General relativity is self-inconsistent

[Guest Zanket]

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.

Yes, I agree to all.

 

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?

Yes to all. All this inelegance is implied by GR’s prediction of black holes and its equation for gravitational acceleration.

 

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

Yep. If their mathematical proof of black holes is correct, then GR is inconsistent. I say the self-inconsistency of GR has been in plain sight this whole time!

 

You keep noting to the effect that “this is flat spacetime!” Here’s something you may not have realized. Put a relativistic rocket in SR’s idealized, gravity-free universe. Somewhere below the rocket is a Rindler horizon. A photon emitted at or below the Rindler horizon will never reach the rocket, assuming the rocket thrusts forever. But, as you keep pointing out, no inertial observer’s light cone is vertical in flat spacetime. So how can a light signal emitted below the horizon ever reach an object in free fall above the rocket, as it seems it must be able to if no inertial observer’s light cone is vertical? The answer is, the object must fall below the rocket before it can receive the signal. And that always happens eventually, since in SR a relativistic rocket eventually passes every object in free fall in front of it.

 

But GR predicts that an escaping particle passing a rocket hovering at r=2M+.0000000000000001ε never falls below the rocket. So the particle can never receive a signal emitted from a point even one femtometer farther in initially, where the rocket’s Rindler horizon is. That contradicts SR, making GR self-inconsistent.

 

What causes GR’s problem? It’s not a problem with the particle’s ability to escape. It’s a problem with GR’s prediction of black holes, namely their horizons. GR did not have to predict those. They are not a requirement in a theory of gravity. Einstein didn’t have any observational evidence as to how nature works at r=2M. He must have guessed, and he guessed something that contradicts the equivalence principle. (In fact we know that he used trial and error to develop GR. He didn’t know that his field equations predicted black holes when he published GR; Schwarzschild showed that later.) Fortunately a small change to the Schwarzschild metric removes this problem and allows self-consistency, while retaining full experimental confirmation.

[Erasmus00]

Zanket, might I suggest your problem is one of the fact that the schwarzchild metric has a COORDINATE singularity at R=2m, not a legitimate singularity.

 

Your argument boils down to this, as far as I can see- GR should reduce, locally to flat space-time. HOWEVER, at the event horizon, it does not reduce to flat space time. Hence, logical contradiction.

 

However, Andrew has shown that, locally, R=2M works exactly like a Rindler Horizon, which IS a flat-spacetime metric. Your problem seems to be that you expect all coordinate systems to reduce the the Minkowsi metric, which is not necessary, they need only be flat.

-Will

[Guest Zanket]

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.

You can completely ignore time dilation, gravitational or otherwise. What a distant observer measures is not important to the question. I’m just asking, can the crew of the rocket experience an increase in their altitude when the gravitational acceleration on their rocket is arbitrarily high? I don’t see how that could be possible.

 

You say that an arbitrarily low thrust is needed for a rocket in orbit to gain altitude, and I agree, but I think a larger gravitational acceleration results in a smaller gain in altitude for a given thrust, right? Then when the gravitational acceleration is arbitrarily high, any given thrust would result in an arbitrarily small gain in altitude, which is effectively no gain.

 

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.

Keep in mind that quantum gravity or whatever is proposed mainly for the region around the central singularity at r=0, not at r=2M. It’s only near the central singularity that GR breaks down. For a body with a sufficiently large mass, a billion years could elapse on the clock of an observer after he falls through r=2M and before he reaches the singularity.

[Guest Zanket]

However, Andrew has shown that, locally, R=2M works exactly like a Rindler Horizon, which IS a flat-spacetime metric. Your problem seems to be that you expect all coordinate systems to reduce the the Minkowsi metric, which is not necessary, they need only be flat.

I don’t know where you got that idea from. Andrew brought up a Rindler horizon just to make a point (he’s been trying to show that GR does not predict black holes, so that GR can be self-consistent). The OP doesn’t mention a Rindler horizon; its case is made without that.

 

The proof in the OP employs an inertial frame X that is falling through a horizon of a black hole. All inertial frames are equivalent, says the relativity principle as stated by Thorne in the OP. The OP shows that SR does not work as expected in X; it does not work the same way in X as it does in an inertial frame that is wholly above a horizon. Your comments don’t resolve that problem.

