What happens to objects pulled in by black hole?

Question

What happens to the particles/elements/objects that sink into the gravitational pull and ultimately go to the interior part of a black hole? If, according to popular notion, it is crushed by the excessive gravitational high density thrust, what is that object reduced to? Which theory provides a basis for calculation of this effect?

Answer 1

If you stick to the theory of general relativity then what happens to the matter is fairly straightforward. As the matter falls inwards it experiences increasing tidal forces. The matter reaches the singularity in a finite (short!) time, and at the singularity it is compressed into a point with zero size and infinite density. Note that nothing special is experienced as the matter crosses the event horizon.

But we expect quantum mechanics to affect what happens. Until recently (more on this below) quantum mechanics was expected to become important only as the matter approached the singularity, and in particular when the matter had been compressed to a size approaching the Planck length. Exactly what happens at such small length scales we don't know because we have no theory of quantum gravity.

The idea of compressing matter to infinite density may seem absurd, but it has a sound theoretical basis. In everyday life matter is held in shape by electrostatic forces, but at the pressures in black holes these forces aren't strong enough to keep the matter in shape and it collapses to form degenerate matter. However even the forces in degenerate matter aren't strong enough to survive a black hole, and the matter will collapse further. At this point we don't know of any other forces capable of resisting further collapse, so when the degenerate matter collapses there is no known force to stop it collapsing completely into a point of infinite density. However because we don't know of any such forces doesn't necessarily mean they don't exist, and a future theory of quantum gravity is expected to introduce new effects that will resist the collapse.

I said above that until recently quantum mechanics was expected to become important only as the matter approached the singularity. In the last few years calculations related to string theory have suggested the black hole may be an even stranger object than we had thought. This is known as the firewall theory, and it suggests that spacetime may end at the event horizon so the black hole doesn't have any inside. Note however that this is a currently a speculative idea and may have no basis in reality.

Answer 2

Here's an answer from a nonspecialist who's reasonably comfortable up till the end and no further of "The Big Black Book" (Misner Thorne and Wheeler) and with a general lay reading other than this.

Although often asked and thought about a great deal by physicists, your question is an excellent one because it leads straight to the edge of our knowledge of physics by way of only fairly elementary concepts: to ideas such as the Black Hole Information Paradox and to grotesque and morbid descriptions of the horrific weird of any hapless spacefarer who happens to be the dropped object: spaghettification.

In recent years, some scientists have proposed that everything falling into a black hole is incinerated at or near the event horizon in a phenomenon known as the Black Hole Firewall. This is by no means mainstream- or even widely accepted: it is postulated to help resolve the black hole information paradox that I talk about below. But from hereon in, I shall discuss what was a fairly mainstream account of what is likely to happen to something falling into a black hole up until about 2012. It is a mainly classical, General Theory of Relativity understanding (which I have already warned you is my limit).

Looking At Things "From Outisde"

Rob's answer is an excellent physical and philosophical summary for the history of a dropped-into-a-blackhole object (let's call it $\mathscr{D}$ for "dropped" or "doomed") as seen from an observer $\mathscr{O}$ a long way off (theoretically infinitely far from the hole). From observer $\mathscr{O}$'s standpoint, the best mathematical description of the falling object $\mathscr{D}$ is through the Schwarzschild geometry, defined by the Schwarzschild metric (first fundamental form), which is a spherically symmetric solution for the metric tensor in the space around a point mass to the Einstein field equations and which was derived by Karl Schwarzschild in 1916 whilst he was being shot at in the WWI battlefields (what a guy!; I nicked the following from Wikipedia BTW to save time typesetting):

$$c^2 {d \tau}^{2} = \left(1 - \frac{r_s}{r} \right) c^2 dt^2 - \left(1-\frac{r_s}{r}\right)^{-1} dr^2 - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right)\tag{1}$$

where:

$\tau$ is the proper (time measured by a clock moving along the same world line with the test particle),

$c$ is the speed of light,

$t$ is the time coordinate (measured by a stationary clock located infinitely far from the massive body),

$r$ is the radial coordinate (measured as the circumference, divided by $2\pi$, of a sphere centered around the massive body),

$θ$ is the colatitude (angle from North, in units of radians),

$φ$ is the longitude (also in radians), and

$r_s$ is the Schwarzschild radius of the massive body, a scale factor which is related to its mass $M$ by $r_s=2 G M/c^2$, where $G$ is the gravitational constant.

So ${\rm d}\tau$ here stands for a short time interval as measured by a clock carried with our Doomed Observer $D$ and ${\rm d} t$ is the corresponding time interval as measured by a clock. Take heed of two things:

1. The careful wording that $r$ is the circumference of a sphere centred on the body divided by $2 \pi$: owing to non-Euclidean nature of general solutions to the Einstein field equations (i.e. the geometry does not fulfil the Euclid parallel postulate), this quantity is NOT the same as the sphere's "radius";
2. When $\mathscr{D}$ approaches the Schwarzschild Horizon (or Event Horizon) $r=r_s$, notice how the co-efficient of $d t^2$ vanishes. This is the description of the phenomenon cited by Rob's answer; that is, from $\mathscr{O}$'s standpoint, the doomed object $\mathscr{D}$ takes an infinite time to reach the horizon. Rob makes a very good point: we (as the observer $\mathscr{O}$) can never see the object crossing the event horizon, so there is no way of testing any model of what happens to anything inside. From our ($\mathscr{O}$'s) standpoint, theories of what goes on inside are not really science, if we accept Popper's Falsifiability Criterion of what science is.

