General relativity is self-inconsistent

[andrewgray]

Let’s be clear that you have not shown here that something can pass outward through r=2M, which is what you have claimed before.

 

Correct. This was the second of two cases that I sited in post number 130

 

The first case was when the balls could escape from the Rindler horizon if they left before the rocket launched. The first case is what shows that there are no literal black holes. Things can only be trapped in a black hole by the time factor, not because they are prohibited from exiting. Both cases show that the balls can stay equidistant in the box.

 

Step back to see the absurdity of what you’re suggesting. Do you really think that the box and the escaping particle can be at rest relative to each other even as the particle passes Earth, where the Earth is at r=1 million light years?

 

Yes, that is absurd. But the Schwarzchild geometry is only Rindler in a local frame. So there is no inconsistency here. I only show that since the Schwarzchild horizon is equivalent to a Rindler horizon in a small inertial neighborhood, that the the balls can stay equidistant for the Schwarzchild case while they remain near the locally defined frame. After they leave the small inertial frame, "all bets are off".

 

So again, there seems to be two cases. One where the two balls are ejected after the rocket launches and are "straddling" the horizon, and one where the two balls are ejected before the rocket is launched, and the inside ball leaves the horizon.

 

Both cases can be extended to the small Rindler neighborhood of a Schwarzchild horizon. So depending on where/when in spacetime you are talking about, the two balls can either straddle the horizon for a while and stay equidistant, or both can exit and they can stay equidistant as they both escape. After they leave the small inertial neighborhood, they no longer stay equidistant of course.

 

Let me be clear here. In the first case, both balls can leave.

 

In the second case, the balls can stay equidistant in the local frame, but after they leave the local frame, the lagging ball cannot ever leave the horizon.

 

The point is that the balls can stay equidistant in the small local inertial frame near a Scwarzchild horizon in both cases for the duration of the frame.

 

Andrew A. Gray

[Qfwfq]

GR does violate SR. It says that SR applies in X, but then does not allow SR to work as expected in X, by not allowing the box and the particle to be at rest relative to each other.

What do you mean by "as expected"? Do you mean "to the first order" or "in the tangent space" or do you mean something else? Surely "as expected" shouldn't mean that you can observe a future event, or "travel" to yesterday? :xx:

 

No, it can also be expressed in terms of widely published predictions, which can be supported without differential geometry. (For example, above I showed mathematically that GR predicts that nothing can pass outward through a horizon, by using the relativistic rocket equations.) Einstein’s words in the equivalence principle are relevant to this discussion, and they explicitly refute you. This is a scientific debate.

I'm sorry but I wasn't following the details of your dicussion of the rocket equation because, as far as I can see, it can show no more than that a body can only approach c asymptotically and I needed no further proof of that. It certainly has an implication about any spacelike 3-surface but, without analyzing the Schwarzschild solution, how can any conclusion be drawn about which ones are spacelike in the Schwarzschild solution? Repetition intended.

 

Your reply says nothing about what you replied to, except for the insistence on Einstein's words. I don't care if, on some occasions, the Parson said 'small' without even specifying 'infinitely' because what he meant was the same as is meant in any proper statement of the equivalence principle for GR. Now, if you want the word of the Parson, from Die Grundlage der allgemeinen Relativitätstheorie, A § 4 (starts 9th page of PDF, excerpts from 11th, p. 779):

Wir werden später sehen, daß die Wahl solcher Koordinaten für endliche Gebiete im allgemeinen nicht möglich ist.

 

.......

.......

.......

 

Auch in dem allgemeinen Falle, daß wir nicht in einem endlichen Gebiete bei passender Koordinatenwahl die Gültigkeit der speziellen Relativitätstheorie herbeiführen können, werden wir an der Auffassung festzuhalten haben, daß die [imath]g_{\sigma\tau}[/imath] das Gravitationsfeld beschreiben.

Wherein the words bolded for emphasis mean "...in general is not possible for finite areas..." and "...that we cannot hold the validity of Special Relativity, by suitable choice of coordinates, in a finite area..." in my translation. You might like to see an independent translation of the whole thing. Acknowledements to Popular for having supplied links to both these PDFs, elsewhere on this forum. The possibility referred to is that of choosing coordinates such that [imath]g_{\sigma\tau}[/imath] be that of SR. :shrug:

 

GR does say this, in the equivalence principle. GR makes no exceptions for where SR applies locally in empty spacetime.

