Halc

Yes, it's a neat coincidence, and memorized since it's a useful fact for such discussions. I imagine it means: Reachable before you die. In an inertial frame, any distance is reachable in finite proper time given enough speed, so with a speedy ship, I can get to the other side of the galaxy (where the dinosaurs were) before I die. I say dinosaurs because some of the most popular ones were around when our solar system was half a galactic orbit prior. In a universe with dark energy, the event horizon is the limit and there is no reaching that or anything beyond, regardless of the power of your ship or your life expectancy.

It's outside Jupiter's Roche limit, so we will remain a large moon and not break up into rings. As Janus points out, the tidal forces would be incredible, and the heat generated from it would more than compensate for lack of sunlight. The oceans I'm guessing would boil away as long as Earth's spin was there to generate the heat. The earthquakes and volcanoes would probably be bad enough that you wouldn't much notice what the weather was trying to add to it. No sun, but still plenty of heat to drive atmospheric convection, so yea, lots of wind but no rain. Nowhere safe to hide. We'd be pretty close to the orbit of IO. Enough for them to interfere? IO is long since tide locked and doesn't get the geothermal tide heating that a spinning Earth would.

It is a current proper distance measured along a line of constant cosmological time. Light emitted from here will never ever reach a comoving object 17 GLY away. There's no hypothetical about that. That's what an event horizon is. Communication is not possible across one. SR does not apply since it only applies to inertial (Minkowskian) frames, and the universe is not Minkowskian. Under SR and the flat spacetime under which it is relevant, yes, light will eventually get to any coordinate in any given inertial frame given enough time. But even in SR and flat spacetime, light will never catch up to a sufficiently distant object that undergoes constant proper acceleration (as does said object 17 GLY away), so in a way, SR rules are sufficient to illustrate the effect. If an object is accelerating away from us at say 1G, it only needs to be a light year away from us and any signal we send will never reach it. A star 16 GLY away (as measured in our inertial frame) need accelerate away at far less than 1G to put us beyond its event horizon. Sorry for slow reply. I don't log on here every day.

Not anything beyond what is now about 16 GLY away, which is where the event horizon currently is, even though we can see it (and they can see our past). I assume the OP was speaking hypothetically, assuming an arbitrarily powerful ship and a totally clear path. The point wasn't an engineering one. Well actually you experience nothing special, per the first postulate of special relativity. In the frame that you're experiencing, you're stationary. But yes, all the galaxies and stuff are moving fast past your and, being things in motion, are length contracted. So you're correct that in an inertial frame, there is no limit to the distance (in a given frame) that one can travel in a human lifetime. Unfortunately the universe is not described by any inertial frame, hence the event horizon limit referenced above. An infinitely powerful ship cannot reach a star 17 GLY away (proper distance along a line of constant cosmic time), even though one could see it (presuming it was a super long lived star that was visible both now and for as long as the attempt takes). But if the distant star system launched a similar ship in our direction today, the two ships could at least meet.

But the numbers were unrealistic, and you never used them anyway, except to attempt an invalid force relation. Using real numbers gives some plausibility to the scenario. Using exact numbers is probably not useful. To illustrate: Theia was (supposedly) 10 times the lunar mass, not 1.5 times. The extra mass either escaped (high speed object, mild glancing hit), in which case where did it go? or it hit more directly and was completely absorbed except for the ejecta that managed to coalesce in orbit. Anything not already in orbit has to be moving at a minimum of around 40000 km/hr relative to Earth, so that's a lower bound to any impact from an external object. Mass and speed do not determine force, and force is not directly relevant to the scenario. Force is different everywhere, and is not one value. Fluid dynamics must be invoked. Yes, it was an off-center hit. A straight on shot would have resulted in ejecta with minimum net angular momentum, and thus no moon that requires it. The offset hit would have significantly altered the spin of both bodies, resulting in an Earth will say a 10 hour (or less) day. How long did it take for the moon to get tide locked? Does the simulation answer that sort of thing? Certainly not until any second moon merged into just the one. You don't seem to contribute after this. All your recent posts seem focused on personal attacks against those trying to help. I suppose I will also be the eventual target of that.

