I am the first in the world

[ben m]

When I lower you all the way to the event horizon, you see the whole evolution of the universe happening in an instant.

Note that Sol said "an observer FALLING through the event horizon". An observer hanging on a rope is under acceleration, in a distinctly non-inertial reference frame, and he knows it. Any observer under this amount of acceleration will expect this type of time distortion---whether they're being accelerated by a rocket, or by a rope holding them just above a black hole.

The distortion is not something special that happens at event horizons.

 

[Vorpal]

But for someone falling through the horizon, at least for a large black hole where tidal forces are small, nothing even slightly unusual happens.

I know that's what they say, but all processes slow and stop on the horizon as viewed from outside.

And that's not at all contradictory.

So imagine I'm lowering you down on a gedanken rope. I lower you down a little, and we compare notes through the comms line wrapped around the rope. ...

Then Sol is held up by the tension of the rope, and so is in an accelerated frame. Even in STR, acceleration gives a one-way horizon attempting to cross which will rip you apart. (Appropriately, near the horizon of a large Schwarzschild black hole locally looks exactly like an acceleration horizon of STR.)

When I lower you all the way to the event horizon, you see the whole evolution of the universe happening in an instant.

If Sol has an idealized body that's both pointlike and lightlike, maybe. Being a static observer the event horizon is a mathematical impossibility. For any remotely realistic body, the tidal forces in this frame will rip him up long before that. But if he's falling inwards instead, then (relative to the stationary frame) he'll be approaching lightspeed near the horizon, so you can think of this as Lorentz contraction keeping the tidal forces on his body finite and small.

Sorry, I didn't look at it. The thing is, mathematics doesn't get this crucial point across. That's why people blithely switch to a different metric I suppose, and do that hop skip and a jump over the end of time without even noticing.

Ok, how about just taking the Schwarzschild geometry in different coordinates?
[latex]\[ds^2 = -dt^2 + \left(dr+\sqrt{\frac{2m}{r}}dt\right)^2 + r^2\left(d\theta^2 + \sin^2\theta d\phi^2\right)\][/latex]
There's indeed an absolute event horizon. No, time doesn't stop there in that frame.

And when all processes slow and stop, you can't measure anything.

You're taking a statement that's true in a particular reference frame and treating as if it was absolute. It isn't. This is the same kind of mistake commonly made in STR, just in a different context.

 

[sol invictus]

I know that's what they say, but all processes slow and stop on the horizon as viewed from outside.

So?

So imagine I'm lowering you down on a gedanken rope. I lower you down a little, and we compare notes through the comms line wrapped around the rope. We note that initially you experience fairly modest gravitational time dilation. Your parallel-mirror light clock doesn't keep pace with mine, a pulsar a few light years away appears to have speeded up, my voice sounds like Pinky and Perky. I lower you further and the gravitational time dilation becomes so dramatic that a minute of your time is an hour of mine. When I lower you to the very near the event horizon you see suns forming flaring and dying in what seems to be minutes, and galaxies racing across the sky then growing dim. When I lower you all the way to the event horizon, you see the whole evolution of the universe happening in an instant.

That's inaccurate for a number of reasons. But in any case, it is not relevant to what you would see if you fell freely through - as ben and Vorpal point out, you're describing an accelerated observer.

Sorry, I didn't look at it. The thing is, mathematics doesn't get this crucial point across. That's why people blithely switch to a different metric I suppose, and do that hop skip and a jump over the end of time without even noticing.

How do you think we know that time slows down at the horizon as viewed by an external observer, Farsight? It's certainly not from observations - it's what the math tells us. And exactly the same math tells us that observers falling through notice nothing unusual. So on what basis do you accept one conclusion and reject the other?

 

[Farsight]

So?

So all processes slow and stop on the horizon. Everything is stopped. Light has stopped. Light isn't moving, it isn't going down any more, so neither are you.

That's inaccurate for a number of reasons. But in any case, it is not relevant to what you would see if you fell freely through - as ben and Vorpal point out, you're describing an accelerated observer.

It doesn't make any difference. Whether you're hanging on a rope or not, the light in your light clock isn't moving. It isn't moving between the mirrors, and it isn't moving downwards either.

How do you think we know that time slows down at the horizon as viewed by an external observer, Farsight? It's certainly not from observations - it's what the math tells us.

