# Why Do Extremal Black Holes Not Radiate?

This might be a "metaphysical" question with no actual physics content and if such is the case then I apologize. Actually, to figure out whether or not it is an actual physics question is one of the major reasons I am posting this question.

Extremal black holes are the black holes that have the lowest possible mass for a given value of charge. If the mass were lower than the extremality value then the singularity would have been a naked singularity.

Now, naively speaking, quantum mechanics provides us with an attractive possibility of a black hole evolving into a naked singularity by the means of losing mass (while preserving charge) via Hawking radiation. But, as we know, when calculated, Hawking radiation ceases to exist precisely when the black hole becomes extremal - stopping the black hole from losing any more mass right at the stage at which it could have evolved into a naked singularity if it were to radiate even a little bit. This all seems extremely beautiful but even more freaky.

What I want to ask is that is there any reasoning due to which a Theoretical Physicist could have argued before actually calculating the explicit expression for Hawking temperature that it must be zero for the extremal black holes? Or do we have to accept this situation as a happy coincidence?

I think it is plausible that some might think that the censorship conjecture might be considered as the underlying reason behind this apparent coincidence. Thus, I would like to clarify that I think the censorship conjecture cannot provide any such reasoning as it has no deeper theoretical grounds but is just a form of wishful thinking. In more concrete words, the censorship conjecture just lays out some expectations that the consequences of all the fundamental laws should fulfill. It doesn't provide us with any understanding of the underlying physics. At best, it could have made the Theoretical Physicist hope that the Hawking temperature better be zero for extremal black holes. But it couldn't have provided rigorous grounds for predicting that it must be zero in the absence of an explicit expression for the Hawking temperature.

PS

I think the fact that extremal black holes do not radiate should be considered as strong circumstantial evidence to the censorship conjecture. But the converse way of thinking the censorship conjecture as the reasoning behind the zero temperature at extremality seems absurd.

Edit

Note that naked singularities are not a problem in themselves. Because, in reality, there exists a well-behaved quantum gravity theory. Thus, it wouldn't be an internal inconsistency of the theory if naked singularities existed. We expect the censorship because otherwise, our experimentally verified calculations about the world that we have done without taking into account the effects of some naked singularities lurking around would have no reason to be verified in the first place. As far as I know, this is all the logic that we have behind the censorship. And it is an anthropic kind of reasoning. There is not any perfectly rigorous way of justifying my use of the word "coincidence" for the described situation but the following two are the reasons behind my use of the word "coincidence":

1. Charged black holes don't really exist in the universe. And thus, our anthropic conjecture shouldn't really be stretched to assert anything about them. But it surprisingly works even if we do such a naive thing.

2. Naively speaking, if we expect the censorship to emerge then also, zero temperature could have occurred at many other places including the extremality. But it happens precisely at the extremality and nowhere else. As if it specifically cared about the censorship. What I mean is that we ought to think of some basic principle of physics that very badly requires the censorship to be followed.

• Why do you want to interpret this as a "happy coincidence", and not as a conformation that GR+ED is indeed consistent and doesn't produce naked singularities? (I don't understand what "coincidence" is supposed to mean here - it's not that we have reason to believe that the Hawking temperature of an extremal black whole was chosen "at random" by something...) Why do you want to argue it doesn't radiate without doing the best you can do to determine its Hawking radiation, namely computing temperature? Haven't you already given the reason we should expect this, namely prevention of singularities? – ACuriousMind Sep 13 '17 at 10:43
• See, Naked Singularities are not a problem in themselves. Because, in reality, there exists a well behaved Quantum Gravity. Thus, it wouldn't be an internal inconsistency of the theory if naked singularities existed. We expect the censorship because otherwise our experimentally verified calculations about the world that we have done without taking into account the effects of some naked singularities lurking around would have no reason to be verified in the first place. Afaik, this is all the logic that we have behind the censorship. And it is anthropic kind of reasoning.... – Dvij Mankad Sep 13 '17 at 11:03
• ... but in this case, it appears that the basic theory is indeed preventing naked singularities to form. There is not any much rigorous way of saying that Hawking temperature is chosen out of many possible values but one can think in two ways to understand the "coincidence". 1. Charged black holes don't really exist in the universe. And thus, our anthropic conjecture shouldn't really be stretched to assert anything about them. But it surprisingly works even if we do such a naive thing. – Dvij Mankad Sep 13 '17 at 11:16
• 2. Naively speaking, if we expect the censorship to emerge then also, zero temperature could have occurred at many other places including the extremality. But it happens precisely at the extremality and no where else. As if it specifically cared about the censorship. What I mean is we ought to think of some basic principle of physics that very badly requires the censorship to be followed. – Dvij Mankad Sep 13 '17 at 11:16
• Cosmic censorship is defined as a hypothesis about classical gravity. CC doesn't have consequences for Hawking radiation and Hawking radiation doesn't have consequences for CC. – Ben Crowell Sep 13 '17 at 23:11

