General relativity is self-inconsistent

[andrewgray]

Zanket,

 

You still may not be getting what I am saying about accelerated coordinates. Take our Rindler flat spacetime metric:

 

[math]-d\tau^2 \;=\; -R^2 dt^2 \;+\; dR^2[/math]

 

To get light geodesics, one sets [math]d\tau = 0[/math].

 

This yields

 

[math]\frac{dt}{dR} \;=\; \frac {1}{R}[/math]

 

This means that as R→0, the slope of light worldlines in these accelerated coordinates goes vertical, i.e., the light cones become vertical. But this is flat spacetime, and we must ask, "so what?"

 

Light cones can go vertical in accelerated coordinate systems and it means nothing.

 

Here we have an example of light cones going vertical in the hyperbolic frame, and nothing is there. This is flat spacetime. If light cones go vertical in an inertial frame, then that is something. But there is no inertial frame with vertical light cones near the worldline R=0 or r=2M.

 

So what I am saying is that light cones become vertical in the accelerated Schwarzchild frames near r=2M, but not in inertial frames near r=2M.

 

That’s because the gravitational acceleration between r=2M and r=2M+ε is arbitrarily high. . .

 

Zanket, the hyperbolic acceleration is arbitrarily high as well, in flat spacetime. It becomes infinite at R=0. So what? There is no inertial frame where the acceleration is infinite. The same is true near r=2M, since I have shown that the gravitational horizon and Rindler horizons are equivalent.

 

Do you think GR’s equation for gravitational acceleration is incorrectly derived from the Schwarzschild metric?

 

No. The expression for constant r acceleration is correct from the Schwarzchild metric. So just like an observer cannot stay at R=0 in flat spacetime, and observer cannot stay at r=2M in curved spacetime. And just like an observer can cross the R=0 worldline in both directions, an observer can cross the r=2M worldline in both directions. It's the same.

 

Do you agree that the equation predicts that an object dropped from rest at r=2M+ε passes an observer hovering at r=2M in the limit at c in the limit?

 

Zanket, what you are asking can be reworded like this:

 

"Do you agree that an object dropped from r=2M+ε passes a lightspeed observer moving at the speed of light?"

 

The question is impossible in itself. No observer can stay along the r=2M worldline, just like no observer can stay along the R=0 worldline in flat spacetime.

 

In the accelerated frame, an object dropped towards the horizon reaches lightspeed. It is possible for accelerated frames to have objects moving at lightspeed. In our rotating frame above, where ρ=1/ω, all stationary objects in the inertial frame move at lightspeed in the rotating frame. These objects have V=c in this frame. So what? Physics is only valid in inertial frames. In the Schwarzchild frame, objects enter the horizon at c, and they leave the horizon at c. This is also true for the hyperbolic frame.

 

But for the hyperbolic case, do we care if something has a velocity of c relative to the hyperbolic frame? No, because this is flat spacetime, and the object's velocity remains sublight in all inertial frames.

 

Then no material object passing upward through r=2M will pass r=2M+ε . . .

 

An object dropping through the horizon will feel no acceleration and have a finite velocity relative to all inertial frames. Thus, time reverse this and the object leaves the horizon with no acceleration and a finite velocity relative to all inertial frames.

 

An object dropping through the horizon will feel no acceleration and have a lightspeed velocity relative to the Schwarzchild frames. Thus, time reverse this and the object leaves the horizon with no acceleration and a lightspeed velocity relative to the Schwarzchild frames.

 

What I need to do is start from the two freefall metrics:

 

[math]-d\tau^2 = -dT^2\; +\; (4M/3)^{2/3} \frac{dR^2}{(R-T)^{2/3}} \;+ \;(2M)^{2/3}(3/2(R-T))^{4/3} d \Omega^2 \;\;\;\; [/math] (Incoming)

 

[math]-d\tau^2 = -dT^2 \;+\; (4M/3)^{2/3} \frac{dR^2}{(T-R)^{2/3}}\; + \;(2M)^{2/3}(3/2(T-R))^{4/3} d \Omega^2 \;\;\;\; [/math] (Outgoing)

 

and confirm that these two satisfy Einstein's equations with no coordinate singularities and no vertical light cones near r=2M. Then for the coupe de grace, calculate the incoming's velocity relative to the outgoing. Will that satisfy your requirements?

