I don't understand how, as a black hole gets smaller and smaller from the excretion of Hawking radiation, it retains its ability to capture photons. I imagine there would be a point in its life cycle where its mass/gravity just isn't enough for it to be able to do so and its body could be revealed, and perhaps gain back some of its previous volume from the lack of gravity being able to hold it together as tightly?

I have no formal education in physics yet.

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    $\begingroup$ If you bend a flexible rod and connect its ends, it would always remain a circle, even if it shrinks. $\endgroup$
    – safesphere
    Commented Nov 2, 2022 at 4:45
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    $\begingroup$ I don't see the point of closing this as duplicate now that this has attracted lots of well-received answers, but a very similar question is here, with also a few answers: Will a black hole eventually turn into a neutron star? $\endgroup$
    – kutschkem
    Commented Nov 3, 2022 at 13:25
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    $\begingroup$ Several comments removed. To provide “a simple answer,” please post an answer, not a comment. $\endgroup$
    – rob
    Commented Nov 3, 2022 at 16:48
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    $\begingroup$ Only point to consider is the necessary timespan involved. A "stellar black hole" would emit hawking radiation at a rate that even pessimistic time estimates make it last ~10^12 years. And one point to think about - a black hole is an encapsulated region of space time. If you use the often used picture of space time being a rubber foil with dents from masses, then a black hole is a bubble formed out of an area of said foil, but twisted shut. Hence we don't know if there is still an object inside. $\endgroup$
    – eagle275
    Commented Nov 7, 2022 at 9:17
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    $\begingroup$ @lxrose Most matter falling into a BH doesn't fall straight in. It first gets swept up into an [accretion disk]( en.wikipedia.org/wiki/Accretion_disk) outside the event horizon. The spewed matter in that space.com article is coming from the accretion disk, not from inside the event horizon. $\endgroup$
    – PM 2Ring
    Commented Nov 15, 2022 at 7:37

5 Answers 5


After writing this answer, I noticed there are a couple alternative explanations that might be interesting to mention, so I'll add them as well.

Explanation 1

What makes something into a black hole isn't exactly how much mass it has, but also how compactified it is. In principle, any amount of mass can form a black hole, as long as you compactify it enough.

The size needed for some amount of mass to form a black hole is know as the Schwarzschild radius. Roughly speaking, if you pick an amount of mass and manage to compress it below the Schwarzschild radius, you'll have a black hole. It is given by a simple expression. Namely, $$R_S = \frac{2 G M}{c^2},$$ where $M$ is the mass, $c$ is the speed of light, and $G$ is Newton's gravitational constant (which sort of measures how intense gravity is). For example, for something with the mass of the Earth, the Schwarzschild radius is roughly $0.88$ cm, while for the Sun it is about $2.9$ km (I must admit I didn't double check the computation, I'm trusting Google on these numbers, but they are pretty much what I remember).

Hence, the black hole stays a black hole while it evaporates because it is shrinking while it is losing mass, and always shrinking enough so that it is always at the correct size.

Explanation 2

The second way of thinking is a bit less familiar. It turns out that black holes aren't really objects, but rather regions in spacetime. In fact, this is so true that black holes are what we call vacuum solutions: there isn't matter anywhere in the spacetime. All of the mass of the black hole is there due to effects of gravity itself. Another way of thinking it is that a black hole is so collapsed that its mass is entirely due to gravitational energy.

It is a bit harder to grasp this concept, but once you get it, the rest is simpler. The black hole stays there because it isn't "made" of anything. There isn't a star just below the event horizon waiting to come out. There is nothing there, but gravity. As it loses mass, gravity weakens and it gets smaller, but there isn't anything behind the horizon to come out.

Edit: the question "What do we mean when we say that black holes aren't made of anything?" later asked for a more technical discussion of parts of this explanation. I suggest checking it out.

Explanation 3

The third explanation might be a bit simpler than the second. Once something falls into a black hole, that's it. It can't come out. Ever. By the very definition of what a black hole is. Hence, as the hole shrinks, there is no way something could come out of the hole to be its "body". That would violate the very meaning of what is a black hole.

This is a simplified answer. Since OP doesn't have formal education in Physics, I might have overlooked a few details and nuances in here, but I did my best to keep the answer as faithful as possible.

