# Extremal black hole with no angular momentum and no electric charge

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

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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 at 23:44
Siva, existing derivations of Hawking temperature of Kerr-Newman and Reissner-Nordström black holes disagree with your statement (that extremal black holes have entropy), as all of them state that the temperature decreases as they approach the extremality boundary. The reason why i state that the black hole has zero temperature is because the initial entropy of the black hole is zero or nearly zero, hence the temperature must be zero too. (at least initially, but is not clear why or how will the black hole regain enough entropy so that the temperature matches the radius) – lurscher Mar 21 at 0:55
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 at 16:41
those holographic degrees of freedom look quite fluffy! – lurscher Mar 27 at 16:49
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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?

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 Here is a link to the same reference on the preprint arXiv. – Qmechanic♦ Mar 20 at 23:36

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.

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 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 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 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 at 6:00 The way I phrase the information loss problem is as a conversion of an initial pure state into a final mixed state, which is at odds with unitarity. It seems that this is exactly the situation you were interested in, is it not? – Daniel Grumiller Mar 22 at 11:49 Regarding the area-law in Einstein gravity: you do not need to assume "stringy microstates" to show that black holes have an entropy, you need them only if you want a microscopic understanding of black hole entropy. If you are happy with a macroscopic description of black hole entropy the four laws of black hole mechanics, which can be proven in Einstein gravity given some natural assumptions, are sufficient to show that entropy is proportional to the area, and Hawking's derivation fixes the proportionality factor to 1/4 (in Planck units). – Daniel Grumiller Mar 22 at 11:59
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