I don't know if I'm a proponent of any particular explanation. A large glancing collision might result in a big molten planet which spins too fast for stability, giving time for all the material to mix and become somewhat one composition. I don't know the dynamics of spinning too fast, nor the possible dynamics of a slow collision, especially if two non-mutually orbiting moons are to collide. It's a little easier to envision if they're orbiting each other, but Roche limits would tear apart both bodies before they coalesced I'd think. Not unreasonable to say that happened. 8 BY is less than the time before the moon turns around and starts approaching Earth again, and peanuts to the amount of time (trillions of years) before it hits Earth. It seems the sun will swallow the pair before then, unless the sun loses sufficient mass to let Earth's orbit get sufficiently far away, and I don't think it will lose enough. Damage deeper than that, yes, but the crater is 3 km, filled with maria (lava). The cooled lava field (the dark stuff we see) is measured at 3 km at worst. Yay for them knowing this. What, do they set up seismic sensors all over the place? I don't know what all Apollo brought with them. I don't think the moon 'hit' us. If there was a collision, the moon was formed by the ejecta of that collision, but the collision was with something not-moon. If this is the case, yes, where did it come from and where did it go? Was the matter entirely absorbed by Earth such that no significant chunk retained escape velocity? What was Earth's orbit before that hit, and why is its orbit so 'undisturbed' now?? That question seems to be one ignored by proponents of the Theia idea. Yes, of course Earth had as many craters as any other object, but like Jupiter, Earth doesn't retain them long, the smaller ones leveled by weather more than heat. You can see a lot of the really large ones (Hudson bay being one), but none as big as the one creating the moon. No crater from that, and it erased all prior craters. Plenty of the current moon craters (the ones on top) are from hits from material that is not 'returning', but rather new unrelated objects. 4 billion years is plenty time to clean up the original bits that flew off that collision, but slow enough to 'come back'. Every single object that retains its craters is pretty much covered with them everywhere. It is just a show of lack of erosion, as studiot points out. If that were true, the debris wouldn't have left 'the moon' in the first place. Earth is a bigger target and larger gravity source, but so is the sun, and I imagine not much of the ejecta found its way there, so I admit that it being bigger isn't a valid argument. But returning material would move too slow to make much of a crater. Most of the big damage seems to be done by things not in Earth orbit. Moon forming within hours. It would almost have to, no? I mean, if it isn't ejected in hours, it's not going to eject. Coalescing in hours? Maybe... Depends on what percentage coalescs into one or a limited number of small objects before somebody decides "yes, that's 'the moon' now". The simulation needs to produce an Earth-moon pair of current mass and spin. They don't just pick random values (except as guesses). So to counter their proposal, you need to run a different simulation with different masses, speeds, and offset, one that produces an outcome more similar to what we see today. So it's not all unknown, since most guesses result in wrong results. If Theia was not totally absorbed (something continued on, not in Earth orbit), then yes, it should be out there. Scars on it? Heck no. Those would be gone for the reason you yourself gave. Melted away. But we don't see any object that fits the bill. I was involved in another topic about this sort of solid anomaly we have at our core. The center of Earth has asymmetrical lumps of different material. Related?

