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So basically they took off three words of what he said to make a super sensationalistic headline? How scientific. there are no black holes with event horizons

by ck2 13y ago

So basically they took off three words of what he said to make a super sensationalistic headline? 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.

VikingCoder 13y ago
That's the thing though - Hawking is saying it's not "inescapable."
- ck2 13y ago
  I think he is saying the gravity is not even enough to have that effect and therefore would be points that are escapable? 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.
  - lmm 13y ago
    No; adding mass doesn't make it less dense (well, it's a singularity, so it's always infinitely dense. But even if it weren't, normally when you add mass to something it becomes more compressed and gets denser, not less dense), just heavier. (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)
    - waps 13y ago
      It's only necessarily infinitely dense from an outside reference frame. Inside you could have anything you'd want according to GR. 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.
      - fennecfoxen 13y ago
        > 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. 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.
      - pdonis 13y ago
        it's impossible to approach the edge of the universe too, despite that border having a physical location. 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.
  - panzi 13y ago
    What surface? A point has no surface.
    - rosser 13y ago
      The event horizon isn't the surface of the black hole. As classically defined, the event horizon is the point at which the black hole's gravity becomes inescapable. That's not at its surface.
      - panzi 13y ago
        Exactly.
      - pdonis 13y ago
        The event horizon isn't the surface of the black hole. As classically defined, the event horizon is the point at which the black hole's gravity becomes inescapable. 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.
  - Steuard 13y ago
    The "surface area" of a black hole is defined precisely as the area of the boundary beyond which light cannot escape, so I don't this angle will work the way you're thinking it might. (It's fairly straightforward to show that that surface does indeed expand as more mass is added, too.) 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.
    - saalweachter 13y ago
      This has always bothered me. 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?
      - rca 13y ago
        What if the signals emitted from inside the horizon do reach your head, but only once your head have crossed the 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)
      - Steuard 13y ago
        Well, your feet aren't ever "missing": as far as classical GR is concerned (even with Hawking's semi-classical quantum corrections included), your experience would be perfectly comfortable and unexceptional as you drifted in. The issue is that in that scenario, your head isn't on an escaping trajectory: the nerve signals from your feet might be moving "outward", but your head will be moving inward, and the signal will only get to your brain after your brain has also crossed the horizon. 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.
lmm 13y ago
By that logic neutron stars should be black holes. But they're not.
- ck2 13y ago
  Sorry I didn't mean the sol sun, I mean a sun with enough mass to become a black hole. 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.
  - venomsnake 13y ago
    That is one of the standard candles - type 2 supernova IIRC - when neutron star in binary system is eating the other star and turns nova very predictably.
    - larubbio 13y ago
      I think it's type 1a. There was a discussion on this recently after a new supernova observation. https://news.ycombinator.com/item?id=7104056 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.
    - splat 13y ago
      The standard candles are Type Ia supernova which are when white dwarfs exceed the Chandrasekhar mass and become neutron stars. It is thought that when neutron stars exceed the maximum mass of a neutron star (which is not well known, by the way) they form a gamma-ray burst.
      - kevinastone 13y ago
        The TOV Limit for those following at home. It's when the star's gravitational pressure exceeds neutron degeneracy pressure and collapses into a black hole. http://en.wikipedia.org/wiki/Tolman–Oppenheimer–Volkoff_limit http://en.wikipedia.org/wiki/Tolman–Oppenheimer–Volkoff_limi...
Crito 13y ago
> 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. 3.8 miles (1.9 mile Schwarzchild radius), to be a bit more precise.
Steuard 13y ago
This isn't really all that sensationalized, based on Hawking's suggestions in question. The presence of an event horizon is quite often used as the defining feature of a black hole. The event horizon is the "point of no return" beyond which gravity is inescapable. If Hawking is claiming there is no definable event horizon (due to turbulence or whatever else), then he's claiming that there is no set of points near a dense, massive object that are inescapable. 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!
- VladRussian2 13y ago
  >The event horizon is the "point of no return" beyond which gravity is inescapable. 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.
anigbrowl 13y ago
In its stead, Hawking’s radical proposal is a much more benign “apparent horizon”, which only temporarily holds matter and energy prisoner before eventually releasing them, albeit in a more garbled form. '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
- codeflo 13y ago
  It's partly a question of definitions. There's an astrological phenomenon that occurs when large masses collapse, and we believe there is one of those at the center of the Milky Way. This exists in nature, we just haven't definitely figured out how they work yet. We currently call those "black holes". If a new theory suggests that these holes are not so black, but really rather "grey holes" or something, we might adopt a new word for those things, or we might change our definition of black hole to match the current theory.
  - summerdown2 13y ago
    > There's an astrological phenomenon that occurs when large masses collapse Astronomical. Astrological would suggest it depends which star sign the masses are in.
    - codeflo 13y ago
      I'm stupid. Of course, thanks.
nathancahill 13y ago
"There is no spoon."

