188

This is really a footnote to the accepted answer. Light cannot escape from an event horizon. But how can you check that light can never escape? You can watch the surface for some time $T$, but all you have proved is that light can't escape in the time $T$. This is what we mean by an apparent horizon, i.e. it is a surface from which light can't escape within ...


50

The expression for the power emitted as Hawking radiation is $$ P = \frac{\hbar c^6}{15360 \pi G^2 M^2} = 3.6\times10^{32} M^{-2}\ \text{W} = -c^2 \frac{dM}{dt},$$ where the term on the far right hand side expresses the rate at which the black hole mass decreases due to the emission of Hawking radiation. You can see that what happens is that the power ...


45

(The answers by Mark H and B.fox were posted while this one was being written. This answer says the same thing in different words, but I went ahead and posted it anyway because sometimes saying the same thing in different words can be helpful.) The key is to appreciate the difference between losing information in practice and losing information in principle....


38

At first many people didn't care much for black holes. But later people showed that they were pretty unavoidable features of the theory of general relativity and that theory made other quite precise predictions that were tested and found good. So when you are told that black holes are required if you have GR and GR looks like the best game in town then it ...


34

There are two ways to approach your question. The first is to explain what Brian Greene means, and the second is to point out that the "particles being swallowed" explanation is a metaphor and isn't actually how the calculation is done. I'll attempt both, but I'm outside my comfort zone so if others can expand or correct what follows please jump in! When a ...


28

To answer this we need to talk a bit about how particles are described in quantum field theory. For every type of particle there is an associated quantum field. So for the electron there is an electron field, for the photon there is a photon field, and so on. These quantum fields occupy all of spacetime i.e. they exist everywhere in space and everywhere in ...


25

The reason for many contradictory statements regarding the nature of virtual particles is that they are often invoked for heuristical explanations of phenomena that arise within the framework of quantum field theory. One then tries to justify those explanations by attributing certain properties to virtual particles they do not actually possess. What ...


24

...why do we trust black hole physics? ... (physics which is derived by combining quantum mechanics and GR such as Hawking Radiation, things relating to the Information Paradox, etc. ) Formally, there isn't quite a reason to because we've not observed these things yet. But that's also perfectly okay as well because that is how science sometimes works: we ...


23

There are a number of equivalent ways to think about Hawking radiation. One is pair creation, as endolith mentions, where the infalling particle has negative total energy and so reduces the mass of the black hole. Another way, perhaps more useful here, involves de Broglie wavelength. If the wavelength of a particle (not just photons, by the way) is greater ...


23

When Dr. Hawking talks about information being destroyed, he is talking about the erasure of all evidence that the information ever existed. In the case of burning a written letter, you could track the trajectory and composition of every smoke particle as the book burned. Since ink and paper generate different kinds of smoke and start at different locations, ...


18

Craig Feinstein asked: Does Stephen Hawking believe that General Relativity is wrong? Here is my answer (I will shift my answer there if some one reopen that question): Stephen Hawking did NOT say that black holes do not exist. Hawking used to think black holes are oblivious. Now he admits (like some other people do) black holes have perfect memory, just ...


16

@JohnDuffield: I can give you both a correct answer in simple terms and the fairy tale, together with references to an explanation how the fairy tale is related to the real thing! The dry facts are that two real particles (e.g., two photons, or an electron and a positron) are created from the energy in the very strong gravitational field near the horizon of ...


15

I don't know about extracting mass-energy from inside the black hole; it seems pretty inefficient to me. Hawking radiation isn't that powerful (as well as still being highly hypothetical), but building a Dyson sphere around a star is better. Remember, black holes, are, well, black. If there was enough Hawking radiation, they would cease to be black and would ...


15

The Newtonian gravitational acceleration for an object of mass $M$ is given by the well known expression: $$ a = \frac{GM}{r^2} \tag{1} $$ And the radius of the event horizon of a black hole is given by: $$ r_s = \frac{2GM}{c^2} \tag{2} $$ Suppose we calculate the Newtonian gravitational acceleration at the event horizon. Let's not worry whether this is ...