 

Andrew A. Gray

[Qfwfq]

a = M / (r^2 * sqrt(1 - (2M / r)))

So apparently you mean in Schwarzschild coordinates, presumably the second time derivative of r. However, it is somewhat ill-defined to call this "the acceleration the observer feels there" as a free falling observer will "feel" no acceleration. What will be felt by an object not pointlike is the gradient and this goes for the box in your OP.

[Guest Zanket]

However, it is somewhat ill-defined to call this "the acceleration the observer feels there" as a free falling observer will "feel" no acceleration. What will be felt by an object not pointlike is the gradient and this goes for the box in your OP.

I didn't suggest otherwise. I said "It returns the initial acceleration that an observer fixed at a given r-coordinate measures for an object dropped from rest there, or the acceleration the observer feels there." An observer who is fixed at a given r-coordinate—like you are now—is not in free fall.

[Guest Zanket]

If light cones go vertical in an inertial frame, then that is something. But there is no inertial frame with vertical light cones near the worldline R=0 or r=2M.

In Penrose’s plot above, the light cones are going vertical in an inertial frame. At the tip of each cone is an inertial observer radiating a signal in all directions. T&W also use a “rain-frame metric”, a metric for an inertial observer falling to r=0, to draw their similar plot.

 

Zanket, the hyperbolic acceleration is arbitrarily high as well, in flat spacetime. It becomes infinite at R=0. So what? There is no inertial frame where the acceleration is infinite. The same is true near r=2M, since I have shown that the gravitational horizon and Rindler horizons are equivalent.

Since you don’t deny the equation, here you’re denying the outcome of experimental tests. The acceleration doesn’t have to become infinite for me to make my case. It need only be arbitrarily high. You’re implying that any ball thrown upward from the ground can reach r=infinity. But no, experimental tests and common experience shows that r_max depends on g at the ground. If the ball is thrown upward with a force less than g then it won’t even reach a higher r-coordinate. Likewise, a material object passing upward through r=2M can reach no higher than r=2M+ε, because g is arbitrarily high between r=2M and r=2M+ε.

 

The question is impossible in itself. No observer can stay along the r=2M worldline, just like no observer can stay along the R=0 worldline in flat spacetime.

I said “at r=2M in the limit”. That is, arbitrarily close to (and above) r=2M. An observer can stay there, so I suggested something possible. This is an experiment that can be conducted in principle. Let the object be dropped from rest at r=(2 + 10^-70)*M. Let the observer be at r=(2 + 10^-7000)*M. GR’s equation predicts that the object will pass the observer at close to c. An object dropped from rest at r=2M+ε passes an observer who is at r=2M in the limit at c in the limit (arbitrarily close to but less than c).

 

In the Schwarzchild frame, objects enter the horizon at c, and they leave the horizon at c. This is also true for the hyperbolic frame.

 

But for the hyperbolic case, do we care if something has a velocity of c relative to the hyperbolic frame? No, because this is flat spacetime, and the object's velocity remains sublight in all inertial frames.

You must care, if you want to refute me. You’re trying to show that the box in the OP can be at rest relative to the particle, which is escaping to r=infinity. But I have shown using GR’s equation that an object passing upward through the horizon comes to rest at no higher than r=2M+ε, where ε is zero in the limit. Just like a ball thrown upward from the ground at a force less than g there does not reach a higher r-coordinate, an object thrown upward through r=2M reaches no higher than r=2M+ε, because the force at which the object is thrown is less than some g between r=2M and r=2M+ε.

 

An object dropping through the horizon will feel no acceleration and have a lightspeed velocity relative to the Schwarzchild frames. Thus, time reverse this and the object leaves the horizon with no acceleration and a lightspeed velocity relative to the Schwarzchild frames.

Yes, a velocity of c in the limit. And an object passing upward through the horizon comes to rest at no higher than r=2M+ε, according to GR’s equation for gravitational acceleration. The object need not feel an acceleration. The r_max for a ball thrown upward from the ground on Earth is dependent on g even though the ball feels no acceleration.

 

What I need to do is start from the two freefall metrics:

...

Will that satisfy your requirements?

