Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The new Jan 2014 Hawking paper (arXiv:1401.5761v1) asserts on page 3:

The absence of event horizons means that there are no black holes - in the sense of regimes from which light can't escape to infinity. There are however apparent horizons which persist for a period of time.

If light can always escape, then by simple geometry the only path by which light can be emitted from very deep within a black hole is straight up.

That is, I am reading Hawking's above statement as an assertion that when an object falls into a black hole, it will see watch its view of the outside world shrink to a smaller and smaller point-like angle. However, critically and in contrast to prior views, that final point will never completely disappear. An object swallowed billions of years ago thus will retain some incredibly tiny solid angle over which a sufficiently vertical photon will eventually reemerge into the outside world, at some vastly lower frequency due to gravity shift.

The size of this solid angle should be directly linked to the apparent temperature of the black hole to an outside observer. Here's why: A very large black hole, say a galactic center, will be like a super-hot but perfectly insulated furnace in which only an incredibly tiny porthole allows an exterior observer to see the interior. Since the apparent temperature will be the average of that tiny porthole over the entire furnace surface, the result is an apparent temperature that is appallingly close to absolute zero.

Conversely, a very small black hole in roughly the mountain-of-mass range will only barely have enough gravity to keep even photons emitted horizontally trapped in a circular orbit.

These would be Hawking's exploding black holes. The solid angle of emission would be nearly hemispherical, meaning that the hole has lost essentially all of its gravitational insulation. I'm not sure quantum mechanics is even needed to describe what happens next, since it would be like suddenly opening the side of a perfectly insulating chamber containing incredible densities of energy. So, boom, big time.

Incidentally, if matter falling into a black hole always retains even a trace of connection to the outside world, I'm pretty sure you can't just dismiss deep conservation laws based on symmetries. Hawking seems to imply something like that when he asserts on page 3 that:

The chaotic collapsed object will radiate deterministically but chaotically.

A "chaotic collapsed object" is not a pure energy object. So when a small antimatter-fed black hole loses its last gravitational insulation, I'm betting the resulting radiation burst signature would differ somehow that of a matter-fed black hole of the same mass.

One think I really like about Hawking's new black hole is that is seems to show a path for getting rid of the extremely paradox-prone distinction between singularity mass and event horizon mass. Here's just one small example: For a galactic center black hole, how exactly does "pump" mass-energy from the singularity back out to the event horizon, assuming that a reasonable and self-consistent location for that horizon can even be defined? If black holes evaporate, you have to do that somehow, and the mechanism is not very clear.

Hawking instead seems to be heading toward some form of reunification of interior of the black hole with the event horizon. In his new scheme, even as the interior becomes astronomically introspective and isolated, it never quite takes the final leap of cutting itself off from the apparent horizon at which the vast majority of photons fall back. My own guess is that a careful re-examination of Kruskal-Szekeres coordinates in terms of quantities that are forced to remain "in touch" with and meaningful to the outside world will provide some insights on that.

I would even wager based on these more symmetric ideas that small black holes need to be a lot "bouncier" than we give them credit for. For example, smaller ones may be nearly time symmetric when they when they form at the center of collapsing stars, and so can be dismantled at that stage almost as easily as they can be created. Hawking again seems to be deeply interested in exploring higher symmetries between collapse and re-radiation:

This suggests that black holes should be redefined as metastable bound states of the gravitational field. It will also mean that the CFT on the boundary of anti-deSitter space will be dual to the whole anti deSitter space, and not merely the region outside the horizon.

My own guess, a prediction I guess, is that the strange bounce seen at the last moment in the cores of collapsing stars is nothing more than a small black hole forming briefly and then just as quickly self-destructing. Why not? Small Hawking black holes already blow up, so this would just be a generalization of that concept to a larger range of masses, albeit with higher levels of time symmetries that make even larger holes unstable. Persistent black holes would form only if a threshold of mass is reached that allows gravitational insulation to dominate and keep the hole stable. The increased insulation would show up inside the holes as reduced solid angles of emission, and outside the holes as abrupt and precipitous drops in the apparent temperature of the surface of the hole. Such "too big to bounce" black holes would exhibit the equivalent of sudden and massive phase changes, transforming within a few milliseconds from very hot objects into very cold ones.

So, my question: Are my above conceptual interpretations of Hawking's paper reasonably accurate?


Related prior questions:

  1. Why does Stephen Hawking say black holes don't exist?

  2. Photon Escape Angle From Black Hole

share|cite|improve this question

An object swallowed billions of years ago thus will retain some incredibly tiny solid angle over which a sufficiently vertical photon will eventually reemerge into the outside world...

For a galactic center black hole, how exactly does "pump" mass-energy from the singularity back out to the event horizon... ?

... the apparent horizon at which the vast majority of photons fall back...

From these and similar statements throughout your question, it appears that you think that Hawking radiation has something to do with mass or photons traveling from inside the black hole to outside of it. Given that the escape velocity below the horizon is greater than the speed of light, massive particles cannot travel out; and given that massless particles must travel at $c$, they can only travel in a direction in which their speed will be $c$, which inside the horizon is always toward the center.

So there's no mechanism for Hawking radiation to result from anything traveling from inside the horizon out of it, and thus all the conjecture about solid angles of emission is a bit nonsensical. Once inside the horizon everything travels toward the singularity, and nothing travels away from it. There's no "incredibly tiny porthole [which] allows an exterior observer to see the interior."

So then, where does the energy for the Hawking radiation come from? It can't come from inside the black hole, since... well, that's black holes for you. What do we have to work with? Assuming an uncharged black hole as is reasonable in our universe, we've only got the angular momentum (which only changes the shape of the singularity, not the horizon) and the gravitational field.

Interestingly, this is where Hawking radiation (as a concept) comes from: the gravitational field is (for now) considered a classical field in our best theory, but if it is treated in a quantum manner, we can use the energy stored in the field to play the usual quantum tricks with virtual particles, which is where Hawking himself started on the problem.

So the idea is that the local gravitational field energy fluctuates enough that a particle pair is produced, and since the pair is produced near the horizon, one falls into the black hole and the other escapes, and the black hole loses that amount of energy permanently from its gravitational field, shrinking the gravitational attraction (and thus the horizon) by a little bit. The energy hasn't "come out of" the black hole because the energy wasn't in the black hole; it was outside the horizon, where the particle pair formed. But the change in the local field propagates outward and eventually an observer at some distance will notice the change - i.e. will "see the black hole shrink."

The local curvature just above the horizon of a small black hole is higher than outside the horizon for a larger black hole, meaning more energy is available and more particle pairs will form, and the black hole evaporates faster, allowing ever increasing access to the gravitational field energy, etc. But at no point in this process is anything coming out of the black hole - rather, the black hole is getting smaller and smaller, such that more and more is "outside" the black hole. The energy for Hawking radiation comes from the gravitational field which already extends beyond and across the horizon, and which effectively (in the classical theory) is the black hole.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.