Scientists May Get Best View Yet of a Black Hole in Action
wired.com
wired.com
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.
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!
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 :)
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.)
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?
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.
- 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
Hawking radiation is a quantum effect, which nobody has ever observed; the reasons for thinking that it exists are purely theoretical.
Also armchair physicist here - my understanding is that since information can never be destroyed, Blackholes must emit something.
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.
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.
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.
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:
http://www.physicsforums.com/showpost.php?p=4285946&postcoun...
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.
> 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.
I'm discussing the simple picture in the blog and its caption. Not a more complex puzzle, or Laws J & K. The cloud of particles is demanded by GR to be splitting in two along the horizon, when all the particles above the horizon are let to be escaping. This "splitting in two" contradicts the equivalence principle and occurs in every arbitrarily short period of time in the life of frame X. There is always an arbitrarily short duration of time available in any LIF, so frame X's size in spacetime is not an issue.
When you talk about "global coordinate chart" and a "lightlike curve" you're being unnecessarily complex, I say. We're talking a simple inertial frame of special relativity here, plus one basic prediction of GR as it relates to the horizon, namely that everything below the horizon moves inexorably inward toward the singularity. This discussion can be much simpler than you're making it. After reading your explanation I shouldn't be left wondering which sentence in the blog is incorrect or doesn't follow from its premises.
The cloud splits in two and that's a "bug". To prove the blog wrong you need to show that the cloud doesn't split in two, that all the cloud's particles can in fact move outward in formation. Can you do that in a way that clearly shows which sentence in the picture's caption is incorrect?
The cloud of particles is demanded by GR to be splitting in two along the horizon
This is wrong; GR "demands" no such thing. The blog post's author has misunderstood GR in this respect. See below.
When you talk about "global coordinate chart" and a "lightlike curve" you're being unnecessarily complex, I say
And I say that you are making a big mistake here, because a proper understanding of those concepts and how they relate to the LIF centered on the horizon is crucial to properly understanding and stating what GR says about this scenario. You'll see several examples of this below.
one basic prediction of GR as it relates to the horizon, namely that everything below the horizon moves inexorably inward toward the singularity
That means everything below the horizon must decrease its r coordinate in the global coordinate chart. It does not mean that everything below the horizon must decrease its x coordinate in the LIF. Remember: the horizon itself is the line t = x in the LIF; i.e., the horizon is moving in the positive x direction in the LIF at the speed of light, and it passes through the origin of the LIF at t = 0, x = 0. So a "cloud" particle that is at some negative x value at t = 0 in the LIF will be inside the horizon, and a "cloud" particle that is at some positive x value at t = 0 in the LIF will be outside the horizon; and it is perfectly possible for the first particle (inside the horizon) to be moving in the positive x direction in the LIF faster than the second particle. And this can be true even if the second particle (the one outside the horizon) is moving at "escape velocity". And it can also be true even though the first particle's r coordinate is decreasing and the second particle's r coordinate is increasing. My previous post gave more details on why those statements are true.
Oh, and one other thing: what does "toward the singularity" mean in the LIF? Which direction is the singularity in? The answer is, it's in the positive t direction--i.e., it's in the future. So any object in the LIF is moving "toward the singularity" as far as the LIF is concerned, since all objects move towards the future. There is no way to tell, from within the LIF, which objects are going to ultimately hit the singularity and which ones are not. That's a global concept.
To prove the blog wrong you need to show that the cloud doesn't split in two, that all the cloud's particles can in fact move outward in formation.
My previous posts (plus all the ones in the earlier HN thread I linked to, plus what I posted on PF) have already done that, multiple times. However, I've given a quick recap above.
This discussion can be much simpler than you're making it.
If only that were true.
Agreed.
> It does not mean that everything below the horizon must decrease its x coordinate in the LIF.
Agreed.
> ...it is perfectly possible for the first particle (inside the horizon) to be moving in the positive x direction in the LIF faster than the second particle.
Agreed.
> And this can be true even if the second particle (the one outside the horizon) is moving at "escape velocity".
Agreed.
> And it can also be true even though the first particle's r coordinate is decreasing and the second particle's r coordinate is increasing.
Disagree. There is no way you could show this for an inertial frame falling in the Earth's atmosphere, like for a skydiver (ignoring air friction). The EP demands that the laws of physics in the skydiver's frame and frame X are the same, so what you say here should be the same for the skydiver. Of course in the skydiver's frame you wouldn't use terms like "global coordinate chart" and a "lightlike curve", as that would be unnecessarily complex.