I don't ask it to make exceptions. As long as inertial is understood to mean locally inertial, and this is meant in the manner I've lost hope of you understanding.

 

The equivalence principle says “anywhere”.

Actually it says “any where-when”.

 

Taylor, Thorne, and Wheeler disagree with you. Note that both Andrew and Erasmus agree that SR applies in X, disagreeing with you. The ice you are skating on broke. You're flailing in the water.

Did they say "SR applies exactly to a spacetime region partly with [imath]r<r_S[/imath] and partly with [imath]r>r_S[/imath]"? And I even wasted so much bandwidth yesterday to add the word exactly which I had left out! :doh: Even Einstein, in what I quote above, doesn't bother to qualify the word Gültigkeit with an adjective to the same meaning (In the negative, note!), he leaves it implicit. Nobody's perfect. Not even the Parson. So you, reading that instead of your quote, might easily think he says that, in general, SR doesn't apply at all, in a finite region.

 

 

You are picking on words, details and the angel's sex without understanding the essence. I don't know why I'm still wasting so much time.

 

Why not just explain where you’re going with the timelike and spacelike reasoning?

Absolutely simple. I'm going from past to future and I'm not going vice versa.

 

It’s reasonable for me to ask when it seems clear that you cannot use it to refute me. I don’t like going down dead end paths, paths that are already refuted; that takes time and leads people to not see the forest for the trees. I suspect your reasoning has something to do with your erroneous belief that “GR does not say this at all”.

Wow, now clearly has become "it seems clear". The only dead-end paths are the timelike ones that go to [imath]r<r_S[/imath]. What you quote as my erroneous belief can pretty much be found in the above quote and without even meaning just a region that extends below the Schwarzschild horizon (not yet known at the time of his writing).

 

It doesn’t help refute the OP. GR says that SR works in X the same as it does in any other inertial frame. The equivalence principle is clear about that. Nothing you can point out about X will refute that.

If that hint doesn't help, maybe you haven't connected it with what it referred to, i. e. the questions you still haven't answered.

[andrewgray]

So let's do a Schwarzchild example, so we can be crystal clear. Again, let a=1 and 4M=1 arbitrarily to keep things ultra simple. Here is the question:

 

Can a ball at (t=99.9, r=2M-(4x10^-86)M) reach (t=100.0, r=2M+(4x10^-86)M)?

 

Well, we have from post number 48

 

that above the horizon,

 

[math]R^2 = r -2M \;\;\; or \;\;\; R^2 = r - \frac{1}{2}[/math]

 

and for below the horizon we have that

 

[math](-R)^2 = 2M - r[/math]

 

So our Rindler coordinates (in a small frame) are:

 

[math](T,R)_{outside} =(100.0,\;\; 1 \times 10^{-43})[/math]

[math](T,R)_{inside} \; = ( \; 99.9, \; -1 \times 10^{-43})[/math]

 

The flat frame coordinates are related above by:

 

[math]x_o=R \sinh T [/math]

[math]x_1=R \cosh T [/math]

 

And below by:

 

[math]x_o=-R \cosh T [/math]

[math]x_1=-R \sinh T [/math]

 

So our flat coordinates become:

 

 

Plotting we get:

 

 

And we see from the slope of an imaginary line between the two points (slope less than 1 in x_o,x₁ plane) that the speed necessary is "warp speed" (greater than c), and the ball cannot escape outward through the horizon to the specified event.

 

Next question:

 

Can a ball at (t=-99.9, r=2M-(4x10^-86)M) reach (t=+100.0, r=2M+(4x10^-86)M)?

 

So our Rindler coordinates (in a small frame) are:

 

[math](T,R)_{outside} =(+100.0,\; +1 \times 10^{-43})[/math]

[math](T,R)_{inside} \; = ( \; -99.9, \; -1 \times 10^{-43})[/math]

 

 

And our flat coordinates become:

 

 

 

And we see from the slope of an imaginary line between the two points (slope greater than 1 in x_o,x₁ plane) that the speed necessary is "sublight", and the ball can escape outward through the horizon to the specified event.

