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

Professor_Fink

planck2 wrote:

Back to calculating the geodesics of the 6 dimensional spherically symmetric spacetime.

Hyper-black holes!!!

By the way, I've just bought 'Quantum Gravity' by Rovelli. Have you read this? I'd be interested in your opinion.

Son Goku

planck2 wrote:

Yes you are correct, R_ab of Zanket's paper is not zero.

I think the case is closed at this point.

Back to calculating the geodesics of the 6 dimensional spherically symmetric spacetime.

What do you research?
(If you don't mind.)

Zanket

Professor_Fink wrote:

How exactly is it an empty claim?

It’s empty because you didn’t support it. You said that something is wrong with my paper without including a supporting explanation (which anyone can do without even reading my paper, to show how meaningless an empty claim is). In the post I’m replying to now, you added some support, but it’s still incomplete.

If the field equations don't hold, then R_ab may be related to T_ab in some different manner, but the relationship musyt exist if mass curves space in _any_ way. You cannot avoid this. The fact that you're paper does not have R_ab=0 for T_ab=0 implies some kind of additional curvature for space.

What do you mean by “additional curvature”? Are you saying that my theory is not one of curved spacetime? Show me your basis for the second sentence. How are you sure that the “relationship musyt exist if mass curves space in _any_ way”, given that Einstein’s field equations are invalid? Why should that relationship still be trusted?

I believe what he is saying is that a full discription of the physics involved is not possible without differential geometry or Clifford algebra. As such the book by T&W is not a full discription of the situation, but rather a gentle introduction to some of the ideas involved.

That may well be the case, in which case I don’t deny that. But in that case, it’s a moot point. This thread is about my paper, not GR as a whole or my understanding of that, and a theory of Schwarzschild geometry need provide nothing more than a metric to be complete. My paper shows that a metric can be derived without differential geometry or Clifford algebra. The scientific method is also clear about that; it does not require a particular method of derivation for an equation.

Setting aside your paper, what evidence do you have for this claim? If anything it opens our eyes to the possible flawed assumptions that can be made.

Two examples:

It has been suggested in this thread that my paper must be wrong because it ...

... does not use differential geometry or Clifford algebra.

... uses thought experiments, which are claimed to always be pop science.

Neither withstands scrutiny. They are signs of unscientific bias.

So what happens if the central starts spinning?

It’s irrelevant to the paper, which handles only Schwarzschild geometry. So I decline to answer. My paper is not incomplete because it handles only Schwarzschild geometry.

Ok, here is an analogy (since you seem fond of analogies).

Great, and I am!

It's a proof that the universe doesn't really exist:
asfdsefqwfasdfcasdcasdcacascasc
asdcasdascasxcasxcasxcasc3423qsx
awsf432tvb 4yt 4bvaswer2t5 q3gfva
34 rewgv ewrg etrhrf wr5y zd5ytbv543

hence the universe doesn't exist!

Prove it's wrong! Show me the exact spot!

It’s wrong because the conclusion does not follow from the “basis”, which, lacking an accompanying explanation, is presumably gibberish. The exact spot where it goes wrong is the conclusion. The gibberish could well mean something intelligible if an explanation accompanied it (so it cannot be said to go wrong at the spot of the gibberish, or else we might not get to an explanation that follows it). Then perhaps the conclusion would not be invalid.

If I was challenging this theory, I’d do something like this:

hence the universe doesn't exist!

This conclusion does not follow from the “basis”. Without an accompanying explanation, the basis is presumably gibberish.

The problem is that there isn't any one spot where it goes wrong, merely the whole arguement is nonsensical. This is the problem (all be it a slightly more extreme version) we are encountering when we try to point out the flaws in your arguement.

I showed that even the gibberish proof above is clearly refutable. Then it should be possible to refute my paper, if it is invalid.

Arguments can be subtle, so as to be challenging to grasp even when valid. I suggest you take a look at the argument in section 7, which also shows an inconsistency of GR. The argument there is more straightforward than the one in section 2, and it depends only on section 1. Nobody in this thread has yet mentioned it.

Son Goku

Zanket wrote:

What do you mean by “additional curvature”? Are you saying that my theory is not one of curved spacetime? Show me your basis for the second sentence. How are you sure that the “relationship musyt exist if mass curves space in _any_ way”, given that Einstein’s field equations are invalid? Why should that relationship still be trusted?

