Can matter really fall through an event horizon?

Question

This question is closely related to Event horizons without singularities from about a year ago (May 2012), which John Rennie answered nicely and persuasively.

My variant of the question is this: Given an existing large-scale black hole and associated event horizon, how does matter manage to fall through the event horizon?

Here's the reference thought experiment I'm using:

Assume two clocks communicating in both directions via radio or laser, with the observer clock kept distant from the black hole, and the falling clock heading towards the event horizon of the black hole.

Each clock "sees" the time units (call them seconds) of the other clock through the radio link, and can express the size of those time units in terms of its own units of time. The observer clock watches the time units of the falling clock grow quickly in length, until at the event horizon the seconds of the falling clock become infinite in length. To the observer clock, it looks as though the falling clock has become suspended in time at the event horizon, since a clock with infinitely long seconds requires an infinite time to do anything.

Now the standard interpretation is that since the falling clock has its own time standard, it sees nothing amiss at the event horizon. In that interpretation, the apparent "freezing" of the falling clock is more-or-less an illusion caused by the falling clock red-shifting out of communication with the rest of the universe. By that interpretation, the observer clock is viewing what amounts to a massively slowed-down recording of the moments right up to the falling clock leaving the visible universe. The remaining, um, "singular" fate of the falling clock is simply hidden from view. It is an appealing scenario, one that "feels right" for interpreting the oddities of infinite time dilation.

My problem is with what the falling clock sees.

As best I can understand it -- and my question is what I'm seeing wrongly -- the falling clock will not see the event horizon as a "no big deal" event. Instead, it will see the time flow of the observer region accelerate very quickly, so that the falling clock can observe and in principle exchange data with events in the very distant future of the observer universe.

A second rather noticeable effect will be that unless the external universe stays very dark indeed, the falling clock will be incinerated by blue shifted radiation before hitting the event horizon proper. Why that is so is not hard to see: If at some point the falling clock measures the observer clock as having seconds that are one billionth the length of its own falling clock seconds, then the frequency of any electromagnetic radiation sent to it from the observer region will also be multiplied in frequency by a billion times. Or from the observer clock perspective, the falling clock has slowed down so severely in time that it begins accumulating energy over very long periods of time.

My biggest problem is that if the falling clock can interact with the future universe, no matter how painfully, its time dilation is necessarily real and observable, and not simply a left-over recording of the last moments of its fall out of this universe. So, if the falling clock is still available to interact with an observer clock a billion years from now, then it is not truly "in the black" yet, just very, very cold and slow -- and still perched very close to, but still not all the way through, the event horizon.

This would mean that regardless of how the black hole formed -- which is a separate question, and one that John Rennie addressed nicely last time -- then once it has formed, external matter and light cannot penetrate its event horizon.

So what is the deal here? Is there something fundamentally wrong with my thought experiment? How exactly does a falling clock move through a region where the seconds are infinite in length? (And one more thought in passing: Does the observer clock also appear to become more distant in space? That might help... maybe?)

Addendum 2013-10-27

Here's the most succinct version of the question I can come up with:

What is the mathematical procedure for calculating the last distant-clock time tag that the falling clock sees as it approaches the event horizon?

The above version keeps the calculation firmly embedded in the time system of the falling clock, avoiding the dangers in statements such as "time flows normally for the falling clock." That assertion is patently true, but since it does not calculate the last time tag seen, it does not answer the question.

@twistor95 left a nice, highly relevant reference to an online article by John Baez on why most physicists now think that a clock falling into a black hole will not see the end of the universe. @Qmechanic noted an an earlier Physics SE question that to be honest I think is the same as mine. (I really did look, but Qmechanic is a lot better at such searches.) What's worrisome is that the answer to that question was the older physics view that the falling clock would see the end of the universe!

I must 'fess up that I exchanged a few emails with John Baez on this topic years before he wrote that piece. What left me baffled at that time was a subtle switch in whose time standard was used. So, in my current variant of this question, I tried as carefully as I could to ask the question in terms of predicting the last time stamp the falling clock would receive. This phrasing shifts focus from "Will he see the end of the universe?" (Baez: no) to what the math predicts. Prediction is, after all, the very essence of what good scientific theory is all about.

The old answer was that the cutoff occurs at infinity, that is, at the end of the external universe. If you accept that cutoff for a clock that is merely approaching the event horizon and has not yet fallen through it, then the idea that you can fall through an event horizon becomes very problematic indeed. See for example @Anixx's accepted "collapsar" answer for the older version of this question.