[CraigD]

I’m just asking, can the crew of the rocket experience an increase in their altitude when the gravitational acceleration on their rocket is arbitrarily high?

Yes

I don’t see how that could be possible.

Consider the case of a primary body consisting of a supermassive black hole of about 1 billion solar masses ([math]2 \times 10^{37} \,\mbox{kg}[/math]).

 

It’s Schwartzchild radius is [math]r_s = \frac{2GM}{c^2} \dot= 2.969 \times 10^{10} \,\mbox{m}[/math] (a bit smaller than orbital radius of Mercury around the Sun.

 

A transfer orbit between a circular orbit of radius [math]r_1 = r_s - 1000 \,\mbox{m}[/math] and one [math]r_2 = r_s + 1000 \,\mbox{m}[/math] requires an initial [math]\Delta V_1 = \sqrt{\frac{MG}{r_1}} \left ( \sqrt{\frac{2 r_2}{r_1 + r_2}} -1 \right ) \dot= 3.571 \, \mbox{m/s}[/math], then, [math]t_H = \pi \sqrt{\frac{(r1+r2)^3}{8u}} \dot= 440 \, \mbox{s}[/math] later, one of [math]\Delta V_2 = \sqrt{\frac{MG}{r_2}} \left ( 1 - \sqrt{\frac{2 r_1}{r_1 + r_2}} \right ) \dot= 3.571 \, \mbox{m/s}[/math]. A 0.1 g accelerating of less than 4 seconds duration at periapsis and apoapsis would be adequate to accomplish this maneuver.

 

Rereading this thread, especially your last post, I get the impression, Zanket, than you are assuming the gravitational force at the event horizon of a black hole ([math]r_s[/math]) is infinite. This is not a prediction of any theory of which I’m aware. In the above example, the acceleration due to gravity at [math]r_s[/math] is about [math]1.514 \times 10^6 \,\mbox{m/s/s}[/math] – about 100,000 gs. The gravitational tidal forces on an object at [math]r_s[/math] is slight, about .01 m/s/s on a 100 m long object, the equivalent of it hanging in a mere .001 g gravitational field.

 

As I hinted at in post #85, the “you can’t cross it” nature of a black hole horizon predicted by General Relativity can be interpreted more as having to do with your passage of time relative to the rest of the universe than a mechanical restriction.

 

The vicinities of Black holes are typically Bad Places To Visit, with accretion disks of hot, high-energy matter eager to convert other matter into high energy photons, plasma jets, and other phenomena of practical discomfort to the crew of a rocket ship, but, were they to find a “clean” black hole of sufficient mass that the tidal forces in the vicinity of its event horizon were not too great, an event horizon-penetrating orbit is theoretically possible – assuming that the “infinite time passing in the outside universe” issue doesn’t entail some dire consequence, such as 12100.

[Qfwfq]

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.

I had gathered more or less this but I disagree about something getting from [imath]r<r_S[/imath] to [imath]r>r_S[/imath]. As you go the opposite way, the direction of decreasing r becomes time, this is why it doesn't even make sense to imagine the box having one side in and one side out.

 

SR predicts that the box can be at rest relative to the escaping particle, without breaking.

I'm unaware of SR saying something specifically about the Schwarzschild metric near the event horizon, so in which sense do you mean that? :confused:

 

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.

The tidal force in an inertial frame is exactly zero, but why should it be negligible in a locally inertial chart? Or in any chart? At a given point of space-time, it depends on the intrinsic curvature, it should be independent of coordinate choice among those for which the coordinate differentials match up with proper length and time.

[Guest Zanket]

Rereading this thread, especially your last post, I get the impression, Zanket, than you are assuming the gravitational force at the event horizon of a black hole ([math]r_s[/math]) is infinite.

Yes that is what I think.

 

This is not a prediction of any theory of which I’m aware. In the above example, the acceleration due to gravity at [math]r_s[/math] is about [math]1.514 \times 10^6 \,\mbox{m/s/s}[/math] – about 100,000 gs.