In the skydiver's LIF under the conditions above, the first particle (the lower particle) would always be moving in the positive x direction in the LIF slower than the second (upper) particle, regardless of the skydiver's speed relative to the Earth, and regardless of the second particle's speed relative to the Earth (i.e. it doesn't need to be escaping).
Then you are disagreeing with the theory of relativity, because what I've said is what the theory of relativity says. See below.
> The EP demands that the laws of physics in the skydiver's frame and frame X are the same
True.
> so what you say here should be the same for the skydiver
False, because there is no law of physics that says the r coordinate has to behave the same in every LIF. The r coordinate is a global coordinate, not a coordinate in the LIF; so as soon as you talk about the r coordinate, you are not just talking about the LIF, you are talking about the relationship between the LIF and a global coordinate chart. And there is no law of physics that says that relationship must be the same for every LIF. In fact that relationship is very different for the skydiver LIF as compared to the LIF that is falling through the horizon of a black hole. So any reasoning you do based on the assumption that that relationship is the same for both is simply wrong.
> Of course in the skydiver's frame you wouldn't use terms like "global coordinate chart"
As soon as you talk about the r coordinate, you are using a global coordinate chart, whether you realize it or not. So by not using such terms, you are failing to understand a key aspect of the scenario.
> as that would be unnecessarily complex.
It's (somewhat) complex, yes, but it's not "unnecessarily" complex. As I've said several times, understanding the proper relationship between the LIF and the global r coordinate is crucial if you want to correctly state what relativity says about this scenario. You and the blog post author have given excellent demonstrations of the mistakes you make if you don't have that understanding.
I'm responding to this separately because it was easier than trying to cram this plus my other responses into one post. If you are going to talk about what is or is not the same in the skydiver LIF and the LIF falling through the black hole's horizon, you have to first make sure the initial conditions are set up the same. Here's how you would do that:
(1) The LIF is in free fall, i.e., the astronaut/skydiver who is at rest in the LIF is freely falling in the gravitational field of some central body.
(2) At time t = 0 in the LIF, the astronaut/skydiver meets an outgoing light ray. (In the LIF falling through the black hole's horizon, this outgoing light ray is the horizon; in the skydiver LIF, it's just whatever outgoing light ray happens to be passing him at t = 0. Within the LIFs, there is no way to distinguish the two.)
(3) At some time t = minus epsilon in the LIF, the astronaut/skydiver releases a probe that flies outward at nearly the speed of light. (This is a key point that I don't think you understand: the initial condition in the LIF is that the relative velocity of the probe and the astronaut/skydiver must be the same. It is not that the probe's initial velocity is escape velocity. "Escape velocity" is a global concept, not a local concept; it has no meaning within the LIF. It so happens that, in the LIF falling through the black hole horizon, the first probe gets launched at a velocity that, globally, is just sufficient for it to escape to infinity, whereas in the skydiver LIF, the probe's initial velocity is way, way more than needed for it to escape; but there's no way to tell that from within the LIF.)
(4) At some time t = plus epsilon in the LIF, the astronaut/skydiver releases a second probe that flies outward at a speed even closer to the speed of light than the first probe.
These conditions are perfectly possible to set up in both LIF's (the skydiver LIF and the LIF falling through the black hole's horizon), and within the LIF's, there is no way to tell which LIF you are in; every observation within the LIF will be the same for both. The second probe will move closer to the first probe (while they are both within the LIF); but the second probe will be falling behind the light ray that passes the astronaut/skydiver at t = 0.
It's true that, once all these objects exit the LIF, things will be very different in the two cases. In the skydiver case, the outgoing light ray will catch up with and pass the first probe. In the black hole horizon case, it won't. But there's no way to tell that from within the LIF.
It's also true that the r coordinates of these objects behave very differently, even within the LIF. In the skydiver case, all three of the objects that are moving outward (the first probe, the light ray, and the second probe) are increasing their r coordinates (and rather rapidly at that). In the black hole case, the first probe has (very slowly) increasing r, the light ray (the horizon) has constant r, and the second probe has (very slowly) decreasing r. But as I said in the other post I made in response to your latest, the r coordinate is a global coordinate, not a coordinate in the LIF; and there is no law of physics that says the relationship between local coordinates within an LIF and global coordinates must be the same for every LIF. In fact, it obviously can't be, because the whole point of the equivalence principle is that LIFs that look the same locally can occur in parts of spacetime that look very different on a global scale.
