Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

A black hole will have a temperature that is a function of the mass, the angular momentum and the electric charge. For a fixed mass, Angular momentum and electric charge are bounded by the extremality condition

$$M^2 - a^2 - Q^2 \gt 0$$

Exactly at the extremality boundary, both entropy and temperature are zero.

Suppose i create a black hole with a spherically symmetric, incoming wavefront of electromagnetic radiation in a pure quantum state (that is, the density matrix satisfies the property $\rho^2 = \rho$). The wavefront is shaped in a way such that the whole energy of the packet will be inside the Schwarzschild radius, which will form an event horizon.

Since the wavepacket is as nearly as pure as it is physically possible to create, the quantum (Von Neumann) entropy is zero or nearly zero. But the formation of the black hole does not create nor destroy entropy, so the black hole must contain zero or nearly zero entropy as well. So the black hole seems to be extremal (it has zero temperature) but it nonetheless does not have any angular momentum (it is formed from a wavefront with zero net polarization over the whole sphere) and it does not have any charge (electromagnetic radiation is neutral).

Question: what "hair" does have a black hole formed from such a pure state, so that it can be extremal and still do not have angular momentum or electric charge (which are the classical hair that we come to expect from classical general relativity)

This question is a mutation of this question, but while that specific question tries to look what input black hole states create specific output (Hawking) radiation states that are far from thermal from a statistical point of view, this question is specific about extremality that is unrelated to angular momentum and charge

share|improve this question
Nice question +1 –  Dilaton Dec 27 '12 at 0:31
It's not clear why the black hole you create in your process has no temperature. Also, extremal black holes do have entropy, unlike what you say in the question. –  Siva Mar 20 '13 at 23:44
Afaik, the temperature goes to zero as a black hole approaches extremality, but not the entropy (which is propotional to the horizon area, a la Bekenstein-Hawking). If not, what happens to the validity of Bek-Hawk area formula for entropy? Entropy at zero temp would depend on the number of ground states (and I feel there ought to be more than one). I could be wrong; I would appreciate it if you can give a reference. –  Siva Mar 21 '13 at 3:22
There are suggestions that the microstates of a black hole look like this; at least this is how I personally visualize the fuzzballs conjecture of Samir D. Mathur. –  Dilaton Mar 27 '13 at 16:41
those holographic degrees of freedom look quite fluffy! –  lurscher Mar 27 '13 at 16:49
show 5 more comments

3 Answers

Since we want its energy density to reside strictly inside the Schwarzschild radius, the question then reduces to "can we construct arbitrarily localized photon pure states"?

This reference suggests that constructing a spatially localized photon state by applying a momentum space shaping factor:

$$|\phi \rangle = \sum_\lambda\int{}\frac{d^3k}{(2\pi)^3}f_\lambda({\bf{k}})a_\lambda^{\dagger}({\bf{k}})|0\rangle $$ results in a configuration space wavepacket that can't have compact support.

This suggests that, even if a horizon were to form, there would be some residual component of the original photon wavefunction outside the horizon.

I think this only partially answers your question in that it suggests that perhaps the direction to look in is - what is the maximum energy density that can be constructed using these spatially localized photon states?

share|improve this answer
Here is a link to the same reference on the preprint arXiv. –  Qmechanic Mar 20 '13 at 23:36
add comment

As pointed out in the comments it is not true that extremal black holes have no entropy. They do have an entropy given by the area of the event horizon (at least in simple theories like Einstein gravity; otherwise use Wald's entropy or generalizations thereof).

The conversion of an initial pure state into a final thermal state that you mentioned in one of your comments is the famous information loss problem.

The information loss problem most likely is solved in the same way as it is solved in condensed matter systems: information is lost for practical purposes, but not in principle.

For comparison, imagine you shine with a pure laser beam on your hand. So the initial state is your hand and the laser beam. For simplicity let me put your hand at zero temperature and assume it is also a pure state initially. The final state will be an approximately thermal state, namely your hand at some finite temperature, which will then radiate away approximately thermal radiation. So as in the black hole case you have an information loss problem - a conversion of a pure initial state into a thermal final state.

The information loss problem is resolved if you place detectors around your hand and measure the outgoing radiation at arbitrary precision and for arbitrary long time. You will find that the spectrum is not exactly thermal, and that the deviations from thermality allow you, in principle, to reconstruct the initial state.

If you do not believe that black holes are profoundly different in this respect then also in black holes information should be lost only for practical purposes but not in principle. So by observing the outgoing "Hawking"-radiation you should be able to reconstruct the initial state. (I put "Hawking" under quotation marks since real Hawking radiation is exactly thermal.)

