How Does Hawking Radiation Affect Black Hole Mass?

In summary: It's a virtual particle, so it doesn't have the same rules and restrictions as real particles do. And it's not like the black hole is losing mass by emitting it, because the mass of the black hole is a function of its total energy, not just its mass. So the black hole can lose energy, and as a result, lose mass, without anything actually being removed from it.
  • #1
Khursed
43
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I've read as much on the subject as I can before getting into the actual math of it, so I need people who know the subject better then I do to answer my question.

The way I understand it, at the quantum level, particle/anti-particle pairs are created all the time and annihilate before they can impact the universe with their presence.

The way hawking Radiation works as I understand it, a pair is created on the edge of the event horizon, and as the pair is created, and before it can annihilate, the anti-particle falls into the black hole, and the positive particle is seen by the rest of the universe as if the black hole had just expelled matter.

Now, given how the uncertainty principle will never allow for any particle to have a known position, how can the black hole only absorb the anti-particle? Why isn't the black hole sometime getting the positive particle and therefore gaining weight? While the universe gets the bad end of the deal and gets an anti-particle?

So in summary, shouldn't by sheer luck a black hole remain stable in regards to hawking radiation? Or is there some artifact of physics, or something similar to the CP violation going on?

If its CP violation, I would imagine that the discrepency would be immensely small given the original scale for the universe, so that explanation seems somewhat weak.

Finally, as far as I understood it, negative particle retains their mass, as far as physics is concerned, there are no particle with negative mass, so what exactly happens for a particle from a pair creation from quantum fluctuation to emit a particle that is absorbed by the black hole to count as lowering its mass?

I mean, ok, so the universe sees a particle that seems to have been emited by the black hole, however, since the black hole itself absorbed something, and as far as I know mass is invariant to the actual polarity of a particle, how does the black hole loses mass?

Is this a new quantum mechanic property I've never read? Redistribution of mass to satisfy math?

I remains puzzled by how the black hole should lose mass because it so happened to gobble up a particle that was too close to its event horizon.

The way I see it, I'm also not sure how a particle created so close to the horizon would have the energy to get away from it?

I can imagine the theory says that the black hole imparts some energy to the particle to escape, but how would such a mechanism work? And why wouldn't it work for normal matter?

That would mean black hole can arbitrarely eject matter from its event horizon thus losing mass? Do we know of any such mechanism?

Anyway, looking forward to some ideas or explanation regarding that one.
 
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  • #2
It doesn't matter which the black hole absorbs. The particle and antiparticle have the same mass, so the net gain to the outside is the same. If the antiparticle is the survivor, it will soon be annihilated and the net gain will be in the form of energy.
 
  • #3
mathman said:
It doesn't matter which the black hole absorbs. The particle and antiparticle have the same mass, so the net gain to the outside is the same. If the antiparticle is the survivor, it will soon be annihilated and the net gain will be in the form of energy.
That doesn't answer the question at all. The OP understands how mass/energy is bestowed upon the outside, but not how mass/energy is removed from the black-hole. I have the same confusion; can anyone explain/clarify?
 
  • #4
The two virtual particles together have zero mass. When one falls in (it doesn't matter which one) the particle on the outside becomes a real particle with positive mass. Since together they had zero mass and now the one on the outside has positive mass, the one on the inside must have negative mass.
 
  • #5
skeptic2 said:
The two virtual particles together have zero mass. When one falls in (it doesn't matter which one) the particle on the outside becomes a real particle with positive mass. Since together they had zero mass and now the one on the outside has positive mass, the one on the inside must have negative mass.
Interesting. I'm okay to accept that, its a good argument... but there isn't any reason that, given the outside particle has become real, the inside one must be real and therefore have standard properties (like positive mass)?
 
  • #6
But where would its mass have come from?
 
  • #7
Sure, that's a great question. But I would ask the same thing about the accepted paradigm, i.e. where did the mass/energy of the hawking radiation come from? For a virtual pair to appear, one has to be an (e.g.) electron, and the other a positron... both of which have positive mass.
 
  • #8
Khursed said:
The way I understand it, at the quantum level, particle/anti-particle pairs are created all the time and annihilate before they can impact the universe with their presence.

There is a better way of thinking about it...

Quantum mechanics says that there is some inherent uncertainty about what the location of a particle is or how much energy and momentum it has. It turns out that if you ask "how many particles are in this box" that's also an number that is fuzzy. It's not that there are twenty particles in the box, it's that the exact number of particles in the box is undefined. It's "around 20-ish."

So in summary, shouldn't by sheer luck a black hole remain stable in regards to hawking radiation?

Both the particle and anti-particle have positive mass.

Also the reason that something like Hawking radiation has to exist is that if you had a black hole suck everything, it would turn into an ultimate heat sink, and you could use it to create a perpetual motion machine.

Is this a new quantum mechanic property I've never read? Redistribution of mass to satisfy math?

The mass of the particle that gets emitted comes from gravitation energy of the black hole.
 
  • #9
skeptic2 said:
The two virtual particles together have zero mass. When one falls in (it doesn't matter which one) the particle on the outside becomes a real particle with positive mass. Since together they had zero mass and now the one on the outside has positive mass, the one on the inside must have negative mass.

That's wrong. "Virtual particles" exist because there is fundamental uncertainty about the amount of energy that a particle pair has, so it could be zero. Once you actually create a "real particle" that subtracts mass from the gravitational field. One falls in, and one escapes.
 
