# Are elementary particles ultimate fate of black holes?

From the "no hair theorem" we know that black holes have only 3 characteristic external observables, mass, electric charge and angular momentum (except the possible exceptions in the higher dimensional theories). These make them very similar to elementary particles. One question naively comes to mind. Is it possible that elementary particles are ultimate nuggets of the final stages of black holes after emitting all the Hawking radiation it could?

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This is indeed a tempting suggestion (see also this paper). However, there is a crucial difference between elementary particles and macroscopic black holes: the latter are described, to a good approximation, by non-quantum (aka classical) physics, while elementary particles are described by quantum physics. The reason for this is simple.

If the classical radius of an object is larger than its Compton wavelength, then a classical description is sufficient. For black holes whose Schwarzschild radius is bigger than the Planck length this is fulfilled. However, for elementary particles this is not fulfilled (e.g. for an electron the "radius" would refer to the classical electron radius, which is about $10^{-13}$cm, whereas its Compton wavelength is about three orders of magnitude larger).

Near the Planck scale your intuition is probably correct, and there is no fundamental difference between black holes and elementary particles - both could be described by certain string excitations.

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Yes, black holes are special kinds of elementary particles. That's how they have to be represented in every consistent quantum theory of gravity. This representation of a black hole becomes especially useful and important for small black holes - whose mass is not much larger than the Planck mass.

And indeed, a black hole evaporates, which is just a form of a decay of a heavy elementary particle, and when it becomes very light, at the end of the Hawking evaporation process, it is literally indistinguishable from a heavy elementary particle that ultimately decays into a few stable elementary particles.

However, a difference that you seem to neglect is that black holes actually carry a large entropy $$S = \frac{A}{4A_0} k_B$$ where $A$ is the area of the black hole's event horizon and $A_0$ is the Planck area $A_0=\hbar G / c^3$. The constant $k_B$ is Boltzmann's constant. This means that there actually exists a huge number of microstates $$N = \exp(S / k_B )$$ and a single black hole, with a fixed value of mass, charges, and spin, is just a macroscopic description of the ensemble of $N$ "microstates". In reality, the black hole carries a huge information - the world distinguishes which of the $N$ microstates is actually present.

It is these "microstates" that are really analogous to types of elementary particles. But the number of particle species that macroscopically look like the black hole of given mass, charges, and spin is not one: instead, it is huge, approximately $N$.

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All this is fine and good. Black holes may behave like elementary particles in string theory, but can the electron, for example, be the last stage of the dissipation of a black hole? –  anna v Feb 27 '11 at 18:07
@anna v I understand what @Luboš Motl said as implying that there is no difference between particle "remnants" and particles emitted as Hawking radiation in the final stage of black hole decay. The planckian black hole just decays into a bunch of particles, probably including some electrons in the final state. –  mmc Feb 28 '11 at 2:39
Dear @anna, yes, the electron is pretty likely to be the last "remainder" of a black hole. Before it becomes an electron, the black hole may be a W boson that decays into an electron and a neutrino. Before it was a W boson, it could have been a much more excited string state. Note that I didn't use the term "string theory" in my answer. There is a confusion in between the lines of your comment: you try to pretend that the answer in string theory and the answer in the real world are 2 things: but they're the same thing. String theory is the real world. –  Luboš Motl Feb 28 '11 at 8:18
"String Theory is the real world". This statement needs justification to those of us who would expect the statement "String Theory is a scientific theory of the real world". There is a large gap between these statements that would require several Stack philosophy questions to get to clarification, I suspect. –  Roy Simpson Feb 28 '11 at 11:21
Fine, Roy. What I meant is that "the answer to the question how XY behaves in the real world" is the same thing as "the answer to the question how XY behaves according to the best theory describing XY et al. that we have," and in the case of the character of black hole microstates, it's string theory. However, the conclusion that black hole microstates are continuously connected to (other) elementary particle species - there's no qualitative difference - doesn't really depend on specific features of string theory. It's a general fact that strings just confirm and make more specific. –  Luboš Motl Mar 1 '11 at 15:20

The short answer is no. Have a look at the wikipedia article on dissipation of black holes.

quote: Unlike most objects, a black hole's temperature increases as it radiates away mass. The rate of temperature increase is exponential, with the most likely endpoint being the dissolution of the black hole in a violent burst of gamma rays.

