General relativity is self-inconsistent

[Qfwfq]

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

Sorry, I did not closely at that post for lack of time; I took a quick look through it now.

 

By definition the Schwarzschild solution is independent of the Schwarzschild t coordinate. The only difference between what you start with in your two examples is the value of t (which I suppose to be Schwarzschild's) as far as I can see. If I haven't missed another difference between the premises, this would appear to rule out the difference between the two conclusions being possible. Mysterious? The first thing that came to mind was if, somehow, in the map between the two charts, the opposite sign of t could perhaps end up becoming equivalent to a time reversal.(?) Oooohps!... Now, thinking of nothing more than the definition of the Rindler coordinates, I already see a faint glow of suspicion as to a possible vague reason why such a trick might happen. I don't have time to work it out more in detail.

 

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

I fail to see your point.

 

Show relevancy please, and if it’s adequate then I’ll answer.

Answer yes or no please, and if your answer is the obviously true one and not the obviously false one, I might get nearer to showing relevancy.

 

It is a matter of that, as any decent book on GR will tell you.

My GR course was an excelent one and I used an excelent book for it. The professors were excelent too.

 

Again I ask: So you think SR is not experimentally confirmed?

I already answered, why insist? SR is excelently confirmed.

 

His words in the OP were handpicked by Thorne.

Show relevancy please.

 

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.

No, that is not by my logic, it is by yours.

 

Have you ever learnt how to find a straight line tangent to a circumference using rule and compass? Must you have an infinitely small rule and an infinitely small compass?

 

You haven't shown that the wording isn't meaningful.

You did not address what I followed it with, which started with a more meaningful statement which I believe was what you were trying to say. You exhibted no valid basis of objection to saying 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. And who said that Einstein, Taylor, Thorne, and Wheeler are all barefaced liars?

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.

Exactly in which sense does the equivalence principle equate X to any other inertial frame? The EQ says nothing about parallel transport, even less about the Schwarzschild r and t coordinates, so it can't be in contradiction of the statement you are attempting to criticize.

 

If you take mod action because I insist on relevancy then I will know that this is...

Who talked about mod action? Me? At the most this thread belongs in Strange Claims Forum. In there, you can insist on anything you like without an adequately scientific basis.

 

...not a scientific site, hence not worth my time.

:bwa:

 

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

I can imagine this only being due to it mentioning the difference between the two space-time regions you were talking about. The mention of something so utterly blasphemous is supposed to make it supportive of your claim, is it? :rolleyes:

 

Follow Will's good advice and get yourself a proper course in GR, perhaps first one in calculus and then the basics of differential geometry.

 

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. :eek2:

Well, you've managed to find a true example of a formal system which is self inconsistent, by adding an axiom that contradicts those to which you added it. Congrats. In this case it actually does contradict the others, it doesn't just simply restrict possibilities, a distinction I was hoping to get across, someday.

 

And you have the chutzpah to tell Will that he should learn logic! :xx:

[andrewgray]

. . . The first thing that came to mind was if, somehow, in the map between the two charts, the opposite sign of t could perhaps end up becoming equivalent to a time reversal.(?) Oooohps!... Now, thinking of nothing more than the definition of the Rindler coordinates, I already see a faint glow of suspicion as to a possible vague reason why such a trick might happen. I don't have time to work it out more in detail.

 

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

 

I fail to see your point.

 

Qfwfq,

 

No, time reversal is not the correct symptom. The symptom is time coordinate singularity.

 

The point is clearly seen when comparing this route: (t=-99.9, r=2M-ε) going to (t=100.0, r=2M+ε) where ε=4x10^-86M

 

in both the Rindler coordinates and the Kruskal-Szekeres coordinates:

 

 

The route out of the horizon in the Kruskal coordinates is shown in green. This is the analogy to the green route in the Rindler coordinates, which was described as "being ejected before the rocket launched" (t=0).

 

The other two routes (black route on left, and red route on right) are warp speed and impossible.

 

See post #156

 

Andrew A. Gray

[Qfwfq]

No, time reversal is not the correct symptom. The symptom is time coordinate singularity.

Well, I had missed something, which is that from the first to the second case only the initail t is of opposite sign, so [imath]\Delta t[/imath] changes. This knocks down the strong objection of t independence but doesn't dissipate my scepticism.

