Black holes

[sol invictus]

The metric doesn't tell me it varies.

The metric defines both the coordinates and the spacetime, and it lets you predict the results of all experiments.

So what happens when you take the lower of the two parallel-mirror light clocks down to the event horizon? Draw us all a picture. In your own time.

Exactly the same as at a black hole horizon. Clocks held at fixed, smaller r run more slowly than clocks held at fixed, larger r, and clocks held at r=r0 freeze. You can see that immediately from the metric - if you know GR, that is.

 

[Farsight]

The metric defines both the coordinates and the spacetime, and it lets you predict the results of all experiments.

OK. So let me ask you to make a prediction...

Exactly the same as at a black hole horizon. Clocks held at fixed, smaller r run more slowly than clocks held at fixed, larger r, and clocks held at r=r0 freeze. You can see that immediately from the metric - if you know GR, that is.

We're cooking on gas now sol. We have a clock at r=r0. It's a light clock. A parallel-mirror light clock. It's frozen. And you're holding it. So what do you see? Tell me what you see, popsicle.

 

[Brian-M]

We're cooking on gas now sol. We have a clock at r=r0. It's a light clock. A parallel-mirror light clock. It's frozen. And you're holding it. So what do you see? Tell me what you see, popsicle.

Since you're "frozen" too the clock appears to behave normally, the same as it would outside the black-hole.

(Yes, I know the question wasn't addressed to me, but the answer was obvious.)

ETA: I suspect it would take infinite (external) time to reach the point where you're completely frozen, so the best you could do is reach a point where motion is reduced to the point where it appears frozen to the external observer, but is still slowly advancing.

 

[sol invictus]

OK. So let me ask you to make a prediction...

We're cooking on gas now sol. We have a clock at r=r0. It's a light clock. A parallel-mirror light clock. It's frozen. And you're holding it. So what do you see? Tell me what you see, popsicle.

What?

OK, I think you're asking the question Brian-M answered. See below. By the way, the answer is also "exactly what you'd see at a black hole horizon, because the metrics are identical there."

 

[sol invictus]

Since you're "frozen" too the clock appears to behave normally, the same as it would outside the black-hole.

(Yes, I know the question wasn't addressed to me, but the answer was obvious.)

That's correct - assuming we don't worry about the infinite g-force required to remain right at the horizon.

 

[Farsight]

Since you're "frozen" too the clock appears to behave normally, the same as it would outside the black-hole.

(Yes, I know the question wasn't addressed to me, but the answer was obvious.)

And wrong I'm afraid Brian. You're frozen. Light is frozen. You don't see anything. You can't think, and you don't know anything about anything.

ETA: I suspect it would take infinite (external) time to reach the point where you're completely frozen, so the best you could do is reach a point where motion is reduced to the point where it appears frozen to the external observer, but is still slowly advancing.

Let's say it would take infinite external time for you to see the clock tick. Have you seen it tick yet? No. Let's wait for a billion years of external time. Have you see it tick yet? No. Let's wait for a trillion years. Have you see it tick yet? No. You never see it tick.

That's correct - assuming we don't worry about the infinite g-force required to remain right at the horizon.

Let's wait for a zillion years. Has Brian seen it tick yet? No.

 

[sol invictus]

And wrong I'm afraid Brian. You're frozen. Light is frozen. You don't see anything. You can't think, and you don't know anything about anything.

Let's say it would take infinite external time for you to see the clock tick. Have you seen it tick yet? No. Let's wait for a billion years of external time. Have you see it tick yet? No. Let's wait for a trillion years. Have you see it tick yet? No. You never see it tick.

Let's wait for a zillion years. Has Brian seen it tick yet? No.

So let's be very clear - is it your position that in the spacetime described by that metric, no observer will ever see a clock at r=r0 tick?

 

[W.D.Clinger]

You're wrong. You DO view a coordinate system as more than an artifact (not artefact, BTW). Except you view one particular coordinate system as being special.

