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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.

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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.

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    $\begingroup$ 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). $\endgroup$
    – Ted Bunn
    Commented Jan 20, 2011 at 20:27
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    $\begingroup$ but if 1kg mass is rotated? $\endgroup$
    – voix
    Commented Feb 19, 2011 at 13:24
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    $\begingroup$ 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. $\endgroup$
    – Ted Bunn
    Commented Feb 19, 2011 at 14:48
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    $\begingroup$ 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? $\endgroup$
    – user1062760
    Commented Feb 18, 2017 at 23:07
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    $\begingroup$ @user1062760 the answer is perfectly "on point". Changing reference frames never changes what actually happens. It is a simple as that. $\endgroup$
    – m4r35n357
    Commented Sep 8, 2019 at 13:05
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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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    $\begingroup$ 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? $\endgroup$
    – shopsinc
    Commented Jan 20, 2011 at 20:39
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    $\begingroup$ @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. $\endgroup$
    – Mark Eichenlaub
    Commented Jan 20, 2011 at 20:50
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    $\begingroup$ @MarkEichenlaub Please explain why the principle of relativity is in fact true, perhaps the principle of relativity is only a reasonable good approximation? $\endgroup$
    – Ms. Molly Stewart-Gallus
    Commented May 8, 2015 at 1:26
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    $\begingroup$ 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? $\endgroup$
    – user1062760
    Commented Feb 18, 2017 at 23:10
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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.

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    $\begingroup$ 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. $\endgroup$
    – Ted Bunn
    Commented Jan 21, 2011 at 15:03
  • $\begingroup$ +1 For adding more realistic context to the answer - obviously a mass accelerated up to near C needs to spend a finite time accelerating. $\endgroup$
    – B T
    Commented Nov 30, 2015 at 22:33
  • $\begingroup$ Best answer here. It also answers this: physics.stackexchange.com/q/708360/226902 $\endgroup$
    – Quillo
    Commented May 12, 2022 at 20:46
  • $\begingroup$ "Light from other regions of space will eventually reach it, unlike a black hole." This is not correct. Light from other regions WILL reach a black hole. What they won't do is escape it. You are also stating the accelerated particles see an event horizon and while this true, it does not make their point of view the same as being part of a black hole . The accelerating Rindler observers are equivalent to being outside a black hole. The only Rindler observer that is remotely equivalent to being part of a black hole has infinite acceleration, but you do not make that clear. $\endgroup$
    – KDP
    Commented Jun 24 at 6:50
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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.

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    $\begingroup$ 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. $\endgroup$
    – Ms. Molly Stewart-Gallus
    Commented May 8, 2015 at 1:28
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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.

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It can be noted that particles (cosmic rays) pass the Earth at close to the speed of light every day. From their point of view, the Earth IS moving at very close to the speed of light, yet clearly the Earth does not turn into a black hole. That is the short answer, but how do we prove that is what the equations of relativity predict, when the object appears to be moving close to the speed of light?

In Newtonian physics, inertial mass and gravitational mass are equal and Newton considered this a mystery as he could find no reason why this should be the case. Things are slightly different in relativity. The gravitational formula is:

$$F = \frac{GMm}{R^2}$$
where M is the active gravitational mass (The attracting mass) and m is the passive inertial mass (which is the proportionality factor between how a particle accelerates relative to the force applied to it). The relativistic equation for force transverse to the motion is:

$$F'_{\perp} = \frac{G(M\gamma^{-2})(m\gamma)}{R^2}$$

where $\gamma$ is the usual relativistic gamma factor equal to $1/\sqrt{1-v^2/c^2}$ where -v is the velocity of the observer orthogonal to the gravitational force vector. The end result is that the observer sees the transverse gravitational force to be weaker by a factor of gamma such that $F_{\perp}' = F/\gamma$. This is in agreement with the Lorentz transformation of transverse force and the old fashioned concept of transverse inertial mass.

Consider the following scenario: A scientist measure the time for a mass to fall from a height L above the surface as $t_0$. He calculates the acceleration as $a_0 = L_0 t_0^{-2} = g \ m/s^2$. To an observer (Anne) moving East to West at velocity such that the gamma factor is $\gamma = 2 $, the time to fall a vertical distance of $L$ is $2 \ t_0$ seconds, due to time dilation. Using the equation of motion, $ L =1/2 a t^2 \rightarrow a = 2L/t^2$, she calculates the acceleration to be $2L_0 t_0^{-2} \gamma ^{-2} = g/4 \ m/s^2 $. In other words, $a =a_0/\gamma$ and the force of gravity on the surface of the Earth appears to be weaker to the passing observer. Objects appear to fall in slow motion.

To Anne, the passive inertial mass of the falling 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 $F = m a = (m_0 \gamma )(a_0 \gamma^{-2}) = F_0 \gamma^{-1}$ in agreement with the result obtained earlier. For a large object like the Earth to become a black hole it has to compressed by a force. I have shown that the transverse compression force is actually weaker from the point of view of an observer that sees the system as moving with relativistic velocity. The Lorentz transformation also tells us the compression force parallel to the motion is the same as the force measured in the rest frame of the system, so overall there is no increased compression force. The proper volume of the Earth remains the same in its rest frame so there is no increase in the mass density, so there no reason to think the Earth should become a black hole as result of being accelerated to relativistic velocities, or equivalently, as a result of an observer passing by at close to the speed of light. It is not predicted by the equations of relativity.

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  • $\begingroup$ I haven't checked all your equations, but it appears you've lost a factor of 1/2 in your "scientist drops a mass from 9.8m above the surface at the North Pole" scenario. $s=\frac12at^2$. $\endgroup$
    – PM 2Ring
    Commented Jun 24 at 9:15
  • $\begingroup$ @PM2Ring Thanks for the correction. Fixed that error in the text now. $\endgroup$
    – KDP
    Commented Jun 24 at 15:26

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