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I'm a big fan of the podcast Astronomy Cast and a while back I was listening to a Q&A episode they did. A listener sent in a question that I found fascinating and have been wondering about ever since.

From the show transcript:

Arunus Gidgowdusk from Lithuania asks: "If you took a one kilogram mass and accelerated it close to the speed of light would it form into a black hole? Would it stay a black hole if you then decreased the speed?"

Dr. Gay, an astrophysicist and one of the hosts, explained that she'd asked a number of her colleagues and that none of them could provide a satisfactory answer. I asked her more recently on Facebook if anyone had come forward with one and she said they had not. So I thought maybe this would be a good place to ask.

David Z
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shopsinc
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    Here is a link from John Baez's site. http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_fast.html – MBN Jan 20 '11 at 22:26

6 Answers6

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The answer is no.

The simplest proof is just the principle of relativity: the laws of physics are the same in all reference frames. So you can look at that 1-kg mass in a reference frame that's moving along with it. In that frame, it's just the same 1-kg mass it always was; it's not a black hole.

Ted Bunn
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    An addendum: It's worth pausing to ask why one might have thought it would form a black hole, and why those reasons are incorrect. Presumably the thought is that a combination of Lorentz contraction and relativistic "mass increase" squeeze the object below its Schwarzschild radius. So what's wrong with that reasoning? The main thing is just that the derivation of the Schwarzschild radius only applies under certain conditions. At the very least, it only applies in the object's rest frame (since it assumes spherical symmetry -- i.e., no preferred direction). – Ted Bunn Jan 20 '11 at 20:27
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    but if 1kg mass is rotated? – voix Feb 19 '11 at 13:24
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    The gravitational field of a rotating 1-kg mass is different from that of a non-rotating mass. I don't remember the details, which are complicated, but the gravitational pull probably does get stronger because the rotational kinetic energy gravitates. If you start with a mass that's larger than its Schwarzschild radius, I don't know whether you can make it turn into a black hole by supplying rotational kinetic energy. – Ted Bunn Feb 19 '11 at 14:48
  • @TedBunn "If you start with a mass that's larger than its Schwarzschild radius, I don't know whether you can make it turn into a black hole by supplying rotational kinetic energy." But even kinetic energy is relative, if I accelerate to the same velocity as your object, your object don't have any kinetic energy relative to me. – Calmarius Apr 22 '13 at 15:59
  • @TedBunn: And yet, light bending will become very peculiar for a very fast mass. Special relativistic aberration will come into play in ultrarelativistic regime which will enhance the otherwise weak lensing. Also, the fast body will feel strongly enhanced tidal forces in its frame, and there will be quite a lot of other interesting things happening. – Alexey Bobrick Jun 04 '14 at 15:17
  • @voix, if you rotate a mass, then you have to provide a lot of energy and pressure to keep it rotating (centrifugal forces). Those in turn will gravitate on theire own, and it will mess up the problem a little bit. – Alexey Bobrick Jun 04 '14 at 15:18
  • @TedBunn Prove the principle of relativity is correct and is not just some invalid assumption. – Molly Stewart-Gallus May 08 '15 at 01:22
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    so in this case we could have two realities simultaneously? One where its a black hole and one where it's not? for an observer in the frame with respect to which the object is moving at C the laws of physics would appear to be violated since the object would be weighing infinite and yet not form a black hole. – user1062760 Feb 13 '16 at 06:28
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    This answer is not on point. Op clearly means looking at the object from a different frame. Why is that so hard to understand what op means? – user1062760 Feb 18 '17 at 23:07
  • So you are saying we simple assume a "no"? – Shing Nov 22 '18 at 00:07
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    @user1062760 the answer is perfectly "on point". Changing reference frames never changes what actually happens. It is a simple as that. – m4r35n357 Sep 08 '19 at 13:05
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    @StevenStewart-Gallus it cannot be proved, it is an assumption. It has been proved valid by a century of experiment (in the case of SR/GR) and many centuries in the case of Newton. If that is not enough to satisfy your objection, nothing will be. – m4r35n357 Sep 08 '19 at 13:12
  • @m4r35n357 that's a statement without explanation that contradicts relativity. If something has cetain mass in my reference frame why it won't act the way it should with that mass? The increase in mass is why its impossible to accelerate an object to c, so its very real – user1062760 Oct 01 '19 at 13:20
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    @user1062760 no, your statement contradicts relativity. An object only has one mass, and it is the same in all reference frames. Your alternative leads to mayhem where each object has an infinite number of masses. No-one can do physics that way! – m4r35n357 Oct 01 '19 at 17:21
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    If a moving object from a given reference point does not generate additional gravity, does kinetic energy not gravitate? If so, how is this compatible with with mass-energy equivalence? – Abdul Moiz Qureshi Feb 13 '20 at 10:15
  • @m4r35n357 you said "Changing reference frames never changes what actually happens." I don't understand. If I enter a supermassive black hole, i observe myself entering the event horizon in finite time. But my friend on earth observes me being continuously redshifted forever as I approach the horizon, and from his frame of reference I never enter the horizon. So, the reference frame does change what happens (or at least what appears to happen). – cowlinator Oct 14 '22 at 23:11
  • @TedBunn (I fully agree) Just to add; spagetti-fication (spelling??) broadly speaking a - extreme version of Lorentz contraction is effectively what happens close to a black hole – William Martens Nov 15 '22 at 15:48
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    @cowlinator just because you ex-friend does not see it happen doesn't mean you aren't geting shredded! – m4r35n357 Nov 25 '22 at 09:52
  • Aside from the OP, I do have a related (General Relativity) angle I would like to ask about - the Stress-Energy Tensor contains momentum terms, so shouldn't relative motion between an observer and mass be reflected in these? Has anyone done a back-of-the-envelope calculation assuming non-zero momentum flow? Do these terms tend to "increase" or "decrease" gravitational effects from the stationary situation? Perhaps I need to consider a question of my own . . . – m4r35n357 Nov 25 '22 at 09:59
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No, a 1kg mass would not turn into a black hole, even if it were zipping past you at very close to the speed of light.