 

Zanket,

 

This is a mathematical Q.E.D.

 

 

Andrew A. Gray

[Guest Zanket]

The first case was when the balls could escape from the Rindler horizon if they left before the rocket launched. The first case is what shows that there are no literal black holes.

I disagree; you have not shown that. You've drawn that, but you haven't supported it with the relativistic rocket equations, as you should be able to if you’re right. When using those equations to show that GR predicts that nothing can pass outward through r=2M, there is no such thing as “before the rocket launched”. In principle a rocket can always be hovering at r=2M+ε. A Rindler horizon exists at r=2M starting from the first moment that r-coordinate exists in empty spacetime.

 

After they leave the small inertial frame, "all bets are off".

Yes, so your second case doesn’t refute the OP. The particle must exit the box on its way to r=infinity, because the box cannot pass outward through the horizon. The only way the particle can exit the box is to move relative to it in X. Then GR requires that the box and the particle move relative to each other in X, in contradiction to SR, which GR says applies in X. Then GR is self-inconsistent.

 

Also note that the upper ball can reach R=infinity only when the proper distance between the balls is infinite. When the proper distance between the balls is finite—as it always is—the upper ball reaches some maximum height in the rocket’s frame, and then falls, eventually falling below the rocket. That’s how the balls can have a two-way communication despite being on either side of a Rindler horizon. The upper ball can receive a signal from the lower ball when the upper ball has fallen below the rocket. The box in the OP is finite-sized, and the particle will never fall below an observer hovering at r=2M+ε.

 

Let me be clear here. In the first case, both balls can leave.

To show this, and refute me, you need to show that the relativistic rocket equations allow something to pass outward through r=2M; you haven’t done that. Can you do that? I have shown otherwise, and my result agrees with the Rindler Horizon site. You haven’t shown any problem with that.

[Guest Zanket]

This is a mathematical Q.E.D.

Even if you were right, it wouldn’t refute the OP. In post 140 I proved using the relativistic rocket equations that nothing can pass outward through r=2M. My result agrees with the Rindler Horizon site.

 

To make my case I needed show in only one way that GR predicts that nothing can pass outward through r=2M. (And I’ve gone beyond the call of duty here since you agree that hundreds of physicists agree with me.) That’s why you need to refute me using those equations. You need to show what mistake I made in post 140. You haven’t done that. To be right, I don’t need to show what mistake, if any, you made above. You can't refute me by showing that GR predicts in some other way that something can pass outward through r=2M.

 

Starting from post 140, when a = 1 then

 

d = sqrt(t^2 + 1) – 1

 

d_photon = t – 1

 

It is easy to see that d is greater than d_photon for any given t. Then it would be mathematically impossible for you to use the relativistic rocket equations to show that something can pass outward through r=2M to reach r=2M+ε, where ε is arbitrarily small. A photon moving radially upward at r=2M never reaches r=2M+ε. In post 156 you used a particular value for ε, in which case it’s easy to be fooled into thinking that something can pass outward through r=2M. But ε can always be smaller than whatever value you assign to it. In the limit, ε is zero, so a photon moving radially upward at r=2M does not increase its radial position, just like T&W say in the OP.

[Guest Zanket]

What do you mean by "as expected"?

I mean “as expected, to match generally accepted predictions of SR”.

 

Now, if you want the word of the Parson, from Die Grundlage der allgemeinen Relativitätstheorie, A § 4 (starts 9th page of PDF, excerpts from 11th, p. 779):Wherein the words bolded for emphasis mean "...in general is not possible for finite areas..." and "...that we cannot hold the validity of Special Relativity, by suitable choice of coordinates, in a finite area..." in my translation.

So you think SR is not experimentally confirmed? That text is not saying that SR has no applicability in a finite-sized inertial frame our real, gravity-endowed universe. It is implying that it is accurate to only an finite number of significant digits in such a frame. Accuracy to an infinite number of significant digits isn’t needed for the proof in the OP to be valid, just like it isn’t needed for SR to be experimentally confirmed.

 

Actually it says “any where-when”.

No, it says “anywhere”. The OP includes the equivalence principle in Einstein’s own words. (He was fluent in English.)

