When You Fall Into a Black Hole, How Long Have You Got?
blogs.scientificamerican.com
blogs.scientificamerican.com
Though it may be possible to design some machines that can do computation in this unidirectional fashion. Consider something like: http://en.wikipedia.org/wiki/Rule_110
Another possibility is finding a really enormous black hole, large enough that stuff can live inside it: http://www.technologyreview.com/view/423608/planets-could-or... Of course, humans are not wired topologically to survive in such an environment, but if there is an energy source and computing restrictions are followed, it may be possible to grow a civilization inside an event horizon.
A 1 trillion solar mass black hole would have an event horizon at about half a light year, enough room for a few solar systems, depending on our good luck.
Now, a brain that was only half across would have some trouble as you describe I think. If you crossed it fast enough maybe no part of the brain would have enough time to realize it was isolated? I suspect if you are falling in faster than anything in your body was moving in the opposite direction nothing would be noticed.
First, I'm not sure it's possible for the brain to be half-way across an event horizon. When an in-falling object nears the event horizon, the horizon "dimples" in response to the changing gravitational field caused by the object's mass.
Second, once inside the horizon, it is true that messages cannot be sent opposite the direction of the singularity. However, your entire brain is moving toward the singularity, and therefore neural communication does not require messages to be sent away from the singularity.
>Things past the event horizon should be able to travel in the direction of it, though they will always turn back before actually crossing it.
According to the best of my understanding, if you're inside the event horizon, every direction you look leads to singularity. There are no paths away from the singularity.
True, but you are not standing still. You are moving. Accelerating, in fact. No messages need to be sent away from the event horizon for the brain to work.
Sure it is. But the brain is falling inward, so nothing has to move outward across the horizon for the brain to continue working. Everything falls inward, but signals traveling from one part of the brain to another might fall inward slightly more slowly, so the part of the brain the signal is going to will "catch up" with the signal, from a global perspective.
Maybe that's where we are already :)
One of the few "no brainers" in quantum physics.
That's the theory anyway. Black holes are disproved using the theory itself and rock-solid logic, here: finbot.wordpress.com/2008/03/05/no-black-holes/
Of course I'm using a throwaway account because people love to assume that any science not coming from an academic source is both false and crackpotish, evidence be damned. (This unscientific behavior that pervades our society is the strangest thing to me. Especially among hackers, who are otherwise highly logical.) It's generally accepted among astrophysicists that black hole theory forms a physical paradox called the information loss paradox, a contradiction with quantum mechanics, a much better tested theory. No black hole has yet been definitely observed (despite media harping to the contrary), and Einstein's theory itself explains all suspected black holes without predicting them to be black holes. Sorry to pop anyone's sci-fi bubble!
The basic mistake in this whole argument is that there is no such thing as an inertial frame or near-inertial frame that crosses an event horizon. An inertial frame outside the event horizon will appear to be fully outside it from within the frame, even if extended.
Event horizons can exist in GR even without black holes, for example, an accelerating observer sees an apparent horizon or Rindler horizon, where signals beyond it will never reach him.
But there is no near-inertial frame that, from within the frame itself, appears to cross the Rindler horizon.
Sure there is. Consider an object falling into the black hole; you can set up a local inertial frame around the event of that object crossing the horizon and it will work just fine even though it straddles the horizon.
What you can't have is a static inertial frame that crosses the horizon; i.e., you can't have one that stays at the same radius, even for an instant. The inertial frame I described above is infalling--i.e., an observer at rest in the frame is falling into the hole.
But there is no near-inertial frame that, from within the frame itself, appears to cross the Rindler horizon.
Yes, there is. This is even easier than the black hole case because you can have a Rindler horizon in flat spacetime, where all inertial frames are global--they cover the entire spacetime, including the region behind the Rindler horizon.
I should note that I do not mean to imply that the "finbot" article linked to is correct; it isn't. As I described above, inertial frames that straddle the horizon can be constructed, and they are perfectly consistent with the equivalence principle.
At this point, my understanding of physics gets hazy so I'm not sure how much this extrapolates to the black hole case.
Actually, the accelerating observer can't "see" the Rindler horizon; light emitted at the Rindler horizon never reaches the accelerating observer (that's the definition of the Rindler horizon). The inertial observer is the one who can actually "see" the horizon, because he passes it.
Physical laws that hold in any inertial frame do not necessarily have to hold when observing from a non-inertial frame (for example ability to signal between any two points).
Any two points within the "accelerating frame" of the accelerating observer can signal to each other. The Rindler horizon is not, strictly speaking, "within" that frame; Rindler coordinates become singular at the Rindler horizon.
