# Evaporating Black Holes?

## Summary:

Does Hawking radiation at the event horizon of a black hole result in the reduction of the mass of the black hole?
Does Hawking radiation at the event horizon of a black hole result in the reduction of the mass of the black hole?

Quantum mechanics requires the creation and destruction of particle-antiparticle pairs throughout space-time. When this effect occurs at the event horizon of a black hole, sometimes one of the newly created particles ends up falling into the black hole while the other particle moves away from the event horizon.* The accumulated particles that move away from the event horizon are called “Hawking radiation.”

When the pair-particle that falls into the black hole is an antiparticle, it is annihilated by combining in the black hole with what is (or was?) a particle. At the same time, its partner the particle moves away from the event horizon. The combined effects of this particle-antiparticle pair’s end-state is to reduce the mass of the black hole.

When the pair-particle that falls into the black hole is a particle, it adds to the mass of the black hole. Its partner the antiparticle moves away from the event horizon and is annihilated when it meets a particle. The combined effects of this particle-antiparticle pair’s end-state is to add to the mass of the black hole.

On the assumption that each of these scenarios is equally likely, would not the net effect of quantum mechanics pair-particle creation at the event horizon be neutral despite the existence of Hawking radiation? If this is true, black holes would not evaporate away due to Hawking radiation.

*On the assumption that the pair-particle that moves away from the event horizon requires an escape velocity to do so, this requires the event horizon itself to have a “thickness”. Consider the location of a quantum mechanics event that occurs at x-y-z-t, and that one pair-particle created at that location has sufficient velocity v to move away from the event horizon. Then consider a location x’-y’-z’-t that is slightly different from x-y-z-t and that one pair-particle created at that different location also has sufficient velocity v to move away from the event horizon. That means that there is a non-zero distance between x-y-z-t and z’-y’-z’-t that defines a non-zero thickness for the event horizon itself.

Ibix
2020 Award
Quantum mechanics requires the creation and destruction of particle-antiparticle pairs throughout space-time. When this effect occurs at the event horizon of a black hole, sometimes one of the newly created particles ends up falling into the black hole while the other particle moves away from the event horizon.*
This explanation is nonsense, by Hawking's own admission. He just couldn't come up with a non-mathematical description of what actually happens. It's also an effect near an event horizon, not at an event horizon, so there's no problem with particles escaping. And event horizons most definitely do not have a thickness, because they are boundaries between two regions.
On the assumption that each of these scenarios is equally likely, would not the net effect of quantum mechanics pair-particle creation at the event horizon be neutral despite the existence of Hawking radiation?
The model you are using is wrong, as noted as above. The actual maths remains beyond me, so I can't really offer a better explanation, although others here may do so. But Hawking radiation always leads to a reduction in mass of the black hole, since it isn't a case of "sometimes a negative energy particle goes in and sometimes it doesn't".

Note that Hawking radiation from stellar mass black holes is blackbody radiation below the temperature of the CMB, so stellar mass black holes are not currently decreasing in mass and will not start to do so for a long, long time. Microscopic black holes should have a much higher temperature and might well evaporate on short timescales.

berkeman
This is a discussion forum, not a game of whack-a-mole. Considerations about the subtleties of the English language aside, I have several questions. First, if the phenomenon occurs near but not at the event horizon, how is the phenomenon different from any other point in space where a quantum mechanic creation of a particle-antiparticle pair occurs? Also, even if the phenomenon occurs near the event horizon, could not one of the particle-antiparticle pairs be drawn into the black hole?

weirdoguy
berkeman
Mentor
This is a discussion forum, not a game of whack-a-mole. Considerations about the subtleties of the English language aside, I have several questions. First, if the phenomenon occurs near but not at the event horizon, how is the phenomenon different from any other point in space where a quantum mechanic creation of a particle-antiparticle pair occurs? Also, even if the phenomenon occurs near the event horizon, could not one of the particle-antiparticle pairs be drawn into the black hole?
Are you sure that you want your thread prefix to stay at "A" (Advanced = Graduate School / PhD level)? I can change it to "I" or "B" if that would help you to get more appropriate responses...

Are you sure that you want your thread prefix to stay at "A" (Advanced = Graduate School / PhD level)? I can change it to "I" or "B" if that would help you to get more appropriate responses...
More responses would be a big plus. Please move it down one notch, and thank you!

berkeman
berkeman
Mentor
More responses would be a big plus. Please move it down one notch, and thank you!
Thank you sir, I'll do that now.

