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

Son Goku

Zanket wrote:

Which Penrose’s Proof? The first one I see involves artificial intelligence; i.e. it is inapplicable to this topic.

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.
But more importantly......

That isn’t so.

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.
There are millions of functions which lead to 1/0. This is useful, for instance in complex analysis. Particularly contour integration and Laurent series where it is totally expected in most of the functions you work with. There is nothing wrong with 1/0.

For instance the integral of 1/{z(z-2)^4} is -iπ/8, because there are 1/0 points at z = 0 + 0i and z = 2 + 0i. If there weren't the integral would be 0.
Integrals like this come up all the time in QFT. 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.
The very area of mathematics needed for your paper.

The observer need only be the particle,

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.

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

“Section 1 shows that directly measured free-fall velocity asymptotes to c in a uniform gravitational field. Then v asymptotes to c locally.

Directly measured free-fall velocity asymptotes to c in a nonuniform gravitational field as measured by shell observers. This says nothing about velocity locally in the particles own frame or about the existence of horizons globally.
Why do you think it does?

Zanket

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

Zanket wrote:

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

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

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).

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.

Son Goku

Zanket wrote:

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?

Of course because to hold themselves at a constant altitude they have to thrust upward at a constant rate, which gets higher and higher as you get closer to the horizon. At the horizon the velocity required is c. After the horizon this concept loses meaning as t and r interchange, i.e. after the horizon the velocity doesn't exist.
Why do you think this means there is no event horizon?
The whole point of the event horizon, in fact its loose definition is that within it you would require a velocity greater than c to stay at a constant r, because r becomes timelike.
Therefore you can't measure the escape velocity within the event horizon using other objects because no object can maintain that velocity.

Just because no observer can attain the escape velocity does not mean it has exceeded c.
It asymptotes to c up to the horizon and then surpasses it. The only difference is that before the horizon material objects can match the escape velocity.

The whole point of the inside of the hole is that nobody can attain the escape velocity.

Professor_Fink

Zanket wrote:

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.

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

Zanket wrote:

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.

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

Zanket wrote:

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.

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

Zanket wrote:

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.

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.

Zanket wrote:

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.

Yes I can.

Velocity for a particle falling from rest at infinite distance under a uniform gravitational field (i.e. constant force = m g):

Potential energy = mgh (for height h)

For a field obeying Newtonian gravitation (i.e. non-uniform) F = G m M / r^2

Porential energy = -GMm/h

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

Zanket

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.

Zanket

Son Goku wrote:

Just because no observer can attain the escape velocity does not mean it has exceeded c.
It asymptotes to c up to the horizon and then surpasses it. The only difference is that before the horizon material objects can match the escape velocity.

You violate the definition of an asymptote. A velocity cannot asymptote to a value and then surpass it. Again I note, the simple meaning of an asymptote applies here: a line that draws increasingly nearer to a curve without ever meeting it. When v asymptotes to c, the line, the asymptote, is v = c. Then the curve of v as a function of r has an asymptote at v = c, in which case v is always less than c, and GR tells us that escape velocity shares the same curve. Hence escape velocity is always less than c, and so there are no event horizons. It doesn’t matter that GR tells us that escape velocity shares the same curve only above the Schwarzschild radius—that is enough to conclude that the curve of escape velocity has an asymptote at v = c, hence it is enough to preclude event horizons. Then I am free to ignore the region r <= the Schwarzschild radius.

Son Goku

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

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.

You violate the definition of an asymptote.

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.

Zanket wrote:

It doesn’t matter that GR tells us that escape velocity shares the same curve only above the Schwarzschild radius—that is enough to conclude that the curve of escape velocity has an asymptote at v = c, hence it is enough to preclude event horizons. Then I am free to ignore the region r <= the Schwarzschild radius.

No your not. Why do you think you are? This is really flawed reasoning.
Escape velocity is a totally Newtonian concept, it is no surprise that it is not defined after the horizon.
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.
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.

This is equivalent to saying that Quantum Mechanics is disproven because angular momentum shouldn't have certain values based on Newtonian mechanics in the Hydrogen Atom.

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

Professor_Fink

Zanket wrote:

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”.