The moon was not tide locked shortly after the Theia impact, nor was it even 'the moon', something it only became after most of the pieces coalesced. There was a likely merger of two moons, but what was that like? Two moons in a similar orbit will set up a horseshoe orbit and never touch. There are several objects that are in the same solar orbit as Earth in such a horseshoe arrangement. So what else? Two moons in mutual tide locked orbit might be small/solid enough to do a soft touch, but only if they're in a tight orbit, so lots of angular momentum. There can be no sustained bulge from tides, but you still get the density asymmetry that we see today. Why is the dense side on the Earth side? Tide will tend to align a linear mass vertically, but with either end up, not necessarily the low density end. All questions, not so much answers. Yes, interesting discussion. NASA says "The Moon’s core is proportionally smaller than other terrestrial bodies' cores. The solid, iron-rich inner core is 149 miles (240 kilometers) in radius. It is surrounded by a liquid iron shell 56 miles (90 kilometers) thick. A partially molten layer with a thickness of 93 miles (150 kilometers) surrounds the iron core." https://science.nasa.gov/moon/facts/ Unsure how they measured this. Yes I am. Having something the size of Mars hit us will do that. There is an alternate (far less consensus) theory that there was no impact, and Earth just turned so fast that it spun off a big gob of material. Or even one week of simple fluid dynamics. I cannot leave a scar on the ocean after the waves have dissipated and currents have mixed any dyes and other evidence of disturbance. That they do. Many of the pieces either escaped or fell to Earth. Many of the small pieces in orbit were cleaned up by the largest lump, which eventually formed one moon, but perhaps two of them for quite a while. Assuming we wait long enough (far far more than the current age of the universe), and assuming unreasonably that the sun will not consume the Earth/moon system, the moon will eventually fall to low enough orbit to break up into rings like Saturn, and then fall as small bits to Earth. Yes, it's currently receding, but only as long as kinetic energy exists to fuel that.

Yes, the biggest evidence is that the surface has similar composition to Earth, which is true of no other planet's moon. The larger impacts impart enough energy to melt vast volumes of rock, and this liquid spreads out and covers significant area. The largest impacts might have created a crater perhaps 3 km deep, but the lingering energy from the event also continues to trigger volcanic activity for some time, and this adds to the maria, and also originates from much deeper than just a few km. I've not seen a consistent explanation for this, but the best explanation seems to be the thickness of the crust. Where the crust is thin and dense, large impacts can melt rocks instead of just eject lighter thick material sort of like a child jumping into a ball pit. Other way. Thicker (and less dense) on the far side, sort of like it's the continent over there, and the near side having thin, more dense crust like our ocean floors. The near side is the dense side, which is why it faces Earth. They're the stuff on the far side actually. The small merged bits are the low density stuff. Yes, the theory is that there were probably two primary moons for a while, both formed from the same event, which eventually did a slow motion collision/merge, with the smaller less dense moon forming the thick crust on what is now the far side. It is unclear if the moon becoming tide locked occurred before or after (or because of) this slow merge. Yes, there would have been shrapnel everywhere, which eventually collected into a limited number of moons whose gravity cleaned up all the little pieces. Much of it of course achieved escape velocity and didn't ever come home. Theia itself? One would think that it would still be around if it survived, so I think the material merged with the pre-Earth to become one planet-moons system, the material of which was composed of matter from both objects. The pre-impact planet was not Earth particularly. It had a different mass, spin, and orbited somewhere other than where Earth is now. The post-imact Earth orbited similar to where it is now, but more eccentric, with a spin of perhaps 10 hours, similar to Jupiter. I don't think it is meaningful to refer to the 'side which struck Earth', or for that matter, refer to the place on Earth where the Theia impact took place. Both bodies were reduced to molten balls, a bunch of smaller molten balls, and a lot of ejecta. A liquid ball does not have a meaningful location of where something occurred on it. There's not going to be a crater or other scar, or a place where cooling takes place faster. Not if you get it hot enough.

Better worded, thanks. The wiki site explains the pressure, due to electrons not being classical particles: "The pressure exerted by the electrons is related to their kinetic energy. The degeneracy pressure is most prominent at low temperatures: If electrons were classical particles, the movement of the electrons would cease at absolute zero and the pressure of the electron gas would vanish. However, since electrons are quantum mechanical particles that obey the Pauli exclusion principle, no two electrons can occupy the same state, and it is not possible for all the electrons to have zero kinetic energy. Instead, the confinement makes the allowed energy levels quantized, and the electrons fill them from the bottom upwards. If many electrons are confined to a small volume, on average the electrons have a large kinetic energy, and a large pressure is exerted." Yes, the degeneracy pressure is what supports the stars, and the effect is related to the Pauli exclusion principle. My point still stands: Such large dense things squashed together by gravity are arguably a single nucleus of some unfathomable element. It certainly isn't a collection of small separate atoms that are just really close to each other. For one, the one large atom has far higher neutron-to-proton ratio than any known element. It's a low-energy solution to the otherwise problem of where to go with all the extra electrons forming the degeneracy pressure.