The principle of relativity is a fundamental idea in physics, and one consequence of it is that we can understand the physics of something that's moving by imagining we're moving alongside it.

For example, you are watching people play pool on a train as it rushes past you. You want to know whether a certain shot that's just been made will sink the 8-ball. You figure it out by imagining you're inside the train and calculating everything you'd expect to happen from that simpler viewpoint where the pool table is stationary. If the 8-ball goes into a certain pocket from that point of view, you can rest assured it will go into the same pocket if you analyze the situation again from your original vantage point on terra firma.

Applying the same principle to the 1kg mass, we see that moving along side it, it just looks like a normal mass, not a black hole. Hence, from another point of view in which it moves close to the speed of light, it still looks like a normal mass, not a black hole.

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Mark Eichenlaub
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    So does it then follow that the relative speed of a mass has no bearing on the gravitational force felt by a nearby stationary mass? That is if a mass flew by me at .1 c, would I feel the same tug as if it flew by at .999c? Would there be some sort of equivalence given the time it takes the object to pass? That is, would the total force felt over time be the same; sort of like how the area covered by an orbit is the same over a given time? – shopsinc Jan 20 '11 at 20:39
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    @shops Your question can't be answered using simply the principle of relativity because it's asking about different types of relative motion. You might try asking it as a separate question on the main site. I don't have a good, concise answer to that question. – Mark Eichenlaub Jan 20 '11 at 20:50
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    @MarkEichenlaub Please explain why the principle of relativity is in fact true, perhaps the principle of relativity is only a reasonable good approximation? – Molly Stewart-Gallus May 08 '15 at 01:26
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    No. Another answer that is completely unnecessary and missed the point. Talk from the viewpoint of the observer. If a non zero mass zips at speed of light it's mass for stationary observer would be infinite so why won't the stationary observer see a blackhole instead? – user1062760 Feb 18 '17 at 23:10
  • @user1062760 see my comment in the accepted answer – m4r35n357 Sep 08 '19 at 13:06
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While good, I think the other answers are currently missing one ingredient, so I'll post this answer.

For particles traveling at constant velocity there is no event horizon, and so they act nothing like a black hole. Light from other regions of space will eventually reach it, unlike a black hole. Further, the forces between atoms in what ever matter constitutes the mass are co-moving and so there is no increased gravitational interaction between them. While the distances between them appear to change to an outside observer (as the mass is accelerated) once it reaches constant velocity they are fixed.

What has not been mentioned in other answers is the effect of acceleration. When a particle is continuously accelerated there is an apparent event horizon. See the relevant Wikipedia page here. So this has some features that we associate with a black hole, however there are still major differences. An object undergoing constant acceleration does indeed behave like it is static in a constant gravitational field. However, in the case of such an object the direction of the equivalent field is constant (and in a constant direction) throughout the object. This is not true for the gravitational field of a black hole, which is spherically symmetric.

