If a particle is being expelled from a black hole and an antiparticle is being driven into it, shouldn't the opposite occur as well and in the same frequencies?

I mean, black holes should emit antiparticle radiation as well and gather energy from it and the black hole energy shouldn't change, right?

What am I missing?

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    $\begingroup$ I think this is a good question - a detail that is missing in the layman explanation of Hawking radiation. My best thinking is that there is a cosmic firewall argument behind it. After all, what ordinarily causes spontaneous particle and antiparticle pairs to annihilate? The net energy of the system is zero. But if the antiparticle passes the firewall, the charge of the system is unbalanced, so the particle freaks out. But I lack a more technical description of freaking out. $\endgroup$ Jul 21, 2013 at 0:10
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    $\begingroup$ physics.stackexchange.com/q/44922 I think that guy in that post has the same question with a better understanding $\endgroup$
    – eJunior
    Jul 23, 2013 at 16:15
  • $\begingroup$ I have found the answer I needed in this article. arxiv.org/abs/1008.3657 $\endgroup$
    – eJunior
    Aug 14, 2013 at 17:01

4 Answers 4


EDIT:The full machinery of quantum field theory in curved space-time is required to derive Hawking radiation properly. In QFT there are various equivalent ways to think about scattering processes. The best and easiest way, due to Feynman diagrams, is to think about virtual particles. These particles are not real, and in this sense they do not exist. They are counterparts of real particles, but they can have any mass they like. They "exist" only as intermediaries between real particles. The popular science explanation of Hawking radiation mostly focuses on this virtual particle picture. What does this have to do with vacuum energy? Well, if you have a field and it has the lowest possible energy (vacuum energy) how many particles there are is not a "natural" question, so to speak. It is something that can change for short periods of time, due to the famous Heisenberg Uncertainty Principle. So these virtual particles are really just fluctuations around the vacuum energy, but note that the negative energy of one of the particles is absolutely necessary in Hawking radiation, because at the end of this process you need to have a situation which is real and is not just a fluctuation. The energy of the vacuum can not be lowered (it is the lowest possible energy the field can have) so the energy has to come from the black hole. The vacuum is not what gives the energy! If there were no vacuum there would be no fields, and no Universe, as we know it, not just no Hawking radiation. :) People usually talk about event horizon and not the ergosphere because when they explain Hawking radiation they focus on the simpler case of a black hole which does not rotate. See this answer about Hawking radiation of rotating, charged black holes.

OLD ANSWER: Black holes do emit all sorts of different particles and antiparticles as a part of Hawking radiation. The loss of mass has nothing to do with whether matter or antimatter falls into a black hole. It has to do with the sign of the energy of the (virtual) particle. Quantum fluctuations create particle-antiparticle pairs near the event horizon, but these are virtual. That means that they can't exist very long and can only be observed indirectly. One of these (anti)particles manages to escape and become a real particle with positive energy. Conservation of energy demands that the (anti)particle which did not escape and has fallen into the black hole has negative energy. The opposite does not happen because if it were to happen the negative energy virtual particle would become a real particle and this is not possible.

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    $\begingroup$ "Black holes do emit all sort of radiation." This is a bit vague... are you talking about only Hawking radiation? My immediate thought was including emission from accreting material, which is not from the black hole. Maybe it's just me though. $\endgroup$
    – Kyle Oman
    Jul 22, 2013 at 16:10
  • $\begingroup$ Yes, you're right. I should have made it more clear. I have edited my answer. Thanks! $\endgroup$
    – Bubble
    Jul 22, 2013 at 18:07
  • $\begingroup$ I have also edited my answer to address question later raised by the OP. $\endgroup$
    – Bubble
    Jul 22, 2013 at 19:03

The following answer is not "rigourous", but it may gives a simple explanation.

Suppose you have a quantum fluctuation just near the horizon, but outside. This quantum fluctuation create 2 particles, one with a negative energy -E, the other with a positive energy +E.

If the 2 particles stay outside the black hole, they have to anihilate themselves in a time time $t \le \frac{\hbar}{E}$

Now, one of the 2 particles may fall inside the black hole, and there are 2 possibilities ; the escaping particle may have a positive or a negative energy. The key point is that there is an asymmetry between these 2 cases.

