They’re the boogeyman of pop astronomy: the ravenous, all consuming black hole, with gravity so intense that you can’t escape it, and lurking in the center of the Milky Way, working to devour the entire galaxy.
Except they’re not that bad.
Read the whole story. Worrying about getting sucked into a black hole is like worrying about getting sucked into the sun — things tend to orbit them, crashing in only occasionally, and once they’ve drawn in all the matter that’s in close proximity, they’re terribly inefficient at gathering more.
Well, gosh. I’m going to scratch that one off my list of nightmare inducers. I’m going to go back to dreaming about parasites and viruses…much scarier stuff.
Has there ever been a person who feared any astronomical phenomenon more than parasites?
::shudder::
Parasites & viruses? For parasites, there’s all the religious evangelizers, but whad’ya mean by viruses?
Yes, drawing in more than everything is inefficient.
Well comets seem to have a track record of being fairly effective at causing all kinds of panic just by appearing in the sky…
Truly god loves black holes. He created a universe with lots of them, and made them the centers of galaxies.
Reminds me of a thought experiment / astronomy tute question: what would happen to the Earth’s orbit if the sun were replaced with a black hole of the same mass at the same location? Answer in rot13: Shpx nyy.
I haven’t read the article yet (shame on me), but the cover of this month’s Scientific American is titled, “Rings of Fire”, that black holes seem to do something more than just “suck” (ie “be a huge gravity well”), they do something to all the matter falling in, to create fire(!) surrounding them completely. After Interstellar‘s attempt at humanizing black holes, seems like SA is rebutting the movie; that they are really nightmare fuel, not the lovey-dovey thingy the movie mistakenly portrayed.
aargh typo again! I always skew up the html tags. italics intended for magzine title only aaarrrrggghhh
Twas Brillig@7
When a black hole gobbles up gas, the gas forms a disk as it has to give its angular momentum to something else before it can fall into the black hole. It becomes dense enough to heat via collisions as it whips around in its death spiral. If you fell into a quiescent black hole, you wouldn’t heat up. (I mean, there’s still good reasons not to, but hot accretion disks aren’t one.)
Other objects in astronomy do that, but the denser the object is, the more kinetic energy things have when they orbit, so the more energy that can be transformed into thermal energy and radiated.
There’s an interesting story by Larry Niven, appropriately titled Neutron Star, in which the protagonist Beowulf Schaeffer makes a close approach to a neutron star. While his (fortunately indestructable monomolecular-hulled) spaceship is tidally locked with the star, he fires his hugely overpowered engines, providing thrust directly toward the star. Due to orbital physics, there is no way this could actually make him impact the star, but it does bring his approach marginally closer, and dramatically increases his velocity at perihelion. He survives the experience by climbing an access tube to the ship’s center of gravity, and hold on for dear life while spaghettification tries to pull his head off. The pilot’s seat ends up crumpled into a flat wad at the nose of the craft.
Becca Stareyes:
If you fell into a quiescent black hole, you wouldn’t fall in. Because if it’s quiescent then there’s no accretion disc of stuff to collide with and reduce your angular momentum. So you’d stay in orbit, just like around a star.
(Unless you gave yourself a lot of Δ-v in exactly the right direction. But that’s not really falling, is it?)
Does anyone actually worry about black holes? Frankly, if you’re going to worry about astronomical events then a big meteor crashing into earth is a lot more realistic, with the oceans being the biggest and most dangerous target.
There was a glob of something…they weren’t sure if it was a star or gas…that went fairly close to the black hole at the center of our galaxy in the past couple of years. Scientist had great expectations of seeing fireworks, but apparently it was sort of a dud.
twas brillig — Read that “Ring of Fire” piece the other night. It’s basically yet another attempt to explain the quantum-gravity paradox, particular for the conservation of information, based on strings/branes. Interesting stuff but speculative. As he points out, there’s no way to study them directly.
Johnny Vector @11:
Assuming you have enough angular momentum to begin with. Which, if you’re falling in, you don’t…
twas brillig @7: Looks like they’re talking about the AMPS firewall. Nice background article here, and if you have an hour to kill, a talk by Leonard Susskind.
Well, a Chicxulub-like asteroid impact might cause some concern. Comet Shoemaker-Levy 9 did a number on Jupiter.
Takes an infinite amount of time for something to fall into a black hole.
I concede that a catastrophic impact event is far more fearsome than any parasite. My squeamish loathing of the latter got the better of my reason (such as it is).
#16 madtom1999:
Nope. It just looks that way from the outside. Because, you know, it’s hard for light to escape from even close to where the escape velocity is the speed of light.
But there’s nothing that slows down in-falling matter – you only get faster.