No I don't.

Yes, you do. As Ziggurat said:

Yet your entire argument is based on the Schwarzchild coordinate system. You keep referring to that one picture in MTW of the Schwarzchild coordinates, and insisting that it represents reality while the others do not.

Light at the event horizon is moving. It's moving along the event horizon.

Where gravitational time dilation is infinite according to observers at a great distance? Like me? I know what, I'll look at that light through my gedanken telescope. Has it moved yet? No. Let's give it half an hour. Has it moved yet? No. How about a year? Has it moved yet? No. And so it goes.

Everything you just said is based upon your implicit assumption that the distant observer is comoving (i.e., sitting still with respect to) the static Schwarzschild coordinates. You used the Schwarzschild metric to calculate the stoppage of time at the event horizon.

Not so. They represent the exact same solution.

Yes so. Because in my gedanken telescope I can also see you. Have you moved yet? No. You don't move, the light doesn't move, and you don't see things happening normally. You don't see anything. Not ever. Your finite proper time takes forever in the real world. The things you thing you'd be able to see are in neverneverland.

You're assuming you can use the Schwarzschild metric to calculate the results of your gedanken experiments. That assumption is justifiable, so long as you're only concerned with what would be observed by an observer who's comoving (sitting still with respect to) the Schwarzschild coordinate system.

You don't understand that your assumptions give special status to the Schwarzschild coordinate system. The only way for an observer located at some finite distance to sit still with respect to Schwarzschild coordinates is for that observer to undergo constant acceleration in the outward radial direction.

That acceleration is measurable in principle. For nearby observers, that acceleration is measurable in practice. Even today, those measurements provide the best direct evidence for the theory of general relativity.

Unaccelerated observers do not sit still with respect to Schwarzschild coordinates. They do sit still with respect to some of the coordinate systems you dismiss with flippant language ("neverneverland") instead of reason.

Do you even get the difference between a coordinate singularity and a genuine singularity?

Yes. You don't understand that you cannot eliminate this c=0 by choosing neverneverland coordinates where a stopped Zig in front of stopped light sees everything happening normally. Don't you get it yet? Nothing happens. Ever.

You're wrong.

Light doesn't stop anywhere within the Schwarzschild manifold (restricted to r > 2m). The Schwarzschild coordinate singularity prevents Schwarzschild coordinates from being used at r=2m.

The physical interpretation of that coordinate singularity is that it would require infinite outward acceleration to prevent the radial coordinate of an observer situated at r=2m from changing over time.

It's a mathematical fact that the Schwarzschild coordinate singularity can be removed by a straightforward transformation of the time coordinate t that replaces that coordinate by the proper time of an unaccelerated (infalling) observer. That coordinate transformation gives us Painlevé-Gullstrand coordinates.

It's a mathematical fact that, as reckoned by an infalling observer at the event horizon, outgoing light makes no outward headway (as measured by the Schwarzschild radial coordinate) against gravity. It is another mathematical fact that, at the event horizon, ingoing light makes rapid inward progress by that same measure. I proved those facts in another thread.

Not at all. Light most definitely moves at the event horizon in Kruskal coordinates. Or in any coordinate system which doesn't have a coordinate singularity.

Light doesn't move in a coordinate system. It moves through space.

The quality of discourse here is not improved by your desire not to understand what others are saying.

In a coordinate system, the movement of light through space is represented by its world line, which will be a null geodesic. At any point on that world line, we can calculate the coordinate velocity of the light.

And if there is a location in space where that light is stopped, you can't make it move by flicking to a coordinate system that pretends that a stopped light-clock isn't stopped just because a stopped observer is sitting in front of it.

It is a mathematical fact that the coordinate velocity of light depends upon the coordinate system. It is another mathematical fact that the coordinate velocity of light is undefined (not "stopped") at coordinate singularities.

