Black holes evaporate (Hawking Radiation) acting as black bodies with the temperature inversely proportional to the mass.

No physical process, be it evaporation or any other "trick", can make a black hole "super-extremal", namely it can't make the square of the angular momentum or the charge too large compared to the mass. As the mass drops due to evaporation the charge and angular momentum must also drop.

Angular momentum can be shed with photons, because photons have spin and can be emitted in a non-radial direction.

Charge is trickier. There are three methods to shed charge:

  1. Accrete oppositely charged particles. This can be stopped, hypothetically at least, by isolating the hole.

  2. Emit charged particles. There are no massless charged particles, so a temperature << 511keV strongly suppresses this positron and electron generation, so making the hole large enough should stop this.

  3. Break down the vacuum with the electric field (requires ~10^18 V/m). Again, making the hole large enough should suppress this because the electric field in the vicinity of a near-extremal charged hole scales as 1/M.

If all three of these are suppressed, we have a new candidate for the longest lived objects possible! So will a large enough black hole evaporate itself toward extremity after which there will be essentially zero Hawking radiation?

  • $\begingroup$ It is a misconception that extremal black holes will not radiate. The existence of superradient modes for charged and rotating black holes allow for the emission of "Hawking" radiation even for extremal black holes. $\endgroup$
    – TimRias
    Jul 9, 2019 at 7:30

2 Answers 2


Short answer: Yes, if an isolated black hole is large enough (supermassive) and has an initial charge comparable to its mass, then it would lose mass through Hawking radiation much quicker than it would lose charge and would eventually reach nearly extreme state. It would still continue to lose mass and charge though at much slower rates and would remain in a near extreme state nearly till the end of its long, but still finite, lifetime, exceeding by many orders of magnitude the lifetime of an uncharged black hole of the same initial mass.

Longer answer: In what follows, we are using Planck units $G=\hbar=c=1$. $Q$ and $M$ are the charge and mass of the black hole, and $e$ and $m$ are the charge and mass of the electron, the lightest charged particle.

First, let us emphasize that in a realistic astrophysical environment free electrons/positrons would quickly neutralize any significant charge that the black hole may possess, so OP's condition 1 makes the situation quite artificial.

If we consider charged particle orbits in the Reissner–Nordström metric, the condition $$ \frac{e Q }{ r_+} > m $$ makes it energetically favorable for a pair of charged particles to form with one escaping to infinity and another falling into the black hole. Here, $r_+$ is the horizon radius, so $Q/r_+$ is the electrostatic potential at the horizon.

If Compton wavelength of an electron is much smaller than $r_+$ then pair production could be described by Schwinger's equations. The rate of pair production would be exponentially suppressed if the maximal field strength is lower than $E_{S}\sim \frac{m^2}e$. Since the field strength at the horizon is $\frac{Q}{r^2_+}$ and for a RN black hole $M\leq r_+ \leq 2M$, a black hole can carry a geometrically significant charge ($Q$ comparable to mass $M$) for a long time only if $$ M > \frac{e}{m^2} \approx 5 \cdot 10^5 M _\odot. $$ This also automatically enforces OP's condition 2. Such a black hole would fall into a SMBH range.

Evolution of charge and mass for such massive isolated black hole has been considered in literature:

  • Hiscock, W. A., & Weems, L. D. (1990). Evolution of charged evaporating black holes. Physical Review D, 41(4), 1142, doi:10.1103/PhysRevD.41.1142.

The rate of charge loss is obtained by integrating the Schwinger pair production rate over the volume near the horizon, while the mass loss is a sum of thermal radiation from massless particles and energy carried away by the charged particles. The resulting system is then integrated numerically. The overall evolution for the system is best illustrated by the following plot:

Figure from Hiscock & Weems 1990

FIG. 2. Evolution paths followed by evaporating charged black holes. The charged-black-hole configuration space is divided into two regions: a “charge dissipation zone” in the upper left, where black holes rapidly discharge, and a “mass dissipation zone” to the lower right, in which evaporation causes the charge-to-mass ratio of the black holes to increase. The boundary area between these two regions is a dissipative attractor, which all charged black holes evolve toward as they evaporate.