Of course, you can quibble based on the fact that there’s no reason to assume a black hole has, you know, some place you could call a surface where you can impact. But once you’re beyond, say, the diameter of an atom, I have no problem saying that you’re essentially done falling in.
On the third hand, given that we suspect black holes cannot be actual singularities as those probably are unphysical, nobody really knows what happens after you get beyond the event horizon. And last I heard, we still don’t have a TOE.
khms:
But if you’re going to be that precise about it, it matters how large it is. If it’s a relatively small one, you’d already be ripped apart by then. Or at least it might not be a very comfortable “landing.”
To reiterate khms: it does NOT take an infinite amount of time to fall into a black hole. It takes an infinite amount of time for an external observer to see the falling object reach the event horizon, due to gravitational time dilation. From the point of view of the falling object, it takes a finite time that can be directly calculated from the mass of the black hole (assuming non-rotation and neutral charge). Even then, the point of interest as per survival is not hitting the centre of the black hole, it’s reaching the point where the tidal force is sufficiently strong to tear you into your component atoms (popularly known as “spaghettification”). Interestingly, for a sufficiently large black hole, you can fall a long way past the event horizon before spaghettification takes place…but you can’t send signals out past the event horizon (one of the glaring flaws in Interstellar‘s understanding of GR).
Artor@10: “Neutron Star” is a neat little story, but it illustrates the cognitive bias of “hard” science fiction fandom in awarding it the 1967 Hugo for Best Short Story for “predicting” the existence of neutron stars and their high tidal forces even though the existence of neutron stars and their gravitational properties had already been shown mathematically way back in 1934. And even then, Niven made a number of major mistakes, for instance he ignored the rotational kick that would have put the ship into a high-g spin and killed its occupant; he misunderstood tidal forces as being at their weakest in the middle of the ship (nope, doesn’t work like that); and it’s also risible that an advanced spacefaring civilisation like the Puppeteers would know nothing about the existence of neutron stars and tidal forces or, even if they had this inexplicable gap in their knowledge, they couldn’t figure it out with unmanned probes. I’m not saying this to put down the story or Niven himself or even hard sf (which is my personal favourite of all the sf subgenres when it’s done well, as in Vernor Vinge, Greg Benford, Nancy Kress, Pat Cadigan) — my real beef is with the lack of self-awareness in hard sf fandom that a lot of their favourite stories are as scientifically shaky as many of the subgenres they scoff at. (And no, I’m not intimating that you are one of these people.) If I were to choose a Niven story, my personal favourite for hard sf would be “Flash Crowd”.
I used to have nightmares about the varieties of fungi that can grow to fill up a given sinus (do NOT google image search that set of ideas, you have been warned), and then I got some topical antibiotic stuff for my sinuses and the dreams went away.
Probably my subconscious telling me what was up, since I also had fewer headaches afterwards, but it wasn’t a fun process.
chrislawson #20:
(I’m about to ask a question from extreme ignorance, sorry in advance)
I’ve heard that said many many times, but wouldn’t that make every asteroid, planet, star, pony or whatever that’s ever been pulled in throughout the life of the black hole stuck just before the event horizon? And in the case of those quiescent black holes with no accretion disks, wouldn’t they still be exceedingly bright with multiple stars stuck eternally above the event horizon?
I’m sure I’m missing something.
Radioactive Elephant @23: As objects approach the event horizon, light from them is redshifted to the point that they ‘wink out’ in a finite time, from the point of few of a distant external observer.
That should be ‘point of view’ rather than ‘point of few’. Or maybe ‘point of phew!’.
No, spacetime doesn’t work that way. In ordinary non-relativistic settings, there are many different paths you could take from A to B, but they are not all the same length nor are they the shortest length (which is a straight line, at 45 degrees if it’s exactly Northwest let’s say). What basically happens when taking relativity into account is that everybody does move through the same (space and time) “distance” no matter what, and consequently you’re either moving more through time or more through space. Each one happens because of the other, since they are linked together.
The point is that time runs differently for you than it does for the person falling in, if you’re more or less “stationary” with respect to the black hole while they are not at all stationary with respect to it (because they’re falling, or they were falling and you’re just now finding out about it). Time is running slower or faster, but you would not say that you are stuck, because it has everything to do with things moving around or with the framework you’re moving in being distorted/dilated by the existence of stuff with mass/energy.
If you’re seeing anything just above the horizon, it’s very dim, not bright. It would be redshifted until eventually you get no more light from them (given however long you’re willing to wait for the next super-long-wavelength photon). But they are not stuck at all. The light which may or may not eventually reach you is some other physical object (the photons) for you to observe wherever you happen to be and however you happen to be moving, and that is not what the star is or where it is or how it is moving.