The way I see it, I’m refuting you by simpler means. I’d like you to refute those simpler means first. If you were able to refute my claims above, I would drop back to showing that a photon passing upward through r=2M cannot reach r=2M+ε, based on your proof that there is a Rindler horizon at r=2M for an observer hovering at r=2M+ε. And if you were able to refute that, I would drop back to pointing out that GR’s equation for gravitational acceleration returns every possible finite rational value for g at r>2M, in which case it would be impossible for the body’s surface to be fixed at r<=2M, unless gravity is weaker below r=2M than above it.

 

Let me just ask now: What does GR predict for g at the surface of a body whose surface is fixed at r=2M? You have been claiming that a body’s surface can be fixed at r<=2M, right? If you are right about that and GR is self-consistent, then it must predict some finite value for g at r=2M, right?

[Qfwfq]

OK I see what you mean now about the observer, but there are the other oddities of Schwarzschild coordinates near the horizon, as have been discussed here. The way to put it is basically that Schwarzschild's r and t cease to have their "usual" meaning, buy the time you reach the horizon; one way of seeing this is by the discussions of the light cones. Indeed your point about the observer at a fixed r, which makes sense for [imath]r > r_s[/imath], shows that this observer is not fixed but accelerating according to locally inertial coordinates.

[Guest Zanket]

OK I see what you mean now about the observer, but there are the other oddities of Schwarzschild coordinates near the horizon, as have been discussed here. The way to put it is basically that Schwarzschild's r and t cease to have their "usual" meaning, buy the time you reach the horizon; one way of seeing this is by the discussions of the light cones. Indeed your point about the observer at a fixed r, which makes sense for [imath]r > r_s[/imath], shows that this observer is not fixed but accelerating according to locally inertial coordinates.

If there’s a refutation or problem stated in there, please elaborate because I’m not seeing it.

[andrewgray]

Zanket,

 

In my opinion, I have shown mathematically that GR predicts that your box "spanning the horizon" can leave the horizon, since I have shown that the Schwarzchild horizon is exactly equivalent to a Rindler horizon. You may still not understand that proof, but it seems to me that all I have left to do is to refute the incorrect traditional logic that says otherwise.

 

What does GR predict for g at the surface of a body whose surface is fixed at r=2M? You have been claiming that a body’s surface can be fixed at r<=2M, right? If you are right about that and GR is self-consistent, then it must predict some finite value for g at r=2M, right?

 

Not quite correct. What I have been saying is that the worldline of a point on the surface can intersect the worldline r=2M-h, then intersect a "higher" worldline r=2M-h+ε. When r<2M, constant r-lines are spacelike, not timelike. Timelike worldlines (for objects) can easily intersect spacelike worldlines, but they cannot follow them (i.e., remain at constant r when r<2M). It is similar to our rotating coordinate system. When ρ>1/ω, constant φ worldlines become impossible. The tangential velocity of the coordinate system becomes greater than c. But freefall worldlines can intersect constant φ worldlines, no problem.

 

So what you are not quite understanding is that a constant acceleration coordinate system always includes a time when the coordinates stop and come from the other direction relative to inertial frames. Thus it is possible to both "enter" and "leave" constant r worldlines. But an object may not remain at constant r (when r<2M) for a finite amount of time.

 

You’re suggesting that the light cones do not become vertical at r=2M.

 

Zanket, add "in any inertial frame". Light cones do not become vertical at r=2M relative to any inertial frame. Light cones do not become vertical in any local inertial frame. Light cones do become vertical relative to the accelerated Schwarzchild coordinates at r=2M.

 

However, I can say that T&W use what they call a “rain-frame metric”, the metric for an inertial observer falling to r=0. They show a plot similar to this one by Penrose (click on it for more info):

 

Can we see that metric? (Zanket, are you going to learn latex?) Are you sure that it is inertial? Because if it is, then your inconsistency is proven if light cones go vertical.

 

Andrew A. Gray

[Guest Zanket]

In my opinion, I have shown mathematically that GR predicts that your box "spanning the horizon" can leave the horizon, since I have shown that the Schwarzchild horizon is exactly equivalent to a Rindler horizon. You may still not understand that proof, but it seems to me that all I have left to do is to refute the incorrect traditional logic that says otherwise.

Let’s be clear that I haven’t denied that you’ve shown that. I’ve said that GR disagrees with it. When GR is self-inconsistent, it can both agree and disagree with your proof. I don’t think you’ve refuted all of the logic that contradicts your proof.

 

But an object may not remain at constant r (when r<2M) for a finite amount of time.