In your paper R_ab doesn't equal zero.
However since T_ab does equal 0 in your paper, your universe has Ricci curvature even when there is no matter around.
Space has innate curvature in your universe and this is something which is not observed.

Professor_Fink

Zanket wrote:

It’s empty because you didn’t support it. You said that something is wrong with my paper without including a supporting explanation (which anyone can do without even reading my paper, to show how meaningless an empty claim is). In the post I’m replying to now, you added some support, but it’s still incomplete.

Yes Zanket, I did support it. g_11 and g_44 in your metric cannot satisfy R_ab=0. See Son Goku's post above to see why this is such a major problem when T_ab=0. It means that you are saying that special relativity doesn't hold in the absence of a gravitational field. There are literally tousands of experiments which verify that R_ab=0 (or very close to it) when T_ab=0. Almost any test of special relativity does this.

Zanket wrote:

What do you mean by “additional curvature”? Are you saying that my theory is not one of curved spacetime? Show me your basis for the second sentence. How are you sure that the “relationship musyt exist if mass curves space in _any_ way”, given that Einstein’s field equations are invalid? Why should that relationship still be trusted?

See Son Goku's reply and/or Plank2's comment about T_ab.

Zanket wrote:

Two examples:

It has been suggested in this thread that my paper must be wrong because it ...

... does not use differential geometry or Clifford algebra.

... uses thought experiments, which are claimed to always be pop science.

When I said "setting aside your paper", I meant it. What evidence, aside from your current disagreement with us, do you have that we have learned GR in a particularly unintuitive way that blinds us to the obvious. That means no references to you or your paper in the reply.

Zanket wrote:

It’s irrelevant to the paper, which handles only Schwarzschild geometry. So I decline to answer. My paper is not incomplete because it handles only Schwarzschild geometry.

Where in the paper do you assume a steady state solution?

Zanket wrote:

It’s wrong because the conclusion does not follow from the “basis”, which, lacking an accompanying explanation, is presumably gibberish. The exact spot where it goes wrong is the conclusion. The gibberish could well mean something intelligible if an explanation accompanied it (so it cannot be said to go wrong at the spot of the gibberish, or else we might not get to an explanation that follows it). Then perhaps the conclusion would not be invalid.

If I was challenging this theory, I’d do something like this:

This conclusion does not follow from the “basis”. Without an accompanying explanation, the basis is presumably gibberish.

You mean like the conclusion doesn't follow? From the physicsforum thread:

Zanket wrote:

Numbered for reference:

1. The equivalence principle tells us that the crew of a rocket, traveling in flat spacetime and where the crew feels a constant acceleration, experiences a uniform gravitational field identical to that experienced locally by an observer on a planet.

2. Special relativity tells us that in principle the crew can traverse between any two points A and B in flat spacetime in an arbitrarily short proper time, where the two points are at rest with respect to each other and the rocket accelerates from rest at point A to the halfway point and then decelerates from the halfway point to rest at point B.

3. Then the crew can observe free-rising objects (such as an object floating stationary at the halfway point) to recede apparently arbitrarily fast—a million c is not out of the question—while causal contact is maintained since the actual velocity is always less than c.

Example: Let the rocket travel from Earth to Andromeda, two million light years away as we measure. (Assume that Earth and Andromeda are at rest with respect to each other and the spacetime between them is flat.) Let a buoy float stationary at the halfway point. Let the half of the trip from the buoy take ten proper years as the crew measures. Then during this half the buoy recedes by one million proper light years in ten proper years, an apparent (not actual) velocity of one hundred thousand c. From the crew's perspective the buoy free-rises in a uniform gravitational field.

4. Then why does general relativity not predict the same possible observation for the observer on the planet?

to which Doc Al replies:

Doc_Al wrote:

I fail to see any argument for your point (#4). #1 has to do with acceleration, while #2 & #3 have to do with speed. Your conclusion (#4) does not follow.

Seems to me that he counters your claim in exactly the same way you've countered mine... but when he does it you reply

Zanket wrote:

You must disagree with #1, 2, or 3 for #4 to not follow. (I added some clarification. Take another look.) Free-rising objects have speed for either the crew or the observer on the planet.

Surely I could say the same about mine, which is clearly gibberish.

Zanket wrote:

Arguments can be subtle, so as to be challenging to grasp even when valid. I suggest you take a look at the argument in section 7, which also shows an inconsistency of GR. The argument there is more straightforward than the one in section 2, and it depends only on section 1. Nobody in this thread has yet mentioned it.