Ibix
2020 Award
This is a discussion forum, not a game of whack-a-mole.
Not sure what you are attempting to communicate here. I addressed one error in your post about the event horizon having thickness (it does not) and attempted to address the problem with the rest (that it appears to be based on an inaccurate pop-sci model of Hawking radiation, albeit one promulgated by Hawking himself).
First, if the phenomenon occurs near but not at the event horizon, how is the phenomenon different from any other point in space where a quantum mechanic creation of a particle-antiparticle pair occurs?
Near the event horizon there's a nearby event horizon, which there isn't when you are far from a black hole. The presence of the horizon is important in this. My point was that emitted particles can always be traced to somewhere above the horizon.
Also, even if the phenomenon occurs near the event horizon, could not one of the particle-antiparticle pairs be drawn into the black hole?
I'll have to refer that question to others who understand the process properly. However, when anything escapes, what escapes has positive energy and the hole reduces in mass. If the particles don't escape (assuming that's a possible outcome), the net change to the hole's mass is zero because both the positive and negative energy parts got absorbed.

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berkeman
My initial reaction to Ibix was incorrect, albeit for communication issues not substantive issues. Scientific knowledge is best served when point and counterpoint are aired without prejudice. Scientific work cannot prove something to be "right", only that it conforms to experimentation. Since this is a gedankenerfahrung, only vision, logic, and math can lead us ever closer to nature's reality.

The problem with substance of the initial response following my initial post was that it was a decisive stance on "what it isn't". The issue of particles and antiparticles being drawn into the black hole is not addressed, which is the crux of the discussion. I welcome point and counterpoint, for anyone interested.

weirdoguy
Ibix
2020 Award
communication issues
You ain't kiddin'. I've read that post three times and I still have absolutely no idea what your problem with my responses is. I think that's my cue to duck out of this conversation.

berkeman and Robert Friz
In my opinion, the problem with this diagram is that only antiparticles are shown to be drawn into the black hole. If this were the case, naturally there would be a net loss of mass as the particles exit as Hawking Radiation. I should have included this diagram in my initial post.

Unless there is a sound physics reason why only antiparticles that result from the creation of a virtual pair are drawn into the black hole, as opposed to a balance being drawn in, I am at a loss why this discussion has for some reached a stopping point.

Ibix
2020 Award
Unless there is a sound physics reason why only antiparticles that result from the creation of a virtual pair are drawn into the black hole, as opposed to a balance being drawn in, I am at a loss why this discussion has for some reached a stopping point.
OK - this I do understand. It doesn't say "antiparticles". It says "negative energy particles". These are not the same thing. Nothing has negative energy under normal circumstances (which is a part of why the negative energy part can't be radiated). This includes anti-particles - they don't have negative energy under normal circumstances, and don't even necessarily have negative energy in the virtual pair-production model.

To the extent that the explanation you are citing is in any way related to the actual model (as noted, even Hawking admits it's a very tenuous connection), the negative energy particle is always absorbed by the black hole, necessarily so. You'll have to wait for someone who understands the maths to try to explain why that is so. I warn you that the answer is probably "either learn the maths or trust us". As you have seen, even Hawking's attempt at a non-mathematical answer (the diagram you quoted) is seriously lacking in explanatory power.

PeterDonis
Mentor
2020 Award
Does Hawking radiation at the event horizon of a black hole result in the reduction of the mass of the black hole?
First, note my strikethrough in the quote above. Hawking radiation is not emitted at the event horizon. It is emitted just outside it. (At least, it is to the extent that the emission can be localized at all.)

With the strikethrough, the answer to the question is yes: Hawking radiation does reduce the mass of the black hole. More precisely, all current models of Hawking radiation make that prediction: I assume you are aware that nobody has actually observed Hawking radiation from a black hole or measured that it reduces the hole's mass.

With all that said, the rest of your OP is inaccurate speculation based on an inaccurate model of how Hawking radiation is produced and of quantum field theory in general. The heuristic of "particle-antiparticle pairs constantly being created and destroyed everywhere in spacetime", while it is popular in pop science discussions, is misleading and causes far more problems than it solves. That applies to trying to understand Hawking radiation as much as anything else.

I suggest that you read and carefully consider the following articles. The first is one of a series of PF Insights articles on "virtual particles" (which is another version of the "particle-antiparticle pairs constantly being created and destroyed everywhere in spacetime" heuristic):

https://www.physicsforums.com/insights/misconceptions-virtual-particles/

The second is a Usenet Physics FAQ article by John Baez (who is now a Science Advisor on PF although he doesn't post here often) which discusses why the "particle-antiparticle pair" heuristic isn't really a good one for understanding Hawking radiation:

https://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html

PeterDonis
Mentor
2020 Award
the problem with this diagram
...is that it's a pop science diagram based on an inaccurate pop science understanding of how Hawking radiation works. You cannot learn actual science from pop science sources.