Arguably, any neutron star is a giant nucleus, continuously changing the number of protons contained, but on a percentage basis, not by much. It is the Pauli exclusion principle which supports their stability, not so much any particular 'balance of forces'. Somehow I think this reply is not what the OP was looking for.

From what I've read on such noted reluctance to mergers of large masses, it seems that the energy/angular-momentum decay needed to merge is mostly dissipated by gravitational drag, the flinging of nearby matter away at high speeds, as is typical of 3 body dynamics when one of the bodies masses much less than the other two. Eventually, the local area is cleaned out and there's nothing left to fling, leaving a fairly stable 2 body orbit, decaying at least by gravitational waves, but that is trivial at that distance. Dark matter is just more [small] mass, just like regular matter. If it gets close it gets flung away and is cleared out like anything else. I don't see how dark matter is therefore any sort of solution to the effect. I don't know if 1 parsec is typical, or can be derived from any math. It would seem to be a function of the masses of the respective black holes and of the density of other matter near them. Clearly black holes do merge since our own galaxy is a collection of dozens of smaller galaxies, many of which contributed their own SMBH. We have but the one, so they've merged at some point. Some of then recent swallowed galaxies may have central masses that are en-route to the center, but none orbits particularly nearby at this time. Sgr-A has plenty of stars orbiting within that 1 parsec radius, so there is available mass to be ejected during any upcoming merger. Notably, in perhaps 10-20 billion years, Sgr-A will be merging with something 10 times its own mass. I wonder how far away you need to be from that for it not to be dangerous.

Using a dish would focus on signals only from a very specific direction, and would not amplify signals from elsewhere. Using a straight solar array would receive signals from all directions, great if you're just after electricity, but awful if you're trying to filter the noise away from the signal.

That seems completely inappropriate. I don't get the bike wheel thing, whose perimeter was never at the axle. There is no coordinate system that ties events in our universe to whatever constitutes a point in another. There's not even units for that. In our universe, the big bang, in an idealized naïve model, might be the event (0). Just one coordinate. No space coordinate because as MigL says, it's everywhere. But beyond that it doesn't work. For instance, there is not coordinate system that gives the absolute location of you. To do that (10 meters west of my mailbox), one must first identify an origin for your coordinate system (the mailbox). This is fine and relative, but it isn't an absolute coordinate because the location of the mailbox hasn't been specified. You can't use the big bang reference as your spatial reference because it lacks a location in space. Not true of any other event, but true of that one. It isn't like Utah where all you need to do to get a letter somewhere is give two coordinates. Three in Hong Kong.

Sorry, this makes no sense. Likewise there are no straight trajectories - every time it will be a moving object aiming to meet a moving object where their gravity-curved trajectories cross. That comment had nothing to do with straight trajectories. It has to do with preventative action against an object whose orbit crosses the orbit of Earth. Of course orbits are curved. Such an object will intersect Earth given enough time. It is officially dangerous. Point is, if you deflect the object while the object is at Earth's orbit, it will return to that location regardless of the new trajectory you send it (unless you impart escape velocity from the sun to it). Orbits are regular visitations to the same locations over and over. So the idea is to deflect the dangerous object when it is well away from Earth order, deflecting into a new orbit that does not intersect Earth's orbit. My comment above stated that doing it at the wrong time accomplishes nothing.

