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

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.

Professor_Fink

Ok, if you think I don't quote the paper to back up my claims, here is a direct quote.

Zanket's Paper wrote:

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

This statement is completely false.

It's like saying that because a ball appears flat on a small enough scale, it must be flat everywhere and is hence actually a plane.

planck2

this may be out of line, but you know joe and son goku, i wouldn't even bother replying to the thread any more, i haven't even read it [zanket's paper] and i'm not going to

Professor_Fink

planck2 wrote:

this may be out of line, but you know joe and son goku, i wouldn't even bother replying to the thread any more, i haven't even read it [zanket's paper] and i'm not going to

Yeah, I know I should stop, but it's sucking me in.

Like watching a car crash.

Zanket

Professor_Fink wrote:

Ok, if you think I don't quote the paper to back up my claims, here is a direct quote.

Great. Was that so hard?

This statement is completely false.

It's like saying that because a ball appears flat on a small enough scale, it must be flat everywhere and is hence actually a plane.

No, it's not like that. The quote does not suggest that g is constant globally, as your ball example suggests. You read that into it. It’s like saying that because the surface of a ball appears flat on a small enough scale, a tiny bug moving on the ball always moves in a region that can be deemed flat.

Let’s examine the quote. First, the basis:

since a gravitational field is everywhere uniform locally

You didn’t contest that. Next, the conclusion:

The particle always falls in a uniform gravitational field

This must be valid when the basis is valid. Nothing here suggests that the “uniform gravitational field” is global. The basis says “locally”, and the definition of “uniform gravitational field” specifies “local”.

Zanket

planck2 wrote:

this may be out of line, but you know joe and son goku, i wouldn't even bother replying to the thread any more, i haven't even read it [zanket's paper] and i'm not going to

By all means depend on hearsay, for it's doubtful you could refute the paper scientifically.

Professor_Fink

Zanket wrote:

This must be valid when the basis is valid. Nothing here suggests that the “uniform gravitational field” is global. The basis says “locally”, and the definition of “uniform gravitational field” specifies “local”.

Since you only talk about how the spacetime appears locally, then you are stuck in a comoving frame. In a comoving frame the particle doesn't fall, it's stationary.

Now I realise that you won't believe me, and don't seem to have the mathematical ability to deal with GR to see why your wrong without having to ask us, and even if some kind of zombie Einstein rose from the grave to challenge you, you'd just take so quote from a pop science book and say he doesn't understand GR. So at this point, I'll take Planck's advice and try to ignore further comments.

But it is like a car wreck, so I might occasionally give in to temptation.

P.S. I haven't read many of the books on intelligent design, but I am familiar with much of the evidence for evolution. I hear what creationists say, but I can dismiss their claims.

There are currently 381,469 preprints on the LANL archive, none of us have read them all. Nor have you. So we are forced to ignore the vast majority papers even some of those that are in our field. Because of this, we rely (at least somewhat) on the judgement of others to some extent in highlighting what is important.

If Planck2 has decided not to read your paper, it should not be taken as an insult. Physicists simply do not have enough time to read every paper, and I for one am certainly not offended when people chose not to read mine. I certainly do not insult them (i.e. "it's doubtful you could refute the paper scientifically"). So attack me all you want, and Son Goku too (sorry S.G.!), but leave Planck alone. He hasn't said anything against you.

planck2

So tell me what do the Penrose -Catrer diagrams for this new metric look like. Is the space time globally hyperbolic.

Zanket

Professor_Fink wrote:

Since you only talk about how the spacetime appears locally, then you are stuck in a comoving frame. In a comoving frame the particle doesn't fall, it's stationary.

There’s no problem here. Section 2 says: “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 “comoving frame” to which you refer can only be the particle’s inertial frame or one comoving with it, so that the particle is stationary relative to it, as you say. But that’s a moot point, because the particle is measuring its velocity relative not to itself, but rather to the objects passing directly by.

You say I am “stuck in a comoving frame”. That’s vague. The particle’s inertial frame applies for the duration of the particle’s fall. It applies globally.

Now I realise that you won't believe me, ...

It’s not a matter of belief. It’s a matter of being able to refute you scientifically. Make a relevant argument I cannot refute and then I’ll believe you.