 

Did they say "SR applies exactly to a spacetime region partly with [imath]r<r_S[/imath] and partly with [imath]r>r_S[/imath]"?

They didn’t have to. Unlike you, they don’t disagree with Einstein, Taylor, Thorne, and Wheeler about whether SR applies to a finite-sized inertial frame falling through a horizon. They know that SR isn’t accurate to an infinite number of significant digits in X, and they know that doesn’t matter.

 

I don't know why I'm still wasting so much time.

I wonder why too. Why not read some of the links to Taylor and Wheeler’s Exploring Black Holes that I gave in the OP? They can resolve your confusion. If you think SR is not a scientific theory because it isn’t falsifiable due to a lack of infinitely small measuring equipment, that might be a clue that you’re missing something. Quit wasting time and go figure out what you’re missing.

 

If that hint doesn't help, maybe you haven't connected it with what it referred to, i. e. the questions you still haven't answered.

When you have adequately shown that the equivalence principle does not equate X with some inertial frame Y that is wholly above the horizon, then I’ll answer the question. Until then, I need not be concerned with what the Schwarzschild metric says about “timelike” and “spacelike” on either side of r=2M. The OP gives ample evidence from reputable sources that SR applies to a finite-sized inertial frame falling through a horizon. Taylor and Wheeler in particular could hardly be clearer about that.

[Qfwfq]

I mean “as expected, to match generally accepted predictions of SR”.

The question was whether these include that you can observe a future event, or "travel" to yesterday. Do these “generally accepted predictions of SR” include this? Answer either yes or no.

 

So you think SR is not experimentally confirmed? That text is not saying that SR has no applicability in a finite-sized inertial frame our real, gravity-endowed universe. It is implying that it is accurate to only an finite number of significant digits in such a frame. Accuracy to an infinite number of significant digits isn’t needed for the proof in the OP to be valid, just like it isn’t needed for SR to be experimentally confirmed.

I said nothing of the sort and I know exactly what that text is saying. It is not a matter of finite or infinite number of significant decimal digits.

 

No, it says “anywhere”. The OP includes the equivalence principle in Einstein’s own words. (He was fluent in English.)

I believe he was also fluent in German, French and Italian. What I quoted were his own words too. Einstein did not always fuss over details and even glossed over them very often; "his own words" is not the same as "what GR says", and GR is about the space-time continuum. You are clearly not reasoning in space-time. Therefore...

They didn’t have to. Unlike you, they don’t disagree with Einstein, Taylor, Thorne, and Wheeler about whether SR applies to a finite-sized inertial frame falling through a horizon.

Neither did Einstein have to, which also means I'm not disagreeing with him or the others. I'm only not letting myself get confused by use of colloquial wording that even they often use to cut short a bit. This kind of shortcut isn't a problem if one understands a proper formulation and does a bit of mental translation but, when the layperson presumes to argue in terms of it and to draw conclusions, it's a source of disaster.

 

Why not read some of the links to Taylor and Wheeler’s Exploring Black Holes that I gave in the OP? They can resolve your confusion. If you think SR is not a scientific theory because it isn’t falsifiable due to a lack of infinitely small measuring equipment, that might be a clue that you’re missing something. Quit wasting time and go figure out what you’re missing.

I don't need to read that book, it's written for the layperson. I'm rather hoping to have a further look at Weinberg's book when I can; I used it when I took my GR course and got passed the test but that was the years around 1990 and lately I've been wanting to have a better look at a few things, although certainly not due to confusion about the equivalence principle and the meaning of locally inertial. I said no such thing about falsifiability due to a lack of infinitely small measuring equipment. There's nothing wrong with SR at all; even with gravity it is perfectly valid to first order.

 

So, why don't you follow that link and have a look at that book? If you can get through the calculus, it might resolve your confusion; if you can't, that might be a clue that you’re missing something.

 

When you have adequately shown that the equivalence principle does not equate X with some inertial frame Y that is wholly above the horizon, then I’ll answer the question.

I do not have to show this, apart from the wording not being meaningful. I can only suppose it to be meant in the sense that the equivalence principle applies to points of space-time in both such regions. Sodann? Where did I say otherwise? I only said that a timelike worldline, for increasing proper time, cannot go from a point-event with [imath]r<r_S[/imath] to one with [imath]r>r_S[/imath] any more that such a worldline could go from one day to the previous. Anything wrong with saying this?

