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What prevents a static black hole from accumulating more charge than its maximum? Is it just simple Coulomb repulsion?

Is the answer the same for rotating black holes?

Edit

What I understand from the answers given so far, is that maximum charge is a moving target. You can add charge to a black hole but Coulomb repulsion guarantees that you will do so in a manner than will increase "maximum charge value". Is this correct?

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Where should the charge come from? Matter is neutral in general. –  Georg Mar 10 '11 at 20:56
    
@Georg: what do you mean? Let's say a very large electron stream hits the black hole. –  Eelvex Mar 10 '11 at 21:00
    
""more charge than its maximum?"" Is there such a maximum? –  Georg Mar 10 '11 at 21:56
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@Georg: yes there is. –  Eelvex Mar 10 '11 at 22:08
    
to comment on your edit, yes, I'd agree with that. Partly because it's a summary of my answer :P –  JBSnorro Mar 11 '11 at 22:46
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6 Answers

up vote 8 down vote accepted

Coulomb repulsion it is. Specifically, if a black hole has a lot of charge, then particles with a high charge-to-mass ratio will be repelled. Anything that falls in will contribute "more mass than charge," heuristically, keeping the charge-to-mass ratio of the black hole from getting too big.

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That's an OK answer, +1, but what if you just shoot the electrons (whose charge/mass ratio exceeds the extremality bound) into the black hole by violence? ;-) I mean by giving them too high a velocity? Will the black hole re-vomit them? Or is the required velocity such that the total (relativistic) mass of the electrons will keep the charge/mass ratio below the allowed limit? –  Luboš Motl Mar 10 '11 at 21:10
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I've never tried to do the calculation, but surely it's got to be the latter. With the right initial conditions, it seems certain that you can get the particle to cross the horizon. If so, then there's no way "re-vomiting" can occur (classically, anyway -- let's leave Hawking radiation out of it). So it's got to be that the mass goes up enough to keep the q/m ratio in the allowed range. –  Ted Bunn Mar 10 '11 at 21:14
    
That "electron gun" is a problem . One accelerates the electrons within the gun, but a similar voltage will buld up outside with opposit direction. Think of so called ion thruster, the ions are neutralized prior to exhaust to avoid such effects. –  Georg Mar 10 '11 at 21:46
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Dear Georg, I am not sure whether I fully understand your argument. You're surely not saying that "electron guns" are completely impossible, are you? Electron guns, under this very name, en.wikipedia.org/wiki/Electron_gun , are a key component of the devices that some people know under the name "television". Can't you just put a small black hole inside the television, turn the television on, and shoot enough electrons to the black hole before it starts to move? Is the electric field inside the TV the problem? –  Luboš Motl Mar 11 '11 at 6:04
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@Peter Shor: That's not quite right. In particular, an electron has so much charge that $Q > M$, (also $L > M^{2}$) meaning that an electron would be a naked singularity if you believed that it was a Kerr-Newman black hole. The charge limit arises from the fact that a sufficiently large charge will eliminate the horizon of the black hole. –  Jerry Schirmer Mar 14 '11 at 18:32
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A slightly different answer to this is "neutralisation". That is the free positive ions around (in gas clouds nearby maybe) will neutralise the charge. This is generally assumed to keep the charge of a Black Hole near to zero in astrophysical contexts.

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The coulomb force cannot be responsible, because given enough energy I can add more charge to the black hole... There was a related question recently: Paradoxical interaction between a massive charged sphere and a point charge To summarized my asnwer there: It isn't actually the Coulomb force that prevents the addition of charge to the black hole beyond its maximum, but by adding more charge more electrostatic binding energy is added. A lot! (since the black hole must have an incredibly chargedensity.) Therefore also its mass increases, which in turn leads to an increased maximum charge.....

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There is actually a nice and simple calculation of gravitational collapse of charged spherical shells, where you can show that Coulomb repulsion is stronger than gravitational attraction if you exceed the critical bound |Q|>M (in convenient units). You find this simple calculation in the lecture notes by Paul Townsend on black holes [see chapter 3, in particular eqs. (3.10)-(3.13)].

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the energy required to bind together N electrons in the sphere, it is actually

$e/R + 2e/R + ... Ne/R = N(N-1)e/2R$

which is quadratic.

