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Space Science

Largest Black Hole Measured 170

Posted by kdawson from the that's-no-moon dept.
porkpickle tips us to a BBC article on the quasar OJ287, a binary object containing largest black hole yet discovered, weighing in at 18 billion times the mass of Sol. Researchers were able to estimate its mass due to the presence of a smaller black hole in orbit around it. When the smaller companion's orbit intersects OJ287's accretion disk, once every 12 years, it triggers a burst of radiation that was detected by the Spitzer Space Telescope. More detail and a diagram are available on the Turku University site.
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Largest Black Hole Measured More

Largest Black Hole Measured

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  • eh? I don't get it? (Score:4, Interesting)

    by Anonymous Coward on Friday January 11, 2008 @11:50AM (#22000620)
    How large can a singularity be?

    I mean, if they used the word "massive" I'd get it. But large?

    • Re:eh? I don't get it? (Score:5, Informative)

      by AmaDaden ( 794446 ) on Friday January 11, 2008 @12:12PM (#22000948)
      It think they are not 100% sure about the whole "a black hole is a singularity" thing.

      quantum mechanics .... does not allow objects to have zero size--so quantum mechanics says the center of a black hole is not a singularity but just a very large mass compressed into the smallest possible volume.
      from http://en.wikipedia.org/wiki/Black_hole [wikipedia.org]

      • Re:eh? I don't get it? (Score:5, Informative)

        by BobGod8 ( 1123841 ) on Friday January 11, 2008 @01:40PM (#22002262)
        Actually it's way more complicated than that. Only non-rotating black holes could ever truly be point masses. Any angular momentum creates complicated tidal effects near the center, resulting in a non-point-mass. Carried further, the "singularity" expands until the point where it would effectively reach the event horizon itself, resulting in a naked singularity, which some calculations have shown can have actual size. Adding further rotation will (to a point), actually change the size of the "singularity". Of course, this is all moot, since that's not at all what the article was talking about, but that's my .02$.

        • Re:eh? I don't get it? (Score:5, Informative)

          by Anonymous Coward on Friday January 11, 2008 @02:54PM (#22003696)
          That's misleading, and I'm guessing you don't really understand what you're describing. A rotating black hole (aka every black hole, to some extent), is still a singularity (no need for quotation marks, it still has zero volume) despite not being a point. It's a ring with zero cross-sectional area, sort of like an infinitely thin thread arranged in a circle.

          Furthermore, this thread is based on quibbling over semantics without really understanding what the author quite validly meant. The "black hole" aspect of a singularity is a description of the effects of its event horizon, which of course scales with mass. A more massive black hole is by definition larger then a less massive black hole. Someone mod this up so this misunderstanding can be cleared up for more people.

    • Re:eh? I don't get it? (Score:5, Informative)

      by pclminion ( 145572 ) on Friday January 11, 2008 @12:15PM (#22000992)
      A black hole has an event horizon. This horizon has a very well-defined size.

    • Re:eh? I don't get it? (Score:5, Informative)

      by moderatorrater ( 1095745 ) on Friday January 11, 2008 @12:31PM (#22001198)
      The event horizon is often considered the size of a black hole since nothing could ever leave that space.

    • Re: (Score:3, Informative)

      by jgarra23 ( 1109651 )
      How large can a singularity be?

      I mean, if they used the word "massive" I'd get it. But large?

      I believe they are measuring the event horizon, not the singularity.

    • Re: (Score:2, Funny)

      by mtmra70 ( 964928 )
      The more important question is, how do you have an unmeasured 'largest black hole'?

  • Wow. (Score:5, Funny)

    by AltGrendel ( 175092 ) <`su.0tixe' `ta' `todhsals-ga'> on Friday January 11, 2008 @11:53AM (#22000692) Homepage
    A binary black hole system.

    Proctologists across the globe swoon!

  • Which one weighs 18 billion times our sun, and which ones weighs 100 million times our sun?

  • Ask slashdot (Score:5, Funny)

    by sm62704 ( 957197 ) on Friday January 11, 2008 @12:10PM (#22000914) Journal
    Is there a theoretical limit to the size of a black hole?

    That was serious, here's the link [uncyclopedia.org] to the non-serious.

