When a black hole forms from gravitational collapse of a dense, heavy object, there are a few different possibilities for how the density goes up at the center of the mass distribution. If the mass distribution stays at some maximum density and builds up into a sphere until it reaches its own Schwarzschild radius (case 1 below), then the "collapsing spherical shell" model would seem to be correct (though that might have additional repercussions as I'll discuss later). However, if all fermions break down and degeneracy pressure disappears, then density would increase without bound at the center of the mass distribution until a singularity state is reached (case 2 below). That's the possibility that interests me.
The critical density for the formation of a black hole as a function of radius is ρ(R) = 3c2/8πGR2. So as long as the density during collapse goes up at a rate faster than inverse-squared, a black hole will form from the inside-out rather than the outside-in.
What are some of the properties of such a black hole? Well, if you're starting from what is essentially the minimum possible size of a black hole, then there are a few things you might run into. For example, black hole evaporation. Consider a black hole so small that when it evaporates, it releases a shell of Hawking radiation with the same mass-energy density that the black hole itself had:
Obviously, this takes place when the volume of the spherical shell and the volume of the inner sphere are equal, which takes place when the outer radius of the spherical shell is 21/3 times the radius of the inner sphere. The evaporation time Tev of a black hole with mass M can be found as Tev = 5120πG2M3/ħc4, so the thickness of the evaporation shell can be found as the product of Tev and c.
Setting the thickness of the shell equal to (21/3 - 1)*R where R = 2GM/c2 allows us to solve easily for this "critical density" black hole:
5120πG2M3/ħc3 = (21/3 - 1)*2GM/c2
2560πGM2/ħc = (21/3 - 1)
M2 = ħc(21/3 - 1)/2560πG
M2 = hc(21/3 - 1)/5120π2G
M = sqrt(hc(21/3 - 1))/32π*sqrt(5G)
A black hole with a mass of 1.237e-10 kg would have an event horizon at 1.8e-37 m, a temperature of 9.9e32 K, and a lifetime of 1.6e-46 seconds. While these values are all outside of Planck limitations, they are only outside of it by a relatively small margin, so they may still be physically meaningful. However, the blackbody spectrum peaks at a wavelength of 2.9e-36 m, meaning that a single photon emitted at the peak of the radiation curve would have an energy 6,000 times greater than the total mass-energy of the black hole, which throws a wrench into things.
At what point does a photon at the peak of a black hole's radiation curve actually have an energy equal to the mass-energy of the black hole? That's not too hard to figure out; setting hc/λ equal to mc2 and using Wien's displacement constant to find the wavelength as a function of temperature gives M = hc/4π*sqrt(bGkB) = 9.67e-9 kg, just under half a Planck mass. Such a black hole would have an evaporation time of 7.6e-41 seconds, about 1400 Planck times.
What if all this is moot because black holes form from the "outside-in" instead? Well, that prompts the question of what happens inside such a black hole as it is formed. This is not a meaningless question; due to the shell theorem, anything outside a particle's radial position is gravitationally invisible to that particle, and so the event horizon does not exist from the perspective of a particle inside the distribution of matter that would become a black hole. Thus, it will continue to obey the same physical laws it would have obeyed if the event horizon had not yet formed. The closer you get to the center of the black hole, the more of the black hole's mass becomes gravitationally invisible, and thus the density at the center of the collapsing object will continue to rise without bound as discussed above:
In case it's not clear: as an event horizon forms around the uniform-density core, the inside of that core fails to notice and continues to collapse, repeating the process by iteration until it forms an arbitrarily-low-mass black hole.
So here's the hypothesis.
Stable black holes form when macroscopic distributions of matter collapse to a density between 7.8e95 and 4.8e99 kg/m3. At this density, a black hole quanta forms. Because of its extremely small size, the quanta's behavior is dominated by quantum effects and it immediately tunnels out of its own event horizon, producing a wavefunction with a mass-energy distribution having an average density on the order of the density of the black hole quanta.
Subsequent collision between the mass-energy distribution of the Hawking radiation and infalling outside matter, however, immediately causes the wavefunction to collapse into a black hole quanta again, due to the high density of the surrounding medium. This process repeats over and over and over again, rapidly transforming the mass distribution into a cloud of constantly-evaporating-and-collapsing black hole particles.
The propagation of this cloud is at first chaotic, but although the quanta themselves are not affected by radiation pressure, the infalling matter is, and so the spherical symmetry of the original mass distribution causes the formation of a spherical shell of black whole quanta. The black hole quanta themselves are just outside their collective Schwarzschild radius.
Why don't the quanta fall in? Well, their evaporation time is so miniscule that they don't have time to be pulled inside. Hawking radiation produced by each of the evaporating quanta falls inside, but since there is nothing inside, the shell theorem means it feels no gravitational force and may escape through to the other side:
It is impossible for a singularity to form at the center of this distribution because the quanta must necessarily evaporate through quantum tunnelling more rapidly than it would be able to interact with anything at the center, and Hawking radiation released from evaporation by quanta inside the Schwarzschild radius escapes freely. Collapse is only possible by interaction with other quanta or with infalling matter outside the would-be event horizon, so the center of the sphere is merely a diffuse photon gas.
Evaporation happens so rapidly that the outward progress of the quanta cloud is necessarily many orders of magnitude faster than infalling matter. Therefore this distribution of black hole quanta expands outward, swallowing up anything it encounters and converting it to more black hole quanta, rather than just sitting still and allowing matter to fall into it as with a conventional black hole. It will continue to grow as it swallows up new matter, but no collective event horizon ever forms. This growth continues until it runs out of matter to swallow up.
At this point, the quanta forming the surface of the shell are still evaporating...but the radiation they produce is affected by the collective gravity of all the remaining black hole quanta. This redshifts the Hawking radiation, reducing the collective power output of the cloud and decreasing the temperature of the Hawking radiation to match the blackbody spectrum of a classical black hole. As with classical objects, however, this is a statistical match, not an absolute match as would be the case with a true classical black hole.
What about when the cloud runs out of matter to swallow up? Won't the black hole quanta evaporate away? Well, yes...but recall that the energy density of the Hawking radiation is on the order of the black hole quanta themselves. Therefore, as the radius of the collective cloud becomes many many orders of magnitude greater than the radius of a black hole quanta, the density of the emitted radiation becomes more uniform, while also becoming cooler:
Because the density at the surface increases while the temperature of the Hawking radiation decreases, the infalling radiation from the CMBR is sufficient to maintain the re-collapse of the black hole quanta. The black hole never really stops growing, it just grows more and more slowly.
This handily solves the black hole information paradox, because the no-hair theorem no longer applies. There is no black hole and no macroscopic event horizon; there is only a spherical shell of black hole quanta which are continually evaporating and re-collapsing. To an outside observer, the cloud will appear to act in many ways like a classical event horizon black hole, but there is no singularity. A Kerr black hole forms from the collective orbit of the many black hole quanta. While this isn't a complete theory of quantum gravity, it's a step in that direction.
So, that's the hypothesis.
Predictions? Well, the math needs to all work out. I'd have to go in and figure out whether the redshift would affect the spectrum properly, narrow down the "rest mass" of each black hole quanta, and so forth. I have to figure out how Hawking emission can be represented as quantum tunneling and then solve the square-well problem to figure out whether that side of things works. There are a lot of qualitative explanations that will need to be rigorously tested to determine whether they are quantitatively sound as well. But I'm optimistic.