> There is no way to tell which LIF you are in
There is a way.
To keep it simple, let's assume the probes are test particles. In the skydiver's frame the second probe will overtake the first probe, given a sufficiently small epsilon. (We can always make that epsilon small enough that it's within the duration of the LIF.) In the astronaut's frame, for the same epsilon, the second probe won't overtake the first probe. The same experiment, different results, violating the equivalence principle.
NO, YOU CANNOT.
Sorry to shout, but not only have I already said this is false (several times if you include the previous thread I linked to), I have linked to a computation that proves it's false. So once again, you are basing your reasoning on a false assumption, and therefore you are naturally reaching false conclusions. (Note that my computation proves something stronger: that the distance required, extrapolated from the LIF, for the outgoing light ray the astronaut/skydiver passes at t = 0 in the LIF to catch the first probe, is much larger than the size of the LIF. If this is true, it must also be true that the second probe can't catch the first probe within the LIF.)
To briefly expand on what the computation I linked to shows: the smaller you make epsilon, the smaller the difference in velocities between the two probes can be (because the first probe has to be launched at escape velocity, and the smaller you make epsilon, the closer escape velocity gets to the velocity of light). And the smaller the velocity difference, the larger the catch-up distance, in the same proportion. So decreasing epsilon increases the catch-up distance extrapolated from the LIF such that the extrapolated catch-up distance remains much larger than the size of the LIF. (Again, my computation proves something stronger: that the catch-up distance required for the outgoing light ray, extrapolated from the LIF, increases as epsilon decreases, such that the catch-up distance remains much larger than the size of the LIF.)
To expand on the expansion just a bit more: remember that, in order for there to be any potential issue to discuss at all, two things must be true: (1) the initial conditions must match in both LIFs; (2) the global prediction of whether or not the second probe catches the first must be different for the two scenarios. Requirement #2 is what forces us to change the initial velocity of the first probe when we change epsilon; requirement #1 is what forces us to change the initial velocity of the first probe in both LIFs when we change epsilon.
(Actually, to expand one more bit, there is a third condition: the LIF size over which we can do the comparison at all must be the smaller of the two LIF sizes. Otherwise there would be no point to the comparison, since we could always just call globally flat spacetime an "LIF" and find some difference between it and an LIF in any curved spacetime by looking at effects that happen outside the range of the curved spacetime LIF.)
the smaller you make epsilon, the smaller the difference in velocities between the two probes can be
s/can be/has to be/
> So decreasing epsilon increases the catch-up distance extrapolated from the LIF such that the extrapolated catch-up distance remains much larger than the size of the LIF.
A fatal problem though: you arbitrarily chose the size of your LIF to make it too small.
The size is determined by the accuracy the experimenter wishes to achieve, e.g. 6 significant digits. When I compare the laws of physics between two LIFs I have even greater freedom to choose the LIF size. I choose both LIFs to be the same size and as large as needed for overtake to happen in the skydiver's frame (e.g. the width of a particle and extending a megaparsec outward into empty space). I choose a black hole massive enough that the tidal force in the astronaut's frame is less than in the skydiver's frame.
Now, when overtake happens in the skydiver's frame but not the astronaut's, I know the difference isn't due to the tidal force, because tidal force has no ability to cause one particle to overtake another (it only stretches and squeezes objects or systems), and the tidal force in the astronaut's frame can't be the reason overtake didn't happen there, because the tidal force is less there (the system of particles was stretched less there, than in the skydiver's frame). Having ruled out the tidal force completely, my result definitely shows a violation of the EP.
> There is no way to tell which LIF you are in
There are ways that allow the LIF to be arbitrarily small. For example, let the probes be tiny rods. Fire them vertically-oriented, the first probe completely above the horizon, the second initially stradding the horizon and partially alongside the first probe. In the skydiver's frame the second probe will be overtaking the first probe, moving outward faster. In the astronaut's frame the second probe, unlike the first probe, cannot reach a higher r-coordinate (lest it be moving outward through the horizon, which GR disallows), hence it won't be overtaking the first probe.
Also I still agree with the picture in the blog. The cloud splits apart at the horizon, in violation of the EP. Go back to your original probes experiment, as test particles. You say the probes can approach each other in the astronaut's frame. I say they can't. When the first probe's r-coordinate is increasing and the second probe's is decreasing, they are receding from each other as measured in any LIF that wholly contains them. You obviously wouldn't be able to devise an experiment to show otherwise in the skydiver's frame, and there's no relevant difference for the astronaut's frame.