Then what do you think happens to a body whose surface falls below r=2M? Do you think its surface falls all the way to r=0? Or do you think it continuously falls and rises unless it rises above r=2M again? Those seem to be the only alternatives.

 

Do you agree that r_max in your diagram in post #62 is affected by the gravitational acceleration g along the object’s path? Do you agree that a rocket on the ground whose maximum thrust is less than g at the tip of the rocket cannot lift off the ground? Do you agree that g at some r between 2M and 2M+ε is higher than any given finite force with which an object can be hurled?

 

Light cones do become vertical relative to the accelerated Schwarzchild coordinates at r=2M.

Yes, and that makes my case. When light cones become vertical relative to the accelerated Schwarzschild coordinates at r=2M, it means that nothing—not even light—can pass outward through r=2M to reach an accelerated observer hovering just above r=2M. Solid evidence from GR to the contrary—like possibly your proof that the Schwarzschild horizon is exactly equivalent to a Rindler horizon—means that GR self-inconsistent.

 

Zanket, add "in any inertial frame". Light cones do not become vertical at r=2M relative to any inertial frame. Light cones do not become vertical in any local inertial frame.

This is contradicted by your previous statement. The escaping particle in the OP can never receive a signal emitted from r<=2M, or else the signal could also reach an accelerated observer hovering above r=2M, in which case light cones do not become vertical relative to the accelerated Schwarzschild coordinates at r=2M.

 

Can we see that metric? (Zanket, are you going to learn latex?) Are you sure that it is inertial?

I’ll post it; I’m away from the book now. I didn’t want to learn latex, but we’ll see :eek_big:. The metric is for inertial observer; T&W call it a “rain-frame” metric to evoke the visualization of a raindrop falling from rest at some r-coordinate, like from a cloud.

[Guest Zanket]

Can we see that metric? (Zanket, are you going to learn latex?)

Looks like I don’t have to learn latex just yet! The metric is in this PDF. Search it for “rain-frame”. I confirmed that this metric is the same as T&W's in Exploring Black Holes, with the exception that with T&W's motion is limited to an equatorial plane. The rain frame is for a raindrop (or other inertial observer) that falls freely from rest at infinity (correction from above). The raindrop is at rest in the frame. T&W derive this metric from the Schwarzschild metric and SR on pg. B-13. The PDF file also shows how to transform the Schwarzschild metric into the rain-frame metric, and emphasizes that "both metrics describe the same spacetime!"

 

Here's a link that has a mathematical explanation for a horizon at r=2M that does not allow the box in the OP to pass outward through it (search the link for "Speeds of light"). The explanation uses a rain frame. I confirmed that the explanation is equivalent to T&W's in Exploring Black Holes, pg. B-14. T&W note that the first equation under the heading "Speeds of light" in that link is equivalent to the rain-frame metric for light (dτ=0) moving radially (dφ=0).

 

T&W say:

 

From Exploring Black Holes, pg. B-13:

 

Why can't we escape from a black hole? Why is travel inside the horizon inevitably a one-way trip to the center? This section treats these questions from the point of view of one special diver: a "raindrop," namely, a plunger that falls freely from rest at a great distance. The basic conclusion, however, is true for all travelers inside the horizon: Inside the horizon you can move only toward the center!

[andrewgray]

Zanket,

 

Here is the rain-tree metric:

 

 

From this it is obvious that this metric has flat 3-space, but not flat spacetime (4-space). Obviously, this would be impossible where spacetime is curved. So this metric is special, but obviously not an inertial coordinate system. When dt=0, the space metric is flat. But for constant θ,φ and constant r=2GM, we have dθ=dφ=dr=0 and:

 

[math]-d \tau^2 = 0[/math]

 

and we see the coordinates are moving at lightspeed (not inertial).

 

 

Zanket, don't forget that I showed that lightcones go vertical in the hyperbolic coordinate system at R=0 in flat spacetime. Vertcal lightcones have little significance in accelerated coordinates! Hyperbolic coordinants represent flat spacetime. There is nothing there! No restrictions on movement and no restrictions on direction. Obviously, one can go anywhere in flat spacetime. And lightcones are going vertical in the accelerated coordinates that represent part of flat spacetime!

 

Then what do you think happens to a body whose surface falls below r=2M? Do you think its surface falls all the way to r=0? Or do you think it continuously falls and rises unless it rises above r=2M again? Those seem to be the only alternatives.

