Why doesn't the vertical light beam get out of a black hole?

[John Duffield]

For this we need a thought experiment: imagine you're on a gedanken planet manning a gedanken laser cannon, and it's pointing straight up. The light doesn't curve round, or slow down as it ascends, or fall down. It goes straight up. Now let's keep you safe in a bubble of artistic licence, and make a few little changes.

Let's make the planet denser and more massive. The laser cannon is still pointing straight up. The light still doesn't curve round, or slow down as it ascends, or fall down. No problem.

Then let's make the planet even denser and more massive. The light still doesn't curve round, or slow down as it ascends, or fall down. Still no problem.

Then let's make the planet denser still and more massive, so much so that we take it to the limit. Let's make that planet so dense and so massive that it's a black hole. At no point did the light ever curve round, or slow down as it ascends, or fall down. But Houston, we have a problem. Because that black hole is black, because the light can't get out. So the $64,000 dollar question is this:

Why can't the light get out?PS: If you're tempted to invoke redshift, remember that when you send a 511keV photon into a black hole, the black hole mass increases by 511keV/c² only. The photon E=hf energy didn't actually increase, conservation of energy applies. And if you're tempted to invoke the waterfall analogy, please note that Einstein never ever modeled a gravitational field as a place where space is falling down.

 

[phinds]

For the same reason that you can throw a rock off of an asteroid hard enough to let it escape the asteroid's gravity field, but you can't throw one off of the Earth and get it to escape the Earth's gravity field because the field is too strong. As soon as your photon is created, it heads backwards towards the singularity.

To expand on that just slightly, ALL world-lines inside a black hole point to the singularity.

 

[rootone]

Why can't the light get out?

Because it would have to travel faster than light

 

[SlowThinker]

And if you're tempted to invoke the waterfall analogy, please note that Einstein never ever modeled a gravitational field as a place where space is falling down.

The waterfall analogy seems to explain why the light can't get out. Einstein was a genius, but only had one life. If he didn't do something, it doesn't mean it's wrong.
What bugs me, perhaps the light can't escape to infinity, but it can show above the horizon? As in, you can throw a stone to *some* height, even if it doesn't escape Earth. The waterfall analogy would explain why the light can't get out even in this case - it's falling from the start.

 

[Grinkle]

I think the OP is having difficulty picturing a world line that at once originates pointing away and perpendicular to the surface of a sphere (the singularity) and also curves back to the sphere.

I have the same difficulty. A photon emitted at some non-perpendicular angle to what one (at least me) pictures as the sphere of the singularity (I know its not really a sphere, its not actually describable, its a singularity) is easy to picture falling back. Its harder to picture a photon going straight up, never moving slower than the speed of light, and still somehow falling straight back down.

I know that I am relying way too much on the throwing-a-rock analogy, and where that analogy breaks down is that a rock is allowed to travel at any speed <c, including zero, relative to any observer. Light is not allowed to do that, so its hard to use that analogy to picture this situation. At least for me.

Edit -

Perhaps what I need to picture is that all perpendicular world lines point inward. As a world lines get closer and closer to perpendicular, they start to look like smaller and smaller radius loops until finally at the limit of perpendicular emission a photon is never actually emitted at all. So a photon can never be emitted absolutely perpendicular to a singularity.

That also can't be right because a singularity has no radius, so defining perpendicular seems problematic, but thinking of it that way at least gives me a picture.

 

[DaveC426913]

As you make the planet denser and denser, time dilation due to gravity creeps up. The universe ages faster outside than inside. At some point, you are in a gravity well so strong that, while you see the light rise in a split second, the universe outside ages and dies before the light reaches the event horizon.

 

[Grinkle]

Dave -

For me, that is not a very satisfying response. The question invokes an image of an observer who can see what happens to the photon, from emission through absorption by the singularity. Having the answer be that there is no such observer is disappointing. I suppose an observer inside the SR of the singularity would also never see the photon fall back to the singularity in finite time from their perspective, so perhaps that is the sadly boring answer?

 

[DaveC426913]

I'm not suggesting the photon falls back to the singularity, I'm suggesting it continues to the event horizon, but that, to an outside observer, it takes an infinite time to get there.