13

The paper pointed out by Daniel's comment gave me a starting point to find more literature on this topic and do further research. After a while, it became clear to me that my question is actually an unsettled (research) question. Therefore, a definitive answer cannot really be expected. Nonetheless, I think it's valuable to provide something of an answer. ...


12

With this answer, I am going to list some of my notes from the paper that @zephyr posted as a comment, http://arxiv.org/abs/0908.1803v1. Using Hawking radiation as a means of mass-to-energy conversion seems, in a word, absurd. The paper, however, addressed exactly that for use in powering a manned spaceship to the stars. Hawking Radiation as a thrust ...


12

Also, the book mentions that a negative-energy particle would appear to an observer inside the black hole as positive. Why? Very roughly speaking and in as simple terms as possible, inside the black hole, gravity is so intense that the time coordinate and one of the spatial coordinates (the radial coordinate) swap "roles". That's one way to see why you ...


12

Yes and no. Electrons - and all other elementary particles - may be viewed as microstates of very tiny black holes. As one considers increasingly heavy elementary particles (e.g. those in the Hagedorn spectrum of string theory), they increasingly morph into black hole microstates. When the elementary particle masses sufficiently surpass the Planck scale, ...


11

As you said, the case of black holes is conceptually totally analogous to the burning books. In principle, the process is reversible, but the probability of the CPT-conjugated process (more accurate a symmetry than just time reversal) is different from the original one because $$ \frac{Prob(A\to B)}{Prob(B^{CPT}\to A^{CPT})} \approx \exp(S_B-S_A ).$$ This is ...


11

Wouldn't this mean that empty space would have infinite energy? So ignoring quantum issues (and in the absence of a complete theory of quantum gravity we have no choice) and staying strictly with the classical approach let's consider the problem. Regardless of what amount of power is radiated by the black hole, that power is removed from the energy of the ...


11

Does this mean that from some point in time black hole just will cease to be a black hole because it won't be massive enough? No, once a black hole forms there's no turning back. It can lose mass via Hawking radiation, but (as far as we know) it cannot stop being a black hole until there's nothing left. There's no theoretical lower mass limit for a black ...


10

The conditions for the existence of the Hawking effect are described in classical terms, i.e you need 1) A Lorentz signature metric 2) A horizon (given, for example, by space flowing into a BH faster than the speed of light, or fluid flowing downstream faster than the speed of sound) 3) Surface gravity at the horizon Those conditions are then applied to ...


10

Yes, black holes are supposedly near-perfect black bodies. They emit thermal radiation called Hawking radiation, which, however, does not originate from beyond the event horizon, but is a consequence of the interaction of the strong gravitational field outside the horizon with the vacuum. The process is sometimes described as the production of 'virtual' ...


10

First, the Unruh and Hawking radiation aren't quite "the same thing". They have a similar origin and the Unruh radiation may be considered a flat space (large black hole) limit of the Hawking radiation. Now, the near-horizon metric of an extremal black hole is $AdS_2\times S^2$ while for a non-extremal one, the $AdS_2$ is replaced by the Rindler space. ...


10

Hawking radiation is a process that's always there when you have an event horizon. With black holes, the strength of this radiation is a function of its size: The heavier the black hole, and thus the bigger the event horizon, the colder the Hawking radiation. While the strength of the Hawking radiation approaches zero as you go to larger black holes, it ...


9

I'm kind of in your boat. Hawking radiation violates almost all of the energy conditions, and a stacked set of apparent horizons is two-way transversible when their area decreases with time. I see no reason why the typical assumptions like cosmic censorship should apply. And if cosmic censorship is gone, and the black hole is two-way transversible, then ...


9

Any non-mathematical answer is obviously going to be an oversimplification, but as long as you're happy with that here is my oversimplification. The temperature of a big black hole is lower than a small black hole because the curvature at the event horizon decreases with radius. It's the curvature, i.e. the bending, that determines the temperature - more ...


9

The black hole initially lost the gravitational energy that was needed to create the pair. The pair-creation model is a bad description of Hawking radiation, which for macroscopic black holes is really photons. The second particle that gets created above the event horizon doesn't have nearly enough energy to escape. It does, however, produce photons above ...


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