Ok, if you are only talking about numbers painted on signs, then my arguement is inapplicable, but we are nolonger dealing with something analogous to your paper. Since you refered to it as an analogy, the obvious choice is to consider observers at different locations, each one trying to enforce the speed limit. As far as I can see, that is the only analogy that makes any kind of sense. In this case, my arguement does hold.

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.

Zanket wrote:

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.

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

Zanket wrote:

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.

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

Zanket wrote:

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.

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.

Zanket wrote:

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

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

Zanket wrote:

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.

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. 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.

Zanket wrote:

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.

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.

Zanket wrote:

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.

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.

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.

Professor_Fink

Zanket wrote:

You violate the definition of an asymptote. A velocity cannot asymptote to a value and then surpass it.

Actually it can. This happens at singularities. You have a function which is analytic everywhere except at these singularities, and so is defined over the domain except at these points.

A very simple example is f(x) = 1/x. This has an assymptote at 0, but is defined on both sides of x=0.

secret_squirrel

Professor_Fink wrote:

So deep in this thread the only people likely to read the discussion are you, me and Son Goku.

Actually Im quite enjoying having 10 years of dust blown off my physics knowledge - carry on lads.

If its not off topic (if it is please delete this) can I check my understanding of your critiques of Zankets Document.

1. Zanket is trying to extrapolate a Newtonian scenario, involving special relativity over to General Relativity by essentially summing over all possible SR scenarios?

2. Zankets use and understanding of Frames of Reference is flawed?

Please correct me if Im wrong - by PM if this is off topic.

Son Goku

secret_squirrel wrote:

1. Zanket is trying to extrapolate a Newtonian scenario, involving special relativity over to General Relativity by essentially summing over all possible SR scenarios?

2. Zankets use and understanding of Frames of Reference is flawed?

Essentially, yes. Number 1 in particular is a very deep flaw.

Professor_Fink

Hi secret squirrel,

Guess I was wrong after all.

secret_squirrel wrote:

Actually Im quite enjoying having 10 years of dust blown off my physics knowledge - carry on lads.

If its not off topic (if it is please delete this) can I check my understanding of your critiques of Zankets Document.

1. Zanket is trying to extrapolate a Newtonian scenario, involving special relativity over to General Relativity by essentially summing over all possible SR scenarios?

2. Zankets use and understanding of Frames of Reference is flawed?

Please correct me if Im wrong - by PM if this is off topic.

Essentially Zanket's paper attempts to show a flaw in the Schwarzchild solution. The Scwarzchild solution is, according to general relativity the only spherically symmetric, assymptotically flat space time, describing the curvature of space around stars, planets and black holes. Zanket attempts to show that this is not a unique solution by making arguements from special relativity only, without resorting to general realtivity. If this were correct it would show some inconsistency in the field equations.

In his paper, Zanket considers the velocity of a passing particle as measured by observers sitting at fixed distances from the central mass. He argues that the particle never has a velocity greater than light for any of these 'shell observers' and hence there is no event horizon (i.e. black holes don't exist).

The major flaw in this reasoning is the fact that a) Zanket tries to extrapolate the entire space-time from the measurements of shell observers, and b) the curvature of the space is completely ignored (see all the comments along the lines of "a non-uniform field is everywhere local"). Neither of these can be done (although Zanket will no doubt claim that I am wrong about this).

Hope this was a clear enough summary.

Zanket

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.

Son Goku

Zanket wrote:

Let’s review my quote you responded to here:
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”.

How can it not? The particle moves on a geodesic.

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

In certain situations you can use the Newtonian concept of Escape velocity as a mental aid. However it is not defined in GR.
What is the escape velocity of at a point outside a Kerr-Newman hole for example?

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.

How does the particle always measuring an escape velocity to be less than mean there is no horizon?
How can you be confident the concept doesn't just break down?
You're doing mathematics on a curved manifold, this kind of stuff doesn't work.

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.

It is c at r=2M, because there the shell observer is light. Why are you excluding light as a shell observer?
What General Relativity then says is that after this there are no shell observers, hence no escape velocity.

Zanket

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

Zanket wrote:

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

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.

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.