 

[Grinkle]

Its the same for an inside observer, right? I think an observer inside an SR can never see a photon reach the SR, they just see it getting more and more red-shifted.

 

[DaveC426913]

Well, in fact, they won't see anything at all. You cannot see a photon unless you intercept it.

 

[Grinkle]

Sigh. Can you make it MORE boring? ;-) You are right, of course.

 

[SlowThinker]

As you make the planet denser and denser, time dilation due to gravity creeps up. The universe ages faster outside than inside. At some point, you are in a gravity well so strong that, while you see the light rise in a split second, the universe outside ages and dies before the light reaches the event horizon.

I feel there is something wrong with this idea of infinite time dilation.
For example, the black hole should evaporate before the end of the universe, so perhaps the singularity doesn't even have enough time to form? This idea seems to be posted quite often, but the knowledgeable people around always say that the singularity does form, and you fall into it very quickly.
I suggest we wait for a mentor before going too wild. Sorry for occupying Peter for so long

 

[DaveC426913]

I feel there is something wrong with this idea of infinite time dilation.
For example, the black hole should evaporate before the end of the universe, so perhaps the singularity doesn't even have enough time to form? This idea seems to be posted quite often, but the knowledgeable people around always say that the singularity does form, and you fall into it very quickly.
I suggest we wait for a mentor before going too wild. Sorry for occupying Peter for so long

All of that is true, so how is it a problem?

 

[SlowThinker]

All of that is true, so how is it a problem?

It means that saying that the beam would get out after infinite time, is wrong.
Something else is preventing the escape.

 

[rootone]

I feel there is something wrong with this idea of infinite time dilation...

Yes a lot of people are very uncomfortable about that, and personally I am in the camp with those who see the 'singularity' as something not physically real, just an indication that our best theories while impressive enough, have limits to their accuracy.

 

[DaveC426913]

It means that saying that the beam would get out after infinite time, is wrong.

Wait. Why?

You have to do is ask what an external observer would see, and separately, what an internal observer would see.

In the split second that the internal observer sees his laser fire, the outside universe has aged and died. Neither observer ever saw the light escape.

 

[SlowThinker]

It means that saying that the beam would get out after infinite time, is wrong.
Something else is preventing the escape.

Wait. Why?

For starters, the inside of a black hole would be a really busy place. Imagine everything that ever fell to that BH, compressed into, say, 20 minutes at most. I'm pretty sure that's not what happens.

You have to do is ask what an external observer would see, and separately, what an internal observer would see.

In the split second that the internal observer sees his laser fire, the outside universe has aged and died. Neither observer ever saw the light escape.

Again, the BH evaporates before the universe ends. So how can the ray not hit the horizon, if that eventually shrinks to 0?

 

[DaveC426913]

For starters, the inside of a black hole would be a really busy place. Imagine everything that ever fell to that BH, compressed into, say, 20 minutes at most. I'm pretty sure that's not what happens.

Yeah, it is. And highly blue-shifted, hard radiation.

Again, the BH evaporates before the universe ends. So how can the ray not hit the horizon, if that eventually shrinks to 0?

Yeah, but the black hole evaporating is certainly a condition in which the thought experiment breaks down. We're talking about the universe ending. The 'light can't escape' claim was not really meant to apply to that edge condition.
Remember, it is a physically impossible scenario, since no observer with any mass could hover inside the event horizon.

 

[PeterDonis]

Let's make that planet so dense and so massive that it's a black hole.

You can't do this. A black hole is not a static object like a planet, that just happens to have an escape velocity of $c$ from its surface. And a black hole's horizon is not a static surface where light just happens to stay put.

If you work out the math using GR, it turns out that it is impossible to have a static object (i.e., an object whose radius does not change with time), like a planet, with a radius smaller than 9/8 of the Schwarzschild radius corresponding to its mass (which is $r_s = GM / c^2$ in conventional units). (This is known as Buchdahl's Theorem.) So at some point in your process of making the planet smaller and denser, it will become unstable and collapse. A black hole will be the object that the collapse process leaves behind.

