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Black holes would presumably evaporate in the long future via Hawking radiation.

However is this inevitable? Or are there any mechanisms that would compensate the lost mass due to Hawking radiation avoiding the evaporation of certain black holes?

It was recently observed that apparenlty black holes increase their mass as the universe expands (https://newatlas.com/physics/dark-energy-black-holes-accelerate-expansion-universe/ ; https://www.space.com/black-holes-expanding-with-universe). However, these results are still preliminary. So I was looking for something more grounded.

Perhaps if a black hole had a long standing and stable accretion disk with a low infall rate all the mass lost could be compensated by the disk radiation and matter falling towards the black hole (assuming that the universe had no CMB radiation that would also infall to the black hole, altering this equilibrium)?

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    $\begingroup$ I frequently wonder if very tiny black holes would be stable due to quantum effects requiring a minimum of energy that gets evaporated. $\endgroup$ Commented Apr 7 at 20:13
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    $\begingroup$ @GyroGearloose Very tiny BHs emit the most energy of any BH, so that shouldn't be a problem. They evaporate in a bang, not a whimper. $\endgroup$ Commented Apr 7 at 21:11
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    $\begingroup$ @LawnmowerMan as far as I am told, there is a thorough lack of theory about that, as gravity and quantum mechanics are no good friends to each other, in the theories we have just now. $\endgroup$ Commented Apr 7 at 21:17
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    $\begingroup$ fun fact: the Hawking radiation is not limited to the black holes as it does not emerge from the black hole itself. Any massive object is eligible. $\endgroup$
    – fraxinus
    Commented Apr 8 at 7:56
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    $\begingroup$ Any BH is stable until the microwave background becomes colder than the BHs surface temperature. Currently, that's any BH heavier than our moon (en.wikipedia.org/wiki/Hawking_radiation). Stellar mass BH have temperatures on the nanokelvin scale, supermassive BH are even colder by the same factor as they are heavier. The universe has a tremendous amount of expanding left to do in order to reduce the background radiation far enough that those become unstable. $\endgroup$ Commented Apr 8 at 15:42

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It's probably inevitable. However, it's not really possible to make any completely definitive statements about Hawking radiation because it's a quantum process involving general relativity, and we don't yet have a consistent theory of quantum gravity. But let's assume that Hawking's derivation is basically correct...

The Hawking radiation temperature of a black hole is inversely proportional to its mass. Even stellar mass BHs are very cold, so they radiate very slowly, and the supermassive BHs at the hearts of most galaxies are even colder.

According to Viktor Toth's Hawking radiation calculator, a 5 solar mass BH, which has a Schwarzschild radius of ~14.770 km, has a Hawking temperature of 1.23374e-8 K and an expected lifetime of 1.44968e69 years.

However, the cosmic microwave background (CMB) radiation is currently 2.73 K, which is over 200 million times hotter than the 5 solar mass BH. So in the present era, black holes are absorbing more energy than they radiate. The CMB cools as the universe expands, but it will take around 1e29 years before the CMB is cooler than even the smallest BHs that can be formed via core-collapse supernovae. And of course those BHs will continue to grow and get colder in the intervening years.

It's currently around 13.7 billion years since the Big Bang, but that's less than an eyeblink compared to 1e29 years. And that's less than an eyeblink compared to 1.4e69 years. So even though black holes won't start losing mass until the universe is more than a billion billion times its current age, the evaporation time is so long that that delay is effectively negligible. ;)

By the time the BH evaporation era begins there won't be much else left in the universe, except maybe a few neutron stars and black dwarfs that haven't yet fallen into a black hole. There definitely won't be any remaining accretion disks. There may still be some stray matter floating around, but if so, it will be extremely diffuse.

Wikipedia has an interesting timeline of the far future, but it includes some speculative things, like proton decay, and it's not completely self-consistent.

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    $\begingroup$ Black holes are likely to exist in populated areas of space since they are the product of star collapses and therefore likely to be where other stars exist. Which means that they'll absorb, over time, all surrounding matter (parts of the galaxies they reside in) and finally fall into the central black holes (in cosmological time scales, presumably, the energy loss through gravitational waves suffices for that). This means they'll grow enormously and cool likely faster than the universe for a long time. $\endgroup$ Commented Apr 8 at 5:53
  • $\begingroup$ Extremely cool video on this (general) topic: youtube.com/watch?v=uD4izuDMUQA $\endgroup$
    – AnoE
    Commented Apr 9 at 14:33
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Extremal black holes have a Hawking temperature of zero and an infinite lifetime in theory.

The lifetime of large black holes is long enough that it's plausible that the universe will end in another way before they evaporate completely.

I think those are the only two options.