The possibility of micro black holes from extra dimensions in some string models still has them dissolving thermodynamically into elementary particles as soon as they are formed.

Edit: Herein I have been replying to the question stated clearly in the last sentence: Is it possible that elementary particles are ultimate nuggets of the final stages of black holes after emitting all the Hawking radiation it could? Not to the different question that people seem to be replying to: "are black holes like elementary particles."

A yes answer to the latter, does not reply to the former, i.e. whether quarks and leptons are the nugget, what is left over, from a black hole. A yes answer to this last would offer the intriguing model of the snake eating its tail, maybe quite probable in some new more encompassing theory, but not foreseen now, at least from the answers given. If after shedding innumerable quarks leptons and photons and entropy on the way, a black hole ends up as an electron (for example) in an identifiable quantum mechanical history.By this last I mean something similar to a decay chain in nuclear cascades.

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the negative specific heat that you quote does not, however, apply to all black holes; it applies to Schwarzschild or Kerr black holes in 4D asymptotically flat spacetime (which astrophysically are the most relevant ones), but not, e.g., to black holes in AdS (at least on the right side of the Hawking-Page phase transition). Also, I am not sure why the concept of temperature would be relevant for the question? –  Daniel Grumiller Feb 27 '11 at 13:39
I am replying to the direct question of whether the elementary particles we know, quarks, leptons I presume can be the end result of dissipating black holes. It is interesting that at the planck scale black holes might have attributes like elementary particles but we are not at that scale? –  anna v Feb 27 '11 at 13:57
No body knows for sure the final stages of black hole evaporation. There may be "likely" scenarios but hardly any mathematically rigorous answer. The final fate of a black hole is certainly an open question. BTW, I am also curious about what temperature has to do with the question? –  user1355 Feb 27 '11 at 14:19
@ sb1 The quote says that due to exponentially increasing temperature the end of the black hole dissipation will be an explosion into gamma rays. If the temperature is not increasing, no explosion? Now the answer by Daniel Grumiller covers the question of how small a black hole can be, imo. Black holes are a classical object. They are too large to be quantum mechanically coherent as one object, imo. I might be wrong, and string theory might pull a rabbit out of the hat, but not with what we know now. –  anna v Feb 27 '11 at 14:40
Black holes don't have any fixed size. At the final stages they are certainly very small to be treated as quantum object. In fact one must apply quantum theory to understand the final stages. –  user1355 Feb 27 '11 at 14:53

The other answers here are fine. Another point that should be stated is that if you believe in the naïve values of the angular momentum, charge and mass of most elementary particles, and plugged them into the Kerr-Nordstrom solution, you would find that almost every (and probably all of them, I just haven't checked) elementary particle would be a naked singularity, not a black hole--the charge and angular momentum of these objects would be too large to support a horizon.

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Why am I bound to use classical theories? If we can describe black holes in terms of a correct quantum gravity theory, the necessary modifications might help to avoid the naked singularity. Whether the suggestion is right or wrong I don't know. What I do know is outright rejection of the idea on the basis of classical GR is not helpful. –  user1355 Feb 27 '11 at 14:14
@sb1: so, you're using the no hair theorem, which is a classical GR theorem to develop intuition for a conclusion, and then rejecting an argument based upon classical GR? It's far more likely that what quantum gravity will do is eliminate singularities than it is that it will change the horizon structure so that you can super-spin and super-charge mini-black holes. Also, what do you suppose we get an intuition from, if not classical GR? Last I checked there was no coherent quantum theory of gravity. –  Jerry Schirmer Feb 27 '11 at 15:50
What's so wrong about using a low energy effective classical theory like GR to build intuition about the nature of black holes. In physics it is always valid to hold certain principles (like N.H.T.) valid in a more general theory, in this case QG. The last time I checked there was an yet to be experimentally confirmed but very sensible quantum theory of gravity called string theory. People are using it to calculate many features regarding black holes. –  user1355 Feb 27 '11 at 16:31
@sb1: I have no problem with that--my problem is selectively using knowledge of General Relativity. There is no model of a black hole with an angular momentum an order of magnitude larger than its mass. So why call the thing a black hole at all? –  Jerry Schirmer Feb 27 '11 at 19:39

