5 ms·

What would the implication of a no-singularity model be? Would it be that black holes just have a comparatively mundane ball of matter inside the event horizon?

pantulis 3y ago
There's an hypothesis for that! https://en.wikipedia.org/wiki/Fuzzball_(string_theory) https://en.wikipedia.org/wiki/Fuzzball_(string_theory)
jonhohle 3y ago
As someone with no formal experience in astronomy (other than a college for kids course in the 80s), that has been my mental model. Just a large enough mass with enough gravitational force to bend light/prevent light from leaving. There’s no hole, just a name describing what was originally observed.
- Novosell 3y ago
  Mind you, black holes require high (infinite) density, but can exist at arbitrary mass. It'll just be smaller.
- Sharlin 3y ago
  This is the 1800s classical, pre-relativistic idea of a "black hole". But it is mostly a coincidence that if you plug an escape velocity equal to the speed of light to Newton's equations, you get a body with a mass and radius equal to the mass and event horizon radius of a black hole. But given the incredible predictive success of general relativity, there's little doubt that actual black holes are not just "dark stars" but fundamentally relativistic objects: regions of spacetime so warped that time becomes spacelike and space becomes timelike. Note that Kerr most definitely does not disagree; the entire point of the Kerr metric is that it's a solution (set of solutions) to the Einstein field equations.
pdonis 3y ago
The kind of "no singularity" model Kerr discusses--where you have an object inside the inner horizon that replaces the ring singularity--is not actually viable. First, most physicists believe the inner horizon itself is not viable, because it is unstable against small perturbations, which any real hole will have (because there is always something falling in, even if it's just the cosmic background radiation). Second, it's not actually possible to replace the entire ring singularity with a non-singular object. However, there is a different type of "no singularity" model in the literature, called a "Bardeen black hole" (among other things), in which there is not only no singularity, but no event horizon. What appears to us as an event horizon is actually just an apparent horizon--a surface where, locally, light cannot escape, but which will not stay that way forever; eventually it will evaporate away (by a process similar to Hawking radiation), and everything that was trapped inside will come back out (though in hugely scrambled form). So there is nowhere in this spacetime that can't send light signals to infinity; it just takes a really long time for the signals to get out from some parts of it.
- hinkley 3y ago
  It has always bugged me that the model is that space time around a black hole is non Euclidean space that light cannot escape from. How does anything approach a sphere that is surrounded by space that is so distorted? Is not everything suspended in the last few (Euclidean) meters before the event horizon?
  - sliken 3y ago
    Yes anything falling in approaches the speed of light as it approaches the event horizon. So to external observers they slow down. However keep in mind that all black holes are growing (and will continue until the cosmic backround radiation gets lower than the rate of hawking radiation. So I think of black holes as onions, everything approaches the event horizon and then (to external observers) get frozen at the edge, until the black hole gets bigger. Also keep in mind that from the perspective of someone falling in, there's nothing particularly noticeable as you cross the event horizon.
    - 3y ago

pantulis 3y ago

There's an hypothesis for that! https://en.wikipedia.org/wiki/Fuzzball_(string_theory) https://en.wikipedia.org/wiki/Fuzzball_(string_theory)

jonhohle 3y ago

As someone with no formal experience in astronomy (other than a college for kids course in the 80s), that has been my mental model. Just a large enough mass with enough gravitational force to bend light/prevent light from leaving. There’s no hole, just a name describing what was originally observed.

Novosell 3y ago
Mind you, black holes require high (infinite) density, but can exist at arbitrary mass. It'll just be smaller.
Sharlin 3y ago
This is the 1800s classical, pre-relativistic idea of a "black hole". But it is mostly a coincidence that if you plug an escape velocity equal to the speed of light to Newton's equations, you get a body with a mass and radius equal to the mass and event horizon radius of a black hole. But given the incredible predictive success of general relativity, there's little doubt that actual black holes are not just "dark stars" but fundamentally relativistic objects: regions of spacetime so warped that time becomes spacelike and space becomes timelike. Note that Kerr most definitely does not disagree; the entire point of the Kerr metric is that it's a solution (set of solutions) to the Einstein field equations.

pdonis 3y ago

The kind of "no singularity" model Kerr discusses--where you have an object inside the inner horizon that replaces the ring singularity--is not actually viable. First, most physicists believe the inner horizon itself is not viable, because it is unstable against small perturbations, which any real hole will have (because there is always something falling in, even if it's just the cosmic background radiation). Second, it's not actually possible to replace the entire ring singularity with a non-singular object. However, there is a different type of "no singularity" model in the literature, called a "Bardeen black hole" (among other things), in which there is not only no singularity, but no event horizon. What appears to us as an event horizon is actually just an apparent horizon--a surface where, locally, light cannot escape, but which will not stay that way forever; eventually it will evaporate away (by a process similar to Hawking radiation), and everything that was trapped inside will come back out (though in hugely scrambled form). So there is nowhere in this spacetime that can't send light signals to infinity; it just takes a really long time for the signals to get out from some parts of it.

hinkley 3y ago
It has always bugged me that the model is that space time around a black hole is non Euclidean space that light cannot escape from. How does anything approach a sphere that is surrounded by space that is so distorted? Is not everything suspended in the last few (Euclidean) meters before the event horizon?
- sliken 3y ago
  Yes anything falling in approaches the speed of light as it approaches the event horizon. So to external observers they slow down. However keep in mind that all black holes are growing (and will continue until the cosmic backround radiation gets lower than the rate of hawking radiation. So I think of black holes as onions, everything approaches the event horizon and then (to external observers) get frozen at the edge, until the black hole gets bigger. Also keep in mind that from the perspective of someone falling in, there's nothing particularly noticeable as you cross the event horizon.
  - 3y ago