So, since the current answer is that the falling clock does not see the end of the universe, the visibility cutoff must necessarily occur for a tag that is well short of the end of the external universe. You cannot assert the one ("no end seen") without implying the other ("some tag will be the last one seen").

So again: If "no end is seen" is the answer, how is the implied final tag calculated?

I will be blunt on one point: As someone with an information technologies background, I see no strong reason to view either the old or the new answer as more persuasive. Untestable code, whether mathematical or programmatic, is always in danger of errors.

Addendum 2013-10-30

My question has been very nicely answered (no end-of-universe is seen!) in this new (2013-10-29) question asked by John Rennie:

Does someone falling into a black hole see the end of the universe?

Michael Brown provided the answer, and John Rennie then iced the cake by providing an additional diagram that shows the actual intersection of the outside time stamp with the falling clock. Beautiful and elegant stuff!

Alas, though, it also means I don't have an answer to check here. @MichaelBrown, if you happen see this and don't mind adding in a link to your other answer as an answer here, I'd be happy to flag your link to close out this question.

Answer 1

I suspect you've fallen foul of of the conceptual problems that frequently beguile budding general relativists. Assuming I have understood your question correctly it boils down to what really happens?, the implication being that the two experiences of time can't be so different. Indeed, you say:

the apparent "freezing" of the falling clock is more-or-less an illusion caused by the falling clock red-shifting out of communication with the rest of the universe.

The differences in the time experienced by the observer and the falling clock are not illusory. the observer really does measure an infinite time for the clock to reach the event horizon, while the clock measures a finite time. As far as the distant observer is concerned, indeed any observer outside the event horizon, the region inside the black hole does not exist in the sense that there is no spacetime coordinate ($t$, $x$, $y$, $z$) that the external observer can assign to points inside the event horizon. We can construct coordinate systems like Kruskal-Szekeres coordinates that are continuous and well behaved across the event horizon, but these coordinate systems do not correspond to measurements real observers would make.

Answer 2

The question, "can matter fall through an event horizon" is troubling because it raises a number of interrelated issues. To even begin to consider these issues, we must first consider whether a black hole can increase in mass.

I think there is general agreement that black holes can grow in mass. Thus, it is pretty much accepted that black holes "grow" by pulling in matter or colliding with other black holes. It has even been demonstrated mathematically that the mass of a black hole determines the diameter of its event horizon which defines the outer limits of what we think of as the black hole.

Also, gravitational waves have recently been detected from a black hole merger and the data suggests that two black holes coalesced into a single larger black hole, minus the release of the predicted solar masses of energy in the form of gravitational waves.

So we understand that black holes can grow, but this begs the question of exactly how does a black hole increase its mass. Conventional theory suggests that incoming matter just passes the event horizon and coalesces with the singularity at the heart of the black hole. But does this make sense?

We have no idea regarding what is the nature of the "space" between the singularity and the event horizon. Yet it is said with some certainty that matter can traverse this space to reach the singularity and add to the mass of the black hole. That is a pretty simple and comforting mental picture that seems to ignore some obvious difficulties. First, there is no proof that all of the mass of a black hole is necessarily concentrated at its singularity. Some mass could be smeared out over the event horizon or even concentrated in the area between the event horizon and the singularity.

Also, the event horizon defines the outer limits of a region that is so time-dilated by gravity that time itself is frozen. How can matter or even energy traverse a space frozen in time? Movement requires velocity or acceleration over time. With no time, how can movement toward the singularity occur within the sphere defined by the event horizon?

Time dilation also rears its ugly head when you consider the time frame of a distant observer. From that perspective, in-falling matter will seem to freeze in time just as it reaches the event horizon. That "frozen time perspective" is explained away by switching to the time frame of the in-falling matter which is not dilated. Many scientists therefore conclude that in-falling matter will simply pass over the event horizon and approach the singularity with no local effect other than shredding of matter due to differential gravity. This seems like wishful thinking to me as it fails to even suggest how matter can exist and move in the space within a black hole.

Worse, what you are left with is the in-coming matter at the outer edge of the event horizon frozen in time from one perspective and moving toward the singularity from another local perspective. This seems to be an unacceptable paradox. Some have argued that the remote "frozen in time" perception is just an image and not physical reality. Trust me, they say, that frozen image will just eventually fade away since it does not represent the actual location of the new matter embedded in the black hole. This seems like more wishful thinking. I think there is a paradox here that suggests crossing the event horizon and achieving "frozen time" is just not possible.