 

If you are measuring "falls and rises" with the r coordinate, then yes, the only two choices are that it collapses to a point (not likely in my opinion, which I will explain if desired), or it can fall and rise in "oscillation" (non-escape) or "explosion" (escape) if the collapse is elastic. This is correct mathematical logic from the Schwarzchild equations.

 

Do you agree that rmax in your diagram in post #62 is affected by the gravitational acceleration g along the object’s path? Do you agree that a rocket on the ground whose maximum thrust is less than g at the tip of the rocket cannot lift off the ground? Do you agree that g at some r between 2M and 2M+ε is higher than any given finite force with which an object can be hurled?

 

If by "cannot", you mean that in the rocket's own proper time, it cannot go somewhere, then I do not agree. If by "cannot" you mean that a Schwarzchild observer will never see it happen, then yes, I agree. The bottom line is this: the accelerated coordinates cannot describe the rocket leaving r=2M because its time coordinate goes to minus infinity. However, near r=2M, the rocket has no restriction on where it can go with finite energy. It is just that the accelerated Schwarzchild coordinates cannot describe this motion just like the Rindler hyperbolic coordinates fail to describe the crossing of R=0 in flat spacetime. It is just that the accelerated coordinates are deficient.

 

Yes, and that makes my case. When light cones become vertical relative to the accelerated Schwarzschild coordinates at r=2M, it means that nothing—not even light—can pass outward through r=2M to reach an accelerated observer hovering just above r=2M. Solid evidence from GR to the contrary—like possibly your proof that the Schwarzschild horizon is exactly equivalent to a Rindler horizon—means that GR self-inconsistent.

 

No, Zanket. lightcones going vertical have little significance in accelerated coordinates, since I have shown that lightcones go vertical in the hyperbolic flat spacetime coordinates. There is nothing there!

 

Light cones do not become vertical in any local inertial frame.

This is contradicted by your previous statement. The escaping particle in the OP can never receive a signal emitted from r<=2M, or else the signal could also reach an accelerated observer hovering above r=2M, in which case light cones do not become vertical relative to the accelerated Schwarzschild coordinates at r=2M.

 

This is not a contradiction. If the escaping particle is freefall escaping, then it can receive a light signal from r<2M. The accelerated observer, however, can never "see" this happen as his coordinates are deficient in time.

 

Zanket, this same scenario can be set up in flat spacetime near R=0, and there is nothing at R=0!

 

This section treats these questions from the point of view of one special diver: a "raindrop," namely, a plunger that falls freely from rest at a great distance.

 

Zanket, if Taylor & Wheeler were using these "rain-tree" coordinates as their "rain-drop" coordinates, then obviously they were incorrect. These coordinates are obviously not like a "freefalling raindrop" since no object can reach the speed of light. No inertial worldline can ever have dτ=0. If some constant 3-space point reaches the speed of light in a coordinate system, then these coordinates do not represent an inertial system at that point.

 

 

Here is a link . . .

 

Zanket,

 

There are several mistakes in this link. Here is one of them:

 

 

If this author thinks that the Schwarzchild velocity of a freefall raindrop goes to zero as it approaches the horizon, then he needs a little help. I already discussed this here:

 

SciForums.com - View Single Post - How to do basic GR maths in a Schwarzschild geometry?

 

The Schwarzchild velocity actually goes to c as the raindrop approaches the horizon. This author needs help. Just like the tangential velocity in cylindrical coordinates is

 

[math]V_\phi = r \frac{d\phi}{dt} [/math] and not just [math]\frac{d\phi}{dt}[/math]

 

The Schwarzchild velocity of a freefall drop is

 

[math]V_r = \frac{1}{(1-2M/r)}\frac{dr}{dt}[/math] and not just [math]\frac{dr}{dt}[/math]

 

Andrew A. Gray

[Guest Zanket]

Zanket, don't forget that I showed that lightcones go vertical in the hyperbolic coordinate system at R=0 in flat spacetime. Vertcal lightcones have little significance in accelerated coordinates! Hyperbolic coordinants represent flat spacetime. There is nothing there! No restrictions on movement and no restrictions on direction.

You are doubting the math of probably hundreds of physicists and thousands of students. That’s okay, I may not be able to refute you, but it should at least make you suspicious of your findings. What I can refute (and have refuted) is your claim that a material object can pass outward through r=2M to go above r=2M+ε, which makes my case. More on that below.