 

The coordinate singularity at zero is what I was thinking of and the fact that the map between charts is two sheets to one; the Schwarzschild chart per se doesn't really distinguish between dt and -dt and it doesn't change by time inversion but there is the matter of physical significance and interpretation. It isn't enough to show the world-line is timelike. The one you show as spacelike clearly can't be a particle's path, neither one way or the opposite. I repeat that I won't be working out the details before doing some other things, the whole point is that it's tricky to follow the matter through the mapping, or to work it out directly, and I so far haven't gone that much into the details of it. I can't yet rule out you being the one that'll monkey the whole community of GR experts but I'm not persuaded either.

[andrewgray]

Qfwfq,

 

Yes, you must remain skeptical until you see for yourself.

 

Also, do you understand this proof now?

 

Other proof

 

 

Andrew A. Gray

[andrewgray]

So finally, I would like to point out that when they talk about black holes on the Discovery Channel or on The Universe Show, and they say that:

 

"Once something goes into a black hole, it never comes out."

 

This is incorrect.. The correct saying would be:

 

"Nothing ever goes into a black hole and nothing ever comes out".

 

This is the result of the time coordinate singularity and the fact that the geometry is completely time reversal symmetric.

 

Andrew A. Gray

[Guest Zanket]

Answer yes or no please, and if your answer is the obviously true one and not the obviously false one, I might get nearer to showing relevancy.

I decline. If it worked that way then anyone could effectively block discussion by asking irrelevant questions, like on bautforum.com. This forum should adopt sciforum.com’s alpha rules, which are fair to both the thread starter and challengers.

 

I already answered, why insist? SR is excelently confirmed.

SR cannot be experimentally confirmed by your logic. You say that inertial frames must be infinitely small. Then no experiment can test SR. There’s no such thing as infinitely small measuring equipment. You can’t have it both ways. If SR is experimentally confirmed using our finite-sized measuring equipment, then inertial frames need not be infinitely small.

 

Show relevancy please.

Relevancy is that Thorne is widely considered to be an expert on black holes. It’s unlikely that he would have quoted Einstein’s words that he knew to be wrong. I’m going to have to trust Einstein, Taylor, Thorne, Wheeler, and the fact that SR is experimentally confirmed, over your position that SR requires infinitely small inertial frames and is therefore not experimentally confirmable.

 

Have you ever learnt how to find a straight line tangent to a circumference using rule and compass? Must you have an infinitely small rule and an infinitely small compass?

No, but that doesn’t show how SR could be experimentally confirmed when inertial frames must be infinitely small.

 

You did not address what I followed it with, which started with a more meaningful statement which I believe was what you were trying to say. You exhibted no valid basis of objection to saying 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. And who said that Einstein, Taylor, Thorne, and Wheeler are all barefaced liars?

...

Exactly in which sense does the equivalence principle equate X to any other inertial frame? The EQ says nothing about parallel transport, even less about the Schwarzschild r and t coordinates, so it can't be in contradiction of the statement you are attempting to criticize.

The EQ equates any small freely falling frame in our real, gravity-endowed universe to an inertial frame in SR’s idealized, gravity-free universe. X is given to be an inertial frame, which Taylor, Thorne, and Wheeler define as a freely falling frame small enough that the tidal force in it is negligible for the purposes of an experiment. Then the EQ equates X to an inertial frame in SR’s idealized, gravity-free universe, and by transitivity (if a=b and b=c, then a=c), X is equivalent to any other inertial frame in our real, gravity-endowed universe. Thorne confirms this in the supporting info in his statement of the relativity principle.

 

The EQ need not say something about parallel transport, or about the Schwarzschild r and t coordinates, to show that your comments are irrelevant. Since X equates to any other inertial frame, SR applies in X. If GR contradicts SR in X then GR is self-inconsistent. You can’t refute the OP by showing that GR contradicts SR in X; you can only help make my case that way.

 

Who talked about mod action? Me?

No, you’ve been good so far, but it is common practice on some other forums for mods to censor threads they disagree with, even when their disagreement is unfounded.

 

Follow Will's good advice and get yourself a proper course in GR, perhaps first one in calculus and then the basics of differential geometry.

First show me a bona fide problem with the OP. Doubtful you’ll do that by disagreeing with all of Einstein, Taylor, Thorne, and Wheeler.

 

Well, you've managed to find a true example of a formal system which is self inconsistent, by adding an axiom that contradicts those to which you added it. Congrats. In this case it actually does contradict the others, it doesn't just simply restrict possibilities, a distinction I was hoping to get across, someday.