The entropy of a black hole is a measure of the number of microstates, where for $N$ degenerate microstates the entropy is $S~=~k~log(N)$, which is associated with gravity. The entropy for large $N$ is determined by the area of the event horizon $S~=~kA/4L_p^2$, where for the Schwarzschild black hole $A~=~16\pi M^2$. The black hole is a system which holds a set of states with energy $E~=~M$ in a degeneracy $g(E)~=~exp(4\pi E^2)$ and the partition function is $$Z(\beta)~=~\sum_E e^{4\pi E^2}e^{-\beta E}.$$ This partition function is divergent for $E~\rightarrow~\infty$. The statistics for the number of degenerate microstates for a black hole is unbounded, and thus the partition function diverges. Black hole entropy is a coarse graining microstate states, which has been accomplished in string theory for large $N$. The horizon area is a summation of these quantum numbers $$A~=~16\pi\alpha_p\sum_{i=1}^Nn_i,$$ for $\alpha_p$ a Planck area. The quantum numbers $n_i$ determine an element of the horizon area. The energy is then counted as $E_n~=~\alpha E_p\sqrt{n}$, for $n~=~\sum_{i=1}^Nn_i$

The degeneracy for $E_n$ is the number of ways $n~>~0$ is a sum of $N$ or less positive integers $n_i$. It is the cardinality of the set of elements $\{n_1,~n_2,~,\dots,~n_m\}$, such that $n~=~\sum_{i=1}^m n_i$ for $1~\le~m~<~N$. The number of ways a positive integer $m$ may be written as a sum of $m$ positive integers is the same problem as computing the number of ways of arranging $n$ balls in $m$ cells in a row. The result is a degeneracy for the energy $E_n$ $$g(E_n)~=~\sum_{m~=~1}^N\left(\matrix{n~-~1\cr m~-~1}\right),$$ for $N~\le~n$. We also have that $m~\le~n$, which cuts the degeneracy further in $$g’(E_n)~=~\sum_{m~=~1}^n\left(\matrix{n~-~1\cr m~-~1}\right).$$ The partition function is a summation of the two degenerate sets, $$Z(\beta)~=~\sum_{n=1}^N\sum{m=1}^n\left(\matrix{n~-~1\cr m~-~1}\right)e^{-E_p\alpha\sqrt{n}}~+~\sum_{n=M+1}^\infty\sum_{m=1}^N\left(\matrix{n~-~1\cr m~-~1}\right)e^{-E_p\alpha\sqrt{n}}.$$ The two portions of the partition functions play a role at $n$ small and $n~>>~N$, and may be computed independently. The convergence occurs for $n~>>N$ with $$Z(\beta)~\simeq~\sum_{n~=~N+1}^\infty(n~-~1)^{N-1}e^{-\beta E_p\alpha\sqrt{n}}$$ This is a convergent partition function. Conversely for a low black hole temperature $n~<<~N$, the degeneracy from the binomial theorem is $g’(E_n)~\simeq~2^{n-1}$ and the black hole entropy is $S~=~k~ln(2^{n-1})$ $=~(n~-~1)ln2$. The area $A~=~16\pi\alpha^2n$ permits us to set $\alpha~=~{1\over 2}\sqrt{{ln2}\over\pi}$.

The above calculation can be looked at according to strings. By holography the horizon is covered by strings which define all the quantum information which entered the black hole. A generating function for stringy density of states computes a function which is similar to the above, and in the holographic setting describes the black hole as a string sphere on a stretched horizon. This part is a bit involved, so I will jump forwards to say that a black hole may be thought of as a statistical state or phase of strings.

These quantum numbers associated with Planck units of area of the event horizon. This is in keeping with the naturalized units of G = [Area]. These quantum numbers can include a range of quantitites, particular mass, angular momentum and electric charge. The horizon exists as a radius

r_\pm~=~m~\pm~sqrt{m^2~-~Q^2~-~J^2}

which corresponds to the outer an inner horizons. With the term in the square root is zero the two horizons meet and the spacelike region between them is “squashed” into an $AdS_2\times S^2$. This is an extremal black hole, which has zero Hawking temperature. In general these charges can be supersymmetric, or supercharges. In the extremal case these charges are at the BPS bound. In this case all the quantum numbers associated with those unit areas define an object which is similar to an elementary particle.

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use "\sim" for $\sim$ –  user346 Feb 28 '11 at 1:22