Not having studied General Relativity, I have sometimes been puzzled by references to the behaviour for "classic" black holes — as they are popularly portrayed — as being true for black holes which are not rotating and which have no charge. I don't understand the role rotation does play, but at least because it has to do with the motion of massive bodies, I can understand why it could play an important role. But I have no such intuition for electrical charge.

How does charge change the behaviour of a black hole, aside from the obvious role of electrostatic force? (It will preferentially attract particles of the opposite charge, obviously.) But it would seem that charge plays a more intriguing role than just this.

The current status of the Wikipedia page on black holes claims that you can theoretically avoid the singularity of a charged black hole. It also describes there that there is a theoretical upper bound on the charge/mass ratio of a black hole: that any would-be black hole exceeding it (which is generally thought to be impossible — see this related question on trying to force saturation of the charge/mass ratio or black holes) would lack an event horizon (and therefore presumably not be a black hole). Why should that be? Furthermore: from this other related question on repulsion of pairs of charged black holes (and from Willie Wong's comment, below), ir seems that the size of the event horizon may change depending on how close it is to being extremal! Why would the event horizon of a highly charged black hole be different than the event horizon of a neutral black hole, of similar mass?

Is there a clear reason for such an interplay is there between electrodynamics (aside from local Lorentz invariance) and general relativity that bring these things about?

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    $\begingroup$ You are mixing two concepts. First we have the assertion that the event horizon of an extremal black hole is "infinitely far away" (a statement which is not really true, but true under certain interpretations). This does not mean that the event horizon is larger. In fact, looking at the horizon area of Reissner-Nordstrom solutions, you see that charging the black hole actually decreases the horizon area (the area radius being, in appropriate units, $r = M + \sqrt{ M^2 - Q^2}$ where $M$ is the mass ans $Q$ is the charge). $\endgroup$ Aug 26, 2011 at 13:04
  • $\begingroup$ @Willie Wong: Alright --- I took the statement of infinite radius size at it's word, and assumed the variation between was continuous, from which I inferred that it increased. I'll edit my question to reflect that. Of course, that doesn't address my base concern of having no idea why the charge should affect the size of the event horizon at all! $\endgroup$ Aug 26, 2011 at 13:35

2 Answers 2

  1. The ability to "avoid the singularity" is generally regarded as a special property of the very special, stationary, exact solutions we know for black hole space-times. It has to do with the analytic continuation of the solution of Einstein's equations beyond the region that one can predict based on causal principles. (Basically, if you only know that there is a black hole, and you only know what goes on outside of the black hole, you cannot predict what happens inside the black hole; if you actually see the black hole form, on the other hand, you may have a chance at making this prediction.) So one shouldn't take that possibility too seriously. (I for one won't bet my life on it and jump into a charged black hole.) In fact, one interpretation of the Strong Cosmic Censorship conjecture is precisely that for generic black holes, the singularity is unavoidable once you entered the event horizon.
  2. The bound on the charge-to-mass ratio is, at the present day, more of an imprecise conjecture than a stated fact. There are several problems with that statement: in a dynamical space-time, mass-energy can radiate. So the definition of the "mass" of a black hole is already problematic. (This is also related to the fact that mass in general relativity cannot be defined locally; though there are a lot of work put into definitions of quasilocal mass.) Similarly, the charge of a black hole, in a general dynamical space-time, is not well-defined. What we do know is that we have a three parameter family of exact, stationary solutions to Einstein's equation depending on $M, A, Q$. Because these solutions are all stationary, the mass and charge are well-defined. Because these solutions are all axisymmetric, angular momentum is well-defined. And within this family we know that were the charge $Q$ to exceed the mass $M$, the formula that gives the expression of the metric tensor still makes sense as a solution to Einstein's equations, but the formula will lead to solutions with no event horizons. So we conjecture that this is a general fact, despite not knowing how to define the mass and the charge of a generic black hole. Now, there are some cases where this conjecture is known to be true for dynamical solutions. For example, in spherical symmetry, if we also assume that we have, in addition to the electromagnetic field, some other "good" uncharged matter fields (this makes the electromagnetic field non-dynamical, but the gravitational field is still dynamical), then we can prove such a statement using the Hawking mass of the black hole. These types of statements are related to Penrose-type inequalities, most of which are conjectural and only a few have been proven to hold generally. (Remark: there's however evidence that one cannot start with a subextremal black hole and then ``supercharge'' it.)
  3. The horizon area of charged black-holes are smaller than that of uncharged ones. (Again, because of the difficulty in definition, interpret the above in terms of the known stationary black holes.) Your last statement in the third paragraph is incorrect.
  4. The answer to your general question about the interplay and the mechanism is: "no one really knows". The problem is that general relativity, unlike classical Newtonian mechanics coupled to electro-magnetism, or even special relativistic mechanics coupled to electro-magnetism, is a highly nonlinear theory. In classical electrodynamics, the linearity of the system allows you to pin-point the contributors to the dynamics: you can say that the total force acting on this particle is a sum of the gravitational force plus the Lorentz force plus this-and-that. In GR, because of the non-linear feedback, it is in general impossible to disentangle the sum of the parts from the whole. (While the equivalence principle tells you that locally in inertial frames stuff behave as it were linear, the sort of questions you were asking necessarily involve long range effects of gravity and electromagnetism.)