It is possible in principle, if you have a really technologically advanced spaceship (one far beyond our current level of technology), to "hover" as close to a black hole's horizon as you like; but even such a spaceship can't "hover" exactly at the horizon. Any ordinary object (i.e., anything with nonzero rest mass--note that light has zero rest mass) that is at the horizon can't stay there; it must be falling inward. So asking what things would look like to someone "hovering" at the hole's horizon, or why light emitted by such a person can't get out, is meaningless, because there can't be any such person; it's physically impossible.

The correct way of thinking about the hole's horizon is this: it is an outgoing null surface, i.e., a surface made up of radially outgoing light rays. Anyone falling through the horizon will see these light rays come at them and pass them as they fall. So in fact, light at the horizon is moving outward--relative to the only possible observers that can exist at the horizon, who must be falling inward. Locally, the light at the horizon is not staying in the same place; it is moving radially outward, just as one would expect.

But because of the curvature of spacetime, this outward moving light stays at the same radial coordinate $r = r_s$ forever. The horizon is a global phenomenon, not a local one; it's not something that you can tell is there just by local observations of the light there. Those local observations will show the light moving radially outward at $c$. But globally, when you look at the entire spacetime, you will see that it is curved in such a way as to keep that light at $r = r_s$ forever.

 

[PeterDonis]

As soon as your photon is created, it heads backwards towards the singularity.

Light emitted from inside the horizon will do this, yes. But even so, this light will still be moving radially outward at $c$, relative to local observers (who will be falling inward, from a global perspective, faster than the light is).

 

[PeterDonis]

What bugs me, perhaps the light can't escape to infinity, but it can show above the horizon?

No. Light emitted at the horizon stays at the horizon forever. It doesn't go up, stop, and then fall back.

The key thing to remember is that spacetime at and inside the horizon is not static. There are no observers who can "hover" at a constant radial coordinate. So it's simply not correct to think of events there the way you would think of things on Earth, where things that are thrown upward at less than escape velocity move upward, stop, and then fall back. Spacetime at and inside the horizon simply doesn't work that way.

 

[PeterDonis]

As you make the planet denser and denser, time dilation due to gravity creeps up. The universe ages faster outside than inside. At some point, you are in a gravity well so strong that, while you see the light rise in a split second, the universe outside ages and dies before the light reaches the event horizon.

No, this is not correct. The spacetime at and inside the horizon is not static; the concept of a "gravity well", which requires a static spacetime, does not apply there. (It does apply outside the horizon, but outside the horizon light can escape, though it can take a long time to do so.)

I'm suggesting it continues to the event horizon, but that, to an outside observer, it takes an infinite time to get there.

This isn't correct, either. As phinds correctly pointed out, a photon emitted inside the horizon falls into the singularity.

Also, since spacetime at and inside the horizon is not static, it can't be described using the Schwarzschild coordinates that are used in the region outside the horizon to assign an infinite time coordinate to events on the horizon. So there is no meaning to the question of how long it takes light to travel inside the horizon "to an outside observer".

 

[SlowThinker]

For starters, the inside of a black hole would be a really busy place. Imagine everything that ever fell to that BH, compressed into, say, 20 minutes at most. I'm pretty sure that's not what happens.

Yeah, it is. And highly blue-shifted, hard radiation.

Would not that mean that as a star gets sucked into the BH, it is turned into a superdense neutron gas? Otherwise all those stars simply would not fit in. I've never heard of such a scenario.

Yeah, but the black hole evaporating is certainly a condition in which the system we're examining is changing over time. And we
we're talking about the universe ending. I'm not sure light can't escape was really meant to apply to that edge condition.

I think it does apply. If you could send any messages out of a black hole, the very idea of an event horizon would be kind of non-sensical. If it was possible in quantum theory, then the current debate of firewalls and information paradoxes would look very differently.

The waterfall model seems to offer a very intuitive, paradox-less explanation of what's going on. Maybe it's even correct

 

[PeterDonis]

the black hole should evaporate before the end of the universe, so perhaps the singularity doesn't even have enough time to form?

Yes, it does. This has been discussed in a number of previous threads; see, for example, these:

https://www.physicsforums.com/threa...gularity-evaporate-before-it-can-form.683755/

https://www.physicsforums.com/threa...-that-evaporates-by-hawking-radiation.688627/

the BH evaporates before the universe ends. So how can the ray not hit the horizon, if that eventually shrinks to 0?