Zanket

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

Whenever a mod threatens to close the thread due to disagreeing with the opinions expressed within

Zanket my main apprehension comes from seeing this thread:
11 page thread involving you and seven research physicists. Three Experimentalists and four Theorists.

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

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

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.

Zanket

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.

Professor_Fink

Zanket,

you cannot neglect curvature along the geodesics (the path along which the particle falls). You have proclaimed, repeatedly, that you can without offering any proof. The onus is very much on you to prove this assumption, not us to disprove it. Adding a reader comment along the lines of "Can curvature along this path be neglected? Yes." is not a proof.

The fact that you ignore this curvature does indeed remove the singularity at the event horizon, because you've replaced a relatively complex space time with a cone. This is because the acceleration is uniform, implying uniform curvature)

In this conical space time, a particle will indeed assymptote to c as it approaches the centre (if dropped from far enough away).

The problem is that this is not how gravity behaves, either is GR or Newtonian theory. If it did, then we would be pulled as strongly towards all solar mass stars as the sun, and would be hurtling towards the centre of the milky way. The known orbits of the planets are not consistent with a constant acceleration due to gravity as a function of radius, but are consistent with roughly a 1/r^2 relationship.

So if you could make that assumption, you would be right in saying that blackholes were impossible, etc., but the simple fact is that you are specifying a subset of spacetimes, which excludes the ones which really occur.

In fact, your assumption explicitly rules out general relativity (and indeed Newtonian gravity) right at the start, since it cannot be made to fit either the 1/r^2 dependence of acceleration on distance in the newtonian theory, or the line element predicted by GR.

I hope I have now made it clear enough for you to see why you are getting different results.

And don't say I disagree with Einstein, or Wheeler or some other big name, because you'd be wrong.

Professor_Fink

Zanket wrote:

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.



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.

Well I see this has degenerated to the level of insulting people that try to point out the errors in your paper.

But the underlined claim is easy to test:

Zanket, could you please give me the geodesics for a radially assymptotically flat space time containing an infinitely long cylinder of mass M per length L, and radius R?

If you can, then I will accept that you have at least a working knowledge of GR, the theory you are attacking. If you cannot, then I will have to assume that you do not really understand the theory.

Problem solved.

Son Goku

Zanket wrote:

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.

That would be impossible for somebody who studies condensed matter physics.

Now show us you understand GR by answering Professor_Fink's question.

To attempt to disprove and move beyond something you must first understand it in the language it was written in.

Zanket

Professor_Fink wrote:

Actually it can. This happens at singularities. You have a function which is analytic everywhere except at these singularities, and so is defined over the domain except at these points.

A very simple example is f(x) = 1/x. This has an assymptote at 0, but is defined on both sides of x=0.

No curve of that function surpasses 0.

Zanket attempts to show that this is not a unique solution by making arguements from special relativity only, without resorting to general realtivity.

No, I don’t do that. I acknowledge that the Schwarzschild metric is a “unique solution” of Einstein’s field equations for Schwarzschild geometry. I’ve made that clear above. It’s immaterial to my paper.

In his paper, Zanket considers the velocity of a passing particle as measured by observers sitting at fixed distances from the central mass.

In section 2 I say: “Let a test particle fall radially from rest at infinity toward a large point mass while measuring the velocity v of objects fixed at each altitude as they pass directly by.” The particle is the measurer.

He argues that the particle never has a velocity greater than light for any of these 'shell observers' and hence there is no event horizon (i.e. black holes don't exist).

No, section 2 doesn’t use that reasoning to preclude event horizons. See that for yourself.

The major flaw in this reasoning is the fact that a) Zanket tries to extrapolate the entire space-time from the measurements of shell observers, ...

Measurements of shell observers are not used.

... and b) the curvature of the space is completely ignored (see all the comments along the lines of "a non-uniform field is everywhere local").

No, spacetime curvature is not ignored, and nowhere do I make the nonsensical comment that “a non-uniform field is everywhere local”. Section 2 says “a gravitational field is everywhere uniform locally”, which is valid.

you cannot neglect curvature along the geodesics (the path along which the particle falls). You have proclaimed, repeatedly, that you can without offering any proof.

