General relativity is self-inconsistent

[Qfwfq]

The supporting info in the OP does. The box cannot pass outward through the horizon, because nothing can do that, according to the definition of a horizon. The particle is given to be escaping to r=infinity, which GR allows. The particle’s radial position is increasing; the box’s cannot. Then they cannot be at rest relative to each other.

Call that logic.

 

GR allows the little ball to escape but not the box, ergo it does not allow the ball to be at rest with the box. You'll have to show me why it does not allow both, as alternative possibilities.

 

I just wanted to remind you that I showed that the curvature tensor for the Schwarzchild metric goes as

Thanks, I wasn't able to follow this whole thread and all of your contributions, but I've been hoping to work out stuff like that, and possibly more, when I have more spare time. I'm at work too ;) and I'm even a weary commuter. I'm wanting to do so for other reasons though; I've already said that as far as this thread goes I simply don't care, I tried to point out one of Zanket's inaccuracies and tidal forces can come into one of the two branches of the logical argument I attempted, but simply aren't necessary.

 

Not only, but it simply doesn't make sense to think of an object straddling the event horizon (even if it did and with enough rigidity to stand whatever tidal forces there may be, you should know the fundamental reason why it can't escape in one piece).

 

Ooohps forgot, there are enough inaccuracies in Zanket's wording and I don't see the point of adding this:

However, the laws of physics apply in the inertial frame, and they do not apply in the accelerated frame.

They have a different form, inertial frames being those for which they have the simplest. Before GR, at least.

 

The very point of general covariance is that even the form doesn't change. The change in form (including Coriolis) is replaced by the change in components of tensors (such as the metric, or also the non-tensor connection) when changing chart. I'm sure you should know that, and saying things inappropriately has only been aggravating the discussion, all along.

[Guest Zanket]

The acceleration required to "stay" at constant R is infinite, yet we can cross these R coordinates in both directions! And this is flat spacetime with no inconsistencies here!

Where have you resolved the fact that your finding here contradicts the relativistic rocket equations (which are derived from the book Gravitation)?

 

A photon that originates below a Rindler horizon cannot reach the accelerated observer for whom that horizon exists. The acceleration required to stay at r=2M is infinite. Then for an observer hovering at r=2M+ε, where ε is infinitely small, there’s a Rindler horizon at r=2M (because then 1/a, the distance to the Rindler horizon as measured in an MCIF of the rocket, is infinitely small). You’ve proven this. The Rindler Horizon site also agrees with this. Then nothing—not even light—can reach r=2M+ε from r=2M, and so the box in the OP cannot escape to r=infinity to be at rest with the particle. “This is flat spacetime” doesn’t resolve this issue, because a Rindler horizon works as advertised even in SR’s idealized, gravity-free universe.

 

You have not addressed this issue that refutes you. The only way you could be right is if the relativistic rocket equations are wrong. Do you agree? Are you saying those equations are wrong? Are you saying that a photon can get to r=infinity from r=2M, while somehow skipping an observer hovering at r=2M+ε? If so, how does the photon skip that observer?

 

Are you in agreement that the above x=1 geodesic looks like this across the horizon in flat spacetime hyperbolic coordinates?

I don’t think so. It seems that to agree I must also agree that the relativistic rocket equations are wrong. But I think those equations are valid.

 

If not, then I give up trying to make you understand, and what is your full name so I can give you credit if I decide to submit a paper using some of your ideas?

Just call me “Zanket”.

 

GR allows the little ball to escape but not the box, ergo it does not allow the ball to be at rest with the box. You'll have to show me why it does not allow both, as alternative possibilities.

I did not say only that GR allows the particle to escape. I said, “The particle is given to be escaping to r=infinity, which GR allows”. The particle is escaping to r=infinity. The box cannot pass outward through the horizon, according to GR. Then GR does not allow the particle to be at rest relative to the box. If that doesn’t make sense to you, perhaps you can suggest an alternative possibility and I can tell you why my logic already excludes it.

[Qfwfq]

I did not say only that GR allows the particle to escape. I said, “The particle is given to be escaping to r=infinity, which GR allows”.

Exactly. The particle is given to be escaping. This is a premise, and the only reason it cannot be at rest with the box that cannot be escaping, due to another premise. These two premises together imply that the one cannot be at rest with the other.

 

This does not contradict SR.

[DryLab]

Zankeet

I've been following this thread here and in other forums where you've posted similar threads. I can't understand your proof, but just for the sake of argument, lets play like you are correct. What then? Do you have some point you would make given that?

[Guest Zanket]

Exactly. The particle is given to be escaping. This is a premise, and the only reason it cannot be at rest with the box that cannot be escaping, due to another premise. These two premises together imply that the one cannot be at rest with the other.

 

This does not contradict SR.

