# What are the final particles emitted from an evaporating black hole?

Hawking radiation predicts that black holes can slowly evaporate through the effective emission of a particle. This particle is a real particle, as in, it is not a black hole itself. I'll write this (a bit tongue-in-cheek) as follows, with $A$ being the emitted particle, and black hole prime being the black hole with slightly reduced mass.

$$\mbox{black hole} \rightarrow A + \mbox{black hole}'$$

There is every reason to think this continues until the black hole stops being a black hole. So we are left with a radiated particle and something else. We don't know much about this process, but I think we can still limit it to a 2-product decay.

$$\mbox{black hole} \rightarrow A + B$$

What could the something else, $B$, be? I don't know much about it, but I know that it is:

1. Not a black hole
2. A physically plausible particle

It might be more realistic to ask what other conditions should we impose on B? I realize this is probably an unsolved problem. Also, won't $A$ be really highly energetic? How energetic? Actually, what is $A$ to begin with?

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This is definitely in the domain of string theory, and it will depend on the compactification details. If you look at model charged black holes you have a better chance of getting an objective answer. Black holes in string theory are continuous with matter--- there is no sharp separation--- the highly excited matter is a thermal black hole, the cold world-sheet is matter. They all decay according to recognizable black-hole like vertex-operator emissions. – Ron Maimon Nov 14 '11 at 9:54

Most of the time while the black hole is evaporating, the "A" particles will be photons (or other mass-less particles like gravitons). The reason for this is that the black hole emits radiation as if it were a black body and the temperature of the black body is inversely proportional to the black hole mass. For black holes with mass equal to the mass of the sun the temperature would be 60 nano-kelvins. So the energy of the particles emitted would have to be extremely low which is why they need to be mass-less particles like photons or gravitons.

Now in our universe (today), if the mass of the black hole is greater than about the mass of the moon, it would actually gain more energy from the cosmic microwave background radiation than it would lose to Hawking radiation since the black body temperature of black hole with a mass greater than the moon would be lower than the CMB black body temperature of 2.7 K. The CMB temperature decreases over time so eventually all black holes will evaporate but it will take an extremely long time for this to happen.

However as the black hole gets smaller and smaller, the black body temperature rises so that eventually the temperature would rise so high that massive particles could be emitted. For a given amount of energy being emitted any particle that has a rest mass below that energy could be the emitted particle. The reason for this is that the Hawking radiation is a result of the vacuum virtual pair creation in the vicinity of the event horizon. Since all particles participate in the vacuum virtual pair creation, any particle that is compatible with that black body energy spectrum will be emitted.

When the mass of the black hole approaches zero, the black body temperature will approach infinity so the evaporation will be faster and faster with more and more energetic particles. So the last two particles that would be emitted would be any 2 particles compatible with the remaining mass of the micro-black hole.

So if, for example, a micro-black hole could be created at the LHC it would evaporate very rapidly into an isotropic spray of particles of all types compatible with the energy of the black hole. That would be the event signature that would be found by the LHC detectors.

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There is no exact model of the endpoint of black hole evaporation. However, I am persuaded that Samir Mathur's "fuzzball" model of black holes is the right one. In a fuzzball, you don't have a singularity or even an event horizon, because the fuzzball physically extends to where the horizon would be found in classical general relativity. When Mathur constructs individual bound states corresponding to the extremal black holes found in string theory, they have this feature.

So black holes are fuzzballs, and fuzzballs are bound states of strings and branes, and fuzzball evaporation results from these bound objects occasionally escaping, not from pair production at a nonexistent event horizon. The endpoint of fuzzball evaporation will be a fuzzball so small it is just an ordinary bound state. At this point FrankH's answer will apply: the black hole will have disintegrated into a spray of ordinary particles. But my point is that we apparently need the fuzzball description of black holes, which is still work in progress, in order to really understand what's going on.

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The Hawking radiation from a black hole and the Unruh effect where an accelerating observer detects black body radiation are intimately related. The "fuzzball" idea doesn't apply to Unruh radiation, does it? In my opinion multiple models can explain the same physics and virtual particles at an even horizon or fuzzballs may just be two equivalent models. – FrankH Mar 5 '12 at 21:28

Our world is a quantum world so everything, including any stage of the Hawking evaporation, may only be predicted probabilistically. At the end of its life, a tiny black hole has a mass comparable to the Planck mass and such a tiny black hole becomes qualitatively the same thing as just another heavy unstable species of an elementary particle.

A question is whether or not you know the precise microstate of such a black hole. If you know it, you may predict the probabilities of different final products accurately from a well-defined compactification of string/M-theory you consider (without string/M-theory, you will clearly be able to make no precise predictions of the quantum gravity phenomena, and this is a textbook example of one). If you don't know the exact microstate is, it is still true that roughly speaking, the small black hole emits a thermal radiation.

However, at the very end of the life of the black hole, its temperature goes up a lot. At the very end, the temperature is close to the Planck temperature (the highest possible temperature that may marginally be talked about in physics) so the decay products may include (with a high probability) very heavy particle species, too. Right before it disappears, a black hole may surely produce a pair of top quarks or even heavier particles. There's still a nonzero probability that it will decay to two photons or anything else that doesn't violate conservation laws.

Actually, the probability is nonzero that a black hole emits another, smaller black hole. It's just very unlikely: such a process is essentially suppressed by $\exp(-S)$ where $S$ is the entropy of the emitted black hole.

For macroscopic black holes, such a factor is zero for all practical purposes. However, if you want to emit black holes that are slightly larger than the minimal (Planckian) black hole, the factor isn't hopelessly tiny and the emission of a small black hole is a possibility. Again, any black hole microstate may always be interpreted as yet another species of an elementary particle. Large black holes have a high entropy and the description in terms of "exponentially many new particle species" becomes contrived. However, for the smallest (Planckian) black holes, the description in terms of new particle species becomes a condition for any accurate description of the black hole's behavior.

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To be completely clear, all black holes are over the plank mass and all elementary particles are under the plank mass? – Alan Rominger Nov 14 '11 at 12:51
You can say it this way. But again, this is a convention. For particles of mass comparable to the Planck mass, both particle physics of the kind we know from light particles as well as effects of gravity we know from black holes are important. There is no qualitative sharp separation of these objects to elementary particles and black hole microstates. The transition is continuous. The transitional regime near the Planck mass is the only regime in which a detailed model of quantum gravity is really needed to make predictions: lighter objects are just QFT and heavier ones are given by GR. – Luboš Motl Nov 14 '11 at 13:17
"There's still a nonzero probability that it will decay to two photons or anything else that doesn't violate conservation laws." It's important to note that these conservation laws don't include all the conservation laws that are normally believed to hold in particle physics. For instance, baryon number is not conserved. There is a nice discussion of this in Wald, p. 413. – Ben Crowell Nov 14 '11 at 16:54