    A Black hole is an impossible object which makes the Universe work. It has the useful property of being "undetectable". It's like when your spouse comes home with a dent in the car, and blames it on an invisible black mass; the dent is proof of the black mass, but you can't, and never will be able to see it with CCTV cameras, but you know it's there. "Dark matter" is an equally undetectable force that causes cars to defy gravity, and hit invisible black holes. Astronomers will tell you that lots of them have spouses with dents in their cars, and can explain this is very technical terms, so you won't be able to understand why it's not possible.
    More there...

    • Re:Ask slashdot (Score:5, Informative)

      by ArcherB ( 796902 ) * on Friday January 11, 2008 @12:59PM (#22001522) Journal

      Is there a theoretical limit to the size of a black hole?
      While I can't give you numbers since I'm going from memory, but there used to be a theoretical limit to black hole size. This was before "Super Massive Black Holes" were discovered in the center of every galaxy. Super Massive Black Holes are much more massive than the previous theoretical limit and were thought to be impossible so many astronomers were claiming that such a thing was couldn't exist while others were saying, "Oh yeah? Then why don't you put down the chalk, professor, and come down to my observatory and tell me what that big-ass black gravity thing is in the middle of our galaxy!" (Of course, they couldn't really see it, but you get the point)

      I think astronomers are reluctant to guess at a size limit now as they don't want another discovery to make them look like asses.

      • Re: (Score:3, Informative)

        by Ambitwistor ( 1041236 )

        While I can't give you numbers since I'm going from memory, but there used to be a theoretical limit to black hole size.

        There has never been a theoretical limit to the size of a generic black hole. (Technically, the observable universe could be in a giant black hole.) But back when people thought the only way a black hole could form was from the collapse of a single star, there was a practical limit on the size of an astrophysical black hole: if it forms from stellar collapse, it can't be more massive than the most massive stars. Everyone recognized that black holes can get larger by swallowing more mass, but it was a l

      • Re: (Score:3, Informative)

        by Cassini2 ( 956052 )

        The Eddington limit [wikipedia.org] appears to limit the size of a star. At one point in time, it was thought that black holes formed from the collapse of stars. Later on, it was concluded that supermassive black holes are very good at feeding on neighboring stars, and thus supermassive black holes could form. The Wikipedia page on Black Hole Parameters [wikipedia.org] has an explanation.

      • There was a post on slashdot I think in the past year on this subject.

        Where scientists found a black hole seemingly larger than it was possible to be.

        I think the difference is, while it is postulated that there are super massives at the center of every galaxy, what is not known is how they are formed.

        As is commonly accepted, a normal black hole is formed by a collapsing star, and that stars have a finite size, that any black hole formed in this manner is restricted.

        If I remember correctly there was also a t

      • Re: (Score:2)

        by barakn ( 641218 )
        Some weirdos think our universe is the inside of a black hole, so no, there's no limit.

  • When it comes to choosing neighbors, (Score:3, Funny)

    by davidsyes ( 765062 ) on Friday January 11, 2008 @12:23PM (#22001100) Homepage Journal
    I pine for Sol, not a massive black hole. Otherwise, we'll have a massive cleanup job? Oh, wait...
  • Seems like /. is going down one of them two holes...

  • Need a better measurement comparison (Score:5, Funny)

    by vjmurphy ( 190266 ) on Friday January 11, 2008 @12:28PM (#22001160) Homepage
    "largest black hole yet discovered, weighing in at 18 billion times the mass of Sol."

    Yes, but how many Twinkies is that?
  • Yes, there is much gravitational distortion around a black-hole,
    which looks like a very light-bright sphere (maybe a little
    physically distorted) to all humans, and within the absence of
    light there is much levity to consider.

    Tell me again, why is it a big black-hole and not a big bright-spot?

    In the absence of levity there is gravity.
    In the absence of gravity there is levity.

  • Tag as Sun!Sol (Score:3)

    by MtlDty ( 711230 ) on Friday January 11, 2008 @01:09PM (#22001670)
    Why do people say 'sol' instead of 'sun'. Is there some fundamental difference, or are they just trying to sound smart?

    • Re: (Score:2, Informative)

      by AbsoluteXyro ( 1048620 )
      We're pretty used to referring to Sol as "the sun" but the truth is, a sun is a thing and there are many of them. It is silly to call ours THE sun, because it clearly isn't. In actually, it is ONE OF the suns. Sol is our sun's Latin name. Similarly, Luna is our moon's Latin name.

