By Eelvex


2011-03-10 20:53:04 8 Comments

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?

6 comments

@Lawrence B. Crowell 2011-03-11 19:19:37

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.

@Jerry Schirmer 2011-03-11 20:00:32

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.

@Carl Brannen 2011-03-12 03:23:22

This is the best answer.

@user346 2011-03-12 08:49:25

... and the only correct answer. +1

@Jerry Schirmer 2011-03-12 18:54:22

That's not the correct formula for the horizon--$r\rightarrow M$ as $Q\rightarrow 0$. It should be $r=M \pm \sqrt{M^{2}-Q^{2}}$

@Lawrence B. Crowell 2011-03-13 03:20:06

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.

@lurscher 2011-03-11 16:55:21

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

@Peter Shor 2011-03-12 13:02:40

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.

@lurscher 2012-06-18 21:07:31

@PeterShor, thanks! and that is why heuristic arguments need to be taken with a grain of salt, specially on these topics

@Daniel Grumiller 2011-03-11 16:40:43

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)].

@JBSnorro 2011-03-11 00:35:33

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.....

@Roy Simpson 2011-03-10 22:40:18

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.

@Zemyla 2019-01-06 12:31:25

And even if there are no free ions around, the Hawking radiation would preferentially have the same charge as the black hole, and thus radiate the black hole's charge away. For instance, if an electron and a positron appear near a negatively charged black hole, it's much more likely that the positron will fall into the event horizon and the electron will escape towards infinity than the other way around.

@Ted Bunn 2011-03-10 21:02:46

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.

@Luboš Motl 2011-03-10 21:10:56

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?

@Ted Bunn 2011-03-10 21:14:21

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.

@Georg 2011-03-10 21:46:44

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.

@Luboš Motl 2011-03-11 06:04:39

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?

@Georg 2011-03-11 10:04:55

Dear Lubos, Eelvex proposed to "Let's say a very large electron stream hits the black hole". What else than an electron gun could be used to do that? An in strictly astrophysical context: any "natural" electron (or ion) gun is under the same restrictions. Matter is neutral in general. Even a beam from a active galaxy is neutral.

@lurscher 2011-03-11 16:50:44

i thought the answer is that since the binding electrostatic energy grows quadratically in the charge, adding more electrons will increase the mass-energy of the black hole more than enough to account for the new charge limit (that it just increased linearly). Is there controversy on this?

@Ted Bunn 2011-03-11 16:54:21

@lurscher: I think that's an equivalent way of saying what Lubos and I were saying. In order to shoot charges into an already-charged black hole, you need to give them enough energy to overcome the repulsion, and that'll add to the mass of the black hole.

@Jerry Schirmer 2011-03-11 19:57:31

@Ted Bunn: but that'll also add to the momentum of the black hole, and when you re-boost to the rest frame of the final state black hole, you won't necessarily see all of that energy anymore (yes, you can usually ignore that, but not if we're adding a charge of order $M$ to the black hole).

@Ted Bunn 2011-03-11 20:41:49

@Jerry Schirmer: I'm not sure I see your point. If it helps any, put two identical electron guns on opposite sides of the black hole. Anyway, it's a fact that there is a maximum Q/M for a Kerr black hole, so if you fire a charge into a black hole that's already near that bound, it must be the case that M goes up by enough to keep Q/M within that bound.

@Jerry Schirmer 2011-03-11 20:53:23

@Ted Bunn: Yes, though I've never seen a particularly satisfying proof of the Third Law of Black hole dynamics, or at least one that's anywhere near as convincing as the proofs of the 0th-2nd Laws. If the mechanism isn't something about the ratio of coulomb force to surface gravity, I don't see why the black hole being extremal is the critical condition here.

@Peter Shor 2011-03-14 17:22:29

@Jerry: From this discussion, it seems to me that the only way to get an extremal charged black hole would be to have the entire mass of the black hole be coming from the energy of Coulumb repulsion. For this, you'd need to build it out of charged particles with zero mass. Electrons are presumably the closest we can come to this, and their mass isn't quite zero (although it's very small compared to the Planck mass, if that's the right scaling). Is this a proof of the 3rd law for charged black holes, or have I got my reasoning wrong somewhere?

@Jerry Schirmer 2011-03-14 18:32:16

@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.

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