Can you turn a black hole inside out with charge?

[Antiphon]

Thought experiment: you locate a modest sized black hole. You then start feeding it protons. Lots and lots of protons.

Will it be able to eat enough protons so that the repulsive force of the excess charge can overcome the inward pull of gravity?

Would there be three different event horizons for protons, electrons and neutrons?

 

[Matterwave]

I think that would violate the cosmic censorship conjecture...since it would be a naked singularity...I think? (Just a conjecture though)

 

[Dickfore]

http://en.wikipedia.org/wiki/Reissner-Nordström_metric

 

[Antiphon]

Thanks for the link. They seem to say it's not possible. So what would actually happen if you started pouring protons into the hole? Would it reach a point where the next proton couldn't go in before the event horizons could diverge enough to uncloth the singularity?

 

[Nabeshin]

Would it reach a point where the next proton couldn't go in before the event horizons could diverge enough to uncloth the singularity?

Indeed. After a certain point the proton wouldn't be absorbed by the black hole because if it did so the hole would have over-extremal charge.

I'm not sure of anywhere this is worked out in detail -- people seem to normally gloss over charged black holes since they're not really of astrophysical significance. One would think that it would be in a paper somewhere, and if anyone knows I'd love to see. A friend of mine is actually engaged in work related to this problem right now, and so far we've found the literature quite sparse.

 

[Bill_K]

Collapsing charged dust was studied extensively back in the 60s and 70s. For example see Plebanski's book, "An Introduction to General Relativity", parts of which are on books.google.com. A recent paper on the subject is 1010.4211v2.pdf on arxiv.

 

[Antiphon]

Ok, so here's a prescription then for ending the universe or triggering a big bang or something.

You find a fairly large black hole and feed just enough charge into it that it can accept maybe another coulomb or two. You then fire charged torpedoes into it from all six cardinal directions at the same time. Each torpedo contain a 1/2 coulomb charge and travels at near the speed of light. The six torpedoes would be causally disconnected from one another and so the hole would be forced to accept all six charges loads.

Next thing you know, naked singularity.

 

[Bill_K]

Will it be able to eat enough protons so that the repulsive force of the excess charge can overcome the inward pull of gravity?

Protons outside the hole will be repelled, yes. The rule still holds that nothing already inside can escape.

Would there be three different event horizons for protons, electrons and neutrons?

Event horizons refer to the behavior of outgoing null geodesics. Here we're talking about the inward trajectories of particles with mass. The electrons and the protons don't even follow geodesics, let alone null ones. A charged black hole still has an event horizon, although at a different radius that depends on both e and m.

As the Wikipedia article points out, there's an upper limit to the e/m ratio for a black hole. Here's how a growing hole avoids exceeding the limit. Incoming protons get repelled more and more. Therefore you can't just drop them, and let them fall in of their own accord, you have to fire them into overcome the repulsion. In doing so, you must give the proton additional energy, and the amount of energy needed for this increases as you approach the limit. Additional energy means additional mass. The net result is that you're forced to add so much mass along with the proton that the hole stays below the e/m limit.

 

[pervect]

Try MTW, pg 910

Some notes
M is the mass of the black hole
Q is its charge.
S is its Angular momentum
a = S/m, i.e. angular momentum / unit mass

If a black hole is initally of an extreme Kerr-Newmann" variety with M^2 = a^2 + Q^2, so that one might fear a change which makes M^2 < a^2 + Q^2 (ed note: by adding charge Q to the black hole without increasing M) and thereby destroy the horizon, one's fears are unfounded. Eq 33.56 demands

M delta M >= a delta a + Q delta Q

i.e that in order to get that charge Q into the black hole, one must also add enough energy to increase M^2 enough to satisfy the inequality.

So, if your hypothetical charged torpedoes don't have the sufficient energy, they'll won't make it into the black hole, they'll be repelled by its positive charge. Contrawise, if they do have sufficient energy, then they'll increase the mass of the BH as well as its charge to maintain the relation M^2 > a^2 + Q^2