The entropy of the black hole (extremal or not) then arises because there are many different microstates that correspond to the same macrostate.

share|improve this answer
You are repeating the area-entropy law as if it were a kind of axiom. What reason do i have to believe that available microstates grow with charge, other than to keep the area-entropy relationship intact? In quantum mechanics, Von Neumann entropy grows with decoherence that produces effective mixed states. Pure states have zero entropy. If you see the definition of Wald entropy, is defined in such a way as precisely the area-entropy law holds. Given that the Wald entropy does not have any other motivation, i find unsatisfactory to assume that it will work just because 'stringy' microstates –  lurscher Mar 22 '13 at 5:54
Also, i'm not uncomfortable with the way you extrapolate that this is the same as the traditional information loss problem. For one, in that problem, you are wondering what happens to non-zero entropy matter when it falls in the black hole, and how that increases the area. This is subtly different, we are wondering what happens when i add almost-zero entropy matter-energy. Neither me or you have any compelling reason to assume that a bunch of coherent light toward self-collapse will decohere in the right amount such as to satisfy the area-entropy law gods. –  lurscher Mar 22 '13 at 5:59
I will -1 but i'm open to revert it if you address the shortcomings of this answer that i'm highlighting –  lurscher Mar 22 '13 at 6:00
Maybe your confusion is the following one: are you claiming that starting with one given microstate the final macrostate that you observe should have zero entropy? This would be like considering the microstate corresponding to all coordinates and momenta of an ideal gas and claiming that the associated macrostate has zero entropy, since you know precisely the mircostate. But there are many other microstates that lead to the same macrostate. Of course, if you have arbitrarily precise information entropy is always zero, but in a coarse grained description it is not. –  Daniel Grumiller Mar 22 '13 at 12:05
The same is true for the example with the laser beam and the hand that I gave. The pure microstate has zero entropy; the point is that there are many other microstates that lead to the same macrostate (defined by the mass, angular momentum and charges of the black hole or in the hand/laser example by the temperature of the hand). If my point is still unclear to you then consider to contact me in private, so that we do not have to spam the comment section. –  Daniel Grumiller Apr 1 '13 at 22:14
show 4 more comments

The debate that seems to be happening right now is what does it mean that '... the formation of the black hole does not create nor destroy entropy, so the black hole must contain zero or nearly zero entropy as well.' This is correct, of course, except that the material we observe with zero or nearly zero entropy is 'Bose-Einstein condensate' (BEC) and BEC brings properties to the event horizon (such as non-compressibility), and indeed to the process of gravitational collapse, that have not previously factored into black-hole theory or yet been addressed.

Pawel O. Mazur and Emil Mottola's seminal paper have tried to address this, but their results are somewhat controversial. [Gravitational Condensate Stars: An Alternative to Black Holes] They found that as a star collapses, in-falling matter sheds its entropy becoming Bose-Einstein condensate (BEC) which imposes constraints on what can and cannot happen at the even-horizon. Some of these constraint exclude necessary conditions for the formation of traditional black-holes (such as the formation of a singularity, infinite space-time curvature, etc.). Their object, though not quite a 'black-hole', will look, act, and feel very similar (tis said), but will instead be a gravitational vacuum condensate star with event-horizon (consisting of BEC), but no singularity. Actually the interior is described to be a segment of de Sitter space.

This solves Hawking's Black-Hole information paradox because rather than having in-falling matter transforming into a pure quantum states completely independent of Hawking-Radiation destroying information about the original quantum state, it is suggested that instead what happens is that all in-falling matter (protons, neutrons, electrons, etc...) transforms instead into a quantum state known as “super-atom” (coherence). No information is lost since Hawking-Radiation is a product of this transformation, not independent of it.

The appeal of this theory is that it provides a much clearer understanding of behaviour at the event-horizon boundary and solves many stability problems. Some of its consequences are also testable [Hawking radiation in a two-component Bose-Einstein condensate (BEC). P.-É. Larré and N. Pavloff]. Furthermore, this theory solves Hawking's 'Black-Hole Information paradox' by establishing the basis for thermodynamic stability. The Gravastar is theorized to have very low amounts of entropy, in contrast with black holes which apparently have a billion times more entropy than the dying star that formed it.

This theory, as original posed, had sight problems with it, but Matt Visser and David Wiltshire were able to solve them by posing a slight variation [Stable Gravastars — an alternative to black holes?] which also provide an alternate explanation for gamma ray bursts. Even with all of this, there is criticism. The problem that still remains is about the creation of a Gravastar; is a collapsing star capable of shedding enough entropy upon implosion to cause a change in quantum state to 'super-atom'? If "yes", Gravastars and Black-holes would appear the same observationally, and produce similar signatures. (Actually I question this since BECs exhibit the ability to significantly slow the speed of light to a crawl. The only question I've asked is whether or not this is completely true. "Are Black-holes and Gravastars observationally identical? No one has yet answered.)

With respect to your question about 'hair', if this theory holds, can such a cold condensate body, a super-atom, indeed have angular momentum or electric charge? In fact the Gravastar is very close to the ideas of Kerr as rotating Black Hole yet solves the 'hair problem' by positing that the super-atom would modulo quantum fluctuations (Kerr/Hairless hybrid?). Can there be magnetic fields in de Sitter space? Pawel Jan Morawiec says 'Yes, no problem!'. He argues that in the Gravastar model, the non-vanishing magnetic field could be present in de Sitter space (by studying massless Dirac fields as an example of a matter field in the de Sitter spacetime in the vicinity of an event horizon), this postulated to be related to the Josephson effect [Physical and geometrical aspects of de sitter interior of a Gravastar].

It's a fascinating theory, but we'll likely have to wait for the jury to weigh in ...

See also:

[Cosmological milestones and Gravastars — topics in General Relativity, Céline Cattën (supervised by Matt Visser)]

[Scattering of atoms on a Bose-Einstein condensate, Uffe V. Poulsen, Klaus Molmer] - where an atomic wavepacket seemingly leaves the condensate before its arrives

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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