While here is a sample evolution of charge and mass through the black hole lifetime:

Figure from Hiscock & Weems 1990

FIG. 7. Mass and charge as functions of time for a black hole with $M= 168 \times 10^{6} M_\odot$ and $(Q/M)^2=0.1$ initially, and $n_\nu=3$. The charge-to-mass ratio of the black hole reaches a maximum at $(Q/M)^2=0.9999$ just as it reaches the attractor. The black hole spends most of its lifetime very close to the extreme Reissner-Nordstrom limit.

We see that a very heavy black hole with a significant initial charge $Q<M$ would first lose most of its “excess” mass ($M-Q$) and then spent most of its lifetime in a near extreme Reissner–Nordström state evolving along the attractor trajectory. The temperature of the black hole of course, never reaches zero, so there is no violation of the third law of thermodynamics.

The total lifetime of such charged black hole is dominated by the initial charge $Q_i$ and could be approximated as: $$ T\simeq \frac{2 \pi^2 \hbar^2}{e^3} \exp \left(\frac{Q_i}{Q_0} \right)= 10^{47} \exp \left(\frac{Q_i}{Q_0} \right) \,\text{yr}, $$ where $Q_0=\frac{\hbar e}{\pi m^2}\approx 1.7\cdot 10^5 M_\odot$, the equation starts being valid for $Q_i> 60\cdot 10^6M_\odot$. This lifetime is exponentially longer than the lifetime of an uncharged black hole, that scales as $M^3$.

  • $\begingroup$ TON 618 is 66 billion solar masses. If it was artificially charged up (ignoring the mass increase of charging it) and isolated, we are talking 10^170000 years or so! Near this "monument to longevity" spacetime is less curved than here on Earth but the electric field would tear matter apart. $\endgroup$ Jul 13, 2019 at 5:54
  • 1
    $\begingroup$ @KevinKostlan: Yes! Also, after I wrote the answer, I had a thought that one would only have to isolate such a black hole only while there is a lot of matter in the universe. If the universe continues to expand exponentially, then after just a few trillions of years there probably would be too few free charges to neutralize such a black hole, especially if the supercivilzation that charged it would also direct it away from matter concentrations. So after that the BH would persist on its own. $\endgroup$
    – A.V.S.
    Jul 13, 2019 at 13:09

This question has some deep aspects to it, and at this time I really do not think there is a definitive answer to it. This touches on issues of quantum information of black holes and the final state quantum process, which approximately is a form of Hawking radiation explosion of a quantum black hole. So I will not pretend to answer this question in any fullness, but I can offer some points to think about.

To start the Boltzmann constant $k = 8.6\times 10^{-5}eV/T$ with the electron mass $m\simeq 5\times 10^5eV$ means this threshold for electron production is a black hole with temperature $T\le 1.7\times 10^{10}K$. The Hawking formula for mass is then easily used to find this minimal threshold for black hole mass is $7.2\times 10^{12}$kg. Then if black hole radiation is purely spontaneous emission, which is the Hawking theory, then for a charged black hole with a mass larger than this it will not be possible to emit an electron that carries off charge. The black hole with even a single electric charge would not be able to emit this charge and then Hawking radiation would carry off mass until the extremal condition is reached. The Reissnor-Nordstrom charge or charge radius $Q=\frac{Ge^2}{4\pi\epsilon_0c^4}$ is about $4.4\times 10^{-65}m$ and thus far smaller than the Planck radius. Similarly the Schwarzschild radius of an electron mass is much smaller. So this points to some possible black hole remnant with the charge and maybe mass of an electron. We are in a way “straining” our ideas about Planck gravity units and quantum gravitation, which are at best incomplete. A black hole with a Planck unit of charge would have about $10^{30}$ units of charge or about $10^{11}$ Coul of charge. That is a fair amount of charge!