Looking At Things from the Doomed Object's Standpoint: Crossing The Horizon

However, there is one observer whom in principle such a theory would be falsifiable for, and that is our hapless friend $\mathscr{D}$. So (if there is no Black Hole Firewall)we can make some good guesses for what happens (from $\mathscr{D}$'s standpoint) just after $\mathscr{D}$ crosses the horizon, at least for some shortish time afterwards. Especially if the black hole is a very, very big one, such as those that astronomers now believe lie at the centres of most galaxies. What about the singularity at $r=r_s$ in the Schwarzschild geometry I hear you ask? Well actually, that is an "artefact" of the co-ordinates: more physically, it is an artefact of constraining ourselves to observe from outside the black hole. There are co-ordinate systems that spirit away this horizon singularity. I'll get to an example of these co-ordinates in a moment, but the main point is they are written to understand things from $\mathscr{D}$'s standpoint. From the $\mathscr{D}$'s standpoint, what happens as they cross the horizon?. Short answer: (firewalls aside) probably not much at all at first, especially if the black hole is a very big one. If the black hole is big enough, regions much bigger than the size of $\mathscr{D}$'s spaceship are roughly "flat", i.e. just like the spacetime around you and me. The "gravity" at this point is not particularly "strong" (I'll also make this point more precise soon) and $\mathscr{D}$ dreys his dreadful weird later on. And, as a geometer, I refuse to believe that, as long as spacetime can be charted in such regions in a way that fulfils the Einstein Field equations, it's a reasonable hunch that they describe $\mathscr{D}$'s physical experience, at least until $\mathscr{D}$ gets too near the singularity. Many physicists will agree with me here, but I and most physicists also believe that Einstein's theory breaks down as $\mathscr{D}$ nears the singularity: when the "gravity gets much stronger" (again, I'll make this clearer further down). But it would seem a bit weird if the breakdown were "sudden" at some (from $\mathscr{D}$'s standpoint) arbitrary horizon.

So now for the co-ordinates. One can actually transform the co-ordinates in (1) (or, more strictly speaking, switch to another chart for the spacetime manifold) to remove the singularity at $r=r_s$. There are several co-ordinate systems that do this: one of the most useful is Kruskal-Szekeres Co-ordinate System. This maximally extends the charting of the Schwarzschild geometry, i.e. it charts the whole of spacetime aside from the singularity. You'll need to look them up as I'm a bit bored with cutting and pasting other people's equations but here is the most excellent Kruskal-Szekeres diagram which again I have nicked from Wikipedia (who nicked it from Misner Thorne and Wheeler) and marked up for our purposes:

The Kruskal-Szekeres Co-ordinates do a nonlinear transformation on the radial $r$ and observer $\mathscr{O}$ time $t$ co-ordinates (let's see whether my Planck to SI units transformation is correct):

$$U=\left(1-\frac{r}{r_s}\right) \exp\left(\frac{r}{2\,r_s}\right)\cosh\left(\frac{c\,t}{2\,r_s}\right)$$ $$U=\left(1-\frac{r}{r_s}\right) \exp\left(\frac{r}{2\,r_s}\right)\sinh\left(\frac{c\,t}{2\,r_s}\right)$$

to give two new co-ordinates $U$ and $V$. Don't worry too much about the details; these qualitative properties are what is important:

1. Each point on the diagram corresponds to a whole 2-dimensional spherical surface concentric with the singularity at some time $t$ as measured by $\mathscr{O}$;
2. The event horizon is two red straight lines $U=\pm V$;
3. The singularity is the top green hyperbola;
4. Contours of constant $\mathscr{O}$-time $t$ are the dashed straight lines through the origin;
5. Contours of constant Schwarzschild radial co-ordinate $r$ are the dashed hyperbolas;
6. All null lines (World lines or valid histories of light rays or, if you like, photons) are straight lines at $\pm 45^o$ with the vertical (horizontal). Light-cones in the KS diagram therefore look exactly like light-cones in special relativity.

There is a slight awkwardness insofar that there are two universes represented on the one diagram: the black hole outside and inside are above the diagonal $U=-V$, everything else is a corresponding "white hole" and its inside and outside.

Let that last comment (5) sink in for a bit. It means we can really use the KS diagram to build intuition about $\mathscr{D}$'s possible histories. I have drawn our doomed spacefarer $\mathscr{D}$ as the yellow dot, and their future lightcone is the upper region bound by the two orange lines. You can therefore see at a snap that $\mathscr{D}$ is doomed, no matter what they may do, to hit the singularity at some time in their future (recall that the sigularity asymptotes to the lines $U=\pm V$, so that the two orange lines will always intersect it, no matter what point within the black hole that $\mathscr{D}$ may be inside.