@chrislawson, 20
I haven’t seen it since it was in theaters, but I don’t recall them ever broadcasting across the event horizon. I know he was broadcasting during his descent, but I thought he cut off once he was past that point. Unless you’re talking about the broadcasts between the two who were inside the EH (I have no idea how that would work, so I can’t comment).
cr @26:
You won’t have to wait long; there will be a ‘last photon’, simply because of the discrete nature of light.
Yes, chrislawson was presumably talking about McConaughey becoming a five-dimensional being that sends signals (by moving the books around) to his daughter in his own past. That’s the silly deus ex machina which tries to put the entire plot (and scenery!) back together after it was so thoroughly shredded; but anyway, that is what is not even remotely consistent with GR. Blatantly so.
It isn’t consistent, that is, assuming the entire Earth/black-hole system isn’t in a closed loop with time going in a circle and all of the same events happening over and over again, which to me seems more depressing than any famine. Also, in that case, there would definitely be no escape (or entrance!) from it and all the bullshit about black holes would be completely unnecessary, so it’s hard to imagine that’s the kind of story they wanted to tell. That isn’t what they were saying, so it was simply wrong.
I know I don’t understand most of the details, but what does that have to do with it? Remember, I’m a Bohmian, so I’m all about teh particles ;) I was thinking you’d see fewer and fewer discrete light-chunks as time goes by, but that doesn’t mean it must go to a rate of literally zero photons/time (for as long as the black hole still exists and hasn’t yet evaporated, if we’re going to pretend like we could wait that long). And I don’t see why the last one to escape, whenever that is, would need to be a short amount of time after the formation of the horizon. Couldn’t they take whatever orbits they want which might ever so gradually end with them being flung out? I mean, sure, it seems like they’d need to be very lucky and finely-tuned to pull that off, but I don’t see how it has anything to do with light being discrete or continuous.
Radioactive Elephant: as Rob Grigjanis says, what happens is that the light from the falling object becomes more and more redshifted to the external observer and eventually will become undetectable — in theory, though, this is a limitation of the detecting technology rather than a true loss of signal. I wonder if anyone has calculated the amount of redshift it would take before the signal is completely indistinguishable from the CBR.
cr @30: I don’t know what photons want, but they’re as much slaves to equations as the rest of us. If someone falling towards a (uncharged, non-rotating) black hole is shining a flashlight, a distant observer would see the luminosity decrease exponentially. If light were continuous, this would never actually reach zero. But imagine the flashlight firing one photon at a time. There will come a point at which one photon will get out, and the next one won’t.
If you’re not convinced, you could read chapters 32 and 33 of Gravitation by Misner, Thorne and Wheeler. Or this explanation by John Baez;
Matrim@27: there is a moment when McConaughey talks about letting a probe fall into the event horizon, and the probe can broadcast its signal to him and he can repeat the signal to an outside observer. This is NOT general relativity. I’m won’t list all of the mistakes in Interstellar because I’d be here all day, but this kind of thing irks me because the filmmakers made such a big deal about how accurate it was, how they’d brought Kip Thorne onboard to advise, and so on, but they were only interested in being (very) superficially scientifically accurate. It’s really just the science fiction equivalent of stupid ghost movies “based on a real story.”
Oops, the site belongs to Baez, but the explanation is by Matt McIrvin.
Rob@32:
Yes, I was talking about classical GR — which as a classical theory predicts infinite redshift over infinite time for an external observer watching a light source fall into a black hole. However, the classical aspects of GR don’t play well with quantum theory, and I hadn’t considered the effect of falling into an event horizon on discrete events such as photon production. I really should get my hands on MTW’s Gravitation, but at nearly $200 a copy, I keep getting put off.
chrislawson: I got mine for $23 brand new. Which kinda dates me…
Anyway, what MTW and McIrvin are talking about isn’t QFT on a curved background; just classical GR with photons. After all, photons predate QFT and GR!
Damn, you just ruined all of the magic for me.
I don’t know which equations you’re talking about to represent the situation. Maybe we’re asking two different questions.
Suppose I drop a flashlight in, and it’s spitting out some number of photons/second. As r goes down, that rate approaches zero. Fair enough. The flashlight has only so many distinct photons to emit as it travels toward the horizon, so it make perfect sense to talk about there being a photon which is the last one to escape, whenever that is. I’m not disputing that.
However, suppose we’re considering how light can orbit outside the horizon. Some of the photons that do eventually escape as the flashlight falls could be trapped in such an orbit and stay that way for some time (and the question is how much time that is for a distant observer). There’s no reason to think it must either fall into the horizon quickly or that it must escape quickly. It’s not as if it must have some simple trajectory that is easy to calculate or is good enough for practical purposes. It just has any old trajectory whatever and that path is however long it is. The equations for that couldn’t depend only on features of the light source in isolation (or its trajectory separately from the photon) but also on features of the whole exact environment the photon is in (curved space and all) after it has left the source.