That said, I don't see what the properties of the accelerating frame have to do with the statement I made that any inertial frame in flat spacetime will include the Rindler horizon, since all inertial frames in flat spacetime are global.
This is essentially the 'no drama' assumption the article is about. It means roughly, that there is no way to detect that you did cross the event horizon. And therefore the brain still works fine.
The larger the black hole, the smaller the tidal forces at the event horizon, so with a large enough black hole, a human body would be fine.
And don't tell me that it curves away because that would be a violation of conservation of momentum.
It also can't travel directly up then back down, because light can't do that - at the moment of reversal it's not traveling at the speed of light.
And the light particle can't evaporate (i.e. red shift into nothingness) by loosing all its energy to the blackhole because that would be a violation of conservation of angular momentum (every photon caries some angular momentum).
So what happens to it?
I should mention that after studying all the physics that I could understand I do not believe black holes exist. (Super massive objects might, like the observed objects in the center of galaxies, but not as black holes.)
No, the light does make it to the horizon.
our math breaks down at the horizon
No, the horizon, and the spacetime inside all the way down to the singularity, can be described perfectly well mathematically.
And what does that have to do with conservaton of momentum? If i throw a rock upwards it curves away rather than leaving the solar system — momentum is still conserved and the rock has much more than a photon.
It's the same phenomenon as what causes gravitational lensing.
It's not moving in a curved line, it's moving straight - from the POV of a distant observer as well.
In order to curve it must have some lateral motion relative to "down".
Ordinary masses bend space nowhere nearly enough. The Sun bends space just barely enough for the effect to be observable in a solar eclipse (it was used as a verification of Einstein's theories) by causing stars behind it to appear ever so slightly out of place when its disc is just about to overlap them. The light from these stars is NOT being bent - it keeps going in a straight line, but the Sun's gravity slightly alters the very idea of straight in its immediate vicinity. The effect is a tiny distorsion, about 0.00048 degrees.
And the Sun is not exactly small.... but it's not dense enough to have a strong effect. Nothing in our everyday experience is. If you could stuff all of its mass into a tiny volume then in its immediate vicinity you could experience this.
The "barrier" you speak of (the event horizon) is already traveling outward at the speed of light. Gravity curves spacetime so much at the horizon that a light beam traveling radially outward stays at the same radius forever. Inside the horizon, spacetime is curved even more strongly, so that a light beam traveling radially outward still gets pulled inward towards the singularity.
The Hairy ball theorem requires it.
Are you somehow claiming that traveling in a straight line away from the black hole leads toward the black hole? If that were the case then gravity would be pulling objects away from the black hole, and toward it, at the same time, which makes little sense.
And if it existed anyway, despite making no sense, that would imply there is a point away from the black hole where the gravity of the black hole is exactly balanced (i.e. you could hover there forever), which would a stunningly huge thing to say.
If you mean which way is the event horizon traveling, it's traveling radially outward. That is a well-defined direction all by itself; there is no "left or right" involved.
The Hairy ball theorem requires it.
If you mean the theorem that says there can't be an everywhere non-vanishing vector field on a sphere, that theorem doesn't apply in this case because the manifold of the black hole spacetime is not a sphere (meaning, topologically it's not a sphere).
Are you somehow claiming that traveling in a straight line away from the black hole leads toward the black hole?
No. I'm saying that inside the horizon, even something that is moving as fast as it can away from the black hole still has a radial coordinate that decreases with time. That's because of the way the spacetime is curved; your intuitions about how "space" works don't work in a region of spacetime that is curved as strongly as the region inside the black hole's horizon.
that would imply there is a point away from the black hole where the gravity of the black hole is exactly balanced
How does it imply this? I don't understand your argument here.
> How does it imply this? I don't understand your argument here.
Emit a photon from just inside the event horizon, traveling directly away from "down".
What does the photon do?
Does it go into orbit? (What gave it the angular momentum to do that?) Does it hit the black hole? (How did it turbn around without ever curving?) Does it just travel forever thinking it's moving away from the black hole, but not actually going anywhere? (i.e. redshift into nothingness, since it's constantly fighting gravity)
Something else?
Tell me what it does from the POV of the photon, not the black hole.
That said, from just inside the black hole - what is "down"? Towards the singularity? Does that even make sense when you'll hit the singularity no matter what direction you travel?
From the POV of the photon, nothing special happens. You get emitted, you move towards the singularity, and then we have no idea what happens next. You might happen to take a longer path, but that's it.
