A Flaw of General Relativity, a New Metric and Cosmological Implications [Technical]

Professor_Fink wrote:

First, you need to specify what reference frame you're talking about at each step.

The reference frame is always clear.

As far as I can see, you are always looking at this from the frame of some central object towards which the particle is falling.

No. For example, section 2 discusses a free-falling particle’s velocity as directly measured at each altitude. Then the frame in which the measurement is done is either the particle’s (directly measuring the velocity of something at an altitude as it passes by) or that of some observer at an altitude (directly measuring the velocity of the particle as it passes by).

And indeed, from such a frame, you do not see an event horizon at the schwarzchild radius. The event horizon comes up when you look at the system the other way around, from the frame of the incoming particle, as time in that reference frame essentially stops at the event horizon.

In GR, in the frame of the infalling particle, nothing special need occur at the event horizon. You could have crossed one while reading this sentence.

There are other mistakes, but this seems to be at the heart of the matter.

I don’t see how you have shown any mistake.

Also, for what its worth, its _very_ easy to prove that the schwarzchild solution is unique.

It's the only vacuum solution which satisfies spherical symmetry and asymptotic flatness.

The derivation can be found at http://en.wikipedia.org/wiki/Schwarzschild_black_hole

Birkhoff's theorem is immaterial to my paper. It doesn’t prove that the Schwarzschild metric is a valid description of Schwarzschild geometry. It proves that it is “unique” only in the context of Einstein’s field equations. My paper shows that those equations lead to an inconsistency. By definition theories are not proved, so nothing proves that GR is valid.

Professor_Fink wrote:

If you have someone sitting at a specific altitude, they are a in a third reference frame and don't measure the same relative velocity between the other two frames.

Agreed. So get rid of the superfluous third frame, that of the central mass.

Any observer passing directly by an object can directly measure its velocity. So either the infalling particle can directly measure the velocity of something sitting at a specific altitude as it passes directly by, or vice versa.

Professor_Fink wrote:

Ok, but then the observer doesn't measure the velocity of the particle towards the central mass, but rather the velocity towards that point. And while the point in question may be static relative to the central mass, it is not the same frame.

The particle directly measures the velocity of something sitting at a specific altitude as it passes directly by, or vice versa.

The frame of the central mass is not mentioned in the paper.

Also, you keep moving the point your looking at since you are dealing with a different reference frame at each altitude. So if you are not measuring the velocity relative to the central mass, then you are measuring it at each point in a different reference frame, which is totally useless.

It is not useless for the velocity to be directly measured as I describe above, even for a succession of such measurements. Section 2 uses such measurements in a valid way to show that GR is inconsistent. Physics is primarily about predicting measurements. Any measurement is fair game.

Taylor and Wheeler devote a section of their book Exploring Black Holes to the same situation as described in section 2, that of a particle falling radially from rest at infinity, its velocity directly measured at each point by observers sitting at a specific altitudes as the particle passes directly by. From the Schwarzschild metric they derive an equation that predicts what any such observer would measure. That such observers each have their own frame is immaterial. You suggest that their analysis and the resulting equation are meaningless. Can you prove that it is?

Professor_Fink wrote:

I have not read the book, and so I cannot comment on the veracity of their analysis. They're smart guys, so I'm guessing that their analysis is probably correct, but I'm not arguing about their analysis, which I must assume is consistent with GR. I'm arguing with yours, which is not.

You said that measurements such as those in section 2 are useless. But you guess that T&W's use of such measurements is probably correct. Then I don't see how you have shown any problem of my paper. To make a compelling argument you need to be more specific. All I'm seeing is "you're wrong" with nothing to support that.

Son Goku wrote:

No I'm saying one must reach a reasonable standard. I have argued with people who have made these comments before, particularly the one in bold, the argument invariably goes nowhere, because they have an inbuilt aversion to rigour.

Your standard is not reasonable. It’s not an aversion to rigor; rather, it’s an aversion to an argument that says the rigor must exceed what is necessary to show an advance of physics. You have not proven that your standard is necessary to show an advance of physics.

And papers aren't written for a specific audience, they written for the entire community or anybody who can understand the mathematics.

Papers and any other text are written for the audience the author chooses. Mine is not written for those who think the rigor must exceed what is necessary to show an advance of physics. For example, mine is not written for those who would scoff at Einstein’s thought experiment showing relativity of simultaneity, simply because it has no math. By “not written for”, I mean that I expect to see no convincing arguments by those people that there is a problem with my paper along those lines.