To summarise: there's still too much we don't know, even with regards to the basic definitions, in general relativity to be able to answer your questions. We don't have a completely satisfactory definition of black holes that can be used locally (as oppose to teleologically), and we don't know what the local definitions of mass or charge should be. We certainly don't have a general description of how black holes should behave under the influence of electric charge. What we do have is a rather limited zoo of examples. This dearth of data points means that a lot of different conjectures can be made to fit those data points It is hard enough to know even which of those conjectures are right, nevermind to try and understand the principles behind such behaviour.

  • $\begingroup$ Despite your description that we do not know how to solve these problems, your answer is quite informative. I did not imagine, for instance, that the mass of a black hole was ill-defined in general. If I understand correctly, this is because one would normally think to infer it from contributions to the "gravitational field" (which is not actually a field, but shrug) by the superposition principle, which as you note does not apply here. I suppose that the black hole itself is a geometrical object, and any effective mass would be a charicature of the nature of the hole itself. (cont'd) $\endgroup$ Aug 26, 2011 at 14:59
  • $\begingroup$ I appreciate your answer as being words of caution, with respect to taking any of the existing (badly disseminated) theory of black holes as any sort of well-grounded fact or generalizable theory; I value knowing the boundaries of the answer. With the understanding that the theory does not extend well beyond these tranquil steady-state solutions, and that they may arise from analytic continuation (i.e. imaginative mathematics), could you elaborate why in those cases, charge has an impact on the event horizon and admits the possibility of avoiding the singularity? $\endgroup$ Aug 26, 2011 at 15:04
  • $\begingroup$ I don't have an answer for the singularity avoidance: all the mathematics points to that the transition from neutral to charged is a discontinuous process inside the black hole. The drastic difference is akin to the difference between the polynomial $x^2$ which has one double root and $x^2 - \epsilon^2$ which has two single roots. There is a (conspiratorial?) algebraic cancellation when you compare Schwarzschild to Reissner-Nordstrom. Even the nature of the singularities are much different: Schwarzschild has a space-like singularity, and RN has a time-like one. $\endgroup$ Aug 26, 2011 at 15:58
  • $\begingroup$ For horizon area, there is a somewhat often used argument. The argument basically goes like this: the "mass parameter" in RN is not the "mass of the black hole". The "mass of the black hole" should precisely be the area-radius of the horizon. The mass parameter, however, is the ADM mass of the space-time. So the correct interpretation of the mathematics should be that "for two blackholes of the same mass (horizon area), the one which is charged contributes more ADM mass". Now the ADM mass is the total energy content of the universe, so should roughly speaking be sum of all mass (contd) $\endgroup$ Aug 26, 2011 at 16:03
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    $\begingroup$ @Willy: this question is not asking about the conjectured swampland bound that the mass of the lightest charged particle is less than its charge. This is about the certain classical bound that the mass of a black hole is always bigger than its charge. $\endgroup$
    – Ron Maimon
    Aug 26, 2011 at 17:46

This is an area where there is a lot of confusion, and a lot of opinion cannot be supported by sound mathematical theory.

Not-analytic continuation

The theory of black holes requires continuation beyond regions which are accessible. These continuations are often called "analytic continuations", but this is ridiculous, because the fields do not have to be analytic in order to do it. Hyperbolic equations don't have general analytic solutions, and generally covariant equations can't have solutions which are analytic in non-analytic coordinate systems, which are in no way bad for GR.

The term I would use is "Minkowski horizon equivalence principle continuation", because it is continuation past a horizon by assuming that the horizon looks like Minkowski space locally, that is, by assuming the equivalence principle includes horizons too. When you reach a horizon which is locally like a Rindler horizon, like special relativity in accelerating coordinates, so that the curvature does not blow up, then you can continue a little ways past the horizon by extending the solution to be locally flat. In this way, you can continue into the interior of a black hole knowing the outside solution.