Consider the time dilation that occurs when an object approaches the speed of light. In that scenario there is no way that you can ever achieve an absolute dilation that freezes time from a remote perspective for the moving object. The reason for this is that you can't possibly move a physical object at the speed of light and therefore freeze time. That feat would require infinite energy, so it just can't happen and there is therefore no time dilation paradox for travel near the speed of light. Why then is it arguably possible to approach and cross the event horizon of a black hole and achieve complete time dilation and frozen time? I know this logic is somewhat specious, but it makes me question the glib assertion of many that the event horizon can be crossed and full time dilation achieved.

So what is the answer here? Can matter and energy cross the event horizon to add to the mass of a black hole? No one knows with absolute certainty. I think it is likely that as matter approaches the black hole, it is smeared over the outer surface of the event horizon to add mass to the black hole and grow the event horizon, without actually crossing the horizon. The same process would apply for Hawking radiation and merging black holes. There will be a lingering image of the in-coming mass adjacent to the black hole before it is absorbed, but this image will not be frozen in time and will eventually dissipate. The process of surface absorption of new matter will also form jets of ejected matter.

The above paragraph is just a guess. It is probably likely that the true answer is actually something more bizarre than we can imagine at this time.

Answer 3

The falling observer cannot observe or exchange information with the very distant future universe. They will continue to receive information from the outside universe up to the point they cross the event horizon and beyond. However, signals they send back will cease to move outwards once they cross the event horizon. Thus the event horizon is a one-way signal membrane.

The falling observer will not see the universe outside in fast-forward as it were. The falling observer will have a well-defined amount of time to receive information prior to reaching the event horizon. This time is not infinite because, even in Schwarzschild coordinates, the light catches up with the falling observer in a finite time. There is then also a very limited amount of time to receive further information from the outside universe once they have crossed the event horizon.

As a challenge, using Schwarzschild coordinates (which is not the easiest way to do it), I showed in https://physics.stackexchange.com/a/396157/43351 that if we consider something signalling to a falling observer from a distance $r_0$, where the observer falls past them from infinity at $t=0$, that the coordinate time interval $\Delta t$ that they can wait and still send a signal that intercepts the falling observer before they reach the event horizon is (in units where $c=1$) $$\Delta t < \ln \left(\frac{4r_s}{r_0 - r_s}\right)r_s + \left( \frac{2}{3}\left(\frac{r_0}{r_s}\right)^{3/2} + 2\left(\frac{r_0}{r_s}\right)^{1/2} - \ln \left| \frac{\sqrt{r_0/r_s} + 1}{\sqrt{r_0/r_s} -1}\right| - \frac{5}{3}\right)r_s - r_0\ .$$ [Note that this calculation was independently confirmed by Pulsar using Kruskal-Szekeres coordinates (easier) https://physics.stackexchange.com/a/396829/43351 ]

Then, following the falling object on the other side of the event horizon one can calculate an additional time increment of $0.28r_s$ that can be added to the expression above, which accounts for the fact that the falling body can still receive signals from the outside universe between when it crosses the event horizon and meeting the singularity.

For the case of $r_0 \gg r_s$ then $$\Delta t \simeq \frac{2}{3}r_0^{3/2}r_s^{-1/2} - r_0\ .$$ But this is just the freefall time for an object falling from infinity to go from $r_0$ to the singularity minus the time it would take light to travel through flat spacetime from $r_0$ to 0. So this value of $\Delta t$ is pretty much what you would expect if the falling observer had travelled through flat spacetime. Thus from the point of view of the falling observer almost nothing strange seems to happen when looking back (radially) at the (distant) outside universe and the temporal extent of the events they can witness is almost unaltered by the spacetime curvature unless those events are also quite close to the black hole themselves.

Thus the cut-off Schwarzschild coordinate time-tag is $$t = \ln \left(\frac{4r_s}{r_0 - r_s}\right)r_s + \left( \frac{2}{3}\left(\frac{r_0}{r_s}\right)^{3/2} + 2\left(\frac{r_0}{r_s}\right)^{1/2} - \ln \left| \frac{\sqrt{r_0/r_s} + 1}{\sqrt{r_0/r_s} -1}\right| - \frac{5}{3}\right)r_s - r_0 + 0.280r_s\, ,$$ where the clock starts as the falling observer passes the position $r_0$ (from infinity).

