What happens at the atomic level when a black hole forms?

Question

As I understand it, during a gravitational collapse, the star stops fusion and gravity overcomes the radiation pressure. At that point, there is a point in space where there is a lot of gravity in a small amount of space which leads to the formation of a black hole? My question is that, right before the formation of the black hole, if there is a "designated" spot in space where the gravity is highest and is where the black hole forms. If so, what happens at the atomic level at that specific spot right before the formation? What specific action of a particle(s) produces a black hole? Is this process the same if the formation is not due to a gravitational collapse of a high mass star (for example a high-energy collision)?

Answer 1

Gravitational collapse. which cretes a black hole, does not happen at a single point in space. A black hole will form if a given mass $M$ is compressed into a radius smaller than its Schwarzschild radius $r_s = \frac{2GM}{c^2}$. This requires an average density equal to the Schwarzschild density

$\displaystyle \rho_s=\frac{3M}{4\pi r_s^3} = \frac{3c^6}{32 \pi G^3M^2}$

Although the Schwarzschild density needed to a create stellar mass black hole is very large, $\rho_s$ is proportional to $M^{-2}$, so the Schwarzschild density for a much larger black hole is surprisingly small. For example, the Schwarzschild radius for a black hole with a mass equal to the Milky Way ($\approx 10^{42}$ kg) is around a quarter of a light year, and the corresponding Schwarzschild density is only around $1$ gramme per cubic metre. At this density, nothing "unusual" happens at an atomic level.

Answer 2

Black holes doesn't quite form at a spot. Informally we often talk about them as black spheres defined by the event horizon, but they are actually extended structures of spacetime curvature: the horizon is not a local property but just like an everyday horizon seems to move as you move. The "real" event horizon that delineates the point-of-no-return certainly exists but is invisible and can show up even if nothing seems amiss to a local observer. (Strictly speaking, the black sphere in pictures is usually the photon sphere at $1.5 R_S$)

When a star implodes, the implosion is both a change in the nearby spacetime structure and compression of the matter residing in the spacetime. The spacetime change is what general relativity likes to study and there are numerous papers modeling the implosion under different assumptions. The key thing is that as the matter contracts the curvature of spacetime increases. This actually makes the amount of volume available inside the star grow. However, the collapse is stronger and matter is spaghettified by curvature-induced forces: in a sense "down" is extending much more than the sideways directions.

But since time and space are dynamic here and light hardly has time to traverse this mess, "when" and "how fast" needs to be argued carefully and mathematically - different observers see different things. An event horizon quietly forms when the mass is dense enough, but to observers riding the stellar matter nothing odd seems to happen - it is just that light from the outside universe is starting to get extremely blueshifted. After a certain (short) time according to this observer the curvature and blueshifts all go to infinity and the singularity is reached. "Meanwhile" on the outside observers see the collapse slow and redshift into the familiar black event horizon.

In short, there is no spot where an observer riding the star sees a black hole form. And it is not the fault of any individual particles the hole forms, just their collective mass.

Now, what happens to the matter? Basically it is subjected to the increasing tidal forces. One paper analysing this argues that spaghettification turns matter into a quark-gluon plasma (after first ripping apart macroscopic objects and then atoms). Eventually the heating gets so large and the rate of change so quick that hydrodynamic models of the plasma fail. It is unclear what happens then, and a bit later (as seen by particles involved) one enters the realm of quantum gravity effects that is even less understood.

In the high energy collision case there is no real collapse, just particles forming a back hole state that then quickly decays by Hawking radiation. That might just be seen as an interaction vertex in a Feynman diagram. Of course, here we start in quantum gravity realm, so it is hard to say much.

Answer 3

In this article they are talking of quark deconfinement and a quark gluon plasma forming before the black hole in supernovas.

This means that the nuclei have disappeared and the protons and neutrons that composed them also disappeared.

What specific action of a particle(s) produces a black hole?

It is not an action of particles, the particles are acted upon by the various macroscopic forces, from pressure to radiation to gravitational collapse, all are macroscopic emergent effects of very large masses, i.e. zillions of particles together..