One small nitpick about a statement in the article; it says:
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.
Technically there may be no "safe distance" at all, depending on how much time you're talking about. The whole galaxy seems to be caught in the spiral of a super massive blackhole, and it's just a matter of time before everything falls into it, although most stars will probably die out before they do. So I guess you could consider that "safe", since the time it takes to fall into the blackhole is longer than the time the object itself exists. Whole galaxies are probably drawn to each other, too, like we are to Andromeda and Andromeda to us.
It deserves mentioning that it is very difficult for things in space to fall into each other. If two bodies aren't on a direct collision course they will orbit or slingshot, not "circle the drain until they fall in." The effects that do promote "falling in" happen on very, very long timescales with respect to anything, including stellar lifecycles (you mentioned this but didn't emphasize the extent to which one timescale utterly dominates the other).
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 :)
Actually there is a way that objects orbiting other objects can shed angular momentum: emit gravitational waves. We've observed this with binary pulsar systems; it's expected that it would also be taking place with objects orbiting the black hole at the center of our galaxy. This is one of those "long timescale" effects in most cases, but for objects close enough to the black hole at the center of our galaxy its time scale might not actually be longer than the lifetime of some of those objects. (That would only be true pretty close to the hole, though.)
The whole galaxy seems to be caught in the spiral of a super massive blackhole, and it's just a matter of time before everything falls into it
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 central supermassive black hole does consume the galaxy, but in fact the rate of consumption is so excruciatingly slow that the galaxy will have "evaporated" long before then. Random processes like collisions with other galaxies will eventually eject nearly all stellar remnants into intergalactic space. The central black hole will only end up consuming about 1% of the galaxy's stellar remnants (as you say, all stars will have died trillions of years before the galaxy evaporates).
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!
Well, I was told by an astrophysicist on /r/askscience, that inside of 1.5x of the event horizon, there are no more stable orbits (you either go hyperbolic, or fall in). Outside of that distance, you can absolutely orbit a black hole just like any other body, with the same chances of outside forces influencing your orbit.
He shouldn't have used the word "stable". Orbits from 1.5x the horizon radius to 3x the horizon radius exist, but they are unstable, like a pencil balanced on its point; any small perturbation will cause the orbiting object to either fall into the black hole, or fly off to infinity. Orbits at 3x the horizon radius or larger are stable in the sense that small perturbations do not do this; the object will just tend to adopt a slightly different orbit.
Since the post I was responding to used the word "safe", I was not including the orbits from 1.5x to 3x the horizon radius, since they are not, IMO, "safe"; you'll need rocket power continuously available to avoid falling into the hole or flying off to infinity due to small perturbations.
(A link to the askscience thread would be nice, btw.)
A question for the physicists on HN: I've heard that black holes emit radiation. But if nothing can go faster than the speed of light, and a black hole's gravity is so strong that not even light can escape, then how can a black hole emit anything?
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
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.
> 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)
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?
It doesn't take any energy to create virtual particles; virtual particle pairs are constantly being created and destroyed everywhere, according to Quantum Field Theory, but when they're created, on average, they have zero net energy: one has positive energy and one has negative energy. (Note that this is a heuristic description and not every quantum field theorist would agree with it. The only really unambiguous way to describe the process is using math; but translating math into everyday language is often difficult because our intuitions don't really match up with what the math is telling us. I'm doing the best I can.)
However, if a virtual particle pair happens to be created just outside a black hole's horizon, the hole's tidal gravity can pull the negative energy particle inside the horizon before it can be annihilated by the positive energy particle. The positive energy particle can then escape. Effectively, this means the positive energy particle's energy is taken from the hole's mass, so the hole's mass decreases slightly.
IanCal and blaze33 both gave good answers, but just to clarify one thing: the radiation we see coming from regions where there are black holes is of the first type: radiation emitted by objects like gas clouds that are falling into the holes, before those objects cross the event horizon. If we leave out quantum effects like Hawking radiation (see below), it's impossible for light, or any kind of radiation, or indeed anything at all, to escape from inside the event horizon of a black hole.
Hawking radiation is a quantum effect, which nobody has ever observed; the reasons for thinking that it exists are purely theoretical.
Isn't the nonexistence of large black holes from cosmic ray collisions proof of black hole evaporation? Or have we not been able to measure that / have reason to doubt micro-black-holes are created by cosmic ray collisions in the first place?
I would say we don't have accurate enough measurements or an accurate enough theoretical understanding to know how many micro-black-holes we should expect to see from cosmic ray collisions, on the assumption that none of them ever evaporate, or to be able to measure how many there actually are, so as to be able to compare the two numbers to see if there's a significant difference. In principle this would certainly be a good experimental test for the existence of black hole evaporation; I just don't think it's a test we can make with any confidence now or in the near future.
You may be referencing two different kind of radiations:
- 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]
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?)