What equation are you using? I’m using the reciprocal of the third equation on the left here. That is GR's equation for gravitational acceleration.

[Guest Zanket]

As you go the opposite way, the direction of decreasing r becomes time, this is why it doesn't even make sense to imagine the box having one side in and one side out.

You disagree with Taylor and Wheeler here:

 

From pg. 2-4 of Exploring Black Holes by Taylor and Wheeler (boldface mine):

 

Our old, comfy, free-float (inertial) frame carries us unharmed to the center of a black hole. Well, unharmed almost to the center! ... No one can stop us from observing a black hole from an unpowered spaceship that drifts freely toward the black hole from a great distance, then plunges more and more rapidly toward the center. Over a short time the spaceship constitutes a "capsule of flat spacetime" hurtling through curved spacetime. It is a free-float frame like any other. Special relativity makes extensive use of such frames, and special relativity continues to describe Nature correctly for an astronaut in a local free-float frame, even as she falls through curved spacetime, through the horizon, and into a black hole.

 

Do you think you agree with them? You also disagree with Einstein in his statement of the equivalence principle in the OP. You’re implying that an inertial frame cannot fall through a horizon.

 

I'm unaware of SR saying something specifically about the Schwarzschild metric near the event horizon, so in which sense do you mean that?

It doesn’t have to say anything about that. GR says something about SR at the horizon. GR says that SR applies locally (in a sufficiently small region) anywhere in empty spacetime (search the OP for “anywhere”). That includes a region straddling a horizon.

 

The tidal force in an inertial frame is exactly zero, but why should it be negligible in a locally inertial chart?

The tidal force in any nonzero-sized frame is nonzero. An inertial frame need not be zero-sized according to Einstein, Taylor, Thorne, and Wheeler in the OP. See the definitions of an inertial frame by Taylor, Thorne, and Wheeler in the OP. See the equivalence principle as stated by Einstein in the OP.

[CraigD]

This is not a prediction of any theory of which I’m aware. In the above example, the acceleration due to gravity at [math]r_s[/math] is about [math]1.514 \times 10^6 \,\mbox{m/s/s}[/math] – about 100,000 gs.

What equation are you using? I’m using the reciprocal of the third equation on the left here. That is GR's equation for gravitational acceleration.

I’m using [math]a = \frac{MG}{r^2}[/math], Newton’s old formula. Based on the equivalence principle, I believe it’s approximately correct when considering only the immediate vicinity of the inertial frame of the ship. It would not agree with the observations of a distant observer.

 

The equation you’re using,[math]a = \frac{M}{r^2 \sqrt{1- \frac{2M}{r}}}[/math], appears to me to combine Newton’s formula with GR’s formula for gravitational time dilation, [math]\frac{t}{t'} = \sqrt{1- \frac{2GM}{r c^2}}[/math] (with mass units chosen to eliminate the [math]\frac{G}{c^2}[/math] constant term). I think this is done to give the acceleration as measured by a distant observer. I’m confused (as it appears he wrote that equation, perhaps andrewgray can clear up my confusion), however, as it appears to me that the equation for acceleration as seen by a distant observer should be [math]a = \frac{M \sqrt{1- \frac{2M}{r}}}{r^2}[/math], that is, that the “slow-ticking” distant observer and “fast-ticking” one near the event horizon have been erroneously exchanged. :confused:

[Guest Zanket]

The equation you’re using,[math]a = \frac{M}{r^2 \sqrt{1- \frac{2M}{r}}}[/math], appears to me to combine Newton’s formula with GR’s formula for gravitational time dilation, [math]\frac{t}{t'} = \sqrt{1- \frac{2GM}{r c^2}}[/math] (with mass units chosen to eliminate the [math]\frac{G}{c^2}[/math] constant term). I think this is done to give the acceleration as measured by a distant observer.

The first equation (the one Andrew and I are using) is GR’s equation for the gravitational acceleration an observer who is at the given r-coordinate experiences. The equation has nothing to do with a distant observer. Taylor and Wheeler also show this equation in their book Exploring Black Holes. The equation is for geometric units (c=G=1).

 

So you can see that GR does predict that gravitational acceleration approaches infinity as r decreases to a limit of 2M.