I'm not. They are locally homotopic in the sense that there is a homeomorphism from a subset M of Schwarzschild spacetime containing the gravitational worldline to a subset O of Minkowskian spacetime containing the world line of an accelerating observer.

Yes, you are, and your second sentence does not change that. You said “yes” to this:

Zanket wrote:

The relativistic rocket equations, when applied to the “gravitationally effected rocket”, will return negligibly inaccurate results in a local frame (the spacetime of which is negligibly curved by definition in my paper).

Then you clearly agreed that the equations of motion for a relativistic rocket are also those for a uniform gravitational field, unless you want to nitpick that negligibly inaccurate is too inaccurate.

Son Goku wrote:

I already have, but you continuously misunderstand what I'm saying.

You have not refuted it. Obviously, if it asymptotes to c locally everywhere, then it asymptotes to c globally. I have not misunderstood what you are saying. Rather, I have refuted your point.

A collection of local results is not a global result.

For some types of local results that apply everywhere, a global result can be concluded. When every state or province has a speed limit of 100 kph or less, can I conclude that the maximum speed limit is 100 kph nationally? Yes I can.

You can't do that. Local results can not be built up to global results on a curved manifold.

Local results are not “built up”. Rather, they apply locally everywhere, allowing a global result to be concluded. Simply declaring that this can’t be done is insufficient; i.e. your argument is not convincing. I have given an even simpler example showing that it can be done, about speed limits above. Can you refute it?

General Relativity is a differential geometric theory. That would be equivalent to attacking Quantum Mechanics without any understanding of Hilbert Space or Operators.

A theory can be refuted by showing as little as a division-by-zero error. Do you disagree?

My point is that (and this is what makes your argument difficult to criticise) is that your entire analysis makes no comments about the event horizon. You're argument is very difficult to understand because you're not actually criticising General Relativity.

An analysis that refutes the existence of an event horizon need not assume that an event horizon exists. In general, an analysis that refutes the existence of something need not assume that it exists. Section 2 does “criticize” GR—it shows that it is flawed, by showing that it both precludes and predicts event horizons, which is inconsistent.

You have basic misconceptions of the concepts involved in General Relativity as well as its natural mathematical foundation. This leads me to believe that you do not fully understand what GR says.

You have not shown how. I have given clear examples against your points. Can you attack them directly instead of just questioning my understanding?

What are you talking about?

There I’m giving a clear-cut example of putting the cart before the horse, to show what you are doing and how to correct it.

You're entire paper involves sellotaping a bunch of local results together in a way that cannot be done.

Obviously not every part of the paper has something to do with local results. You’re generalizing. The only sections you’ve specifically mentioned are 1 and 2, and for those I’ve refuted your points. Can you refute my example about speed limits above? Until you can, I’m unlikely to be convinced that a local result that applies everywhere can never be used to conclude a global result.

You still haven't grasped it though. And globally is a more advanced differential geometric term than worldline is. Worldline is a basic Special Relativistic term, where as Globally is quite technical.

You missed my point, which is that you’re being overly technical. You can make your points far more simply, and I request that you try to do that for the sake of this argument. For example, we don’t need differential geometry to discuss my example about speed limits. Einstein didn’t need differential geometry in his thought experiments; that would have been overkill.

A particles velocity measured by a shell observer as it falls from past infinity (for simplicity) will never be measured as being higher than c.
Now, how does this show that there is no horizon when the shell observers only measure in the region r > 2M, which is an open set?

Great! This is the way to argue against the paper.

That the velocity will “never be measured as being higher than c” by shell observers is irrelevant. That “shell observers only measure in the region r > 2M” is irrelevant. What is relevant is that the velocity asymptotes to c. Then escape velocity also asymptotes to c (because, as section 2 notes, GR says that “above the Schwarzschild radius the particle passes each altitude at a directly measured velocity equal to the escape velocity there”), in which case escape velocity is always less than c, and then there are no horizons.

I do not need to assume that a horizon exists at r = 2M. I do not need to analyze the region r <= 2M. The simple meaning of an asymptote applies here: a line not touched by a curve. Then when escape velocity asymptotes to c, it is never c; it does not touch the line v = c. Eq. 4 in fig. 2 shows clearly that GR also predicts that escape velocity does not asymptote to c. Rather, there it has a limit of c at r = 2M.