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    $\begingroup$ I came from the question that you linked in question 2. I have to ask, what do you mean exactly with explanation 2? When a black hole forms due to the collapse of a stellar mass, the mass doesn't just disappear once the event horizon forms. Your explanation might be interpreted as the mass not being inside the black hole. A Schwarzschild black hole could have no mass inside but those would not be found in a real universe. Could you elaborate? $\endgroup$ Commented Nov 16, 2022 at 22:57
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    $\begingroup$ @AccidentalTaylorExpansion Since OP doesn't have a formal education in Physics, I simplified things a bit on explanation 2. It applies for an eternal black hole, which hasn't been formed by stellar collapse (and is an extremely good approximation for other cases as well). Strictly speaking, an external observer will never see a black hole form by stellar collapse (the star matter will take infinitely long to reach the event horizon, and hence it never truly becomes a black hole, strictly speaking), so OP's idea of the body of the black hole coming out won't happen because it hasn't formed yet $\endgroup$ Commented Nov 16, 2022 at 23:05
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    $\begingroup$ As I remarked at the bottom of the answer, I overlooked some nuances. This is the main one. It is worth adding that an external observer will still observe Hawking radiation, even though the observer never sees the actual black hole forming. In fact, Hawking's original derivation was carried out for this sort of spacetime, and only a bit later were the calculations for an eternal Schwarzschild black hole carried out . $\endgroup$ Commented Nov 16, 2022 at 23:06
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    $\begingroup$ @WilliamMartens It is a simplification, but it is accurate in the sense that $G$ is what couples spacetime curvature to matter. $G$ is a constant, but it also has some physical meaning. We can understand it as a measure of the strength of gravity in the sense that if $G$ was smaller, matter would curve spacetime less. This is a trick often useful to check whether an equation is correct. Note, for example, that in the limit $G \to 0$ one gets that the Schwarzschild radius vanishes, because the matter is not able to affect spacetime so much. We can't, of course, change the value of $G$, but + $\endgroup$ Commented Jan 7, 2023 at 18:48
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    $\begingroup$ these arguments are often helpful in fundamental Physics as a manner of understanding the meaning of our equations. Similarly, even though $c$ is a constant, it is often useful to check the limit $c \to \infty$ in Relativity to see whether we recover the Newtonian expressions, which are valid when all involved velocities are much smaller than $c$. As so on. $\endgroup$ Commented Jan 7, 2023 at 18:50

This is not a competing answer but rather a supplementary one.

You say "capture photons" as if gravity is so strong that photons cannot escape, like the pop science idea that "escape velocity is faster than light speed." To me, this is the wrong picture. General relativity tells us that what we call "gravity" is not a force; rather, the presence of mass and energy changes the rules of geometry in the regions near them in ways that make the trajectories of objects curve. You don't fall towards Earth because of a force pulling. You fall because paths are bent in such a way that they progress towards the Earth.

You can think of it like space and time are a big sheet of graph paper, which in empty space is flat and with lines at 90°, but near a mass like the Earth are curved. They curve more and more the closer you get to Earth, but before they diverge too much from being straight, you would hit Earth's surface. Now, imagine you crammed all the matter in the Earth into a much smaller space at its center, so the lines could continue curving. At some point they would curve so much, the geometry would change so much, that outward is no longer a possible direction. You could not move further away from the center by changing direction or adding speed, any more than right now you could travel backwards in time by traveling in a certain direction or getting to a certain speed (not even 88 mph). "Backwards in time" is simply not an available direction.

A black hole is simply an object whose mass is compressed so much that the curvature can get close enough to the center to reach this critical stage (Schwartzchild radius) where outward is no longer possible. But there is no minimum size. Earth could become a black hole if it were compressed as described to under 1 cm. The reason in reality we do not observe "small" black holes is the only known process capable of compressing matter to those extremes, is the collapse of stars. And this can only happen with a certain minimum mass.

(It is speculated that cosmic ray particles crashing into the atmosphere at extremely high energies may create black holes with the mass of a few atoms, which would Hawking evaporate in nanoseconds. But this has not been proven.)