Oh dear, Zanket. Your arguement is not too subtle for me, it is simply incorrect. There is a big difference.

planck2

Professor_Fink wrote:

Hyper-black holes!!!

By the way, I've just bought 'Quantum Gravity' by Rovelli. Have you read this? I'd be interested in your opinion.

no i haven't read it. I do have Quantum Gravity in 2+1 dimensions by Steven Carlip, he has worked with Danny Birmingham, but I haven't had the time to read it yet.

Professor_Fink

planck2 wrote:

no i haven't read it. I do have Quantum Gravity in 2+1 dimensions by Steven Carlip, he has worked with Danny Birmingham, but I haven't had the time to read it yet.

Quantum gravity is something I'm finding myself more and more drawn to. Maybe I'll look to that if I get bored with quantum info.

Zanket

Son Goku wrote:

Zanket, you said we learned Relativity in a less intuitive way.

To be clear, I said that it seems that way. I cannot prove it.

I'm saying that this is ridiculous, because GR deals with curved spacetimes, which humans have no intuition of.

Curved spacetime is a paradigm. That’s why humans don’t have an intuition of it—it’s an invention by Einstein. Relativity can be expressed without mentioning “curved spacetime”, by instead employing the tidal force, say, or some other perfect substitute. Your statement here is an example of how you may have learned relativity in a less intuitive way.

They present a toned down version of the Theory for interested students (secondary school to lower undergraduate) willing to take what they know of calculus and see it applied to GR. None of it is incorrect, however it cannot be used to attack GR itself, as some of its concepts don't generalise correctly.

My paper shows that the book can be referenced in an attack of GR. You haven’t refuted it. An attack on GR need not attack its field equations directly or anything outside of Schwarzschild geometry.

Isn’t it time at long last to stop making these baseless arguments? You need to prove your points, or at least give a strong basis for them, and not just claim them. For example, prove to me that “it cannot be used to attack GR itself”. Show me that any argument that attacks GR and references that book leads to incorrect results. Is this a scientific forum or not?

For example, I couldn't imagine learning QFT and then suddenly stopping when I see a "contradiction" in the implications of the Klein-Gordon Field and using this to proclaim QFT was flawed. Then on top of that I'm looking at the Klein-Gordon Field presented in an easy manner that doesn't get into anything fully and only calculates the simplest quantities.

Whether the source gets “into anything fully” is irrelevant. All that would matter is whether the contradiction is valid.

Here is what I think of your claim:
The escape velocity as observed by an infalling observer doesn't even asymptote to c, as it actually hits c at a certain r.
The simple fact of the matter is that the escape velocity doesn't asymptote to c in General Reltivity, it reaches c at a certain surface.
This you cannot contend with.

I can contend with this, and do. My paper shows that GR is inconsistent about that. Can you prove that a theory that predicts something cannot contradict itself about that prediction? Of course not.

Summarized, all you’re saying here is “You cannot show that GR is inconsistent”. That’s not a scientific position. You ignore my basis for an inconsistency in section 2.

Next is the claim that because the observer never observes the escape velocity to be greater than c it therefore never is greater than c.

The conclusion there is close to that of section 2, but the basis is incorrect. Had you quoted the paper then you could not have misrepresented it.

Therefore past the horizon this number is undefined and is not a sensible quantity to attempt to measure.

The paper already has a reader comment that covers this:

Reader: At and below an event horizon no object can be fixed at an altitude, so your analysis fails.

Author: That puts the cart before the horse. When v always asymptotes to c then so does escape velocity, in which case escape velocity is always less than c and then there are no event horizons. Since this was inferred by means general relativity allows, the theory cannot demand otherwise without being inconsistent.

Professor_Fink

Zanket wrote:

Curved spacetime is a paradigm. That’s why humans don’t have an intuition of it—it’s an invention by Einstein. Relativity can be expressed without mentioning “curved spacetime”, by instead employing the tidal force, say, or some other perfect substitute. Your statement here is an example of how you may have learned relativity in a less intuitive way.

No, no, no. This is complete rubbish. How can a force take the place of time dilation, for example?

You, yourself, describe what you come up with as a metric. This makes no sense if you do away with the manifold (i.e. if you throw out the notion of spacetime).

Zanket wrote:

Isn’t it time at long last to stop making these baseless arguments? You need to prove your points, or at least give a strong basis for them, and not just claim them. For example, prove to me that “it cannot be used to attack GR itself”. Show me that any argument that attacks GR and references that book leads to incorrect results. Is this a scientific forum or not?

