The well-known fable of an astronaut sending signals out to an external observer while falling toward an event horizon states that the time lapse between such signals becomes greater even if in the astronaut sends them out periodically (as judged in his inertial frame). When viewed from earth and weighing time-dilation due to the gravitational field of the collapsing star, how is it possible that a black hole can form in finite time (for any external observer) if it takes an infinite amount of time to "see" events occurring at the event horizon?
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For an observer on the surface of the collapsing star, the black hole forms rather quickly. – Alfred Centauri Mar 05 '14 at 23:30
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Right... but how does that permit black-holes forming for any external observer in a finite amount of time? – Mar 06 '14 at 00:07
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This is straying into metaphysics I'm afraid (If a tree falls in a forest...). Either the horizon is in spacetime or it isn't. If it is, some worldlines intersect the horizon and end on the singularity regardless of any external observer, i.e., it objectively exists. – Alfred Centauri Mar 06 '14 at 00:22
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1It is not metaphysics at all... it's clearly physics as astronomers claim the existence (in the here and now) of black-holes. Furthermore the event horizon is a valid boundary according to GR.. So physics predicts a certain object- we should be able to observe it- quite unlike your tree – Mar 06 '14 at 00:31
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Or is physics now free of backing up theory with experimental data? – Mar 06 '14 at 00:34
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2http://physics.stackexchange.com/q/21319/ http://physics.stackexchange.com/q/102202/ – jpm Mar 06 '14 at 00:35
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The accepted answer is unacceptable... particularly "Such collapsars possibly can become BHs for a short time due to quantum fluctuations and thus emit hawking radiation." but quantum fluctuations happen in an observably finite time.. it's the same problem again... (the other link is to my question for some reason) – Mar 06 '14 at 00:55
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1"it's clearly physics as astronomers claim the existence (in the here and now)" the "here and now" is just an event in spacetime, not all of spacetime. – Alfred Centauri Mar 06 '14 at 01:07
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What's with the editing Nick Stauner? – Mar 06 '14 at 01:09
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Sorry Alfred- something odd just happened to my post- the "here and now" indicates observable events... if an event is unobservable then how can it be included in physical law? – Mar 06 '14 at 01:25
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It takes 6 to form. – dfg Mar 06 '14 at 02:28
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"punctuation, non-redundancy[, grammar, brevity]" – Nick Stauner Mar 06 '14 at 03:14
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No more rollbacks, please. If there are any more I will be locking the question. – David Z Mar 06 '14 at 18:27
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I'd like to rollback to an era before you were born... but alas that would invoke casualty violations... – Mar 07 '14 at 07:44
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Similar questions have cropped up on this site many times, and the debate surrounding them is usually fractious because people misunderstand each other's use of words like exist.
One of the lessons of General Relativity is that any observer has to choose a locally convenient coordinate system that may not be globally convenient. We on Earth (quite sensibly) choose time as measured on our clocks and distance as measured by our rulers, and these coordinates are known as the Schwarzschild coordinates (strictly speaking they are shell coordinates, but the difference at the orbital distance of the Earth from the Sun is negligable). Locally our coordinates work very well, but when the central body is a black hole the coordinates become increasingly curved as you approach the event horizon and at the event horizon they fail completely resulting in a coordinate singularity.
I've made this sound like a mathematical nicety, but it's quite real. Remember that by the time coordinate I mean the time we measure on our clocks, and that means there is a singularity in our measurements of time at the event horizion. This is why it takes an infinite time for anything to reach the event horizon, let alone cross it.
The question is whether it is therefore correct to say that: the event horizon never forms. It is quite true that you and I and everyone outside the black hole will never measure the time the event horizon forms, because it would take an infinite time. However there are lots of coordinate systems that have no singularity at the horizon, such as Gullstrand-Painlevé, Eddington-Finkelstein and Kruskal-Szekeres coordinates. The trouble is that these coordinates are somewhat abstract and do not coincide with the experience of any human observer. However since such coordinates exist, physicists tend to be quite comfortable stating that black holes form even if human experimenters could never observe it.