The maths tells you that gravitational time dilation occurs, and that maths is backed up by experiment and observation. We allow for it in the GPS clock adjustment. But what the maths doesn't tell you that it isn't time that slows down, it's motion. Including the motion of light in your light clock that you use to determine the thing you call time t.

And exactly the same math tells us that observers falling through notice nothing unusual.

It isn't exactly the same maths. It's further maths that ignores the fact that both you and your clock are both stopped, and blunders on as if nothing unusual happens to conjure up a hop skip and a jump over the end of time. See the Schwarzschild chart on the left here. Look carefully at that conveniently truncated peak. Also see the wiki article which refers to the "coordinate artifact", and note this:

"In 1939 Howard Robertson showed that a free falling observer descending in the Scwharzschild metric would cross the r = r_s singularity in a finite amount of proper time even though this would take an infinite amount of time in terms of coordinate time t."

So on what basis do you accept one conclusion and reject the other?

Look a bit further down on the wiki article. See the bit that says:

"However due to the obscurity of the journals in which the papers of Lemaître and Synge were published their conclusions went unnoticed, with many of the major players in the field including Einstein believing that singularity at the Schwarzschild radius was physical."

As ever, I'm with Einstein. My interpretation of general relativity adheres to the original. The "modern interpretation" that you, Vorpal, and benm have been taught, doesn't. There's no other way to say this: in certain important respects, it's wrong. And so relativity remains the Cinderella of modern physics.

 

[sol invictus]

It isn't exactly the same maths. It's further maths that ignores the fact that both you and your clock are both stopped, and blunders on as if nothing unusual happens to conjure up a hop skip and a jump over the end of time.

You're directly contradicting every textbook on general relativity. This is a skeptic's forum, so you must provide evidence for such a radical claim. Show us which step in the math is incorrect.

I'll help you start. The defining principle of general relativity is the principle of equivalence, which states that observers in freely falling observatories cannot detect the presence of gravitational fields except through tidal effects. Hence, to show that "nothing unusual happens to a freely falling observer" it's necessary and sufficient to show that tidal forces are weak at the horizon of a large black hole. That's rather trivial to do with modern techniques - but you claim it's wrong, so find the mistake.

As ever, I'm with Einstein. My interpretation of general relativity adheres to the original. The "modern interpretation" that you, Vorpal, and benm have been taught, doesn't. There's no other way to say this: in certain important respects, it's wrong. And so relativity remains the Cinderella of modern physics.

Einstein didn't understand the Schwarzschild solution, that's true. Neither did anyone else until the 50s.

 

[Farsight]

Note that Sol said "an observer FALLING through the event horizon". An observer hanging on a rope is under acceleration, in a distinctly non-inertial reference frame, and he knows it. Any observer under this amount of acceleration will expect this type of time distortion---whether they're being accelerated by a rocket, or by a rope holding them just above a black hole. The distortion is not something special that happens at event horizons.

It doesn't matter ben. A reference frame isn't something that actually exists, it's isn't something that you're actually "in". It's just something we use to describe the uniformity or not of our measurements of space and time, sometimes when in motion, sometimes when not. We do these measurement using the motion of light, and at the event horizon the coordinate speed of light is zero.

Remember what I said about the coordinate speed of light being the actual speed of light? The cordinate speed of light varies in a non-inertial reference frame, such as the room you're in. We now have super-accurate optical clocks that lose synchronisation when separated by only a foot of vertical elevation. You can simplify them to parallel-mirror light clocks, whereupon you can work out that this is what's happening:

|-------------|
|-------------|

Now imagine the lower clock is at the event horizon. This is what's happening:

|-------------|
|-------------|

At the event horizon it isn't nothing unusual happens, it's nothing happens. There are no more events. The light isn't moving any more. You can't measure space, you can't measure time, you can't see, or think, and your reference frame and coordinate system have utterly collapsed.

Think it through.

Vorpal: sorry, I have to go.

 

[Farsight]

You're directly contradicting every textbook on general relativity. This is a skeptic's forum, so you must provide evidence for such a radical claim. Show us which step in the math is incorrect.

No. I've already told you twice that the maths doesn't show you this. I've given you a good explanation, now deal with the argument instead of trying to play the inscrutable mathematics card.

Einstein didn't understand the Schwarzschild solution, that's true. Neither did anyone else until the 50s.