Also armchair physicist here - my understanding is that since information can never be destroyed, Blackholes must emit something.
I think there's pretty general agreement among physicists that we will eventually confirm that black holes emit Hawking radiation. However, that by itself doesn't help us to choose between all the various proposed resolutions of the information paradox.
The simplest answer: it's the region immediately above the black hole that these reports refer to, not the black hole itself (its event horizon and below there).
If only black holes weren't illogical! [1] Any decent software developer (i.e. highly logical thinker) who knows basic info about black holes can look at the picture there to quickly see that black holes are inconsistent with the core postulate of general relativity, the equivalence principle. They're a bug! (which Einstein spent a decade trying to find, or some other way that nature could prevent black holes he thought didn't "smell right", and no that's not my blog) [2] is previous discussion showing no bona fide flaw in the argument. I've not seen anyone refute that simple picture, which should be easy if it's incorrect.
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).
Here's another relevant discussion on HN that you conveniently forgot to link to, which does explain in detail the flaw in your argument (the discussion you did link to does too, but not in detail):
You've linked to that before, but there's nothing there that refutes the blog (not my argument). If there's a flaw it should be a sentence or two of logic, not multiple paragraphs that summarize as "it's not that simple". The blog shows that inertial frames falling through a horizon are that simple (after all, the equivalence principle demands that), and Taylor and Wheeler agree there. Extraordinary evidence is not a requirement per the scientific method. "Extraordinary" is not a scientific concept.
I understand that you don't agree that what I posted refutes the blog, but I'm not trying to convince you, and I'm not going to rehash the argument here. I'm just linking to that discussion for the record, so others reading this thread will understand that your claims and the blog's claims about black holes are not undisputed.
(not my argument).
You may not have written the blog post, but you are claiming it's correct, so it is "your" argument in the only sense that matters here. If you're not willing to own the argument when it's challenged, then you shouldn't be linking to it.
If there's a flaw it should be a sentence or two of logic, not multiple paragraphs that summarize as "it's not that simple"
That's not the summary of what I said. The summary of what I said is this: the blog post's claims about what the theory of relativity actually says are not correct. So the blog post is not refuting the actual theory of relativity; it's refuting a straw man version of the theory that the blog post's author has constructed in his own head.
Taylor and Wheeler agree there
Agree with what? That there can be an inertial frame that falls through the horizon? Yes, of course. But that does not mean Taylor and Wheeler agree with the blog post's claims about black holes. The fact that there can be an inertial frame that falls through the horizon does not mean the blog post's claims about the details of how such frames work are correct. I went into detail about what's wrong with them in the thread I linked to.
What sentence in that picture's caption is wrong, or what doesn't follow from its premises? Asking for extraordinary evidence isn't a scientific thing. Valid logic is all the evidence needed.
Yes it's my argument in that way. I'll challenge a refutation or accept it, but that's difficult when the counter argument is a vague wall of text. If I were to try to summarize what you think is wrong I couldn't do it. I can't decipher your points to see how they attempt to refute any particular sentence in the blog. Please be way more clear and short and to the point, and I'll agree, or disagree with my reasoning. (For example check out Millstone's reasoning above. He/she disagrees that an inertial frame can cross the horizon. One short sentence is enough to summarize!)
> the blog post's claims about what the theory of relativity actually says are not correct
Anyone can say that; be more specific. What the blog says in that way sources from experts in relativity. Its statement of the equivalence principle, for example, is Kip Thorne's.
> Agree with what?
They agree that frame X in the blog post is validly defined and validly used in the thought experiment. Their quote in the blog makes that clear.
I can't decipher your points to see how they refutes any particular sentence in the blog.
That's because the blog post itself is not very well written, and doesn't state correctly what relativity actually says.
Please be way more clear and short and to the point
I'll give it another shot below.
They agree that frame X in the blog post is validly defined and validly used in the thought experiment. Their quote in the blog makes that clear.
It does no such thing. Taylor and Wheeler do not say that the blog post's "Law J" and "Law K" are correct statements of what relativity actually says. They're not; they're not even stated precisely enough to have a definite meaning that can be compared with what relativity says.
Rather than try to go through the blog post in detail, let me try stating two simpler puzzles that bring out what seem to me to be the essential points. Here's the setting for both puzzles: an astronaut is falling through the event horizon of a black hole. Just before reaching the horizon, he launches a probe outward at escape velocity, which will be just a smidgen less than the speed of light.
Now consider the astronaut's local inertial frame (LIF) as he crosses the horizon: i.e., the origin (t = 0, x = 0) of this frame is the event at which the astronaut crosses the horizon, and the time axis of the frame is the astronaut's worldline as he falls in. According to relativity, the following are all true statements:
(1) The horizon is an outward-moving lightlike curve that passes through the frame's origin; i.e., it is the line t = x in that frame.