A Rindler horizon is vertical sheet in time for the accelerated observer. There is another kind of horizon which is locally a horizontal sheet. The Rindler horizon is formed by the spatial light cone going to the right in a 2d space-time diagram where time is vertical. If you tip this cone by 90 degrees clockwise so that it goes into the past, this is called a Cauchy horizon. A Cauchy horizon is a limit to predictability, because it is where new light rays come from infinity. This Cauchy horizon in question is the limit of the future development of a past pointing half-hyperbola which is just the 90 degree rotated trajectory of an accelearated observer.

Cauchy horizons are also locally flat, so one may continue past them by assuming the equivalence principle, just as for Rindler horizons. The issue is that when one does this, there is always new information coming in from someplace that you haven't seen before, so it is difficult to predict what happens, at least in classical GR.

But the full continuation is not in principle any more difficult than the continuation past the event horizon. The equivalence principle should hold at all horizons. But the results are strange, and require interpretation

Charged black hole solutions

I will restrict myself to 4 dimensions here, and to classical General Relativity. In string theory, there is a dilaton and the charges are d-form fields, so the "charged" black holes are extended objects in higher dimensions, sometimes with very different interior structure.

The charged black hole is nice because it is still spherically symmetric, and yet looks like the generic case of an arbitrary rotating black hole. The reason charge and rotation look alike is Kaluza Klein theory--- a charged black hole can be thought of as moving in a small circular fifth dimension. This is similar to rotation.

But the solution in GR is just as simple as the Schwartschild solution: for general mass M and charge Q:

$ ds^2 = - f(r)dt^2 + {dr^2\over f(r)} + r^2 dS_2^2$


$f(r)= 1-{2m\over R} + {Q^2\over R^2}$

The important thing is that there are now two radii where f is zero, and these are the two horizons. The outer horizon is continuously deformable to the Schwartschild horizon, and is an Event horizon. The inner horizon is a Cauchy horizon. As M=Q, the black hole is extremal, because the two zeros of f(r) have the same value of r, so you get a double root. Although the two horizons are collided in the r coordinates, they are not at the same point! The radius r breaks down, because the distance between the horizons approaches a finite limit at extremality.

For $M<Q$ there is no solution to GR without a naked singularity. Such solutions are absurd, they cannot be reached by continuous deformation of the uncharged black hole, and are forbidden by the Censorship conjecture.

The black hole singularity is at r=0, past both horizons, and trajectories in the interior of the Cauchy horizons are repelled by it. So it is impossible for a massive object to hit the singularity. The Cauchy horizon itself is an outgrowth of the singularity, it is the first light rays from the singularitly

The structure: outer event horizon- inner Cauchy horizon - timelike singularity is generic. It is the situation for rotating/charged/rotating-charged black holes. But it is considered unstable. I will discuss why in the next section, and why I think these arguments are bogus.

Bad arguments that the Cauchy Horizon is the singularity

There are, in the literature, arguments that the Cauchy horizon becomes the singularity in realistic solutions. These arguments are hand-wavy, and based on the intuition that everything should be mushed somewhere. In my opinion they are completely wrong.

The arguments are based on the following properties:

  • Cauchy horizons are unstable: you can see that it is possible to deform the solution to get whatever you like there, because you can start emitting arbitrary gravitational waves from the singularity, and these will go out and influence up to the Cauchy horizon.
  • The Cauchy horizon has a property that it can see the entire future development of the black hole exterior, in principle. When you approach a Cauchy horizon, you see the entire future development of your past, which includes the entire future development of the outer region.

The second point is very important, because it means that photons which come to you from outside are infinitely blue-shifted at the Cauchy horizon, leading people to expect a wall of hard-radiation to appear at this point under generic perturbations, which will make a singularity.

I don't buy these arguments, because the black hole is not eternal. It is not clear to me that a decaying black hole has the same Cauchy horizon structure as a regular black hole. It is also not clear to me why the singularity can't send out mollifying fields to make the Cauchy horizon non-singular rather than singular.

But most damningly, we have models of near extremal black holes in d-branes, slightly off from extremality, and these are shiny and thermodynamically nearly time-reversible. There are arguments by Gubser that d-branes will absorb other d-branes irreversibly by going off-diagonal in the non-commutative description, that is, by producing strings between the branes, but I don't buy these arguments at all.

I think that one should take the story in rotating/charged black holes at face value.

Traversable Cauchy horizon implies you can leave a black hole

If you do take the story at face value, objects going into a black hole will come out pretty much unchanged, except for possibly a little singeing when traversing the hard-radiation at the Cauchy horizon. These objects go in, turn around past the Cauchy horizon, and go out.