 

And you have the chutzpah to tell Will that he should learn logic! :xx:

Yes, I easily did exactly what Erasmus said was impossible. It’s an example of a self-inconsistent theory, so your :xx: is groundless.

 

If GR reduced to nothing but SR in X (i.e. locally), then GR could be self-consistent. But the OP shows that GR overrides SR’s predictions in X; otherwise the box and the particle could be at rest relative to each other, in which case the box would be passing outward through the horizon, in contradiction to the definition of a horizon. GR does not reduce to nothing but SR in X, like Erasmus claims. GR also requires that a system in X adhere to GR’s prediction for escape velocity, a concept that SR does not feature.

[Guest Zanket]

So finally, I would like to point out that when they talk about black holes on the Discovery Channel or on The Universe Show, and they say that:

 

"Once something goes into a black hole, it never comes out."

 

This is incorrect.. The correct saying would be:

 

"Nothing ever goes into a black hole and nothing ever comes out".

 

This is the result of the time coordinate singularity and the fact that the geometry is completely time reversal symmetric.

But if you reverse time so that clocks run backwards, it's a white hole, not a black hole. With black holes, time marches forward only. You can't say something about a black hole by changing it to a white hole. Your logic is analogous to saying, "A black car would not be black if it was white, therefore it's not black."

[andrewgray]

Zanket,

 

I can feel your frustration. Sorry about that. You have helped me gain a much better understanding of accelerated coordinates and I now thoroughly understand the Schwarzchild horizon. So believe me, that is appreciated.

 

So I guess I wish I could hear you say, "I now understand what you have been saying Andrew, but perhaps I disagree".

 

But I just don't think you understand yet, so since you have helped me, I guess I am not ready to give up.

 

So here is where I believe that we are. Tell me which step you have a problem with:

 

1) The Schwarzchild geometry reduces to a Rindler geometry in a small neighborhood overlapping the horizon.

2) The Rindler geometry is equivalent to a flat geometry.

3) Therefore, the trajectories of the paths of objects near the horizon is the reverse mapping from:

 

[math]\text{flat spacetime route} \right \text{Rindler route} \right \text {Schwarzchild route}[/math]

 

which was shown in some of the previous posts.

 

Now I understand that you have said that all you need to show is just one inconsistency.

 

OK. So now with all this new knowledge in hand, we go back and reference your original post:

 

GR predicts that the inertial frame X can fall through the horizon of a black hole.

 

Correct.

 

GR says that SR applies in X.

 

Correct.

 

But GR contradicts SR by not allowing the box, which straddles the horizon, to be at rest relative to the particle, which is above the horizon and escaping to r=infinity.

 

Zanket, this is where I showed, in both cases that it can. For the case that you typically think about, it is here at this link.

 

For a Rindler Horizon in flat spacetime, the ball inside can stay equidistant from the outside ball forever, with the horizon in-between the two balls. For the Schwarzchild horizon, the ball inside can stay equidistant from the outside ball for however long the balls are in the local inertial frame. What fooled you into thinking that there was an inconsistency is that even though the outside ball has increasing r coordinates, and the inside ball has decreasing r coordinates, they still stay equidistant in the inertial frame, as shown in the link. We are only concerned with the balls staying equidistant between, say r=2M-2ε and r=2M+2ε. In other words, in the case you usually think about, the inside ball is going nearly lightspeed outward, the horizon is moving at lightspeed outward, and the outside ball is moving at nearly lightspeed outward. For the duration of the inertial frame, the two balls stay equidistant, with the horizon gaining just slightly on the outside ball, and slightly leaving the inside ball behind. But the two balls remain equidistant with the horizon in-between. Remember, a small inertial frame in spacetime has time limitations as well as space limitations.

 

GR predicts that the box cannot pass outward through the horizon.

 

Not Correct. This is the other case. For this case, which I am not completely convinced that you understand, the two balls both just leave the horizon directly. Shown here in the second part of this link. What you do not understand here is that the inside ball can never reach certain events (t, r=2M+ε), but it can reach other events (t', r=2M+ε). The horizon separates the inside ball from a set of events outside the horizon, but there is a set of events outside the horizon that the inside ball can get to, as shown in the link. So finally, one must say that for the inside ball, there exists a Schwarzchild horizon relative to a certain set of events outside r=2M. But there is a set of events outside r=2M that the inside ball can make it to.

 

Andrew A. Gray

 

I hope this helps.