In Hawking's original model of an evaporating black hole, the ray hits the singularity before the horizon shrinks to zero and the hole vanishes.

There is a lot of contention still in this area; most physicsts seem to think that Hawking's original model is not correct, but there is no good consensus on what should replace it. In some candidate models, there isn't ever a true event horizon at all; there are only "apparent horizons", surfaces where light, locally, is trapped for some period of time. But these apparent horizons eventually disappear and the light gets out again, and there are never any singularities or any regions from which light can never escape at all. (This doesn't mean it's not risky to fall into these regions, though; strange quantum gravity stuff is going on there that probably will destroy your body if you ever fall in, and it won't be much consolation to you that the light from the process of your destruction will eventually escape again.)

 

[PeterDonis]

Yeah, it is. And highly blue-shifted, hard radiation.

Not inside the black hole, if you're free-falling towards the singularity. Such observers see radiation falling in from the outside universe redshifted, and the redshift factor increase as they fall towards the singularity.

It's possible to see radiation from the outside universe blueshifted inside the horizon, if you accelerate very hard in an outward radial direction. But that will also greatly decrease the proper time it takes by your clock for you to hit the singularity.

 

[phinds]

Light emitted from inside the horizon will do this, yes. But even so, this light will still be moving radially outward at $c$, relative to local observers (who will be falling inward, from a global perspective, faster than the light is).

Thanks for that correction, Peter, but now I'm confused. You use a global perspective for the light but "global perspective" includes space-time outside the EH, does it not? And I thought, and your posts seem to confirm, that "global perspective" as used outside the EH is not valid inside the EH.

Let me be more specific. A photon has been emitted radially outward and naturally moves at c but its world-line points to the singularity. Does an infalling observer who is "next to" the photon as it is created speed past the photon on his way to the singularity?

 

[PeterDonis]

Would not that mean that as a star gets sucked into the BH, it is turned into a superdense neutron gas?

Matter falling into the hole will tend to become denser as it falls, yes. But it also will be subjected to increasing tidal stretching in the radial direction as it gets near the singularity. We don't actually have a very good understanding of what will happen to it at that point.

Otherwise all those stars simply would not fit in.

They don't have to; in the classical model of a black hole, infalling objects hit the singularity and are destroyed fairly quickly. They certainly don't remain forever in the region inside the horizon.

(There is another wrinkle here as well, but I'll refrain from going into it, since this thread is already complicated enough.)

 

[rootone]

I think that what goes on inside the event horizon of a BH, might as well be described as 'Here be Dragons'.
Hopefully we'll come up with something more believable one day,

 

[PeterDonis]

You use a global perspective for the light but "global perspective" includes space-time outside the EH, does it not? And I thought, and your posts seem to confirm, that "global perspective" as used outside the EH is not valid inside the EH.

It depends on which global perspective you use. Schwarzschild coordinates won't work; the ones valid outside the EH aren't valid inside the EH, as you say. But other coordinate charts work fine both outside and inside the EH; for example, Painleve coordinates, which are the ones I was implicitly using in the passage you refer to.

A photon has been emitted radially outward and naturally moves at c but its world-line points to the singularity. Does an infalling observer who is "next to" the photon as it is created speed past the photon on his way to the singularity?

Just to clarify: the infalling observer is falling radially inward, and the photon is emitted radially outward, correct? In that case, of course the infalling observer will "speed past" the photon; they're moving in opposite directions. The photon's worldline "points toward the singularity" in spacetime, but so does the infalling observer's worldline; and the directions in which they "point towards the singularity" are different. The singularity is not just in one "direction" in spacetime when you're inside the horizon; it's in the future no matter which direction in spacetime you are pointing (as long as it's a timelike or null direction).

 

[PeterDonis]

think that what goes on inside the event horizon of a BH, might as well be described as 'Here be Dragons'.

The classical model is perfectly consistent and well-defined, though it is certainly counterintuitive. Its only problem is that it can't be right , at least not when you get close enough to the singularity for the spacetime curvature to be large enough for quantum gravity effects to become important.