      • Re: (Score:3, Insightful)

        by kalirion ( 728907 )
        It's equally silly to say The White House when there are plenty of white houses around, no?
        • One would rarely find themselves in a situation where they need to distinguish one white house from millions of extremely similar white houses. It is for this reason that naming that particular building "The White House" works. If everyone lived in a white house with nearly identical construction, and each of these houses were regularly referred to as a "White House", naming one of them "The White House" wouldn't be of much use.

      • Re: (Score:2, Informative)

        by illogique ( 598061 )
        why many suns? it's more like they are many stars and our star is name the sun!
        similarly, the moon is the name of the Earth natural satellite
        • A star with planets is a sun. A rock of sufficient mass orbiting a planet is a moon. Our sun is not the only sun, and our moon is not the only moon. Astronomers and other scientists, who study several suns and several moons, typically choose to specify our sun by calling it Sol, and our moon by calling it Luna. "The moon" and "the sun" is far too ambiguous when you are dealing with millions of suns and moons.
      • We're pretty used to referring to Sol as "the sun" but the truth is, a sun is a thing and there are many of them. It is silly to call ours THE sun, because it clearly isn't. In actually, it is ONE OF the suns. Sol is our sun's Latin name. Similarly, Luna is our moon's Latin name.

        Which sucks for us speaking Spanish (and most languages derived from Latin) since the literal translation for "Sun" is "Sol" and for "Moon" is "Luna" :)

    • Re: (Score:2)

      by pembo13 ( 770295 )
      sun is 'common noun' ; sol is 'proper noun'

  • gridwars (Score:3, Informative)

    by doti ( 966971 ) on Friday January 11, 2008 @01:10PM (#22001690) Homepage
    This story makes me want to play gridwars2 [marune.de] again.

    And again, and again...

  • Question about gravity (Score:5, Interesting)

    by caywen ( 942955 ) on Friday January 11, 2008 @01:14PM (#22001754)
    One question I have about gravity and black holes is this: If nothing can escape the event horizon, how can gravity escape it? In other words, would objects outside the event horizon ever feel the pull of gravity from that which is inside the event horizon?

    • Re: (Score:2, Insightful)

      by fred fleenblat ( 463628 )
      gravity *waves* cannot escape the event horizon, so presumably something like a starquake of the singularity cannot be detected. however, the gravitational field around the black hole is/was established before stuff falls in so as far as the rest of the universe is concerned the black hole has normal gravity. there's some weird effects like frame dragging though. check wikipedia for some explanations. IANAP.

    • Re: (Score:3, Interesting)

      by Anonymous Coward
      Gravitational pull isn't something that is being radiated out of bodies. Just changes of it.

      (In fact if the singularity somehow disappeared magically the outside world wouldn't detect it since the signal of black hole disappearing wouldn't escape from the gravitational well.)

    • Re:Question about gravity (Score:4, Informative)

      by Ambitwistor ( 1041236 ) on Friday January 11, 2008 @02:09PM (#22002816)
      Other people have answered your question (radiation cannot escape from inside the horizon, but it can still generate a static external field), but here [ucr.edu] is a FAQ with more detail, including the quantum picture.

    • Re: (Score:3, Insightful)

      by JohnFluxx ( 413620 )
      One hypothesis of gravity is that it is an exchange of 'gravitons'. If this hypothesis is indeed correct, then it does indeed make sense to ask how these gravitons can escape a black hole. And I don't know the answer to that.

      But the most commonly accepted theory is that heavy objects cause the fabric of spacetime to bend under its mass - like a heavy ball placed on rubber sheet.
      With this image, it is spacetime that bends so there's no meaningful question for how gravity 'escapes' from it.

      • Re: (Score:3, Informative)

        by Ambitwistor ( 1041236 )

        One hypothesis of gravity is that it is an exchange of 'gravitons'. If this hypothesis is indeed correct, then it does indeed make sense to ask how these gravitons can escape a black hole. And I don't know the answer to that.

        Static gravitational fields are mediated by virtual gravitons, which can travel at any speed, including faster than light. However, you cannot use them to transmit information, i.e., changes in the field from inside the horizon.

        With this image, it is spacetime that bends so there's no meaningful question for how gravity 'escapes' from it.

        Right. Classically you can see that the exterior field does not depend on the interior field, and that gravitational radiation generated inside the hole can't get out.