We needn't refer to r-coordinates at all. Here is a simple law of physics (I just devised) for an LIF that you seemingly disagree with: Two free test particles each moving toward destinations beyond opposite sides of the LIF recede from each other.
No, I didn't. I didn't specify the size of the LIF at all, in terms of a standard unit like meters. I only specified it in terms of the mass of the black hole; its size has to be much smaller than M (or GM/c^2 in conventional units). That's not an arbitrary choice; it's required by the physics. See below.
> GR doesn't tell us the maximum size of a LIF, nor could it.
This is wrong as you state it. GR does set an upper limit on the size of the LIF, in terms of the spacetime curvature that is present, because if the LIF gets too large, the effects of spacetime curvature on the metric coefficients becomes equal in size to the metric coefficients themselves. At that point you can't have an LIF no matter what your measurement accuracy is, because the whole point of an LIF is that the corrections to the metric coefficients due to curvature are much smaller than the metric coefficients themselves.
In the particular case under discussion, for distances from the origin of the LIF that are of the same order as the mass of the black hole (GM/c^2 in conventional units), the curvature corrections to the metric coefficients are of order 1, i.e., the same size as the metric coefficients themselves. That's all my computation was showing, and it's enough to set an upper limit on the size of the LIF. In other words, the mass of the black hole determines the spacetime curvature at the horizon, and that determines the maximum possible size of an LIF falling through the horizon.
It's true that, by increasing the black hole mass M, you increase the maximum size of the LIF falling through the horizon; but I already took that into account in my analysis, as I said above.
> It's determined by the accuracy the experimenter wishes to achieve, e.g. 6 significant digits
This will determine an upper limit on the size of the LIF, as long as it's smaller than the limit I described above. But also, the more accurate the measurements, the smaller the upper limit on the size of the LIF. I'm not sure you realize this.
> When I compare the laws of physics between two LIFs I have even greater freedom to choose the LIF size.
No, you have exactly the freedom I specified, and no more: the LIF size over which you can do the comparison is the smaller of the two maximum sizes. So we have two possibilities in the case under discussion:
(1) The maximum size of the black hole LIF is larger than the maximum size of the skydiver LIF (which will be based on the spacetime curvature in the skydiver's vicinity). In that case, obviously the "catch-up distance" extrapolated from the LIF will be much larger than both LIF sizes.
(2) The maximum size of the skydiver LIF is larger than the maximum size of the black hole LIF. In that case, the "catch-up distance" will be much larger than the size of the black hole LIF; but if the skydiver LIF is large enough, that distance could be smaller than the size of the skydiver LIF. But the range of the comparison, for purposes of the equivalence principle, will be the size of the black hole LIF, since that's the smaller of the two, so there's no violation of the equivalence principle since there's no difference between the two cases within that range of comparison.
> I choose a black hole massive enough that the tidal force in the astronaut's frame is less than in the skydiver's frame.
And that just means the black hole LIF size will be larger than the skydiver's LIF size, which is the first of the two cases I discussed just above. In that case, since I've proved that the overtake can't happen within the black hole LIF size, it obviously can't happen within the skydiver LIF size either, since the latter is smaller.
> Now, when overtake happens in the skydiver's frame but not the astronaut's
Which it doesn't; see above. Once again, you are basing your reasoning on a false premise, so you're reaching a false conclusion.
If you wanted to make things more interesting, you could have chosen a black hole with tidal force in the astronaut's frame larger than in the skydiver's frame, i.e., the second of the two cases I discussed above. Then it would at least be possible that an overtake would happen within the skydiver's LIF; but, as I noted above, that doesn't violate the EP, because the comparison can only be done over the smaller of the two LIF sizes.
> let the probes be tiny rods. Fire them vertically-oriented, the first probe completely above the horizon, the second initially stradding the horizon and partially alongside the first probe. In the skydiver's frame the second probe will continue to overtake the first probe. In the astronaut's frame the second probe cannot continue to overtake the first probe, lest it be moving outward through the horizon, which GR disallows.