... even if some kind of zombie Einstein rose from the grave to challenge you, you'd just take so quote from a pop science book and say he doesn't understand GR.

Good one. But I haven’t said you or anyone else here doesn’t understand GR. And the only quotes I’ve used from books were said by Hawking, Taylor, Thorne, and Wheeler, i.e. from the top echelon of relativity physicists. And the quotes were specific. You have suggested that in books they sometimes all say the exact opposite of what is valid, even all of them about the same thing, which is not convincing.

If Planck2 has decided not to read your paper, it should not be taken as an insult.

His comment is obviously meant to be rude to me, and not just to inform us that he had decided to not read the paper.

Zanket

planck2 wrote:

So tell me what do the Penrose -Catrer diagrams for this new metric look like. Is the space time globally hyperbolic.

Papers aren't refuted by posing problems to the author. Mine speaks for itself. Can you refute it directly?

Son Goku

Zanket, would you agree that it is possible that you cannot understand GR without understanding the language it is written in?

Would you also agree that it is possible that GR cannot be expressed using the mathematics that you are employing?

Zanket

Son Goku wrote:

Zanket, would you agree that it is possible that you cannot understand GR without understanding the language it is written in?

No, I disagree. GR can be understood in the mind, with visualizations, the way Einstein understood it before he expressed it mathematically and with thought experiments. The language GR was originally written in, math, is just one way to express what’s in the mind. It works in reverse too: the math can be used to create visualizations (plots, say, or even stories) that help one to understand GR. That is what laymen’s books and introductions to GR do, and when they do it correctly, what those visualizations convey is no less valid than the math. They are simply written in a different language than GR was originally written in.

Would you also agree that it is possible that GR cannot be expressed using the mathematics that you are employing?

For the whole theory, yes, I agree. But my paper does not express the whole of GR. The only equation of it that it derives is eq. 8, for gravitational time dilation / length contraction. And it handles only Schwarzschild geometry (the simplest curved spacetime geometry) and not, say, Kerr geometry.

Had you asked whether it is possible that a theory of gravity for Schwarzschild geometry cannot be expressed using the mathematics that I employ, I would disagree, if only because my paper contradicts that. It shows that a metric for Schwarzschild geometry can be derived by only analyzing a free-falling test particle, and takes valid shortcuts to reduce the complexity of the math to that of simple algebra. For example, it shows that the new metric is experimentally confirmed by showing that the results it returns differ from those of the Schwarzschild metric only beyond the number of significant digits in the tests. While a mathematical purist would balk at that technique, there is nothing scientifically wrong with it.

planck2

i haven't tried to refute it. I am just wondering if you have worked them out.

Zanket

planck2 wrote:

i haven't tried to refute it. I am just wondering if you have worked them out.

No, I haven't.

planck2

ok then

Son Goku

If General Relativity is flawed, how do you obtain the dynamics of your theory?

If you throw out the Einstein-Hilbert action, what is the new action and what difference does it make to the dynamics of your theory?

planck2

Son Goku wrote:

If General Relativity is flawed, how do you obtain the dynamics of your theory?

If you throw out the Einstein-Hilbert action, what is the new action and what difference does it make to the dynamics of your theory?

Yes, I am wondering about this too. How do you get the energy-momentum-stress tensor for the theory? Is it divergence free?

And I also don't understand how you can have two metrics. Only one is allowed. Are the geodesics timelike or spacelike? Is casaulity preserved?

planck2

and another thing, neither of your metrics are asymptotically flat. you also use different sign conventions in both metrics, which is inconsistent and you also argue that your knowledge of general relativity is irrelevant in regard to some point of the arguement. Funny because I would think that someone who was trying to dismiss GR would know it inside out.

Professor_Fink

I'm also unclear how Zanket's theory is supposed to prevent a physical singularity at r=0. While he has no longer got an event horizon, he seems to be left with a naked singularity.

The reason I ask is because one of his main criticisms of GR is that it 'breaks down' at the physical singularity.

Zanket

Son Goku wrote:

If you throw out the Einstein-Hilbert action, what is the new action

The Einstein-Hilbert action is used to derive Einstein’s field equations. Then I have nothing to say about that, because the new metric for Schwarzschild geometry that is derived in the paper is not derived from field equations.