I am at a loss why this discussion has for some reached a stopping point.
Because your entire mental model of how Hawking radiation works (based on particle-antiparticle pairs) is wrong. You will not make any progress by asking for particular rebuttals of particular claims based on that mental model. The only way you will make progress is by dropping that mental model and starting from scratch. The article by John Baez that I linked to might provide a better heuristic basis for you to use to start from scratch. After you have read it, feel free to ask questions about particular items; the article is quite brief and leaves a lot of details for the reader to investigate further.

Thanks, Ibix, for your recent reply! This is something I can work with. However, being slow-witted, it usually takes me a number of weeks of contemplation sitting cross-legged to understand and move forward. With thumbs and forefingers touching one another, my rumination will cover at least these points:

1. Peter Donis's response which includes further discussion on "virtual particles" with helpful links. This may clear up my misconceptions.
2. Why positive energy particles appear, both in the above diagram and in related forum texts, to be on a one-way street away from the black hole.
3. What is the difference between an antiparticle and a negative energy entity.

Again, thanks everyone for the stimulating discussion!

PeterDonis
Mentor
2020 Award
my rumination will cover at least these points
Point #1 is good, yes (with the further advice I gave in post #13 just now).

Points #2 and #3 are not good; further thinking along those lines will just waste your time, since the entire model on which those points are based is wrong, as I said in post #13.

Point #1 is good, yes (with the further advice I gave in post #13 just now).

Points #2 and #3 are not good; further thinking along those lines will just waste your time, since the entire model on which those points are based is wrong, as I said in post #13.
Thanks. Rumination will continue however. More is learned from dead ends than the science community often recognizes -- and this DOES NOT diminish from your advise. I need to better understand the WHYs.

PeterDonis
Mentor
2020 Award
[Edit: removed incorrect comment on a previous post.]

To the extent that the explanation you are citing is in any way related to the actual model (as noted, even Hawking admits it's a very tenuous connection), the negative energy particle is always absorbed by the black hole, necessarily so. You'll have to wait for someone who understands the maths to try to explain why that is so.
Unfortunately for that heuristic model, the explanation basically amounts to "drop that heuristic model and learn a better model". One of the (many) weaknesses of the "particle-antiparticle pair" heuristic model is that it doesn't really have a good answer to why the negative energy particle has to be the one that falls in. There are various hand-waving explanations that sound plausible, but because the entire model doesn't really have a solid basis to begin with, those explanations don't either.

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phinds and Ibix
PeterDonis
Mentor
2020 Award
I need to better understand the WHYs.
See my response to @Ibix just now. As I told him, one of the main issues with the "particle-antiparticle pair" heuristic model is that, because it doesn't have a solid basis to begin with, when people try to ask logical questions to probe the details of the model, as you have, there aren't good answers. The only real solution is to drop the model and learn a better one.

See my response to @Ibix just now. As I told him, one of the main issues with the "particle-antiparticle pair" heuristic model is that, because it doesn't have a solid basis to begin with, when people try to ask logical questions to probe the details of the model, as you have, there aren't good answers. The only real solution is to drop the model and learn a better one.
I was an IT manager at 3M before retiring. I am very, very good at hand waving.

Ibix and phinds
PeterDonis
Mentor
2020 Award
I was an IT manager at 3M before retiring. I am very, very good at hand waving.
LOL

Ibix
2020 Award
Actually, in the pop science heuristic model being used (which is misleading and wrong), they pretty much are.
So Hawking radiation is just particles, no anti-particles? Edit: response to removed part of a comment.

Baez's explanation that you linked is fascinating. He seems to be saying that a distant observer sees macrostates that correspond to multiple microstates that can't be distinguished even in principle, because some of the necessary information crossed the horizon. Therefore the black hole has entropy and hence temperature and hence radiates. And this can be shown rigorously using the Bogoliubov transformations, which somehow relate two observers' determination of the vacuum state. I think (?).

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256bits
256bits
Gold Member
both the positive and negative energy parts got absorbed.
I always wondered with these explanations of Hawking radiation, the negative and positive energy is in relation to what, or what type of energy.
The explanations commonly given seem to make one infer that there is such a thing as an absolute energy.

For any massive body, and not a black hole,
Gravitationally speaking, ( localized Newtonian should apply ) then then two particles formed at the same radius some distance apart, have the same energy of this type.
On the other hand, two particles formed at different radii, since they have to be separated by a Δh, the one at radius h+ Δh, has more gravitational energy than the one at h- Δh. If the h + Δh, has enough kinetic energy to escape from the massive body, the h - Δh definitely has not. Believing in mass conservation, and conservation of momentum, both particles have the same mass, and the same velocity wrt a location of formation; but if we use the cm of the massive body as a reference, the two particles do not the same velocity wrt to the centre of mass of the massive body - hence the escaping particle, h + Δh has more energy than the h - Δh particle. Both particles have positive energy in some frame of reference, yet in another frame of reference, one particle can have a positive energy and the other a negative energy.
From a viewer outside the system, there is no total mass or energy gain, but the massive body has lost mass and energy related to the escaping particle h + Δh, or if one prefers by the Δh difference in radii of formation.