## Not charge, rotation scares the theorist

In practice, most people don't worry about the case of a charged black hole getting so charged as to turning into a naked singularity but rather about the fact that infalling matter will make it rotate with such a speed that the horizon "rotates at the speed of light", and it then becomes a naked singularity (an extremal/overextremal black hole) this way.

This is because in astrophysical black holes, we expect the charge-to-mass ratio to be close to zero, certainly absolutely negligibly as to the extremal charge-to-mass ratio. Mass which feeds black hole is almost always quasi-neutral, and if an overcharge does come about, it actually attracts more of the opposite sign and drives away the same sign of charge, so thinking about extremally charged black holes seems to be more of a mental exercise.

However, what is certainly not a mental exercise is the issue of rotation. Isolated stationary rotating black holes (without charge) are uniquely described by the Kerr metric, and we see that if the ratio of the angular momentum to mass squared (plus some physical constant factors) $Jc/(G M^2)$ is larger than 1, the black hole becomes a naked singularity in very much the same way it happens with the charged black hole. (The black hole is extremal when we have exactly $Jc/(G M^2) =1$.)

But if you run some real astrophysical numbers, you will see that for instance for the Milky Way this ratio of angular momentum to mass squared is of orders $\sim 10$. Supermassive black holes at the galactic centers got most of their mass exactly from their host galaxies and this would make us naively guess that most supermassive black holes must be naked singularities.

## Avoiding nakedness

A more sophisticated analysis shows that even though there is a lot of angular momentum in the galaxy, galactic black holes most probably manage to avoid the over-rotation and throughout cosmic history attain an equilibrium rotation rate which is a tad below the critical rate.

This is because particles with large angular momentum do not just fall into a black hole. Technically speaking they are driven away by the centrifugal barrier, normally speaking they just "fly by" and are not attracted enough to the black hole. Think of any other process which would bring in matter into a black hole and this tendency of angular momentum to just fly away will apply there too.

In fact, if particles didn't loose angular momentum in an accretion disk, a black hole would be a rather underwhelming matter gobbler. In an accretion disk, the particles slowly loose angular momentum, spiral down to the inner edge of the accretion disk, and then fall into the black hole with whatever angular momentum they are left with. But the funny thing is that the larger the black hole rotation, the smaller the angular momentum of particles on the accretion-disk inner edge! And guess what is the angular momentum of this inner-edge matter as you approach critical rotation? Yes, you guessed it, exactly zero.

This is a remarkable patter which shows up all over the place - the black hole fights with a stronger and stronger resilience the attempts of its surrounding to convert it to an extremal black hole. Some of these arguments led Kip Thorne to famously estimate in 1974 that an astrophysical black hole would saturate at an angular momentum which is at $0.998$ the critical value, today astrophysicist estimate that this equilibrium value is probably even a little bit smaller.

## Why Do Extremal Black Holes Not Radiate?

I took a slight detour to directly answer your question, but I feel it was needed to address all the issue you mention in the body of your post. Now let's get to it.