    • I'm taking a wild guess here (a little knowledge is a lot dangerous). Forces are understood by the Standard Model of physics to be implemented via mediating particles [wikipedia.org]. That is, a force between two particles is felt when those two particles exchange a mediating particle. The photon is considered the mediating particle of the electromagnetic force, and the graviton is hypothesized to be the mediating particle of the gravitational force.

      However, the mediating particles themselves are not affected by the forc

      • Re: (Score:3, Interesting)

        by Ambitwistor ( 1041236 )

        However, the mediating particles themselves are not affected by the force they mediate. Otherwise the universe would disappear up its own arse.
        Hence, gravity is not affected by gravity.

        Actually, most mediating particles are affected by the force they mediate, including gluons, the hypothetical gravitons, and IIRC the W bosons.

        In gauge theory, a non-Abelian gauge group will in general lead to a nonlinear Yang-Mills theory with self-interacting fields, in contrast to the linear Abelian theory of electrodynamics.

        Because gluons, the mediator of the strong nuclear force, themselves carry strong ("color") charge, it's possible for them to bind to each other. (See glueballs [wikipedia.org] in quantum chromody

        • However, the mediating particles themselves are not affected by the force they mediate. Otherwise the universe would disappear up its own arse.
          Hence, gravity is not affected by gravity.

          Actually, most mediating particles are affected by the force they mediate, including gluons, the hypothetical gravitons, and IIRC the W bosons.

          In gauge theory, a non-Abelian gauge group will in general lead to a nonlinear Yang-Mills theory with self-interacting fields, in contrast to the linear Abelian theory of electrodynamics.

          Because gluons, the mediator of the strong nuclear force, themselves carry strong ("color") charge, it's possible for them to bind to each other. (See glueballs [wikipedia.org] in quantum chromodynamics.)

          Similarly, gravity gravitates: gravitons interact with each other, because they have energy and anything with energy gravitates. This idea holds even in classical general relativity: gravitational fields themselves gravitate. Analogously to QCD glueballs, general relativity can have gravitational geons [wikipedia.org], which are regions of gravitational field which hold themselves together under their own gravity. (You might think that a vacuum black hole has that property too, but I'm talking about purely non-singular field configurations.)

          Pffft. I can put together a bunch of big made-up words to sound smart too.

          'Because, as everyone knows, the gluons gravitate towards gravitons, due to Abelian gauge theory and how it acts upon Bosun-W quark neutrons, regardless of the non-self-interactivity of Yang-Mills theory in relation to the Gass-Black theory of rock.'

          See? It's a perfectly cromulent way to sound smart, but I'm on to you!

    • One question I have about gravity and black holes is this: If nothing can escape the event horizon, how can gravity escape it? In other words, would objects outside the event horizon ever feel the pull of gravity from that which is inside the event horizon?

      Here's an even better question. If I used my magic obliterator to magically make the sun disappear, would Earth go flying off into space at the same moment or would it continue to orbit the missing sun for the 8 minutes it would take the last rays of light to reach us? Me not being a scientist, I would think immediately but I'm wrong. Knowledgeable people say the Earth would continue for those 8 minutes because nothing can communicate faster than the speed of light, violating causality and everything. This

      • Re: (Score:3, Informative)

        by Ambitwistor ( 1041236 )

        If I used my magic obliterator to magically make the sun disappear, would Earth go flying off into space at the same moment or would it continue to orbit the missing sun for the 8 minutes it would take the last rays of light to reach us?

        The latter.

        This is where they say gravitons come in as a particle that conveys gravity which doesn't make any sense.

        Why doesn't it make any sense? Photons are particles which convey electric and magnetic forces, do you have a problem with them too?

        Anyway, you don't need to appeal to graviton particles to answer the above question. Even in classical general relativity, the answer is still "8 minutes later", since that's how long for gravitational waves of spacetime curvature, traveling at the speed of light, take to reach the Earth.

        • Why doesn't it make any sense? Photons are particles which convey electric and magnetic forces, do you have a problem with them too?

          Hey, I'm not saying the science is wrong, I'm just saying I don't understand it. Photons I can understand. They are created at light sources, can be diffracted through gases, bounce off of solids, you can direct beams of light with a mirror, focus them with a lens, split them out into individual colors with prisms, etc. All of that seems reasonable enough. But now when you say that gravity is transmitted by a particle, just how does that happen? An atom of hydrogen is radiating graviton particles? Do gravi

          • Hey, I'm not saying the science is wrong, I'm just saying I don't understand it.