But maybe I am confused about it still. Does this make sense?
No worries, Nibiru is gonna smear us long before any black hole gets the chance to.
cr @37: If a photon is emitted at exactly the event horizon, where the escape velocity is exactly c, it can stick around until the black hole grows a bit by swallowing something*. Then it would fall in. Emitted just above the event horizon, it escapes; below, it falls in.
*This is all ‘classical’ stuff, so no Hawking radiation, but yer average stellar or larger black hole will almost certainly be swallowing up stuff faster than it radiates for a long time to come.
Thanks Rob Grigjanis, Consciousness Razor, and chrislawson. I hadn’t thought about the redshift, well not the point of dimming it to the point of becoming undetectable anyway. that makes sense, and I knew it’d be something silly I was missing.
Consciousness Razor, I understood the first part of your #26, I knew it wasn’t literally “eternally stuck” for what ever was actually experiencing the approach to the black hole. The question was purely from the perspective of an outside observer. Sorry for not specifying that. But yeah, red shift. Awesome! Thanks again.
My question has to do with falling into a black hole, from the victim’s point of view.
It seems to me that if the escape velocity of the hole is greater than the speed of light, an object falling from infinity is going to approach the speed of light. (Assuming you line up right and aim, and maybe goose the engines so as to save time.) Dropping straight in is going to get relativistic. I assume that the black hole cannot accelerate you past the speed of light, but if your mass is increasing, the gravity will have more to pull on. Time, to you, falling, will seem to pass normally inside your ship, but the outside universe will appear to be moving faster as you approach lightspeed, so the black hole will appear to arrive very quickly.
Anyhow, what happens when mass increases due to gravity?
And wouldn’t a black hole seem to jump right at you?
Or does falling straight in not get you up near lightspeed?
None of the above. Once you’re close enough you will of course be accelerating near c,* but your mass doesn’t increase. That would involve getting more atoms to make up your body or your spaceship, which I say you don’t get to have. You have enough atoms.
And it’s not jumping at you any more than the Sun is jumping at you.** If you found you were no longer in a stable orbit here and you were falling straight into the Sun, it’s exactly the same story: your speed toward it would increase as you got closer, just like Newton would have said with his inverse square law. That’s not sudden or jumpy at all. Indeed, the two of us are also gravitationally attracting each other now, but there’s no jumping or any funny business like that. Getting much, much closer would not make a drastic difference either, since we’re so tiny (or since gravity is so weak, depending on how you look at it).
*Relative to the horizon. Keep in mind that right now you and I are both already moving near c in some reference frame or another. We’re not staying put, nor is anything else. Of course, it’s not as if we’re using any fuel to accelerate ourselves toward it or away from it (side to side, around it, etc.), nor are we near any massive object that’s “pulling” on us gravitationally to do the work of the rocket fuel. However, if you wanted to accelerate yourself a lot out of the state of motion you’re already in now, that is what will cost you. Or it simply can’t be done if it’s exactly c.
**Indeed, we are almost certainly falling into the galaxy’s central black hole, or maybe Andromeda’s or both. It’s just going to take quite a while. But if we get to start from infinity, either of those are more than close to enough to count as concrete examples.
Menyambal @41:
Yes, an observer ‘hovering’ just outside the event horizon will measure your speed as almost the speed of light as you pass them. Of course, an observer very far from the black hole will see your speed increase at first, then decrease to zero near the horizon :)
Not sure what you mean. Your mass doesn’t increase; the force of gravity and tidal forces increase.
What appears to be passing faster to you is the the time away from the black hole. But this is fleeting, and when you cross the event horizon*, photons from outside become garbled.
*Assuming the black hole is big enough to not have torn you apart before you hit the event horizon.
Thanks.
I thought an object’s mass increased as it approached light speed.That is what keeps us from hitting lightspeed, I thought. An object gets more massive, and nothing has enough energy to accelerate that increased mass.
But, if that mass is being pulled by a black hole, the gravity of it will have more pull on that increased mass. Which maybe means the mass speeds up, gets more massive, and gets pulled even harder.
There may be something about falling that alters the case – I vaguely recall something from Feynman.
I will have to think about what happens after crossing the event horizon. All I was thinking of was approaching it.
Menyambal @44: OK, you’re probably referring to the so-called relativistic mass, which is a somewhat outdated (and frowned upon by many, including Einstein and Wheeler) notion. It’s defined as
M = m(1 – v²/c²)^(-1/2)
where m is the rest mass, v the speed. So as v increases, so does M, to ∞ as v approaches c. As Einstein wrote;
That would be me.