Is there a kind of black hole which is so massive that not even that radiation can escape, or do all black holes emit some kind of radiation? (In fact, do they emit radiation proportional to their size?)
There are two reasons why you'd say black holes emit radiation. An interesting one and a very interesting one (warning, other peoples scales may be calibrated differently to mine).
1) Black holes accelerate things massively, so they're travelling at an astonishing speed before they "enter". This can result in huge amounts of x-rays being emitted due to heating things to millions of degrees (well beyond white hot!). It's not the black hole itself, but the black hole is certainly to blame. http://hyperphysics.phy-astr.gsu.edu/hbase/astro/blkbin.html
2) The weirder one. Hawking radiation: http://en.wikipedia.org/wiki/Hawking_radiation
Hawking radiation happens when a pair of virtual particles "pop" into existence near the event horizon. Normally these pairs annihilate quickly, but if it happens near the event horizon it's possible for one of the particles to fall in and the other to escape. This results in a loss of mass of the black hole (told you it was weird) so could be considered to be the black hole emitting radiation.
How does the energy to create the virtual particles come from the black hole (which it has to in order for the accounting to work: -2+1=-1)? Is it a "Quantum Field Theory doesn't care about the event horizon" type thing?
Hawking radiation is a quantum effect, which nobody has ever observed; the reasons for thinking that it exists are purely theoretical.
- the one emitted by the heated matter falling in / spinning around the black hole (the quasar referenced in the article [1])
- the Hawking radiation. Basically a pair of particle/antiparticle randomly appears near the black hole event horizon, one of the particle falls into the black hole while the other one escapes. The black hole would eventually evaporate if you waited long enough. [2]
[1] http://en.wikipedia.org/wiki/Quasar [2] http://en.wikipedia.org/wiki/Hawking_radiation
Also armchair physicist here - my understanding is that since information can never be destroyed, Blackholes must emit something.
The question does not currently have a definitive answer. Although current mathematical analysis has in falling matter being dismantled at the sub-atomic level as it undergoes the tidal stresses associated with gravity. Basically if you were standing at the event horizon the pull on your feet would be several billion times the pull on your head.
The confounding factor is that if you're falling into a black hole the acceleration can get your velocity to nearly light speed, and at that velocity your perception of time slows, to the point of nearly stopping. So while people watching you fall in might see a burst of xrays as your physical being converted into energy, "you" might perceive nothing at all, simply that time stopped (which is really not something you can perceive) followed by your non-existence (which depending on your theology either has you returning you energy to the entropy of the universe or a visit with your deity and/or anti-deity if there a judgement step.)
Most theories do not currently postulate a 'far side' of a black hole, mostly because "hole" is a metaphor rather than a physical description of the object. In theory its really just a point where the numbers go out of whack because the equations have a divide by zero error there. This too is what fascinates a lot of people, the universe sets up this problem where it gets to divide by zero. A bit of calculus, a bit of fudging with infinities of both the positive and negative variety, and your guess is as good as anyone else's at this point.
Fun to think about though.
This is not always the case. If the black hole is massive enough there won't be strong tidal force at the event horizon to disintegrate objects.
I don't know GR enough to judge your other statements. However, when discussing the time slowing down close by a black hole I suspect one must take into account of the strong gravity field and not just the velocity of the falling object.
This is true for a black hole with mass a few times the mass of the Sun, the sort we expect to be formed by the gravitational collapse of stars. However, a much larger black hole would have much less tidal gravity at the horizon. Some of the supermassive black holes that are believed to be at the centers of quasars would have less tidal gravity at their horizon than you feel on the surface of the Earth.
The confounding factor is that if you're falling into a black hole the acceleration can get your velocity to nearly light speed
Velocity is relative. You will be moving at the speed of light relative to observers who are "hovering" just outside the hole's horizon; but other people falling into the hole just ahead of or behind you could have much smaller velocities relative to you, even well after you cross the horizon (depending on how close they were to you to start with).
and at that velocity your perception of time slows, to the point of nearly stopping
Your perception of time would be normal; you would notice nothing unusual in the behavior of clocks you carried with you, even well after you fell inside the horizon (assuming the tidal gravity was bearable--see above).
Also, "time" as you're using it here is relative; there is no absolute notion of "perception of time".
while people watching you fall in might see a burst of xrays as your physical being converted into energy, "you" might perceive nothing at all
They would only see this if it happened outside the horizon, which it might if the hole's tidal gravity was large enough outside the horizon. But in this case, while it would be true that you would perceive nothing at all, that would simply be because you would be turned into x-rays and destroyed; it would have nothing to do with any change in your "perception of time" due to relativity (see above).
Once you reach the horizon, even if you emit x-rays, nobody outside the hole will ever see them, since light emitted at or inside the horizon can never get back out. But your "perception of time" will continue just fine, assuming again that tidal gravity is bearable (see above).
Most theories do not currently postulate a 'far side' of a black hole
By "far side" do you mean a region inside the event horizon? If so, you are wrong; our current theories most certainly do predict (not postulate, it's not an assumption, it's derived as a theorem) that there is spacetime inside the horizons of black holes.
In theory its really just a point where the numbers go out of whack because the equations have a divide by zero error there.
This is only true in a particular system of coordinates; it is not true as a statement about the actual physics. That is, there is no actual problem with spacetime at the horizon; all physical quantities are perfectly finite there. The "divide by zero error" is a purely mathematical problem with one coordinate chart, which can be fixed simply by using different coordinates.