For each additional electron shoot into the black hole we will have to add energy to it that is quadratic in the current charge (proportional to N, the existing charge), while the charge will just increase linearly (by one)

so the (additional) mass-energy of the black-hole will grow quadratically in the charge

since the maximum charge of a black hole is some function of the mass of the black hole, this implies that this function grow no quicker than a square root of the mass

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For an extremal black hole, the charge is proportional to the mass. I believe your calculation is incorrect because it's assuming that the radius of the sphere is fixed, while the radius of a black hole grows proportional to the square root of its mass. –  Peter Shor Mar 12 '11 at 13:02
    
@PeterShor, thanks! and that is why heuristic arguments need to be taken with a grain of salt, specially on these topics –  lurscher Jun 18 '12 at 21:07
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There is a way of seeing this more explicitly with the Reissner-Nordstrom (RN) metric $$ ds^2~=~-F(r)dt^2~+~F(r)^{-1}dr^1~+~r^2d\Omega^2 $$ where the $F(r)~=~1~-~r_0/r~+~(Q/r)^2$, $r_0~=~2GM$ and $Q$ the charge in length units. The metric has two critical points $$ r_\pm~=~\frac{r_0}{2}~\pm~\frac{r_0}{2}\sqrt{\frac{4Q^2}{r_0^2}} $$ These are the outer and inner horizons for $r_+$ and $r_-$ respectively. The region between them is a spacelike trapping region, similar to the interior of a Schwarzschild solution. The extremal condition on the black hole is where $r_+~=~r_-$ which is where the spacelike region between the outer and inner horizons has been “removed,” or in a more subtle way mapped into the spacetime $AdS_2\times S^2$.

From the metric components we then compute the Christoffel symbols in the usual straight forwards, though tedious, manner. The most salient of the connection terms is $$ {\Gamma^r}_{tt}~=~F(r)\frac{r_0r~-~2Q^2}{2r^3} $$ which gives the geodesic equation $$ \frac{d^2r}{ds^2}~+~{\Gamma^r}_{tt}U^tU^t~=~0. $$ Far from the black hole We have that $U^t~\simeq~1$ and so $ds~\simeq~dt$ and this is a Newton second law type of equation $$ \frac{d^2r}{dt^2}~+~F(r)\frac{r_0r~-~2Q^2}{2r^3}~=~0, $$ where for $Q~=~0$ recovers Newton's second law for gravitation.

Now consider the extremal case. The connection term is then $$ {\Gamma^r}_{tt}~=~\frac{1}{2}\Big(1~-~\frac{r_0}{r}~+~\frac{r_0^2}{4r^2}\Big)\Big(\frac{r_0}{r^2}~-~\frac{r_0}{2r^3}\Big) $$ which tells us that a neutral particle is still attracted into the black hole. Then we consider a charged particle

The field strength 2-form and tensor components is $$ F~=~\frac{Q}{r^2}dt\wedge dr $$ The geodesic equation is no longer zero, but there is a driving force $F~=~F(r)r_0/2r^2$. With this Newtonian approximation the total force on the particle can be seen to be zero near the horizon. So for the extremal black hole a charge near the horizon will experience no net force.

Other connection terms are also nonzero. An important one is ${\Gamma^\theta}_{r\theta}~=~-1/r$ . For the extremal case the radial acceleration of a charge near the horizon approaches zero, but the angular component remains. Hence if there is a small $U^\theta$ this will move the charged particle off the radial path and ultimately away from the black hole. This in effect prevents the overcharging of a black hole.

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Can we do calculations across the horizon using a singular coordinate system? I think this has to be re-run in Kerr-Schild coordinates or something like that, and that'll give $A_{a}$ an $r$ component. –  Jerry Schirmer Mar 11 '11 at 20:00
    
This is the best answer. –  Carl Brannen Mar 12 '11 at 3:23
    
... and the only correct answer. +1 –  user346 Mar 12 '11 at 8:49
    
That's not the correct formula for the horizon--$r\rightarrow M$ as $Q\rightarrow 0$. It should be $r=M \pm \sqrt{M^{2}-Q^{2}}$ –  Jerry Schirmer Mar 12 '11 at 18:54
    
The radial acceleration on a charged particle is zero for the extremal case. Any deviation from radial motion results in an acceleration away from the BH. Jerry, the equation is defined in a number of different ways. You are right and after I was about 2/3 into this back of envelope calculation I realized your formula (or definition), but kept with what I did. The 4 and 2 there comes from the quadratic equation. It should be absorbed into the Q for similicity. –  Lawrence B. Crowell Mar 13 '11 at 3:20
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