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Consider an observer at a significant distance away from a collapsing star.

As such when a singularity is born at the core of the star the observer would never see it grow anymore than the infinitesimal point it is at the moment of its creation.

This is because the gravitational time dilation at its event horizon would mean any infalling matter would take infinite time to cross the horizon with respect to observers time.

Keep in mind the observer is away from the star and not inside it as answers to similar questions had falsely assumed that otherwise. So the time dilation for the two would be different.

If this is true no Black hole should technically exist for people at earth as the time required for a blackhole to grow from singularity to things like supermassive blackhole at galactic centers would take infinite time from earth's perspective.

Finally this should mean there are only two ways for them exist

  1. If all matter or the universe itself came out of black hole as only in that case we would have equal dilation leading to their formation in finite time

  2. Universe existed since near forever (eternal) .

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marked as duplicate by tparker, Qmechanic Dec 28 '16 at 6:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Possible duplicate of How can anything ever fall into a black hole as seen from an outside observer? $\endgroup$ – tparker Dec 27 '16 at 23:37
  • $\begingroup$ Please do not flag for duplicate if the similar question have answers that completely fails to understand the question. Having a answer doesn't means it's correct. $\endgroup$ – user1062760 Dec 28 '16 at 18:41
  • $\begingroup$ Those other question seem to understand the question to me $\endgroup$ – tparker Dec 28 '16 at 18:46
  • $\begingroup$ @tparker well if you notice the answers to one of them make the wrong assumption that observer is inside Black hole while that question itself States "distant observer". The other one has a answer that does not exactly clarify and is found unacceptable by some as you can see in the comments it also has drastically less upvotes to another answer of it which makes unsaid assumptions again. $\endgroup$ – user1062760 Dec 28 '16 at 19:01
  • $\begingroup$ Please edit your question if you think this is not a duplicate, making explicit what about your question is different to the other ones already asked. If you are merely not satisfied with the answers to the previous questions, you might consider offering a bounty on them requesting additional clarification instead. $\endgroup$ – ACuriousMind Dec 29 '16 at 3:27
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That's why they are called "black", it's not a theoretical basis for an argument that they do not exist.

1) They are generally assumed to exist from the gravitational distortions. That is astronomers observe matter orbiting them. The gravity effects are felt outside the event horizon.
2) Additionally when matter approaches the event horizon the gravitational stress can cause some kind of emissions, which get some time dilation, but can escape.
3) Hawking radiation could under theoretical circumstances be detectable, I don;t think it's been done. This is when spontaneously formed virtual particle pairs in free space fail to anihilate as one is captured by the event horizon in the other's frame of reference. 5) Apparently it's also theorised that black hole surfaces are in fact not actually black, but encode some kind of holographic data related to the matter consumed.

Time dilation at the event horizon is not infinite as light can move in "orbits". Light moves, time ticks.

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  • $\begingroup$ How does this relates to the question? I'm asking a theoretical question not one pertaining to practical observation methods $\endgroup$ – user1062760 Dec 28 '16 at 18:39
  • $\begingroup$ OK, well my answer covered both the possibilities; for what I believe to be sound reasoning. If a conclusion is flying in the face of consensus, it's an indication something is probably wrong. Unless you are in fact a thoretical physiscist working on the "wavefront" in this area The answer you probably want is the last para. $\endgroup$ – JMLCarter Dec 28 '16 at 18:57

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