Einstein did not derive his field equations from the Einstein-Hilbert action; only Hilbert did. Einstein’s “derivation” (actually a trial-and-error method) is scientifically valid, so the Einstein-Hilbert action is superfluous.

If General Relativity is flawed, how do you obtain the dynamics of your theory?

The dynamics of my theory are completely given by the new metric. Taylor and Wheeler say:

By T&W; google for it:

Every (nonquantum) feature of spacetime around this kind of black hole [a Schwarzschild object] is described or implied by the Schwarzschild metric. This one expression tells it all! [italics theirs]

Likewise, the new metric should also describe or imply every nonquantum feature of spacetime around a Schwarzschild object. (Section 6 shows that the new metric is experimentally confirmed to all significant digits. So the Schwarzschild metric approximates the new metric in weak gravity. And those tests cover the vast majority of experimental tests of general relativity.) Keep in mind that my paper handles only Schwarzschild geometry.

Zanket

planck2 wrote:

How do you get the energy-momentum-stress tensor for the theory? Is it divergence free?

I have nothing to say about that, because the new metric for Schwarzschild geometry that is derived in the paper is not derived from field equations. See also my reply to Son Goku immediately above.

And I also don't understand how you can have two metrics. Only one is allowed.

Only one metric is derived in the paper. It is given in timelike and spacelike forms, just like the Schwarzschild metric is (see reference 10 in the paper for an online reference).

Are the geodesics timelike or spacelike? Is casaulity preserved?

I could take a stab at these, but I decline.

and another thing, neither of your metrics are asymptotically flat.

The spacetime predicted by the new metric is asymptotically flat at a great distance from the center of gravitational attraction, just like it is for the Schwarzschild metric, as shown by fig. 3. The Schwarzschild metric approximates the new metric in weak gravity; see section 6 for experimental confirmation of that.

you also use different sign conventions in both metrics, which is inconsistent ...

Then how do you explain that the Schwarzschild metric uses the same signage for its own timelike and spacelike forms? (see reference 10 in the paper)

and you also argue that your knowledge of general relativity is irrelevant in regard to some point of the arguement. Funny because I would think that someone who was trying to dismiss GR would know it inside out.

It isn’t necessary for someone to fully know a theory in order to “dismiss” it. Not only am I proof of that, but also there is nothing in the realm of science to suggest otherwise. A theory can be felled by knowing only its Achilles heel.

Zanket

Professor_Fink wrote:

I'm also unclear how Zanket's theory is supposed to prevent a physical singularity at r=0. While he has no longer got an event horizon, he seems to be left with a naked singularity.

Section 6 says, “Unlike the Schwarzschild metric, the new metric is compatible with quantum mechanics since it precludes singularities (r = 0 is invalid) and nothing need fall to r = 0 when the escape velocity is always less than c”. Singularities are not a requirement in a theory of gravity. They are required by the Schwarzschild metric because its interpretation demands that objects below the Schwarzschild radius must fall all the way to r=0. In a theory of gravity in which the escape velocity is always less than c (in which case escape is always possible) and there is otherwise good reason to believe that r=0 is invalid, like mine, there need not be singularities.

The reason I ask is because one of his main criticisms of GR is that it 'breaks down' at the physical singularity.

To be clear, I wouldn’t call it a main criticism since I mention it only tangentially in a reader comment. It isn’t a basis for the inconsistencies of GR that the paper shows.

Professor_Fink

Zanket wrote:

Section 6 says, “Unlike the Schwarzschild metric, the new metric is compatible with quantum mechanics since it precludes singularities (r = 0 is invalid) and nothing need fall to r = 0 when the escape velocity is always less than c”. Singularities are not a requirement in a theory of gravity. They are required by the Schwarzschild metric because its interpretation demands that objects below the Schwarzschild radius must fall all the way to r=0. In a theory of gravity in which the escape velocity is always less than c (in which case escape is always possible) and there is otherwise good reason to believe that r=0 is invalid, like mine, there need not be singularities.