 

Until then, I need not be concerned with what the Schwarzschild metric says about “timelike” and “spacelike” on either side of r=2M.

If you refuse to be concerned with that then end of the discussion.

 

If instead you shall be concerned with it, it is enough to understand that, for any point-event with [imath]r>r_S[/imath], there are timelike worldlines starting from it that reach ones having even larger values of r, unlike in the [imath]r<r_S[/imath] case. This is what makes the difference between the two space-time regions you are talking about. Does that help? Need more hints? You only have to ask...

[andrewgray]

Zanket,

 

In your #140 post, your diagram:

 

 

shows a dotted line for t>0, but it does not show anything for t<0. I am in agreement with you that nothing can pass "outward through the horizon" from t>0. We are in complete agreement here. But you must show where the horizon is for t<0. You must do this because the Schwarzchild horizon exists for "all time".

 

Also, in both the Schwarzchild and Rindler cases, the transformation

 

[math]t \right t'+C[/math]

 

has absolutely no effect on the metric. So since the metric remains unchanged, we are free to set our time scale to an arbitrary "starting constant" C.

 

So I still believe that I have shown that both the Rindler and the Schwarzchild case remain consistent in that the ball can remain stationary relative to the box (while in the local inertial frame), in both cases that I have presented. So in my opinion, the only thing left to do is to make you understand.

 

So, you must show us your post #140 plot with the horizon shown for t<0, in a way that you understand with your "rocket equations".

 

Andrew A. Gray

 

Thanx again for the interesting discussion. I have decided not to give up.

[andrewgray]

. . . for any point-event with [imath]r>r_S[/imath], there are timelike worldlines starting from it that reach ones having even larger values of r, unlike in the [imath]r<r_S[/imath] case.

 

Qfwfq,

 

In post #156 I showed that there is a timelike worldline from (t=-99.9, r=2M-ε) going to (t=100.0, r=2M+ε) where ε=4x10^-86M. This was done by using a locally flat Rindler frame.

 

This is the Schwarzchild analogy of making it out of the horizon by being ejected "before the rocket launches".

 

So there are timelike worldlines from r<r_s to greater r. It's just that the Schwarzchild singular coordinate t goes crazy during the trip. But that does not stop the trip.

 

Andrew A. Gray

[Guest Zanket]

The question was whether these include that you can observe a future event, or "travel" to yesterday. Do these “generally accepted predictions of SR” include this? Answer either yes or no.

Show relevancy please, and if it’s adequate then I’ll answer. The only generally accepted prediction of SR that matters to the OP is that two objects each anywhere in an inertial frame can be let at rest relative to each other.

 

I said nothing of the sort and I know exactly what that text is saying. It is not a matter of finite or infinite number of significant decimal digits.

It is a matter of that, as any decent book on GR will tell you. Taylor, Thorne, and Wheeler all define an inertial frame such that it is small enough that the tidal force in it is negligible, and not such that it is infinitely small. Again I ask: So you think SR is not experimentally confirmed?

 

What I quoted were his own words too.

His words in the OP were handpicked by Thorne. And Taylor and Wheeler agree with them that an inertial frame need not be infinitely small.

 

I said no such thing about falsifiability due to a lack of infinitely small measuring equipment.

You didn’t have to. By your logic, SR is not a scientific theory. A theory is not scientific if it isn’t falsifiable. A theory isn’t falsifiable if it can’t be experimentally tested. It can’t be experimentally tested if infinitely small measuring equipment is required. Infinitely small measuring equipment is required when inertial frames must be infinitely small.

 

The reality is that SR is experimentally confirmed to a certain number of significant digits. SR is valid to a certain number of significant digits in any given finite-sized inertial frame.

 

I do not have to show this, apart from the wording not being meaningful.

You haven't shown that the wording isn't meaningful. I do not agree that Einstein, Taylor, Thorne, and Wheeler are all barefaced liars. I won’t answer your question unless you adequately establish relevancy.

