# The True Nature of Hawking Radiation

cbd1
With the theory of Hawking Radiation, people generally say to imagine that the black hole has a temperature, so it must radiate heat. But, this really is not how it is said to work.

The means by which the black hole evaporates is not that particles are actually coming out of it, or that it is radiating heat. But rather, the black hole is losing mass by taking in anti-particles from virtual particle pairs. So it is not that mass leaves the black hole, but that the black hole sucks up anti-mass. What comes away from the hole is not actually anything extracted from the hole, but a particle that was created outside of the hole. So what has happened is there has been an exchange, the black hole's taking in anti-mass allows for the mass of the new particles Is this basically correct, actually more accurate than saying that something is coming "out of" the black hole?

On a side note, why again is it that there is not an equal balance. You would think that there would be a 50% chance that the positive particle go in the hole and the anti-particle to fly away. That is, why would it not balance, half the time taking in the normal particle and half the time taking in the other?

Homework Helper
Hi cbd1!
… the black hole is losing mass by taking in anti-particles from virtual particle pairs. So it is not that mass leaves the black hole, but that the black hole sucks up anti-mass.

No, there's no such thing as anti-mass.

It's all mass!

It makes no difference to the Hawking process whether (eg) the electron or the positron is "emitted" … either way, the black hole loses mass.

cbd1
But it is not losing mass from within the black hole. Technically, material is not ejected from within the black hole. Is this not correct?

{~}
Inside the black hole the separated is annihilated as it would if it collided with its partner. The annihilation energy is absorbed by the particle outside the event horrizon giving it enough K.E. to escape.

This is essentially the same as the black hole ejecting mass. Mass and energy are equivalent but in this case the masses are so small and randomized its better to think of the same way as other annihilation reactions, hot particle radiation.

Thats my understanding but I am by no means an astrophysicist.

FunkyDwarf
The energy for these particles comes from the gravitational field which is linked to the mass.

{~}
The energy for these particles comes from the gravitational field which is linked to the mass.

Could you please expand on that?

imaloonru
The idea is that particle/anti-particle pairs are being created all of the time. This process seems to violate conservation of energy and mass, since two particles are suddenly being created from nothing. However, this is allowed by the uncertainty principle, as long as that energy is "given back" right away when the two new particles colide and annihilate each other.

A funny thing happens at the event horizon of a black hole. Crossing the event horizon ensures that no particle can escape the black hole's grasp. When a particle and an anti-particle are spontaneously created so that one is on one side of the event horizon, and the other on the other side, one of the particles is doomed to parish into the black hole, while the other may escape if it's heading in the right direction. Uh oh. Now they ARE violating conservation since the uncertainty principle no longer applies. So where does that energy come from? Who pays? The black hole does. It loses a small amount of energy.

This is why it is called radiation. The black hole appears to be giving off particles (from pair production) while losing energy. Not really evaporation, although it's a convenient analogy. Hope this helps!

Zarqon
From the explanations so far I'm still not clear on what the exact physical process is that makes the black hole lose mass from normal particles.

If it is ok to assume that a black hole is formed consisting of normal matter, then when it happens that the anti-matter part of the pair goes into the hole, then I can sort of see how mass can dissappear from the hole. However, when the normal-matter part of the pair goes in (which seems to be an equally likely process) it seems naively to increase the mass of the black hole rather than decrease it.

Obviously this is not the case, but what exactly happens to a normal-matter particle that goes in?

imaloonru
Zarqon:

It's important to keep in mind that anti-matter particles do not have "negative mass", they have a positive mass, like any other normal particle. So when an anti-particle enters the black hole, it adds to the black hole's mass. Same for normal particles. The part where it looses energy comes from the particle that gets away.

That particle that just escaped, did so illegally. According to conservation of energy, it shouldn't even exist! So how did it come into existence? It "borrowed" energy from the vacuum; a process that generally involves giving that energy back right away before the universe notices it's gone (this is a very hand-wavy explanation). So once that other particle (or anti-particle) escapes, someone has to pay for it. That someone is the black hole.

Does this help?

Bartleby50
As i understand it, the particle that falls into the Black Hole always has negative energy, cause it is entangled with its antiparticle, which can only be measured on positive energy.
is this in anyway close to correct ?

Homework Helper
As i understand it, the particle that falls into the Black Hole always has negative energy, cause it is entangled with its antiparticle, which can only be measured on positive energy.
is this in anyway close to correct ?

Mass is energy, energy is mass.

No such thing as negative mass, no such thing as negative energy.

Both particles have positive energy, "borrowed" for a short time.

(but this is all just maths, isn't it? i mean the reality is that a particle is created just outside the event horizon, and the black hole simultaneously loses an equal amount of energy-or-mass?)