Suppose a theory contains the following statements:

- Birds can fly, but only above water.
- Birds cannot fly at all.

Let a refutation of that theory show a picture of a bird flying above land. Is the refutation invalid because it ignores the “only above water” clause? Of course not. Rather, that clause is irrelevant, in the same way some of your clauses are. And yes, a simple bird theory example like this can make a logical, applicable point.

Professor_Fink wrote:

Either you don't understand me or you are deliberately misinterpreting me. Either way, it is hardly worth pusruing this point as it will only lead to you saying that a theory can never be proved.

I would not deliberately misinterpret you; I don’t play that game. It’s hard to understand you when you make contradictory or at least vague statements like “GR can't be proved, you are correct, at least in some sense, but you are missing the point that it can be proved (and has been proved) ...” Whatever proof you are talking about, it doesn’t affect my paper that you have shown. You haven’t shown proof that GR is consistent, for example.

This is complete drivel. If someone attains different results by being more rigorous, it is because you made a mistake in your analysis (i.e. some hidden assumption).

Someone who shows a “mistake” in my analysis by, say, claiming simply that thought experiments are not rigorous because they lack math, is not making a rigorous argument, because the claim is made without basis; the argument is not loophole free. And I say that such claim suggests a false requirement that one must be overly rigorous, because a thought experiment can be rigorous, and there is no need to be more than loophole free.

Nobody in this thread attained different results than me by being “more rigorous”. Rather different results were stated without a basis given, or else I have pointed out a loophole. That is, the arguments against the paper are not rigorous. I give an example of that below.

Let me quote the first few lines of section 2 for you:

Okay, but let’s be clear that you change the subject here midstream. You previously quoted and attacked section 1, yet here, in response to my quote saying that your attack does not apply to section 1, you quote section 2.

This clearly applies the 'uniform gravitational field' to the whole space. Which is wrong. Acceleration due to gravity changes with distance, even in a classical picture, and so you are building your arguement on faulty foundations. You cannot simply apply the equivalence to a rocket at every point and be considering the same rocket.

Well, it’s a test particle, not a rocket. In some cases one can use results of a uniform gravitational field to make a conclusion about the “whole space”. The situation in section 2 is one of those cases. A simple example: A nonuniform gravitational field is everywhere uniform locally. Then, given that the particle falls in a uniform gravitational field, I can conclude that it falls in a nonuniform field as well. Can you disprove that? Notice that it is irrelevant to my conclusion that “Acceleration due to gravity changes with distance”. That is likewise irrelevant in section 2.

(This is an example of where I have found a loophole in your argument, showing that it is not rigorous.)

I think I've adequately showed this to be blatently false.

Not so. You previously quoted section 1, which has nothing to do with a nonuniform field as you suggested. Your argument above, which I refuted by finding a loophole, is about section 2.

You're quoting pop sci at me (which I think is explicitly forbidden in threads marked technical, but which I'm willing to ignore). The trouble with this sort of thing is that peoples comments are taken out of context. I assume what they are refering to is the fact that you cannot get rid of a physical singularity by making a coordinate transformation.

I reject the notion that it’s pop science. I don’t believe that four top-notch physicists would say or quote the clause “break down” when it’s really the opposite. I don’t believe, for example, that Hawking would say that “the Einstein equations and all the known laws of physics break down there” when he really believes that they work fine there. Not everything that physicists say outside of a paper is automatically pop science. I see lots of papers on Google Scholar that use “break down” or “breakdown” in the same context. Are they all pop science too?

You've also completely misunderstood the point. You cannot get from r=0 out. Why do you think you should be able to? Why would that make it a point of failure. The reason the escape velocity is undefined is because there is no escape velocity.

You make a logical mistake here. When the “escape velocity is undefined” at r=0, you cannot conclude that I “cannot get from r=0 out”. What is your basis for that? GR says nothing about whether one can escape from r=0. So the real question is, why do you think I should not be able to? I’ve not said that I should be able to; I’ve said only that the escape velocity is undefined there.

The central singularity is a failure point because GR interprets that all objects at and below the Schwarzschild radius must fall to r=0, where its own equations, like for escape velocity, do not apply.

Almost every line of every post I have written on this topic as been refuting your claims.

