# When we are talking about black hole evaporation - what exactly happens?

According to Wikipedia:

Hawking radiation reduces the mass and energy of black holes and is therefore also known as black hole evaporation. Because of this, black holes that do not gain mass through other means are expected to shrink and ultimately vanish.

My question is about this shrinking and vanishing part. My (school level at best - I supposed now they teach this stuff better) understanding is that a black hole in order to be a black hole needs to be extremely dense or extremely massive. Does this mean that from some point in time black hole just will cease to be a black hole because it won't be massive enough?

The question suggested as duplicate - An explanation of Hawking Radiation - is actually about the physical nature of the Hawking radiation itself, so, though related, it's still different.

• Possible duplicate of An explanation of Hawking Radiation – PM 2Ring Jul 20 '19 at 14:20
• @PM2Ring I've updated my question with explanation why it's not a duplicate. – shabunc Jul 20 '19 at 14:24
• Oh, ok. It's still worthwhile reading John's answer there; you're unlikely to find a better explanation of Hawking radiation for lay readers. – PM 2Ring Jul 20 '19 at 14:30
• FWIW, there is no theoretical lower mass limit for a black hole, but we don't know of any processes with enough energy & pressure to crush a small amount of matter to smaller than its Schwarzschild radius, except possibly during the early phases of the Big Bang. – PM 2Ring Jul 20 '19 at 14:55
• – Qmechanic Jul 20 '19 at 16:44

Does this mean that from some point in time black hole just will cease to be a black hole because it won't be massive enough?

No, once a black hole forms there's no turning back. It can lose mass via Hawking radiation, but (as far as we know) it cannot stop being a black hole until there's nothing left. There's no theoretical lower mass limit for a black hole. There is a possibility that right near the very end of the evaporation process that some quantum effect creates a stable remnant, but we need a proper theory of Quantum Gravity (which unites General Relativity with Quantum theory) to answer questions like that, and we don't yet have such a theory.

As the Wikipedia article explains, Hawking radiation is a very slow process for black holes with the mass of a typical star, and it's very cold, around a billionth of a degree above absolute zero. So it's very difficult to observe, even if you were close to the black hole. The evaporation rate gets faster and the temperature increases as the mass of the black hole gets smaller, but currently the universe is too warm for an isolated stellar black hole to lose mass: it gains far more energy from the Cosmic Microwave Background (CMB) radiation than what it emits as Hawking radiation.

Hawking radiation is a process that's always there when you have an event horizon. With black holes, the strength of this radiation is a function of its size: The heavier the black hole, and thus the bigger the event horizon, the colder the Hawking radiation.

While the strength of the Hawking radiation approaches zero as you go to larger black holes, it never actually becomes zero. So, in a sense, black holes are never truly black. They always radiate a bit, and they always slowly loose weight due to that radiation.

So, if you isolate a black hole from any incoming radiation, it will slowly shrink, and by shrinking it will become brighter, so it will shrink more rapidly in a self-amplifying process. This self-amplification is so strong, that any sufficiently small black hole looses all its mass within a finite time.

Wikipedia says:

So, for instance, a 1-second-life black hole has a mass of $$2.28×10^5kg$$, equivalent to an energy of $$2.05×10^{22}J$$ that could be released by $$5×10^6$$ megatons of TNT. The initial power is $$6.84×10^{21}W$$.

You see, a 300 ton heavy black hole is not black at all. Saying that it's white-hot is a severe understatement. It's so extremely bright that you just see a huge explosion that far exceeds the destructive power of all the worlds nuclear warheads taken together... And all this radiation is coming out of an object of subatomic size!

So, yes, black holes cease to be black as they shrink. Their Hawking radiation gives them the appearance of a perfectly black, more or less hot object. Big black holes are cooler than the cosmic microwave background, appearing as black as we can imagine. But smaller black holes glow with Hawking radiation. As the black hole shrinks, this glow goes all the way from a dim, reddish glow, over bright white light, brutally bright ultraviolet and deadly intensive X-rays to the destructive brightness of a nuclear warhead.

