# Tag Info

158

This is really a footnote to Adobe's 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 a ...

47

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 ...

31

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 ...

23

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 ...

21

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 Compton wavelength. If the wavelength of a particle (not just photons, by the way) is greater than ...

21

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 ...

19

...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 ...

17

A few things: 1)Just because an observer crossing the event horizon doesn't necessarily feel ill effects AT THE TIME OF CROSSING the horizon, it doesn't mean that they won't inevitably end up at the singularity, where there will be plenty of ill effects--all timelike curves that cross the horizon end up at the singularity in a finite amount of proper time. ...

16

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 ...

12

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 ...

12

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 ...

11

To add to Rory's answer- The ability to radiate particles in a random, statistical way, is in a deep sense identical to an object having the property we know as "temperature." So, black holes have a temperature. It has a particular formula that is inversely proportional to the mass of the black hole. If you set that temperature equal to the current ...

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 ...

9

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 ...

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 ...

8

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 ...

8

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 ...

8

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' ...

8

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, ...

8

Unlike most objects, a black hole's temperature increases as it radiates away mass. The rate of temperature increase is exponential, with the most likely endpoint being the dissolution of the black hole in a violent burst of gamma rays. A complete description of this dissolution requires a model of quantum gravity, however, as it occurs when the black hole ...

8

Hawking radiation is such a miniscule effect we can be sure we'll never detect it for a real astrophysical black hole. The Wikipedia article gives some numbers: For a black hole of the mass of the Sun, the power emitted in Hawking radiation amounts to $9\times10^{-29}\ \mathrm{W}$. Even if all this energy were converted to visible-light photons ($4\times10^{... 8 Any black body in space radiates and ends up very cold, might even crystallize. The law of increasing entropy holds for closed systems, in this case the whole system: "all the radiation that left the black body + the black body itself" microstates. In the sense that a black hole behaves as a black body the same holds true, it cannot be considered a ... 7 Hawking radiation is a very robust prediction. It comes simply from applying quantum field theory in the curved space-time near the event horizon. It's also part of the synthesis called "black hole thermodynamics", for which string theory provides an explanation in terms of the statistical mechanics of microstates. In the S-matrix of quantum gravity, if ... 7 I assume you're asking how a black hole can evaporate due to Hawking radiation. The answer is that the Hawking radiation does not come from the event horizon, but instead comes from a region just outside the event horizon so time has not stopped at its position. If you were to watch a black hole form then evaporate, you would never see an event horizon form.... 7 In the case of Hawking radiation, the direct answer is "no, there is no direct test, nor can we imagine one with anything like current technology." But it is not some wild speculation made in vacuum. The extremely closely related Unruh effect can be derived from basic quantum field theory on a curved spacetime, and many QFT and GR texts have at least an ... 7 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 ... 7 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. ... 7 Black holes are basically neutron stars with such a gravitational force that even light cannot escape from it. A black hole is a mathematical solution. A neutron star over the critical mass gets so dense that it forms larger and larger time dilation relative to the outside universe thus we get to see what happens on short time scales. But what causes ... 7 Perhaps the most direct example in particle physics is$J/\Psi$(or any other meson not including up and down quarks) production. The meson has a valence content of$c\bar{c}\$, so it represent a pair of particles knocked on-shell from the nucleon sea. The reaction is not exactly analogous because it requires a rather large input of kinetic energy (as does ...

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