It does contradict SR. The “another premise” you mention is that the box straddles the horizon. But the reason the box cannot pass outward through the horizon is because of a prediction of GR, a prediction that contradicts a prediction of SR in X. Above I said: “In SR you can put any two objects anywhere in an inertial frame and let them be at rest relative to each other”. That is not the case in X. The box cannot be put anywhere in X and let be at rest relative to the particle. To be let at rest relative to the particle the box must be wholly within a certain part of the frame, namely the part of it that is above r=2M. The clauses “anywhere” (predicted by SR) and “certain part” (predicted by GR) contradict, which means that GR contradicts SR in X, where GR says that SR applies. Then GR is self-inconsistent; it contradicts itself.

[Guest Zanket]

I can't understand your proof, but just for the sake of argument, lets play like you are correct. What then? Do you have some point you would make given that?

No.

[andrewgray]

Zanket,

 

Are you familiar with the Kruskal Coordinates for the Schwarzchild geometry?

 

Here is the transformation:

 

 

And here is what they look like:

 

 

See Kruskal-Szekeres Coordinates

Look familiar? Yes, very similar to the hyperbolic coordinates. For r<2M, the r coordinate is spacelike, so one cannot "stay" at constant r. Zanket, you say that nothing can reach r=2M+ε from r<2M.

 

This is absolutely true when referring to regions II and III of spacetime in the diagram above. Likewise for these regions in hyperbolic coordinates. But it is not true for light originating in region IV. For this geodesic, the path would look like this:

 

 

And for the hyperbolic coordinates like this:

 

 

The point is that, there are certain regions of spacetime from which light can reach r=2M+ε from r<2M. Just not from the regions that you are thinking about.

 

Andrew A. Gray

[Guest Zanket]

The point is that, there are certain regions of spacetime from which light can reach r=2M+ε from r<2M. Just not from the regions that you are thinking about.

That supports my case that GR is self-inconsistent. I need only show that GR is self-inconsistent in one way. I need not consider a naked singularity, or regions of spacetime other than those needed to make my case. I don't have to show that GR is self-inconsistent in more than one way. If GR is self-inconsistent in one way, it is self-inconsistent, period.

[andrewgray]

Yes,

 

But Zanket, if your "inconsistency arguments" apply to GR, then they apply to SR as well, since there is a 1-to-1 analogy going here. This 1-to-1 correspondence comes from the equivalence principle being built into GR mathematically. Therefore, since SR is not inconsistent according to your logic, then GR must not be inconsistent either. Zanket, if you insist that your logic is correct, then you must insist that SR is inconsistent as well. Are you saying that SR is inconsistent?

 

 

Andrew A. Gray

[Guest Zanket]

Are you saying that SR is inconsistent?

No.

 

But Zanket, if your "inconsistency arguments" apply to GR, then they apply to SR as well, since there is a 1-to-1 analogy going here. This 1-to-1 correspondence comes from the equivalence principle being built into GR mathematically. Therefore, since SR is not inconsistent according to your logic, then GR must not be inconsistent either. Zanket, if you insist that your logic is correct, then you must insist that SR is inconsistent as well.

I disagree. Nothing about SR or the equivalence principle prevents GR from contradicting SR or the principle. The proof in the OP shows that GR violates the principle. Einstein made a mistake somewhere when he incorporated the principle into GR. You’re assuming that he didn’t, but my proof refutes you. A proof that GR violates the principle overrides an assumption that GR does not violate it, which is what you’re assuming above.

 

The proof shows that GR contradicts SR where GR says that SR applies. Where does GR say that SR applies? The equivalence principle tells us. Then the proof shows that GR violates the principle. Can GR be self-consistent but just violate the principle? No, because a postulate of GR is that the principle applies (see the postulate here.)

[andrewgray]

OK,

 

One more angle to try to get you to understand.

 

There are two cases for your box with two balls in them spanning the horizon. Here is the spacetime diagram for the first:

 

 

This diagram applies for both the hyperbolic (flat spacetime) diagram, and the Kruskal (curved spacetime) diagram. In this case, the box initially spans the horizon for t<0 and T<0, (before the rocket is launched at t=0), and we see that the balls can remain equidistant in the inertial frame, and that there is no problem for the box to exit the horizon with both balls in it.

 

The next case is this:

 

 

This is the case where the box initially spans the horizon for t>0 and T>0, (after the rocket is launched). It is amazing to see that the "outside ball" can get to arbitrary large R (escape to infinity), but the "inside ball" must remain at R<0 for all time. However, the balls can still remain equidistant in their small local inertial frame! We can also see that the "inside ball" moves to more and more decreasing R coordinates for all time, yet it still remains equidistant to the "outside ball" (r→∞) in the small inertial frame for all time!

 

This particular diagram is flat spacetime. Is it consistent?

