How scientific.
there are no black holes with event horizons
If you take the mass of a sun and squeeze it into a few miles wide, you are definitely going to get inescapable gravity no matter what you want to call it.
Now, it may be fair to label this as sensationalism anyway. The source here is a brief, conceptual talk by Hawking at a conference (and an associated writeup): it's not as if he's proven a theorem (as he did years ago when he first demonstrated Hawking radiation, for instance). But honestly, it may be newsworthy that Hawking is even considering this sort of resolution to the current "firewall" debate. It would be one heck of a change in perspective!
the event horizon is frequently treated as "uncrossable" from inside to outside which is really different from "inescapable". The later means that whatever speed you have on or below the event horizon, you can't reach infinity. The former is just an impression by an outside observer because in his observation the time has stopped on the event horizon, while a stone thrown up from below the event horizon would cross the event horizon just fine on the way up and on the way down and would return back down successfully in its proper time (the point of "inescapability" is that the stone would always return). While above the horizon, the stone can interact with other stuff there and result of this interaction can be observed outside (doesn't mean on practice by us today or tomorrow :).
Many models seems to treat the event horizon as "uncrossable". For example quantum information disappearance - matter goes in, evaporates as Hawking. Yet, just for example, when Hawking radiation decreases the mass of the black hole it causes the shrink of the event horizon and thus whatever "stones"/photons on their way up were stuck (for external observer) in the stopped time of the horizon become free - doesn't mean though that we can observe them on practice as getting out of that gravitational well does take time (again in our time) and redshifts them into oblivion. Like proverbial "the check is in the mail". It is the reason why we can't really observe Hawking radiation until we develop technology to observe light with extremely long wavelength and i just don't have the numbers right now on whether 13B years is enough for the radiation originating right above the event horizon to get out of that well and reach the interstellar space.
This doesn't agree with my understanding of GR. A stone "thrown up" from within a black hole interior cannot cross the event horizon in any reference frame - it cannot even get closer to it.
Look at the future light cones within the black hole interior, e.g. in the illustration at http://en.wikipedia.org/wiki/Eddington–Finkelstein_coordinat.... The future light cone of every event within the EH is skewed so far that even light rays directed outwards are drawn closer to the singularity.
The Schwarzschild radius of the mass of the observable Universe is 10B light years. So, several billions years ago, when observable Universe had 10B radius, it would be a black hole (though i think that in less expanded space of the earlier Universe the constants like "c" had different values and thus that Schwarzschild radius was less). We can imagine it in another way - increase 125 times (the observable Universe has 40+B light years radius) the amount of matter, ie. galaxies, stars, etc... inside the 10B radius ball around us, and you'd get the black hole with 10B light years Schwarzschild radius (and i don't think we would ever notice the change - only with time the galaxies's movement will be affected). Obviously, the gravitational field on the surface of that imaginary 10B radius sphere and inside it would increase somewhat (like 125 times on the surface, 125 times 0 pretty much 0) - not even close though to any values to affect space curvature or to prevent anything from crossing it from inside. Of course, anything that would cross it from inside would return back eventually.
No, it can't; the "acceleration due to gravity", which is the acceleration required to "hover" at a constant altitude above the horizon, diverges as the horizon is approached.
What can be made as small as desired by making the hole's mass large enough is tidal gravity at the horizon.
several billions years ago, when observable Universe had 10B radius, it would be a black hole
No, it wouldn't. A black hole is a stationary spacetime. The spacetime of the universe is not stationary. The spacetime model that describes the universe is very different from the spacetime model that describes a black hole; the fact that you can plug the mass of the observable universe into the Schwarzschild radius formula and get a number out does not mean that number has any physical meaning for the universe.
exactly. There is no stationary spacetime in our Universe. Black hole is pretty artificial model where pure mathematical artifacts of singularity at the horizon is taken for the real thing.
> the fact that you can plug the mass of the observable universe into the Schwarzschild radius formula and get a number out does not mean that number has any physical meaning for the universe.
Taking a big chunk of space and calculating escape speed from its gravitational field - how is that not a physical meaning? At what specific size of the chunk you think it becomes not meaningful?
