Hawking Radiation doesn't make any sense?

Question

Ok, this will probably be an elementary question to you physicists out there but:

Hawking radiation is supposedly radiation that comes out of a black hole. It is theorized by Hawking that due to the Heinsburg uncertainty principle, a pair of particles - a "particle" and an "antiparticle" can spontaneously appear in front of a black hole. Hawking theorized that if the "antiparticle" falls into the black hole and the "particle" escapes, then the black hole loses mass and this is how over time the black hole slowly evaporates.

What my question is that, wouldn't over time a similar number of "particles" fall into the black hole as "antiparticles" therefore maintaining an equilibrium therefore not allowing the black hole to lose any mass? (In fact, wouldn't a higher number of "particles" fall into the black hole than "antiparticles" due to the fact that "particles" have mass and therefore wouldn't the black hole gain mass with Hawking radiation overtime if more "antiparticles" were emitted?)

I'm sure it makes perfect sense in Hawking's proof but I'm young and haven't yet dwelled beyond school mathematics.

Answer 1

You have the right idea, but your specifics are wrong.



It's not a "particle/antiparticle" pair; rather, it's an "existence/antiexistence" pair. Quantum mechanics allows particles to pop into existence "ex nihilo" (from nothing), but only for an extreeeeeeeemely short time (the so-called Planck time). They pop into existence along with another "antiexistence particle". They then pop right back out of existence, UNLESS the "antiexistence particle" meets up with an actually existent particle and annihilates it.



Now, suppose one of the existence/antiexistence pairs pops up on a black hole's event horizon:



If the existence particle is inside the event horizon, it gets sucked in. However, the antiexistence particle does not (we can assume) meet up with any existing particle, so they both pop right back out of existence after an extreeeeeeeeemely short time.



Suppose instead that the antiexistence particle is inside the event horizon. If the existence particle outside the event horizon has sufficient energy, it can escape the black hole's gravitation. The antiexistence particle, being inside the black hole, will of course be "suck in" to the singularity and will reduce the mass of the black hole by annihilating mass exactly equal to the particle that escaped.



Btw, this is called "evaporation" of a black hole. Obviously, I haven't given you any sort of proof, just a description of the idea. If you want to research it more, get into a graduate course on quantum mechanics.

Answer 2

You have the theory correct, but some of your facts are wrong. Antiparticles have the same mass as their paired particle.



Quantum mechanics states that energy and matter are in a continual dance of the energy collapsing (for lack of a better word) to create a particle and antiparticle pair, and then the particles smashing into each other again to turn back into energy. This "dance" is happening throughout the universe, even in what we consider empty space (vacuum fluctuations).



Energy, particles, and antiparticles are continuously being created and destroyed inside of the black hole. The only way for a black hole to loose mass is when this occurs near the event horizon (the edge where gravity becomes to strong for light to escape from the black hole). When energy near the edge creates a particle/antiparticle pair, quantum mechanics allows for one (or both) to materialize just beyond the event horizon and escape the black hole, taking its mass with it.



This is a very, very slow process by our time-line. In the course of a lifetime, there will not be a measurable difference in the mass of a black hole. When speaking of a cosmological time-line (hundreds of thousands to several million years - a blink of an eye to the universe), a black hole can theoretically dissipate into nothing.

Answer 3

The particle that escapes from the black hole, no matter whether it is the electron or the positron, carries off a rest energy that is above the ground for the gravity field at the event horizon where the pair of antiparticles appeared.



The particle that falls into the black hole, again no matter which it is, has less energy than the gravitational ground at the event horizon. So the black hole is constantly soaking up particles that have less energy than the vacuum they were made in; i.e., negative energy particles.

Answer 4

Anti particles have mass. So even if it's the anti-particle the leaves the black hole, the black hole still gets reduced.



Many of the particles that are radiated are in fact photons. The antiparticle for a photon is a photon. And both photons can be emitted, i think.

Answer 5

Good question. I'm by no means an expert on this, but this is my guess:



1st, antiparticles (http://en.wikipedia.org/wiki/Antimatter) DO have mass. Just because they have a cool name doesn't mean they're not matter. They just happen to have the property that when it collides with ordinary matter, they annihilate eachother. So its perfectly reasonable for the gravity of a black whole to pull in antimatter.



2. (This is where I'm guessing), I think two particles you speak of are radiating from the black hole itself-- so even if the matter particle you speak of is the one that gets sucked in, the black hole does not grow at all, because the particle came from the black hole to begin with.

Answer 6

No, the problem is the energy balance. These virtual particles are created and destroyed quickly, so the energy balance is usual zero - no energy is created or destroyed.



When the black hole absorbs one of these particles, it breaks the symmetry (symmetry is a very important thing in quantum physics). As the energy balance has to be kept zero (as otherwise the universe would slowly gain energy from nowhere), the energy has to be spent by the black hole, which is guilty of disturbing the balance.

Answer 7

it relatively is because of fact of capability conservation. The particle which escapes is created and given an get away velocity on the cost of the bh capability (or mass, as you like) it quite is consequently decreased. there's a somewhat common intuitive answer to the exciting question posed via Deltafan', which desires no SCiFi like pseudo techniques. What facilitates the separation and the materialization of the digital pair created from the vacuum interior the region of the black hollow horizon is the gradient of the gravity container. Now this gradient varies as a million/(length of the horizon), it quite is nearly a million/(mass of the black hollow) and consequently that's exceedingly better for a microscopic black hollow than for a macroscopic one (yet in a infinitesimal area of area however!) , thereby taking into consideration speedy production energetic radiation and rot. approximately detrimental mass debris.. the two you communicate approximately that a pair of constructive mass debris (entire 2m) is created close to the horizon on the cost of the bh capability. One escapes to infinity, the different one falls lower back in==> loss of mass = m or you communicate approximately that a pair, one with >0 and one with <0 mass is created (entire m-m =0) at no cost now, m escapes to infinity, -m falls lower back interior the bh ==> loss of mass = m So that's precisely equivalent. No SF, no secret. truthfully detrimental hundreds are a historic coincidence that would desire to be put in the museum now.

Answer 8

It doesn't matter which particle escapes.

Both have mass. Both will annihilate one of their anti-particles.

What doesn't 'make sense' is the appearance of the particle

pair in the first place, but theory does hold that it happens.

Answer 9

Aliens excluded, no one can correctly answer as no Earthling has all the facts. The theories put forth by Hawking and others are the best guesses as to what is happening according to what we see. We will not be able to do better even if we could get closer as our whole Solar system would be drawn into the black hole.

Answer 10

It's photons that are the most important Hawking Radiation, and they don't have anti-particles.



Also, both anti-matter and matter have positive mass.