by ck2 13y ago

VikingCoder 13y ago

That's the thing though - Hawking is saying it's not "inescapable."

ck2 13y ago
I think he is saying the gravity is not even enough to have that effect and therefore would be points that are escapable? 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.
- lmm 13y ago
  No; adding mass doesn't make it less dense (well, it's a singularity, so it's always infinitely dense. But even if it weren't, normally when you add mass to something it becomes more compressed and gets denser, not less dense), just heavier. (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)
  - waps 13y ago
    It's only necessarily infinitely dense from an outside reference frame. Inside you could have anything you'd want according to GR. 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.
    - fennecfoxen 13y ago
      > 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. 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.
    - pdonis 13y ago
      it's impossible to approach the edge of the universe too, despite that border having a physical location. 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.
- panzi 13y ago
  What surface? A point has no surface.
  - rosser 13y ago
    The event horizon isn't the surface of the black hole. As classically defined, the event horizon is the point at which the black hole's gravity becomes inescapable. That's not at its surface.
    - panzi 13y ago
      Exactly.
    - pdonis 13y ago
      The event horizon isn't the surface of the black hole. As classically defined, the event horizon is the point at which the black hole's gravity becomes inescapable. 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.
- Steuard 13y ago
  The "surface area" of a black hole is defined precisely as the area of the boundary beyond which light cannot escape, so I don't this angle will work the way you're thinking it might. (It's fairly straightforward to show that that surface does indeed expand as more mass is added, too.) 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.
  - saalweachter 13y ago
    This has always bothered me. 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?
    - rca 13y ago
      What if the signals emitted from inside the horizon do reach your head, but only once your head have crossed the 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)
    - Steuard 13y ago
      Well, your feet aren't ever "missing": as far as classical GR is concerned (even with Hawking's semi-classical quantum corrections included), your experience would be perfectly comfortable and unexceptional as you drifted in. The issue is that in that scenario, your head isn't on an escaping trajectory: the nerve signals from your feet might be moving "outward", but your head will be moving inward, and the signal will only get to your brain after your brain has also crossed the horizon. 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.

lmm 13y ago

By that logic neutron stars should be black holes. But they're not.

ck2 13y ago
Sorry I didn't mean the sol sun, I mean a sun with enough mass to become a black hole. 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.
- venomsnake 13y ago
  That is one of the standard candles - type 2 supernova IIRC - when neutron star in binary system is eating the other star and turns nova very predictably.
  - larubbio 13y ago
    I think it's type 1a. There was a discussion on this recently after a new supernova observation. https://news.ycombinator.com/item?id=7104056 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.
  - splat 13y ago
    The standard candles are Type Ia supernova which are when white dwarfs exceed the Chandrasekhar mass and become neutron stars. It is thought that when neutron stars exceed the maximum mass of a neutron star (which is not well known, by the way) they form a gamma-ray burst.
    - kevinastone 13y ago
      The TOV Limit for those following at home. It's when the star's gravitational pressure exceeds neutron degeneracy pressure and collapses into a black hole. http://en.wikipedia.org/wiki/Tolman–Oppenheimer–Volkoff_limit http://en.wikipedia.org/wiki/Tolman–Oppenheimer–Volkoff_limi...

Crito 13y ago

> 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. 3.8 miles (1.9 mile Schwarzchild radius), to be a bit more precise.

Steuard 13y ago

This isn't really all that sensationalized, based on Hawking's suggestions in question. The presence of an event horizon is quite often used as the defining feature of a black hole. The event horizon is the "point of no return" beyond which gravity is inescapable. If Hawking is claiming there is no definable event horizon (due to turbulence or whatever else), then he's claiming that there is no set of points near a dense, massive object that are inescapable. 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!

VladRussian2 13y ago
>The event horizon is the "point of no return" beyond which gravity is inescapable. 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.

anigbrowl 13y ago

In its stead, Hawking’s radical proposal is a much more benign “apparent horizon”, which only temporarily holds matter and energy prisoner before eventually releasing them, albeit in a more garbled form. '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

codeflo 13y ago
It's partly a question of definitions. There's an astrological phenomenon that occurs when large masses collapse, and we believe there is one of those at the center of the Milky Way. This exists in nature, we just haven't definitely figured out how they work yet. We currently call those "black holes". If a new theory suggests that these holes are not so black, but really rather "grey holes" or something, we might adopt a new word for those things, or we might change our definition of black hole to match the current theory.
- summerdown2 13y ago
  > There's an astrological phenomenon that occurs when large masses collapse Astronomical. Astrological would suggest it depends which star sign the masses are in.
  - codeflo 13y ago
    I'm stupid. Of course, thanks.

nathancahill 13y ago

"There is no spoon."