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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.

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But, as John Baez points out, can this burst of light from the future universe history ever get into your past light cone? –  twistor59 Sep 30 '13 at 7:52
    
Related: physics.stackexchange.com/q/21319/2451 and links therein. –  Qmechanic Sep 30 '13 at 15:24
    
twistor59, Qmechanic, thanks. I exchanged emails with John Baez on this topic years before he wrote that piece. What left me baffled was a subtle switch in whose time standard was used. That's why I've tried to phrase this variant of the question as carefully as I could in terms of predicting the time stamps that would be seen. That shifts the focus from "Will he see the end of the universe?" (Baez), to the presumably more calculable question of "What is the last time tag he sees?" It implies a finite cutoff point, but on what parameters is that cutoff based? I don't think I've seen that. –  Terry Bollinger Oct 27 '13 at 1:37
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4 Answers 4

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.

Re your question about interacting with the future universe, it's important to distinguish between a shell observer who hovers at a fixed distance from the black hole and a freely falling observer who falls in. Freely falling observers cross the event horizon at the speed of light, so light falling in from the rest of the universe can't catch them. If you hover just outside the event horizon then I haven't seen a calculation of this but I suspect that you're correct and it would get very uncomfortable.

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There's nothing I can object to here as a matter of technical fact, but I disagree with your interpretation. 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. I think the key here is your word "correspond." You seem to have in mind a certain notion of correspondence that you feel is natural. I don't agree that this particular notion of correspondence (which you have only defined implicitly) is natural or [...] –  Ben Crowell Sep 30 '13 at 15:15
    
[...] unique. A distant observer who collects data from the region near the horizon is making indirect inferences about what he believes to be happening there. He can choose to apply various corrections for effects such as time dilation, or not to apply them. He can choose one notion of simultaneity or another. He can infer that matter has "really" "already" fallen in, or not. –  Ben Crowell Sep 30 '13 at 15:19
    
Ben Crowell, thanks for both comments. John Rennie, your reiteration of the standard explanation is appreciated, but try this: since the "freely falling observer" clock undergoes real time dilation where the ratio of its seconds to those of most clocks in the embedding universe becomes infinite, what is the exact objective meaning of "freely falling"? That is, if the falling clock sees a span equal to the entire history of the outside universe going by before it hits the EH, is it really "freely falling" at $v=c$, or does it require actual infinite external time to fall in? What does it see? –  Terry Bollinger Oct 13 '13 at 0:48
    
When you ask "what does it see" I assume you mean what would a camera carried by the infalling observer record. If so, there have been various calculations of this over the years, though surprisingly Googling hasn't found anything. If you can find a copy of William Kaufmann's book The Cosmic Frontiers of General Relativity this describes the results (but not the details) of the calculation. In my edition it starts on page 122. As you approach the event horizon your field of view narrows to a point. This is because you are now moving at the speed of light. –  John Rennie Oct 13 '13 at 9:22
    
Sadly, you do not see the end of the universe :-) –  John Rennie Oct 13 '13 at 9:22
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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)

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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?

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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.

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