There are many problems with this picture. The continuation of the solution for a particle which does a full circuit goes past an event horizon, as many cauchy horizons as it wants to (by choosing whether to leave going "up in t" or "down in t" (remember that t is a spatial coordinate inside the black hole) and then out the event horizon going the other way, the way only Hawking radiation can be said to come.

This picture is crazy, because the continuation goes on producing more sheets, just like analytic continuation (but it's not analytic continuation). Each traversing path puts you in another universe, with another direction of time. The result is total nonsense.

This is why people are adament that one has to throw away interior charged solutions. These would naively lead to information loss, other universes, all that nonsense. In my opinion, the correct solution is to do some cutting and pasting and identify the outgoing and incoming paths. I tried a little, but I never found the right pasting.

If you know the pasting, you can predict what will come out of a charged/rotating black hole, mostly, up to a little randomization which becomes more random as the black hole becomes neutral/non-rotating. This type of idea means that black holes are not black at all, not just slightly thermally non-black, but shiny-mirror non-black.

This idea has exactly one supporter, which is me. I give a reasonable probability to this possibility because the counterarguments about a singular Cauchy horizon are all weak. Perhaps I should take solace in the observation that that mysterious anonymous author of the Wikipedia article also harbors suspicions, but I somehow don't think his opinion is independent.

Heuristic Justification for Pasting

The reason I think pasting is necessary is because of the following classical paradox, which I haven't found in the literature. If you throw a mass into a maximally extended charged black hole, and let it go past the horizons and come out, it comes out in another patch. But the black hole mass went up in your patch after absorption, and the mass of the black hole in the other patch went down after emission. But this means that they no longer fit together as continuations of each other.

So when the object comes out of the other sheet black hole, by consistency, it seems that something must come out of your sheet black hole. But the problem is that coming out can happen in either time direction, depending on how the infalling stuff steers itself, whether it decides to come out going towards positive or towards negative t, and so it means that objects can come out time-reflected, and the only way to make sense of that is to do a full identification with CPT gluing, so matter thrown in can come out as reflected motion-reversed antimatter. This is doubly crazy. So I am not sure if it is true, but I think it is. This might explain some astronomical puzzles of anomalous antimatter production because some astrophysical blackholes are measured to rotate at 99% of extremality.

  • $\begingroup$ Your claim that one may always continue past a Cauchy horizon is not supported by mathematical theory. In fact, it flies against known theorems about inextendibility of spacetimes past Cauchy horizons. For example: arxiv.org/abs/gr-qc/0307013 arxiv.org/abs/gr-qc/0509106 $\endgroup$ Aug 26, 2011 at 18:38
  • $\begingroup$ I hope to think a bit about the local transition you describe at the event horizon, so that I can at least understand what is at issue with continuation. But your description begins to clarify why one would expect solutions with $M < Q$ to be naked singularities. However, is there a reason why the charge should have the effect of rotation through a fourth spatial dimension? (I've heard of Kaluza-Klein theory: are you saying that, in any case, the solutions will be formally identical to a rotation in such a theory?) What does the L or ω look like, as a wedge product, in that case? $\endgroup$ Aug 26, 2011 at 19:39
  • $\begingroup$ @willy: I said that you can continue past a locally_flat Cauchy horizon, and that the instability results are no good for establishing lack of continuation. The paper you linked does not show that you can't continue past a Cauchy horizon. It shows that the metric only gets a discontinuous derivative there. So what. That's exactly like the electric field discontinuously changing in the presence of a charged plate, charges can cross the plate. The result of this paper support the position I hold. I find it weird that you would put a paper as a counterexample to a claim without reading it. $\endgroup$
    – Ron Maimon
    Aug 26, 2011 at 19:43
  • $\begingroup$ @Neil--- charge is linear momentum in the extra dimension, which only resembles rotation if you think of somehow making the extra dimension an angular variable, and the black hole a spinning line extended object, which doesn't work at all. So this is just a heuristic justification. There is no L or $\omega$ in this picture. you don't need this picture to work out the mathematical properties of spinning or charged holes, and to see that they are qualitatively the same, and qualitatively different from the neutral unspinning case. $\endgroup$
    – Ron Maimon
    Aug 26, 2011 at 19:46
  • $\begingroup$ @Willy: I think the confusion results from the idea that the continuation is somehow an analytic continuation. As I explained in the answer, the continuation does not depend on analyticity at all, nor can it. $\endgroup$
    – Ron Maimon
    Aug 26, 2011 at 19:49

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