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Hawking theorized the evaporation of black holes. Photons are pictured to slowly radiate their mass away. The higher the mass the longer it takes (due to smaller tidal forces).

Which makes one question, how can the inside matter, composed of normal matter (so no anti-matter), be transformed into photons only? Doesn't this contradict all conservation laws in particle physics? Are there reactions in particle physics that transform all involved particles into photons, except matter anti-matter reactions?

Put differently, how can quarks and electrons annihilate? They both posses the same amount of plus and minus charge, but I'm not aware of processes in which they are destroyed, mutually or separately. Not even by the negative energy wave falling in from the horizon, which destroys mass but not the charges.

Most black holes form from stellar matter, excluding neutrinos, or collapsing neutron stars, containing up and down quarks only (in the ratio 1:2). If we consider the neutron black hole, you would think that their masses and charges can only be annihilated by negative energy anti up and down quarks. But what happens to the negative energy up and down quarks?

To put it differently one last time. Say we have a hole made up out of equal amounts electrons, protons, neutrons, and neutrinos, the same particles present shortly after the big bang. If the hole is evaporated, are they gone? If so, does that mean that particles can be annihilated without their anti-matter counterpart?

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    $\begingroup$ See the information paradox. $\endgroup$
    – MrQ
    Feb 24 at 10:23
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    $\begingroup$ Nothing escapes from inside the event horizon. Hawking radiation is emitted from outside the event horizon. Nothing inside the horizon can affect anything outside the horizon. See physics.stackexchange.com/a/252236/123208 $\endgroup$
    – PM 2Ring
    Feb 24 at 11:53
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    $\begingroup$ Why should you need antimatter to make photons? Hawking radiation isn't due to annihilation; it's related to the Unruh effect. $\endgroup$
    – J.G.
    Feb 24 at 13:10
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    $\begingroup$ I suppose it comes down to whether black holes conserve baryon number, lepton number etc., which this says they don't, while admitting that claim needs a citation. $\endgroup$
    – J.G.
    Feb 24 at 13:24
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    $\begingroup$ @Felicia see the answer by user566 here physics.stackexchange.com/questions/7290/… $\endgroup$
    – anna v
    Feb 24 at 13:54

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Hawking radiation is not only in the form of photons. Fermions, scalars, other spin-1 particles, and even gravitons also contribute (for example, see https://www.sciencedirect.com/science/article/pii/S0370269311001559). So electrons and positrons are also emitted via Hawking radiation.

The origin of photons emitted during Hawking radiation is not an annihilation process between particles and anti-particles. Rather, the existence of the horizon changes the structure of the vacuum state experienced by an observer far away from the black hole (relative to what the vacuum would look like without a horizon), such that the observer sees a thermal bath of particles. The driving factor is the horizon, not interactions between particles. Exactly how this happens is difficult to explain without math.

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    – Buzz
    Feb 25 at 2:34
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how can the inside matter, composed of normal matter (so no anti-matter), be transformed into photons only?

Firstly, I need to make some disclaimers. We don't have a fully-working quantum gravity (QG) theory, so we don't know exactly what happens to matter when it reaches the core of a black hole. And we can't be certain that Hawking radiation is real without a proper QG theory: Hawking's calculations involve a semi-classical approximation that "bolts on" quantum corrections to the purely classical GR equations, and we need QG to justify that procedure.

However, what goes on inside a black hole isn't actually relevant to the rest of the universe. As I mentioned in a comment, the matter and energy inside the event horizon (EH) cannot affect the outside universe in any way. All events on or inside the EH are in the future light-cone of the observer, and of course events must be in the past light-cone to have an effect on the present.

As I said here, the gravitational field of a black hole is sometimes described as a "fossil field". All matter & energy falling into the black hole modifies the spacetime curvature as it approaches the event horizon. And once it crosses the event horizon it can no longer change the spacetime curvature outside the horizon, so those curvature changes are preserved (until something else comes along to add its own curvature changes).

The same reasoning applies to the electromagnetic charge of the infalling matter and its effect on the electromagnetic field in the vicinity of the BH. It does not apply to the strong or weak nuclear "charges" because those interactions have finite range; only gravitation and electromagnetism are preserved due to their infinite range. This gives rise to the no-hair theorem:

The no-hair theorem states that all black hole solutions of the Einstein–Maxwell equations of gravitation and electromagnetism in general relativity can be completely characterized by only three externally observable classical parameters: mass, electric charge, and angular momentum. — Misner, Charles W.; Thorne, Kip S.; Wheeler, John Archibald (1973). Gravitation

The result was quickly generalized to the cases of charged or spinning black holes. There is still no rigorous mathematical proof of a general no-hair theorem, and mathematicians refer to it as the no-hair conjecture.

So when particles fall towards a BH they "pump" energy into the spacetime curvature and EM field of the BH. We don't know or care what happens to those particles once they cross the EH because in our frame that's always in the future.