Perhaps if a black hole had a long standing and stable accretion disk with a low infall rate all the mass lost could be compensated by the disk radiation and matter falling towards the black hole

If you have an external source of matter to stave off the evaporation, you are better off tossing it in as soon as possible since it will reduce the evaporation rate.

If you're imagining a closed system where the radiation is captured and falls back in, that won't work because of the second law of thermodynamics. In the long term everything near the hole will end up at a uniform temperature and radiating outward at that temperature, and it will be the Hawking temperature.

It was recently observed that apparently black holes increase their mass as the universe expands

That's very unlikely to be true. I wrote this answer and this answer and this answer about it.

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    $\begingroup$ Extreme Kerr loses energy and angular momentum via superradiance (IIRC the lifetime of extreme Kerr is about double of Schwarzschild bh with the same mass). Extreme RN would lose charge via Schwinger pair production, the lifetime of a very large RN bh (with any geometrically significant charge) would be exponentially larger than that of Schw. bh, but still finite $\endgroup$
    – A.V.S.
    Commented Apr 7 at 19:02
  • $\begingroup$ @benrg "If you're imagining a closed system where the radiation is captured and falls back in, that won't work because of the second law of thermodynamics. In the long term everything near the hole will end up at a uniform temperature and radiating outward at that temperature, and it will be the Hawking temperature" but if the black hole evaporates through hawking radiation and then that radiation returns to the black hole, shouldn't that count as a net zero loss for the black hole? $\endgroup$
    – vengaq
    Commented Apr 12 at 10:19
  • $\begingroup$ @A.V.S. then, there are no extremal black holes with an indefinitely long stable lifetime? I know Wikipedia isn't the best source, but here (en.wikipedia.org/wiki/Extremal_black_hole) it says that extremal black holes do not emit hawking radiation and are stable. Is it wrong then? $\endgroup$
    – vengaq
    Commented Apr 12 at 10:28
  • $\begingroup$ @vengaq The claim in Wikipedia refers to extremal black holes in supersymmetric theories specifically. Our Universe is not supersymmetric. Plus it is impossible (from thermodynamics viewpoint) to achieve perfect extremality. $\endgroup$
    – A.V.S.
    Commented Apr 12 at 15:01
  • $\begingroup$ But in this question it appears that they do not radiate in a universe like ours as well, sulersymmetry is not mentioned ( physics.stackexchange.com/questions/356955/…). Also why is it impossible according to thermodynamics? Because they would inevitably loss energy in some way? Would this happen if they were completely isolated (so for instance superradiance could be avoided)? @A.V.S. $\endgroup$
    – vengaq
    Commented Apr 12 at 22:22
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This is not realizable (easily) I think in our Univese:

If you have a (either a sufficiently large) blackhole on a (or a sufficiently small) compact manifold for your universe, then the hawking radiation it emits should in some critical time $T$ be expected to collide again and be re-absorbed by the blackhole. As long as the rate of re-absorption is greater than or equal to the rate of evaporation this would be an eternally stable blackhole.

If I had a better command on the mathematics of GR I would've liked to describe this. I don't think you even need to know any QFT just, you can assume the blackhole emits photons according to the blackbody spectrum uniformly for all directions on its surface, so this becomes a pure Geometry + probability type of problem.

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  • $\begingroup$ Would a de Sitter universe with its horizon allow for this? @SidharthGhoshal $\endgroup$
    – vengaq
    Commented Apr 12 at 9:59
  • $\begingroup$ Tbh I have no idea, my GR is poor. I looked up De Sitter space in wikipedia: I noticed in the "definition" section: en.wikipedia.org/wiki/De_Sitter_space#Definition at the very LAST line it is said $dS_n \cong \mathbb{R} \times S^{n-1}$. That copy of $\mathbb{R}$ certainly results in a manifold that is NOT bounded so without understanding more I would probably say no (my argument being: the hawking radiation which is emitted parallel to the copy of $\mathbb{R}$ is never going to end up back inside your black hole again) $\endgroup$ Commented Apr 12 at 14:19
  • $\begingroup$ I might in light of this have to edit my "compact condition" to quite literally "closed and bounded". $\endgroup$ Commented Apr 12 at 14:22
  • $\begingroup$ @SidharthGoshal mmmh but in dS spacetimes in accelerating universes the cosmological horizon woul emit radiation similarly to Hawking radiation in black holes, and because the Poincarré recurrence theorem then eventually a black hole should be formed $\endgroup$
    – vengaq
    Commented Apr 12 at 21:41
  • $\begingroup$ Completely unrelated to your response: is that copy of $\mathbb{R}$ in DeSitter space supposed to be identified with time? $\endgroup$ Commented Apr 13 at 6:54
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According to current theories, all black holes must evaporate over time, as the laws of physics as we currently know them really won't let anything last truly forever. That said, it is possible to have black holes that evaporate so slowly that on human timescales, it appears they don't evaporate at all.

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