The subextremal, extremal, and overextremal black holes should be understood as completely different space-times which are related only by some weird singular transform. Consider a simple black hole metric without charge or rotation. Then the case $M=0$ is a part of the parametrized family of metrics even though it is completely disparate in terms of causal structure and general nature. It is just flat space-time, whereas any $M > 0$ is a black hole with a true inoperable singularity in its center and a horizon, the "causal divisor" standing between very different regions of space-time.

You can understand the various extremal/overextremal limits similarly. They formally belong to the same parametrized family of metrics, but if you take a look on their coordinate-free properties, they are nothing alike. What stands between them is usually called a topological change" in the sense we would talk about a topological change when one pierces a ball to make a torus.

But the problem isn't only topological, the convergence to the extremal black hole is funny in all sorts of senses. Consider for instance the various physically important circular orbits such as the marginally bound or marginally stable orbits. These all converge to the same coordinate radius (the radius of the extremal horizon) but if you measure what is happening to their proper distance, this one actually diverges as we are reaching the extremal black hole!!

And now we are getting to the Hawking radiation. From the arguments I have given you previously it is obvious that we can only expect the extremal black hole to correspond to some weird limit of the subextremal radiation. In fact, the radiation of an extremal black hole, by direct computation, is simply zero and can thus not be assigned any temperature at all! The fact that we assign a temperature to the extremal black hole is by continuous extension by taking a limit from the subextremal case. So the question is more why does the temperature converge to zero when we are spinning up (or overcharging) a black hole almost to the extremal limit.

The intuitive (and slightly inaccurate) picture for Hawking radiation is that a particle-antiparticle pair is created near the horizon, one of them falls through it, and another one of the particles escapes to infinity to add to the Hawking radiation. The only speck of intuition for the temperature going to zero is the fact that as we approach the extreme black hole, the external horizon is getting "weaker and weaker" because a second "repulsive" horizon is approaching from below and weakening the particle-antiparticle separation effect. Since reaching an extremal (degenerate) horizon will always be associated with the weakening of particle-antiparticle separation, we can intuitively guess that the limiting radiation and thus temperature must be zero.

• Thanks for your wonderful and elaborate answer. So, one of the things that I gather from your answer is that extremal black holes are limiting cases of all sorts of sequences. But I feel that there should be some all-encompassing single underlying "explanation" behind these wonderful properties of extremal BHs. Maybe what I expect is some sort of a fundamental reasoning that validates the censorship and thus demands that a specific theory of gravity must respect this general principle. – Dvij Mankad Sep 17 '17 at 19:32
• In a pure analogy, SR teaches us that why nothing must travel faster than light. But if we hadn't known SR but had only known the funny ways in which all sorts of different mechanisms devised to make superluminal signaling fails, we would have been curious about an all-encompassing explanation for all these individual failures which sound like the pieces of the same story. I feel something similar should be going on here with the role of SR replaced by the fundamental reasoning that validates the censorship and that of the result that nothing can go faster than light by the censorship. – Dvij Mankad Sep 17 '17 at 19:33
• Well, this is a more complicated and "soft" question you might think. Claiming cosmic censorship is simply convenient because it makes relativity "scientifically complete" in the sense that observers sharing data in "our" part of the space-time will always obtain perfectly valid scientific predictions from relativity about any observation they can make. In a scientific sense of making predictions, testing them, and sharing the results for reproduction, the interior of the black hole is entirely irrelevant, in some particular sense "scientifically nonexistent". – Void Sep 17 '17 at 21:19
• The curvature singularities arising in relativity can then be understood as just a mathematical nuisance of the theory while "what we need" from the theory is perfectly ok. From my point of view, I see no more true reasoning for cosmic censorship, only anecdotal evidence. But this is not sufficient fundamentally and can only be resolved by finding a theory where the interior of black holes and irrelevant regions are "quotiented out" automatically. Such a theory would be a holographic theory of gravity, but we do not have one yet. – Void Sep 17 '17 at 21:22
• @Kagaratsch No, the condition really is $Jc/(GM^2)=1$, or in geometrized units $a = M$. Some people define the gravitational radius $r_{\rm g} = GM/c^2$, which is different from the Schwarzschild radius by a factor of 2 - so that might be the problem in your formulas. – Void Mar 5 at 8:33