The universe as a whole is not stationary, nor even close to being so; but portions of the universe are very close to being stationary. The solar system, for example. Black holes do not have to be exactly stationary; if they are as close to being stationary as the solar system, that's plenty close enough.
Black hole is pretty artificial model
The exactly spherically symmetric solution is an idealization, yes; but there are plenty of numerical simulations that show that non-symmetric spacetimes still form event horizons.
where pure mathematical artifacts of singularity at the horizon is taken for the real thing.
There is no physical singularity at the horizon. Some coordinate charts have a coordinate singularity at the horizon, but that's easily fixed by just using a different coordinate chart. The only physical singularity is at r = 0.
Taking a big chunk of space and calculating escape speed from its gravitational field
Escape to where? You can't escape from the universe as a whole. The concept of "escape speed" has no meaning for the universe as a whole.
take a 1B light years radius ball, populate it with density of our Milky Way - that ball will have 1B Schwarzschild radius. Calculation of gravitational potential (escape speed) from it to the rest of the Universe makes sense, doesn't it?. And due to this gravitational potential it will be bona fide black hole from the point of view of the rest of the Universe.
Yes, but that's not the same as escaping from the universe as a whole.
And due to this gravitational potential it will be bona fide black hole from the point of view of the rest of the Universe.
So what? What does that have to do with assigning a gravitational potential to the universe as a whole?
The Schwarzschild radius does have the property that, if you plug it into the classical equation for escape velocity, you get the speed of light. That doesn't mean arbitrary classical analogies (like the "thrown stone" example) hold.
In particular, consider that you can escape the earth while never achieving escape velocity. Just keep firing your rockets to counteract gravity. But if this were possible with black holes, then they would not be very interesting!
> The various local effects like time dilation, light path curving, etc... are defined by the value of gravitation force, not gravitational potential.
These are not local effects! Locally, there is no time dilation, and no curvature of light. I see my watch tick at the same rate, no matter where I am, because my watch and my eyes are in the same local reference frame.
In order to see effects like time dilation or curvature of light, you must compare events separated in spacetime. For example, we can look at the paths of light emitted by distant stars. And since the light had to get to our eyes, we have to account for the entirety of the path that it took. So the time dilation I observe for an event depends on the entirety of spacetime along the path from the event to me, not just the local curvature for the event.
In mathematical language, if you want to move a vector from one point to another on a curved manifold, you can't just specify the start and end point. The path you take affects the resulting vector.
> The potential defines the fact that anything originating at or below the horizon would never escape completely the gravitational field, i.e. never reach the infinity.
So objects originating outside the event horizon reach infinity in a finite time? Huh?
Nor mine. You're correct, any object in the interior of the hole, even one that is moving radially outward at the speed of light, gets closer to the singularity (i.e., its r coordinate decreases) with time, which means it can't get closer to the horizon (that would require increasing r).
I always wondered if a black hole could become completely visible again if it collected enough mass to spread its surface area out so that gravity would become thin enough for light to escape again. Kind of like how only the surface of a pond sometimes freezes.
(Of course, singularities don't make any sense - which is part of what Hawking's talking about. But that's the classical model of a black hole)
One certainty is that the space inside a black hole is bigger than the size of the black hole observed from outside of it. So it's entirely possible that you could have a solar system or even an entire universe on the inside. According to GR the space inside a black hole should pretty much behave like the rest of the universe, as long as the black hole is growing. As for "not getting out", well it's impossible to approach the edge of the universe too, despite that border having a physical location. But it's expanding at light speed, so ... no way to get there.
So density of a black hole is an entirely different concept from the outside and the inside. Seen from the outside it is infinite. Seen from the inside it is probably not infinite.
That is not a certainty at all. The entire premise of Hawking's paper is that the object we call a 'black hole' is just a really degenerate dense lump that slowly turns incoming matter/energy into background radiation with as much entropy as possible. And that's it. No fancy Time Lord bigger-on-the-inside science proposed here.
No, it doesn't. There is no edge to the universe. Our best current model is that the universe is spatially finite, but even if it turns out to be spatially finite, it is unbounded--the spatially finite model is a 3-sphere; like the surface of the Earth but with one more dimension. The surface of the Earth is finite, but it has no edge; a spatially finite universe would be the same way but with one more dimension.