 

If you are measuring "falls and rises" with the r coordinate, then yes, the only two choices are that it collapses to a point (not likely in my opinion, which I will explain if desired), or it can fall and rise in "oscillation" (non-escape) or "explosion" (escape) if the collapse is elastic. This is correct mathematical logic from the Schwarzchild equations.

Sounds a little absurd, doesn’t it? The surface could never again reach r=2M+ε, if only because the gravitational acceleration between r=2M and r=2M+ε is arbitrarily high, which would slow the surface’s r-coordinate velocity to zero before it reaches r=2M+ε. More on that below.

 

If by "cannot", you mean that in the rocket's own proper time, it cannot go somewhere, then I do not agree.

That’s incorrect. A rocket with maximum thrust less than 1g cannot leave the ground (increase its radial position) when the gravitational acceleration at the tip of the rocket is 1g. Are you really asserting that a 0.5g force is stronger than a 1g force? By your thinking it would be effortless to lift a dumbbell.

 

However, near r=2M, the rocket has no restriction on where it can go with finite energy.

I disagree. A rocket with a maximum thrust less than the gravitational acceleration on it cannot increase its radial position. This is Physics 101! Still, I’ll give references:

 

From How a Rocket Flies:

 

Liftoff: At liftoff, the thrust of the rocket engine is greater than the weight of the rocket, lifting the rocket into the air.

 

Where the weight of the rocket is defined as ...

 

From Rocket Weight:

 

Weight is the force generated by the gravitational attraction on the rocket.

 

Then the thrust of the rocket must be greater than the gravitational acceleration on it, or else it will not increase its radial position. Gravitational acceleration slows the ascent of objects in free fall too, or else a ball thrown up would never fall down. The arbitrarily high gravitational acceleration between r=2M and r=2M+ε would slow the r-coordinate velocity of any material object to zero before it reaches r=2M+ε.

 

This is not a contradiction. If the escaping particle is freefall escaping, then it can receive a light signal from r<2M. The accelerated observer, however, can never "see" this happen as his coordinates are deficient in time.

This defies logic. This accelerated observer can be directly in-between the emission point and the escaping particle. In that case tell me how the signal can get from the emission point to the escaping particle without the accelerated observer being able to receive it too; this observer is right on the path of the signal! I’m not asking for a mathematical answer. I’m asking, how can the accelerated observer be bathed in photons from the signal as the signal travels to the escaping particle, without the accelerated observer being able to receive the signal? If you think the accelerated observer will not be bathed in photons from the signal as the signal travels to the escaping particle, then tell me how the photons will leapfrog the accelerated observer, who is right on the path of the signal. Your answer cannot have anything to do with time, because we’re assuming that the accelerated observer hovers indefinitely, including at the moment the signal passes the observer, which it must if it will ever reach a higher r-coordinate. If the signal never reaches the accelerated observer on her clock, then the signal never reaches that r-coordinate in anyone’s frame.

 

Zanket, this same scenario can be set up in flat spacetime near R=0, and there is nothing at R=0!

Such comments have no effect on me! Your comment would make sense if GR was self-consistent. But GR can say whatever it wants. It can contradict you. It does contradict you. Your proof of a Rindler horizon at r=2M for an accelerated observer at r=2M+ε shows this contradiction. The meaning of a Rindler horizon is that nothing—not even light—can get from the horizon to the accelerated observer (I showed this with a chart above). GR is not so weird that a photon that cannot reach r=2M+ε from r<=2M can somehow reach a particle above r=2M+ε. The photon does not somehow leapfrog r=2M+ε; it has to pass through it to get above it. What reference can you give to show that a photon emitted below a Rindler horizon can somehow bypass the accelerated observer to get above that observer? I've given a reference that shows that a photon emitted from r=2M would asymptote to r=2M+ε; it would forever be below r=2M+ε and so it could never reach the escaping particle above r=2M+ε.

 

Zanket, if Taylor & Wheeler were using these "rain-tree" coordinates as their "rain-drop" coordinates, then obviously they were incorrect. These coordinates are obviously not like a "freefalling raindrop" since no object can reach the speed of light. No inertial worldline can ever have dτ=0.