The photon is traveling radially outward; that means it is traveling in a direction opposite to a line pointing "down". At least, that's the way it's traveling spatially. In spacetime, a photon at the horizon is traveling along a curve of constant radial coordinate r (and constant angular coordinates theta, phi if you include them). A photon inside the horizon is traveling along a curve of decreasing radial coordinate r; inside the horizon even outgoing null curves (the worldlines of outgoing light beams) have decreasing r.
Emit a photon from just inside the event horizon, traveling directly away from "down". What does the photon do?
According to an observer that is falling inward just inside the event horizon, the photon moves radially outward at the speed of light.
Does it go into orbit?
No. As you say, it has no angular momentum, but that's not the primary reason; the primary reason is that there are no "orbits" inside the horizon. In fact, there are no "orbits" inside a radius of 3/2 the horizon radius; at that radius, a photon can orbit the hole in a circular orbit.
Does it hit the black hole?
Eventually, yes.
(How did it turn around without ever curving?)
It didn't. The spacetime itself is curved inside the horizon to such an extent that even a photon traveling radially outward ends up hitting the singularity.
Once again, your intuitions about how "space" works and how things "travel in space" break down inside the horizon. You are thinking of a point with radius r < 2M, i.e., just inside the horizon, as a "place in space". It isn't. It's more correct to think of it as a "moment of time". The horizon itself is also not a place in space; it's more correct to think of it as an outgoing light beam.
Does it just travel forever thinking it's moving away from the black hole, but not actually going anywhere?
This is one way (but probably not the best way) to visualize what a photon exactly at the horizon does; it is moving radially outward at the speed of light, but because of the curvature of spacetime at the horizon it stays at the horizon.
(i.e. redshift into nothingness, since it's constantly fighting gravity)
"Redshift" is relative; you have to specify what observer is receiving the photon and measuring its frequency.
Tell me what it does from the POV of the photon, not the black hole.
There is no such thing as "the POV of the photon"; photons don't have "rest frames" in the usual sense of that term. I said above what the photon does from the POV of an observer falling into the hole.
They don't have to for your brain and nervous system to continue functioning. Suppose you are falling feet first towards the singularity, and your feet send a nerve signal to your head. From your viewpoint, the signal travels up your body normally. But from the "god's-eye" viewpoint, the nerve signal from your feet to your head just falls towards the singularity more slowly than your feet and head do, so your feet fall away from the signal and your head catches up with it.
If I can see the floor, then light from the floor is passing the horizon and entering my eye.
No, it isn't. Consider a ray of light emitted by the floor just below the horizon, and suppose that your eye is just above the horizon when it is emitted. By the time that ray of light reaches your eye, your eye will have fallen below the horizon. So your eye will see the ray of light just fine.
If you are trying to "hover" at a constant altitude above the black hole's horizon, then yes, the acceleration you would feel increases without bound the closer to the horizon you try to hover. But if you're falling into the hole you clearly aren't "hovering"; it is perfectly possible to fall in while feeling only 1 g by adjusting your rocket power properly.
The tidal gravity at the horizon might be very large, or very small, depending on the mass of the hole. If it's large enough, yes, tidal forces would tear you apart by the time you reached the horizon. But that would happen even if your center of mass was in free fall.
I guess my question is how fast you'd be traveling by the time you got near/in the event horizon.
Relative to what? There is no absolute answer to "how fast" you are going; velocity is relative.
In order for that to happen you would have to travel faster than light.
Huh? The light ray and you are moving towards each other. Why would you have to travel faster than light to reach it?
Eg: The light ray came from the floor 1 foot below the event horizon, and your eyes are 3 feet away from it.
No, this is incorrect. You are thinking of the event horizon as a "place in space". It's not. It's an outgoing light ray. And "locations" with a given radius inside the horizon are not places in space either; they are moments of time.
Here's how to unpack what I just said in more detail. What kind of curve is a curve of constant r (and constant theta, phi if we include the angular coordinates)? Our ordinary intuition about how "space" works says that a curve of constant r is timelike; that is, it is a possible worldline for an object. This means that an object can "stay in the same place" by staying at constant r for all time.
If a black hole is present, however, this ordinary intuition is only valid outside the event horizon. Outside the horizon, curves of constant r are timelike. However, the event horizon is a curve of constant r, and it is not timelike: it's null. It's the path of an outgoing light ray. A light ray always moves at the speed of light; it can never "stay in the same place". So you can't think of the event horizon as "a place in space" the way you can think of a curve of constant r outside the horizon.
Inside the horizon, a curve of constant r is spacelike. The physical meaning of a spacelike curve is that it is the set of all events that happen "at the same time" for some observer. So a curve of constant r inside the horizon is "a moment of time" rather than "a place in space".