[andrewgray]

Zanket,

 

Now we are getting somewhere. All I need to do is show that there is no discontinuity at r=2M, and I believe that we are done.

 

But first we must clear up some pure GR concepts. In GR, there is no gravitational acceleration. Observers who free fall in curved spacetime feel no acceleration. This is the equivalence principle. So take a minute and think about this. No gravitational acceleration in GR.

 

Now, consider the astronaut at constant r somewhere above the horizon. There is no gravitational acceleration. The acceleration that he feels is from the rocket engines pressing the floor against his feet. It is just the opposite of how you are thinking. You are thinking that the inertial frames are accelerating past the rocket, "being pulled in by gravity". This is incorrect according to GR. There is no gravitational acceleration in GR. Only the curvature tensor. Acceleration is only properly defined relative to inertial frames. If you want to talk about an acceleration, you must define which inertial frame it is relative to.

 

So in General Relativity, we have the curvature tensor. With this quantity, we can tell all about inertial frames at a certain point. From these inertial frames, acceleration is defined.

 

A quick calculation of the Riemann curvature tensor (see Gravitation p360-362, 821) yields:

 

 

 

The payoff of this calculation: according to these equations, none of the components of Riemann in the explorer's orthonormal frame become infinite at the gravitational radius. The tidal forces the traveler feels as he approaches r=2M are finite; they do not tear him apart-at least not when the mass M is sufficiently great, because at r=2M the typical non-zero component [imath]R_{\hat \alpha \hat \beta \hat \gamma \hat \delta} [/imath] of the curvature tensor is of the order 1/M². The gravitational radius is a perfectly well behaved, nonsingular region of spacetime, and nothing there can prevent the explorer from falling on inward.

 

So you see, we have shown that, according to GR, forces at r=2M are finite, and that spacetime is continuous since the Riemann curvature tensor is continuous. Thus according to GR, there are no discontinuities at r=2M. Finally, this means that we can extend our r=2M+ε concepts all the way to and through r=2M due to mathematical theorems on continuity. This means that trajectories remain time reversal symmetric up out of the horizon and back inward.

 

So where is the mistake?

 

It is here:

 

 

 

Here, MTW claim that since the coefficients of dt² and dr² change sign at the horizon, then r becomes timelike and t becomes spacelike. MTW say, "the further decrease in r represents the passage of time." Then, according to them, since time is unidirectional, r can only decrease.

 

But we have already shown a counterexample to these concepts here:

 

 

 

 

 

 

Here we have shown that the coefficient of dt² changes sign and that it is φ that seems to have to decrease. But it is false to claim that t is not timelike since this is flat spacetime. It is just that we are using accelerated coordinates. Timelike vs. spacelike in accelerated coordinates is meaningless, as proven by this counter-example. You must be talking about an inertial frame before you can identify the timelike and spacelike variables!

 

Finally, we have proven that Misner, Thorne, and Wheeler (as well as thousands of conformist followers) have made a great mistake and that in the neighborhood of a gravitational horizon the equivalence principle and GR actually predict that a "spanning" rectangle can leave without a problem. Again, thank you Zanket for pointing this out and increasing my understanding.

 

 

Andrew A. Gray

 

 

P.S. There are conformists throughout science. Do not be so surprised as the sheep fall in line.

[Qfwfq]

Zanket you are attributing things to me which I have not said, which just shows me your misunderstandings all the more. The words of mine that you quote do not contradict what you quote from Taylor and Wheeler. I did not say something can't fall through the event horizon, beware of paralogism.

 

It doesn’t have to say anything about that. GR says something about SR at the horizon. GR says that SR applies locally (in a sufficiently small region) anywhere in empty spacetime (search the OP for “anywhere”). That includes a region straddling a horizon.

Now if you want a reasonable discussion, choose your wording properly and pay attention to mine, because we are getting totally lost here.

