Confused about Hawking Radiation

[PhDnotForMe]

Hawking Radiation says that an antiparticle and particle spontaneously appear and annihilate each other everywhere, but at the horizon, if the antiparticle appearing inside the horizon and the particle appears outside, then the antiparticle is sucked in and the particle is emitted outward. Hawking radiation would be the particles being emitted away. The antiparticles are sucked in and, if the black hole is small enough, all the antiparticles will eventually completely annihilate the black hole.

What I am confused about is why would the black hole lose any mass at all. This relies on the premise that more often than not, the antiparticle will be the particle that appears on the inside of the horizon while the (regular) particle will be the particle to appear on the outside.
Intuitively, it would seem to me that 50% of the time, the antiparticle would appear on the inside while the other 50% of the time the (regular) particle would be the one to appear on the inside.
Why is it that the antiparticle is assumed to be the one appearing inside the horizon more than the (regular) particle?

 

[Nugatory]

Hawking Radiation says that an antiparticle and particle spontaneously appear and annihilate each other everywhere, but at the horizon, if the antiparticle appearing inside the horizon and the particle appears outside, then the antiparticle is sucked in and the particle is emitted outward. Hawking radiation would be the particles being emitted away. The antiparticles are sucked in and, if the black hole is small enough, all the antiparticles will eventually completely annihilate the black hole.

You will read many non-technical descriptions of Hawking radiation that say something like this, but that doesn't make it right. It's not exactly wrong, but it is so dumbed-down as to be completely worthless for understanding the subject at any depth.

There's no substitute for reading and understanding the real thing, but you may also find this explanation somewhat helpful.

This topic has been discussed in many previous threads, and ineed what I just wrote above is almost a verbatim quote from a half-dozen older posts.

 

[Dale]

This relies on the premise that more often than not, the antiparticle will be the particle that appears on the inside of the horizon while the (regular) particle will be the particle to appear on the outside.

As @Nugatory said this is not actually the basis of the prediction. However, from your comment I get the impression that you believe that antiparticles have negative mass. They do not, antiparticles have the same mass but opposite charge. So even in the “pop sci” presentation your objection is a little off.

 

[PhDnotForMe]

As @Nugatory said this is not actually the basis of the prediction. However, from your comment I get the impression that you believe that antiparticles have negative mass. They do not, antiparticles have the same mass but opposite charge. So even in the “pop sci” presentation your objection is a little off.

I don't think they have negative mass. I think that when an antiparticle and particle come together they annihilate each other resulting in a zero mass system and dissipated energy. This was the reasoning I used above. Is this incorrect?

 

[Nugatory]

I don't think they have negative mass. I think that when an antiparticle and particle come together they annihilate each other resulting in a zero mass system and dissipated energy. This was the reasoning I used above. Is this incorrect?

It is generally correct that annihilation turns a particle and an antiparticle into energy... but that energy contributes to the "mass" of the black hole just as much as the particles did. You start with positive mass-energy, you end up with the same amount of mass-energy in a different form. And in the case of a black hole it doesn't matter what form the infalling mass-energy takes because it all ends up at the singularity anyways.

 

[phinds]

I don't think they have negative mass. I think that when an antiparticle and particle come together they annihilate each other resulting in a zero mass system and dissipated energy.

Sure, but that has nothing to do with Hawking Radiation. As has already been pointed out, the "virtual particle pair" is a heuristic that is very misleading, as Hawking himself has said. It's just the best he could come up with to describe in English something that really can only be described in the math.

 

[Dale]

I don't think they have negative mass. I think that when an antiparticle and particle come together they annihilate each other resulting in a zero mass system and dissipated energy.

Well, the system has the same mass before and after. This is confusing, but the mass of a system is generally more than the sum of the masses of its parts. In particular, a photon has no mass, but a pair of photons moving in different directions generally does have mass. When antimatter anhilates with matter the system of photons has the same mass as the system of the original particles.

This was the reasoning I used above. Is this incorrect?

I don’t know. Given your explanation I am not sure why you think that more antimatter needs to go inside in order to reduce the black hole mass.

 

[stevebd1]

What I am confused about is why would the black hole lose any mass at all. This relies on the premise that more often than not, the antiparticle will be the particle that appears on the inside of the horizon while the (regular) particle will be the particle to appear on the outside.
Intuitively, it would seem to me that 50% of the time, the antiparticle would appear on the inside while the other 50% of the time the (regular) particle would be the one to appear on the inside.
Why is it that the antiparticle is assumed to be the one appearing inside the horizon more than the (regular) particle?

The first law of thermodynamics states that energy can neither be destroyed or created, hence there are two scenarios when virtual particles are pulled apart near the event horizon. In the case of say an electron and an anti-electron (a positron) being pulled apart, if the positron was to be pulled into the bh and the electron to escape, then according to the first law, something would have to 'give', as the electron is now a part of our spacetime outside the event horizon, the bh would have to shrink in size by the equivalent mass. If the electron fell into the bh and the positron escaped, the positron would probably encounter another electron very quickly and both would annihilate resulting in gamma ray photons with energy equal to the combined mass of the electron/positron. This energy is now a part of our universe and again, because of the first law, the black hole reduces in mass. It's probably also worth pointing out that this is not necessarily a local effect and could be taking place some distance from the horizon (possibly near the photon sphere).