The velocity (both speed and direction) of an object like an asteroid is very frame dependent. So you seem to envision the frame of the sun say, where the asteroid is on a path not towards Earth, but one that will cross Earth's orbit exactly when Earth gets there. You hit it straight on in a direction opposite its motion in that frame. It slows, and the mostly unaltered paths still cross, but the objects are at that point in a different times. Now do that from Earth frame (the frame from which the rocket was launched). In that frame, Earth is stationary and the the asteroid is heading straight for us. If we hit it straight on in that frame, it will slow, but still get here a little bit later. Point is, a straight bullet shot fired from Earth isn't going to divert it. In that frame, to get it to miss, you need to apply lateral momentum to it. This involves sending the rocket on a curved path, wasting fuel compared to the straight path. And only the fuel expended for the lateral motion will effect the asteroid in a deflecting way. So you need a lot of fuel. One batch to get there (all unusable for deflection) and a whole separate batch to apply laterally to the thing. All this kind of presumes flat inertial motion of both Earth and asteroid, which is accurate if the thing is pretty close, but the idea is to get it when it isn't so close since it takes less effort to divert something far away. A smaller deflection is needed to effect a miss. So maybe our computers predict this collision on some prior orbit and we can manage to hit the thing slightly on some prior pass. Idea is to not hit it when it crosses Earth orbit since no defection there will prevent it from returning to Earth orbit repeatedly.

Speed is not a vector. You seem to be saying that moving 2.8 meters in 2 seconds is 1 m/sec. That's simple arithmetic that you're getting wrong. You also are still labeling a speed in meters where you say v3h=0.7m Do I have to point this out every post? The vertical component of each velocity vector is 1 and -1 per second respectively, not 0.7. Ditto with the h component, which is the speed of the road which you said is 1, regardless of whether ped2 knows about it or not. I tire or repeating this. You seem bent on just asserting the same wrong numbers over and over no matter how many times the mistakes are identified. You seem to be doing this deliberately, perhaps in order to argue some mathematically inconsistent point that you think has some unfathomable bearing on relativity. It doesn't. Relativity doesn't support your numbers at all, nor does it give different numbers for this scenario than does Newtonian physics.

I understand what you seem to be trying to say. I am trying to help you use consistent terminology to do so. You've been saying ped1 is stationary and immediately saying he has a nonzero speed. That's a contradiction, and it has to be fixed. Motion is relative, and ped2 is your designated reference. If motion is specified relative to anything other than ped2, you need to explicitly say this, else the statement is not even wrong. It renders further discussion impossible. Your scenario seems to take place at an airport with moving sidewalks. Ped2 is standing outside of one of them, and ped1 is standing on the sidewalk. Sometimes the sidewalk is on (cases 2 & 3) and sometimes off (case 1). The points in space (A-F) are all locations relative to Ped2, not relative to anything else. Oh, that's much better. You added 'relative to the road', which is the alternate frame reference that was missing before. Now it's not a contradiction. This part is still contradictory. You said it takes 2 seconds to do the two diagonals, one second apiece. Now you assert a different time of 2.8 s. That's a contradiction. It takes 2 seconds to go 2.8 meters at 1.4 m/sec which is the speed ped1 is moving relative to ped2. You also get units wrong. You multiply 1.4 seconds by two, which should get 2.8 seconds, but you call it 2.8 meters. I suspect the unit thing is just a typo. This is still wrong. Ped1 moved from A to D in one second, and that distance is 1.4 m, so his speed relative to Ped2 is 1.4 m./sec, not 1. You did not show your calculation of from where this 1 m/s comes. It really doesn't matter if he's aware of how ped1 is accomplishing his motion. Point is, he sees him 1.4m away after 1 second, and his speed is thus 1.4 m/sec, regardless of what the road is doing. You seem to think it does matter, and that a ped3 doing the same thing with a jetpack, but always being at the same location as ped1, is somehow moving at a different speed relative to Ped2. This is wrong. If you insist otherwise, well then perhaps this topic needs to be moved to strange claims section. This is really hard to parse, but apparently you've imposed a speed limit of 1, and the road moves incredibly slow. I think this is an attempt to an analogy of using natural units for speed rather than m/sec, but given your inability to describe simple Newtonian motion above, I think I won't go there and take your description as worded. "The pedestrian one is 0.7m wide" : What does that mean? He is now on a somewhat narrower super-slow moving sidewalk? Legal, but why? The road speed cannot be 0.1m since meters is not a unit of speed. You perhaps mean 0.1m/sec, so it moves perceptibly fast now. I think a typical airport sidewalk goes about 2 m/sec, but we're in the slow and safe airport today. Why the speed limit? Relative to what? Ped2? What if ped2 decides to walk left? Can nobody put a box on the sidewalk now? What prevents it? How very non Newtonian. The abrupt change of velocity at point D is straight downward (south), so the jet pack should be pointing south, not to the southeast as drawn. Better to just put a wall at D and have him bounce off it. No jet pack needed at all then. He can just be an inertial object in freefall the whole time.