  • #10
twofish-quant said:
Once you actually create a "real particle" that subtracts mass from the gravitational field. One falls in, and one escapes.
Youre saying: some amount of energy is taken from the gravitational field of the black-hole, half is sent back into the black-hole, and half escapes, yes?
Cool. I'm trying to understand how exactly the energy exchange is made; when you say, 'energy is taken from the g-field' you mean, from the acceleration of the trapped particle towards the singularity? i.e. the g-field has to do work to accelerate that particle?
 
  • #11
zhermes said:
Cool. I'm trying to understand how exactly the energy exchange is made; when you say, 'energy is taken from the g-field' you mean, from the acceleration of the trapped particle towards the singularity? i.e. the g-field has to do work to accelerate that particle?

I have to think about exactly how the energy exchange happens. However, one nice thing about physical principles like conservation of mass-energy is that you know that *somehow* the energy for the particles has to come somewhere, even if it isn't immediate obvious where and how.
 
  • #12
twofish-quant said:
That's wrong. "Virtual particles" exist because there is fundamental uncertainty about the amount of energy that a particle pair has, so it could be zero. Once you actually create a "real particle" that subtracts mass from the gravitational field. One falls in, and one escapes.

What happens to the particle that falls in? Does it become real too? What is its mass?
 
  • #13
Another problem I have with it, why would such a particle be able to escape anyway?

I mean we're talking at the edge of an event horizon with an escape velocity of slightly below light speed, how does any particle get away from there? Unless its a photon or a neutrino that's not pointing toward the black hole?

Anything else with mass, I have a hard time imagining it having enough momentum to get away from the black hole.

But then again, as far as I understand there has never been any proof of hawking radiation, so it could be a wrong assumption.
 
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  • #14
Khursed said:
The way hawking Radiation works as I understand it, a pair is created on the edge of the event horizon, and as the pair is created, and before it can annihilate, the anti-particle falls into the black hole, and the positive particle is seen by the rest of the universe as if the black hole had just expelled matter.

Now, given how the uncertainty principle will never allow for any particle to have a known position, how can the black hole only absorb the anti-particle? Why isn't the black hole sometime getting the positive particle and therefore gaining weight? While the universe gets the bad end of the deal and gets an anti-particle?

Khursed said:
Another problem I have with it, why would such a particle be able to escape anyway?

See the post

https://www.physicsforums.com/showthread.php?p=620350#post620350,

and the link to Steve Carlip's explanation that I give in this post.
 
  • #15
zhermes said:
Youre saying: some amount of energy is taken from the gravitational field of the black-hole, half is sent back into the black-hole, and half escapes, yes?
Cool. I'm trying to understand how exactly the energy exchange is made; when you say, 'energy is taken from the g-field' you mean, from the acceleration of the trapped particle towards the singularity? i.e. the g-field has to do work to accelerate that particle?

sorry to bring an old thread to the top, but I've been confused about the exact same thing as the OP, and luckily i found this thread via a search before unnecessarily starting my own. at any rate, after much thought of how particle-antiparticle pair creation near an event horizon can result in Hawking Radiation (BH evaporation), the most sensible explanation i could comtemplate in my own head was more or less what was stated above. if the gravitational field near the event horizon gives up a certain quantity of mass/energy in order to create the particle-antiparticle pair, then the BH's mass/energy must decrease by that same quantity. but the BH must gain some (but not all) of that mass/energy back when it swallows either one of the two particles previously created. that's the only way i can visualize hawking radiation as the actual decrease in the mass/energy of a BH. now granted, i don't expect anyone to be able to definitively confirm or deny my analysis - after all, we won't really know the specifics of the energy exchange until we start to actually witness Hawking Radiation. but i would like to hear from others on whether they generally agree or disagree with this notion of BH evaporation...

also, this blurb is a bit off-topic, but its somewhat related, so i thought i would postulate it here. i don't know if its possible, but i was wondering if the quantities of energy involved in the minute fluctuations of the BH's gravitational field that create these particle-antiparticle pairs in the first place could be small enough that neither particle of the pair is endowed with enough energy to escape the BH. if this scenario is possible (and i have no idea if it is or isn't), then it would seem that not all particle-antiparticle pair-creating fluctuations in the gravitatonal field result in Hakwing Radiation. that is, if the entire quantity of mass/energy given up by the BH during the creation of a particle-antiparticle pair is swallowed back up by the BH, then its mass/energy is conserved, and no mass/energy is radiated away in the form of Hakwing Radiation.
 
  • #16
It was my understanding that the anti particle would annihilate with another particle in the black hole lowering its mass through the destruction of matter. Although, matter is converted into energy in a black hole anyways.
 

Related to How Does Hawking Radiation Affect Black Hole Mass?

What is Hawking radiation?

Hawking radiation is a theoretical concept in physics proposed by scientist Stephen Hawking. It suggests that black holes emit radiation due to quantum effects at the event horizon, causing them to gradually lose mass over time.

How does Hawking radiation solve the "information paradox"?

The information paradox is a problem in physics that arises when considering what happens to the information of matter that falls into a black hole. Hawking radiation proposes that this information is encoded in the radiation emitted by black holes, solving the paradox and preserving the laws of quantum mechanics.

Why is Hawking radiation important in understanding black holes?

Hawking radiation plays a crucial role in our understanding of black holes as it provides a mechanism for them to eventually evaporate and disappear. It also helps reconcile the laws of thermodynamics with the existence of black holes.

Can Hawking radiation be detected?

As of now, Hawking radiation has not been directly observed or detected. This is due to its extremely low intensity and the fact that black holes emit it in all directions. However, scientists are working on ways to potentially detect it in the future.

What are the implications of Hawking radiation for the fate of the universe?

Hawking radiation suggests that black holes will eventually completely evaporate, resulting in the release of all the energy and information they contain. This could have significant implications for the future of the universe and the ultimate fate of black holes.

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