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If a 1kg mass was accelerated close to the speed of light would it turn into a black hole?

Imagine, a rod of length L is moving with velocity approaching the speed of light with respect to a human observer on Earth. Due to Lorentz contraction, the rod will observed to be very short. And since all laws of physic hold true in every frame of references, the gravitation law which state the force acting is inversely proportional to square of distance between them, will be acting between various part of the rod. Now, as velocity approach c , L will approach 0. This should cause an enormous gravitational force enough to form a black hole. Isn't this suggesting a black hole can be formed when the object velocity comes near the speed of light?

According to the rod's frame of reference, it will see itself as stationary and a man who observe it is moving, so according to the rod, the human must be a black hole. Isn't this a paradox?

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    Not really a duplicate. This question doesn't mention the "mass increase" but instead only mentions the length contraction. – John Rennie May 16 '12 at 06:01
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    @JohnRennie: But the answers to the first completely answer the second, this is a dup, you don't want to make people repeat text verbatim. – Ron Maimon May 16 '12 at 08:03
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    I don't think the answers to the previous question actually answer the question. They all just basically say "it would contradict the principle of relativity" and that's no answer. I was tempted to post saying "it's because the Riemann tensor is co-ordinate invarient" but that's a glib answer as well. What I would like to see are some calculations showing me why a black hole can't form (obviously it can't!). I've so far given it one five minute tea break's thought but without any significant progress - maybe my lunch hour will be enough :-) – John Rennie May 16 '12 at 08:44
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    Fundamentally the problem is that you are trying to apply simultaneously Newtonian gravity, special relativity, and general relativity. The paradox basically says that your assumption that all three can be applied to the same system at the same time is daft. – Willie Wong May 16 '12 at 11:39
  • Well thanks a bunch, you closed the question six minutes after I'd put a lot of effort into answering it! – John Rennie May 16 '12 at 14:18
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    BTW--@John, if you can't raise enough interest to re-open this question, you can flag it to be considered for merging with the possible duplicate. Or you can just leave your answer here: the question will still show up in search, people can still vote on it, and the OP can still accept your answer if (s)he is happy with it. My take is that this is the better version of the question but that they are duplicates. – dmckee --- ex-moderator kitten May 16 '12 at 14:38
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    Generally speaking (i.e. not just this question), in the spirit of SE regulations to avoid duplicates, OP (and potential answerers) are encouraged to try to detect duplicates by doing site searches before posting, cf. the faq. If a new question is a duplicate, it seems most logical if new answer are placed at the original post in order not to promote the new duplicate unnecessarily. – Qmechanic May 17 '12 at 15:03
  • Good answer is this one: https://physics.stackexchange.com/a/3465/226902 – Quillo Feb 08 '23 at 23:24
  • Moreover, length contraction is such that nothing happens to the rod: it proper length is unchanged (the same for its proper density, so no collapse) https://physics.stackexchange.com/a/270063/226902 – Quillo Feb 08 '23 at 23:33

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OK, having devoted my lunch hour to this (the sacrifices I make for Physics!) I have an answer for you. I'm not sure this is the best possible answer, so if anyone can improve on it please jump in.

Firstly you're absolutely correct to say that the density of your object increases as it Lorentz contracts. This isn't an illusion: the RHIC observes this every day. Note how the illustrations on the RHIC page I've linked to show the colliding nuclei flattened into disks. However the contracted object can't form a black hole because this violates one of the principles of relativity i.e. the presence or otherwise of the black hole could be used to tell who was moving and who was standing still. So what's going on?

The paradox arises from your assumption that it's the mass/density of the object that determines whether it will be a black hole or not, because this isn't true, or rather it's only true in special cases. The Einstein equation that gives us the curvature, and therefore whether a black hole will form, is:

$$G_{\alpha\beta} = 8\pi T_{\alpha\beta}$$

$G_{\alpha\beta}$ is the Einstein tensor that describes the curvature, while $T_{\alpha\beta}$ is the stress-energy tensor. So it's not the mass/density of the object that determines the curvature, it's the stress-energy tensor.

There's a shortcut here, because the stress-energy tensor is an invarient i.e. it is the same in all co-ordinate systems. That means the stress-energy tensor we observe is the same as the stress-energy tensor observed in the rest frame of your test object. So if the test object doesn't form a black hole in it's rest frame it will not form a black hole in any other co-ordinate system, even the one you describe in which the object is moving at almost the speed of light.

However it's at this point that I run out of steam a bit, which is why I think there's scope for this answer to be improved. It would be nice to give an intuitive feel for what the stress-energy tensor is, and why it doesn't change when we see the object moving at almost the speed of light. We normally write the stress-energy tensor as a 4 x 4 matrix, and with a few approximations about your test object the tensor only has one non-zero value, $T_{00}$, which is indeed the density. If we write the stress-energy tensor in our frame, where the object is moving, our value for $T_{00}$ will increase as the density increases, and if nothing else changed this would eventually form a black hole. However in our frame the other entries in the matrix are no longer zero. The changes in the other entries balance out the change in the density, so when we plug our stress-energy tensor into the Einstein equation we get the same curvature as in the test object's rest frame. No black hole!

John Rennie
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    The intuitive stuff: the time-space components are the momentum, and in the limit you are discussing, they are equal in magnitude to the time-time component. Two parallel pencils of light (or two masses highly boosted in the same direction) don't attract--- they repel by gravito-magnetic force to balance the attraction. This can be understood from SR (since if you boost two attracting objects perpendicular to the line of attraction, the time dilation makes the time to collision go to infinity). The same is true in EM, where two stationary charges boosted to near c don't repel (E and B cancel). – Ron Maimon May 17 '12 at 01:49
  • If a boat starts moving ,would it sink due to increase in relativistic energy?because in that case wouldn't the stress energy change? – Paul Jun 26 '15 at 05:29
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    To add to the answer... In the special case of a sphere of radius $R$ rather than a rod of length $L$. A black hole is formed when a mass is completely enclosed in its Schwarzschild radius $R_S$. Since at rest, the sphere is not a black hole: $R> R_S$. When the sphere is accelerated, one of its directions will shrink to arbitrarily small length scales, but the two directions perpendicular to the motion will stay unchanged, this even in the limit of a flat disk, the object will still be outside its Schwarzschild radius: not a black hole. @WizardOfMenlo – Andrea Jan 18 '16 at 15:42
  • +1 For your lost lunch hour. This answer is better than those in the duplicated question. – Reinstate Monica -- notmaynard Apr 07 '17 at 20:12
  • After reading several dozens of answers to this question this is by far the best. Specifically the invariance of the stress energy tensor really says it all. Thank you 10 years later. – Peter Moore Jul 29 '22 at 16:35