 

 

Andrew A. Gray

[Guest Zanket]

In this case, the box initially spans the horizon for t<0 and T<0, (before the rocket is launched at t=0), and we see that the balls can remain equidistant in the inertial frame, and that there is no problem for the box to exit the horizon with both balls in it.

What I see is diagrams that have boxes drawn on them where you drew them. You contradict the two sites I gave in this post, regarding a Rindler horizon. I’m not going to believe your placement of the boxes over those two sites. I’m not going to believe any analysis that does not directly refute those sites. I’ve plotted the path of a photon emitted at a Rindler horizon using the relativistic rocket equations, to see for myself that the photon cannot get to r=2M+ε from r=2M. I showed the plot above.

 

Above I said:

 

You have not addressed this issue that refutes you. The only way you could be right is if the relativistic rocket equations are wrong. Do you agree? Are you saying those equations are wrong? Are you saying that a photon can get to r=infinity from r=2M, while somehow skipping an observer hovering at r=2M+ε? If so, how does the photon skip that observer?

What are your answers? Why should I trust your diagrams when you ignore these questions? Why should I trust your diagrams when I can plot a photon’s path using simple math to see that you are refuted, while confirming the findings at the two reputable sites I gave above?

 

In an idealized, gravity-free universe, one of two balls at rest relative to each other in a common inertial frame can rise above a relativistic rocket to any height, while the other ball, initially below the crew’s Rindler horizon, never reaches the crew. In no way does that indicate a problem with the proof in the OP. The balls can communicate with each other because a signal sent from the lower ball can reach the upper ball when that ball is below the crew. The upper ball will always fall below the crew eventually. In our real, gravity-endowed universe, however, a ball can forever rise above a crew hovering at r=2M+ε. The ball need never fall below the crew.

[Qfwfq]

It does contradict SR. The “another premise” you mention is that the box straddles the horizon. But the reason the box cannot pass outward through the horizon is because of a prediction of GR, a prediction that contradicts a prediction of SR in X. Above I said: “In SR you can put any two objects anywhere in an inertial frame and let them be at rest relative to each other”.

SR, without further assumptions, does not forbid two bodies to be and remain at rest with each other. The two further premises we are discussing, taken together, go against this. It is not incompatible with SR as this, by itself, makes no determination. Can you get fined for not going through a green light?

 

Again, you are also neglecting the fact that [imath]dr[/imath] becomes a null intervall at the horizon ([imath]r=r_S[/imath]) and a timelike one inside it ([imath]r<r_S[/imath]), which completely counters the whole scenario, but your refutation can be made void even regardless of this. It simply doesn't contradict SR. If you let go of an apple in lack of gravity, it doesn't fall. According to SR the apple could remain forever in mid-air, simply because SR supposes nothing about gravity which can be considered a force like any other one. In GR, SR is viewed as the limit of zero gravitation, in which the extra axioms become nil. The addition of these changes certain predictions but this doesn't cause self-inconsistence. According to your kind of paralogism, even the apple not remaining where you let go of it would be a refutation of GR.

 

The box cannot be put anywhere in X and let be at rest relative to the particle. To be let at rest relative to the particle the box must be wholly within a certain part of the frame, namely the part of it that is above r=2M. The clauses “anywhere” (predicted by SR) and “certain part” (predicted by GR) contradict, which means that GR contradicts SR in X, where GR says that SR applies. Then GR is self-inconsistent; it contradicts itself.

Suppose A and B to be two regions of spacetime as follows: A is in your bedroom during December 31, 2007 and B is in your bedroom during January 1, 2008. Can a body get from B to A? Is this impossible in GR but possible in SR?

[andrewgray]

I’m not going to believe your placement of the boxes . . .

 

Zanket,

 

Of course you cannot believe what you do not comprehend. These diagrams have addressed your logic, and directly.

 

Suggestion: Get someone you can trust to explain these diagrams so you can understand.

 

Suggestion: I believe that I have shown that your logic shows an inconsistency, but not the one you are portraying.

 

So good luck to you. ;)

 

Andrew A. Gray

[Guest Zanket]

SR, without further assumptions, does not forbid two bodies to be and remain at rest with each other. The two further premises we are discussing, taken together, go against this. It is not incompatible with SR as this, by itself, makes no determination.

SR allows two bodies each anywhere in a common inertial frame to be at rest relative to each other. GR forbids that in X, where GR says that SR applies. Then GR contradicts itself. According to GR, to let be at rest relative to the particle the box must be wholly above the horizon. GR does not allow SR to work in X as SR advertises, even though GR says that SR applies in X.

 

Can you get fined for not going through a green light?

The question indicates no problem with the proof in the OP. If SR said you cannot get fined for not going through a green light, and GR overrode that, by saying you will where SR applies, then GR would be self-inconsistent.