Below are the metrics from you're paper:

dT^2 = ((r / (r + R)) * dt^2) - (dr^2 / (r / (r + R))) - (r^2 * do^2)
ds^2 = (-(r / (r + R)) * dt^2) + (dr^2 / (r / (r + R))) + (r^2 * do^2)

Taking r =0 I get dT^2 = -infinity*dr^2 and ds^2 = infinity*dr^2
This is a singularity Zanket!

Professor_Fink

Or are you saying that nothing can ever reach r=0?

If that is the case you have a major problem. How can the spacetime even arise. How does the central mass reach that position? What about if the mass is zero? This gives R=0 and you're back to flat space. This is fine. But what happens if you have vacuum fluctuations? Well, you'll get spontaneous productions of electron/positron pairs, etc. Consider what happens to one of these. They occupy different positions in space and each will have your proposed space time in its immediate vicinity. As such they can never anihilate, as they should do, and the vacuum starts pumping out particles from nothing.

This is interesting because it violates conservation of energy, but also because it means that matter would be produced all around us all the time, at a constant rate. It isn't!

Last but not least, it would prevent the formation of BECs, since no two particles could be made occupy the same space. Unfortunately for your theory, BECs can and do exist. So your theory does not agree with all known experiments, let alone full blown quantum field theory.

Zanket

Professor_Fink wrote:

Or are you saying that nothing can ever reach r=0?

Yes.

If that is the case you have a major problem. How can the spacetime even arise. How does the central mass reach that position?

In the new cosmological model proposed in section 7, spacetime always existed; it never arose. There is no requirement that a cosmological model have a beginning of spacetime. There is no direct observational evidence that our universe had a beginning.

What about if the mass is zero? This gives R=0 and you're back to flat space. This is fine. But what happens if you have vacuum fluctuations?

If the mass is zero, neither r nor R applies. These apply to only material objects.

Unfortunately for your theory, BECs can and do exist. So your theory does not agree with all known experiments, let alone full blown quantum field theory.

The only place that GR is incompatible with quantum mechanics is at a singularity. Then my theory need not predict BECs or the observations that quantum mechanics predicts.

planck2

now you are digging an even bigger hole, saying that what quantum mech predicts cannot effect your theory.

your spacetime is not asymptotically flat because the line element blows up at r=infinity.
Also it can't be because you have the spherical solid angle in the line element and not the surface metric of a 2-sphere

The action principle is important because this can tell you the dynamics of the theory , i.e. how particles move in the spacetime providing you with predictions you can match with experiment.

the sign usage is not the same, if the line element is +--- , LL convention, then you have spacelike for <0 and timelike for >0.this is how the different signs occur.

I disagree with you absolutely 100%. Why dont you write down the geodesics, and show me that causality is preserved.

planck2

If your metric is as I presume contains r^2*(dtheta^2 +sin^2(theta)*dphi^2). All I do to get it is replace -2M (-R in your language) in the Schwarzschild metric with 2M or R and swap g00 and g11.

Zanket

planck2 wrote:

now you are digging an even bigger hole, saying that what quantum mech predicts cannot effect your theory.

How so? Quantum mechanics doesn’t contradict the Schwarzschild metric, except at a singularity. Then it does not at all contradict the new metric, which precludes singularities.

your spacetime is not asymptotically flat because the line element blows up at r=infinity.
Also it can't be because you have the spherical solid angle in the line element and not the surface metric of a 2-sphere

I gave you a reference for the Schwarzschild metric. Did you look at it? The placement of do (the increment of a spherical solid angle) in the new metric is unchanged from that of the Schwarzschild metric. There is no difference between the new metric and the Schwarzschild metric that supports your “not asymptotically flat” argument. Fig. 3 shows that they converge as r increases.

The action principle is important because this can tell you the dynamics of the theory , i.e. how particles move in the spacetime providing you with predictions you can match with experiment.

As I noted above yesterday, the new metric alone tells one everything about the (nonquantum) dynamics for Schwarzschild geometry, just as the Schwarzschild metric does in GR. The vast majority of experimental tests of general relativity are tests of the Schwarzschild metric, not tests directly of the Einstein field equations or the Einstein-Hilbert action principle.

the sign usage is not the same, if the line element is +--- , LL convention, then you have spacelike for <0 and timelike for >0.