The chart (coordinate system) is not the manifold. Whether light is moving is a property of the manifold and its metric, not the chart. Your argument rests upon the exalted status you bestow upon Schwarzschild coordinates, compounded by the errors you make when you treat an undefined quantity as a definite zero.

Do not presume to lecture me. You know far less than you think you do.

Ditto.

As for the artefact, I'm John Duffield, I live in Poole in the UK. Pleased to meet you. And you are?

Okay, John Duffield of Poole: Do not presume to lecture us. You haven't done your homework.

So a beam of light directed straight up doesn't get out because it travels sideways?

No. The very question shows how lost you are.

At the event horizon, outwardly directed light has to travel at the speed of light just to remain at r=2m. In effect, space is falling into the black hole just as fast as light can travel through that space. That's the river model we've already discussed in this thread.

At the event horizon, inwardly directed light diminishes its radial r coordinate twice as fast as it would in flat spacetime. You can examine the calculation in another thread.

And no mention of the coordinate speed of light varying with gravitational potential? Or the speed of light? Or the infinite gravitational time dilation according to distant observers?

Had you done your homework, you'd know that all of those things are taken into account by the calculations.

I have made no fundamental mistake, I've plumbed the fundamentals.

You have made many fundamental mistakes, including those noted above. You repeat those mistakes because you refuse to do any homework.

The metric doesn't tell me it varies. Experiment tells me that.

Experiment tells you the general theory of relativity is correct or nearly so. The general theory of relativity tells you the metric is correct and can be used to calculate.

You aren't listening to what experiment tells you. You aren't listening to what the general theory of relativity tells you.

If everybody gives their real names you get a more civilised discussion. If it's only me, I've got the moral high ground.

ETA:

So let's be very clear - is it your position that in the spacetime described by that metric, no observer will ever see a clock at r=r0 tick?

That's an excellent question. I look forward to Farsight's answer.​

 

[Farsight]

So let's be very clear - is it your position that in the spacetime described by that metric, no observer will ever see a clock at r=r0 tick?

Not quite. I don't mean to nitpick, but it's more fundamental, wherein motion is given priority over time. More Raum und Bewegung rather than Raum und Zeit. My position is that spacetime is a static "all times view" that combines space with motion through it, that the metric is derived from the motion of light, that the observer is located in space rather than spacetime, and that when he's situated at a location where we as distant observers say the event horizon is located, then he never sees a clock tick.

 

[Farsight]

Everything you just said is based upon your implicit assumption that the distant observer is comoving (i.e., sitting still with respect to) the static Schwarzschild coordinates. You used the Schwarzschild metric to calculate the stoppage of time at the event horizon.

No it isn't. Now look at my answer to sol above. The stoppage of time does not feature. It's based on the hard scientific evidence that the speed of light varies with gravitational potential, and a clear understanding that we define the second and the metre using the motion of light and so always deem the local speed of light to be 299,792,458 m/s, that clocks clock up regular cyclic motion rather than "the flow of time", and by reading the original Einstein to understand that that gravitational time dilation is the result of a reduced rate of motion caused in turn by a concentration of energy "conditioning" the surrounding space. From that I say that if gravitational time dilation goes infinite, the rate of motion goes to zero, and I test this by asking why a vertical light beam, which does not curve or fall back or decelerate, does not escape a black hole. Answer: because it's stopped.

I have to go I'm afraid, otherwise I'm in trouble with the wife this Saturday night. I'll look at the rest of your post tomorrow. Meanwhile take care with your other thread. It's a commendable effort, you are perhaps not beyond redemption, but remember this:

"Spacetime manifolds are mathematical objects. We can prove their mathematical existence follows from the laws of logic and the axioms of mathematics, but we cannot use mathematics alone to prove a spacetime manifold accurately describes the physical universe. Whether something exists in a physical sense is a question for science, not mathematics".

And please note that I've mentioned vacuum impedance before now. Make sure you check that out.

Catch you later.

