# Will a giant ball of protons form a black hole?

Suppose you have enough energy and resources to put together (in a momentarily static configuration in which they are all at rest at the same time) as many protons as you want to form a "proton star". This ball will last a split second before the protons repulsion disperses it back. But if made of enough protons, it will momentarily have enough mass to form a black hole (or am I wrong and there is no upper mass limit for this situation? So the question is, will a black hole form or not? My doubt is that it is always argued that a star collapses into a black hole because there is no other force in nature to counterbalance it, however notice that in this case, the repulsive force of of the charged protons is always larger than the implosive force due to gravity.

• The mass-energy of the energy you use to "smash them together" will offset the repelling force of the charge, so why wouldn't it form a black hole? Generating microscopic black holes in accelerators is being discussed as a real technique of exploring quantum gravity, so there is nothing special about your giant proton ball, except that it's overkill. Two protons will probably be enough. – CuriousOne Jun 21 '16 at 23:00
• First of all, protons are not a good example to use. Protons are compound particles and even at the energy range of the LHC we aren't actually colliding protons but their constituents (quarks and gluons). Heavy ion collisions at LHC, RHIC (etc.) are forming quark-gluon plasmas at high energies, in which the electrostatic force is of less and less importance as the energy increases (the mass-energy is being converted into ever more particles which are, on average, ever less charged). That leaves the case of extremal black holes, which are still compact and don't break up. – CuriousOne Jun 21 '16 at 23:27
• @WetSavannaAnimalakaRodVance: Yes, that was my point. We put two heavy ions with a net charge of something like 160 in and end up with a quark-gluon plasma with $175MeV$ per constituent. The ingoing energy is something like $13 TeV * 82$ (lead), so there should be an effective number of something like $13TeV*82/175MeV\approx$ 6million constituents in the plasma (I have no idea if I am making a fool of myself here, or not :-)). The net charge (164) per constituent is therefor pretty small... Is particle meaningful in this sense? Is a "phonon" at 2300K a "particle"? I honestly don't know. :-) – CuriousOne Jun 22 '16 at 1:56
• Obligatory XKCD reference what-if.xkcd.com/140 – Aron Jun 22 '16 at 3:31
• @Aron: I wrote about ten protons in a 3m ball because I am not sure what scale we are talking about if not something of the size of a neutron star plus... or a Planck scale accelerator. – CuriousOne Jun 22 '16 at 4:02

This question has already been answered by a Randall Monroe of XKCD and Dr. Cindy Keeler of the Niels Bohr Institute.

It forms a Naked Singularity. Which is an infinitely dense object, from which light can escape.

You Reissner–Nordström metric for this question, as opposed to the more well known Schwarzschild metric.

Overall, you will get a blackhole, but it would be a blackhole without an event horizon (a naked singularity). Naked Singularities are forbidden by General Relativity, which I assume is the model of framework we want to work within for this question.

• Actually, although I love xkcd and, like anywhere else, Mr. Munroe clearly has done his research, I think there's a practicality that all four of us, you, I , he and Dr. Keeler have overlooked, as most elegantly explained in Lewis's answer: think about it: the proposed scenario is almost exactly what happens in a supernova. – WetSavannaAnimal Jun 23 '16 at 2:46
• This answer is wrong. The Coulomb self-energy contributes to the mass of the black hole. When you include it, the hole is (very) subextremal. The What If article correctly points this out (a couple of paragraphs below the paragraph that mentions naked singularities). – benrg Jul 3 '16 at 3:23

I think this is one of those situations where it is cleanest to think relativistically and concentrate on gravity as spacetime geometry, not gravity as a force.

If the stress energy tensor's components reach a such a magnitude that an horizon forms, then the mass/energy inside the horizon is doomed to stay inside evermore. This is a question of spacetime curvature, the geometry becomes such that nothing can escape the horizon unless somehow it travels backwards in time. Forces, no matter how big, cannot change this result once the horizon forms; although they will change the trajectories of the sources in the stress energy tensor, yet you've posed your question in a way that the assumption that the protons are converging fast enough to reach this point is implicit. The critical speed / energy of convergence to achieve this assumption is finite.

You also make the point

the repulsive force of of the charged protons is always larger than the implosive force due to gravity.

Again, this is indeed true in Newtonian gravity where we think of gravity as a force. In Newtonian gravity space does not deviate from Euclidean geometry, so once we're tilting local light cones enough that they are contained within a horizon, then we're well beyond Newtonian validity. Again, you simply need to give the protons enough kinetic energy to come condensed enough to form a horizon, then you have one mightily charged black hole and whose steady state is described by the Reissner–Nordström metric.