 

Now, your reply about the equivalence principle does not address my point about what sense it makes to say that one side of the box can be outside the horizon "at the same time" as the other side is inside. It's a subtle matter of reasoning on world lines, but even if we axiomatically suppose a similar thing about a case which does make sense it is trivial to resolve the apparent contradiction in your OP. Suppose that no object, or part of, can cross the border from Liechtenstein to Switzerland in that sense (the Prince's guards grab it with an unopposable grip, as soon as it comes in from Switzerland), and start with what I said about the two possibilities: A) The guards drag the whole thing in and the Prince will keep it all. B) You can keep at the most what is still in Switzerland, at the cost of chopping it off. Obviously there is no contradiction about the particle inside the box but still on the Swiss side. In the case of Schwarzschild radius you invoke SR, applied locally, as contradicting this, are you really convinced?

 

The tidal force in any nonzero-sized frame is nonzero. An inertial frame need not be zero-sized according to Einstein, Taylor, Thorne, and Wheeler in the OP. See the definitions of an inertial frame by Taylor, Thorne, and Wheeler in the OP. See the equivalence principle as stated by Einstein in the OP.

Again you fail to address my query about your point. Notice that I was distinguishing between inertial coordinates and locally inertial ones. And don't worry about teaching me the basics of GR, I've had very good teaching of it, although it was more than 15 years ago and I've had to shake a few cobwebs out of some subtleties. Concerning the equivalence principle as stated by Einstein, I have read it in Die Grundlagen der allgemeiner Relativitätstheorie.

 

I’m using [math]a = \frac{MG}{r^2}[/math], Newton’s old formula. Based on the equivalence principle, I believe it’s approximately correct when considering only the immediate vicinity of the inertial frame of the ship.

:umno: If this were so even down at the Schwarzschild radius, the orbital mechanics could not be qualitatively different at radii up to about 3 times [imath]r_S[/imath], GR could not even explain Mercury's perihelion shift. In any case one must specify according to which coordinates.

 

But first we must clear up some pure GR concepts. In GR, there is no gravitational acceleration. Observers who free fall in curved spacetime feel no acceleration. This is the equivalence principle. So take a minute and think about this. No gravitational acceleration in GR.

I think part of your problem in getting across to Zanket is that you keep repeating a fundamental fact, the very equivalence principle, but try to infer something not consequential to it.

 

It is tautologigal that there is no gravitational acceleration according to inertial coordinates. For other coordinates there is and, of course, according to coordinates that are only locally inertial the acceleration is zero at the specified point but not in a whole neighborhood of it.

[Guest Zanket]

All I need to do is show that there is no discontinuity at r=2M, and I believe that we are done.

You need to do more than that. Look again at the list at the bottom of my post 79. Like I said there, I’m not contesting the first item on the list, your claim that T&W & others made a mathematical mistake. But your thinking that “we are done” is refuted by the last three items on the list. You would need to resolve those before you could be right. I believe those items are irrefutable. For example, for you to be right, a photon moving upward would have to be able to skip some r-coordinates. Like, on its way from below r=2M to r=2M+2ε, it would skip r=2M+ε so that an observer hovering there could not receive it. But no experimental test shows that light skips some r-coordinates on its way upward.

 

But first we must clear up some pure GR concepts. In GR, there is no gravitational acceleration. Observers who free fall in curved spacetime feel no acceleration. This is the equivalence principle. So take a minute and think about this. No gravitational acceleration in GR.

 

Now, consider the astronaut at constant r somewhere above the horizon. There is no gravitational acceleration. The acceleration that he feels is from the rocket engines pressing the floor against his feet. It is just the opposite of how you are thinking.

It’s not the opposite of how I am thinking. In GR, there is a gravitational acceleration, and observers in free fall feel no acceleration. There is no contradiction there. The reason the astronaut fixed at some r (and fixed along some radius) does not increase her radial position despite feeling an acceleration is because the acceleration she feels precisely counteracts the gravitational acceleration at that r (indeed she’d have to have adjusted her acceleration to precisely counteract the gravitational acceleration at that r; otherwise she’d not have stayed fixed at that r). Nowhere did I suggest that an observer in free fall feels an acceleration. If there was no gravitational acceleration, then we’d be in free fall right now, if we were even alive. The Earth would just be dust floating in space then, because it would not have coalesced into a ball due to gravitational acceleration. GR’s equation for gravitational acceleration tells the astronaut fixed at some r the initial acceleration she will measure for an object dropped from rest there. The equation also tells the dropped object what initial acceleration it can expect to measure for the astronaut. Gravity is not a “pull down” force in GR. Nevertheless it coalesced free-floating particles to create the Earth, even as no particle felt an acceleration as it freely fell.