I do not neglect spacetime curvature, and I have not proclaimed that. What I did say is that spacetime curvature does not affect the conclusion in the analogy I gave you, that refutes your claim that a local result that applies everywhere cannot be used to conclude a global result. Nor does spacetime curvature affect the conclusion in section 2 that “v always asymptotes to c; i.e. as long as the particle falls”.

The basis I gave is simple. If it is refutable, then refuting it should be easy. Instead you take me out of context. Why?

The onus is very much on you to prove this assumption, not us to disprove it.

My paper already provides bases for its claims. Until someone refutes it, I have met the burden of proof. I have also given bases for all of my refutations in this thread.

Adding a reader comment along the lines of "Can curvature along this path be neglected? Yes." is not a proof.

No such reader comment ever existed in the paper.

The fact that you ignore this curvature does indeed remove the singularity at the event horizon, because you've replaced a relatively complex space time with a cone. This is because the acceleration is uniform, implying uniform curvature)

I do not ignore spacetime curvature anywhere in the paper. The gravitational acceleration along the particle’s fall as a whole in section 2 is nonuniform.

Well I see this has degenerated to the level of insulting people that try to point out the errors in your paper.

Funny. So it’s bad for me to call a spade a spade, but it’s okay for you to take me out of context, repeatedly spinning my words rather than bother to quote me or the paper.

Zanket, could you please give me the geodesics for a radially assymptotically flat space time containing an infinitely long cylinder of mass M per length L, and radius R?

Papers aren’t refutable by posing story problems to the author. But I’ll answer this one. In asymptotically flat spacetime, a geodesic is an all but straight line. The cylinder is immaterial because the spacetime is given to be asymptotically flat.

Zanket

Son Goku wrote:

How can it not? The particle moves on a geodesic.

How do you think the fact that the “particle moves on a geodesic” shows that the fact that spacetime is curved along the particle’s path necessarily affects the conclusion?

Let’s analyze section 2:

Zanket wrote:

Section 1 shows that directly measured free-fall velocity asymptotes to c in a uniform gravitational field. Then v asymptotes to c locally.

The second sentence is valid, given the definitions in the paper.

Zanket wrote:

The particle always falls in a uniform gravitational field since a gravitational field is everywhere uniform locally.

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

Zanket wrote:

Then v always asymptotes to c; i.e. as long as the particle falls.

When the other statements are valid then this one must also be valid. Then spacetime curvature neither affects the conclusion nor was ignored. The degree of spacetime curvature in the nonuniform gravitational field can be as negligible or as great as imaginable, yet v will still always asymptote to c; i.e. as long as the particle falls.

Son Goku wrote:

In certain situations you can use the Newtonian concept of Escape velocity as a mental aid. However it is not defined in GR.

Someone better tell that to Taylor and Wheeler, who say, on pg. 2-22 of the link I gave above, “Here we elbow Newton aside and give the relativistic answer: The maximum escape velocity is the speed of light”. Oh yes, I know that it’s pop science; everything in their books is a lie. And NASA is forced to use Newtonian mechanics when they launch a satellite.

How does the particle always measuring an escape velocity to be less than mean there is no horizon?

When the escape velocity is less than c along the particle’s entire fall, then event horizons are precluded, because they are a surface where the escape velocity is c.

How can you be confident the concept doesn't just break down?

Because the logic is simple, seemingly undeniable (without loophole), and has not been refuted.

You're doing mathematics on a curved manifold, this kind of stuff doesn't work.

You haven’t proven that “this kind of stuff doesn't work”. I’ve made simple arguments in the paper. They should be simple to refute, if you are correct here.

It is c at r=2M, because there the shell observer is light. Why are you excluding light as a shell observer?

Why are you contradicting yourself? You said above that escape velocity is “not defined in GR”. Now you’re telling me what its value is at r=2M in the context of GR.

You put the cart before the horse here. Section 2 shows that event horizons, surfaces where the escape velocity is c, are precluded in a theory consistent with section 1, which was inferred by means GR allows. GR cannot force an inconsistency. Then I need not assume that the escape velocity is c at r=2M. As section 2 puts it, “General relativity’s prediction of event horizons does not take precedence over logic precluding that prediction”.

