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What actually does happen to an object after it crosses the Event Horizon of a BH? Does it-

  • lose the properties of matter and become a new, undiscovered type of matter?
  • does it become negative matter?
  • does it get completely converted into energy? (if so, what type of energy?)
  • or does something else happen?

Actually, is there any theory on this question, or is it opinion based?

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  • $\begingroup$ @ACuriousJim that should be an answer $\endgroup$ – David Z♦ Apr 18 '15 at 17:28
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We don't have any adequate theory to describe what happens to matter inside the event horizon (and some theories even say that matter never actually crosses the event horizon). But let me address you specific questions:

1) It's possible that matter entering a black hole is converted into a new form of matter.

2) I have no idea what negative matter would be. Either way, no, I fairly positive that whatever this negative matter is, that's not what it becomes.

3) No, since the black hole has a mass, we can be reasonably sure that matter that falls into it keeps its mass. While mass is technically a form of energy, the mass that falls in doesn't get converted to pure energy.

4) Yes, something else probably does happen. As I mentioned, we have no real idea what that something is, we can only guess and approximate.

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According the old classical general relativity, The event horizon can be approximated by flat spacetime so nothing happens when it crosses. But space and time coordinate interchanged for an outside observer, so its energy and momentum are interchanged for an outside observer. However, firewall is a new phenomena considering quantum effects, it says the event horizon is replaced by firewall

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  • $\begingroup$ I think that is wrong saying things like: "...so its energy and momentum are interchanged." Momentum and energy now is a pure relativity thing. For a outside far enough observer this is true, and is because that we have a event horizon. For the free falling observer this is not true. The energy and the momentum still pretty the same when cross the event horizon. After all don't make any sense to talk about a event horizon for a free falling observer. Event horizon is only a thing that is seen by the outside observer. $\endgroup$ – Nogueira Apr 18 '15 at 17:38
  • $\begingroup$ Only in the outside observer! This is my point. For an free fall observer the energy and momentum do not interchange. This interchange is only a result of a "bad" coordinate system. $\endgroup$ – Nogueira Apr 20 '15 at 19:14
  • $\begingroup$ However, on a second thought, i will appreciate if you can prove it with Kruskal coordinate $\endgroup$ – Haolin Lu Apr 21 '15 at 16:35
  • $\begingroup$ Everyone wins when the truth is achieved ;). You don't need go so far. Is simple an application of equivalence principle. For a free falling observer, locally, are an inertial observer who see things locally with minkowski metric. Energy and momentum assigned for a neighbour free fall object is simple as any other problem of minkowski metric. The distortion o space and time is in order of Schwarzschild's radius $r_s$ and $\frac{r_s}{c}$ respectively. $\endgroup$ – Nogueira Apr 21 '15 at 19:49
  • $\begingroup$ By this considerations of typical length and time, we can make sense of word as "neighbour" and "locally". Actually, in spirit of general relativity, we can't assign an absolute meaning of energy and momentum so easy. We have a lot of attempts as stress–energy–momentum pseudotensor. Note that here the energy and momentum in question is the energy and momentum due gravitational field $\endgroup$ – Nogueira Apr 21 '15 at 19:55
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In general relativity it is believed that the Schwartz Child radius is approximately Ricci-flat spacetime, meaning that there is little tidal force or at least no amount that should be surprising. A smaller black holes have a greater tidal force because the Schwartz Child radius is closer to the singularity than in a bigger one.

In this theory falling in doesn't seem special to the person who is falling in until they get close enough to the singularity to start feeling extreme tidal forces, enough to induce stretching extreme enough to be called "spaghettification".

To an outside observer, the person falling in would never hit the Schwartz Child radius, they would slow down and flatten as they approached asymptotically to the edge of the black hole. They would eventually disappear from view as the red-shift stretched the wavelength of the light reflecting off of their body to enormous sizes.

What interests me the most is that inside the Schwartz Child the "proper time" (time experienced by the observer) mathematically switches with the distance of the observer from the black hole, so space and time switch; of course this is only a mathematical expression of what happens and it doesn't imply that one would feel anything odd.

There are also theories that state that the outside observer would witness the person falling in being thermalized, radiated out like infrared light off of a heat source. But it is usually also believed that the person falling in will not experience being thermalized, and will go on for a bit longer to be spaghettified. So it is theorized that there is a complete disconnect between the two observed events depending on the observer, but there is no way for them to communicate once one of them is passed the Schwartz Child radius so there is no real contradiction there.

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