When you include quantum effects, we don't have a good theory at this point to predict what happens; it depends on how the "black hole information paradox" is finally resolved. However, in terms of what will happen to you in practical terms, not much changes from the above: you will still most likely get torn apart by either tidal gravity or some kind of quantum "firewall". The details of what happens after that won't make much difference to you in practical terms.
See
Though we think of them as cosmic vacuum cleaners, black holes are actually just like any other massive body, such as a star. This means other objects can safely orbit them, until they get within a particular distance and pass what’s known as the event horizon, after which there is no escaping being sucked in.
Technically, you can't "safely orbit" a black hole if you're any closer to it than three times the horizon radius. That is, there are no stable free-falling orbits, like orbits around a planet or star, closer than that. If you want to get closer than that to the event horizon without falling in, you will need to continuously fire your rockets to hold yourself at altitude against the hole's gravity.
We are used to circulating fluid inevitably falling into the center of a drain, but this only happens in our daily lives because viscosity allows the water to shed angular momentum (about the drain) to its surroundings. No angular momentum transfer = no falling into the center, and a galaxy doesn't have a gigantic porcelain fixture anchored to the central black hole to which stars can transfer their angular momentum :)
That's not necessarily true. From far away, a black hole is just like any other object with the same mass; there's no extra ingredient to a black hole's gravity that makes things more likely to fall into it from far away.
The Milky Way is fated for just such a collision with Andromeda in a few billion years, and there's a chance our solar system will be ejected!
That black holes are a bug doesn't mean we can't explain observations. Gravitational redshift still exists to explain massive objects emitting no discernible light (might be only one photon every year, away from Earth).
https://news.ycombinator.com/item?id=5233894
Click on "parent" to see the post of yours that I was responding to.
This picture shows the event horizon as having constant position in a free-falling frame. But the event horizon accelerates in all nearby inertial frames, exactly like how the surface of the earth accelerates (towards you) in every inertial frame near the earth. So the line representing the event horizon cannot be straight: it must curve. It must be hyperbolic. That is the key feature that the article gets wrong.
> Then law J shows that a locally inertial frame relative to which the particle is at rest can’t even in principle extend below the horizon
The horizon forms a hyperbola in any such inertial frame. Because the horizon accelerates, whether a spatial point is below the horizon depends on time.
If you have a test particle moving with constant acceleration, can you reach it with a signal at some point after it leaves? In Newtonian physics you can: just send the signal fast enough. But in relativity, test particles moving with constant acceleration can be "uncatchable" even though they never reach c. See Rindler coordinates for more.
This illustrates why an outgoing test particle cannot cross the event horizon. Picture the event horizon as a hyperbola, and the test particle as an asymptote: it can never intersect it.
> a test of the laws of physics can distinguish X from an inertial frame in an idealized, gravity-free universe
Well duh. There are no inertial frames that cross the horizon. Technically any two separated points in our frame will be accelerating with respect to each other. We approximate this acceleration as zero, but this approximation can break down near the speed of light, as things tend to do.
No, it doesn't. The event horizon is an outgoing null geodesic; it has zero proper acceleration. It is certainly not anything like the surface of the Earth.
The horizon forms a hyperbola in any such inertial frame
No, it doesn't. The horizon is an outgoing lightlike surface, so in any local inertial frame that contains it, it will be a straight line moving up and to the right at 45 degrees. (See my post in response to fargolime upthread for a more detailed description.)
in relativity, test particles moving with constant acceleration can be "uncatchable" even though they never reach c. See Rindler coordinates for more.
It's true that the Rindler horizon gives a good flat spacetime analogy for many key features of the event horizon of a black hole. But you appear to have things backwards: the object moving at constant acceleration is not the Rindler horizon; it is "uncatchable" by the Rindler horizon! In other words, the Rindler horizon is a light ray moving in the same direction as the accelerating object, but which never quite catches up with it (because the test object has just enough of a head start).
This illustrates why an outgoing test particle cannot cross the event horizon.
No, it doesn't. The reason an outgoing test particle inside the horizon can't cross it is simple: the horizon is moving outward at the speed of light, and nothing can go faster than light.
There are no inertial frames that cross the horizon
This is not correct; there are, just as Taylor and Wheeler say. There are plenty of wrong statements in the blog post, but this is not one of them.
The line represents the horizon at a single moment in the life of frame X, during which it has a specific position relative to X and is not accelerating relative to X.
> If you have a test particle moving with constant acceleration, ...
The test particles are defined as freely falling. The frame X is defined as inertial. No free particle can accelerate in an inertial frame (at least not measurably), including in relativity. Check out Taylor and Wheeler's quote at the bottom of the blog post: "Keys, coins, and coffee cups continue to move in straight lines with constant speed in such a local free-float frame." According to you, those things would not move at constant speed; rather they'd be "moving with constant acceleration".
> Well duh. There are no inertial frames that cross the horizon.
This again disagrees with Taylor and Wheeler's quote. They are clear that not only can an inertial frame cross a horizon, but also that frame is "a free-float frame like any other".
I didn't respond to everything you said because it hinges on the same disagreement. Before we could get further in the discussion you would need to accept Taylor and Wheeler there, or show how I'm misunderstanding the conflict between what you said and their quote that seems to explicitly disagree with you.
Astronomers know that as recently as a few hundred years
ago, the Milky Way’s central supermassive black hole was
producing much more radiation.