(oh, and entanglement is irrelevant … for the infalling particle, existence is futile ! )

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Bartleby50

"So the black hole can absorb the negative-energy particle from a vacuum fluctuation without violating the uncertainty principle, leaving its positive-energy partner free to escape to infinity."

The entanglement explains for me, why its always the "negative-energy" particle, that is swallowed by the Black Hole.

Zarqon
Zarqon:

It's important to keep in mind that anti-matter particles do not have "negative mass", they have a positive mass, like any other normal particle. So when an anti-particle enters the black hole, it adds to the black hole's mass. Same for normal particles. The part where it looses energy comes from the particle that gets away.

That particle that just escaped, did so illegally. According to conservation of energy, it shouldn't even exist! So how did it come into existence? It "borrowed" energy from the vacuum; a process that generally involves giving that energy back right away before the universe notices it's gone (this is a very hand-wavy explanation). So once that other particle (or anti-particle) escapes, someone has to pay for it. That someone is the black hole.

Does this help?

I'm afraid I consider these types of reasonings not as explanations, but rather as "absence of violations" if you understand the distinction I'm talking about.

Just because there there technically exists a possibility to take the energy from the black hole and restore conservation and general happiness to the universe, doesn't mean that it's explained how this process happens.

I understand what you mean about the anti-particle also contributing to the total mass/energy of the black hole, but I don't understand the physical process inside the black hole after a particle (or anti-particle) is sent in, that somehow causes a decrease in mass/energy.

Let me try an example.

Let's say an electron/positron pair is created, by random chance it happens that the positron is sent away from the hole and the electron is sent into the hole. Since an outside observer sees the positron leave the black hole, he would say the hole is losing energy. However, let's consider the electron that went in. Once inside, the electron from the pair production is indistinguishable from any other electron that would go into the black hole through other means than pair production and so it should contribute to it's mass/energy. The question is: what is the physical meachnism that causes the decrease from this seemingly normal electron?

Naty1
Kip Thorne, BLACK HOLES AND TIME WARPS, PGS 435-440

Hawking concluded that a black hole behaves precisely as though its horizon has a finite temperature. There are several different ways to picture black hole evaporation...all acknowledge vacuum fluctuations as the ultimate source of the outflowing radiation...the simplist is that...tidal gravity pulls the virtual photons apart and the one that escapes carries away the enrgy tyhat the hole's tidal gravity gave it.

As soon as one is separated from its virtual partner, it becomes "real" (observable)...but only to an accelerating (stationary) observer outside the horizon.

Since black holes are still considerably colder than our universe, they radiate nothing but do absorb energy (radiation) from the universe and slowly grow. So nobody has ever seen Hawking radiation.

If we could hang a thermometer on a long string and dangle it just outside the horizon of a black hole, it would register incredibleby high perhaps approaching infinite temperature. On the other hand a free falling observer falling towards the black hole registers no such increase in temperature and passes harmlessly thrugh the mathematical horizon without any immediate effect except increasing gravitational acceleration and tidal effects...So thermal and quantum radiation become two sides of the same coin near a horizon.
Leonard Susskind, THE BLACK HOLE WAR, almost an exact quote with some omissions...

The black hole would be losing energy, rather than mass, IF it were hotter than the surrounding universe...

Wikipedia describes it this way:

Emission process
Hawking radiation is required by the Unruh effect and the equivalence principle applied to black hole horizons. Close to the event horizon of a black hole, a local observer must accelerate to keep from falling in. An accelerating observer sees a thermal bath of particles that pop out of the local acceleration horizon, turn around, and free-fall back in. The condition of local thermal equilibrium implies that the consistent extension of this local thermal bath has a finite temperature at infinity, which implies that some of these particles emitted by the horizon are not reabsorbed and become outgoing Hawking radiation

Hawking radiation can also be described in terms of strings, quantum tunneling, and entropy/information...the description one uses depends on the theory one chooses to employ.

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Gold Member
The 'negative' energy that falls into a black hole is a balancing act, not unlike the imaginary charges that arise in electronic circuits. It is a convenient way of expressing the observational consequences. When you mediate physical processes using bosons, you are forced to acknowledge this kind of particle. We never directly detect such particles and they probably never exist in a naked, detectable state.

I just thought the intense gravity from the black hole is slowly making the mass get converted into energy and then the energy is converted into space.

andyzoom
id have said the intense gravity pulls the particles in, accelerating until they get close to the center of the black hole by which point they have reached lightspeed and turn to energy, just like ice turns to water at 0'c. As matter cannot go faster than that they get released in a beam...could it be so simple?