A body's mass increases when it aquires velocity. It approaches infinity when the velocity approaches the speed of light. This will not result though in the formation of a black hole.
A micro black hole in reach for observation is speculated to exist in a theory that tries to unify the three basis forces with gravity. Extra curled up space dimensions are postulated and these dimensions can make gravity vary in a non-$\frac{1}{r^2}$ way. Because of this the mass needed to produce a micro black hole is in reach for experimental probing. See this article for more details.
One can read that two colliding protons can (according to the theory) form a micro black hole. It's obvious that two protons with a high enough relative velocity can have a mass equivalent of a micro black hole. But if the protons can't form black holes at high enough velocities, how can a black hole be formed after they have collided? How emerges a black hole spacetime in this case?

  • $\begingroup$ @safesphere I get you! Thanks! All momentum (although the total is zero) is conversed in energy, so...Which is also obvious from a quark perspective. Due to the collision the six valence quarks get into an excited high energy state which translates in positive curvature(?). I like the citatation of Mencken, by the way... $\endgroup$ Commented Apr 17, 2021 at 16:54

1 Answer 1


"A body's mass increases when it acquires velocity" is false. Velocity is relative, so any apparent increase in mass is entirely in the eyes of the observer moving relative to the body, i.e. it's an artifact of the coordinate system. That's why modern relativity textbooks generally avoid "relativistic mass" and only talk about "invariant mass" or "proper mass" (the rest mass of the object).

So a single proton moving very fast relative to you won't form a black hole, any more than you would form a black hole because you're moving very fast relative to the proton.

However, two protons moving in opposite directions do have an invariant kinetic energy associated with the whole system. That is, every observer can agree that the total system of protons has energy greater than the sum of the rest masses -- no matter what you consider to be "at rest", one or both of the protons is going to be moving.

One could imagine that if the protons collide this kinetic energy could end up concentrated in a small enough region to form a black hole. Whether this could in fact happen in the real world is another question, and probably requires a quantum theory of gravity (which we don't have!) to answer for sure. Certainly it has never been observed.

  • $\begingroup$ See also this question $\endgroup$
    – Eric Smith
    Commented Apr 16, 2021 at 22:14
  • $\begingroup$ @safesphere : The OP and I are discussing protons, i.e. hydrogen nuclei, not "photons". So your objections are starting from an incorrect premise. $\endgroup$
    – Eric Smith
    Commented Apr 17, 2021 at 15:57
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    $\begingroup$ Protons do have mass, and so they do curve spacetime. And the stress-energy tensor for a system of two protons in relative motion is not the same as the stress-energy tensor for two protons at rest relative to one another. It's certainly true that the effect is complicated -- it's not as simple as plugging the "mass equivalent" of the kinetic energy into Newton's formula. $\endgroup$
    – Eric Smith
    Commented Apr 17, 2021 at 16:21
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    $\begingroup$ If the particles are moving relative to one another then the stress energy tensor, and hence the curvature, is objectively different than in the case where they are at rest relative to one another. See also Tolman, Ehrenfest, and Podolsky "On the Gravitational Field Produced by Light", Phys. Rev. 37, 602, wherein they show that photons propagating in the same direction have no attraction to one another, but photons propagating in opposite directions do attract. $\endgroup$
    – Eric Smith
    Commented Apr 17, 2021 at 21:51
  • $\begingroup$ @safesphere said “you need a collision to convert the kinetic energy to the energy of a different type whose curvature is not offset by the gravitomagnetism of momentum” which is not exactly correct. You need a collision, as Eric Smith mentioned, to get the energy into a small region, but conversion into a different form of energy is not needed. In the center of momentum frame the momentum of the two are equal and opposite. So the gravitomagnetic terms you mention are naturally canceled in a “collision” even without any conversion. Eric’s argument is valid. $\endgroup$
    – Dale
    Commented Apr 18, 2021 at 16:45

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