Einstein thought the singularity at the Schwarzschild radius was physical. I'm with Einstein. You're not, and you're attempting to pooh-pooh Einstein by saying he didn't understand it? Come on sol, you can do better than that. Think this thing through.

Now I really must go. Again, apologies Vorpal.

 

[sol invictus]

No. I've already told you twice that the maths doesn't show you this. I've given you a good explanation, now deal with the argument instead of trying to play the inscrutable mathematics card.

Now you're claiming "the maths doesn't show you this"? So are you saying I'm lying? Are you also denying that this is what all the textbooks say? Do you think they're lying too? If so, you're either delusional or ignorant and extraordinarily arrogant. If not, you think the math in those textbooks is wrong. So, tell us what's wrong with it. If you can't, why should anyone pay any attention to you?

As for your "explanation", no one disputes that time slows down at the horizon as viewed by an external observer. Time also slows down for a moving object as viewed by a stationary observer. Does this mean that something changes in the rest frame of the object? Nope - and the same goes at the horizon.

Einstein thought the singularity at the Schwarzschild radius was physical. I'm with Einstein. You're not, and you're attempting to pooh-pooh Einstein by saying he didn't understand it? Come on sol, you can do better than that. Think this thing through.

I don't know for sure what Einstein thought, and you're obviously not a reliable source. My understanding is that he thought the Schw. solution was simply unphysical. And yes - he was wrong.

 

[ben m]

We do these measurement using the motion of light, and at the event horizon the coordinate speed of light is zero.

The coordinate speed of light according to whom, Farsight? You can't throw three different observers into a problem---observers who, remember, will disagree on that speed---and declare that the speed "actually" has some special value.

At the event horizon it isn't nothing unusual happens, it's nothing happens. There are no more events. The light isn't moving any more. You can't measure space, you can't measure time, you can't see, or think, and your reference frame and coordinate system have utterly collapsed.

For any observer accelerating very strongly, your reference frame and coordinate system have "collapsed". That's true if you are accelerating in flat space (and picking up speed w/r/t reference objects). It's also accelerating near a black hole (and hovering just above the event horizon).

For an inertial observer, your local space is approximately flat, your local light speed is c, Newton's Laws hold, etc.. That's true if you are drifting in (globally) flat space, it's true if you are falling inertially and just outside an event horizon, it's true if you are falling inertially and just inside the event horizon.

Your idea that "your coordinate system does something special at the horizon" is flatly wrong. It's not subtly wrong, it's straightforwardly wrong; and it's been derived explicitly in many places that you have access to; you have found no error in the derivations, you apparently just don't like the answer because your hobby-horse suggested a different one.

 

[Vorpal]

The maths tells you that gravitational time dilation occurs, and that maths is backed up by experiment and observation. We allow for it in the GPS clock adjustment. But what the maths doesn't tell you that it isn't time that slows down, it's motion. Including the motion of light in your light clock that you use to determine the thing you call time t.

But t isn't the thing I would call time if I was either suspended over a black hole or falling toward in. Actually, no one could naturally call the Schwarzschild t coordinate time except a stationary observer at infinity.

Say I'm in completely ordinary Minkowski spacetime and I have a big rocket tuned to always give me a constant acceleration a in some particular direction. Around me are a bunch of people in astronaut suits. What would I see? I'd see them going toward a horizon away from me, slowdown, get redshifted, with their images hovering above just above it, with the redshift factor diverging to infinity. What would I observe if I tied a rope around one, preventing them from going away from me? I'd observe that the closer they are to the horizon, the more tension is in the rope, with the required force also diverging.

This is basic STR requiring absolutely nothing of GTR. And it works out very analogously. What's important to note here is that despite all my observations of them slowing down or redshifting, the people go on their merry away. Because as far they are concerned, there's no special surface where I see them slowing down. Time goes on--for them.

In Schwarzschild spacetime is that though stationary observers are accelerated, stationary observer at infinity becomes inertial. That's possible because of spacetime curvature.

"In 1939 Howard Robertson showed that a free falling observer descending in the Scwharzschild metric would cross the r = r_s singularity in a finite amount of proper time even though this would take an infinite amount of time in terms of coordinate time t."

For those interested in a little historical diversion, as far as I'm aware of the first to seriously consider that "collapse goes on" in the formation of a black hole (rather than a "God-given" Schwarzschild geometry) were Oppenheimer and Snyder, also in 1939,
Phys. Rev. 56, 455–459 (1939)
immediately following the Oppenheimer-Volkoff result about neutron degeneracy pressure being unable to stop black hole formation.