Incorrect. Someone brought up this case in the PhysicsForums thread I linked to earlier. My detailed answer is there. A quick summary: the second probe will overtake the first probe, within the range of comparison between LIFs (i.e., the smaller of the two LIF sizes). Only outside that range of comparison will the difference between the cases become apparent. And this will be true even though the second probe is indeed falling through the horizon, while the first probe is moving outward at escape velocity. (The upper end of the second probe will be moving faster than escape velocity at its initial altitude, since by hypothesis it is moving in the positive x direction in the LIF faster than the first probe, so the r coordinate of its upper end will be initially increasing, even though the r coordinate of its center of mass is decreasing, which is what is necessary for it to be falling through the horizon.)
> You say the probes can approach each other in the astronaut's frame. I say they can't. When the first probe's r-coordinate is increasing and the second probe's is decreasing, they are receding from each other as measured in any LIF that wholly contains them.
This is wrong and I've repeatedly explained why it's wrong. If you can't understand what I've already explained and why it's true, I'm sorry, that's the best you're going to get. I'm not going to teach you a course in relativity.
> We needn't refer to r-coordinates at all. Here is a simple law of physics (I just devised) for an LIF that you seemingly disagree with: Two free test particles each moving toward destinations beyond opposite sides of the LIF recede from each other.
Why do you think I would disagree with this? I don't. It's perfectly consistent with everything I've been saying, and with GR.
However, this claim is not equivalent to the claim I suspect you have in the back of your mind. I suspect the claim you have in the back of your mind is this: the first probe is moving upward from the horizon, while the second is moving downward from the horizon, therefore they are moving in opposite directions, therefore they must be receding from each other. The problem is that "upward from the horizon" and "downward from the horizon" are not the same as "moving toward destinations beyond opposite sides of the LIF".
First, "upward from the horizon" and "downward from the horizon" are global concepts, not local ones; they are defined in terms of the global r coordinate. You keep on making this mistake even though I've repeatedly explained why it's wrong, and even though you think you are not referring to r coordinates at all.
Second, the horizon is not a place in space. It's an outgoing light ray. And within the LIF, both probes are moving in the same spatial direction as that outgoing light ray! (The first probe is ahead of the light ray, and the second is behind it.) So within the LIF, the two probes are not moving in "opposite directions". So your argument, once again, is based on a false premise, so naturally you are reaching false conclusions.
http://www.physicsforums.com/showpost.php?p=4694521&postcoun...
The post explains the key features of the diagram that illustrate what I've been saying in this thread.
I'd have to think about this for a while, like weeks, so I'll move on to lower hanging fruit for now.
> So within the LIF, the two probes are not moving in "opposite directions".
Agreed.
> Why do you think I would disagree with this? I don't.
You do. The first probe moves toward infinity. The second probe toward the singularity. They each move toward destinations beyond opposite sides of the LIF. Then, as measured in that LIF, they recede from each other. But you disagreed here:
> The second probe will move closer to the first probe (while they are both within the LIF);
Again, you wouldn't be able to show that the probes approach each other in the skydiver's frame, given the first probe moving toward higher r-coordinates and the second probe moving toward lower r-coordinates. You should be able to do that if you are correct.
> You keep on making this mistake even though I've repeatedly explained why it's wrong, and even though you think you are not referring to r coordinates at all.
GR places additional conditions on what happens in a LIF, when it predicts that everything below a horizon falls toward the singularity. To be a falsifiable scientific theory we must be able to set up a thought experiment that accounts for that prediction within an LIF straddling the horizon. So what I do in that regard cannot be a mistake, as long as I'm careful that nothing is measured outside of the LIF. For example, it's okay to have an experiment in an LIF that says "fire a particle toward Vega", even though Vega is outside the LIF, as long as the measurements take place wholly within the LIF.
Globally, yes. Within the LIF, it's impossible to tell, just from the way either probe is moving within the LIF, what is going to happen to it after it leaves the LIF.
Also, to the extent that "toward infinity" and "toward the singularity" can be defined within the LIF, they aren't the directions you think they are--more precisely, the latter isn't. See below.
They each move toward destinations beyond opposite sides of the LIF.
This is wrong and I've already explained why it's wrong, but I'll recap again since apparently you aren't reading my posts very carefully. The singularity is not in the negative x direction in the LIF; it's in the positive t direction, i.e., in the future. Infinity is more or less in the positive x direction, yes. But that means that both probes, as far as you can tell within the LIF, are moving in both directions: toward the singularity (in the positive t direction) and toward infinity (in the positive x direction). There's no way to tell, from within the LIF, where the probes will end up.