[Guest Zanket]

Andrew,

 

There’s no frustration on my part. Just scientific thinking.

 

I already refuted both of your cases, and you haven’t refuted me in return. A brief rehash:

 

Your first case is refuted here.

 

Your second case is refuted here.

 

I pointed out that the only way you can show that GR does not predict black holes is to show that the relativistic rocket equations allow something to pass outward through r=2M. You must use that method because I have used those equations here and here to show that GR predicts black holes (in agreement with the Rindler Horizon site), and I need only show that GR predicts black holes in one way to refute your claim to the contrary. You replied:

 

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".

I pointed out here that t is defined as elapsed time. It can be less than zero only when clocks run backwards. But clocks always run forwards for a black hole, and a black hole is what the OP is about. So I need not show where the horizon is for t<0. I need not be concerned with a white hole or any other case where clocks run backwards. You’ve not refuted that, which leaves us with me having refuted your claim that GR does not predict black holes.

 

Of course I appreciate your input too and I wish you Godspeed if I’m wrong. But I’d be wary of putting forth an argument that contends that hundreds of physicists all made the same mathematical mistake. If there is a problem with GR intrinsically, it is almost certainly very subtle. The problem I point out with GR in the OP is so subtle that people cannot see it even when they’re staring right at it. For example, Erasmus is insistent that GR cannot be self-inconsistent when GR reduces to SR locally. He cannot see that GR can reduce to SR locally and contradict SR locally.

[andrewgray]

Zanket,

 

OK, the elapsed time is t for the rocket equations.

 

But no Schwarzchild horizon before t=0?

 

So is t=0 the creation of the black hole?

 

Isn't [math]t \right t'+C[/math] a valid transformation, such that we put our time "starting point" at any arbitrary time?

 

Aren't the Schwarzchild metric and the Rindler metric independent of t, thus independent of the start time?

 

That is, we can put any time as t=0, and consider any time before that (t<0)???

 

If t=0 has your meaning, doesn't it have to be the creation at the big bang?

 

Andrew A. Gray

[Qfwfq]

I decline. If it worked that way then anyone could effectively block discussion by asking irrelevant questions, like on bautforum.com.

Zanket, I'm really not trying to block the discussion with irrelevant questions, you're wasting far more time by refusing. I would agree with the intent of those alpha rules but they would have to solve the problem of someone insisting on relevance only to beat around the bush in the face of argument which corners them, as well as possible attempts at shifting the onus of proof. It's a question of the type with a yes or no answer and it's implicit that you could answer "I don't know" so it's not by answering that you'll let the discussion be blocked. You could complain only about long contorted rambling or of being forced into it. However, when the problem is due to your misunderstanding of what you presume to criticize (which is a bona fide problem with the OP), it is not up to me to teach you differential geometry and a proper understanding of GR when you only refuse to accept it.

 

Now, I'm not going to discuss the matter considering either answer as possible and the answer to the question is obviously "No." so I'll discuss the matter if you don't try to say otherwise. What's more you have already wasted my sparse online time and I'll scarcely be here in the next couple of weeks, so you may as well think very, very, very carefully before replying and perhaps do a bit of reading.

 

Anyway you don't comply with that alpha rule in:

Relevancy is that Thorne is widely considered to be an expert on black holes. It’s unlikely that he would have quoted Einstein’s words that he knew to be wrong. I’m going to have to trust Einstein, Taylor, Thorne, Wheeler, and the fact that SR is experimentally confirmed, over your position that SR requires infinitely small inertial frames and is therefore not experimentally confirmable.

Call that relevance! Thorne is widely considered to be an expert on black holes. Therfore those words in your quote of him, which you claim to be Einstein's own without support, must definitely override direct references to das Original, Einstein's very paper on GR, and a translation of it by Princeton University Press. Hats off to your relevance. On page 154 of the translation, right at the beginning of A, § 4 you find his own first statement of the equivalence principle:

For infinitely small four dimensional regions the theory of relativity in the restricted sense is appropriate, if the coordinates are suitably chosen.

So, Einstein says "infinitely small four dimensional regions" but Thorne quotes him saying "small, freely falling reference frame anywhere...". Therefore Einstein is inconsistent with Einstein. Therfore Einstein is self inconsistent. Therefore GR, being based on Einstein, is self-inconsistent. How do you like that logic?

 

No, but that doesn’t show how SR could be experimentally confirmed when inertial frames must be infinitely small.

Yes it does.