John Rennie
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Doesn't this remove the role of experimental observation when it comes to validating theory? In particular if a theory predicts a certain thing then how isn't the validity of that theory based upon data (taken by us humans, as we have no data from other intelligent creatures that pertains to GR) that evidences that very thing? – Mar 06 '14 at 21:29
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@jaskey13: GR predicts thing we cannot possibly observe. However it also predicts things that we can observe, and we observe that GR makes the correct predictions. If we always find GR to be correct about the things we can observe then is it safe to assume it's also correct about the things we can't observe? That's a tricky question, and even amongst physicists view differ. However I suspect most us believe that GR is reliable even when we can't confirm what it predicts. – John Rennie Mar 07 '14 at 07:10
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That is not in the true spirit of the scientific method- nor Occams Razor..... In particular a theory that makes predictions that cannot ever be proven by experiment is not the simplest theory... As a) if predicts things unprovable by a method of experimentation and b) it includes within it extraneous things that are outside of experimental data – Mar 07 '14 at 07:27
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@jaskey13: the universe didn't consult us before it decided to exist – John Rennie Mar 07 '14 at 07:47
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and how does a universe "decide" to exist? Is there a conscious principle I'm not aware of? – Mar 07 '14 at 07:53
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@jaskey13: No, we have theories that describe the universe (the bits we can see) very well. But those theories predict some parts of the universe are inherently unobservable. That doesn't mean we've stopped trying. For example some of the firewell theories predict spacetime simply ends at an event horizon. – John Rennie Mar 07 '14 at 07:54
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So back to my question... how long does it take for the horizon form? Can it form in finite time for an external observer? – Mar 07 '14 at 07:59
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1@jaskey13: No. For an external observer the event horizon never forms. The best we see is an apparent horizon. – John Rennie Mar 07 '14 at 08:00
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But an apparent horizon is based on the idea that there is not enough time (referenced by an external observer) for signals to propagate.... If the universe does not "crunch" then there will always be enough time... Right? – Mar 07 '14 at 08:03
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@jaskey13: No (I seem to starting all my answers with "No" :-). The whole point of infinity is you can never reach it. When we say the event horizon takes an infinite time to form we mean there will never be enough time to observe it. – John Rennie Mar 07 '14 at 08:18
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Can you explain to me how the idea of an apparent horizon does not assume that there is a constraint on the time coordinate for an external observer- in particular, given enough time in any frame the apparent horizon shrinks, correct? So we must still see a collapsing object yet no barrier? – Mar 15 '14 at 02:56
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1The time coordinate in Gulstrand-Painleve is the observer’s proper time and so it’s easy to interpret why it’s is regular at the horizon – bapowell Apr 29 '20 at 12:01
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It's a matter of what you mean by "see". Even for a distant observer, it will take a small amount of time for the gravitational redshift effect to become essentially infinite. If your collapsing gas star redshifts to the point where it won't emit a single photon in the age of the universe, it may not have yet technically "redshifted to zero", but it has functionally redshifted to zero as far as experiment is concerned.
The actual picture seen by an external observer is consistent with the general picture described by a naïve interpretation -- a black object that is not emitting anything other than (what is, for macroscopic holes, a vanishingly small amount of) Hawking radiation, that will absorb objects that enters it. I should also add that the apparent horizon of a collapsing star will move outward at a speed faster than the speed of light, which will solve the infinite redshift issue, as well.