 

Again, you are also neglecting the fact that [imath]dr[/imath] becomes a null intervall at the horizon ([imath]r=r_S[/imath]) and a timelike one inside it ([imath]r<r_S[/imath]), which completely counters the whole scenario, but your refutation can be made void even regardless of this.

This does not show a problem with the proof in the OP. GR allows the inertial frame X to fall through a horizon, and says that SR applies in X. Then to show that GR is self-inconsistent I need only show that GR does not allow SR to work in X as SR advertises. Which I do. No other prediction of GR would refute that. Consider the following theory about birds:

 

Some birds can fly.

No birds can fly.

 

Cleary this theory is self-inconsistent. If it also made this statement:

 

[imath]dr[/imath] becomes a null intervall at the horizon ([imath]r=r_S[/imath]) and a timelike one inside it ([imath]r<r_S[/imath])

 

then the theory would still be self-inconsistent. Regardless of any additional statement, the theory would be self-consistent.

 

It simply doesn't contradict SR. If you let go of an apple in lack of gravity, it doesn't fall. According to SR the apple could remain forever in mid-air, simply because SR supposes nothing about gravity which can be considered a force like any other one. In GR, SR is viewed as the limit of zero gravitation, in which the extra axioms become nil. The addition of these changes certain predictions but this doesn't cause self-inconsistence.

This does not show a problem with the proof in the OP. You have not shown here that GR does not contradict SR in X. The only way you can refute me is to show that GR allows SR to work in X as SR advertises. You have not shown here that GR allows the box and the particle to be at rest relative to each other, as SR allows.

 

Suppose A and B to be two regions of spacetime as follows: A is in your bedroom during December 31, 2007 and B is in your bedroom during January 1, 2008. Can a body get from B to A? Is this impossible in GR but possible in SR?

How is this relevant to the discussion?

[Guest Zanket]

These diagrams have addressed your logic, and directly.

Then why can’t you use the relativistic rocket equations to refute me? Those equations work for plotting the geodesic of a photon moving radially upward. When I plot such a photon, I see that the geodesic disagrees with your diagrams.

 

Suggestion: Get someone you can trust to explain these diagrams so you can understand.

Even if someone did that, I would still be left wondering why the relativistic rocket equations refute the diagrams. The equations are so much simpler than your diagrams. There’s no reason I shouldn’t trust them; you haven’t given me any reason not to. If you had said something like “the equations are wrong because ...” then I might have more faith in your diagrams. But you’ve never directly addressed why your diagrams contradict those equations.

 

So good luck to you. ;)

And to you. Thank you for helping me to understand the Rindler horizon, for showing me that the directly measured radial distance between r=2M and some higher r is finite, and for the excellent discussion in general.

[Qfwfq]

If SR said you cannot get fined for not going through a green light, and GR overrode that, by saying you will where SR applies, then GR would be self-inconsistent.

The green light does not forbid you from going through. If for some reason you don't go through, at the most you are annoying the driver behind you but you aren't violating the green light.

 

If it also made this statement:

 

[imath]dr[/imath] becomes a null intervall at the horizon ([imath]r=r_S[/imath]) and a timelike one inside it ([imath]r<r_S[/imath])

 

then the theory would still be self-inconsistent. Regardless of any additional statement, the theory would be self-consistent.

You really don't realize how this makes it meaningless to say that the box straddles the horizon. You also fail to realize the difference between saying "Some birds can fly." and SR not forbiding birds from flying, and also that between saying "No birds can fly." and GR predicting that a particular bird cannot fly.

 

Likewise, SR describes dynamics of bodies in terms of their masses but makes no assumption about these having the effect called gavity. It therefore does not forbid the apple from remaining where you let go of it. GR does, by saying that the frame at rest with the ground isn't an inertial one, as it would be according to SR. Why is this not inconsistent?

 

How is this relevant to the discussion?

Perhaps I should have asked a different question first: "Can your chest of drawers have one side in A and the other in B?" Even more directly: "Can the bed be in A and the chair in B, and can these be at rest with each other?". A and B are two contiguous regions of spacetime, two parts of a single region that are separated by the boundary defined as New Year's midnight. Now I was replying to you having said:

“In SR you can put any two objects anywhere in an inertial frame and let them be at rest relative to each other”.

 

........................

 

To be let at rest relative to the particle the box must be wholly within a certain part of the frame, namely the part of it that is above r=2M. The clauses “anywhere” (predicted by SR) and “certain part” (predicted by GR) contradict, which means that GR contradicts SR in X, where GR says that SR applies.

You still seem unaware that the "perts of the frame" you are talking about share something very fundamental with the regions A and B I defined regarding the bedroom. In a region of spacetime that straddles the event horizon, which directions are spacelike and which are timelike?