You have not proven that GR is self-consistent; you’ve only alluded to a proof. And I have otherwise refuted all of your refutations of my claims.

Please show me where I have, even once, invoked Birkoff's theorem.

You have not invoked it by name. But Birkoff's theorem is the only proof that “the schwarzchild solution is unique” that I know of. If that’s not what you meant, then what did you mean?

No, as I have repeatedly said, you have not shown this because you proof is seriously flawed.

You haven’t shown how it is flawed, that I have not refuted.

Oh, come on!

Come on, what? I still don’t know what your point is there. Please enlighten me.

I can. Different parts of the nation are at different heights (like mountains and valleys which have slightly different gravitational fields) and different lattitudes (which are revolving at different speeds, due to the earths rotation) as such, if I stand in certain places, I can see cars breaking the speen limit, even though for an observer standing at the side of the road (just like your shell observers) the cars appear to be sticking to the speed limit. So putting together all the observations from observers very near the car does not explain why I see a different velocity, and so does not give a global description.

My analogy refers only to speed limits (like the value on the signs), not the speeds of the cars.

This is exactly the same mistake you are making in your paper.

Section 2 explicitly refers to only directly measured velocities. So the “mistake” doesn’t apply to it.

Son Goku wrote:

Zanket, physics is not searching things on google. Roger Penrose is one of the world's best General Relativists. He made several contributions to GR during the 60s.

Why digress? Again I ask, which Penrose’s Proof?

Alternatively, just give me any proof that my example above doesn’t work, the example that shows that a local result that applies everywhere can be used to conclude a global result, refuting your claim to the contrary.

It is so.
You have said that a theory can be proven incorrect by a single division by zero error. This shows a gross misunderstanding of mathematics as used in physics.
It also shows a gross misunderstanding of 1/0, which mearly implies a topological gap.

If the division by zero is not problematic, then it is not a division-by-zero error, i.e. the kind that leads to untenable false results or absurdities. Division by zero is not okay or tenable in every case.

There is nothing wrong with 1/0. It cannot be used to refute a theory. That you
think it can shows you have no knowledge of analysis.

I didn’t say “1/0”. I said “division-by-zero error”, which can be used to refute a theory.

Wait a second. The method in your paper only works for shell observers, it doesn't work for geodesic observers like the particle.
i.e. your method only works for people who don't fall into the hole. What the particle experiences is completely different to what shell observers predict.

What do you think “the method” is? If I want the particle to measure the velocity v of objects fixed at each altitude as they pass directly by, that’s my prerogative. It’s my thought experiment. What shell observers predict is immaterial. They don’t measure anything in my thought experiment

The particle doesn't see itself as moving at all. v doesn't asymptote at all to him.

The particle doesn’t have to see itself moving. v, the velocity the particle measures of objects fixed at each altitude as they pass directly by, does asymptote to c, for the reasons given in the paper. Can you refute it?

Note that in fig. 1, v asymptotes to c in the frame of the gantry, the observer in free fall.

And a series of shell observers who follow the particle as it falls, don't give the same results as the particle, because they have a boosted frame.
...
Directly measured free-fall velocity asymptotes to c in a nonuniform gravitational field as measured by shell observers.

Shell observers don’t measure anything in my thought experiment. Even if they did measure something, their results would be immaterial.

This says nothing about velocity locally in the particles own frame or about the existence of horizons globally.
Why do you think it does?

I don't suggest that it does.

Professor_Fink wrote:

Ok. Speed limits. Given the different reference frames mentioned in my original post about your analogy, different observers interpret the speed limits differently. The speed (magnitude of velocity) of some object (car, whatever) measured by someone beside the sign will be different from someone who is observing from the top of a nearby hill. So your analogy completely neglects the curvature of the space (ie hills and valleys which will correspond to different reference frames due to the slight differences in the earths gravitstional attraction).

Above I emphasized to you that my analogy refers only to speed limits (like the value on the signs), not the speeds of the cars. Those were my exact words, and their meaning is clear. Yet here you again mention the speed of a “car, whatever”, again taking the analogy out of context. Different observers do not interpret the speed limits differently. Every observer (distant, local, moving, or whatever) sees the same speed limit on the signs. For example, if the posted speed limit is 100 kph, every observer sees the value “100” on the sign. Nobody sees “50”, say, simply because the sign reads “100”, not “50”.