 

[Ziggurat]

So a beam of light directed straight up doesn't get out because it travels sideways?

How exactly do you define which way a beam of light points? That's not a trivial question. I suspect you haven't actually really thought about the answer.

And no mention of the coordinate speed of light varying with gravitational potential?

Note that hilighted word: the coordinate speed of light varies with gravitational potential in Schwarzchild coordinates. It does not vary in Kruskal coordinates. You are favoring one set of coordinates over every other set.

Or the speed of light? Or the infinite gravitational time dilation according to distant observers?

I can get infinite time dilation for distant observers even without gravity. That really is no problem at all.

Let's tease this one out. Let's say our vertical emitter is a small distance above the event horizon. It employs the hyperfine transition, an electromagnetic spin-flip "process" that emits light, which travels straight up. We see this light as redshifted. Now move the emitter to a point closer to the event horizon. We still see the light, but now it's even more redshifted, and we know that the emission process is running at a slower rate. We repeat this, and with some care, we now detect shortwave radio waves. Then longwave, and so on. At what point do those electromagnetic waves suddenly turn sideways, and at what speed do they then move with respect to distant observers? And with respect to the emission process?

Every single thing you describe here applies to the event horizon created by acceleration.

I have made no fundamental mistake, I've plumbed the fundamentals.

You have made many fundamental mistakes. Including a number below, which I will describe to you..

In flat spacetime the speed of light through the universe is uniform. An observer moving through the universe doesn't change that. Look, there he is out in space. Can you see him? Yes. Now look just behind him? See that black cone? Can you see his Rindler horizon? No. You see plain old stars. There is no gravitational lensing surrounding a black region. This isn't in the same league as a black hole.

You are wrong on many, many counts. First off, you can't ever see a horizon, not even for a black hole. Second, the Rindler horizon isn't a cone, it's a plane. Third, the stars you see behind your accelerating observer are the exact same stars that he sees (though he sees them redshifted). The light that you now see was emitted long ago, before they crossed the Rindler horizon. The Rindler horizon moves at the speed of light in your non-accelerating frame, and so you won't see light from stars on the other side of the Rindler horizon until you've crossed the Rindler horizon yourself. And that's exactly what will happen with a black hole. Drop a probe in and fall in after it, and you will NEVER lose sight of the probe. In fact, if you pick the right trajectory, you won't even see it get red shifted. And lastly, you're even wrong about lensing: light WILL bend around a Rindler horizon, in the accelerating reference frame. The horizon doesn't have the same shape, so it won't the the same sort of lensing, but you'll get equivalent bending of the trajectory of light.

You may have heard about Rindler horizons, but you clearly don't understand them.

 

[Reality Check]

Not quite. ...

You did not answer sol invictus's question.
Here is the metric:

I've already given you an example. Instead of "knocking it down", you ignored it. Here it is again:

The coordinate speed of light is zero at r=r0. Clocks (including light clocks) run slower and slower if they are held at fixed, smaller and smaller r. In fact, this metric has exactly the properties you keep referring to in the black hole.

Do you know what this metric describes, Farsight? It's not a black hole.

The actual question is

So let's be very clear - is it your position that in the spacetime described by that metric, no observer will ever see a clock at r=r0 tick?

Or better yet: At what rate will a clock at r=0 be observed by an external observer to be ticking?

 

[sol invictus]

Not quite. I don't mean to nitpick, but it's more fundamental, wherein motion is given priority over time. More Raum und Bewegung rather than Raum und Zeit. My position is that spacetime is a static "all times view" that combines space with motion through it, that the metric is derived from the motion of light, that the observer is located in space rather than spacetime, and that when he's situated at a location where we as distant observers say the event horizon is located, then he never sees a clock tick.

It's a yes or no question, Farsight. What's the answer?

And while you're at it, is it your position that locations r<r0 (behind the horizon) don't exist, can be ignored, or in some other way aren't relevant?