For a particle to be real, its energy has to be positive, but relatively to the time coordinate. With an evolution variable $\tau$, this can be write $\frac{dt}{d\tau}>0$

When the horizon is being crossed (by the infalling particle), we may consider, that there is a change in the nature of the time and radial space coordinate. The time-like coordinate becomes a space-like coordinate, and the radial space coordinate becomes a time-like coordinate.

More precisely, if, outside the black hole, the coordinate are(in units $c=1$) : $z=r+it$, then the "coordinates" inside the black hole are $z \rightarrow z'\sim -iz$ So, $$z'=r'+it'\sim-i(r+it)=(t-ir)$$

So, $t'\sim-r$ and $x'\sim r$

For an escaping particle, the energy must be positive relatively to $t$, so $ E=\frac{dt}{d\tau} >0$, but for the infalling particle the "energy" must be positive relatively to $t'$, that is = $\frac{dt'}{d\tau}>0$, which is "equivalent" to $-\frac{dr}{d\tau}>0$.

But the last expression means only that the particle is an infalling particle, which was our hypothesis. We could say also, for the infalling particle, that the "outside" energy $-E$ becomes a "inside" momentum.

  • $\begingroup$ According to your logic the infalling particle dont really have negative energy, just is a particle that is going back in time and being expelled from the BH in the form of the radiation? Because T become negative also? $\endgroup$
    – eJunior
    Aug 14, 2013 at 17:11
  • $\begingroup$ As it is precised at the end of the answer, the "inside" radial space dimension "corresponds" to a outside time, so, $-E$ is just an "inside" momentum. There is no "back" in time process. The radiation corresponds to the escaping particle (of "outside" positive energy $E$) $\endgroup$
    – Trimok
    Aug 14, 2013 at 18:12
  • $\begingroup$ 'Back in time' as in the particle being pulled away by gravity instead of pushed. But why does this reverse in radial space dimension would only expel radiation? Wouldnt it affect all matter crossing the horizon? $\endgroup$
    – eJunior
    Aug 14, 2013 at 19:26

It is called energy conservation.

When a real particle departs from the field of the black hole, energy is taken away to infinity, and the hole is depleted by that delta(E), no matter how soft the particle leaving is. Eventually it is possible that all the mass of the black hole will be depleted to the point of not being able to have the strength of the gravitational field that makes it a black hole .

Now if you are worried that the universe is in danger of losing its black hole population don’t be. Hawking radiation only has a noticeable effect on black holes of around 10^12 kg 1. or smaller. This is because a black hole must be starved to death. That is the matter-energy it puts out must be greater than the matter-energy it takes in. Since black holes mass is inversely proportional to its temperature and the amount of matter-energy that it radiates is dependent on its temperature, then as the mass of the black hole goes up the amount of matter-energy that it emits goes down. So for larger back holes such as one about the mass of our sun their temperature is very low about 10^-7 K and so they have very low radiation. And since they receive light from stars, cosmic back ground radiation, and even matter, in the form of dust, planets, and stars they continue to grow. It will only be at some future date when all stars have gone out, and matter is out of reach of large black holes and when the temperature of the universe is less than that of the black hole itself that the larger black holes will begin to evaporate.

Edit after comments:

The above replies to the title.

In the content you ask:

If a particle is being expelled from the Black Hole and an antiparticle is being driven into it, shouldn't the opposite occur as well and in the same frequencies?


I mean Black holes should emit antiparticle radiation as well and gather energy from it and the black hole energy shouldn't change, right?


Whether it is a particle or an antiparticle, the energy it carries, mass and kinetic part, is always positive for a real particle. Once a particle/antiparticle leaves the black hole it is real and carries away energy depleting the store of energy of the black hole.

What am I missing?

That antiparticles are anti only in their quantum numbers versus their corresponding particles, not in the energy/mass. The electron and the positron both have the same positive mass.