What General Relativity then says is that after this there are no shell observers, hence no escape velocity.

Yes, now we agree. So escape velocity is defined in GR, just only for r>=2M.

That would be impossible for somebody who studies condensed matter physics.

No, that would be possible. This is ironic. You threaten to close the thread because you think I don’t meet some scientific standard. Yet you repeatedly try to disprove me using unscientific baseless opinions, like this one.

Why not be scientific instead? Can you support ZapperZ’s belief that the frame of a ball falling toward a planet in a uniform gravitational field cannot be inertial?

To attempt to disprove and move beyond something you must first understand it in the language it was written in.

No, I already refuted this conjecture above. A theory can be refuted as simply as pointing out a division by zero error. You did not disprove that. Your attempt to do so erroneously omitted the “error” qualifier.

Professor_Fink

Zanket wrote:

Papers aren’t refutable by posing story problems to the author. But I’ll answer this one. In asymptotically flat spacetime, a geodesic is an all but straight line. The cylinder is immaterial because the spacetime is given to be asymptotically flat.

Actually, it's a very useful problem. But you got the answer completely wrong. Assymptotically flat means flat as r->infinity, and curvature does matter in the presence of a central mass.

Also, should you wish to retry, you can just give me the line element. Wordy answers are not necessary, or welcome. Just something along the lines of:

ds = f(theta,phi,r)

I just want to know the function f(...). Answer that and I will accept that you at least know enough general relativity to be able to ask google for the answer, which would be a start, at least.

Professor_Fink

Zanket wrote:

And NASA is forced to use Newtonian mechanics when they launch a satellite.

Actually if its low earth orbit, it makes little difference. For GPS, etc., you have to make the GR corrections for the clock timings (GPS works by recieving the time simultaneously from a number of different satellites). Also you need relativistic corrections if you want to rendez-vous with some far off object.

Oh, and feel free to tell me I know nothing about NASA...:rolleyes:

Zanket

Professor_Fink wrote:

Actually, it's a very useful problem. But you got the answer completely wrong. Assymptotically flat means flat as r->infinity, and curvature does matter in the presence of a central mass.

Let’s review your question:

Zanket, could you please give me the geodesics for a radially assymptotically flat space time containing an infinitely long cylinder of mass M per length L, and radius R?

This can easily be misconstrued. If you wanted the geodesics in the immediately vicinity of the cylinder, as opposed to just in asymptotically flat spacetime, then the question should be worded something like: “please give me the geodesics in the immediate vicinity of an infinitely long cylinder of mass ...?” You don’t need to mention “asymptotically flat spacetime”; that the spacetime far from the cylinder is asymptotically flat can be assumed because the cylinder is the only object mentioned, and the spacetime far from the cylinder is immaterial anyway because you want the geodesics in the immediately vicinity of the cylinder.

Also, should you wish to retry, you can just give me the line element.

Nah, I decline to spend more time on this. My overall knowledge of GR is irrelevant here. My paper already shows a flaw of GR, using simple logic, and it has not been refuted. My paper already derives a new metric for Schwarzschild geometry that is shown to be experimentally confirmed to all significant digits, and that has not been refuted. Your question is outside the scope of my paper, and my refusal to answer your question does not show a flaw of my paper. So there’s no incentive for me to spend more time on this.

Keep in mind: I’m not here to convince you or anyone else. I’m here to see if anyone can refute my paper. Anyone who thinks that my paper is invalid because I don’t answer your question doesn’t convince me, for that is unscientific reasoning. Even if the thread is closed because I don’t answer your question, I’ve lost nothing.

Actually if its low earth orbit, it makes little difference. For GPS, etc., you have to make the GR corrections for the clock timings (GPS works by recieving the time simultaneously from a number of different satellites).

Agreed. My point was only that NASA is not forced to use Newtonian mechanics when they want to predict escape velocity. They can use GR, which is easy because the equation for escape velocity is shared by both theories for an r-coordinate as low as the Schwarzschild radius.