OK. And all this is 'as observed by' ped2, which means ped2 is by definition stationary, and is with ped1 the whole time. The road moves to the right (as the arrow in the picture shows). Both pedestrians are at location A the whole time, since you said they were stationary. Locations A-F are coordinate locations relative to ped2, our definer of the frame. The locations are not painted on the road as I had earlier guessed. No it isn't. You said he was stationary. If the road is transporting him anywhere, he isn't stationary. You're being inconsistent. If you want him to ride with the moving road, then don't say he's stationary. As I said, ped2 seems to be there to define the frame. He cannot remain at (0,0,0,0) since after 1 second he will be at (1,0,0,0). But he starts at that event in all systems. I am presuming location A is at x=y=0. Apparently you've assigned the x to vertical (across the road, to B) and y horizontal (along with it), towards C,E and such. This is a contradiction. Ped1 is stationary relative to ped2 or he's moving. You have to pick one or the other. Scenario 3: OK. He's moving faster than 1 m/sec then because he's getting to D (1.41 away) in only one second. Your OP description said he was moving at 1/sec, not 1.4 That last line is wrong. You show time 2 above in the E coordinate, not 2.84. 2.84 is not computed from the above values as you assert. Maybe you should fix all these problems before moving on to the silly parts. You can't. You said the max speed was 1 (harsh cops, I tell ya), and then immediately suggest going faster than that.

This whole topic seems to be related to Galilean relativity and Newtonian physics. There is no relativity theory going on in any of this. I am assuming that points A through F are painted points marked on the sides of the path/road. If you mean otherwise, it isn't indicated. How does ped1 move from A to E if he's not moving? This is a contradiction. Given the pic, the road is moving to the left, and spot E painted on the road moves left to the pedestrian over 2 seconds. Anything other than that and the ped is not stationary. I take it ped2 is always with ped1 in this scenario? If not, then ped1 would be moving, and you say he isn't, as observed by ped2. So ped2 stays at point A, the road is stationary, and ped1 takes this diagonal path to D, then E. The distance is under 2.83, but close enough. Speed is as you say. If ped2 is not staying at A, then none of the above works. Ped2 is your reference frame. You have an orange line going from A to E but that's apparently not ped2, it's just a line. No it isn't. You have ped1 stationary in system 2, but you also contradict that. I am presuming that we're using the frame of the road for system 3. The speed values make no sense otherwise. Thus, since ped2 defines the frame, ped2 stays at location A of the road. Yes, it takes 2.83 seconds to go 2.83 meters at 1 m/sec. This is news? It isn't a difference in observed time, it is a difference in path taken. ped1 goes straight across the meter path, or he takes a diagonal, a longer path. All observers will measure 2.83 seconds in this 3rd scenario. Changing observers (frames) will change the measured speed and distance, but it won't change the 2.83 seconds. Not at these speeds anyway. A car might go by and measure ped1 moving at 100 m/sec relative to the car. That guy will still see it take him 2.83 seconds to go from A to E.