 

You are thinking that the inertial frames are accelerating past the rocket, "being pulled in by gravity". This is incorrect according to GR.

An inertial frame does not need to be pulled to accelerate past the rocket. It accelerates past the rocket because the rocket is noninertially accelerating. Inertial frames do accelerate past the rocket. To see that, just drop a ball and measure its acceleration relative to you, a noninertially accelerating observer.

 

Acceleration is only properly defined relative to inertial frames. If you want to talk about an acceleration, you must define which inertial frame it is relative to.

Agreed. The acceleration you feel now is best measured by an observer in a momentarily comoving inertial frame (MCIF), like it says at the relativistic rocket site: “The acceleration of the rocket must be measured at any given instant in a non-accelerating frame of reference travelling at the same instantaneous speed as the rocket”. The acceleration you feel now, your acceleration as measured in an MCIF for you (like your acceleration as measured by a measuring device once you drop it), is predicted by GR’s equation for gravitational acceleration.

 

So you see, we have shown that, according to GR, forces at r=2M are finite, and that spacetime is continuous since the Riemann curvature tensor is continuous. Thus according to GR, there are no discontinuities at r=2M.

That the tidal force is finite there does not mean that the gravitational acceleration is finite there. Those are different animals. You are disagreeing with the predictions of GR’s equation for gravitational acceleration. The fact that the equation stops applying at r=2M is the “sort of discontinuity” to which I agreed. You used the equation in your proof. Are you now rejecting your proof, or are you just trying to have it both ways?

 

Finally, this means that we can extend our r=2M+ε concepts all the way to and through r=2M due to mathematical theorems on continuity. This means that trajectories remain time reversal symmetric up out of the horizon and back inward.

It doesn’t mean that. If it meant that, then an object could increase its radial position despite an arbitrarily high gravitational acceleration on it. If an object could do that, there’d be no decent explanation for why we are held to the Earth now, fixed at some r-coordinate. We would be in free fall if not for the Earth pressing against us; just jump off a cliff to see that. The force with which the Earth presses against us is predicted by GR’s equation for gravitational acceleration. If the equation predicted an arbitrarily high gravitational acceleration at our r-coordinate then we’d be falling, for the Earth could not counteract such acceleration (which would be higher than any given finite value for the force at which the Earth presses against us). We’d be falling along with the Earth’s surface, moving toward r=0. But apparently you think otherwise. You think that an object’s ability to increase its radial position is independent of gravitational acceleration. You think we could rise off the Earth’s surface without regard to gravitational acceleration. What do you think is holding us to the r-coordinate of our chairs? Why doesn’t the force of the Earth’s surface pressing against us push us to a higher r-coordinate?

 

Finally, we have proven that Misner, Thorne, and Wheeler (as well as thousands of conformist followers) have made a great mistake and that in the neighborhood of a gravitational horizon the equivalence principle and GR actually predict that a "spanning" rectangle can leave without a problem.

I say this only matter-of-factly: I think yours is a classic case of not seeing the forest for the trees. I pointed out some crippling flaws in your thinking, not only here but also in the list in post 79. You’re ignoring them. I suggest you step back and think about those. (But I am always willing to be wrong if you can refute me.)

 

P.S. There are conformists throughout science. Do not be so surprised as the sheep fall in line.

I agree!

[Guest Zanket]

I did not say something can't fall through the event horizon, beware of paralogism.

You said:

 

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

To any reasonable person, this is suggesting a problem with the idea of something straddling the horizon. And that suggestion is all I addressed. Had you not been coy and elaborated on what the “main trouble” was, instead of continuing with “But suppose we forget about this and just take it as it is”, then I need not have been confused by your comment.

 

Now, your reply about the equivalence principle does not address my point about what sense it makes to say that one side of the box can be outside the horizon "at the same time" as the other side is inside.