So a light ray that is emitted "1 foot below the horizon" is really emitted "1 foot to the future of the horizon". Or, since light travels about 1 foot per nanosecond, the light from your feet is emitted 1 nanosecond to the future of the horizon. That light can't travel "upward" to meet your feet; that would mean traveling into the past. Instead, your feet have to fall through the horizon to meet the light.
This is not correct. You will only see the portion of the universe's history that is in your past light cone; there will be a lot of the life of the universe that you will never see if you fall into a black hole, because you will be destroyed in the singularity at the center before light from most of the universe has a chance to reach you.
Edit: some corrections.
It's like a swimmer going over a waterfall. The water is like spacetime being dragged towards the black hole, and the swimmer is an object moving within spacetime. If the waterfall is falling faster than the person can possibly swim, the person can never get back up the waterfall.
For more info on this "river model" of spacetime, see:
http://jila.colorado.edu/~ajsh/insidebh/waterfall.html
http://arxiv.org/abs/gr-qc/0411060
Relatedly: Since spacetime can "flow" faster than the speed of light, you might think this allows FTL travel. Indeed, this idea is called the "Alcubierre drive".
It's a good question, because it exposes some subtle assumptions being made.
The string breaks, because you cannot have an infinitely strong wire in a universe that allows black holes. If we could have such a wire, black holes would not be formed, since the wire's material would be able to resist being crushed into a black hole. Therefore collapsing stars would compress into this material rather than into black holes, and there would be no black holes.
Once inside the event horizon, every path leads to the singularity. You cannot get further away from it, and you cannot orbit it (or otherwise keep distance constant).
It doesn't seem impractical to be able to establish an orbit inside that radius, though, even a highly elliptical one, provided this orbit never slings you back out. It just seems if your orbit decays even slightly, you'll never be able to recover, leading to the inevitable conclusion of being pulled directly into the singularity.
(Note: this applies to an electrically neutral non-rotating black hole.)
For example, why can't it be the case that there are two kinds of matter, one that can be used to build infinitely strong wire, and one that can't, with stars being made of the latter? Or why can't it be the case that the crushing process doesn't take matter through the infinitely strong configuration on the way to black hole configuration?
There's a good argument to be made that this isn't necessarily the case. I was thinking that all particles are interchangeable at sufficiently high temperatures (i.e. they can switch from one particle to another so long as energy is conserved). I overlooked the fact that stars don't get hot enough.
> Or why can't it be the case that the crushing process doesn't take matter through the infinitely strong configuration on the way to black hole configuration?
If the "infinitely strong" configuration got crushed into a black hole, it wasn't infinitely strong.
> For example, why can't it be the case that there are two kinds of matter, one that can be used to build infinitely strong wire, and one that can't, with stars being made of the latter?
How would the infinitely strong wire be held together? Electromagnetic force? Strong force? Weak force? Those forces (which are the only ones we know about apart from gravity) travel at c. If part of the wire is in, while part is out, we will necessarily break the bonds holding the material together if we attempt to pull the wire clear. If nothing else, this breaking will occur because those bonds rely on force carriers traveling at c, and so somewhere along the wire those force carriers will not see the items they are supposed to bind together.
Let us say we hold the wire so that a portion of it is past the event horizon. The bit of the wire at the event horizon, if held stationary, is traveling at c because light is also stationary there. Whatever bond existed at that point is no more; the force carriers can't make it to the particles they're supposed to bind together.
About your other questions ( and being a bit sloppy with my definitions): Inside the black hole the radial coordinate becomes timelike, that means the singularity is in the future of every particle which has fallen into the black hole. And you can not orbit around the singularity any more than you can orbit around next Thursday.
* the gravitational acceleration at the event horizon can be overcome by your rocket engines
* the tidal forces at the event horizon don't kill you
You'd be in free fall, and that's a perfectly nice reference frame. Within that frame, your gravitational sensors would observe certain doom in every direction. This sphere of doom would get smaller until you were ripped apart from the tidal stress.
Well, that's how reality actually works. Statements about reality are either true or false. How many physicists would even agree with that? Therein lies my distrust of modern physics, or at least part of it.
I'm no fan of Popper, though.
The history of science (Kuhn, etc) shows that this is just not how it works.
I'm no fan of Kuhn either, though. He also probably wouldn't agree with my litmus test of "Statements about reality are either true or false."
I'm no physicist, but if gravity were to act on a human body with that much force, wouldn't the force alone kill you? I imagine the force of the pull would be stronger than the force holding together the atoms that make up the tissue of your body, and you would disintegrate into a stream of particles accelerating toward the singularity. Am I way off here? If a smaller object approached an object of such great mass that not even light can escape its gravitational pull, wouldn't the smaller mass's particles inevitably be funneled into a single-file line until they are absorbed into the singularity?