If you wished to convert that into a timestamp measured in the proper time coordinate at a stationary position $r_0$ then you would have to use the usual $(1 - r_s/r_0)^{1/2}$ time-dilation correction.

It would be interesting to do some calculations for light on non-radial trajectories.

Answer 4

Well, this is by no means settled, but that's ok (meaning any answer is likely going to be seen as controversial until further observation provide more data).

John Rennie's answer doesn't appear to address whether or not singularities actually form in the first place; he does suggests that few skeptics would argue this, but in fact a few have (David Hilbert, for example, argued infinities cannot physically exist in nature, but merely as metaphysical constructs, and so would likely argue against a singularity with infinite space/time curvature. Likewise Pawel Mazur and Emil Mottola in "Gravitational Condensate Stars: An Alternative to Black Holes" have assumed the same limitation as Hilbert and explored the theoretical consequence and come up with the 'Gravastar', a theoretical construct with properties similar to black-holes, and mathematically no less real (given that black-holes are only ever observed indirectly by the effect they have on surrounding space ). Notwithstanding John Rennie's suggestion about singularities, he's otherwise made it pretty clear, if singularities do exist, they cannot do so without an event horizon. That still doesn't address event horizons without singularities though.

Consider, as your falling clock falls toward an event horizon it gains speed, and loses thermal energy (else poses problems for the black-hole due to the 2nd law of thermodynamics). As it falls more quickly, because its time slows, the rate at which it loses energy will appear to decrease. It will appear to go the speed of light at the event-horizon but never appear to cross over, or go faster, because its time appears stopped. It should also not have any energy. Because black-holes are not thought to break the 2nd law of thermodynamics it is generally believed necessary that all energy is radiated away through thermal or Hawking radiation before the event Horizon is reached, but for the sake of your question lets suppose this doesn't matter. The point is that once it reaches the event-horizon the rate of energy loss must appear to be zero for, as you point out, this clock's time period compared to the remote clock observing, will be infinitely long.

Matter whose energy content, whose rate of energy energy emission said to be effectively zero is recognized to be Bose-Einstein condensate - which raises a question about the nature of the event-horizon. Paul R. Anderson, Roberto Balbinot, Alessandro Fabbri, Renaud Parentani and others have shown (Hawking radiation correlations in Bose Einstein condensates using quantum field theory in curved space) that a Bose-Einstein event horizon will produce Hawking radiation, and this has been observed (Hawking radiation in a two-component Bose-Einstein condensate, P.-É. Larré, N. Pavloff)

Although some have argued that 'infinite time' is an illusion caused by the falling clock red-shifting out of communication with the rest of the universe, it doesn't need to be mere illusion, but may actually be (well, as close as we can get at least) if all energy is radiated away prior to reaching the event-horizon causing the formation of Bose-Einstein condensate whose surface is the event-horizon itself. This means that as the clock's period appears to slow it does so because all it's energy is being lost which suggests that not only is space time, but also that energy is as well - not a surprising result given super-string theory which posits matter and energy are one and the same.

By that interpretation, the observer clock is viewing what amounts to a massively slowed-down recording of the moments right up to the falling clock leaving the visible universe where the rate of emission of energy is directly related to the energy still contained in the clock as it falls.

This means then, with respect to your original question that; you can have event-horizons without singularities, albeit Bose-Einstein condensate ones known as 'Gravastars'; matter cannot actually fall through an event horizon, since the EH effectively represents a zero-energy boundary; and that infinite space time curvature and singularities need not exist for there to be event-horizons (Hilbert would be so pleased).

It also means that the falling clock is still available to interact with an observer clock a billion years from now, that it is not truly "in the black" but to do so would require it to be able to absorb energy. The property of Bose-Einstein condensate with respect to Hawking radiation has been studied, both near and on the event-horizon itself, and although it is not impossible for this to happen, it is not very likely. Matter near but not touching a Bose-Einstein event-horizon can still radiate away energy, but matter in physical contact with it must become part of the event-horizon itself (which is fascinating too because Bose-Einstein event-horizons have also been shown to have fractal dimensions)