The question is whether those quantum gravity effects are only important near the singularity, or whether they become important sooner--at the horizon, or even before the horizon is reached/formed. That's the question to which we don't (yet) have a good answer.

 

[phinds]

The photon's worldline "points toward the singularity" in spacetime, but so does the infalling observer's worldline; and the directions in which they "point towards the singularity" are different. The singularity is not just in one "direction" in spacetime when you're inside the horizon; it's in the future no matter which direction in spacetime you are pointing (as long as it's a timelike or null direction).

Yeah, I think this is the point where my meager understanding breaks down, but I appreciate your input.

 

[John Duffield]

Thanks for the replies everybody. I have to go to work, so I can't respond properly now. But meanwhile take a look at this and this and this and this. I and a few other guys kicked this one around a year or so ago and we were able to "ask the profs" and get input from a number of names. It was an interesting experience.

 

[SlowThinker]

I and a few other guys kicked this one around a year or so ago and we were able to "ask the profs" and get input from a number of names. It was an interesting experience.

What exactly is the "this idea" and how do the linked articles relate to it? Is there a record of said discussion?

It's not very nice to post a link to tens of pages without a specific question. There is a good chance most of us already know what's written there.

 

[John Duffield]

Re the earlier posts posts:

Phinds: sorry, but at no stage in the gedankenexperiment did the vertical light beam start slowing down as it ascended. The ascending photon does not slow down like an asteroid.

Rootone : IMHO because it would have to travel faster than light is kind of correct. But maybe not the way you think.

Grinkle: the idea that all paths lead back to the event horizon is a fairy tale I'm afraid. I could shine my laser straight down, and you could shine yours up my beam.

DaveC426913: Hi.

Slowthinker: we have some good evidence that there are black holes out there, whether they contain point singularities is a separate issue.

Peter: I beg to differ about the claim that it's impossible to have a static object. I'll come back to that once we've established why the light can't get out. Meanwhile please note that "the curvature of spacetime" relates to the tidal force and the second derivative of potential, the force of gravity relates to the first derivative of potential, and gravitational potential can be related to the coordinate speed of light. Hence light emitted at the horizon stays at the horizon forever. It doesn't go up, stop, and then fall back. As for matter falling into the hole, I suspect it doesn't even make it as far as the event horizon, but that's one for another day.

Re the above:

Slowthinker: it isn't some idea, it's being able to read the Einstein digital papers. And then following it up and finding out that the vertical light beam doesn't slow down. Because instead, it speeds up. Here's what Don Koks said:

"...light speeds up as it ascends from floor to ceiling (it doesn't slow down, as apparently quoted on your discussion site), and it slows down as it descends from ceiling to floor; it's not like a ball that slows on the way up and goes faster on the way down. Light travels faster near the ceiling than near the floor. But where -you- are, you always measure it to travel at c, because no matter where you place yourself, the mechanism that runs the clock you're using to measure the light's speed will speed up or slow down precisely in step with what the light is doing".

Don Koks is the editor of the PhysicsFAQ website which is sometimes called "Baez". See this article written by him to replace this one.

 

[phinds]

Phinds: sorry, but at no stage in the gedankenexperiment did the vertical light beam start slowing down as it ascended. The ascending photon does not slow down like an asteroid.

You must have misread or misunderstood my post. At no point did I say or even suggest that it slows down. What I said was that it heads toward the singularity.

 

[PeterDonis]

I beg to differ about the claim that it's impossible to have a static object.

I didn't say it was impossible to have a static object. I said that spacetime inside a black hole's horizon is not static. That is a straightforward conclusion from the metric of said spacetime, and is not in any way controversial. (In fact, the black hole interior is not even stationary, which is a stronger condition than not being static. See below.)

please note that "the curvature of spacetime" relates to the tidal force

The curvature of spacetime is tidal gravity (not "force"--tidal gravity is not a force). They are the same thing, physically.

and the second derivative of potential

Only in a stationary region of spacetime, where a "potential" can be defined.

the force of gravity relates to the first derivative of potential

Gravity is not a force in GR, and the "acceleration due to gravity"--which is better expressed as the proper acceleration required to remain at the same spatial location in a stationary spacetime--is what relates to the first derivative of the potential, which, again, can only be defined in a stationary spacetime.