 

The equivalence principle equates X to any other inertial frame, in which case an attempt to show that X is different than any other inertial frame cannot be used to refute the OP. If GR predicts that X is different than any other inertial frame then it contradicts its own equivalence principle, hence is self-inconsistent. You can’t use a self-inconsistency of GR against a proof of that.

 

Sodann?

English please.

 

I only said that a timelike worldline, for increasing proper time, cannot go from a point-event with [imath]r<r_S[/imath] to one with [imath]r>r_S[/imath] any more that such a worldline could go from one day to the previous. Anything wrong with saying this?

Yes. The equivalence principle equates X to any other inertial frame, in which case an attempt to show that X is different than any other inertial frame cannot be used to refute the OP.

 

If you refuse to be concerned with that then end of the discussion.

Fine with me. If you take mod action because I insist on relevancy then I will know that this is not a scientific site, hence not worth my time.

 

If instead you shall be concerned with it, it is enough to understand that, for any point-event with [imath]r>r_S[/imath], there are timelike worldlines starting from it that reach ones having even larger values of r, unlike in the [imath]r<r_S[/imath] case. This is what makes the difference between the two space-time regions you are talking about. Does that help?

Here it can only help to make my case that GR is self-inconsistent.

[Guest Zanket]

I am in agreement with you that nothing can pass "outward through the horizon" from t>0. We are in complete agreement here. But you must show where the horizon is for t<0. You must do this because the Schwarzchild horizon exists for "all time".

t is the time elapsed on an observer’s clock. There’s no such thing as negative time elapsed.

[Erasmus00]

Zanket, might I make a suggestion: take the time to learn GR. GR is formulated in terms of the calculus of curved spaces, and in this calculus it is simply impossible for small frames of reference to not approach flat. Hence, ANY 3+1 dimensional surface will approach SR for a small enough area. The mathematical language of GR GUARANTEES its consistency. For a discussion of this, see any text on the geometry calculus of curved surfaces.

 

Now, since the consistency is a mathematical certainty- any "thought experiment" that shows otherwise must be wrong. This is because the consistency of the theory is mathematically guaranteed. So your approach would be more constructive if you approached the problem as a student: why is this wrong?

 

Now, might I suggest somethings that may help- 1. the event horizon is only singular in schwarzschild coordinates. Perhaps some of your confusion could be remedied by looking at your thought experiment in different coordinates. Most of the analysis in this thread has been in terms of Schwarzchild/Rindler coordinates (Personally, I believe the fact that the schwarzchild metric reduces to a Rindler horizon PROVES the consistency of GR).

 

2. At the very least, learn how to interpret a spacetime diagram. Andrew Grey has made a large effort to discuss your "paradox" but repeatedly you refuse to put the same kind of effort into understanding his presentation.

-Will

[Guest Zanket]

GR is formulated in terms of the calculus of curved spaces, and in this calculus it is simply impossible for small frames of reference to not approach flat. Hence, ANY 3+1 dimensional surface will approach SR for a small enough area. The mathematical language of GR GUARANTEES its consistency.

No, its consistency is not guaranteed that way. You’re assuming that GR cannot say something else that contradicts SR. That’s a mistake in your logic.

 

Now, since the consistency is a mathematical certainty- any "thought experiment" that shows otherwise must be wrong.

If you were right and know so much, it should be easy for you to point out a flaw in my simple proof. Why have you been unable to do that?

 

... (Personally, I believe the fact that the schwarzchild metric reduces to a Rindler horizon PROVES the consistency of GR).

Since my proof refutes that, your approach would be more constructive if you approached the problem as a student: why is this wrong?

 

Andrew Grey has made a large effort to discuss your "paradox" but repeatedly you refuse to put the same kind of effort into understanding his presentation.

Ah, so you agree with Andrew that GR does not predict black holes? :doh: Had you put the same kind of effort into understanding his presentation as I have, you would know that he agrees that GR is self-inconsistent if it predicts black holes.

[Erasmus00]

No, its consistency is not guaranteed that way. You’re assuming that GR cannot say something else that contradicts SR. That’s a mistake in your logic.

 

It cannot- the entire SR theory is encompassed in the flat metric. GR must reduce to SR locally because of the language it is formulated in- therefore it is self consistent.

 

Since my proof refutes that, your approach would be more constructive if you approached the problem as a student: why is this wrong?