 

 

The EQ equates any small freely falling frame in our real, gravity-endowed universe to an inertial frame in SR’s idealized, gravity-free universe. X is given to be an inertial frame, which Taylor, Thorne, and Wheeler define as a freely falling frame small enough that the tidal force in it is negligible for the purposes of an experiment. Then the EQ equates X to an inertial frame in SR’s idealized, gravity-free universe, and by transitivity (if a=b and b=c, then a=c), X is equivalent to any other inertial frame in our real, gravity-endowed universe. Thorne confirms this in the supporting info in his statement of the relativity principle.

You still haven't given a meaning to the word equate in there although it was already obvious you meant it as some equivalence relation, therefore compliant with the transitive property. But what exact equivalence relation? It has to do with parallel trasport.

 

Yes, I easily did exactly what Erasmus said was impossible. It’s an example of a self-inconsistent theory, so your :) is groundless.

My :eek2: was about your chutzpah, regardless of whether Will did say such a thing. Get the difference between that example and the argument you uphold for self-inconsistence of GR.

 

My time here is up, see you soon and apologies for ant typos.

[Guest Zanket]

Zanket, I'm really not trying to block the discussion with irrelevant questions, you're wasting far more time by refusing. ...

Your question just leads to obfuscation. The beauty of “SR applies locally”, GR’s postulate, is that locally SR, and SR alone, can be used to make predictions. Your question implies that more than SR is required to make predictions locally, but that contradicts GR’s postulate, so there’s no point in answering it.

 

... You could complain only about long contorted rambling or of being forced into it. However, when the problem is due to your misunderstanding of what you presume to criticize (which is a bona fide problem with the OP), it is not up to me to teach you differential geometry and a proper understanding of GR when you only refuse to accept it.

But you haven’t shown a misunderstanding of GR on my part.

 

On page 154 of the translation, right at the beginning of A, § 4 you find his own first statement of the equivalence principle:So, Einstein says "infinitely small four dimensional regions" but Thorne quotes him saying "small, freely falling reference frame anywhere...". Therefore Einstein is inconsistent with Einstein. Therfore Einstein is self inconsistent. Therefore GR, being based on Einstein, is self-inconsistent. How do you like that logic?

It makes sense when you realize that Einstein is really saying that SR is more accurate the smaller the frame, as many references (including some in the OP) explain.

 

Yes it does.

No, it doesn’t. Sorry, you can’t test a theory that requires infinitely small frames with finitely-sized measuring equipment; the equipment is too large.

 

You still haven't given a meaning to the word equate in there although it was already obvious you meant it as some equivalence relation, therefore compliant with the transitive property. But what exact equivalence relation?

Whatever! When people can’t find a bona fide fault with an idea in these forums, they tend to question the meaning of words like “equal”.

[Guest Zanket]

OK, the elapsed time is t for the rocket equations.

 

But no Schwarzchild horizon before t=0?

 

So is t=0 the creation of the black hole?

The t=0 is the moment an object is dropped by a hovering observer. The t is the time interval in the dropped object’s frame since being dropped.

 

...That is, we can put any time as t=0, and consider any time before that (t<0)???

That’s like starting the stopwatch before the race starts. Doesn’t seem to be any benefit to that.

 

If t=0 has your meaning, doesn't it have to be the creation at the big bang?

No. Let the black hole exist. Let someone hovering above and arbitrarily close to its horizon drop an object. Start the dropped object’s clock at t=0 when it’s dropped. At t=0, let a light pulse be emitted directly upward from the horizon. The equations show that in the dropped object’s frame the pulse never reaches the hoverer; rather it asymptotes to the hoverer’s position, staying below the hoverer. The hoverer is arbitrarily close to the horizon, so to never reach the hoverer the pulse must stay precisely at the horizon, just like the books say.

[andrewgray]

Zanket,

 

Glad to see you are back. It has been interesting. I actually have written a paper and will be submitting it for publication. It is your last chance to get some credit. The publisher does need at least a first initial and contact info. I respect your privacy, so don't say I didn't try to give credit where credit is due if you rightfully decline.

 

OK, onward. Here is the Rindler diagram of your scenario.

 

 

The hovering Schwarzschild observer is at r=2.0000032M (purple). At t=T=0, he drops an object (black arrow pointed in T direction), headed towards the horizon. At the same time, a red laser is fired pointing outward from the horizon. So, I agree so far with everything that you are saying. However, t=-M and T=-.001M are still defined in this diagram, and the horizon (r=2M) is still there at these "negative-valued" times.