Since the region at and inside the horizon of a black hole is not stationary, none of the above even makes sense there.

gravitational potential can be related to the coordinate speed of light

Only in a stationary region of spacetime, where "potential" has meaning. And even there, the coordinate speed of light has no physical meaning; it's not something anyone can actually measure. But that's irrelevant for this discussion, because...

Hence light emitted at the horizon stays at the horizon forever. It doesn't go up, stop, and then fall back.

No, this is not the reason light stays at the horizon--because the horizon, and the region inside it, is not stationary, so none of the concepts above are meaningful there.

As for matter falling into the hole, I suspect it doesn't even make it as far as the event horizon, but that's one for another day.

This is wrong too, as many, many previous threads here on PF have discussed. If you continue to post these incorrect claims, you will receive a warning.

Here's what Don Koks said

I like the Usenet Physics FAQ, and it's often a good source, but that doesn't mean it's always right, or that its authors always are. The coordinate speed of light, which is what "speeds up" as a light ray moves upward in the gravitational potential field in a stationary spacetime, and "slows down" as the light moves downward, has no physical meaning. Don Koks knows this, because he adds the qualification that nobody actually measures it; anyone actually measuring, locally, the speed of light will find it to be $c$. He attributes this to the measuring device "speeding up" or "slowing down" exactly in sync with the light itself; but that is attributing a physical meaning to the coordinate speed of light that it simply doesn't have.

This is a good illustration of why, when push comes to shove, we don't use pop science articles, even good ones like the Usenet Physics FAQ, as references here on PF; we only use peer-reviewed scientific papers or textbooks, and even then we only use those in which the physics is set out rigorously in math, not heuristically in ordinary language. Ordinary language is simply too imprecise to rely on if you really want to understand the physics.

 

[PeterDonis]

Here's what Don Koks said

To expand on this issue a bit more, I'll bring up something which I'm surprised you didn't bring up, since it is a better argument for your case than what you quoted. This is from the updated article in the Usenet Physics FAQ by Koks that you linked to:

"That the speed of light depends on position when measured by a non-inertial observer is a fact routinely used by laser gyroscopes that form the core of some inertial navigation systems. These gyroscopes send light around a closed loop, and if the loop rotates, an observer riding on the loop will measure light to travel more slowly when it traverses the loop in one direction than when it traverses the loop in the opposite direction."

What Koks is referring to here is the Sagnac Effect, and it is indeed real. However, the quote above subtly misdescribes what is actually observed. What is actually observed is that, if light beams are sent around a rotating ring in opposite directions, the counter-rotating beam will arrive back at the source in a shorter time than the co-rotating beam, and the difference in times is related to the angular velocity of rotation.

In other words, what is directly measured is not that "light travels more slowly" in the co-rotating beam than in the counter-rotating beam. If light speed detectors were placed at various points in the ring, to measure the speed of the light beams passing them, they would measure the beams to be moving at $c$. What is directly measure is the difference in travel times of the two beams; and that cannot be due to a change in the speed of light, because we can measure that to be the same locally. Instead, it is due to the way the geometry of spacetime works. A heuristic way of describing this is to say that, relative to an inertial frame, the light source is moving towards the counter-rotating beam and away from the co-rotating beam--in other words, it's because the rotation of the ring breaks the symmetry between the two different directions in space around the ring.

Since the difference in travel times is an invariant, it must still be present if we switch to a non-inertial frame in which the ring is at rest. In this frame, the coordinate speed of light is indeed different in the two directions; but once again, that coordinate speed has no physical meaning, as we can show by putting light speed detectors at various points around the ring and measuring the speeds of both beams to be $c$. So the difference in travel times has to be due to something else--once again, to a breaking of the symmetry between the two opposite spatial directions around the ring.

How else could we test this? Here's one way: put a sensor on the ring that detects the frequency of incoming light from some distant object at a fixed position (relative to an inertial frame). If the ring is not rotating, it can be placed so that the sensor continuously detects that light, and reports a constant frequency, the same as the (known) frequency of emission. If we now start the ring rotating, the sensor will, first of all, only detect the light intermittently, once per rotation; and each time it detects the light, it will measure it to have a different frequency--either blueshifted or redshifted, depending on the direction of the incoming light beam relative to the ring. The frequency shift will be directly related to the angular velocity of rotation of the ring, and will give a direct measurement of the breaking of the symmetry of space in the frame in which the ring is at rest. We could make this even stronger by having two incoming beams, coming in from opposite directions, and measuring that one beam's frequency shift is exactly equal in magnitude and opposite in direction to the other beam's frequency shift when the ring is rotating.