My reply did address that. GR says that an inertial frame can fall through a horizon (see supporting info I gave you above). X is given to be such a frame. Then the box can be on both sides of the horizon simultaneously according to clocks throughout X and at rest with respect to it.

 

Suppose that no object, or part of, can cross the border from Liechtenstein to Switzerland in that sense (the Prince's guards grab it with an unopposable grip, as soon as it comes in from Switzerland), and start with what I said about the two possibilities: ...

Since it “comes in from” Switzerland, I’ll assume you meant that it crosses the border between Switzerland and Liechtenstein, from Switzerland. See, I pay attention to your wording.

 

A) The guards drag the whole thing in and the Prince will keep it all. :shrug: You can keep at the most what is still in Switzerland, at the cost of chopping it off. Obviously there is no contradiction about the particle inside the box but still on the Swiss side. In the case of Schwarzschild radius you invoke SR, applied locally, as contradicting this, are you really convinced?

GR allows an inertial frame to fall through a horizon, and X is given to be such a frame. SR applies to an inertial frame. If A occurs, then the box cannot be at rest with respect to the escaping particle, in violation of SR, which allows any two objects to be at rest relative to each other in an inertial frame. If B occurs, then the frame is not inertial, because the tidal force is the only force that exists in X that could break the box, and the tidal force is negligible in an inertial frame by the definition of an inertial frame. Then if A and B are the only possibilities, GR is self-inconsistent.

 

Again you fail to address my query about your point. Notice that I was distinguishing between inertial coordinates and locally inertial ones.

If a “locally inertial chart” is different than an inertial frame, then it’s probably overkill for this discussion. X is given to be an inertial frame. If you agree that GR allows an inertial frame to fall through a horizon, then all that matters beyond that point is what SR predicts for an inertial frame. Do you agree that GR allows an inertial frame to fall through a horizon? Do you agree that SR applies to an inertial frame?

[andrewgray]

Zanket,

 

I am at work so I cannot address all of your logic right now, but I will soon. But first we must resolve this acceleration issue.

 

An inertial frame does not need to be pulled to accelerate past the rocket. It accelerates past the rocket because the rocket is noninertially accelerating. Inertial frames do accelerate past the rocket. To see that, just drop a ball and measure its acceleration relative to you, a noninertially accelerating observer.

 

Zanket, inertial frames do not accelerate past the rocket. A frame is "accelerating" if and only if an accelerometer detects acceleration in the frame. In the freefall inertial frame momentarily at rest by the rocket, the accelerometer measures zero. Agreed?

 

Therefore, the inertial frame is not accelerating past the rocket. It is moving past the rocket with constant velocity. Only the rocket is accelerating since its accelerometer is measuring something finite. Agreed?

 

 

Andrew A. Gray

[CraigD]

The equation you’re using,[math]a = \frac{M}{r^2 \sqrt{1- \frac{2M}{r}}}[/math], appears to me to combine Newton’s formula with GR’s formula for gravitational time dilation, [math]\frac{t}{t'} = \sqrt{1- \frac{2GM}{r c^2}}[/math] (with mass units chosen to eliminate the [math]\frac{G}{c^2}[/math] constant term). I think this is done to give the acceleration as measured by a distant observer. I’m confused (as it appears he wrote that equation, perhaps andrewgray can clear up my confusion), however, as it appears to me that the equation for acceleration as seen by a distant observer should be [math]a = \frac{M \sqrt{1- \frac{2M}{r}}}{r^2}[/math], that is, that the “slow-ticking” distant observer and “fast-ticking” one near the event horizon have been erroneously exchanged. :confused:

The first equation (the one Andrew and I are using) is GR’s equation for the gravitational acceleration an observer who is at the given r-coordinate experiences. The equation has nothing to do with a distant observer. Taylor and Wheeler also show this equation in their book Exploring Black Holes.

If I’m to clear up my confusion for myself, I certainly must acquaint myself with this equation – and there seems no surer way than reading Taylor and Wheeler’s book. My academic acquaintance with GR’s field equations was all of about a weeks worth, 25 years ago, and have largely fled my awareness :(. The first 2 chapters of the book are available online – if they’re not enough, the book appears available only in print, so I could be a while accessing it.