 

I have limited my responses to this thread to what I feel others have not said. You were refuted early on by Andrew Grey.

 

Ah, so you agree with Andrew that GR does not predict black holes? :doh: Had you put the same kind of effort into understanding his presentation as I have, you would know that he agrees that GR is self-inconsistent if it predicts black holes.

 

I agree with Andrew that the traditional analysis in Schwarzchild coordinates is flawed, and that there are proper time paths that move out of the singularity. That there must be is evidenced in the time reversal symmetry of the problem. Also, this is obvious in any set of coordinates that don't go singular at the horizon. For instance, see Kruskal-Szekeres coordinates.

 

Where Andrew and I may disagree (although I'm not sure what his stance is) is whether or not these mathematical solutions are physical or not. Under time reversal, the two hyperbolic time branches flip (just as with the Rindler metric) and the black hole becomes a "white hole." I'm unconvinced the white hole solutions are physically relevant. However,self consistency is a mathematical concern, so whether the solutions are physical, they certainly exist mathematically.

 

Once more, I recommend study- study any coordinates that cover the event horizon.

-Will

[Guest Zanket]

GR must reduce to SR locally because of the language it is formulated in- therefore it is self consistent.

That’s a non sequitur. Yes, GR reduces to SR locally. But in no way does that prevent GR from saying something else that contradicts SR locally. Then the fact that GR reduces to SR locally does not prove that GR is self-consistent.

 

I have limited my responses to this thread to what I feel others have not said. You were refuted early on by Andrew Grey.

Nice try!

 

I'm unconvinced the white hole solutions are physically relevant. However,self consistency is a mathematical concern, so whether the solutions are physical, they certainly exist mathematically.

The proof of GR’s self-inconsistency is made using a black hole, not a white hole. If GR is self-inconsistent in one way, it is self-inconsistent, period. Whatever GR says about white holes is irrelevant.

 

Once more, I recommend study- study any coordinates that cover the event horizon.

I recommend you study logic.

[Erasmus00]

That’s a non sequitur. Yes, GR reduces to SR locally. But in no way does that prevent GR from saying something else that contradicts SR locally.

 

Think about this for a second. If SR reduces to GR locally then, locally, THEY ARE THE SAME THEORY. Locally, SR=GR. Hence, GR can only be inconsistent with SR locally if SR is inconsistent.

 

 

The proof of GR’s self-inconsistency is made using a black hole, not a white hole. If GR is self-inconsistent in one way, it is self-inconsistent, period. Whatever GR says about white holes is irrelevant.

 

Black/White holes are different regions OF THE SAME SOLUTION! This is what Andrew has been trying to show you with the diagrams! Again, you should learn how to read a spacetime diagram. I find your arrogant refusal to actually study what Andrew has spent time showing you beyond rude. This is my last post in this thread- you may continue tilting at windmills all you like.

-Will

[Guest Zanket]

If SR reduces to GR locally then, locally, THEY ARE THE SAME THEORY. Locally, SR=GR. Hence, GR can only be inconsistent with SR locally if SR is inconsistent.

No, GR can reduce to SR locally and still contradict SR locally, simply by including an additional element that contradicts SR locally.

 

Watch how easy it is to create such a theory:

 

Zanket’s Self-Inconsistent Theory

GR applies globally.

The speed of light is always 5c.

 

My theory mathematically reduces to SR locally, and locally my theory = SR, yet it contradicts SR locally. :shrug: I have proven your conclusion to be false. By your logic, it’s impossible to create a self-inconsistent theory.

 

Black/White holes are different regions OF THE SAME SOLUTION!

So what? What statement in the proof in the OP does that refute? Answer: none, because the proof in the OP is based on a black hole, and not a white hole.

 

Again, you should learn how to read a spacetime diagram.

You should learn logic. If someone tried to refute Zanket’s Self-Inconsistent Theory, you’d be saying “The speed of light is 5c for white holes! No self-inconsistency!”, ignoring that the speed of light is not 5c for other regions.

 

I find your arrogant refusal to actually study what Andrew has spent time showing you beyond rude.

I have addressed all of his posts to the point of showing that they do not refute the OP. Doing no more than that is not rude; it’s scientific.