Seen from the outside it is infinite.
No, it isn't. Seen from the outside, the hole has a finite mass in a finite volume, so its density is finite. (The finite volume an outside observer would assign, based on the surface area of the hole's horizon, is not the "actual" volume of the hole; but it works well enough as an "effective volume", the volume that the hole occupies from the standpoint of the rest of the universe.) The singularity can be viewed as infinitely dense, but the singularity does not occupy the entire black hole; it only occupies the center at r = 0.
But the horizon isn't a point; it is in fact a surface (a 3-surface in 4-dimensional spacetime). Whether you want to call that the "surface" of the black hole is a matter of terminology; the hole is not a solid object--it's not really an "object" at all in the ordinary sense of the term, so it doesn't have a "surface" in the ordinary sense of the term.
It's pretty neat, actually: if you had a black hole containing the mass of a whole galaxy (which isn't unthinkable, especially in the distant future), local space-time at the event horizon would seem completely ordinary (at least as far as we understand from classical relativity and semi-classical quantum corrections like Hawking radiation). You wouldn't have the foggiest idea that you had crossed that fateful line... until you caught on to the fact that you weren't actually making any progress when you tried to turn back.
I am not a point mass. If I can cross the event horizon and not know it, then there is a time when my feet are on one side and my head on the other. If signals cannot cross the event, how would I not notice my missing feet? If signals from my doomed feet can reach my not yet doomed head, then how is it the event horizon?
Basically, you can change your reference point and consider that the signal is not moving, but your head is still moving towards the signal. This way, you're still seeing your legs. (Note that I have no idea what I'm talking about)
Now, if you were wearing some sort of rocket collar that fired powerfully just before your head crossed the horizon and carried it safely away from the black hole (or even just held it there), you're right: you wouldn't ever receive that signal from your feet. But I'm pretty sure that would be the least of your worries! (And you would notice your missing feet. And torso.) Roughly speaking, the only non-orbital trajectories close to an event horizon that don't fall straight in must be moving pretty close to light speed as seen by infalling observers.
So in your example, for a large black hole, you will experience no difference in the rate of tingling as you cross the event horizon. This is what the principle of relativity tells us: you are in free fall, so the laws of physics must appear the same to you.
This is a little bit weird, I know!
It's wrong. Time dilation is not something you observe yourself to experience; it's something another observer observes you to be experiencing. "Time" is relative, so you can feel your own time to be ticking away perfectly normally while it appears to be ticking much more slowly to someone else.
As your feet approach the event horizon, the tingling will get slower and slower, because time passes much faster for your head than for your feet.
No, this is not correct. As you fall toward and through the horizon, signals travel between your head and your feet just fine, and you don't notice any difference in how fast they travel. But, as I posted upthread, if your foot emits a signal towards your head when it's below the horizon and your head is above it, then by the time the signal reaches your head, your head will have fallen below the horizon.
BTW, it's not hard to convince yourself that masses and springs in a centrifuge behave exactly like those near a large body. (In Newtonian mechanics even.) How are you sure you know the difference?
If your feet are below the horizon and your head is above it, then you must be falling through the horizon; it's impossible to remain at a constant altitude at the horizon, even for an instant. So a signal emitted by your foot would still reach your head; but by the time it reaches your head, your head will have fallen below the horizon.
3.8 miles (1.9 mile Schwarzchild radius), to be a bit more precise.
'No black holes' is good enough for me. Now, Hawking's been chipping away at these ideas for a long time - remember his essay 'black holes ain't so black' - but this is indeed a radical departure. Cue fooljhardy adventurers dashing off in search of black holes and an outbreak of highly similar sci-fi plots. /opens Final Draft
Astronomical. Astrological would suggest it depends which star sign the masses are in.
Neutron stars are in a sense "failed" black holes because of not enough mass. I wonder if they could eat more mass and achieve that status though.
https://news.ycombinator.com/item?id=7104056
Basically it's unknown if it is a two star system where the neutron consumes the partner, or a three star system where two neutron stars collide head on.
http://en.wikipedia.org/wiki/Tolman–Oppenheimer–Volkoff_limi...