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Do they look like singularities? As I understand it, any object that crosses a certain density threshold, to a point where light cannot escape its gravitational

Do they look like singularities? As I understand it, any object that crosses a certain density threshold, to a point where light cannot escape its gravitational pull, is effectively a black-hole (even if the mathematical model for them is more purist, only described by a handful of parameters). I don't think they need to be infinitely dense to explain our observations. You could say that we do observe a singularity, not in the centre of the black-hole but in its event horizon. But technically that's just an infinity in the maths not a physical singularity, in the sense that if you were there it would just seem like normal space.

chasil 1y ago
I understand that black holes are described by precisely three parameters: mass, spin, and charge. I am assuming that these are the only way to describe anything within an event horizon.
- raattgift 1y ago
  The key parameter is mass. The no-hair conjecture says that a Kerr-Newman black hole (KN BH: stationary, eternal) has only position, linear momentum, angular momentum, and electromagnetic charge. 3 components of linear momentum and three of position can be removed by keeping the KN BH at the origin of a system of coordinates. The KN BH doesn't evolve with time so we can remove two related time components as well. This leaves us with 3 components of angular momentum ("spin"), electromagnetic charge (because by definition the KN BH is immersed in an electromagnetic field), and mass. Schwarzschild BHs are a special case of KN BH where spin and electromagnetic charge vanish. But we could add other fields with charges, and make those charges more complicated thanks to interactions among the matter fields. That's not a Kerr-Newman black hole any more, though. In practice the other standard model fields don't really make much difference: the charges will tend to neutralize before gravitation is relevant, and won't build up around the BH itself. KN BHs are in that sense electromagnetically quasi-neutral. Things falling into a KN BH cause a perturbation that decays quickly away, changing the mass, and possibly spin, of the BH held at the coordinate origin. A change in electromagnetic charge will probably "reach out" and capture a charged particle in order to neutralize. Electromagnetic attraction is much stronger than gravitational attraction, while also being long range. However neutralization is not necessarily instant (the closest proton might be several light-minutes away, for example) or completely matched (oops, two protons are electromagnetically nudged into an infalling trajectory in response to the small negative charge, so when the second one arrives there'll be a small positive charge for a bit until a further electron is pulled in...). So quasi-neutral. These parameters are only about the horizon. It says nothing about whether the mass was a bunch of individual protons and electrons or a bunch of neutral hydrogen or a bunch of heavier molecules. (The same "says nothing" also exists classically without reference to particles: start with a Schwarzschild BH and drop in a uniform spherical shell of mass M or two concentric uniform spherical shells of mass M/2 each, and after some "balding" time we cannot tell whether our increased-mass Schwarzschild BH had one or two shells dropped into it). More prosaically, "no hair" is a statement about the stability of black holes in the face of small perturbations. If you throw something (a star, a big gravitational wave) into a black hole and wait a bit, does the resulting configuration still look like a black hole in the sense that the trajectories one grinds out of the Kerr-Newman metric (with adjusted mass, spin and charge parameters) accurately represent the trajectories around the new configuration? There is a lot of literature on black hole stability justifying a "yes" answer.
myrmidon 1y ago
This is not actually a density threshold; the more mass you have, the less density you need/get withing the event horizon, meaning bigger black hole => less average mass density ("Schwarzschild density"). And this Schwarzschild density is comparable to the mass density of neutron stars (or atomic nuclei) for small black holes, and can be as low as the density of water (!!) for the supermassive ones.
- oersted 1y ago
  Right, makes sense, it's all about it causing a sufficiently "steep" space curvature and/or over a long enough distance, to offset the speed of light trying to escape. Of course, high density is not the only way to have that effect. Thanks for the insight!

chasil 1y ago

I understand that black holes are described by precisely three parameters: mass, spin, and charge. I am assuming that these are the only way to describe anything within an event horizon.

raattgift 1y ago
The key parameter is mass. The no-hair conjecture says that a Kerr-Newman black hole (KN BH: stationary, eternal) has only position, linear momentum, angular momentum, and electromagnetic charge. 3 components of linear momentum and three of position can be removed by keeping the KN BH at the origin of a system of coordinates. The KN BH doesn't evolve with time so we can remove two related time components as well. This leaves us with 3 components of angular momentum ("spin"), electromagnetic charge (because by definition the KN BH is immersed in an electromagnetic field), and mass. Schwarzschild BHs are a special case of KN BH where spin and electromagnetic charge vanish. But we could add other fields with charges, and make those charges more complicated thanks to interactions among the matter fields. That's not a Kerr-Newman black hole any more, though. In practice the other standard model fields don't really make much difference: the charges will tend to neutralize before gravitation is relevant, and won't build up around the BH itself. KN BHs are in that sense electromagnetically quasi-neutral. Things falling into a KN BH cause a perturbation that decays quickly away, changing the mass, and possibly spin, of the BH held at the coordinate origin. A change in electromagnetic charge will probably "reach out" and capture a charged particle in order to neutralize. Electromagnetic attraction is much stronger than gravitational attraction, while also being long range. However neutralization is not necessarily instant (the closest proton might be several light-minutes away, for example) or completely matched (oops, two protons are electromagnetically nudged into an infalling trajectory in response to the small negative charge, so when the second one arrives there'll be a small positive charge for a bit until a further electron is pulled in...). So quasi-neutral. These parameters are only about the horizon. It says nothing about whether the mass was a bunch of individual protons and electrons or a bunch of neutral hydrogen or a bunch of heavier molecules. (The same "says nothing" also exists classically without reference to particles: start with a Schwarzschild BH and drop in a uniform spherical shell of mass M or two concentric uniform spherical shells of mass M/2 each, and after some "balding" time we cannot tell whether our increased-mass Schwarzschild BH had one or two shells dropped into it). More prosaically, "no hair" is a statement about the stability of black holes in the face of small perturbations. If you throw something (a star, a big gravitational wave) into a black hole and wait a bit, does the resulting configuration still look like a black hole in the sense that the trajectories one grinds out of the Kerr-Newman metric (with adjusted mass, spin and charge parameters) accurately represent the trajectories around the new configuration? There is a lot of literature on black hole stability justifying a "yes" answer.

myrmidon 1y ago

This is not actually a density threshold; the more mass you have, the less density you need/get withing the event horizon, meaning bigger black hole => less average mass density ("Schwarzschild density"). And this Schwarzschild density is comparable to the mass density of neutron stars (or atomic nuclei) for small black holes, and can be as low as the density of water (!!) for the supermassive ones.

oersted 1y ago
Right, makes sense, it's all about it causing a sufficiently "steep" space curvature and/or over a long enough distance, to offset the speed of light trying to escape. Of course, high density is not the only way to have that effect. Thanks for the insight!