics, black holes are trouble:

Approximately everything in nature rotates. Including black holes. Schwarzschild blockholes do not rotate. Rotating black holes are much more complicated and don't necessarily shield their singularity behind an event horizon.

Rotating black holes are described by Kerr geometries and have more than one event horizon, but still have their singularities hidden from anyone outside behind their inner horizon.

pardon a maybe stupid quesiton but a few comments say this and i wonder.

why does it matter that it is not 'visible' for anyone?

"Visible" in this case means "able to influence in any way".

If a problem is not able to influence anything, even in theory, then by definition, it cannot possible influence any testable predictions we have.

We don't actually know if a black hole has an inside. Some theories/hypotheses say spacetime just stops at the event horizon.

This is like saying the Riemann zeta function is only defined for real numbers. You can always extend the singularity mathematically by incorporating new axioms.

My point is, it’s not super meaningful to argue whether a black hole has an inside.

I don't mean the coordinate singularity, I mean there is no more spacetime after that.

Why would that happen? In a different coordinate system (Kruskal–Szekeres coordinates) nothing special happens at the event horizon at all.

Why wouldn't it happen? From the perspective of anyone outside the black hole, nothing can ever enter it. What if that's just true?

Have you considered that the apparent event horizon is not uniform for all observers?

So does spacetime exist in some frames of reference but not others because those frames disagree on the radius of the apparent event horizon?

Also note that in general an event horizon doesn’t require a singularity.

All observers outside the black hole agree that nothing ever enters the black hole though.

And all observers on the event horizon agree that there is no event horizon :)

What I’m trying to say is that there is nothing special about the region of space near the event horizon.

> 1. I must admit to a lack of sufficient GR education to feel confident in this, but I think that one of the issues that made physicists unwilling to accept the idea of black holes when they were first postulated was that there ended up being a division by zero in the mathematics.

Well, the Ricci curvature scalar blows up to infinity, which is obviously unphysical.

Why is it unphysical?

The curvature scalar can be physically evaluated by measuring the volume of a small ball of freely falling test particles and comparing to its volume in flat spacetime.

https://en.wikipedia.org/wiki/Scalar_curvature#Relation_betw...

Okay — why is it unphysical such a ball has unbounded curvature?

If this unbounded state were asymptotic then it wouldn't be unphysical, but a freely falling particle reaches the singularity in finite time. A freely falling observer crossing the event horizon could in principle perform this measurement as they approached the singularity, and general relativity could no longer describe their measurements.

Okay — you’re still not explaining what actually breaks, just repeating they could measure over and over.

Please say what you specifically believe is unphysical about the situation — what trajectory reaches the singularity in finite time and why specifically is that unphysical?

My understanding is that you have a cusp singularity that is actually an infinite spike, ie, distance to the singularity is unbounded; that is, no matter how small a circle/sphere around the singularity, you have an infinite diameter. And so you will need to be much more explicit about where the problem lies.

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