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.


6 Answers 6


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? $\endgroup$ 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

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. $\endgroup$ 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? $\endgroup$ 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? $\endgroup$ Commented Feb 18, 2017 at 23:10

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

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. $\endgroup$ Commented May 8, 2015 at 1:28

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."


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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