Professor_Fink

Zanket wrote:

This can easily be misconstrued. If you wanted the geodesics in the immediately vicinity of the cylinder, as opposed to just in asymptotically flat spacetime, then the question should be worded something like: “please give me the geodesics in the immediate vicinity of an infinitely long cylinder of mass ...?” You don’t need to mention “asymptotically flat spacetime”; that the spacetime far from the cylinder is asymptotically flat can be assumed because the cylinder is the only object mentioned, and the spacetime far from the cylinder is immaterial anyway because you want the geodesics in the immediately vicinity of the cylinder.

No, the question was precise. You only misconstrued it because you are not familiar with the language of relativity. I have to mention assymptotic flatness, as this cannot necessarily be assumed.

Also it is almost word for word a question from my finals paper in general relativity.

Zanket wrote:

Nah, I decline to spend more time on this. My overall knowledge of GR is irrelevant here. My paper already shows a flaw of GR, using simple logic, and it has not been refuted. My paper already derives a new metric for Schwarzschild geometry that is shown to be experimentally confirmed to all significant digits, and that has not been refuted. Your question is outside the scope of my paper, and my refusal to answer your question does not show a flaw of my paper. So there’s no incentive for me to spend more time on this.

Zanket, you are trying to show an error in a theory you clearly do not understand and are making assumptions that are not warrented (i.e. that the effect of curvature along the geodesic does not affect your conclusions).

As it stands, what you say implies a space time that is essentially a cone (although the curvature could change with theta/phi if you are only considering an inward falling particle). You say that this is wrong, but it isn't. You _have_ restricted yourself to unnatural spacetimes where the force of gravity is the same at any distance from the central mass.

Zanket wrote:

Keep in mind: I’m not here to convince you or anyone else. I’m here to see if anyone can refute my paper. Anyone who thinks that my paper is invalid because I don’t answer your question doesn’t convince me, for that is unscientific reasoning. Even if the thread is closed because I don’t answer your question, I’ve lost nothing.

We _have_ refuted you at every corner, you just keep saying that we haven't. And it seems more than clear that this is all an ego trip anyway.

Zanket wrote:

Agreed. My point was only that NASA is not forced to use Newtonian mechanics when they want to predict escape velocity. They can use GR, which is easy because the equation for escape velocity is shared by both theories for an r-coordinate as low as the Schwarzschild radius.

The Schwarzchild radius for an earth mass is << 6400km. Actually its about 9mm.

Zanket

Professor_Fink wrote:

I have to mention assymptotic flatness, as this cannot necessarily be assumed.

Go ahead and mention it, but put it in the correct spot in the problem.

Also it is almost word for word a question from my finals paper in general relativity.

If so, then it was badly worded there too. Considering how many times you’ve taken me out of context in this thread, you may well have done that for the story problem too. You said “...give me the geodesics for a radially assymptotically flat space time containing...” Then the spacetime is asymptotically flat regardless what it contains. I don’t even need GR to know that; I need only knowledge of grammar. Then I can assume that you want the geodesics far from the cylinder. I thought it was a trick question.

Zanket, you are trying to show an error in a theory you clearly do not understand and are making assumptions that are not warrented (i.e. that the effect of curvature along the geodesic does not affect your conclusions).

Yes, the spacetime curvature does not affect the conclusions I specified above. My logic is simple, and I went through it statement by statement above. And it’s only a few statements. It should be easy to refute directly rather than just claim without basis that I don’t understand the theory.

You _have_ restricted yourself to unnatural spacetimes where the force of gravity is the same at any distance from the central mass.

No. Nothing in section 2 suggests that. You’re making it up. If you were right, you could quote the paper, but you can’t. The particle in section 2 falls through curved spacetime, where g increases as the particle falls; nothing in section 2 suggests otherwise.

It's a safe bet that your next post, if any, will still not quote the paper. You’ll just keep making your specious claims about the paper.

We _have_ refuted you at every corner, you just keep saying that we haven't.

Your “refutations” have mostly been of things I’ve not said or suggested. Otherwise I refuted them.

And it seems more than clear that this is all an ego trip anyway.

Yes, but not mine.

The Schwarzchild radius for an earth mass is << 6400km. Actually its about 9mm.

I didn’t suggest otherwise.