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    $\begingroup$ Thank you so much for your time and consideration to add to the previous comment and extend my understanding! $\endgroup$
    – levi shell
    Commented Nov 2, 2022 at 4:40
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    $\begingroup$ This should also give a minimum "size" for elementary particles which would constitute black holes themselves if they weren't fuzzy or spatially finite. $\endgroup$ Commented Nov 2, 2022 at 17:51
  • $\begingroup$ Maybe it's not so pop science problem? The warp field equation is known. Any way to calculate if that can escape a black hole or not, or is that just too difficult still? $\endgroup$
    – Joshua
    Commented Nov 2, 2022 at 18:27
  • $\begingroup$ @Joshua Pop-sci isn't the best description (indeed, the original idea of a black hole comes from before general relativity, and makes do with newtonian mechanics, where gravity is a force - of course, we know that's not accurate, but neither is GR, really). It's mostly that pop-sci has trouble getting into GR and QM properly. The Alcubierre metric is mathematically possible, but there's no indication it's actually possible in the real world (and I don't mean practicality). But the gist of it is that if you can escape a black hole by shaping spacetime, you can travel backwards in time too $\endgroup$
    – Luaan
    Commented Nov 4, 2022 at 7:02
  • $\begingroup$ One caveat re your 4th paragraph: Dust collapse is one other mode of BH formation, but it has been observed (a few years ago) only once, in Sagitarrius A. (As about half of all stars are in binary pairs, and nearly all have some rotation, the existence of more than 90 BH's has been observed indirectly, by the elliptical orbit maintained by the surviving star in such pairs.) $\endgroup$
    – Edouard
    Commented Nov 8, 2022 at 19:28

I take it, that this question is actually about how tiny black holes actually behave. That the aim is to get some kind of intuition about what happens. As such, this answer focuses on describing the last seconds of a black hole. The question gets answered, but the context is also illuminated.

Actually, the black hole does not stay black at all. It evaporates. The more mass it loses to Hawking radiation, the hotter the Hawking radiation becomes, and the faster it evaporates. This is a self-amplifying process that reaches infinite power within a finite time. In the last second of the life of the "black" hole, the Hawking radiation becomes so fierce that it carries away more energy than all the nuclear arsenals on this planet , all taken together, could deliver. Magnitudes of more energy! The radiation will consist of ever harder gamma rays, and at some points electrons/positrons, (anti-)protons, (anti-)neutrons, and other (anti-)particles will start being emitted with ever growing numbers. And all of this radiation is coming out of a tiny sphere comparable to the size of a proton!

As such, "capturing photons" is not really a description of the evaporating black hole anymore. It's a freaking source of photons. Nevertheless, this is all Hawking radiation, and does not tell you any more about the inside of the event horizon than any other Hawking photon does that is emitted in the earlier, cooler stages of evaporation.

And anyways, if something does not interact with a photon, the photon cannot tell you anything about it. So, even though the event horizon becomes too small to interact with incoming radiation appreciatively, it does not reveal anything about its innards. The event horizon still remains the impenetrable shroud that gobbles up anything of sizes comparable to the Schwarzchild radius of the black hole and smaller.