A second thing that is obvious from the KS diagram is that their past light cone (that part of the lower region bounded by the orange rays that is also in "our universe") cannot wholly contain any constant $\mathscr{O}$ time $t$ section of our universe. There is a part of our universe defined by $t=t_0$ for all finite $t_0$ outside the light cone, for the all the dashed lines (constant $t$ contours) intersect the past light cone edge somewhere. This tells us that $\mathscr{D}$ does not see the end of the universe as they fall through the horizon, even though, from $\mathscr{O}$'s standpoint, they take an infinite time to do so.

A third thing to note is an elegant description of the phenomenon cited by Rob's Answer. Witness that World Lines (histories) of a "stationary" observer (someone hovering at some radius $r>r_s$ outside the black hole) is an hyperbola (dashed curve) on the diagram that asymptotes to, yet never crosses, the event horizon. The $\mathscr{O}$ time $t$ is $\frac{2\,r_s}{c} \operatorname{arctanh}\left(\frac{V}{U}\right)$ and becomes infinite as $U\to V$.

Our hapless spacefarer, aside from seeing some interesting optical effects if they look over their shoulder, hasn't noticed anything much yet.

Looking At Things from the Doomed Object's Standpoint: Spaghettification

This will eventually change. Think of all the points in $\mathscr{D}$'s body separately. They freefall at first, i.e. follow spacetime geodesics. They can do this because, for sometime after crossing the horizon, at a given $\tau$ (from $\mathscr{D}$'s standpoint) the points on each geodesic are pretty much exactly where they would naturally be in the freefalling body. A region wherein special relativity holds is much bigger than $\mathscr{D}$ and their spaceship.

However, as the "gravity gets stronger" and $\mathscr{D}$ draws near the singularity, eventually a region around $\mathscr{D}$ small enough for special relativity to hold in gets smaller than $\mathscr{D}$ him or herself. This means that for different parts of their body, the spacetime geodeics diverge more than the distance between the unstressed parts of $\mathscr{D}$'s body. $\mathscr{D}$'s body's tensile and compressive strength at first resists this tendency, and the points deviate from their geodesics because they are forced to by bone and sinew and flesh. Just as the Earth must push up with a force that you call "your weight" on you to hold you up and hold you back from following spacetime geodesics (which would have you falling towards the Earth's centre at acceleration $g$ relative to the Earth's surface), so too do the connective mechanisms in $\mathscr{D}$'s body at first prevent different bits from following their momentarily comoving freefall frames. As $\mathscr{D}$'s leg "wants" to follow a certain geodesic, and $\mathscr{D}$'s torso an ever more nonparallel one, the tensile / compressive (depending on $\mathscr{D}$'s orientiation) stress between the two gets bigger and bigger. I think you can see where this is going. Before the black hole firewal was mooted, I think many if not most physicists thought that GTR would hold good well long enough to describe a scene with as much gore as even Alfred Hitchcock could swallow. The spacefarer's body is eventually torn asunder, being drawn out along radial lines and squashed in the orthogonal direction. This grotesque and gruesome fate is called "spaghettification". The following diagram, from the Wiki "spaghettification" article, probably does better than my words.

The Limits of GTR and the Black Hole Information Paradox

OK, we've told a tale to sate Alfred Hitchcock's bloodlust, and our hapless spacefarer is dead, but what's the ultimate fate of the matter of their body? This is where we reach the limit of our knowledge of physics. Many physicists, especially those like me who only have a classical GTR knowledge, will tell you that GTR holds outside some smallish region near the green hyperbola in the KS diagram, but that somehow GTR must break down and be replaced by something that makes it altogether compatible with quantum mechanics.

Ultimately, is thought a black hole evapourates in Hawking Radiation, and the question of how the information content of what goes into a black hole and what comes out and even whether that information content is the same is the substance of this discussion, and some future theory of quantum gravity is sought to answer these questions. Many physicists believe that information is conserved, i.e. the mapping between the quantum state of the World at any time and that at any other time is a one-to-one map and reversible. That is, in principle, if you know the full quantum state of the World at any time, you can foretell what it will be at any other and you can also say what the state was whence it came at any past time. "Nature does not forget how She got to Her state" at any time is a colloquial way of saying this. Some physicists, though (although I think it is fair to say as yet a minority) believe that this principle may be violated as a black hole swallows matter and evapourates. Now I'm likely as out of my depth as you are, so I would suggest reading the Wiki Page on the Black Hole Information Paradox for further details.

Answer 3

If the event horizon is really a one-way surface, the answer is that whatever happens inside the event horizon is irrelevant to our universe. An object that crosses the event horizon will appear to an outside observer to take an infinite amount of time to do so, and no model of the interior of a black hole can ever be tested.

Whether the event horizon is really a one-way surface is an open question that has received a lot of discussion, usually in the context that predictions for what happen inside the event horizon don't make very much sense.