The point of all this is that, instead of blindly saying that "light travels faster or slower in a gravitational potential" based on one measurement, we should be looking at all the measurements we can make, and coming up with a single scheme of description that covers all of them. That's how scientific theories work. And "the speed of light changes" is not a workable single scheme that accounts for all of the experiments. The geometry of spacetime, with the speed of light always locally measured to be $c$, is a workable single scheme. That's why we prefer it here on PF.

 

[Max Hofbauer]

Nothing shot vertically upwards has a chance to escape a g-field... in the long run.
To illustrate this we need another Gedanken Experiment:

In a universe holding only the Gedanken-Planet, but no other celestial
bodies, you would never be able to shoot so far away that gravity could
be overridden by dark energy...

In the case of a black hole, it seems perfectly possible for a beam of light
to escape, only to fall back later. In order for us to see that beam, it would
have to reach all the way to our retinas, but we are too far away.

Maybe anyone, bound to be gobbled up by a black hole, can witness a bright
dot in the very center of it.

 

[PeterDonis]

In a universe holding only the Gedanken-Planet, but no other celestial
bodies, you would never be able to shoot so far away that gravity could
be overridden by dark energy...

For an object shot upward to escape the Gedanken-Planet, it doesn't need to have the Gedanken-Planet's gravity overridden by dark energy. It just needs to be shot upward fast enough that the Gedanken-Planet's gravity will never slow it down to zero speed and make it fall back again.

In the case of a black hole, it seems perfectly possible for a beam of light
to escape, only to fall back later

It may seem that way to you, but it's not correct. Light emitted radially outward at a black hole's horizon stays at the horizon forever; it doesn't move upward and then fall back. (Light emitted radially outward inside a black hole's horizon falls inward continuously.)

In order for us to see that beam, it would
have to reach all the way to our retinas, but we are too far away.

Incorrect. You would not be able to see light from a black hole's horizon even if you were "hovering" just a very small distance above the horizon. The only way to see a black hole's horizon or what's inside it is to fall in, in which case you yourself can never escape back out.

Maybe anyone, bound to be gobbled up by a black hole, can witness a bright
dot in the very center of it.

Even if you fall inside the hole, you will never see what's at the "center", because there is no "center" in the usual sense. The singularity at $r = 0$ is in your future if you fall in; it's not a place in space, it's a moment of time. The only way to reach it is to move into the future, which you can't help doing; and once you reach it, you cease to exist.

 

[Grinkle]

once you reach it, you cease to exist

Or at least we can say we don't know / have no model for what happens "after time ends" (sic) for particles. Playing on words here, hope I don't come off as just trying to sound clever, but I think the logic is sound - if everything at the singularity does not exist, then the singularity is either a consequential phenomena somehow made of nothing or it also does not exist.

I appreciate the insight that reaching the singularity is really describing a particular tick of the clock more than a particular increment forward in distance.

 

[PeterDonis]

Or at least we can say we don't know / have no model for what happens "after time ends" (sic) for particles.

More precisely, we don't know that the classical model I described, which says that things that reach the singularity cease to exist, is correct; in fact most physicists think it isn't. But we don't have a good model to replace it; we don't have a good theory of quantum gravity (yet), and that's what we would need to know what sort of model would replace the classical model in situations where the classical model predicts a singularity.

if everything at the singularity does not exist

That's not what the classical model says. It says that things that reach the singularity cease to exist. Everything up until that point exists just fine.

As I said above, most physicists believe the classical model is not correct in this situation, but that's not really because of the "cease to exist" part. It's because, even before anything reaches the singularity, according to the classical model, it will encounter spacetime curvatures that are so large that we expect quantum gravity effects to become important and the classical theory to break down. And, as I said above, we don't know what those effects are.

 

[greswd]

Farsight is that you?