As ever, I'm with Einstein. My interpretation of general relativity adheres to the original. The "modern interpretation" that you, Vorpal, and benm have been taught, doesn't. There's no other way to say this: in certain important respects, it's wrong. And so relativity remains the Cinderella of modern physics.

It's the same theory. The mathematics for studying its implications have greatly evolved (compared to today, differential geometry was barely in infacy when Einstein used to formulate GTR).

No. I've already told you twice that the maths doesn't show you this. I've given you a good explanation, now deal with the argument instead of trying to play the inscrutable mathematics card.

It's not inscrutable mathematics. It's basic calculus. Take Einstein's original paper for GTR, which tells you how to calculate geodesics, and the Schwarzschild metric. The radial freefall of an observer is very simple, and it doesn't stop at the horizon. (There's a neat little coincidence that's interesting in itself: for Schwarzschild r-coordinate vs proper time of an radially freefalling observer, the relationship between them is exactly the same as that of radius vs time in Newtonian gravity.)

Most of the other claims in this thread are only marginally more complicated to derive. There's no curvature singularity at the horizon. There's no reason for the manifold to be cut off there.

Einstein thought the singularity at the Schwarzschild radius was physical. I'm with Einstein.

And he was wrong. So what? Einstein made some mistakes, though even your own source (for whatever wiki's worth) even implied he wasn't even aware of Lemaître and Robertson's work.

Your argument is a bit like saying that kinetic energy being relevant to collisions is a nonsensical "modern interpretation" of Newtonian physics because Newton himself thought it was crap.

 

[alfa1]

The Information and the Universe

Conclusions:

In my view Susskind’s first mistake: he is talking about bits of information, but today information is measured in qubits.
I do not agree with Susskind-Hawking: the information is located on the ‘area’ of the universe. In my view the information of a system depends of the energy of the system, we need energy to change the state of the qubit.
Susskind is talking about Hawking’s thought experiment; there is one bit per Planck area. This is wrong because inside a Planck volume the Planck energy EP = 1.9561 × 10^9 J.
Today there is more information inside the universe then on the edge of the universe.
The photons from the edge of the universe can not record all the changing information from the whole Universe. All the qubits like photons, electrons, atoms…are generating a lot of information.
Information is lost in the quantum system trough interaction with an observer.
The information within the quantum states cannot entirely be retained once the system has collapsed from a superposition of states into a specific state. That lost information can never be retained, as opposed to classical theory where information transformations can be reversed to obtain a history of the previous states of the system. That’s why in my view Susskind, Hawking, Gerard 't Hooft…theory is wrong.
When a measurement of any type is made to a quantum system, decoherence breaks down and the wave function collapses into a single state.
Susskind is talking about the information from a book; to have a correct discussion we can not mixed bits with qubits, to describe a system we have to stay at the quantum level.
I calculated precisely the lower bound , the lowest number of qubits after the Big Bang:
min Iuniverse = 15.392 × 10^61 qubits.

http://adrianferent.blogspot.com/

 

[Reality Check]

In my view Susskind’s first mistake: he is talking about bits of information, but today information is measured in qubits.
...

In my view your first mistake: You seem ignorant of the actual topic of this thread.
The topic is what happens to bits of information as they fall into a black hole. Stephen Hawkings asked this at a conference in 1981 and seemed to show that they were lost. The consensus today is that they are not.

Your ignorance also leads you to think that the topic involves the universe, not black holes.

 

[Travis]

Okay....about not seeing something fall into the black hole.

Spaceship B is watching Spaceship A fall into the black hole. From what I'm reading here Spaceship B will never really see Spaceship A disappear into it.

What if Spaceship B then leaves, the crew is put into cryostasis and a million years later it returns to the same viewing point. Would it still see Spaceship A down there on the edge of the event horizon?

 

[sol invictus]

Okay....about not seeing something fall into the black hole.

Spaceship B is watching Spaceship A fall into the black hole. From what I'm reading here Spaceship B will never really see Spaceship A disappear into it.

What if Spaceship B then leaves, the crew is put into cryostasis and a million years later it returns to the same viewing point. Would it still see Spaceship A down there on the edge of the event horizon?

