Hawking radiation and black hole evaporation...?

Question

While I'm on the subject of theories I have a problem with... Hawking and I do not disagree on much, but if you know what the Casimir Effect or Hawking radiation are, you'll catch my drift here.

I'm not necessarily saying he's incorrect, but I have a problem with the concept. Or perhaps just a question that needs answering.

Why would Hawking radiation cause a black hole to evaporate? That seems to imply that antiparticles would fall inside the event horizon more frequently than the massive particle. Why would this be the case? Why would a black hole lose or gain any net mass as a result of spontaneous virtual particles? Or, on the contrary, wouldn't a particle with mass be more likely to be the one that passes the event horizon? Why wouldn't Hawking radiation result in black holes constantly getting heavier by shooting out streams of negative mass particles?

Answer 1

I don't quite understand the Casimir Effect, but I thought the concept of Hawking Radiation was quite simple. According to Quantum Field Theory, ordinary space is filled with 'vacuum fluctuations' in electromagnetic fields, which consist of pairs of photons being produced at one event and recombining at another. Such pairs violate conservation of energy, but if they last less than t = (h-bar)/(Energy produced), they violate no physical laws. So while the law of conservation of energy on a large scale works, it is constantly being violated on small scales. This happens just outside of the event horizon of a black hole too. So imagine a pair of photons is produced there, and one of them falls into the black hole before it annihilates with its pair. The left over one can now freely propagate through space, but the energy it carries violates the laws of physics. Since the black hole absorbed the other "virtual photon," it must pay for its partner's energy my losing some of its own. And as we all know, energy is the same as mass. Therefore, this is how black holes lose mass. If the black hole is losing more mass to this radiation than it gains from swallownig gas, dust, planets, etc., then it will eventually evaporate. Turns out that the time it takes for this to happen is proportional to the mass of the black hole (M) cubed, M^3. So bigger black holes would take longer than the current age of the universe to evaporate, while microblackholes may take a few million years. Does that help explain Hawking Radiation a bit better?



EDIT: There really is no simple or intuitive way to explain why it loses mass. Its in the mathematics of it all. I guess that extra particle has energy that should not exist, and the black hole loses energy/mass in order to balance out the Universe. But why doesn't the fallen photon add to the mass of the black hole? I don't know.



EDIT: In response to 'Harry' below, I was referring to energy conservation. This is a 'virtual photon' that is created, but energy still must be conserved if it exists for a time longer than what I mentioned. The photon is not conserved, its energy is.

Answer 2

If we accept that all particles are surrounded by a cloud of virtual particles whose existence is very short, then some interesting possibilities arise.



A strong electric field will separate the positively and negatively charged particles during the moment that they exist. If separated far enough they will not recombine and we will describe the event as pair creation.



Similarly a strong enough gravitatinal field will have a similar effect. Only this time the particle separation depends upon the spontaneous momentum of the particles rather than their charge. One of the particles will descend into the event horizon and the other will take off into space.



By this means the bh slowly evaporates, the rate of evaporation being said to increase the smaller the black hole.



All this is theory at present as Hawking radiation has never been positively observed.

Answer 3

Hmmmm.... in order for a charged particle to be emitted, a lot more energy needs to be lost by the black hole at once than is the case with a photon. Hawking radiation is therefor photons, not leptons. It's light with a black body spectrum. The anti-particle of a photon is... a photon.



You might want to understand first what you criticize. As of now you are talking absolute rubbish. The question is... is it intentional as in "troll" or do you really not know better?



By the way.. you are aware that the temperature of Hawking radiation for a typical black hole is MUCH lower than the temperature of the CMB? Right? It follows that the CMB actually "feeds" black holes and will continue to do so for eons. Net black hole evaporation will not be happening for a long, long time.



Misconceptions... that's all you got, so far.



:-)

Answer 4

Google vacuum energy The gravity of the black hole puts space-time in tension. Think of it like the elastic energy in a rubber sheet. From time to time that energy is released creating a particle - anti particle pair at the Event Horizon The formation of a pair is required by Conservation of Matter/Energy. One of the pair falls into the BH, the other particle escapes Conservation of Matter/Energy requires that the matter (particle that excaped) Be compensated for by a loss mass by the BH so that the total mass/energy in the universe is a constant

Answer 5

Particles and anitparticles both have the same mass, they just have opposing charges and spin.



In response to SVAL's observation, photons do not need to be conserved. Fermions are conserved, but not bosons (such as photons) I can create as many photons as I like at will by turning on my hallway light.

Answer 6

There are enough simplifications and assumptions in the theory to put it on shaky ground. I understand there is an experimental lab going into orbit in a few years designed to look for evidence of the effect.

Answer 7

It is an ENTROPY problem...and temperature is in the solution....



see here for a description and explanation...



.http://nrumiano.free.fr/Estars/bh_thermo.html

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