  • $\begingroup$ The only problem is that you dont explain the process, only that something would take place. How can a real particle depart from the EH? I dont see that working in the conventional explanation of Hawking Radiation, is it a tunneling effect? Why Hawking Radiation wouldnt be something generated from energy just barely outside the EH (and the BH wouldnt then lose energy)? Thank you! $\endgroup$
    – eJunior
    Aug 13, 2013 at 21:00
  • $\begingroup$ Have a look at this mathematical model of the black hole which shows the ergosphere en.wikipedia.org/wiki/Ergosphere . The ergosphere is a part of the field/entity that is defined as a black hole. Energy depleted from the ergosphere is energy depleted from the black hole. I like @Trimok 's non rigorous explanation. To really dig in to this stuff one needs quantum field theory. $\endgroup$
    – anna v
    Aug 14, 2013 at 3:22
  • $\begingroup$ I have looked at the documents you have showed. And In my mind if the depleted energy came from the ergosphere it would make perfect sense. But then the BH would not die if the energy didnt came from the EH, right? I was a little confused about the terminology when I had made the question, and I think I have a better understanding now. What I really meant by the BH 'gathering energy' is that if the deplated energy came from the ergosphere, then positive energy 'virtual particles' would also happen to enter the EH (increasing the EH in cost of the ergosphere via the same kind of radiation). $\endgroup$
    – eJunior
    Aug 14, 2013 at 15:38
  • $\begingroup$ The point is that the ergosphere is part of the black hole energy system. There is no ergosphere if there is no black hole. Any energy taken from the ergosphere diminishes the energy of the black hole. $\endgroup$
    – anna v
    Aug 14, 2013 at 15:50
  • $\begingroup$ I agree with you, but then the BH wouldnt have a 'expiration date' right? Because trough this process the EH wouldnt diminish forever till extinction, like Hawking concluded that in some point the EH would dissipate. I think Hawking Radiation is somehow related to something called 'killing vector field' and that in theory he said it should generate negative-energy inside of the EH and somehow possibilitate the positive energy to get out the EH leaving only its negative par inside. $\endgroup$
    – eJunior
    Aug 22, 2013 at 14:12

I was the one who posted the question... But I'm having HUGE problems with the website as I cannot comment anything and I cannot log in into eMagus account for some reason (everytime I login into eMagus it automatically logs into eJunior). So I cannot reply to any answer unless I post it as an answer as well... Sorry.

I have done some research and the problem I'm having with your answers are this: There is no such thing as a negative energy particle; it is impossible to exist. Both the antiparticle and particle are positive in energy.

They 'draw' energy from the vacuum... And normally they would collide and release the energy back in the vacuum. I suppose that Hawking radiation theory has nothing to do with negative energy particles... Its origins must be from the vacuum energy drained comiong from the black hole so no matter if the particle or antiparticle are erradiated, the 'energy drained' comes from the microparticles from the vacuum of the event horizon.

But what I don't understand is why the energy vacuum radiation drained comes from the event horizon and NOT from the ergosphere of the black hole?

In my mind it is the vacuum from the ergosphere that should provide energy to the radiation and not the vacuum from the black hole itself. If just barely two virtual particles from the liminar of the ergosphere are created and one fall into the event horizon, the energy still comes from the ergosphere, right?

And even if it were true that is draining energy from the ergosphere, what I already have trouble understanding, shouldn't the ergosphere emmit Hawking radiation also and on a much bigger scale than anything that could possibly come from the event horizon?

I'm trying to merge the two accounts to comment properly, thank you!

  • $\begingroup$ And so.. if hipotetically there was no vacuum energy outside the EH... , should be no hawking radiation coming from it. $\endgroup$
    – eJunior
    Jul 22, 2013 at 17:43
  • $\begingroup$ I have added text to my answer to address your concerns. $\endgroup$
    – Bubble
    Jul 22, 2013 at 18:53
  • $\begingroup$ Man i wish i could comment other people answers. Im honestly thinking about creating another question just for that. But anyways thanks bubble for answering. I will feedback on your answers right way. $\endgroup$
    – eJunior
    Jul 22, 2013 at 19:45
  • $\begingroup$ @Bubbles As i see the energy of the vacuum is constantly lowering due to entropy and because of universe expansion, correct me if its not (actually im not a physicist). In my understanding virtual particles gather from the energy of the vacuum (draining energy from it)... the particle and anti-particle obliterate and the energy comes back to the vacuum. There is no 'negative energy particle' involved.. i tought it would be an aliteration of this draining energy process. $\endgroup$
    – eJunior
    Jul 22, 2013 at 19:53
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    $\begingroup$ And by the Universe being destroyed I literally mean that a lowering of vacuum energy would be favorable by nature and new types of particles would be formed which would cause vacuum decay which would in time destroy the Universe: en.wikipedia.org/wiki/Vacuum_decay At the end you need to end with a real process, no virtual particles being stable. You start with a black hole, and you end with a particle and a slightly smaller black hole. This is a real process. $\endgroup$
    – Bubble
    Jul 22, 2013 at 20:24

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