The is EXACTLY the mistake you are making in your paper. You are taking these locally measured velocities and throwing out any notion that the space may not be flat.

Section 2 of the paper does not throw out “out any notion that the space may not be flat”. Rather, that spacetime is curved along the particle’s path does not affect the conclusion, just as it doesn’t affect the conclusion in the second analogy I gave; more on that below.

The entire derivation of the field equations gives testament to this.

Einstein’s field equations are not derived, as I note in the paper (and give a reference for) near the bottom of section 3. The equivalence principle and special relativity stand alone, independent of the field equations. The field equations are supposedly compatible with them, but there is no proof of that. Even for derived field equations that match Einstein’s, my paper would show that they are not derived correctly.

Not if the thought experiment was massively flawed by their complete lack of understanding of the theory they claim to be debunking.

No, even for a hypothetical massively flawed thought experiment, an argument that claims simply that thought experiments are not rigorous because they lack math, is not making a rigorous argument, because the claim is made without basis; the argument is not loophole free. Such argument does not even refer to a specific thought experiment.

You were arguing against the merits of rigor. I was countering your argument

No I did not argue against the merits of rigor. Above I said, “I don’t think there’s something inherently wrong with any level of rigor”. I was arguing against a nonrigorous argument that “suggests a false requirement that one must be overly rigorous”.

No, I'm not. You claimed that you only used those equations in section 1, and hence claimed that my criticism didn't apply to section 2, both of which are blatently untrue.

Nowhere did I claim that I “only used those equations in section 1”. Quote me otherwise.

Your criticism initially quoted only section 1, saying, for example, “We are not talking about a uniform gravitational field”. But section 1 is only about a gravitational field that is uniform. When I called you on that, you switched to section 2. You did change the subject midstream. If you were talking about section 2 initially then you should have quoted section 2 initially.

so if the particle is let drop it reaches different velocities as its potential energy is converted to kinetic.

That the particle “reaches different velocities” does not refute my conclusion that the particle “falls in a nonuniform field as well”. My conclusion clearly does not claim or imply that the particle cannot reach different velocities. That spacetime is curved along the particle’s path does not affect my conclusion.

I have to say, you have taken so many of my comments out of context that it seems it may be a deliberate method of arguing. I’ll give you the benefit of the doubt, but I respectfully ask that you read my comments more closely before commenting.

Son Goku wrote:

You are ignoring the issue of the curvature of spacetime in a paper attempting to disprove General Relativity?
General Relativity is all about the curvature of spacetime.

Let’s review my quote you responded to here:

Zanket wrote:

Rather, that spacetime is curved along the particle’s path does not affect the conclusion

Am I “ignoring the issue of the curvature of spacetime”? No. Instead it is obvious that I explicitly address that issue, saying that it “does not affect the conclusion”.

Son Goku wrote:

Assymptote's are defined within open sets.
It asymptotes up to the event horizon. However in general, looking at the whole space there is no asymptote. I'm trying to get you to see this.

I added a reader comment to the paper for this:

Reader: The velocity v asymptotes to c only as low as the Schwarzschild radius, the event horizon.

Author: The analysis above shows that v always asymptotes to c; i.e. as long as the particle falls. General relativity cannot demand otherwise without being inconsistent.

Neither you nor GR can force an inconsistency.

Escape velocity is a totally Newtonian concept, it is no surprise that it is not defined after the horizon.

Escape velocity is not solely a Newtonian concept. It is obviously also an Einsteinian concept; otherwise GR would not define it anywhere.

You are using intuitive notions of speed, such as escape velocities to a manifold where the concept can only be used up to a certain point.
And no it is not okay because your analysis only goes up to that point.

The analysis in section 2 shows that, in a theory consistent with section 1, neither v nor escape velocity can asymptote to c just up to a point beyond which the particle keeps falling.

Also remember that at the horizon the escape velocity is c. It only asymptotes up to the horizon. If you include the horizon, it doesn't asymptote.

That puts the cart before the horse. If v always asymptoted to c then so would escape velocity (as section 2 shows), in which case escape velocity would always be less than c and then there would be no event horizons. You haven’t refuted my (section 2’s) basis for that. Your attempt to do so above was refuted.