The reference I gave you for the Schwarzschild metric shows that the sign usage is the same.

I disagree with you absolutely 100%.

You also disagree with Taylor and Wheeler, so I don’t find your argument convincing.

Why dont you write down the geodesics, and show me that causality is preserved.

Nah, I still decline. The objectives that the paper meets are enough.

If your metric is as I presume contains r^2*(dtheta^2 +sin^2(theta)*dphi^2). All I do to get it is replace -2M (-R in your language) in the Schwarzschild metric with 2M or R and swap g00 and g11.

What is your point? Section 5 clearly gives the difference between the Schwarzschild metric and the new metric.

Son Goku

Zanket wrote:

As I noted above yesterday, the new metric alone tells one everything about the (nonquantum) dynamics for Schwarzschild geometry, just as the Schwarzschild metric does in GR. The vast majority of experimental tests of general relativity are tests of the Schwarzschild metric, not tests directly of the Einstein field equations or the Einstein-Hilbert action principle.

You're missing the point, a set of field equations are useless without an action.

For instance solve Maxwell's Equations in a given scenerio. This will give you the E and B fields.
However without the action of a nonrelativistic particle, this tells you nothing about the motion of things in response to those fields.

To put it in plain english, in your theory matter tells spacetime how to curve, but curved spacetime doesn't tell matter how to move.

What are your new dynamics?

planck2

no i don't disagree with Taylor and Wheeler, the sign convention is +---, Landau and Lif****z. I have it right here in front of me, along with Gravitation by MTW and General Relativity by Wald.

the point is i know how to get your new metric and it doesn't predict anything.

clearly they don't converge as r goes to infinity.



you are missing the fundamental point which is that the EH action allows one to predict the equations of motion for the particle which can be tested against experiment. Therefore I need an action to determine the theoretical equations of motion so as to test them against experiment. You don't provide one. So how am I to believe you?

the increment of solid angle is not used any where in the Schwarzschild metric, the metric for the surface of 2-sphere of radius r is though.

planck2

your theory doesn't need to predict BEC's or any of QM's predictions because neither does Einstein's.

Zanket

Son Goku wrote:

You're missing the point, a set of field equations are useless without an action.
...
What are your new dynamics?

You’re missing the point that the new metric is all that is needed to predict the results of experimental tests; i.e. the dynamics. Experimental tests of general relativity do not directly test the field equations or the Einstein-Hilbert action. Instead they test the metrics. I gave a reference for that by T&W above.

My paper does not derive its metric from field equations or from an action principle. There is no scientific requirement that it do so.

Son Goku

Zanket wrote:

You’re missing the point that the new metric is all that is needed to predict the results of experimental tests; i.e. the dynamics. Experimental tests of general relativity do not directly test the field equations or the Einstein-Hilbert action. Instead they test the metrics. I gave a reference for that by T&W above.

My paper does not derive its metric from field equations or from an action principle. There is no scientific requirement that it do so.

Yes, I know. I'm not saying you have to derive your equation from an action.
Obviously since you are taking away General Relativity you are also taking away it's dynamical content.
What generates the dynamics in your theory?
What is your new rule for the response of matter to curvature?

The metric on its own is useless, even the Schwarschild metric is.

For instance what use is a potential in Newtonian Mechanics without F=ma?

planck2

yes you can test the experiment from the metric, but one can do it from the action too. However, doing so from the metric has a very limited power, doing so from the action is more powerful.

planck2

there is no similarity between the two metrics at large r, why do you belive there is. show me that there is with refering to the graph. Do it with plain mathematics taking the limit as r goes to infinity

planck2

and another point Einstein came up with his field equations before Schwarzscild came up his vacuum solution.
So you must have a set of field equations which this metric satisfies as a vacuum solution, which can also be derived from an action and do as I described above.

Zanket

planck2 wrote:

no i don't disagree with Taylor and Wheeler, the sign convention is +---, Landau and Lif****z. I have it right here in front of me, along with Gravitation by MTW and General Relativity by Wald.

the point is i know how to get your new metric and it doesn't predict anything.

clearly they don't converge as r goes to infinity.