The link says, “For light, dτ=0”. It’s light that’s moving at the speed of light. There’s no problem there. The proof shows that a photon emitted radially upward by the raindrop crossing r=2M stays at r=2M. Since light has the highest speed (and nothing can skip an r-coordinate to reach one beyond it, as you have suggested), the proof shows that nothing can pass outward through r=2M. The proof in the link is equivalent to T&W’s.

 

If this author thinks that the Schwarzchild velocity of a freefall raindrop goes to zero as it approaches the horizon, then he needs a little help.

The link says “as measured by the bookkeeper”. The bookkeeper is an observer fixed at an r-coordinate a great distance from the body. There’s no problem there. The link says, “However, the bookkeeper does not physically measure the speed directly. He translates data relayed by the shell observer into Schwarzschild values and compute the speed. The result is only an accounting entry.” T&W agree.

 

It seems that the crux of our argument has come down to these points:

 

• You think that T&W and many others made a mathematical mistake in their proofs that show that nothing can pass outward through r=2M.

 

• You think that the gravitational acceleration on a material object—regardless of its degree—cannot prevent the object from increasing its radial position. Like, you think that a rocket with a maximum thrust less than the gravitational acceleration on it can rise to a higher r-coordinate.

 

• You think that a photon can travel from r<=2M to any higher r-coordinate without reaching an observer hovering directly between those r-coordinates.

 

• You disagree with the published prediction for a Rindler horizon. Although I gave a link that shows—based on your proof of a Rindler horizon at r=2M for an observer hovering at r=2M+ε—that a photon emitted radially upward from r=2M would asymptote to r=2M+ε (effective staying at r=2M since ε is zero in the limit), you disagree with that and think that the photon can reach any higher r-coordinate (while somehow never reaching the hovering observer).

I’m not contesting the first one, but I’d like proof or at least some good references on the bottom three points, especially since I've given references to the contrary. I don’t think you’ve given a sufficient explanation for any of those points.

[Qfwfq]

If there’s a refutation or problem stated in there, please elaborate because I’m not seeing it.

You are the one claiming to refute GR and I still don't see the point of your refutation, so perhaps you would have to elaborate it.

 

First of all I disagree with Andrew that the box straddling the horizon could get back out. The main trouble is, first, the meaning of saying that the box straddles the horizon. But suppose we forget about this and just take it as it is. Certainly the parts of the box at [imath]r<r_S[/imath] can't reach the outside. There are two possibilities: A) the box doesn't break, so the parts initially at [imath]r>r_S[/imath] are dragged in as well, or B) these parts can be pulled away at the cost of being torn off the rest. Where is the contradiction?

[CraigD]

A rocket with a maximum thrust less than the gravitational acceleration on it cannot increase its radial position. This is Physics 101!

Although not directly relevant to a discussion of General Relativity, requiring classical mechanics only, this is a statement of a misconception common and serious enough to, IMHO, warrant correction.

 

It is true only for a rocket with no nonzero component of its velocity tangent to the line between it and its primary body (horizontal) – that is, a rocket moving only straight up or down.

 

A rocket in orbit can decrease or increase its radial position (altitude), and make its orbit more or less circular, with an arbitrarily low thrust. For example, a 1 kg rocket in a circular orbit of 2 Earth radii (1.276e7m) experiences about 2.448 Newtons of gravitational attraction force, and 2.448 m/s/s of acceleration. It can transfer to a circular orbit 1000 meters greater by having its horizontal velocity increased by about 0.2892 m/s, then decreased about 2716 seconds later by about the same amount at the high point (apoapsis) of its transfer orbit (a Hohmann transfer orbit). If can be given this change in velocity by a force of 0.2892 N applied for 1 s, 0.02892 N for 10 s, etc. In short, the change in altitude can be made arbitrarily large, and the force arbitrarily small, by increasing the time required for the total transfer maneuver.

 

Intuitively, the small thrusts necessary to increase the altitude of spacecraft not resisted by a thick atmosphere is evidenced by the small size of the typical orbital maneuvering thruster rocket motor, compared to the large motors necessary to lift it vertically from a launch pad on the ground.

[Guest Zanket]

First of all I disagree with Andrew that the box straddling the horizon could get back out.

He’s trying to show that GR’s prediction that the box cannot get back out is a result of a mistake in a widely accepted mathematical proof that is based on GR’s equations. He thinks that GR is valid.

 

The main trouble is, first, the meaning of saying that the box straddles the horizon.