(You could, of course, define "toward the singularity" as "decreasing r coordinate" and "toward infinity" as "increasing r coordinate"; but as I have already shown multiple times, the respective probes satisfy those definitions while both moving in the positive x direction within the LIF; and as I've also said multiple times, the r coordinate is irrelevant to the LIF since it's global, not local.)
you wouldn't be able to show that the probes approach each other in the skydiver's frame
Wrong; I already have. But if it will help, I am working on a second spacetime diagram that shows the skydiver's LIF, to complement the one showing the black hole LIF that I already posted on PhysicsForums (see the link I posted in this thread). What that diagram will show is that the only difference between the two LIFs is how the lines of constant r coordinate appear; everything else is identical.
given the first probe moving toward higher r-coordinates and the second probe moving toward lower r-coordinates
This isn't true in the skydiver's LIF; it's only true in the black hole LIF. As I've already pointed out. And which is irrelevant to the equivalence principle anyway, since the r coordinate is global, not local. How many times are you going to repeat the same erroneous statements?
GR places additional conditions on what happens in a LIF, when it predicts that everything below a horizon falls toward the singularity
No, it doesn't. You can give a definition of what "toward the singularity" means in a given LIF (I did it above for the black hole LIF--note that it isn't what you think it is); but you can't restrict what happens within a given LIF based on which direction "toward the singularity" is. All a global definition like "toward the singularity" does is tell you the relationship between local conditions in the LIF and some global condition. That can help you to determine what the right local conditions are in an LIF to model some desired global condition; but it can't ever tell you that a given local condition that would be permissible in special relativity is not permissible in an LIF.
we must be able to set up a thought experiment that accounts for that prediction within an LIF straddling the horizon.
Yes, and I've already shown, multiple times, how the thought experiment we've been discussing does this.
For example, it's okay to have an experiment in an LIF that says "fire a particle toward Vega", even though Vega is outside the LIF, as long as the measurements take place wholly within the LIF.
Sure; and the way you would model that experiment is to determine what condition in the LIF corresponds to the global condition "fire a particle toward Vega".
Similarly, in the thought experiment under discussion, the way we model it in the LIF crossing the horizon is to determine what local condition in that LIF corresponds to "the first probe is fired at escape velocity just outside the horizon", and "the second probe is fired at a higher velocity, relative to the astronaut, than the first probe, but just inside the horizon". And then we have to check that the two probes' trajectories, in the LIF, that we come up with meet the criteria of "first probe is moving toward infinity" and "second probe is moving toward the singularity". And I've already shown, multiple times, how all these conditions are met, and how your claims that they are not met are based on an erroneous understanding of how the global conditions translate into local conditions in the LIF.
At this point, I'm tempted to conclude that you're not really trying to understand my posts, but just looking for things to disagree with based on the erroneous straw-man version of relativity that you have in your head. You keep on repeating the same errors, even though I've already shown why they're errors, and you keep on missing the things I'm telling you about how relativity actually models this scenario, that address the concerns you're raising. I really think you need to step back and take an honest look at what I've been saying.
http://www.physicsforums.com/showpost.php?p=4695258&postcoun...
Note that everything is identical to the other diagram except for the lines of constant r.
I've given this a lot of thought to try to see where our major disagreement is. It's hard to tell from your walls of text. I now think it's this: You disagree that any prediction for a global frame can affect experiments in the LIF. Therefore you ignore such predictions and draw spacetime diagrams as one would in SR. Of course when you do that you find no difference between two LIFs. I say you're not seeing the forest for the trees.
Forget about black holes and light rays and especially complex GR terminology. Focus on the simpler skydiver's frame. Let the analogous prediction for the global frame be this: any particle below 1 km above sea level must fall inexorably toward r = 0. Let your first probe be launched just above the 1 km mark. It doesn't need to be escaping, just always moving away from the Earth during our experiment. Let the second probe be launched just below the 1 km mark. As measured in any LIF containing both particles, they'll recede from each other. As measured in a global frame, the first probe is moving toward a destination beyond the side of the LIF that's facing away from the Earth, while the second probe is moving toward a destination beyond the opposite side of the LIF that's facing toward the Earth.
For the record I still believe my argument using megaparsec-sized frames was sound. Your counterargument didn't explicitly show that any sentence in mine was false or didn't follow from its premises. Also I still don't know from this discussion how the argument in the blog isn't sound.
I still don't know from this discussion how the argument in the blog isn't sound.
Sigh. Here are the basics boiled down as much as I can. I'll base my comments on what you've said in this thread, since what you've said here is a lot more coherent to me than the blog post itself is. I'll give just bare statements, with no supporting argument; I've already given the supporting arguments many, many times in this thread.