To try to reduce this common illogical contention, I changed a statement in section 2, from “According to general relativity, for any r above the Schwarzschild radius, v equals the escape velocity there” to “According to general relativity, when v is less than c at an r-coordinate r, it equals the escape velocity there”. The second statement is just as valid, but may reduce the frequency of arguments that insist that an event horizon exists despite logic which shows otherwise and takes precedence.

I don't understand why you think it is something amazing or profound that escape velocity never exceeds c above the horizon.

Nowhere did I suggest anything like that. Section 2 shows that escape velocity cannot be c anywhere in a theory consistent with section 1 (which was inferred by means GR allows), hence a theory consistent with section 1 precludes event horizons.

Professor_Fink wrote:

Either way, it is the mistake you are making. You are just throwing out information about the gravitational field by saying "a non-uniform field is locally uniform everywhere". You are removing all the curvature and simply imposing a constant radial force. This is not how gravity behaves.

What you paraphrased of mine there is valid. You have not shown that I am “removing all the curvature and simply imposing a constant radial force”.

Yes, it does affect it, and this is where you are getting errors.

You have not shown how it affects the conclusion of the analogy. I repeat the analogy in question here for reference, since you didn’t quote it:

Zanket wrote:

A simple example: A nonuniform gravitational field is everywhere uniform locally. Then, given that the particle falls in a uniform gravitational field, I can conclude that it falls in a nonuniform field as well.

The conclusion is, “it falls in a nonuniform field as well”. How does spacetime curvature affect that conclusion? It obviously does not affect it; i.e. it does not change the conclusion.

Professor_Fink wrote:

I beg to differ. The equations are derived from the equivalence principle and the least action principle (the latter holds in Newtonian physics too).

Then you disagree with Taylor and Wheeler. It doesn’t matter; whatever derivation the field equations have does not prove that they are correctly derived from general relativity’s standalone components.

The arguement is that the tought experiment is massively flawed. The fact that it lacks rigor is just an obstacle in seeing the flaws. Its much harder to point to a specific line in a very hand-wavy paper than in a rigorous one.

Basically you’re saying that it’s so flawed that it’s hard to show it’s flawed. That’s a weak argument. Anybody could say that without even reading my paper.

With pleasure:

"Your point is irrelevant because section 1, the only section being discussed in the quote you gave, has nothing to do with Schwarzschild’s solution. That section references only a uniform gravitational field." – Zanket

What you quoted obviously in no way indicates that I “only used those equations in section 1”. More on that below.

No I didn't. First you started telling me that my arguement only applies to section 1, which is demonstrably, as I have shown, false.

I didn’t say that your argument only applied to section 1. Please quote me.

Section 2 contains the problems, and when I point this out you claim I'm changing the subject. So deep in this thread the only people likely to read the discussion are you, me and Son Goku. I'm trying to be helpful, and I have nothing to gain by shooting down arguements if they are correct. The problem is that yours aren't.

You referred to section 2 for the first time only midstream, effectively changing the subject. When you refer only to section 1 initially then I cannot be expected to know that you meant section 2.

You completely neglect curvature. You claim that doing so makes no difference. What I presented is a counter example. A very simple case in which your assumptions are clearly shown to be false.

I do not “completely neglect curvature”. The spacetime of a nonuniform gravitational field is curved. Then the analogy features curved spacetime. You have not shown that its conclusion is false. The analogy shows simply that a local result that applies everywhere can be used to conclude a global result, refuting your claim to the contrary.

Zanket, I have absolutely nothing to gain by taking your comments out of context. I have nothing to prove here. I am only trying to help you. There are clear mistakes in the paper, and I am trying to point these out, so that you can learn from them.

That’s good to know. But why then do you so often and obviously take them out of context? Maybe you just can’t see that you’re doing that. Your “With pleasure” example above is a prime example. Clearly, nothing at all in the quote of mine you gave indicates I “only used those equations in section 1”. And yet it is your direct “proof” of that.

If I have taken comments out of the context you intended it is only because I am unclear on what you mean by many of your statements. Your style of writing is very opaque and it is difficult to try to extract a coherent arguement.

Let’s take this quote of mine for example: “Your point is irrelevant because section 1, the only section being discussed in the quote you gave, has nothing to do with Schwarzschild’s solution.” Please tell me how you gathered from that quote that I “only used those equations in section 1”? My quote seems pointed to me. I don’t see how it could have been made any simpler or clearer. And yet you took it out of context.