I notice you still haven't addressed the problem with R_ab!=0 in your paper. It is you that are making the baseless claims here.

Zanket wrote:

Whether the source gets “into anything fully” is irrelevant. All that would matter is whether the contradiction is valid.

It's not valid because R_ab!=0 which runs counter to observed phenomena.

Zanket wrote:

Can you prove that a theory that predicts something cannot contradict itself about that prediction? Of course not.

For many theories you can indeed show that they are self consistant.

Zanket wrote:

Summarized, all you’re saying here is “You cannot show that GR is inconsistent”. That’s not a scientific position. You ignore my basis for an inconsistency in section 2.

We haven't ignored it, half the posts in this thread have been refuting its very existance.

Zanket wrote:

The conclusion there is close to that of section 2, but the basis is incorrect. Had you quoted the paper then you could not have misrepresented it.


The paper already has a reader comment that covers this:

Reader: At and below an event horizon no object can be fixed at an altitude, so your analysis fails.

Author: That puts the cart before the horse. When v always asymptotes to c then so does escape velocity, in which case escape velocity is always less than c and then there are no event horizons. Since this was inferred by means general relativity allows, the theory cannot demand otherwise without being inconsistent.

v doesn't assymptote to c.

Son Goku

Zanket, answer this:
In you paper R_ab != 0 when T_ab = 0.
This is not observed. How do you account for this?

Son Goku

Also, as Professor_Fink said, the escape velocity doesn't asymptote to c, it hits c.
There is an r where light is fixed at that r. Therefore the escape velocity here is the speed of light.

The paper already has a reader comment that covers this:

Reader: At and below an event horizon no object can be fixed at an altitude, so your analysis fails.

Author: That puts the cart before the horse. When v always asymptotes to c then so does escape velocity, in which case escape velocity is always less than c and then there are no event horizons. Since this was inferred by means general relativity allows, the theory cannot demand otherwise without being inconsistent.

That is useless Zanket. There is an r where the escape velocity is c. So the escape velocity is always less than or equal to c.
You still don't understand that even if what you were saying was true, it doesn't even matter.
GR is different from Newtonian physics, below r = 2M escape velocity is undefined. Remember in GR there are instances where Newtonain Energy is undefined.

All of this stuff about the escape velocity being less than c doesn't matter.
A falling observer will observe objects at constant r travelling at greater and greater velocities.
At a certain point the object at constant r will be light itself. After this point there is no objects which can maintain a constant r.

There are no timelike paths out of a black hole, therefore there is no such thing as an escape velocity.

I still don't understand where this contradiction is.

Zanket

planck2 wrote:

First of all I would like to say that you agree with special relaitvity don't you.

Yes, SR is held as valid in the paper.

Gravitational fields are not globally uniform which is what you are saying.

No, you’re reading that into the paper, but I admit that it’s confusing. A uniform gravitational field is defined as local in the paper; a gravitational field is not globally uniform. To avoid this contention in the future, I changed it to:

From Zanket’s paper:
The particle falls through a succession of locally uniform gravitational fields since a nonuniform gravitational field is everywhere uniform locally.

This seems to cover the rest of your post to which I’m now replying. I move on to the next one.

planck2 wrote:

In your paper you wrote
...
I don't know why you put this in because you're saying exactly what the reader says.

The reader is suggesting that I used SR in curved spacetime, where it does not apply. I am denying that it was used in curved spacetime. The reader’s comment is a common knee-jerk one that is made without basis, which is why it doesn’t have one.

And technically speaking this is not true, SR can be applied to any region big or small where the tidal forces are small enough to be neglected.

That is why my paper and T&W define “flat spacetime” as the spacetime of a frame throughout which the tidal force is negligible.

You also write
...
Where is there?, here, there and everywhere.

“There” is where v is less than c. To be more clear on that, I changed “when” to “where”.

Just because a particle has a velocity v at some point in a uniform gravitational field doesn't mean that it has the escape velocity or that the escape velocity is v. It needs to have a velocity >= to that given by the field.

For my statement “According to general relativity, where v is less than c it equals the escape velocity there” I give an online reference from T&W. Do you still disagree with them, now that you know that my paper doesn’t feature a globally uniform gravitational field?

In the case of a globally uniform gravitational field - providing you still contend that curvature =field stength - the components of the Riemann tensor are all equivalent and constants and R_abcd*R^(abcd) = h, h is a constant. So there is no infinite curvature, which is what a black hole is.