 

In the meanwhile, if anybodies is able and willing, could they directly address this question, which summarizes my current confusion?

Per classical gravity [math]\frac{d^2r}{d^2t} = \frac{GM}{r^2}[/math], where [math]t[/math] is time as measured by an observer distance [math]r[/math] (as observed by itself or a distant observer) from a massive body mass [math]M[/math]

Per General Relativity [math]\frac{dt_0}{dt} = \sqrt{1 - \frac{2GM}{rc^2}}[/math], where [math]t_0[/math] is time as measured by a distant observer, and [math]c[/math] is the speed of light.

Would not [math] \frac{d^2r}{d^2t_0} = \frac{GM\sqrt{1 - \frac{2GM}{rc^2}}}{r^2} [/math], supporting the popular science description that, from the point-of-view of a distant observer, an object falling toward a black hole appears never to reach its event horizon?

 

The text excerpts posted in this thread seem to tantalizingly support my idea of the “impossibility” of an object passing an event horizon being due to time dilation ([math] \frac{dt_0}{dt}[/math]), not a local mechanical effect, but without an adequate understanding of the full text, I can’t reach a confident conclusion.

[Guest Zanket]

In the freefall inertial frame momentarily at rest by the rocket, the accelerometer measures zero. Agreed?

Yes, assuming that the accelerometer is the typical kind designed to be attached to a noninertially accelerating frame, like the shock stickers used on Mythbusters. An accelerometer for attaching to an inertial frame is possible. An accelerometer just measures the acceleration of something moving relative to the accelerometer. Either an inertial (gravitationally accelerating) or noninertially accelerating measuring device can do that, and actually the inertial version would be more accurate in principle. More on that below.

 

Zanket, inertial frames do not accelerate past the rocket. A frame is "accelerating" if and only if an accelerometer detects acceleration in the frame. ... Agreed?

No. How would an accelerometer attached to the rocket work? The rocket is at rest in the accelerometer's frame, so how does the accelerometer register acceleration? What is accelerating in the accelerometer’s frame? The accelerometer would have to measure the change in velocity of an object in free fall (or at least an object mostly free to not noninertially accelerate, like a ball rolling in a tube), some object within the device. The accelerometer effectively measures the acceleration of an inertial frame relative to the rocket. So inertial frames do accelerate relative to the rocket. No one at rest with respect to those inertial frames need feel an acceleration.

 

I imagine that the shock stickers used on Mythbusters have a ball bearing or something similar inside a compartment, where if the ball bearing accelerates above a certain value relative to the compartment, then the compartment hits the ball bearing with enough force to cause the compartment wall to break, releasing an indicator dye into the compartment. The ball bearing would represent an object in free fall.

 

T&W’s example for using GR’s equation for gravitational acceleration is a robot at a fixed r-coordinate measuring the initial acceleration of tool dropped by the robot. And Physics 101 describes g in terms of the acceleration of a falling object.

 

Ideally the rocket’s or robot’s acceleration is measured by an inertial observer in an MCIF, like the relativistic rocket site describes. Then there are no problems with time dilation along the measuring equipment or physical deformation of the measuring equipment, that occur in a frame that is noninertially accelerating.

 

An inertial frame can accelerate gravitationally relative to some noninertially accelerating object passing by. A relativistic rocket in SR's idealized, gravity-free universe experiences the equivalent of a uniform gravitational field, so inertial objects passing by, or dropped within the rocket, can measure "gravitational" acceleration by measuring the acceleration of the rocket.

 

Therefore, the inertial frame is not accelerating past the rocket. It is moving past the rocket with constant velocity. Only the rocket is accelerating since its accelerometer is measuring something finite. Agreed?

No. If the inertial frame were moving past the rocket with constant velocity, the rocket would be inertial; it would be in free fall. If inertial frames moved past the rocket with constant velocity, an accelerometer attached to the rocket would register zero acceleration. The accelerometer has to measure the acceleration of something, and it can’t be the rocket, because the rocket is at rest in the accelerometer's frame. What do you think is accelerating in the accelerometer’s frame, if not an object in free fall (or at least an object mostly free to not noninertially accelerate) within the device?