- You have claimed that, within the astronaut LIF, the directions "toward the singularity" and "toward infinity" are opposite directions. That's wrong; they're not.
- You have claimed that, since "toward r = 0" and "toward infinity" are opposite directions in the skydiver LIF, they must also be opposite directions in any LIF. (You made the claim about the astronaut LIF, but the way you made the claim makes it clear that you would make the same claim about any LIF whatsoever). That's wrong; see above.
- You have claimed that, if we choose the mass of the black hole appropriately, we can make it so that, if the two probes are initially converging, the second will catch up to the first within the astronaut LIF. That's wrong; we can't.
- You have claimed that, if we can set up the skydiver LIF so that the second probe catches the first within that LIF, that somehow invalidates the equivalence principle, even if the second probe can't catch the first within the astronaut LIF. That's wrong; it doesn't.
- You have claimed that, if the first probe has an increasing global r coordinate while the second has a decreasing global r coordinate, the two probes can't be converging within the astronaut LIF. That's wrong; they can.
- You have claimed that, since the two probes must be diverging in the skydiver LIF if their r coordinates behave as above, they must also be diverging in any LIF if their r coordinates behave as above. (Again, you made the claim about the astronaut LIF, but the way you made the claim makes it clear that you would make the same claim about any LIF whatsoever). That's wrong; see above.
And just for comparison, I'll briefly summarize what GR actually says about the astronaut scenario; again, just bare statements since I've already given the supporting arguments many times in this thread:
- As the initial conditions are given, the two probes will converge within the astronaut LIF.
- Within the astronaut LIF, the black hole's horizon is a light ray moving in the positive x direction, which passes the astronaut halfway between him launching the first and second probes. So clearly the second probe will remain below the horizon, even based on observations solely within the LIF, since it starts out behind the light ray and is moving slower.
- Within the astronaut LIF, the horizon will (slowly) catch up to the first probe. However, the distance it would take for the horizon to catch up, based on its "closure rate" within the astronaut LIF, is much, much larger than the size of the LIF. This holds regardless of the mass of the black hole. And if the horizon can't catch the first probe within the LIF, the second probe certainly can't either.
- Once the probes and the horizon have exited the astronaut LIF, tidal gravity can no longer be neglected; and tidal gravity will cause the horizon to stop catching up with the first probe and start falling behind it. Eventually, the first probe will escape to infinity.
Two free test particles each moving toward destinations beyond opposite sides of a LIF, as measured in a global frame containing the LIF, recede from each other as measured in that LIF.
This law is fine, but it doesn't apply in the case under discussion, because the two probes are not moving "toward destinations beyond opposite sides" of the LIF in question. I've repeatedly explained why not.
You disagree that any prediction for a global frame can affect experiments in the LIF.
No; the problem is that you are incorrectly translating predictions in a global frame into predictions for experiments in the LIF. I've repeatedly explained why your translations are incorrect. One example of an incorrect translation is that you think the two probes are moving "toward destinations beyond opposite sides" of the LIF. The correct translation is that the singularity is in the positive t direction in the LIF, while infinity is in the positive x direction; these are not opposite sides of the LIF.
Forget about black holes and light rays and especially complex GR terminology.
In other words, forget about the fact that the relationship between global parameters and local parameters is different for different LIFs, which is the whole point under discussion. The relationship between global and local is different for an LIF falling through a black hole horizon than it is for a "skydiver" frame free-falling at 1 km above the Earth's surface.
Let the analogous prediction for the global frame be this: any particle below 1 km above sea level must fall inexorably toward r = 0. Let your first probe be launched just above the 1 km mark. It doesn't need to be escaping, just always moving away from the Earth during our experiment. Let the second probe be launched just below the 1 km mark. As measured in any LIF containing both particles, they'll recede from each other.
If you set up the initial conditions that way, sure. But that already makes this experiment different from the one posed in the blog post, because the initial conditions in that one were that, as measured in the LIF, the probes were approaching each other.
As measured in a global frame, the first probe is moving toward a destination beyond the side of the LIF that's facing away from the Earth, while the second probe is moving toward a destination beyond the opposite side of the LIF that's facing toward the Earth.