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    $\begingroup$ Note that all astrophysical black holes (that is, with masses comparable to stellar masses or greater) have Hawking temperatures colder than the cosmic microwave background, and are therefore net absorbers of radiation. No stellar-mass black hole has evaporated during the lifetime of our universe. $\endgroup$
    – rob
    Commented Nov 3, 2022 at 16:52
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    $\begingroup$ @rob True. But these are decidedly not in the category of microscopic black holes which shrink by emitting hawking radiation, and which are so tiny that they have trouble interacting with incoming light. As such, they are simply off-topic in the context of this question. $\endgroup$ Commented Nov 3, 2022 at 17:45
  • $\begingroup$ @cmaster-reinstatemonica I would want to argue that a black hole is black because that's it's very nature - it cannot be seen (that is, you can't actually "see" a black hole) Note; what you can see, however - is the effects the black hole do to it's "neighbors" so to say. that is, well - curving the way a ray/beam of light so it looks like a "water drop" or small "magnification" - this is what you can see, which is not(Directly) the black-hole, but(Indirectly I guess) It is; just to clarify what I mean by see - is with naked eye. Black holes can be detected trough other means $\endgroup$ Commented Nov 5, 2022 at 13:38
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    $\begingroup$ @WilliamMartens The shroud of the event horizon is so perfect that it does remain "black" in a sense; the Hawking radiation originates from its direct surrounding. However, a) because the event horizon is such a perfect shroud, all we can interact with is the lingering effects that infalling matters has left in the curved spacetime. For all intents and purposes, you can equate the BH with its horizon. And b) the Hawking radiation is caused by the presence of the event horizon. And in that sense, it is fair to say that the Hawking radiation is emitted by the BH. $\endgroup$ Commented Nov 5, 2022 at 21:02
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    $\begingroup$ @WilliamMartens As to the color of the BH's Hawking radiation, it is perfectly thermal. I.e. the BH does indeed behave like a perfectly black body with a temperature. For observable black holes, that temperature is way below the cosmic microwave background radiation. But for a BH so tiny that it can evaporate, the temperature goes up to infinity as its death draws nearer. As such, the maximum radiation intensity frequency sweeps through the entire spectrum, from microwaves over infrared, visible light, ultra violet, x-rays and gamma rays all the way to particle radiation. $\endgroup$ Commented Nov 5, 2022 at 21:09

As simplified answer to your question is this:

  • It is possible that they stay black holes because only large black holes can exist.

A longer discussion.

  • Remember that Physics is an experimental or observational science first of all. Theories that are not supported by experiments or observations may be interesting thought experiments at best.
  • We have never seen a "small" black hole and can not really know if they can, will or ever have existed. With small I here mean a black hole that loses more in mass over time than it gains (the very basis of your question).
  • What we have found out is that at small scale of stuff, the classic mechanics break down and are better described by other theories, often based on quantuum effects. The so called Schwartschild radius is built on classical mechanics and probably does not hold up for small things, and hence is probably not relevant for "small" black holes.
  • Basically a lot of our theories, makes assumptions and speculates around how small black holes could be created and how they would behave. History has shown that sometimes theories get it right, sometimes wrong. Until proven by experiments the only thing we can do with the theories is to consider them as brain games. You may want to learn the theories, but stay away from believing that they are reality, until experimentally shown to be at least possible.

Some other supplementary points to amplify other answers.

A black hole isn't really a hole in the sense we use on earth. Its a name, a label. So let's not get hung up on the name that we use. So I'll call it a BH here.

General relativity tells us that any matter or energy (same thing really for this) bends spacetime. Meaning it bends the geometry of space itself (among other things).

If you pack enough "stuff" in any space, it bends space enough that there is a region of space where anything whatsoever cannot avoid moving inwards. Literally all directions, every which way, are so twisted they * all * point inwards. There isn't any direction which lets you go outward, at all.

What this means is, anything whatsoever within that part of space, can only ever go more "into" the center of that part of space, never ever outward. We say that "all futures (the future for any object or energy) points inward".

Its in that sense it is like having some kind of "black hole". Its a part of space where no light can emerge from, and like an infinitely deep hole, whatever falls over the "edge" is lost from view forever. You can't ever get near enough the edge to see through inside that part of space, because light itself also falls inward.

General relativity gets complicated when you ask what counts as "stuff". Because matter and energy count as the same thing, enough energy in a small enough space, creates this BH effect as well. This is why some answers say that enough gravitational energy alone, can create and sustain a BH, with or without matter.

As a BH incredibly slowly loses energy via Hawkings radiation, the amount of "stuff" within it, shrinks. If it was spread out equally everywhere it might be a problem. But it isn't. Its moved to the center and so its now an infinite or near infinitely dense amount of stuff in an infinitely small part of the hidden off area of space. The boundary can shrink, and there's still always enough stuff at the center to maintain a BH effect, albeit for a smaller volume of space.

  • $\begingroup$ Uh, one request for clarification: In what sense is a black hole "not a hole in the sense that we use on earth"? What about the holes in, say, a block of swiss cheese? A lot of them seem to be as spherically-outlined as the black holes depicted in, say, the Nobel Prize Committee's graphics, put together at the time of its award to Roger Penrose for his work on BH's. $\endgroup$
    – Edouard
    Commented Dec 13, 2022 at 19:58

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