Son Goku wrote:

Zanket I'm considering closing this thread if you continue with comments like this. All you have done is used a local result and proclaimed that it works as a global result. The fact that you have used it is not proof that it is valid. Unless you make a turn around in the way you argue I will close this.

Whenever a mod threatens to close the thread due to disagreeing with the opinions expressed within, which is a no-no in my book, I stop responding to the other posts to focus on that. To do otherwise just wastes my time. (Try to put yourself in my position to see how.) I know from experience that it won’t matter what case I present. Simply because you disagree you will close the thread. The thread is no longer about scientific discussion, but rather about ego. So I am considering no longer responding to this thread.

What I have done is shown, i.e. with basis, that a local result that applies everywhere can be used to conclude a global result. I have not simply “used a local result and proclaimed that it works as a global result”. (italics mine)

But it won’t matter that I have given a basis, and it won’t matter that you cannot refute it. Experience shows that you will still close the thread. It's convenient being the mod in the discussion, eh?

Let’s dissect my analogy, which is:

Zanket wrote:

A nonuniform gravitational field is everywhere uniform locally. Then, given that the particle falls in a uniform gravitational field, I can conclude that it falls in a nonuniform field as well; i.e. across any range of r-coordinates.

First, drop any assumption that a particle falls in a nonuniform field. We need not assume that, even though it might seem obvious. Then take the first statement:

Zanket wrote:

A nonuniform gravitational field is everywhere uniform locally.

Hard to argue with that, given the definitions in the paper.

Zanket wrote:

Then, given that the particle falls in a uniform gravitational field, ...

This is a given, so no arguing with it. And, given the first statement, this is a local result that applies everywhere.

Now the conclusion:

Zanket wrote:

... it falls in a nonuniform field as well; i.e. across any range of r-coordinates

This conclusion follows unequivocally from the first two statements. Then I have shown that a local result that applies everywhere can be used to conclude a global result, refuting your claim to the contrary.

Son Goku wrote:

Half of your explanations just amount to telling us you didn't do something or that something works.

I do that only where it is appropriate. It’s just been appropriate often.

Let’s take an example:

Son Goku wrote:

I don't understand why you think it is something amazing or profound that escape velocity never exceeds c above the horizon.

Zanket wrote:

Nowhere did I suggest anything like that.

Is my statement true? Yes. The first use of the words “amazing” or “profound” in this thread are yours. Nowhere have I used a similar word. You misrepresented me.

He is saying that the flaws are entire paragraphs so it is hard to point to any specific area and say what is wrong with it.

Anyone can say that without even reading the paragraphs, so it’s a pointless argument. And it’s a ridiculous argument. So we can have whole theories that nobody can find a flaw of, because the whole thing is wrong? Give me a break. As a mod, you should be rebuking those who make such arguments, instead of the opposite.

Son Goku wrote:

Never once did you manage to fully understand their arguements, particularly ZapperZ's one about your use of the equivalence principle.

In post 62, page 5 of that thread ZapperZ expresses the belief that the frame of a ball falling toward a planet in a uniform gravitational field cannot be inertial, simply because the ball is accelerating relative to the planet. That contradicts not only Einstein and the equivalence principle, but also the definition of an inertial frame by Taylor, Thorne, and Wheeler. When I asked him to clarify his belief, he stopped responding and soon thereafter closed the thread.

ZapperZ may be great at whatever type of physics he does for a living, but it seems he doesn’t have even a layman’s understanding of general relativity. (And he’s certainly a poor mod.)

I would never close something because I disagree with it. This is a technical thread and you are not making a technical standard.

That’s unconvincing from someone who believes that thought experiments are insufficient to advance physics by themselves. You are expressing the bias of a mathematical purist, no more. Since you don’t support your claims, your opinion that I am not “making a technical standard” is indistinguishable from simple disagreement. It’s just a lame excuse for when you close the thread.

You should learn differential geometry and then go back and attempt to understand Einstein's theory, not attack a water downed version from a text designed for interested secondary school students.

What’s really happening is that you can’t refute my points and refutations of your points, so you consistently ignore them in favor of cheap shots like referencing some other forum and claiming without basis that I don’t understand the theory. Then I don’t expect it to have any affect on your position that my paper uses only a bit of high school algebra to derive a new metric that is shown to be confirmed to all significant digits by all experimental tests of Schwarzschild geometry, which I'd have to be a miracle worker to do without understanding the theory.