# Tag Info

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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. ...

13

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 ...

6

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 ...

6

Such a process is forbidden by energy conservation: the proton is the lightest baryon (that is the lightest bound state of three quarks). hawking radiation finds it's energy by reducing the energy of the black hole, but there is not lighter baryon state for the proton to go to. Baryon number violating proton decay processes are theorized, but have not been ...

5

We are already living in a nearly empty de Sitter space - the cosmological constant already represents 73% of the energy density in the Universe - and the Universe won't experience any qualitative change in the future: the percentage will just approach 100%. However, once the space may be approximated as an empty de Sitter space, all moments of time are ...

5

Q1 - The black hole spacetime sees actual particles in the "static" frame because the diminishing curvature across the black hole spacetime continuously connects the static frame at infinity - which is really static and corresponds to freely falling, static observers - with the static observers who otherwise live in a strong gravitational field near the ...

5

There's a topology change in the spacetime when you complete the $M\rightarrow 0$ limit--The Minkowski spacetime is non-singular, while the Schwarzschild spacetime contains a spacetime singularity for any (positive) value of $M$. You could also look at it in terms of the total energy available for a small mass BH is going to be miniscule, and you would ...

5

Actually working out a rigorous prediction for Hawking radiation, you need to solve the equations for the relevant QFT semi-classically--treating the metric of spacetime as a substitute for the Minkowski metric usually used in QFT. This is similar to deriving the quantum mechanics of the Hydrogen atom without paying attention to the fact that photons exist ...

5

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 ...

5

Radiation normally contains subtle correlations. For all practical purposes you can't use it, but it's there. Hawking radiation is, according to the theory, perfectly thermal and does not contain any more information than the temperature itself. The problem is that then the process of black hole evaporation is not reversible, in principle. Unlike all other ...

5

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 ...

5

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 ...

4

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 ...

4

I don't have a full answer to your question, but your question did make me realize that the Hawking radiation must be carrying away angular momentum from a rotating Kerr black hole. If it only carried away energy (mass) the Kerr black hole would become extremal as it lost mass but kept the same angular momentum. This, I think, proves that there has to be ...

4

I think your main question is how to reconcile the infinite time-to-reach-horizon needed for a particle created just outside the horizon with the fact that virtual pairs only exist for a short time before re-annihilation? As you've correctly pointed out, the virtual pair picture is only a heuristic (as Hawking said in his original paper), a more ...

4

Physicists are tasked to construct models of reality that are internally consistent and that have predictive power. Working on black holes and Hawking radiation fits this bill. The growing theoretical framework on black hole thermodynamics is absolutely essential to eliminate any internal inconsistencies that seemed to affect thermodynamics when applied to ...

4

First of all, in lower dimensions (2+1 and 1+1) the gravity is much simpler. This is because in 3d curvature tensor is completely defined by Ricci tensor (and metric at a given point) while in 2d curvature tensor is completely defined by scalar curvature. This means that there are no purely gravitational dynamical degrees of freedom, in particular no ...

4

Imagine how a black hole is formed. According to the most prevalent theory most black holes start out as supermassive stars - Much larger then our sun. As they run out of fuel to support their own weight they go through a series of implosions. The shock waves from those implosions throw off the outer shells of material into space while the core collapses ...

3

First of all I do not think that conservation of information is an established statement. It seems to be an open problem still as far as black holes go. Even if true, it is a different type of conservation, analogous to the unitarity requirements of a system of functions or phase space considerations. From the conclusions of a paper by Hawking : In ...

3

Yes, P will observe a thermal radiation that is locally identical to the Hawking radiation and it is called the Unruh radiation. Its temperature is $a/2\pi$ in the $\hbar=c=k_B=1$ units. http://en.wikipedia.org/wiki/Unruh_radiation A few more words about the relationship of Unruh and Hawking radiation. Unruh found his derivation after Hawking found ...

3

It is simply not true that a non-vanishing stress tensor is incompatible with asymptotic flatness. The Schwarzschild spacetime is asymptotically flat, period. The semiclassical Hawking-calculation does not in any way change this background, unless you consider effects from backreactions. Once you take into account backreactions the Hawking radiation does ...

3

Dear D-brane, indeed, a uniform thermal radiation would curve the Universe. Even if one doesn't immerse the black hole in a thermal bath, the outgoing Hawking radiation may violate the asymptotically flat conditions at any finite time, although just mildly. However, an evaporating black hole that is not surrounded in the thermal bath ultimately evaporates ...

3

Well, here is a simple scenario : The quantum field theoretic vacuum is seething with pair production of virtual particles everywhere. Take one such virtual pair at the horizon of the black hole. One part of the pair if it is going down is grabbed by the hole and disappears while the other is with its high momentum runs away from the horizon, on shell. ...

3

The lower the black hole mass, the higher the temperature of the Hawking radiation. This prediction is based on approximations that seem warranted as long as the black hole has mass $M >> M_\text{Planck}$. No-one knows what precise phenomena occur when the black hole mass approaches $M_\text{Planck}$. Drawing conclusions on the $M \to 0$ limit based on ...

3

My previous answer is beside the point now that the question has been edited. There is a simpler example of a question of this type which has been analyzed in great detail in the literature, and that involves an electric charge which is stationary in a gravitational field. Since the power radiated is nonzero if a charge is accelerated one might, by the ...

3

Black hole has no knowledge of what kind of matter went in. As long as it went in it disappeared in the black hole. To create particles, it needs to fulfill conservation laws, which is not easy and black hole produces generally photons. However, take into account that evaporation of a black hole is extremely slow. For a black hole of reasonable size this ...

3

Nice question! I'll try to put together a few ideas into what might be a valid answer, but this could be wrong. The event horizon is a lightlike surface. Therefore no proper time passes at the horizon, and any photon emitted exactly at the horizon would remain exactly at the horizon until the moment when evaporation was complete, after which it would fly ...

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I remember attending a seminar by Unruh a few months ago and the same question arised. As far as I remember, he enfasized that in these hydrodynamic analogs of black holes, the flow is not quantized, it is a classical fluid, and everything is classical and that the dumb hole behaves like a quantum amplifier emitting quantum noise from the Horizon. ...

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