Answer 5

These answers giving Hilbert's (and I believe Einstein's) viewpoint(s) are interesting, but perhaps there is a simpler way of expressing the issue: if time "stops" at the event horizon, then the local timelines actually truncate at a finite time before anything actually reaches the event horizon.
Taking this a little further, this would imply that the event horizon never quite forms. this is the same as saying that, in external time, the black hole never fully forms. This would remove any mystery about the growth of a black hole: the near-black-hole within a certain radius is simply tends towards its limiting mass, and growth merely adds a contimuum of new layers around it. Clearly, this description as worded applies only to a Schwarzschild hole, but the extension to Kerr-Newman holes is conceptually straightforward.
If this represents reality, the levels of Hawking Radiation would be pretty-much unaffected; I'm not certain what (if any) impact the non-existence of the event horizon would have on the paradoxes of that processes. I realise that this post is rather late in the day; nevertheless, better-informed critique would be welcome.

PS - addendum following Terry Bollinger's (TB's) initial response: If I understand it correclty, the final clause of TB's response may be the nail in the coffin of my argument - i.e.: "the absolute horizon starts as an interior point". If I read this correclty, it implies that the size of the event horizon is expanded by the presence of mass outside the horizon. If this interpretation is correct it would allow a "real" event horizon to arise in finite time (albeit this assumes that the analyses that gives rise to this result is properly time-constrained - see below).

However, TB also talks about "physicists' view that only the infalling viewer counts"; certainly it should be possible for the infalling viewer to come up with an accurate assessment of what it observes - indeed, the equivalence of all properly analysed viewpoints is one of Einstein's own tenets. On the other hand, this equivalence means that any analysis from the viewpoint of the infalling viewer that cannot be reconciled with outside viewpoints is meaningless to the outside viewer - and very probably meaningless in aggregate. Thus the infalling viewer counts only as long as its existence is meaningful; thus, it cannot be meaningful to analyse this viewpoint beyond the time that the viewer ceases to exist in an externally-consistent model. I regard any analysis that ignores this as failing the most basic credibility test.

Answer 6

The old answer was that the cutoff occurs at infinity, that is, at the end of the external universe.

It seems that you treat "infinity" as a particular moment in time. However, "the end of the universe at infinity" simply means: it will never happen. There will never be an end to the universe in such case, therefore nobody will be able to see it, regardless if they are falling into the black hole endlessly or not (and if they live forever or not).

So, going back to your original question "Can matter really fall through an event horizon?", if it takes finite time than the answer is a decided "yes", otherwise the answer is "no". The end of the universe (presuming it happens "at infinity") has nothing to do with it.

Answer 7

I have considered this same question, and I think it is quite insightful, and probably has profound ramifications. If observer A falls toward a black hole, they appear to an external observer B to never actually pass the event horizon, separating the external observer from the causally disconnected "inside" of the black hole. The external observer B never sees observer A cross the event horizon, so as far as I can tell, at any time for observer B it should be possible for A to turn around and reunite with A by expending some finite amount of rocket fuel to ascend back out of the gravitational field. Evidently, although observer A measures finite proper time to reach the "event horizon", only the proper time prior to that is actually physical from observer B's perspective. So any information carried by observer A is always available, in principle, to observer B.

Answer 8

Are you considering Swartzschield space (eternal unchanging black hole) or evaporating black hole?

In eternal unchanging black hole, in Kruskal-Szekeres coordinates you cross the event horizon in the infinite future (by coordinate time).

Since real black holes should evaporate, this means you will not cross the event horizon before the black hole evaporates, but I know of no good mathematical model of what should happen in this case.

Back to the eternal solution. As you approach the BH, you find yourself in the far future by the coordinate time, but you cannot receive information from the faraway world beyond some point in time because you are quickly moving towards the BH.

If at the last moment you change direction from the BH, you will quickly find yourself in far future and receive all the information sent to you all this time.

If you still fall on the black hole... Well, after infinite coordinate time (but finite proper time) you cross the horizon (remember no evaporating BH exists infinite coordinate time!).

Even though in the outside universe passed infinite time, your last received information belongs to some finite moment.

Answer 9

It is striking how people are trying to apply the theories outside of their domain. Your question asks about the Schwartzshield's solution to the General Relativity equations, which is unchanging and eternal. Yet, it is known, that any black hole has a finite lifetime defined by its evaporation.

What result are you expecting to derive by trying to draw any conclusions from a purely mathematical solution and extrapolating its conditions to the timespans to which it is known the model is unapplicable?