It's a good question. The truth is, the whole "never see anything fall into a black hole" statement is rather misleading. What you'd really see as you watch the ship approach the horizon (I'm going to assume this is a black hole in otherwise empty space) is that the light emitted by the ship's windows gets more and more shifted towards the red end of the spectrum, and at the same time fainter and fainter. If it was falling straight in, after a short time you'd cease to see it at all as the light went infra-red and very faint.

If you then did a very, very, very precise measurement with the right instruments, you'd detect an extremely faint spectrum of thermal (at a very cold temperature) radiation emitted almost uniformly by the horizon of the black hole (that's the famous Hawking radiation). The ship falling in would affect that radiation, but the effects dissipate extremely quickly - they decrease exponentially with a time constant set by the light-crossing time of the hole. For a hole with the mass of the sun, that means after a small fraction of a second the effects of the ship falling in would be totally undetectable.

So no - you wouldn't see it there.

 

[Farsight]

Okay....about not seeing something fall into the black hole. Spaceship B is watching Spaceship A fall into the black hole. From what I'm reading here Spaceship B will never really see Spaceship A disappear into it. What if Spaceship B then leaves, the crew is put into cryostasis and a million years later it returns to the same viewing point. Would it still see Spaceship A down there on the edge of the event horizon?

Not practically, they won't be actually able to see it because the light can't escape. But that's where it will be. See post #25 above. A free falling observer would cross the r = r_s singularity in a "finite amount of proper time", but this takes infinite coordinate time. So he never actually crosses the event horizon. Not in a million years. Not in a billion years. Not ever.

 

[Farsight]

And that's not at all contradictory.

No, it isn't contradictory, it's crystal clear. All processes slow and stop on the horizon as viewed from outside. So a parallel-mirror light clock stops. And light stops. It stops moving back and forth between the mirrors. It stops moving, full stop. So it isn't moving down any more.

Then Sol is held up by the tension of the rope, and so is in an accelerated frame. Even in STR, acceleration gives a one-way horizon attempting to cross which will rip you apart. (Appropriately, near the horizon of a large Schwarzschild black hole locally looks exactly like an acceleration horizon of STR.)

It doesn't matter. Make it a supermassive black hole so you don't have to worry about the tidal gradient. And take note: this is a place where light has stopped. The coordinate speed of light is zero. Aim a photon at the black hole and it doesn't get blue-shifted to some infinite frequency as it crosses th event horizon. Instead, it stops.

If Sol has an idealized body that's both pointlike and lightlike, maybe. Being a static observer the event horizon is a mathematical impossibility. For any remotely realistic body, the tidal forces in this frame will rip him up long before that. But if he's falling inwards instead, then (relative to the stationary frame) he'll be approaching lightspeed near the horizon, so you can think of this as Lorentz contraction keeping the tidal forces on his body finite and small.

And what's lightspeed at the horizon? Remember what i said about the coordinate speed of light above. It's zero.

Ok, how about just taking the Schwarzschild geometry in different coordinates?
[latex]\[ds^2 = -dt^2 + \left(dr+\sqrt{\frac{2m}{r}}dt\right)^2 + r^2\left(d\theta^2 + \sin^2\theta d\phi^2\right)\][/latex]
There's indeed an absolute event horizon. No, time doesn't stop there in that frame.

There is no time in that frame. In that frame, you're keeping time on a stopped clock. And you're stopped too. The two don't cancel each other out so that "you don't notice anything unusual". Everything has ground to a halt, light doesn't move, nor do nerve impulses in your head, so you don't notice anything.

You're taking a statement that's true in a particular reference frame and treating as if it was absolute. It isn't. This is the same kind of mistake commonly made in STR, just in a different context.

I'm not making a mistake here Vorpal. In the STR scenario, it's like you travelling at the speed of light. We all know you can't actually do this, but think it through. If you were travelliung at c, you'd be totally time-dilated, you don't notice anything any more. You might imagine that "in your frame" everything is carrying on as normal, but it isn't. And in the real world, everybody sheds a tear, because poor old Vorpal is out there in space frozen and insensible, forever.

 

[Vorpal]

Re: #36, but highly relevant to #37.
That doesn't make sense even in STR, Farsight. Take the Rindler chart of Minkowski spacetime, with the coordinate time corresponding to a uniformly accelerated observer seeing inertial (freefalling) objects. The observations made in this frame are in a fair analogy to what a stationary observer in Schwarzschild spacetime would: the slowing, the redshift, the non-crossing, even the thermal radiation. And yet the obviously objects cross the horizon in their own frames.