It's not easy to get an intuitive picture of what happens in and around a BH. Our intuitions aren't very good at dealing with curved spacetime. ;) But here's an analogous situation in flat spacetime that may be helpful.

Imagine we set off an H-bomb at location X at noon on Tuesday. If we try to measure the energy of the blast on the previous day, we won't measure anything, no matter how close we get to X, how big the bomb is, or how sensitive our instruments are. The blast energy simply doesn't travel backwards in time.

Similarly, energy from events inside the EH would have to travel backwards in time to affect events outside the EH.

I should mention that the EH is observer-dependent. The "official" EH at the Schwarzschild radius is the horizon of the Schwarzschild observer. That observer is the limiting case of an observer free-falling towards the BH with zero velocity, so they're at an infinite distance. The horizon for any observer free-falling towards a BH is always below them, until they hit the singularity. This relativity of horizon location means that different observers will measure different numbers of particles in the vicinity of the BH, and is ultimately what gives rise to Hawking radiation, as John Rennie explains here. The mathematics used to perform the Hawking radiation calculations involves the Bogoliubov transformation, which is a bit above my paygrade. ;)

In summary, the Hawking radiation is produced from the energy stored in the gravitational & electromagnetic field around the BH, and the types of particles that originally caused those stresses and strains in those fields is irrelevant.

Now, according to the Bekenstein bound, the entropy of a BH is proptional to the surface area of the EH. So the information regarding the particles that formed the BH isn't exactly lost, but it's not exactly accessible either. As far as I know, Hawking radiation is supposed to be perfectly thermal, so we can't decode that information from the Hawking radiation.


BTW, you may enjoy playing with the Hawking radiation calculator. It makes a few approximations, but it's generally pretty good.

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  • $\begingroup$ "the matter and energy inside the event horizon (EH) cannot affect the outside universe in any way." What is it that produces the gravitational field around the black hole (which seems to affect the outside universe)? $\endgroup$
    – robjohn
    Feb 25 at 9:38
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    $\begingroup$ @robjohn Ghosts. ;) The spacetime curvature outside the EH was produced by matter before it crossed the EH. In the frame of a distant observer, it takes infinite time for anything to reach the horizon. So for such observers the black hole is still in the process of forming. Bear in mind that the Schwarzschild solution (which is a vacuum solution) is only an approximate model to a real astrophysical black hole, and real black holes tend to have a lot of spin, which makes things more complicated. $\endgroup$
    – PM 2Ring
    Feb 25 at 10:15
  • $\begingroup$ "We don't know or care what happens to those particles once they cross the EH because in our frame that's always in the future." Imagine though you fall in. The fall to the center takes little proper time (about the Schwarzschild radius divided by c). Before reaching the center, the hole could be evaporated. There will be an empty space again and you will be gone, while a faraway observer sees it all happening over a large time. Will you reappear again in the future? I mean, the particles you are made of and that fall in? $\endgroup$
    – Felicia
    Feb 25 at 20:52
  • $\begingroup$ @Felicia I don't think so, because almost all Hawking radiation must be photons, only a tiny fraction of the BH energy gets emitted as particles with non-zero mass. Electrons & positrons don't get produced much until the mass drops below ~4E-17 solar masses. $\endgroup$
    – PM 2Ring
    Feb 25 at 21:36
  • $\begingroup$ If the location of the EH is observer-dependent, does it truly prevent infomation escaping? I could imagine a scenario where an infalling observer who is still outside the classical EH observes an event inside it - because said event is outside of where they observe the EH to be - and then re-transmits information about that event to the outside world. $\endgroup$
    – Skyler
    Feb 25 at 21:53
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Black-holes are peculiar but nevertheless bona-fide quantum states of a given quantum field theory and therefore are expected to respect conservation laws: charge conservation, in particular. In a Einstein-Maxwell theory, if you build a black hole by throwing in only charged particles, it would have to radiate this charge away. A variation of this reasoning was used to come up with the "weak gravity conjecture", which roughly states that gravity is the weakest force. More precisely, it states that in a physical theory with charged particles coupled to gravity, you need to a have at least one particle with "more charge than mass" (in suitable units), for otherwise charged black-holes would be able to radiate all the charge away before fully evaporating.

Other symmetries of the standard model like Baryon number, Lepton number (which relate to matter/anti-matter asymmetry) are believed to be only approximate. But in grand unified theories, B-L (baryon minus lepton number) is usually taken to be conserved: this would constitute a global symmetry. And indeed, in quantum gravity, it is also believed that no global symmetry should exist, by similar sort of black-hole evaporation arguments. A rigorous demonstration of this has been given in certain space-times with boundary conditions such that quantum gravity is tractable in there (with negative cosmological constant), see Harlow-Ooguri. Quantum gravitational effects should therefore arise to break the B-L global symmetry at the Planck scale (or before).