The paper calls them 'black stars' and seems to posit a complete lack of event horizon, and that the matter all piles up just outside where the EH would be. This is very similar to Schmelzer's alternative generalization of LET (relativity, but denying both premises of SR). https://arxiv.org/pdf/gr-qc/0001095 He calls the 'frozen stars' since all the matter piles up outside the nonexistent EH. It's a presentist interpretation of relativity, and it equate absolute time with coordinate time. Just my thoughts. The alternative is that the matter gets 'inside' and is somehow prevented from going 'forward' which is the same as positing that if you put enough force on matter, you can push it into the past where there's more room.

For instance, yellow-orange laser light has a frequency of about 5x10**14 Hz. The binary light frequency is equivalently 1x10**110001 Hz. How this makes a difference is of course a mystery, but adding mirrors to the situation is just going to add to the delays since the path is less straight. Yea, about 6 to 45 minutes depending on how far away Mars is at the time. 6-45 minutes later is not 'before it was sent'. Zero justification for the suggestion otherwise has been given.

The collapsing star is a dynamic state, and quite different from a free-falling object. There's dense stuff, and then not even Pauli-exclusion can support it. OK, so there's even more squish, at least at first. A spatial dimension rotates and is replaced by time, and that time dimension is bounded. What was the time dimension rotates out to a spatial one, one with nearly infinite extension at that. Lots of new room to spread out, but the causal light cones don't allow arbitrary travel down this big space, so I cannot say the compression ends. An example of the space available inside a black hole, Sgr-A and the black hole at the center of Andromeda share a common singularity. They're the same black hole, a region of 4D spacetime bounded by a 3D event horizon hypersurface, and in that case, the same (connected) hypersurface. It's only in a slice of coordinate space at a given time that the one object has multiple cross sections, manifesting as a pair of black holes to us, for now. None of the above is particularly an answer to the question of if there is compressed matter in an established black hole. In coordinate time, yes, it's very dense, but that's more like length contraction than pressure. None of the matter actually reaches the event horizon in coordinate time, and yes, in that state, it (the original collapsing star matter) is very much under compression. The singularity condition exists, but isn't described, precisely because the physics there is singular, sort of like asking what the perspective of a photon is like. Got a link? That sounds like pop nonsense. Is it peer reviewed? That makes more sense. Still, to be matter, it has to persist, no? Agree to all. There's also a naked singularity. You can for instance just keep dropping electrons into a black hole until the charge is more than the gravity and no more (isolated) electrons can be added by any means. Similar issues if the infalling matter adds too much angular momentum. A given mass can only have so much of that. These are examples of frame independent singularities not obscured by the coordinate singularity of the EH. Ditto with the LLM answers, which is just massaged google results. Anyway, thanks for the post. Good informative stuff in there.