We have to of course ponder whether that can happen, for charge is a quantum number and correlated with information. Such a black hole might have Bekenstein bound issues. It would tend to imply a Planck mass black hole, with one qubit or at least very few qubits of information is composed with a large number of charges. Much the same could be said of Kerr black holes, where we may think of high angular momentum as converging to harmonic oscillators in the Holstein–Primakoff transformation. This means there is a large number of states that have no contribution to the entropy of the black hole. Carrol, Johnson and Randall showed how the interior region between the inner and outer horizons of a Kerr black hole are projected into and $AdS_2$ that is equivalent to the conformal field theory $CFT_1$. Extremal black holes in general have zero (classically) or near-zero (quantum effects) temperature. Entropy with $r_+=r_-$ is $S = A/4\ell_p^2$ $= \pi r_+^2/\ell_p^2$ We are then faced with some funny questions. In particular, does this quantum information “escape” into this $AdS_2$ to be never seen again, or does it reappear? If there is no temperature then is this quantum information sealed away forever?

There are a number of possible mechanisms though that can still operate. The first is that while a black hole of mass above this threshold will not readily emit an electron mass, it really means the quantum probability is reduced. So a black hole that has a Planck unit of mass and a Planck unit of charge, thus being extremal, has zero temperature classically, but quantum mechanically things are a bit different. The exact quantum mechanical state is probably some form of condensate, which has a finite small temperature, not entirely zero, and this means there is some quantum probability this Planck mass/charge black hole could erupt due to a quantum emission that is not spontaneous. Remember Einstein's coefficients, where radiation fields have a spontaneous part and a superradiant or stimulated emission part. If excited atoms are sending out photons with a wavelength longer than the mutual spacing between these atoms then it is statistically probable that photons are emitted in exactly the same state, such as polarization etc. The Hawking radiation emitted by a very near extremal black hole will have a very long wavelength as $T~\rightarrow~0$, which means the emitted radiation near the horizon has a long wavelength relative to Planck units of horizon area. This means that stimulated emission of radiation is possible. This would give rise to superradiance physics.

For an observer far from a black hole the Hawking radiation appears thermal and random. However, for an observer very close to the horizon there is a torrent of out flowing radiation. The time dilation effect means an observer, or some probe very close to the horizon, Hawking radiation emitted over a long period as measured in the far exterior occurs in a much shorter period of time. This may be seen in how the accelerated frame transforms a vacuum into a distribution of particles. We would then have for an extremal or very near extremal black hole the case where very feeble radiation would be observed by a distant observer, while the accelerated observer close to the horizon might observe a stimulated emission effect. The near horizon observer would observe highly correlated radiation, while the distant observer might not, as photons would appear with different timing data. These photons are entangled in time, but the observation of different timing by the distant observer would demolish this entanglement. So the radiation may appear random. A special type of experimental system would be needed to insure against this, similar that in https://arxiv.org/abs/1209.4191. This though does suggest that even an extremal black hole will emit radiation, though it may do so very slowly. Further, a complete extremal condition may not be possible, just as the third law of thermodynamics prevent zero temperature. A Bose-Einstein condensate is as close as one can get to absolute zero for $N$ bosons.

The stretched horizon of a black hole is a 2-space plus time manifold. Two dimensional spaces are spaces that can be both symplectic and Riemannian. A 1-form $\omega = \omega_idx^i$ defines $\Omega = d\omega$ so that $\Omega_{ij} = \partial_i\omega_j- \partial_j\omega_i$ with $\Omega_{ij} = -\Omega_{ji}$. The 2 dimensional space can also be a Riemannian geometry. The symplectic manifold as a pseudo-complex manifold is $\mathbb CP^1 \sim \mathbb S^2$. The stretched horizon of a black hole is such a manifold.

Quantum coherent states are a classical-like subset of states in Hilbert space. They have symplectic structure. This is then an idea way to think of the stretched horizon, which has quantum modes. These of course transition into states away from the black hole. These modes are in effect “frozen gravitons” with their two polarization degrees of freedom given by the $(\theta, \phi)$ for the metric of the 2-sphere. All the other modes are red shifted away and the degrees of freedom that compose a black hole, at least from the perspective of a distant exterior observer are gravitational harmonic oscillators. This means the black-brane or D2-brane of the black hole is quantum mechanical! In M-theory these D-branes and dual NS-branes are treated as classical. However, this classical behavior for the black brane or D2-brane may really be a theory of coherent states.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.