            Ok, but I don't know what's so different about gravitons that makes you not understand them, if you can understand photons.

            Photons I can understand. They are created at light sources, can be diffracted through gases, bounce off of solids, you can direct beams of light with a mirror, focus them with a lens, split them out into individual colors with prisms, etc.

            You can do a lot of those things with gravitons, too. You can diffract and refract them, scatter them, etc. Things are somewhat different though because electromagnetic fields can both attract and repel while gravitational fields only attract, so gravity interacts with matter differently than does electromagnetism.

            But now when you say that gravity is transmitted by a particle, just how does that happen?

            The same way that electromagnetism is transmitted by a particle: ma

            • Re: (Score:2)

              by TexVex ( 669445 )

              is your conceptual problem with gravity vs. electromagnetism

              I think the confusion is over the difference between virtual and non-virtual particles.

              If you put a thick enough sheet of lead between two masses, the sheet will block photons emitted from one mass from reaching the other. But that sheet of lead can't block the exchange of virtual photons, so the two masses can still exert force on each other through electric or magnetic fields.

              A gravity field is analogous to a magnetic field. Assuming grav

          • Some people think gravity travels faster than c [metaresearch.org].

            Larry

            • Well, some people are wrong. Tom van Flandern is a well known crackpot from Usenet, sci.physics.relativity. Numerous posters there explained his mistakes there, if you want to search Google Groups. It turns out that you can apply his argument for FTL gravity to similarly "prove" that light itself travels faster than light in Maxwellian electromagnetism. Both conclusions are based on the same error ignoring the source velocity depedence of the field.

  • That's incredible! (Score:5, Informative)

    by renfrow ( 232180 ) on Friday January 11, 2008 @01:16PM (#22001794) Homepage
    Using this illustration [astro.utu.fi] and my trusty piece of paper straight edge, I estimate the long axis of the orbit to be 21000 AU and the minor axis to be 16000 AU. Using Ramunjan's Approximation [wikipedia.org] for the circumference of the elliptical orbit and converting to light years [glyphweb.com], I guesstimate the circumference of the orbit to be ~1.99 (call it 2) light years.

    For a 12 year orbital period this means that the orbiting black hole is AVERAGING 1/6c (~49965km/sec, call it 50k km/sec)... meaning at periquaserion it's really booking! Much faster than The Dash!

    Tom.

  • And so... (Score:4, Funny)

    by Cleon ( 471197 ) <<moc.oohay> <ta> <24noelc>> on Friday January 11, 2008 @01:20PM (#22001878) Homepage
    I think this finally means that we have a definition for the SI unit "fuck-ton."
  • I always thought they'd get stuck because the black hole would suck in the tape before they even got half way around it. Then it dawned on me that they probably used stiff measuring tape.

    Those Turku University guys are pretty smart. :-)

  • I see they used the term "Sol" (Score:3, Funny)

    by not already in use ( 972294 ) on Friday January 11, 2008 @04:51PM (#22005814)
    For those of you who don't know, the term "Sol" means "A whale's vagina."
  • ...sounds like a girl I dated in high school.

    Ugh, some memories SHOULD be repressed.
  • I once did some calculations about black hole gravity gradients, and calculated that you could end up with a black hole with a gravity at it's "surface" (Schwarzschild radius) of less than earth's gravity (and that the diameter would be approximately one light-month).
    As I remember it, I figured that the mass of that black hole would be just a couple of GigaSols (it was a LONG time ago, so I no longer have the exact numbers). If my calculations (and memory) were correct, then this extra-super-massive bla
    • The surface gravity at the event horizon of a black hole is always infinite: it requires infinite thrust to hover at a horizon. (Technically: proper acceleration for a stationary observer diverges at the Schwarzschild radius.) I think you were doing a Newtonian calculation, not a relativistic calculation.

      (The tidal force, however, is finite at the horizon, and of course you experience no weight if you're freely falling through the horizon.)
      • No. It doesn't take infinite thrust to hover at the Horizon. It takes infinite energy to escape from the black hole, once you're at the horizon. It would also take infinite energy to slow yourself to a stop at the event horizon (if you arrived in a freefall from a roughly infinite distance, you should be accelerated to apx. Light speed by the time you get there), but if you managed to bring yourself to a stop at the event horizon of a large-enough black hole, you'd need very little energy to just stay at th

        • No. It doesn't take infinite thrust to hover at the Horizon.

