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

Question

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.

Answer 1

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.

Answer 2

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!

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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.

Answer 3

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.)

This answer is based on an artificial mix of two different theories, classical general relativity and quantum phyiscs, 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.