No, because you don't feel this "force" of gravity; you are in free fall, just like astronauts on the International Space Station.
Your body will be subject to tidal gravity because it has a finite size; the effects you talk about in the rest of your post (your body's atoms being pulled apart, etc.) are due to tidal gravity, not "the force of gravity" per se. But for a large enough black hole, tidal gravity is small enough to be bearable at the horizon.
A constant gravitational field doesn't change anything - everything is accelerating the same amount, in the same direction. What can cause problems is the difference between the gravitational field at two different points (for example, when a planet gets too close to a star, it can be torn apart due to the force exerted on the near side being much greater than the force exerted on the far side). But this isn't the same as the overall strength of the field, and a sufficiently large black hole could have a very survivable field gradient at the event horizon.
They will if there's enough of a gradient and the observers are not ideal points.
Many years ago, Larry Niven wrote a story (http://en.wikipedia.org/wiki/Neutron_Star_%28short_story%29 ) where tidal forces killed observers trying to orbit a neutron star.
The tidal forces and the x-rays resulting from said forces are pretty much lethally intense.
Tidal gravity is equivalent to spacetime curvature, so it's there whether or not there is matter falling into the hole; the term "tidal force" is sometimes used to refer just to that fact. But X-rays won't be produced unless enough matter is falling in and being acted on by the tidal force. If there were no large quantities of matter falling in, a very small object like, say, an astronaut in a rocket ship, could fall in and not encounter any X-rays, but the tidal force would still be there (and could potentially pull his ship and him apart). I wasn't sure if that distinction was clear.
In the ordinary universe this often plays out on small moons in a low orbit around very large planets such as Jupiter.
A definition (http://en.wikipedia.org/wiki/Tidal_force): "The term tidal force is used to describe the forces due to tidal acceleration." You can't have acceleration without an object being accelerated.
Not trying to split hairs here, but to try and establish if it's possible to have a force that's not acting on anything.
This is one way of looking at it, but it's not very general. The more general way of looking at it is to look at two nearby "test objects" (meaning objects so small that they don't affect the gravitational field enough to make a difference), both in free fall, that start out at rest relative to each other. If they don't stay at rest relative to each other, then tidal gravity is present.
The term "tidal force", on the more general view, refers to forces felt by objects that are not allowed to follow the free-fall paths that tidal gravity is trying to make them follow. For example, consider a small piece of the Earth on the side facing the Moon, and another small piece on the side opposite the Moon. If these two pieces of matter were in independent free-fall orbits, they would separate due to the tidal gravity produced by the Moon. Since the two pieces of matter are not independent--they are both attached to the Earth--they both feel a "tidal force" because they are being prevented by the rest of the Earth from following the "normal" free-fall orbit they would follow due to the Moon's gravity if they were not attached to the Earth.
if it's possible to have a force that's not acting on anything.
If you don't like using the term "force" when there's no object feeling the force, then you can just use the term "tidal gravity", as I did above, to refer to the underlying spacetime curvature that, when objects are present to feel the force, causes "tidal force". So tidal gravity can be present even when there is no object feeling tidal force due to the tidal gravity.
Therefore, what really happens? No one knows for sure. If the no-drama principle correct (which is implied by Einstein's theory of gravity), in that you cannot feel gravity when you are in free-fall, you pass through the event horizon just fine, and then die when you are torn apart by tidal forces. (Of course, the tidal forces kick in outside the event horizon in smaller black holes - but that's beside the point). If the no-drama principle is correct, then there is a fundamental and insoluble contradiction in our understanding of physics. That's the point of the article.
falling into it, it has enough degrees of freedom to put you into any kind of virtual reality, so you might experience falling into a singularity or somehow travelling into an "asymptotically flat spacetime" but I'm certain that the classical picture of a black hole is completely wrong.
There is plenty of reason to believe that, if a horizon does form, the region of spacetime on the other side will be spacetime, not something completely different.
There are some (speculative) reasons to believe that quantum effects might possibly prevent horizons from ever forming, but that would mean there is no "other side".
This looks like a mixture of misunderstanding the classical picture of a black hole, and misunderstanding how the classical picture is altered by quantum effects.
Classically, stuff that falls inside the black hole does not "stack up" around the horizon; it just falls in, past the horizon and on to the singularity, where it is destroyed.
There is no established theory of how quantum effects alter the classical picture, but there is at least one "semi-classical" picture of how Hawking radiation is produced (the one that Hawking originally came up with) that doesn't require any "grinding up" of infalling matter at the horizon.