I don’t deny that, but my paper doesn’t feature a globally uniform gravitational field.

Yes you are correct, R_ab of Zanket's paper is not zero.

I assume that this is refuted, since my paper does not feature a globally uniform gravitational field. If not, please give your basis.

Zanket

Son Goku wrote:

In your paper R_ab doesn't equal zero.
However since T_ab does equal 0 in your paper, your universe has Ricci curvature even when there is no matter around.

What is your basis? I reviewed the posts above and see no basis for this from anyone. The paper refutes Einstein’s field equations. Then I find any argument based on his field equations to be suspect. Why do you think the argument should be trusted when it’s based on invalid field equations?

Space has innate curvature in your universe and this is something which is not observed.

Section 6 shows that the new metric is fully experimentally confirmed. It shows that there is no discrepancy (outside of the margin of error) between what the new metric predicts and observations made to date. Can you refute it scientifically instead of just deny it without basis?

The new metric does not predict “innate curvature” as you describe it. At the limit of r = infinity, where there is effectively “no matter around”, it predicts that spacetime is flat.

Zanket, answer this:
In you paper R_ab != 0 when T_ab = 0.
This is not observed. How do you account for this?

The new metric is fully experimentally confirmed. No observation contradicts it. Then regardless how you translate the new metric into R_ab and T_ab or anything else, it agrees with observations. If observations are your only basis, then your argument is refuted and your question is invalid (unless and until you can refute section 6).

That is useless Zanket. There is an r where the escape velocity is c. So the escape velocity is always less than or equal to c.

Section 2 concludes that “escape velocity is always less than c”, which is different.

You still don't understand that even if what you were saying was true, it doesn't even matter.
GR is different from Newtonian physics, below r = 2M escape velocity is undefined.

It does matter, when GR is inconsistent about its prediction for escape velocity. GR says or implies both of the following:

- Escape velocity is always less than c (including at r <= 2M).
- Escape velocity is c at r = 2M, and undefined below that.

Do you see a contradiction there? Section 2 shows that the first statement is inferable from GR.

All of this stuff about the escape velocity being less than c doesn't matter.
A falling observer will observe objects at constant r travelling at greater and greater velocities.
At a certain point the object at constant r will be light itself. After this point there is no objects which can maintain a constant r.

Section 2 shows that GR contradicts itself on that, and you have not refuted that argument. The point after which “there is no objects which can maintain a constant r” is where the escape velocity is c, but this is denied by the inference from GR that “escape velocity is always less than c (including at r <= 2M)”. You need to address that to refute the contradiction. You can’t refute it by referencing only one side of the contradiction, namely “escape velocity is c at r = 2M, and undefined below that”.

I still don't understand where this contradiction is.

If it’s still not clear to you, please tell me.

Professor_Fink

Zanket wrote:

What is your basis? I reviewed the posts above and see no basis for this from anyone. The paper refutes Einstein’s field equations. Then I find any argument based on his field equations to be suspect. Why do you think the argument should be trusted when it’s based on invalid field equations?

Because our arguement is not based on the field equations. Think of T_ab as a measure of whether or not there is matter, and R_ab as a measure of whether space is curved. The metric must be a function of R_ab. Surely you can see this.

If matter curves space in some way (any way), then R_ab = f(T_ab), for some function f. Do you still follow?

Now independent of f, we do expect R_ab=0 when T_ab=0 for special relativity to be correct in the absence of mass.

Your paper has T_ab=0 since it claims to be a vacuum solution. However R_ab is not equal to zero in your paper, since your g_11 and g_44 don't satisfy this. As I said earlier g_11 and g_44 are the dt^2 and dr^2 components.

Zanket wrote:

Section 6 shows that the new metric is fully experimentally confirmed. It shows that there is no discrepancy (outside of the margin of error) between what the new metric predicts and observations made to date. Can you refute it scientifically instead of just deny it without basis?

No, it just makes the claim that certain observstions match the metric (although you say all do, but you don't prove that).

Your metric implies R_ab!=0 for T_ab=0. This however has been shown to be untrue by many experimental confermations of special relativity.

Zanket wrote:

The new metric does not predict “innate curvature” as you describe it. At the limit of r = infinity, where there is effectively “no matter around”, it predicts that spacetime is flat.

That is due to a cancelation effect between the inate curvature of space you predict, and the way a particular arrangement of mass curves it. All this proves is that when you put the two together you get an assymptotically flat space. I am saying that if you have no mass, then you don't have flat space (according to your metric).