          Yes, it does. This is not a hard calculation, although it takes a while unless you already have the Christoffel symbols handy.

          I could rederive it for you if you really wanted, but in general relativity the answer works out to be:

          g = GM/r^2 / sqrt(1-2GM/(rc^2))

          where g is the "surface gravity" experienced by a stationary observer, r is the Schwarzschild distance, G is Newton's constant, and c is the speed of light.

          This expression diverges as r->2GM/c^2, which is the Schwarzschild radius, s

        • By the way, an 18 billion solar mass black hole has a Schwarzschild radius of 0.00561932487 lightyears, or 355.36426 AU (calculation [google.com]).

          • units 0.00561932487years days
            * 2.0524146
            Which is roughly my calculation (I just forgot to update my post). Way back when I calculate 3 light-months as being the size that would have a surface gravity of 1G. I just don't remember the mass needed... It now looks like something near 800billion solar masses would do the trick. (just a guestimate, not a calculation).

            • Way back when I calculate 3 light-months as being the size that would have a surface gravity of 1G. I just don't remember the mass needed... It now looks like something near 800billion solar masses would do the trick.
              As I just explained, the surface gravity of a black hole is always infinite. Unless now you're talking about non-black holes now.
              • If the surface gravity of a black hole is always infinite, then how does hawkings radiation word? It depends on the fact that the gravity gradient at the event horizon is much larger for small black holes than it is for large ones. I think that you're confusing the fact that equations diverges (which rather implies a problem with the chosen coordinate system) with the actual gravity that would be experienced.

                • If the surface gravity of a black hole is always infinite, then how does hawkings radiation word? It depends on the fact that the gravity gradient at the event horizon is much larger for small black holes than it is for large ones.

                  The gravity gradient (tidal force) at the event horizon is finite, and decreases for large black holes. However, the surface gravity (proper acceleration of a static observer) is infinite.

                  I think that you're confusing the fact that equations diverges (which rather implies a problem with the chosen coordinate system) with the actual gravity that would be experienced.

                  No. The expression I gave is the proper acceleration, which is a coordinate independent invariant.

                  Think about it: if you could hover at the horizon with a finite acceleration g, then with any acceleration a>g, you could escape the horizon. But you cannot escape an event horizon.

                  Here's a sketch of the calculation l

                • I've been looking for a calculation online which goes through in detail the calculation I sketched, so I don't have to rederive it myself. I found this page [mathpages.com]. The author derives it just with the line elements of the metric, without using the Christoffel symbols as I did.

                  The formula I gave appears below the text which reads, "Substituting this expression for dr into the above formula gives the proper local acceleration of a stationary observer", except he's using geometric units in which G=c=1. The precedi
                  • So, what I hear you saying is that the limit of the value of the acceleration as you approach the horizon is finite, but the the value at the horizon is infinite (and the coordinate system fundamentally changes once you enter)(?)

                    • So, what I hear you saying is that the limit of the value of the acceleration as you approach the horizon is finite, but the the value at the horizon is infinite (and the coordinate system fundamentally changes once you enter)(?)

                      No.

                      Schwarzschild coordinates fundamentally change once you enter the horizon, but not all coordinate systems do (see, e.g., Kruskal-Szekeres coordinates), and the statement I made about the proper acceleration is independent of any coordinate system, by definition.

                      As you approach the horizon, the proper acceleration diverges — increases without bound — has no finite limit.

                      (This is true for static observers, i.e., ones who are hovering at a fixed location. Freely falling observers will experien

                    • So, the gravitational curve that one would normally associate with orbital mechanics and acceleration is (roughly) as I calculated, but other relativistic effects mean that the acceleration needed to stay stationary diverges?

                    • So, the gravitational curve that one would normally associate with orbital mechanics and acceleration is (roughly) as I calculated, but other relativistic effects mean that the acceleration needed to stay stationary diverges?

                      I don't know if I would put it that way.

                      Fundamentally, what is going on is that spacetime inside a black hole becomes so curved that it is no longer stationary. Stationary observers cannot exist within a black hole. The horizon is the limiting case: light can remain stationary there, but no (timelike, massive) observer can. An observer near the horizon will have to boost harder and harder to remain stationary, because he's fighting the spacetime curvature. At the horizon, he loses, because only light

                    • I'm wondering if it's appropriate to simply say that spacetime, inside of a black hole is so bent, that it's better to describe it as 'just warped'?

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