 

[Brian-M]

And wrong I'm afraid Brian. You're frozen. Light is frozen. You don't see anything. You can't think, and you don't know anything about anything.

Let's say it would take infinite external time for you to see the clock tick. Have you seen it tick yet? No. Let's wait for a billion years of external time. Have you see it tick yet? No. Let's wait for a trillion years. Have you see it tick yet? No. You never see it tick.

But from my point of view it's ticking at the usual rate, it's just that the rest of the universe is moving infinitely fast.

I don't think you can ever get to the point where you're completely frozen to an outside observer, just to a point indistinguishable to being completely frozen to an outside observer.

Let's wait for a zillion years. Has Brian seen it tick yet? No.

Possibly. But if I'm still falling into the event horizon, the next tick will take longer.

Of course, that's assuming I'm not exactly at the event horizon yet, which would take forever for me to reach from the viewpoint of an external observer.

Which makes the question about how long it'd take for me to see the clock tick from an external observer's point of view once I get there meaningless, because from an external observer's point of view I can never quite get there to begin with.

 

[Ziggurat]

But from my point of view it's ticking at the usual rate, it's just that the rest of the universe is moving infinitely fast.

Actually, if you're in an infalling reference frame, the rest of the universe is NOT moving particularly fast, even as you cross the event horizon.

 

[Brian-M]

Actually, if you're in an infalling reference frame, the rest of the universe is NOT moving particularly fast, even as you cross the event horizon.

Ah, that makes sense. You'd have to be standing still to experience the blue-shift. My mistake.

 

[Farsight]

It's a yes or no question, Farsight. What's the answer?

No, it isn't a yes or no answer. Here's your question again:

"So let's be very clear - is it your position that in the spacetime described by that metric, no observer will ever see a clock at r=r0 tick?"

The metric isn't something that actually exists, and whilst you employ it to describe spacetime, that doesn't actually exist either. What actually exists is space and light moving through it. Or not, as the case may be. And when it isn't moving, the observer sees nothing.

And while you're at it, is it your position that locations r<r0 (behind the horizon) don't exist, can be ignored, or in some other way aren't relevant?

No. They don't exist as locations per se, in that where light doesn't move we can't define space or time. But it's very interesting because here we have a massive concentration of energy in a strange primal form, perhaps something like the early universe. It's "solid space and no space too", and there may be some parallels with a photon BEC.

 

[Farsight]

But from my point of view it's ticking at the usual rate, it's just that the rest of the universe is moving infinitely fast.

And you haven't seen that infinitely-fast universe motion happen yet. You never ever do.

I don't think you can ever get to the point where you're completely frozen to an outside observer, just to a point indistinguishable to being completely frozen to an outside observer.

Maybe. But if I wait a year, a thousand years, a million years, at some point I give up on somebody trying to flog me a dead parrot by claiming you aren't completely frozen.

Possibly. But if I'm still falling into the event horizon, the next tick will take longer. Of course, that's assuming I'm not exactly at the event horizon yet, which would take forever for me to reach from the viewpoint of an external observer.

You need to think about being at the event horizon first, then come back to the falling. Note that you only fall if the speed of light at your feet is lower than it is at your head. When the speed of light is zero, it can't get any lower, so when you're at the event horizon, you don't fall down. In practice of course your feet would be at the event horizon and your head wouldn't be, so things would get messy. But let's use a bit of artistic licence and say that you are entirely at the event horizon.

Which makes the question about how long it'd take for me to see the clock tick from an external observer's point of view once I get there meaningless, because from an external observer's point of view I can never quite get there to begin with.

It isn't meaningless, that external observer's point of view isn't something you should ignore. If he reckons it takes you forever to get to the event horizon, you still never actually cross it. Thus crossing the event horizon in finite proper time by your clock is a never-neverland myth.

 

[W.D.Clinger]

And while you're at it, is it your position that locations r<r0 (behind the horizon) don't exist, can be ignored, or in some other way aren't relevant?