Zanket wrote:

The new metric is fully experimentally confirmed. No observation contradicts it. Then regardless how you translate the new metric into R_ab and T_ab or anything else, it agrees with observations. If observations are your only basis, then your argument is refuted and your question is invalid (unless and until you can refute section 6).

It is not fully confirmed by experiment as we have already pointed out.

Son Goku

Zanket wrote:

What is your basis? I reviewed the posts above and see no basis for this from anyone. The paper refutes Einstein’s field equations. Then I find any argument based on his field equations to be suspect. Why do you think the argument should be trusted when it’s based on invalid field equations?

What?
R_ab != 0 when T_ab = 0. This has nothing to do with the Field Equation.
You have Ricci Curvature when there is no Stress-Energy. This is not observed.
I can calculate R_ab for you and show that it is not zero.
If you say this is a baseless claim the thread will be locked because you are obviously not dealing with a serious problem of your theory.
Again this has nothing to do with presuming the Field Equation is correct.

The new metric does not predict “innate curvature” as you describe it. At the limit of r = infinity, where there is effectively “no matter around”, it predicts that spacetime is flat.

That is correct. However the simple fact of the matter is you have Ricci curvature where there is no matter. This is not observed.

It does matter, when GR is inconsistent about its prediction for escape velocity. GR says or implies both of the following:

- Escape velocity is always less than c (including at r <= 2M).
- Escape velocity is c at r = 2M, and undefined below that.

No it doesn't. It only says the former.
From your paper:

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.

At r = 2M that "object" is light. How is the escape velocity always less than c? I can see no part of your paper that shows how it is always less than c.
The only way you arrive at this is by asserting this:

Then the equations of motion for a relativistic rocket in section 8 are also those for a uniform gravitational field.

Which is wrong. One of Einstein's insight was that his own equivalence principle was totally local. If your read transcripts of one of his journals, you'll see this statement underlined.
First of all the Schwarschild metric is not a uniform gravitational field. Your conclusions would be correct except for the fact that the Schwarschild solution is a quiltwork of local areas where what you claim holds. However the manner in which the smoothly overlap ruins this conclusion. Results based on the relativistic rocket cannot be used because of this.

If you want the proof that this is so I will give it to you.

planck2

R_ab= 2*R^3/((r+R)^3*r^2), modulo a sign of course because I used -+++.

planck2

Zanket wrote:

For my statement “According to general relativity, where v is less than c it equals the escape velocity there” I give an online reference from T&W. Do you still disagree with them, now that you know that my paper doesn’t feature a globally uniform gravitational field?

I disagree with your interpretation of what they are saying. The citation is Chapter 2 page 22, correct? I read it and I understand it.

planck2 wrote:

In the case of a globally uniform gravitational field - providing you still contend that curvature =field stength - the components of the Riemann tensor are all equivalent and constants and R_abcd*R^(abcd) = h, h is a constant. So there is no infinite curvature, which is what a black hole is.

Zanket wrote:

I don’t deny that, but my paper doesn’t feature a globally uniform gravitational field.

planck2 wrote:

Yes you are correct, R_ab of Zanket's paper is not zero.

Zanket wrote:

I assume that this is refuted, since my paper does not feature a globally uniform gravitational field. If not, please give your basis.

What does a globally uniform gravitiational field have to do with R_ab being zero? I was merely stating that a uniform gravitational field has a constant R_abcd*R^abcd. The Riemann tensor will have components which are equal and independent of r and still R_ab might not be zero
Any even you say that your paper does not deal with uniform gravitational fields, but R_ab is still nonzero. so how is the claim R_ab non zero refuted by your gravity field being not globally uniform.

Zanket

Professor_Fink wrote:

Yes Zanket, I did support it. g_11 and g_44 in your metric cannot satisfy R_ab=0.

You did not support it either then or here. (You do a little better below; more on that later.) You said “it does not satisfy R_ab=0”. Well, how did you conclude that? You don’t say. My paper does not mention R_ab, so you must have some other way to determine that. Anybody could make the same claim without even reading my paper. The only reference you made to my paper is “it”.

See Son Goku's post above to see why this is such a major problem when T_ab=0. It means that you are saying that special relativity doesn't hold in the absence of a gravitational field. There are literally tousands of experiments which verify that R_ab=0 (or very close to it) when T_ab=0. Almost any test of special relativity does this.