But all the time, it's just the Hawking radiation that you see. The singularity (or whatever happens to be within a black hole) remains shrouded behind the event horizon until the black hole has lost all its mass.

• Well, it's either possible to light to escape the black whole it's not, there's either an event horizon or there isn't. Are you claiming that at some point there isn't event horizon? Otherwise it's still a black hole. – shabunc Jul 20 '19 at 18:12
• @shabunc The Hawking radiation comes from just outside the event horizon. Nothing has to actually cross over from inside the event horizon. As I said in this answer on Astronomy, 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. – PM 2Ring Jul 20 '19 at 18:22
• @shabunc I suppose that wording can be a bit confusing. But the tiny black hole is still technically a black hole, with a proper event horizon. It's just that a lot of radiation is being emitted in the vicinity just outside the event horizon. – PM 2Ring Jul 20 '19 at 18:29
• @shabunc While the black hole itself, and the singularity within (or whatever there is, we really don't know) remains shrouded behind the black veil of the event horizon, you cannot describe a small black hole without its Hawking radiation. It's just an integral part of what a black hole is. That hawking radiation can be seen, and it definitely gives the black hole a non-black appearance to an outside observer. – cmaster - reinstate monica Jul 20 '19 at 18:36
• @shabunc Yes, the black hole remains a black hole all the way down to its full evaporation (which looks like a violent explosion). It does not change its nature when it grows tiny, but it sure does change how it looks like (from pitch black, over red glow, to bright white, etc.). I find it important to point this out, because it's so contrary to the very name of the black hole, and the way people think about black holes. As long as someone thinks black holes look black, well, they have not really understood black holes. – cmaster - reinstate monica Jul 20 '19 at 19:25

First, if we ignore quantum effects like Hawking radiation, then there would not be any limit to how small a black hole can be. Classical general relativity allows black-hole solutions with arbitrarily small mass $$M>0$$, and the corresponding Schwarzschild radius (for a non-rotating black hole, which is the simplest case) is $$R=2GM/c^2$$. If we take $$M$$ to be the mass of the earth, then $$R$$ comes out to be roughly one centimeter. If we take $$M$$ to be the mass of a large mountain, then $$R$$ comes out to be less than the radius of an atom (but more than the radius of a proton). Even though it's tiny, it's still a black hole — at least if we ignore quantum effects like Hawking radiation.

Exactly how quantum effects change this picture is not yet understood, so I don't think we can definitively say when an evaporating black hole ceases to be a black hole. However, we have good reason to think that classical general relativity will remain a good approximation to the spacetime geometry as long as the mass of the black hole is much larger than the Planck mass $$\sqrt{\hbar c/G}$$, which is a small fraction of a milligram. In particular, we have good reason to be confident that an evaporating black hole that starts with a typical stellar mass (or larger) will still be a black hole after it shrinks to earth-mass proportions, and presumably even after it shrinks to mountain-mass (subatomic) proportions.

(Note that this would take much, much longer than the current age of the universe, and even that's only if the black hole is radiating more than it's consuming, which is not likely in a universe filled with cosmic background radiation.)

This answer is based on an artificial mix of two different theories, classical general relativity and quantum physics, that we don't quite know how to combine yet. We have good reason to think that at some point, where both general-relativistic and quantum effects have competing magnitudes, the classical concept of spacetime will somehow break down. This must at least happen near the "singularity" that classical general relativity predicts inside a black hole, and for the entirety of any black hole that is not much larger than the Planck mass. Exactly what happens under those conditions is not yet known. However, as long as we only consider situations that are not that extreme, basing answers on the "artificial mix of two different theories" is a reasonable thing to do. Reasonable doesn't necessarily mean correct... just reasonable.