The equations are how everybody knows. No links were provided, so I googled the question and the first 8 hits (NASA, Smithsonian, various you-tubes, reddit) all suggest matter is compressed without bound. Much of this list of bad hits is due to my search terms of "black hole infinite density". First correct answer came from of all places Quora, a site known for severe wrongness of replies. Question was: Do black holes have infinite density? Answer by Toth: "The equations that describe some of the simplest black hole solutions, including the Schwarzschild black hole are (drum roll, please)… equations of general relativity in the vacuum. Yes, that’s right. The vacuum. There is no matter. The density is zero everywhere. The Schwarzschild solution is the simplest, spherically symmetric, static vacuum solution of Einstein’s field equations." Next hit was probably the most respectable forum I can name. https://physics.stackexchange.com/questions/246061/are-black-holes-very-dense-matter-or-empty Rennie (I think) replies specifically about the Schwarzschild metric, which wasn't technically the question: "The archetypal black hole is a mathematical object discovered by Karl Schwarzschild in 1915 - the Schwarzschild metric. The curious thing about this object is that it contains no matter. Technically it is a vacuum solution to Einstein's equations. There is a parameter in the Schwarzschild metric that looks like a mass, but this is actually the ADM mass i.e. it is a mass associated with the overall geometry." The Kerr metric is also a vacuum solution, which differs only by a nonzero angular momentum. There is an Oppenheimer Snyder metric that is an 'unrealistically simplified' solution to the formation of a black hole, but it fails to describe conditions at the singularity. I was hoping at least for some indication of the whole compression vs. tension distinction. None of these metrics seem to include Hawking radiation, so they describe black holes that exist for infinite coordinate time. Rennie continues: "[Observers falling with the star collapse] see the singularity form in a finite (short!) time, but ... the Oppenheimer-Snyder metric becomes singular at the singularity, and that means it cannot describe what happens there. So we cannot tell what happens to the matter at the centre of the black hole. This isn't just because the OS metric is a simplified model, we expect that even the most sophisticated description of a collapse will have the same problem. The whole point of a singularity is that our equations become singular there and cannot describe what happens. All this means that there is no answer to your question, but hopefully I've given you a better idea of the physics involved. In particular matter doesn't mysteriously cease to exist in some magical way as a black hole forms." So my post seems to be based on information about static metrics (Schwarzschild, Kerr, others), the geometry of which shows an end to time and no matter at all, but neither do those metrics show the end to the matter that made them since these kinds of black holes are not 'made'. They exist for eternity. So Op-Sny is probably a better metric despite being 'unrealistically simplified'. A coordinate system that isn't singular at the event horizon (like Kruskal–Szekeres coordinates) shows worldlines of infalling particles just ending in time at the singularity, not persisting with the other matter persisting there. The worldline of compressed matter would not end, but only join all together with the worldlines of other particles. As you (as an observer) fall into one, tidal forces pull you apart, not compress you. This doesn't stop at the EH. So compression ever happens, then the naive description would be when you smack into that physical singularity there where everything else has gathered. None of the metrics describe that. At best they just don't answer the question at all, and on those grounds, I am reneging on the authoritarian tone of my prior replies without suggesting that the 'high density matter' description is a better description. Learned stuff today, which makes this a win topic. I hope we all have. Yes, I've seen places that compute that mass. A moon mass is still going to take an awfully long time to radiate away at CMB temperatures. Infinite time actually, at least until the CMB radiation stops adding mass as fast as HR bleeds it off.

Yes. A black hole near end of life has almost no mass remaining. Gravitons as in gravitational waves, not any sort of force carrier. Gravitational waves carry information about changes to spacetime geometry, and an evaporating black hole is such a thing, so it has to generate such waves, whether or not those waves can be broken down into quanta. Light is energy. Any radiation reduces the mass of the thing radiating it. Light also has momentum. For small black holes, sure, but for larger ones, the odds of something like a positron escaping is incredibly low. Most would fall right back in due to gravity. Gravity can't pull back light if it's going in the correct direction (straight up). There's no matter in a black hole. A Schwarzchild black hole is a vacuum solution. Nothing gets squished in there. Things falling in actually get pulled apart. The singularity is not a location in space where there is matter squished together unreasonably. It's a line/plane/fuzzy region where time just ends. This comment suggests dense material in there somewhere. This is a misconception. Yes, better. Energy & mass are equivalent. The mass doesn't exit the black hole, but is created outside by separation of virtual particles, with the one with negative energy falling in and adding that much negative energy to the BH. The vast majority of the time, both virtual particles are thus pulled in, netting zero energy to the BH. The odds of one escaping becomes larger with the small holes. I don't know where the limit is, and what it means for mass to not be able to support an EH. I think a unified theory would really help give real answers to this. My statement of 'a few grams' might be way off, but classically there is no minimum mass, and at sufficiently low mass, the radiation becomes significant enough to qualify as an explosion. There is still nothing actually from inside the black hole escaping. There is no matter in there.

The end might produce a few grams of actual matter in a brief burst of radiation. "Explode" makes it sound bigger than a wink. The vast majority of the original mass radiates away as massless particles, mostly photons and gravitons, neither of which is a building material for stars.