The consequences certainly sound dire. But section 6 shows not only that no experiment conflicts with my new metric, but also that any experiment that does conflict with it, and where the gravity is weaker than the experiment I reference there (and the only experiment I need reference, the test of the Schwarzschild metric having the strongest gravity to date), must also conflict with the Schwarzschild metric. That includes any experiment of SR. Nobody has refuted section 6.

So you’re making a claim that is not only unsupported, but also it contradicts section 6, which has not been refuted. Then I find your argument to be unconvincing.

When I said "setting aside your paper", I meant it. What evidence, aside from your current disagreement with us, do you have that we have learned GR in a particularly unintuitive way that blinds us to the obvious. That means no references to you or your paper in the reply.

To be clear, I said that it seems that way (the way I said, not your twisted version). I cannot prove it. The only knowledge I have of y’all is this thread. So you’re asking that I base my claim on nothing, which makes no sense, so I decline. To set aside my paper, change any specific reference to my paper in my examples to papers like mine in general. They will still apply.

Where in the paper do you assume a steady state solution?

The conventions section at the top says “This paper handles only Schwarzschild geometry”.

You mean like the conclusion doesn't follow? From the physicsforum thread:
...
to which Doc Al replies:
...
Seems to me that he counters your claim in exactly the same way you've countered mine... but when he does it you reply
...
Surely I could say the same about mine, which is clearly gibberish.

Yes, you could reply to my refutation and say something like, “Oh, I forgot to include an explanation that translates the gibberish into something meaningful. Take another look.” But unless you do something like that, the refutation reasonably holds. The burden of proof would be on you to do it.

Oh dear, Zanket. Your arguement is not too subtle for me, it is simply incorrect.

Not that you have shown. This is an empty claim.

No, no, no. This is complete rubbish. How can a force take the place of time dilation, for example?

Your question doesn’t seem applicable. I said: “Relativity can be expressed without mentioning "curved spacetime", by instead employing the tidal force, say, or some other perfect substitute”. In the conventions section of the paper I give support for that: “Thorne notes that "spacetime curvature and tidal gravity [herein called tidal force] must be precisely the same thing, expressed in different languages".” T&W repeat that quote, and give their support to it, in their book I reference.

T&W also say:

From T&W; google for it:

The constant, ever-present "force of gravity" that we experience on Earth is gone, eliminated as we step into a free-float frame. What remains of "gravity"? Only curvature of spacetime remains. What is this curvature? Nothing but tidal acceleration. Curvature is tidal acceleration and tidal acceleration is curvature.

Professor_Fink wrote:

You, yourself, describe what you come up with as a metric. This makes no sense if you do away with the manifold (i.e. if you throw out the notion of spacetime).

I didn’t say “throw out”, and I said “curved spacetime”. The tidal force is a perfect substitute for spacetime curvature. T&W say:

From T&W; google for it:

[Reader:] You keep talking about "curvature" of spacetime. What is curvature?

[T&W:] The word curvature is an analogy, a visual way of extending ideas about three-dimensional space to the four dimensions of spacetime. (italics theirs)

I notice you still haven't addressed the problem with R_ab!=0 in your paper.

It’s a problem only claimed, not supported, and it contradicts section 6.

It's not valid because R_ab!=0 which runs counter to observed phenomena.

What you say here has no relation to the quote you gave. But in any case, section 6 refutes this argument.

For many theories you can indeed show that they are self consistant.

I said “cannot”, not “does not”.

We haven't ignored it, half the posts in this thread have been refuting its very existance.

Nobody has refuted it. Excepting those I have not yet replied to, show me a “refutation” that I did not refute or show to be baseless or incompletely supported. You can’t do it.

v doesn't assymptote to c.

Not that you have shown. This is an empty claim.

Your paper has T_ab=0 since it claims to be a vacuum solution. However R_ab is not equal to zero in your paper, since your g_11 and g_44 don't satisfy this. As I said earlier g_11 and g_44 are the dt^2 and dr^2 components.

Now this is a little better. I see some support for your claim here, but it’s incomplete. What is your basis for “your g_11 and g_44 don't satisfy this”? You don’t say. My paper does not mention g_11 or g_44, so you must have some other way to determine that. What is it?

No, it just makes the claim that certain observstions match the metric (although you say all do, but you don't prove that).

Section 6 does prove that all experimental tests of the Schwarzschild metric also confirm the new metric to all significant digits. Your simple denial of it does not convince me. You need to do better than that to be convincing.