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    $\begingroup$ Classical black holes obey charge conservation, though, without the need to invoke quantum mechanics at all. The no-hair theorem has charge right there as one of the valid bulk parameters. That's very different from things like baryon number, which would involve distinguishing a black hole that formed from stellar collapse from one that formed due to geons or some sort of exotic collapse of radiation or whatever. $\endgroup$ Feb 24 at 16:46
  • $\begingroup$ A hole is formed out of matter. All this matter is gone after the evaporation. On the horizon, virtual particles turn real. Particle-antiparticle pairs (matter as well as gauge particles) radiate away and their negative energy parts fall to the inside. Can The positron negative energy states can annihilate all electrons, their mass as well as their charge. But what happens to negative energy electron states? Can they annihilate up-quarks (+2/3), or maybe two positrons (+2) annihilating three ups (also +2)? $\endgroup$
    – Felicia
    Feb 24 at 17:26
  • $\begingroup$ @Felicia if you read the answers and links, baryon number, lepton number are not conserved quantities within the horizon, at the present understanding of the theories. $\endgroup$
    – anna v
    Feb 25 at 5:16
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We do not know

We have never seen a black hole up close. The black holes we are reasonably certain exist, are very far away.

Hawking radiation is a theoretical feature of very small black holes. The black holes we think we have seen are very big. Their Hawking radiation levels would be undetectable, even if we were up close.

Meanwhile, we have theories. The problem is that we have many such theories and they do not fit together. In particular, general relativity and quantum mechanics aren't on speaking terms at all.

General relativity is our best theory for strong gravity. Quantum mechanics is our best theory for elemental particle behavior. Both are important to black holes and Hawking radiation.

As long as we do not have a working theory uniting the two, we cannot say we understand black holes at all. Anything we say are educated guesses at best.

The following guesses doesn't fit together:

  1. Black holes radiate, and will eventually lose all their mass to radiation.
  2. This radiation does not reflect the particles used to make the black hole.
  3. The total number of baryons and leptons are conserved.

As you are saying, these guesses can't all be true simultaneously.

picop's answer states that some people believe number 3 is wrong, baryon and lepton conservation might not be absolute.

My personal less educated guess is that number 2 is wrong, Hawking radiation will reflect the particles used to create it. Everything is conserved. My reasoning is from the viewpoint of any in-falling particle, the black hole will evaporate before the particle reaches the singularity --- meaning the singularity will never form in the first place.

Another alternative is the "baby universe" theory. Here the mass collected by a black hole will blow up as a Big Bang in a new universe. One could argue that the baryons and leptons lost to our universe are conserved in the new universe.

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    $\begingroup$ Here are two difficulties for making your guess work. (1) Assuming we can trust general relativity outside the event horizon, no one outside the black hole can measure the baryon or lepton number of a black hole, by the no-hair theorem. Quantum gravity would need to hide a lot of information inside the BH. (2) In GR, matter crossing the event horizon of a solar mass black hole will hit the singularity in order microseconds, while the timescale to evaporate is much larger than the current age of the Universe. It is a big ask for a theory of quantum gravity to reconcile these two timescales. $\endgroup$
    – Andrew
    Feb 25 at 13:29
  • $\begingroup$ @Andrew “matter crossing the event horizon of a solar mass black hole will hit the singularity in order microseconds, while the timescale to evaporate is much larger than the current age of the Universe” - Your objection is incorrect. In the coordinates of a free falling observer, the black hole evaporates before he crosses the horizon. And in the coordinates, in which a black hole takes longer than the age of the universe to evaporate, a falling matter never crosses the horizon. There is nothing to reconcile here for QG, because GR already reconciles these timescales with no contradiction.. $\endgroup$
    – safesphere
    Mar 7 at 10:41
  • $\begingroup$ @safesphere You're probably right that the numerical values I used are not directly comparable. I'm not convinced the point isn't still valid, though. Do you have a reference for this claim? In the coordinates of a free falling observer, the black hole evaporates before he crosses the horizon. $\endgroup$
    – Andrew
    Mar 7 at 12:59
  • $\begingroup$ @Andrew your objection is correct. The phrase “the coordinates of a free falling observer” is ambiguous. It doesn’t have a standard meaning, and if it did it would probably be something like Gullstrand Painleve coordinates, where your statement is true. Regardless of the coordinates used, there are worldlines that reach the singularity after a finite proper time. Such world lines will not escape if the black hole evaporates $\endgroup$
    – Dale
    Mar 8 at 17:48
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    $\begingroup$ @Dale Thanks. That is what I thought. Safesphere's comment made me wonder if there was something I hadn't understood about what BH evaporation looks like from the perspective of an infalling observer. $\endgroup$
    – Andrew
    Mar 8 at 17:50

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