Zo the Relativist
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I strongly disagree. I did the calculations several times, and I'm positive it takes 6 to form. – dfg Mar 06 '14 at 03:42
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1@dfg: I think I saw you at the coffeeshop last week: http://bitsocialmedia.com/wp-content/uploads/2013/07/Internet-Troll.jpg – Zo the Relativist Mar 06 '14 at 03:54
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Doesn't the superluminal speed of formation of an apparent event horizon imply that gravitational influence has propagated superluminally as well? – Mar 06 '14 at 23:01
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And also a collapsing star will not just all sudden collapse to a point that it cannot emit radiation within the known time span of the universe... In fact the "collapsing" will simply appear to decelerate to an outside observer.... In any given time span signals will emerge... Those time spans just become further apart.. Unless you propose an end date to the universe? – Mar 06 '14 at 23:09
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@jaskey13: you can't transmit information from point A to point B by having an apparent horizon expand, so there is no causality or influence violation. And the spacetime is the thing doing the radiation emission, and that happens in the neighborhood of the horizon, not at the horizon, so there is no transmission of information out of the interior. Anything that falls into the hole will redshift to within an arbitrary tolerance of zero intensity in microseconds. – Zo the Relativist Mar 07 '14 at 00:29
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Then how we say even an apparent horizon forms if no information about it's forming is allowed for any observer in contact with us (the ones who make theoretical laws)? – Mar 07 '14 at 00:45
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and you say "fall into the hole" again like it is an event that can be referenced from experimental data and not just theory – Mar 07 '14 at 00:46
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but spacetime itself is still theoretically under Lorentz invariation- that being it does not somehow change superluminally- else the graviton cannot be a fluctuation of the space-time "field" right? – Mar 07 '14 at 00:49
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And while I suspect the reactions can be much shorter than micro-seconds- they still must be dilated from our reference frame... so even nanosecond reactions and femtosecond reactions contribute to our observation of this collapsing object... of course such interactions are witnessed in a dilated time in our frame... – Mar 07 '14 at 00:58
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1@jaskey13: I can walk you through the calculations, but I suspect that you don't have the technical capacity to really get much from them. Suffice it to say that the real-world experience of a distant observer would be to see an object disappear due to redshifting. Even if they don't see a "crossing" event, they WILL see the falling object disappear relative to any possible observation, and this will happen in finite time relative to the shell observer. – Zo the Relativist Mar 07 '14 at 01:28
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Please show me the calculation that shows a falling object disappear in finite time for an external observer...And watch any coordinate that "goes to infinity" – Mar 07 '14 at 01:54
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The short answer is infinity.
The elastic body model, resulting from the work of Milo Wolff and Gabriel LaFreniere and Jeff Yee, says that the elementary particle is a pulsating soliton in a medium that is non-linear. The non-linear medium has an absolute density (analogous to the curvature of space-time), instead of a purely abstract amplitude, and Hook's law applies in it - it reacts to compression with opposing force - this is something that is missing in Einstein's GTR, and this is his biggest mistake in this matter. What he said about GTR, saying that the GTR is not full.
Well, the biggest absurdity arising from this theory - black holes - is just a consequence of not taking into account the internal pressure of space, which is an obvious component of every classic elastic body.
The collapsing star first breaks the pressure inside the atoms, then inside the neutrons, until it disappears in singularity. However, most thinkers do not take into account the fact that the same pressure that prevents atoms and neutrons from collapsing will also stop the entire star from collapsing. This is because both the individual material particles and the gravitational field around the star consist of the density of the same substance - space-time at Einstein, flexible space at Wolff, Aether at LaFreniere and Jeff Yee.
Also we must remember the atomic and electromagnetic forces are much more powerful than gravitational forces.
Noting by Haramein, then by Gabriel LaFreniere (probably also by others :)) that the energy contained in the volume of the proton / electron is enough to make it a black hole, points to an important issue - well, according to the model of the elastic medium, each the material particle is a kind of black hole - looking at how much energy it has. It arises where, within a given area, the density of Ether exceeds a certain critical value. However, this will not lead to the collapse of this area, but to its transformation into a soliton - a creature that can sustain its own existence. What feeds him? Waves around it, called by scientists quantum fluctuations, or vacuum energy, relict radiation, etc.
Vacuum has its energy, some have called it background relikt radiation others have the lowest energy state. Vacuum thickens to form a temporary unstable soliton - what others have called quantum fluctuations.
That is why the equation of the black hole needed to include the vacuum energy it contained. For the same reason, Jeff Yee had to put a constant density of ether in the equation of photon and electron energy.
This is remarkably coincident with Tesla's belief that matter 'absorbs' the Ether to exist.
So the same mechanism that causes matter to exist also excludes black holes. Aether density will always take energy, bend the paths of other densities and waves, but it will never collapse because it will have internal pressure. This is a direct consequence of the characteristics of a classical elastic medium.
The same mechanism contained in the elastic body model is responsible for the process of absorption and emission of an electromagnetic wave by an atom (a soliton made from of other solitons). Soliton aims to return to its ground state - stable state.
Einstein did not suspect that his confusing space-time could be replaced by something so simple. However, physicists who perceive the wave nature of matter are easier because the physical wave, despite the trumpeting, requires a medium.
The unequivocal experimental proof of the wave (frequency) nature of matter is the electron image recorded in 2007. This image was recorded by a team of Swedish scientists from the University of Lund. The image turned out to be a soliton image in line with the proposals of Milo Wolf and Gabriel LaFreniere published in the years 1998 - 2002:
Sorry for My English.
Regards.
user1785960
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