Applying this to particle formation near the event horizon of a black hole ....

PeterDonis
Mentor
2020 Award
So Hawking radiation is just particles, no anti-particles?
No. In principle any particle or antiparticle can be emitted as Hawking radiation. As a black hole gets close to final evaporation, meaning its temperature gets very high, its Hawking radiation would be expected to contain roughly equal numbers of various particles and antiparticles.

I see that I misread your previous post on this point; I'll edit mine to remove the comment you responded to.

Ibix
PeterDonis
Mentor
2020 Award
He seems to be saying that a distant observer sees macrostates that correspond to multiple microstates that can't be distinguished even in principle, because some of the necessary information crossed the horizon.
I think this is one reasonable heuristic way of describing what is going on, yes.

On a side note, your description also leads to a good quick heuristic for why the black hole information paradox is a problem: if the hole finally evaporates, so there is no horizon any more and no region inside it any more, all the information stored in those microstates that fell below the horizon has to come back out, otherwise unitarity is violated. The question is how it comes out.

Ibix
PeterDonis
Mentor
2020 Award
The explanations commonly given seem to make one infer that there is such a thing as an absolute energy.
There isn't, quite, but there sort of is. Unpacking this leads to one reasonable heuristic way of describing what is going on with Hawking radiation (which is not quite the same as the one @Ibix gave a few posts ago). (Note that this unpacking, as far as I can see, does not justify the speculations in the rest of your post; those appear to me to be based on the incorrect model using particle-antiparticle pairs.)

For a quantum field in flat Minkowski spacetime, which state of the field is the "ground state" or "vacuum state"--the state of lowest energy, which by definition has exactly zero quanta of the field--is the same for all inertial observers. (One way this is sometimes expressed is to say that the vacuum state of a quantum field is Lorentz invariant.) This can also be viewed as saying that there is an "absolute energy" for the quantum field, at least for inertial observers--the Lorentz invariant vacuum state having zero energy for those observers is an absolute statement.

However, for accelerated observers, this is not the case: different accelerated observers will not agree on which state of the quantum field is the vacuum state, and will not agree with the inertial observers either. For example, if the quantum field is in the state that is the vacuum state for inertial observers, this state will appear to an accelerated observer as a thermal state with temperature ##T = 1 / a##, where ##a## is the observer's proper acceleration (and I have left out a bunch of physical constants like ##c## and ##\hbar## by assuming units where they are equal to 1). Thus, the accelerated observer will observe radiation, which can be viewed as coming from the vicinity of that observer's Rindler horizon. This is called the Unruh effect.

One of the analogies that originally led to the hypothesis of Hawking radiation is the analogy between the Rindler horizon of an accelerated observer and the event horizon of a black hole. To an observer who is "hovering" sufficiently close to a black hole's horizon that the effects of spacetime curvature due to the hole can be ignored locally--and who must be accelerated in order to stay there--the hole's event horizon is the observer's Rindler horizon (in that local patch of spacetime). So, by analogy with the Unruh effect, an observer "hovering" at a finite altitude above a black hole should observe radiation coming from the vicinity of the hole's horizon, just as the Unruh radiation observed by an accelerated observer in flat spacetime appears to come from the vicinity of that observer's Rindler horizon.

That analogy, by itself, might help to explain why an observer sufficiently close to a black hole's horizon would see radiation coming from the vicinity of the horizon. But it doesn't explain why an observer far away from the hole would see such radiation, or why such radiation would carry away actual energy from the hole and decrease its mass. For those things, we need to take into account the global geometry of the spacetime, i.e., the spacetime curvature due to the hole. That curvature means, roughly speaking, that if the quantum field is in a state which is a vacuum state with respect to radiation flowing into the hole, it will not be a vacuum state with respect to radiation flowing out of the hole. (This is related to the description in terms of observers in the far past and observers in the far future in the Baez article; "far past" corresponds to "radiation flowing into the hole", since that radiation will have been emitted far away in the far past, and "far future" corresponds to "radiation flowing out of the hole", since that radiation will be received far away in the far future.)

In short, one heuristic way of viewing Hawking radiation is that it is a manifestation of there being different notions of "vacuum state" for different observers in a curved spacetime.

256bits