Imagine you have a particle accelerator that you can crank up to arbitrarily high energies. Because of General Relativity, the particles get heavier and heavier as you dump more energy into them. Will these particles ever become a black hole from this relativistic mass? If so, what will an observer particle moving parallel to the black hole, just below that speed, see? What I'm getting at is, kinetic energy is relative. So therefore relativistic mass is as well. So would these particles be black holes to some observers but not others?

  • $\begingroup$ Might get better quality answers in Physics.SE $\endgroup$ – tuomas Jul 8 '19 at 19:52
  • 4
    $\begingroup$ physics.stackexchange.com/questions/3436/… seems like a place where you might find an answer, although I didn't find any of the answers quite as clear as I'd like $\endgroup$ – Steve Linton Jul 8 '19 at 20:09
  • 2
    $\begingroup$ Relativistic mass is a deprecated concept because it can be misleading and confusing. This is one of those situations where it's misleading. ;) Please see physics.stackexchange.com/questions/133376/… $\endgroup$ – PM 2Ring Jul 9 '19 at 7:22
  • 1
    $\begingroup$ However, if you collide a pair of objects with insanely high relative kinetic energy, then you could create a black hole. "All" you need to do is to make sure you have enough energy in a small enough radius. $\endgroup$ – PM 2Ring Jul 9 '19 at 7:37
  • 1
    $\begingroup$ @Carl For example, a black hole of radius 1 fermi (about the size of a proton) has a mass a little over 673 million metric tons. So if you get all the energy obtained by annihilating 337 million tons of antimatter with an equal amount of normal matter and somehow convert that into the KE of a pair of protons in a head-on collision, then when the protons collide you'll have all that energy within the required Schwarzschild radius. I think. ;) The protons may radiate some of that energy away before they collide. $\endgroup$ – PM 2Ring Jul 9 '19 at 17:27

Relativistic mass increase is inertial, not gravitational, so no, you couldn't make a black hole by accelerating a particle. What happens is that the more energy you use to accelerate the particle, the more its inertial mass increases, but the more its inertial mass increases, the harder it is to accelerate. There comes a point near the speed of light when virtually all the energy you pump into the system is manifested as mass increase rather than velocity, and you can't make the particle go any faster. The best idea, if you want a really fast particle with huge amounts of energy, much more than could be produced by an accelerator, is to use a cosmic ray. Cosmic rays are fast protons or atomic nuclei which have been accelerated by natural events such as supernova explosions to much higher energies than you could get in an accelerator.

Cosmic rays have been used for particle experiments, but the snag is that they are random events, so you don't know exactly where to put your particle detectors to intercept the best ones. To give you an idea of how great their energies can sometimes be, the most powerful cosmic ray particles have the kinetic energy of a rifle bullet, yet the atomic nucleus which carries this energy is so minute that it cannot be seen in even the most powerful electron microscope!

|improve this answer|||||
  • $\begingroup$ I thought we had no way to tell inertial mass and gravitational mass apart? physlink.com/Education/AskExperts/ae305.cfm $\endgroup$ – Ryan_L Jul 9 '19 at 0:02
  • 5
    $\begingroup$ "There comes a point near the speed of light when virtually all the energy you pump into the system is manifested as mass increase rather than velocity, and you can't make the particle go any faster." That's not quite correct. There is no point where velocity increase stops. As you accelerate an ultra-relativistic body, its velocity, momentum, and kinetic energy continue to increase; the velocity increases asymptotically towards c, the momentum & KE increase without bounds. $\endgroup$ – PM 2Ring Jul 9 '19 at 7:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.