I can’t explain that. From the reference by Taylor and Wheeler above:



This shows that the sign usage is the same as the new metric. It shows that the only difference between the metrics is the difference between eq. 8 and eq. 9 in the paper, the curves of which fig. 3 shows converge as r increases.

you are missing the fundamental point which is that the EH action allows one to predict the equations of motion for the particle which can be tested against experiment. Therefore I need an action to determine the theoretical equations of motion so as to test them against experiment. You don't provide one. So how am I to believe you?

The metric is the equation of motion. It is all that is needed to predict the motion. One does not need anything that sources the metric. See my reply to Son Goku above. And note that the EH action is not employed in Einstein’s original valid version of GR, showing that it is superfluous.

the increment of solid angle is not used any where in the Schwarzschild metric, the metric for the surface of 2-sphere of radius r is though.

That contradicts T&W and this site:

From Schwarzschild geometry:

Schwarzschild's geometry is described by the metric (in units where the speed of light is one, c = 1)

ds^2 = –((1 – (rs / r)) * dt^2) + ((1 – (rs / r))^-1 * dr^2) + (r^2 * do^2)

[Reformatted for clarity. rs = the Schwarzschild radius.]

The quantity ds denotes the invariant spacetime interval, an absolute measure of the distance between two events in space and time, t is a 'universal' time coordinate, r is the circumferential radius, defined so that the circumference of a sphere at radius r is 2 pi r, and do is an interval of spherical solid angle. (boldface mine)

The Schwarzschild metric from the site above matches the spacelike metric by T&W.

planck2

Zanket wrote:

I can’t explain that. From the reference by Taylor and Wheeler above:



This shows that the sign usage is the same as the new metric. It shows that the only difference between the metrics is the difference between eq. 8 and eq. 9 in the paper, the curves of which fig. 3 shows converge as r increases.

Zanket wrote:

to get those signs T&W use the LL convention (+---) and set ds^2 to be > 0 for timelike and < 0 for spacelike, they also set theta equal to pi/2, so particles move in the equatorial plane.

Zanket wrote:

The metric is the equation of motion. It is all that is needed to predict the motion. One does not need anything that sources the metric.

Zanket wrote:

the metric is not an equation of motion, it describes how the spacetime is curved and how certain geodisics behave. The geodesics or equations of motion can be dervived using the Euler-Lagrange formalism taking the line element as the lagrangian. Or it can be done another way using an action principle which gives the field equations

Zanket wrote:

That contradicts T&W and this site:



The Schwarzschild metric from the site above matches the spacelike metric by T&W.

I am afraid it does not they do not refer to "do" only dphi which is an increment of angle on a circle

planck2

i know how you got the difference in the two metrics, but the point is this if your metric is in fact asymptotically flat then how come the scalar curvature is non zero

Zanket

planck2 wrote:

your theory doesn't need to predict BEC's or any of QM's predictions because neither does Einstein's.

Nice to have confirmation on that, thanks.

yes you can test the experiment from the metric, but one can do it from the action too. However, doing so from the metric has a very limited power, doing so from the action is more powerful.

I don’t deny that “doing so from the action is more powerful”, but I don’t see how the metric has a “very limited power”, given that T&W say:

From T&W; google for it:

Further investigation has shown that the Schwarzschild metric gives a complete description of spacetime external to a spherically symmetric, nonspinning, uncharged massive body (and everywhere around a black hole but at its central crunch point). Every (nonquantum) feature of spacetime around this kind of black hole is described or implied by the Schwarzschild metric. This one expression tells it all! (italics theirs)

Then nothing else is needed to make predictions of GR for Schwarzschild geometry. That is powerful indeed. Elsewhere they say:

From T&W; google for it:

The metric helps to answer every scientific question about (nonquantum) features of spacetime surrounding a black hole, every possible question about trajectories of light and satellites around the black hole as well around more familiar centers of attraction such as Earth and Sun.

It’s extra clear from this that my theory needs no more than a metric.

