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Reading this comment makes me ask what does that actually mean that

In fact, if you are close enough to $c$, the mass-equivalent of your kinetic energy may be enough to make you appear as a black hole to an external observer

? Does it mean that any object travelling at a near the speed of light will act as a black hole even if its mass isn't contained within the necessary Schwartzschild Radius? If yes, then what would cause the regular properties of the black hole (such as event horizon as point of no return etc) to appear just because the object is moving fast?

Or what does "appear as a black hole" mean in this case?

Qmechanic
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Alma Do
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    The comment is completely wrong. – G. Smith Jun 17 '19 at 15:13
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    Duplicate of If a mass moves close to the speed of light, does it turn into a black hole? – G. Smith Jun 17 '19 at 15:28
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    @G.Smith Not quite a duplicate: the other question asks if it would become a black hole. This one asks if it would act as a black hole for the external observer. To the former, the answer is obviously "no", because kinetic energy is relative. To the latter, the answer happens to be "yes", the object will appear as a black hole to the observer. (Alas, I cannot find the source where I have seen it demonstrated at the moment.) – Eth Jun 17 '19 at 16:37
  • @Eth What specifically does it mean for an object to "act as a black hole for the external observer"? – probably_someone Jun 17 '19 at 16:40
  • @probably_someone An object from which even light cannot escape to reach the observer. Unfortunately, I don't remember enough about the demonstration to give more details about what exactly it would look like... – Eth Jun 17 '19 at 16:43
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    @Eth but even with my (very limited) physics knowledge this statement then appears to not be true: clearly the light will be able to reach an observer is emitted from the object itself. If I accelerate a lightbulb to a .99c it will still be a lightbulb casting light towards me at a constant speed ... – Alma Do Jun 17 '19 at 16:47
  • @AlmaDo That's precisely why I remember reading it, because it was so counter-intuitive. I should have saved it somewhere :/ – Eth Jun 17 '19 at 16:49
  • @Eth The metric of spacetime is Lorentz-invariant - the object affects trajectories in the "same" way (up to a coordinate transformation) in every frame. If you were talking about the object being redshifted as it moves away from you, therefore being "darker", you can make the opposite claim when it's moving toward you and blueshifted. – probably_someone Jun 17 '19 at 17:13
  • @Eth Were you thinking of the apparent horizon behind the accelerated object? https://physics.stackexchange.com/a/3465/127931 – JMac Jun 17 '19 at 17:16
  • I think @Eth is talking about the situation mentioned here. However, if a body is a black hole in any frame, then it's a black hole in all frames. OTOH, if you do collide a pair of objects with sufficient KE you can make a black hole, but we're talking about a lot of energy! – PM 2Ring Jun 17 '19 at 17:46
  • I was just about to answer your question on escaping a black hole at 1m/s when you deleted it. If you undelete the question I'll answer it. – John Rennie Jul 17 '19 at 17:00
  • @JohnRennie my oversight was spotted in the comment already (you'll probably see it); I googled that and found out about the spacetime curvature there – Alma Do Jul 17 '19 at 17:05
  • @AlmaDo as you wish, though I think it's a good question (and the downvote was unfair). The answer is not at all obvious. – John Rennie Jul 17 '19 at 17:06

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