Sure, because this LIF is at a radial coordinate that is way, way, way larger (about a million times larger) than the Schwarzschild radius corresponding to the mass of the Earth. So the relationship between directions within the LIF and directions defined globally is very, very different than it would be in an LIF that was free-falling just at the Schwarzschild radius. As long as you continue to ignore this huge difference, even though I've explained it multiple times and given you links to a pair of spacetime diagrams that illustrate it, you will continue to make the same mistakes you've been making.
For the record I still believe my argument using megaparsec-sized frames was sound. Your counterargument didn't explicitly show that any sentence in mine was false or didn't follow from its premises.
That's because I'm not going to hold your hand when making arguments; I expect you to use your intelligence. Your argument, as I remember it, assumed that you can independently adjust the size of the LIF and the initial velocities of the probes. You can't do that, because both of those things depend on the mass of the black hole. The dependence of LIF size on the mass of the hole should be obvious. The probe initial velocities depend on the mass of the hole because that affects what the escape velocity is at the point that the first probe is launched. So there is only one adjustable parameter: the mass of the hole. And my calculation showed how the LIF size is much, much smaller than the distance it would take for the second probe to catch the first probe, regardless of the mass of the hole.
(Edit: Having looked back, I see you also claimed, as far as I can understand your argument, that if the "catch-up" happens within the skydiver LIF, even if it doesn't happen within the astronaut LIF, that somehow invalidates the equivalence principle. That's wrong, and I explained why: the smaller of the two LIF sizes determines the range of comparison. If the "catch-up" happens within the skydiver LIF but not within the astronaut LIF, it must be because the skydiver LIF is the larger of the two.)
On thinking this over, I realized that even in this "skydiver" LIF, you can, in fact, set up initial conditions so that the two probes are converging, even though one probe's r coordinate is increasing and the other's is decreasing. Here's how:
At time t = minus epsilon in the LIF, the "skydiver", who is at rest in the LIF, launches the first probe. At that instant, his downward velocity, relative to an observer who is "hovering" at constant global radial coordinate r, is v1. That means that, in the skydiver LIF, an object with constant r that passes through coordinates x = 0, t = minus epsilon has velocity v1 in the positive x direction at that instant. So the skydiver launches the first probe in the positive x direction with velocity v1 + a, where a is some small constant; that means the first probe's r coordinate is increasing.
At time t = 0 in the LIF, the skydiver passes the 1 km mark.
At time t = plus epsilon in the LIF, the skydiver launches the second probe. At that instant, his downward velocity, relative to an observer who is "hovering" at constant global radial coordinate r, is v2, and v2 > v1 (because, relative to observers who are "hovering" at constant r, the skydiver is accelerating downward). So, relative to the LIF, an object with constant r that passes through coordinates x = 0, t = plus epsilon has velocity v2 in the positive x direction at that instant. So the skydiver launches the second probe in the positive x direction with velocity v2 - b, where b is some small constant; that means the second probe's r coordinate is decreasing.
Now all we have to do is choose a and b so that v2 - b > v1 + a; i.e., the velocity of the second probe, relative to the skydiver (and therefore relative to the LIF) is larger than that of the first probe. (This is always possible because v2 > v1, as above.) That means the two probes will be converging, not diverging; and yet the first probe's r coordinate is increasing while the second probe's r coordinate is decreasing.
Notice the key facts that make the above possible:
(1) The skydiver is falling downwards, with respect to the Earth, with nonzero velocity. That means an object can be moving in the positive x direction in the LIF but still be falling downwards, as long as it's falling slower, with respect to the Earth, than the skydiver.
(2) The skydiver's downward velocity, relative to observers "hovering" at constant r, increases as he falls. That is what makes v2 > v1, and thus "makes room" for the second probe's velocity, relative to the LIF, to be larger than the first probe's, so that the two converge.
You may object: but that's tidal gravity, isn't it? No, it isn't; it's just downward acceleration. The above argument holds even if the skydiver's downward acceleration, relative to observers "hovering" at constant r, does not change (which of course it can't within the LIF, since tidal gravity is by definition negligible within the LIF).
Note also that none of the above changes what I've said before; all the things I said about how the relationship between global and local is very different for the astronaut LIF as compared to the skydiver LIF are still true. But perhaps the above will help show how, even in a highly non-relativistic case (all the velocities in the above example are very small compared to the speed of light), the relationship between the global r coordinate and the local coordinates in a free-falling LIF is not quite what you might think it is.
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 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.
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.
Astronomers know that as recently as a few hundred years
ago, the Milky Way’s central supermassive black hole was
producing much more radiation.
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.
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.
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.
See
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.