Agreed?

 

Andrew

[Guest Zanket]

The publisher does need at least a first initial and contact info.

You can call me Zan Ket, or Z. Ket, email [email protected].

 

However, t=-M and T=-.001M are still defined in this diagram, and the horizon (r=2M) is still there at these "negative-valued" times.

Agreed?

I see that in the diagram. What is the significance? What about it would change the fact that the light pulse never reaches the hoverer (thus the definition of "black hole" is satisfied)?

 

Think about this: in the dropped object's frame the pulse approaches the hoverer, but in the hoverer's frame the pulse remains the same distance away, albeit undetectable. How can that be? It's because in the hoverer's frame, time is stopped at the horizon (so nothing's moving there or even emanating photons), whereas in the dropped object's frame, the object's clock runs normally at the horizon.

[Qfwfq]

The beauty of “SR applies locally”, GR’s postulate, is that locally SR, and SR alone, can be used to make predictions.

You are omitting the words "to the first order".

 

Your question implies that more than SR is required to make predictions locally, but that contradicts GR’s postulate, so there’s no point in answering it.

:confused: And I thought the connection was a fundamental and necessary part of the formalism of GR. I even thought it was essential in defining the intrinsic curvature, the very difference between SR and GR with gravity. I really didn't realize that, au contraire, it contradicts the principle of equivalence.

 

But you haven’t shown a misunderstanding of GR on my part.

As well as misunderstanding a proper formulation of GR, you've been unable to understand how I've been showing it all along. I'm not saying you know zero, only that your understanding isn't enough to sort out the inconsistence that you claim there to be.

 

It makes sense when you realize that Einstein is really saying that SR is more accurate the smaller the frame

Thank yo' for your most exquisite enlightenment young man, and good luck to you.

 

I realized now from the quote that I had linked to the original instead of the translation. :angel: Sorry, I replaced the link in that post and repeat it here:

http://www.alberteinstein.info/gallery/pdf/CP6Doc30_English_pp146-200.pdf

 

Now despite so brilliantly giving me the above clarification, you immediately go on to tell me:

Sorry, you can’t test a theory that requires infinitely small frames with finitely-sized measuring equipment; the equipment is too large.

You can't measure the derivative of a function that isn't linear --not absolutely exactly with a finite number of measurements-- unless you can suppose that function to be smooth enough. Do you understand calculus at all, let alone differential geometry? Physicists do this all the time and not just for special or general relativity. No serious physicist has ever made perfect measurements or claimed to have done so. Does this mean we have never tested any whatsoever theory of physics?

 

In any case, this dispute is ludicrous simply because SR has been confirmed in circumstances in which the effect of gravitation is totally unimportant. It would be less simple to confirm the principle of equivalence near the event horizon of Cygnus X-1 but this hasn't caused any impossibility of confirming SR for lack of infinitely small equipment.

 

Whatever! When people can’t find a bona fide fault with an idea in these forums, they tend to question the meaning of words like “equal”.

Especially when one of the many faults in your arguments is in the use of the word. Obfuscation indeed!

 

The equations show that in the dropped object’s frame the pulse never reaches the hoverer; rather it asymptotes to the hoverer’s position, staying below the hoverer. The hoverer is arbitrarily close to the horizon, so to never reach the hoverer the pulse must stay precisely at the horizon, just like the books say.

Given that it stays exactly at the horizon, it can't asymptote to the hoverer’s position. A plain contradiction.

[Guest Zanket]

No serious physicist has ever made perfect measurements or claimed to have done so. Does this mean we have never tested any whatsoever theory of physics?

 

In any case, this dispute is ludicrous simply because SR has been confirmed in circumstances in which the effect of gravitation is totally unimportant. It would be less simple to confirm the principle of equivalence near the event horizon of Cygnus X-1 but this hasn't caused any impossibility of confirming SR for lack of infinitely small equipment.

You’re arguing against yourself here, not me. Your contention has been that because SR requires infinitely small frames, SR does not apply in X. Now you’re suggesting otherwise. Either SR cannot be confirmed in X, or it can. Which is it?

 

Given that it stays exactly at the horizon, it can't asymptote to the hoverer’s position. A plain contradiction.

Seeing as how both of these predictions are those of GR, it would be GR’s contradiction, not mine. The contradiction is explained away in post #185 at “think about this”.