Of course once the particle stops accelerating the apparent horizon disappears.

Joe Fitzsimons
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    True, but the apparent horizon in this situation is very different from a black hole horizon. In the case of the accelerating particle, the stuff that's "behind" the apparent horizon is far away from the particle -- just the opposite of the black hole's event horizon. That is, heuristically, a black hole's event horizon says that once you're sufficiently close to the black hole you can't get far away, whereas the accelerated particle's horizon says that once you're sufficiently far from the particle you can't get close. – Ted Bunn Jan 21 '11 at 15:03
  • +1 For adding more realistic context to the answer - obviously a mass accelerated up to near C needs to spend a finite time accelerating. – B T Nov 30 '15 at 22:33
  • Best answer here. It also answers this: https://physics.stackexchange.com/q/708360/226902 – Quillo May 12 '22 at 20:46
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I am presuming the idea is the 1kg mass will length contract to below the Planck length. It is either that or the relativistic energy (mass) $E~=~\gamma mc^2$ would be so large it would gravitationally implode. The question though can be thought of according to what would happen to an observer on the mass. The question could be turned around: Would the universe implode? If a mass $M$ passes by a smaller mass $m~<<~M$ then one might think that $M$ could become a black hole and the small mass $m$ if close enough would become trapped in the black hole. However, from the frame of the big mass $M$ the small mass is not a black hole. This is a contradiction.

A ultra-relativistic mass will behave similar to a gravity wave as it passes another reference point. This Aichelburg-Sexl ultraboost has a plane wave pulse of spacetime. The relativistic mass will result in a gravity wave pulse as detected by a stationary observer. So there is a gravitational implication to such extreme relativistic boosts.

Lawrence B. Crowell
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    This answer seems like it could be useful in the future but as of now it is a bit jargony. Also, it assumes that the principle of relativity is in fact true which it probably is but is still something that everybody should be rigorous about. – Molly Stewart-Gallus May 08 '15 at 01:28
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While answers that say something along the lines of : "if an object is not a black hole in one reference frame then it is not a black hole in any reference frame" are technically correct, they are not satisfying as they do not explain why the alleged increase in mass of the test object does not eventually gravitationally collapse the moving object.

The simplest explanation is that not only does mass and length change at relativistic speeds, but other factors like time dilation and force transformation play a part. In relativity, transverse force is reduced by a factor of gamma when the test object is moving relative to the observer. This is true for any transverse force. Consider the following scenario: A scientist drops a mass from 9.8m above the surface at the North Pole. It takes $t_0 = 1$ second to fall by the scientist's measurements. He calculates the acceleration $a_0 = L_0 t_0^{-2} = 9.8 \ m/s^2$. An advanced alien happens to be flying past from East to West at $v/c = \sqrt{3/4}$ such that the gamma factor is $\gamma = 2. $ Due to time dilation, he measures the time to fall a vertical distance of $L$ as 2 seconds and calculates the acceleration to be $L_0 t_0^{-2} \gamma ^{-2} = 9.8/4 \ m/s^2 $. In other words, the force of gravity on the surface of the Earth appears to be weaker to the passing Alien. Objects appear to fall in slow motion.

Since the discussion implicitly introduced the concept of relativistic mass, I will stay with that and say the mass of the test object has increased by a factor of gamma. We can now calculate the force acting on the test object with rest mass of $m_0$ and acceleration $a_0$ in the rest frame as $F = m a = (M_0 \gamma )(a_0 \gamma^{-2}) = F_0 \gamma^{-1}$ as stated earlier. If we consider an object on the surface of the Earth as part of the Earth's crust, the gravitational force acting on it is weaker according to the passing observer and there is no danger of the Earth crust collapsing inward, no matter how fast the observer is passing.

Another factor to ponder is that in Newtonian physics, transverse and parallel inertial mass, momentum mass, gravitational source and target mass, kinetic mass are all the same thing, but in relativity they are not. This is what Einstein was alluding to when he stated "It is not good to introduce the concept of the mass of a moving body for which no clear definition can be given."

KDP
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None of the existing answers mention one of the most salient points, which is that Newtonian gravity is only an approximation valid at low speeds. When relativistic considerations enter (as they certainly do for a mass moving close to light speeds) then the relativistic theory of gravity, i.e. general relativity, must be used.

The source of gravity in GR is not just mass but the stress energy tensor, which includes terms for energy, momentum, and pressure. In a sense the momentum terms for a moving object cancel out the extra kinetic energy from motion.

Eric Smith
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