Proper time is what an actual infalling clock would directly measure. I don't know why you discount it as unreal. If anything, it has far more fundamental than any coordinate because it's completely observer-independent. You're making an huge conceptual mistake.

 

[steenkh]

I'm not making a mistake here Vorpal. In the STR scenario, it's like you travelling at the speed of light. We all know you can't actually do this, but think it through. If you were travelliung at c, you'd be totally time-dilated, you don't notice anything any more. You might imagine that "in your frame" everything is carrying on as normal, but it isn't. And in the real world, everybody sheds a tear, because poor old Vorpal is out there in space frozen and insensible, forever.

You write as if there are two worlds: the traveller's and the real world. As I understand it, they are both equally real. The traveller would see our world recede at the speed of light, and for him, our time will stop.

 

[sol invictus]

Farsight, understanding just a little about the metric in post #17 - which is the Rindler chart of Minkowski spacetime that Vorpal just mentioned - would probably help you realize your mistake. That metric has a horizon that is exactly like the horizon of a large black hole. Time stops at that horizon in exactly the same way, the redshift goes to infinity in the same way, you'd have to accelerate to keep from falling in, etc.

And yet, the horizon of that metric is unquestionably a coordinate artifact - I can choose a different set of coordinates on the same spacetime such that any point I like is on the horizon. So the assertion that objects never "actually" cross such horizons is not even logically self-consistent - it's just wrong.

 

[Farsight]

Now you're claiming "the maths doesn't show you this"? So are you saying I'm lying?

No, I'm saying you don't understand this.

Are you also denying that this is what all the textbooks say? Do you think they're lying too? If so, you're either delusional or ignorant and extraordinarily arrogant.

Don't thump your "good book" at me, sol. Deal with the argument and the logic.

If not, you think the math in those textbooks is wrong. So, tell us what's wrong with it. If you can't, why should anyone pay any attention to you?

I've told you before, don't try hiding behind the maths, not with me. Look at this expression:

[latex]t_0 = t_f \sqrt{1 - \frac{2GM}{rc^2}} = t_f \sqrt{1 - \frac{r_0}{r}}[/latex]

t₀ is the proper time between events A and B for a slow-ticking observer within the gravitational field,
t_f is the coordinate time between events A and B for a fast-ticking observer at an arbitrarily large distance from the massive object...
r is the radial coordinate of the observer...
r₀ is the Schwarzschild radius...

It's a simple expression, and it tells you that when r is very very large, when you're a long way from the black hole, t₀ tends to t_f. It's the infalling observer's clock rate expressed as a fraction of the distant observer's clock rate. I'm the distant observer, you're the other guy. As you fall, when you run the numbers for ever-smaller values of r expressed as a multiple of r₀, you can see the way t₀ trends.

When r is 4, t₀ = .866
When r is 3, t₀ = .816
When r is 2, t₀ = .707
When r is 1.5, t₀ = .577
When r is 1.1, t₀ = .301
When r is 1.01, t₀ = .099
When r is 1.00, t₀ = 0

At the event horizon, your clock rate, expressed as a fraction of mine, is zero. Like you said, all processes slow and stop on the horizon as viewed from outside. And your light clock is a parallel-mirror light clock, remember. With a clock rate of zero. I'll wait a million years. Has your clock ticked yet? No. I'll wait a billion years? Has your clock ticked yet? No.

Has the penny dropped yet?

As for your "explanation", no one disputes that time slows down at the horizon as viewed by an external observer. Time also slows down for a moving object as viewed by a stationary observer. Does this mean that something changes in the rest frame of the object? Nope - and the same goes at the horizon.

Are you being deliberately obtuse? Read post #25 again, and this time pay attention: what the maths doesn't tell you that it isn't time that slows down, it's motion. Including the motion of light in your light clock that you use to determine the thing you call time t. You've got a clock rate of zero. See what I told Vorpal. You're like the gedanken observer travelling at c. You observe nothing.

I don't know for sure what Einstein thought, and you're obviously not a reliable source. My understanding is that he thought the Schw. solution was simply unphysical. And yes - he was wrong.

You know how you get all those cranks and crackpots who know b*ggar all physics and who insist that Einstein was wrong? Well, take a look at which side of the fence you're on.