No. They don't exist as locations per se, in that where light doesn't move we can't define space or time.

I'm sorry, but many speakers of English use "no" when affirming a negative question, and your "don't exist as locations per se" suggests you may be doing that here.

Please clarify what you mean by your "No" that I highlighted. Here are three things you might have meant by that word:

1. No, sol invictus did not describe your position, and you don't intend to let us know what position you take on the question sol invictus asked.
2. No, those locations don't exist, can be ignored, and/or aren't relevant.
3. No, it's not the case that those locations don't exist. In other words, those locations do exist and cannot be ignored.

But it's very interesting because here we have a massive concentration of energy in a strange primal form, perhaps something like the early universe. It's "solid space and no space too", and there may be some parallels with a photon BEC.

What massive concentration of energy?

I think you're confused. The two questions that sol invictus asked were not questions about black holes. He was asking about the event horizon and coordinate singularity that appear within a particular metric that's very similar to the Schwarzschild metric on which you base your arguments (even as you deny that your arguments are based on that metric).

Here's the metric that pertains to the question sol invictus was asking:

[latex]$ds^2 = -(r-r_0) dt^2 + dr^2/(r-r_0) + dy^2 + dz^2$[/latex]

The coordinate speed of light is zero at r=r0. Clocks (including light clocks) run slower and slower if they are held at fixed, smaller and smaller r. In fact, this metric has exactly the properties you keep referring to in the black hole.

Do you know what this metric describes, Farsight? It's not a black hole.

Here is the first question, which you still have not answered:

So let's be very clear - is it your position that in the spacetime described by that metric, no observer will ever see a clock at r=r0 tick?

Here is the second question, which you answered with the "No" I asked you to clarify above:

And while you're at it, is it your position that locations r<r0 (behind the horizon) don't exist, can be ignored, or in some other way aren't relevant?

 

[Farsight]

Everything you just said is based upon your implicit assumption that the distant observer is comoving (i.e., sitting still with respect to) the static Schwarzschild coordinates. You used the Schwarzschild metric to calculate the stoppage of time at the event horizon.

No it isn't. Everything I say is based on the hard scientific evidence that the speed of light varies with gravitational potential, like Einstein said repeatedly.

You're assuming you can use the Schwarzschild metric to calculate the results of your gedanken experiments. That assumption is justifiable, so long as you're only concerned with what would be observed by an observer who's comoving (sitting still with respect to) the Schwarzschild coordinate system.

I'm concerned with what actually happens in this real universe.

You don't understand that your assumptions give special status to the Schwarzschild coordinate system. The only way for an observer located at some finite distance to sit still with respect to Schwarzschild coordinates is for that observer to undergo constant acceleration in the outward radial direction.

That acceleration is measurable in principle. For nearby observers, that acceleration is measurable in practice. Even today, those measurements provide the best direct evidence for the theory of general relativity.

Unaccelerated observers do not sit still with respect to Schwarzschild coordinates. They do sit still with respect to some of the coordinate systems you dismiss with flippant language ("neverneverland") instead of reason.

There's black holes out there, Clinger, like a very big very massive dark thing at the centre of our galaxy. And we ain't falling into it. That'll do.

You're wrong. Light doesn't stop anywhere within the Schwarzschild manifold (restricted to r > 2m). The Schwarzschild coordinate singularity prevents Schwarzschild coordinates from being used at r=2m.

Light stops in space. All coordinate systems are artefacts, remember? They don't actually exist.

The physical interpretation of that coordinate singularity is that it would require infinite outward acceleration to prevent the radial coordinate of an observer situated at r=2m from changing over time.

It's wrong on two counts. The speed of light is zero, it can't go any lower, so there's no more falling. And the observer can't exceed the speed of light, so he can't accelerate.

