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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