Some Afternotes

The black hole apparently doesn't needfully end up charged. Apparently (this is well outside my knowledge), according to a comment by theoretical physicist Physics SE User Lewis Miller:

The scaling of the electromagnetic force is irrelevant because long before the protons reach the density suggested here the protons will beta decay into neutrons and positrons. The positrons, being much less massive will disperse and the remaining neutrons will form the black hole. This is not too much different from what happens in a supernova of a very large star.

As I said- outside my knowledge, but certainly seems sound and intuitively sensible to me. One of the things that I like about this comment and Lewis's answer is that they show we can clearly bring observations and knowledge of real astronomical objects to the discussion of the OP's question.

Also see the XKCD article cited in User Aron's answer.

If a Reissner–Nordström black hole has a large enough amount of positive charge then General Relativity foretells a naked singularity. Whether or not classical gravity can be trusted to this level is moot. However, note that this does not happen if the charge is assembled as in the OP's question; the large amount of energy needed to assemble the black hole in the first place contributes to mass energy $M$ of the black hole such that $r_s \gg 2\,r_q$ (in the notation of the Wikipedia Reissner-Nordström metric article); $r_s$ being the Schwarzschild radius and $r_q$ a radius term resulting from the Coulombic repulsion (this is the quantitative description of my qualitative statement that "once we're tilting local light cones enough that they are contained within a horizon, then we're well beyond Newtonian validity"). The condition $r_s \leq 2\,r_q$ is the condition for a naked singularity.

• How does this "nothing ever leaves the event horizon" thing play out with Hawking radiation? This absolutely must be discussed if we're talking about black holes the mass of two protons. – John Dvorak Jun 22 '16 at 7:44
• @JanDvorak It doesn't; the amount of protons converging is not specified in the OPs question and I took it to mean considerably more than two protons. I agree these quantum aspects need to be discussed for a full analysis of a problem like this; I'm simply trying to show that, classically, the Newtonian notion of electromagnetic force's always overwhelming gravity is not correct and what the fundamental flaw is. – WetSavannaAnimal Jun 22 '16 at 7:54
• "you simply need to give the protons enough kinetic energy to come condensed enough to form a horizon" smbc-comics.com/index.php?id=1984 – Robin Ekman Jun 22 '16 at 12:00
• @RobinEkman LOL. I swear I've never seen that one before in my life: very funny, thanks :) I'll have to following SMBC now! – WetSavannaAnimal Jun 22 '16 at 12:35
• Thanks for incorporating my comment in your expanded answer +1. – Lewis Miller Jun 23 '16 at 15:31

The strong force is responsible for neutron stars not becoming black holes. Although attractive at nuclear densities, it becomes repulsive at somewhat higher densities. This repulsion is many times greater than the electromagnetic repulsion between protons at similar densities. Yet we know that if a star is too large, it's supernova event will lead to a black hole rather than a neutron star. For star masses this large gravity crushes the strong force. Similarly, a proton star of this mass (if somehow assembled) will also be crushed into a black hole.

• the strong force does not scale like 1/r^2, thus at some point it will not be strong enough to stop gravity, which you can make as large as you like. However the electrostatic force cannot be cancelled this way. – Wolphram jonny Jun 21 '16 at 23:44
• @Wolphramjonny: At high energies the electromagnetic force doesn't scale like that, either, you need to start calculating this with quantum electrodynamics (and at slightly higher energies with electroweak theory). What really happens is that new particles are being formed. The semi-classical electron is already a super-extremal black hole, I believe, which shows that one can't apply the theory naively and get useful results. – CuriousOne Jun 21 '16 at 23:47
• Some good observational points here, +1. Can you say anything about the scaling of the electromagnetic force?: you would seem to have the background and these points are a pretty necessary complement to my answer, which is made by someone who lacks particle physics knowledge. – WetSavannaAnimal Jun 22 '16 at 1:07
• The scaling of the electromagnetic force is irrelevant because long before the protons reach the density suggested here the protons will beta decay into neutrons and positrons. The positrons, being much less massive will disperse and the remaining neutrons will form the black hole. This is not too much different from what happens in a supernova of a very large star. – Lewis Miller Jun 22 '16 at 1:54
• Even gravity would do some very weird things in a proton star. Remember, gravity increases with energy density, the force fields of the strong/electro-weak would contribute to the energy density, resulting in more gravity. With electro-weak I would hazard that, the gravity would always be weaker, but with with the strong field, the limit range might just mean gravity wins out... – Aron Jun 23 '16 at 2:58