(2) The probe's worldline starts at t = some value just a little bit less than zero, x = 0, and moves in the positive x direction at just a bit less than the speed of light.
(3) Therefore, in this LIF, the horizon is moving outward faster than the probe.
The first puzzle is simple: if all of the above are true, how can the probe ever escape? Won't the horizon catch it? (Meaning, won't it end up below the horizon, not escaping to infinity?)
The resolution of this puzzle is equally simple: the size of the LIF is much, much smaller than the predicted distance, extrapolated from within the LIF, that it will take for the horizon to catch the probe. Therefore, within the LIF, the probe remains outside the horizon. And once we're outside the LIF, tidal gravity is not negligible, and it will "pull" the horizon down with respect to the probe, so the horizon will never catch the probe.
The calculations underlying what I just said about the size of the LIF vs. the distance required for the horizon to catch the probe are on PhysicsForums here:
The second puzzle is a bit more subtle. Consider the following statements, which are also true according to relativity:
(4) With respect to a global coordinate chart describing the black hole spacetime, the horizon is at a constant radial coordinate r.
(5) With respect to the same global coordinate chart, the probe's radial coordinate r is increasing.
(6) However, the distance from the probe to the horizon, in the astronaut's LIF, is decreasing, at least while the probe and horizon remain within the LIF. (This is obvious from statement #3 given earlier.)
The second puzzle then is, how can #5 and #6 be reconciled? How can the probe have increasing r when it's getting closer to the horizon, which is at constant r?
The best way I know of to see the resolution this second puzzle is to actually draw a spacetime diagram of the LIF, with curves of constant r correctly drawn. If you do that, you will see that the curves of constant r are drawn in such a way that the probe's worldline has increasing r even though its distance from the horizon, in terms of the LIF's x coordinate, decreases. I can't draw such a diagram here, but the key fact that makes this true can be seen by considering statement #4 above, combined with statement #1: the horizon is a curve of constant r, but in the LIF, it is a 45 degree line moving up and to the right! (It's the line t = x in the LIF, per #1.) A curve of constant r lying just outside the horizon will slope up and to the right, a bit more vertical than 45 degrees (which makes it timelike); whereas a curve of constant r just inside the horizon will slope up and to the right a bit more horizontal than 45 degrees (which makes it spacelike). The slope of the lines of constant r that pass through the probe's worldline, which also slopes up and to the right at almost 45 degrees (since the probe is moving outward at almost the speed of light) turn out to be sloped a little closer to 45 degrees than the probe's worldline is; so the probe ends up crossing curves with gradually increasing values of r.
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.
the event horizon accelerates in all nearby inertial frame
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.
> 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
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.
Wow, what a near miss. Would that have been visible at all? To think that if human technology had kicked off a couple centuries earlier we might have observed a black hole eating something already, that the answers to our questions about black holes could be common knowledge... wow.
If you read Stephen Hawking's book "A Brief History of Time" you will see that a variant on this question is what started him on the path of reasoning about black holes in the first place.
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.
> Basically if you were standing at the event horizon the pull on your feet would be several billion times the pull on your head.
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.
if you were standing at the event horizon the pull on your feet would be several billion times the pull on your head.
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.
Sigh. And who are you going to inspire to study physics with those dry facts? :-) I wasn't going for depth here, shire could just goto google scholar or a decent physics library and follow the citation references off of Dr. Hawking's work if they want to pursue this in depth. I was shooting for a bit more pizazz than that :-).
It is a solid point though that the tidal forces on the event horizon of a really big black hole would be minimal. You'd be just as dead though. But I stand by my assertion that "Something pedantically accurate like 'nothing alive'" would be pretty boring and not really convey the interesting aspects of gravitational theory that result in singularities on what most folks consider "normal" space-time.
I was shooting for a bit more pizazz than that :-).
Pizazz is fine, but not at the expense of truth.
You'd be just as dead though
Eventually, yes. Not at the horizon.
singularities on what most folks consider "normal" space-time.
There is a singularity at the "center" of the black hole, yes. (I put "center" in scare-quotes because a black hole doesn't have a "center" in the usual sense; but we don't have a better word for it.) But not at the horizon.
According to classical general relativity (i.e., without taking any quantum effects into account), if you fall inside the hole's event horizon, you will at some point get torn apart and destroyed by tidal gravity. (If the hole is small enough, that could happen even before you reach the horizon). Eventually even the atoms and subatomic particles that you were composed of will be destroyed in the singularity inside the hole.
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.
To be completely honest - There is no straight answer. One modern answer was that particles falling into a black hole would appear to be destroyed for outside observer (in a vortex of fire) while the observer would be scrambled into a holographic representation (keep in mind ashes would be a highly energetic representation of same observer).
well yesterday some were talking about tracking a plane lost in earth and today i read about tracking a gas cloud in space!And supposedly the latter seems more like a soccer penalty shot!
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.