I don’t see any trouble there. GR predicts than an object can fall through the horizon. Then it must be able to straddle it at least temporarily, as the object itself reckons.

 

Certainly the parts of the box at [imath]r<r_S[/imath] can't reach the outside.

I agree that GR predicts that.

 

There are two possibilities: A) the box doesn't break, so the parts initially at [imath]r>r_S[/imath] are dragged in as well, or B) these parts can be pulled away at the cost of being torn off the rest. Where is the contradiction?

Look at the proof in the OP. If A or B are the only possibilities, then SR does not apply in X like GR says it does. (SR predicts that the box can be at rest relative to the escaping particle, without breaking. In SR, any two objects can be at rest relative to each other, and the tidal force in an inertial frame is negligible so no object in the frame need break.) Then GR contradicts itself. That’s a self-inconsistency, a fatal flaw for a theory.

[Guest Zanket]

It is true only for a rocket with no nonzero component of its velocity tangent to the line between it and its primary body (horizontal) – that is, a rocket moving only straight up or down.

 

A rocket in orbit can decrease or increase its radial position (altitude), and make its orbit more or less circular, with an arbitrarily low thrust.

I agree, and those are good points. Thanks for bringing that up and explaining that.

 

Can a rocket increase its radial position when the gravitational acceleration on it is arbitrarily high?

 

GR predicts that gravitational acceleration approaches infinity as the r-coordinate decreases to approach a limit of 2M. I’m claiming that this means that no material object can pass outward through r=2M, whether it’s a thrusting rocket or an object in free fall. Can you refute that?

[andrewgray]

So Zanket,

 

Let's once again get back to basics. In this picture:

 

 

You agreed that an object could "freefall escape" from r=2M+ε with an initial velocity V_o and return again with the same velocity -V_o that it had when it started its freefall escape. And you agreed that this was still true for all ε as ε→0. So you are thus agreeing that an object can escape from r=2M+ε while feeling no acceleration. So are you suggesting that an object can escape from, say r=2M+.0000000000000001ε feeling no acceleration, then magically, a femtometer farther in, the "escape acceleration" is infinite?

 

And, are you suggesting that just 1 femtometer farther in, the trajectory magically becomes no longer time reversal symmetric?

 

That is, are you suggesting that there is some sort of discontinuity at r=2M?

 

Andrew A. Gray

 

P.S. We do agree on this: Either thousands of researchers are wrong or GR is inconsistent.

[CraigD]

Can a rocket increase its radial position when the gravitational acceleration on it is arbitrarily high?

In the formalism of classical mechanical, yes – the required [math]\Delta v[/math] for orbital transfers increases as [math]\sqrt{M}[/math] where [math]M[/math] is the mass of the primary.

 

In the formalism of Relativity, gravitational time dilation is given by [math]\frac{t'}{t} = \sqrt{1 - \frac{r_0}{r}}[/math], where [math]t'[/math] is a time duration measured by an observer at distance [math]r[/math] from the primary’s center of mass, [math]t[/math] as measured by a distant observer, and [math]r_0[/math] the Schwarzschild radius of the primary. This is further complicated by an orbiting body’s velocity time dilation – I’ve never done or followed the derivation, but according to the linked wikipedia article above, this limits possible circular orbits to [math]r \ge \frac32 r_0[/math].

 

For lesser radii, relativistic time dilation given by these formulae is complex-valued and physically nonsensical. My interpretation of what an observer on a body orbiting a primary in an elliptical orbit with low-point (periapsis) [math]r \dot< \frac32 r_0[/math] experience is that he perceives his orbit as normal, but upon reaching an altitude [math]r \dot> \frac32 r_0[/math], discovers that an infinite (to be mathematically precise, an indeterminate and imaginary) amount of time has passed in the outside universe. The cosmological consequences of this are, I think, beyond the scope of this post.

I’m claiming that this means that no material object can pass outward through r=2M, whether it’s a thrusting rocket or an object in free fall. Can you refute that?

As noted above, in a formal sense, yes. In a practical sense, considering but not rigorously using the applicable formalisms, no. Practically, I’m comfortable terming “impossible” an event taking an infinite amount of time as perceived by a distant observer.

 

In terms of what I suspect but can’t prove, I think Relativity fails to correctly describe scenarios such as the above, being superceded by an as yet unknown theory of “quantum gravity”, or some as yet unnamed theory.