I am afraid it does not they do not refer to "do" only dphi which is an increment of angle on a circle

I can’t refute this. I don’t know why I chose to use “do”, as in the link above, instead of dphi, used by T&W, my main reference. Either way, you have forced me to change the paper. Congratulations, and thank you. I have changed “do” to dphi throughout the paper, including the meaning (which is now an “increment of an angle in a plane through a center of gravitational attraction”), and re-checked the meanings of all the symbols in my metric to verify that they are now all equivalent to the symbols used in the Schwarzschild metric by T&W in the image above. Now I can support my claim that the Schwarzschild metric and my metric differ only by the difference between eqs. 8 and 9 in the paper.

I would like to acknowledge you in the paper. Is that okay with you?

there is no similarity between the two metrics at large r, why do you belive there is. show me that there is with refering to the graph. Do it with plain mathematics taking the limit as r goes to infinity

Now that I use dphi, the Schwarzschild metric in the image above and the new metric in the paper differ only by the difference between eqs. 8 and 9 in the paper (that is how the new metric is derived, by swapping eq. 8 for eq. 9), a difference shown by fig. 3, which shows that the curves converge as r increases.

More mathematically, the right-hand side of eq. 9 is equivalent to sqrt(1 - (R / (r + R))). Comparing that the right-hand side of eq. 8, sqrt(1 - (R / r)), it is easy to see that the effect of R in the denominator of eq. 9, the only difference between the two equations, diminishes as r increases. Then the curves given by the equations must converge as r increases.

An equivalent explanation is given in section 6.

and another point Einstein came up with his field equations before Schwarzscild came up his vacuum solution.
So you must have a set of field equations which this metric satisfies as a vacuum solution, which can also be derived from an action and do as I described above.

That’s a non sequitur. The order of those events does not show a scientific requirement for field equations. I already have reader comments in the paper for this:

Reader: A metric must be derived from field equations.

Author: The scientific method lets any type of equation be presented without derivation. Then no particular method of derivation is required.

Reader: Without new field equations, your theory is worthless.

Author: Field equations are not required. A metric makes falsifiable predictions.

to get those signs T&W use the LL convention (+---) and set ds^2 to be > 0 for timelike and < 0 for spacelike, they also set theta equal to pi/2, so particles move in the equatorial plane.

OK, thanks for pointing that out. It might not be good for my paper to be limited to a plane. I’ll think about changing that, or at least clarifying it.

the metric is not an equation of motion, it describes how the spacetime is curved and how certain geodisics behave. The geodesics or equations of motion can be dervived using the Euler-Lagrange formalism taking the metric as the action. Or it can be done another way using an action principle which gives the field equations

Given that it answers “every possible question about trajectories of light and satellites around the black hole as well around more familiar centers of attraction such as Earth and Sun”, it sure seems to meet the definition of an equation of motion, given by Wikipedia as “equations that describe the behavior of a system (e.g., the motion of a particle under an influence of a force) as a function of time”. How do you explain that it doesn't meet that definition?

i know how you got the difference in the two metrics, but the point is this if your metric is in fact asymptotically flat then how come the scalar curvature is non zero

Do you still think there is a problem, now that I changed the paper? If yes, then how can the Schwarzschild metric be asymptotically flat yet mine not, when the difference between the two metrics approaches nothing as r goes to infinity?

Zanket

Son Goku wrote:

Yes, I know. I'm not saying you have to derive your equation from an action.
Obviously since you are taking away General Relativity you are also taking away it's dynamical content.
What generates the dynamics in your theory?
What is your new rule for the response of matter to curvature?

The metric on its own is useless, even the Schwarschild metric is.

For instance what use is a potential in Newtonian Mechanics without F=ma?

Given that the Schwarzschild metric and the new metric in the paper make falsifiable predictions on their own, then they cannot be useless, for they serve a prime focus of physics, predicting observations.

A theory of gravity need not show what “generates the dynamics” or show a “rule for the response of matter to curvature”. It is enough to predict observations.

Suppose beings on some other planet have a metric that approximates the Schwarzschild metric for the tests done so far on Earth, plus it accurately predicts phenomena that GR fails to predict (like stars that accelerate away), but it doesn’t have a “rule for the response of matter to curvature”. Is it useless?

Suppose someone creates an equation that consistently and accurately predicts the path of hurricanes, but it has no basis other than it just “felt right” to its author. Should science reject it? Can science even invalidate it?