It's a mathematical fact that the Schwarzschild coordinate singularity can be removed by a straightforward transormation of the time coordinate t that replaces that coordinate by the proper time of an unaccelerated (infalling) observer. That coordinate transformation gives us Painlevé-Gullstrand coordinates.

Your mathematical fact is fiction. The time coordinate t is derived from motion, and your straightforward transformation blindly ignores it. Look hard at a clock, Clinger. It doesn't measure the flow of time, it clocks up some kind of regular cyclic motion, and displays you an accumulated total called "the time". When that clock stops, a mathematical transformation won't make it start again. You can't make the clock start by stopping yourself.

It's a mathematical fact that, as reckoned by an infalling observer at the event horizon, outgoing light makes no outward headway (as measured by the Schwarzschild radial coordinate) against gravity. It is another mathematical fact that, at the event horizon, ingoing light makes rapid inward progress by that same measure. I proved those facts in another thread.

Spare me your "mathematical fact", I deal in scientific fact. And I'm not seeing downward sunbeams travelling faster than their upward reflections.

The quality of discourse here is not improved by your desire not to understand what others are saying.

I understand what you're saying. I understand why it's wrong.

In a coordinate system, the movement of light through space is represented by its world line, which will be a null geodesic. At any point on that world line, we can calculate the coordinate velocity of the light.

In the real world, we see that the speed of light varies with gravitational potential. We can also see it curve. And we all know that coordinate systems, and worldlines, and null geodesics are all artefacts of mathematical abstraction.

It is a mathematical fact that the coordinate velocity of light depends upon the coordinate system. It is another mathematical fact that the coordinate velocity of light is undefined (not "stopped") at coordinate singularities.

It is a physical fact that the speed of light varies with gravitational potential. Optical clocks go slower if they're lower. That doesn't depend on the coordinate system.

The chart (coordinate system) is not the manifold. Whether light is moving is a property of the manifold and its metric, not the chart. Your argument rests upon the exalted status you bestow upon Schwarzschild coordinates, compounded by the errors you make when you treat an undefined quantity as a definite zero.

Whether light is moving is a property of space, not the manifold and its metric. Your denial of scientific fact rests upon the exalted staus you bestow on mathematical abstraction.

Okay, John Duffield of Poole: Do not presume to lecture us. You haven't done your homework.

I will hold your nose to the grindstone of hard scientific evidence, Clinger. Don't try to evade it with specious demands for "homework".

No. The very question shows how lost you are. At the event horizon, outwardly directed light has to travel at the speed of light just to remain at r=2m. In effect, space is falling into the black hole just as fast as light can travel through that space. That's the river model we've already discussed in this thread.

And it's baloney. A black hole isn't sucking in the surrounding space like some aether wind. And nor is the sky falling in.

At the event horizon, inwardly directed light diminishes its radial r coordinate twice as fast as it would in flat spacetime. You can examine the calculation in another thread.

Maybe I'll weigh in and rain on your parade there. We'll see. But meanwhile, we can see that light moves slower where gravitational potential is lower, but at the event horizon it's suddenly going twice as fast as when it started? And lookie here, the light coming down from the sun is going faster than its upward reflection. No Clinger. I hate to break it to you, but it's a fantasy I'm afraid.

Had you done your homework, you'd know that all of those things are taken into account by the calculations.

Your calculations are wrong from the off. Your axioms are flawed.

You have made many fundamental mistakes, including those noted above. You repeat those mistakes because you refuse to do any homework.

I'm the one plumbing the fundamentals here, not you.

Experiment tells you the general theory of relativity is correct or nearly so. The general theory of relativity tells you the metric is correct and can be used to calculate.

Einstein tells you the speed of light varies with gravitational potential. The general relativity you're employing is a corrupted version of the original. Think for yourself, don't let MTW do your thinking for you. And no, I don't have a copy.

You aren't listening to what experiment tells you. You aren't listening to what the general theory of relativity tells you.

Oh yes I am. And I'm listening to Einstein. And the hard scientific evidence. You aren't.