Your metric implies R_ab!=0 for T_ab=0. This however has been shown to be untrue by many experimental confermations of special relativity.

Section 6 shows that if my metric doesn’t agree with SR, then neither does the Schwarzschild metric.

I am saying that if you have no mass, then you don't have flat space (according to your metric).

The “proof” of this is incomplete so far, and contradicts section 6, which has not been refuted. I don’t find it to be convincing.

It is not fully confirmed by experiment as we have already pointed out.

Previous claims to that effect were just as empty as this one.

Zanket

Son Goku wrote:

What?
R_ab != 0 when T_ab = 0. This has nothing to do with the Field Equation.
You have Ricci Curvature when there is no Stress-Energy. This is not observed.
I can calculate R_ab for you and show that it is not zero.
If you say this is a baseless claim the thread will be locked because you are obviously not dealing with a serious problem of your theory.

Go ahead and lock the thread then. It will just prove that this forum is not about scientific discussion. Since none of you are being very scientific here, I lose nothing.

Your “I can calculate R_ab for you and show that it is not zero” is as close as you’ve come to giving a basis, and of course it’s incomplete. Up to that, you have only claimed it without support. In a scientific argument, one supports their claims up front. As the mod, you should be rebuking those who make empty claims, rather than ignoring those and even making empty claims yourself.

I wrote a whole paper backing up my claims. I’ve backed up every claim I’ve made in this thread, and stand ready to give further support when asked. That’s how reasonable people participate in a scientific discussion.

Sorry, an unsupported claim does not show a “serious problem” with my paper. There may be one, but you’re only claiming it, not showing it. That doesn’t convince me. It wouldn’t convince any reasonable person.

What happened when planck2 kept repeating ad naseum that my metric is not asymptotically flat? That was a baseless claim he eventually recanted. I noted that the paper clearly shows otherwise; I gave a basis for my refutation. But that didn't stop planck2, and you didn't stop him. Why didn't you threaten to lock the thread then, by the same logic you're displaying here? Had he given a basis then maybe I could have pointed out his error, and resolved the issue earlier. Were you to give a basis for your claim then maybe I could do that for yours too.

Then you say:

That is correct. However the simple fact of the matter is you have Ricci curvature where there is no matter. This is not observed.

And when you said this before, I pointed out to you that section 6 refutes this, in which case it refutes your whole argument re R_ab. Section 6 shows that no observation contradicts my metric, that does not also contradict the Schwarzschild metric. But did you respond to that? No, you don’t even mention it here. That’s not how reasonable people participate in a scientific discussion.

When I point out simple logic that refutes you, then, if you were reasonable and scientific, you would check it out and say either “oh yeah, I guess you must be right” or “there’s a problem with this statement ...” One or the other. But instead you ignore what I point out to you and just repeat your claim.

You also ride on others’ “refutations” (like ZapperZ’s misunderstanding) and then won’t support them yourself when asked. You also twist my words, and let others repeatedly do so. You also threaten to lock the thread if I point out valid things, like a claim that is baseless, or someone twisting my words. Your arguing and moderation style seems designed to feed your ego. I’m not here to play those games. I’m here to have a scientific discussion. If you can’t do that, and won’t let me do that, then by all means lock the thread.

No it doesn't. It only says the former.

I assume you mean the latter.

From your paper:

At r = 2M that "object" is light. How is the escape velocity always less than c? I can see no part of your paper that shows how it is always less than c.

It seems that you simply ignore the basis. You say:

The only way you arrive at this is by asserting this:

No, that is not the only way I arrive at that. (And that is a conclusion, not an assertion.) Many sentences compose the basis of the conclusion in section 2, not just one. Do you really not see them? There’s no sense in me repeating the basis, since you ignore it.

Zanket&#8217 wrote: »

Then the equations of motion for a relativistic rocket in section 8 are also those for a uniform gravitational field.

Which is wrong. One of Einstein's insight was that his own equivalence principle was totally local.

Which agrees with a uniform gravitational field being correctly defined in the paper as a local thing. So there’s no problem identified here.

First of all the Schwarschild metric is not a uniform gravitational field.

Why do you say things like this? My paper has no such quote. It does not say or imply this in any way.

Your conclusions would be correct except for the fact that the Schwarschild solution is a quiltwork of local areas where what you claim holds. However the manner in which the smoothly overlap ruins this conclusion. Results based on the relativistic rocket cannot be used because of this.

The equivalence principle is clear that SR, which includes the relativistic rocket equations, can be used in every locally uniform part of a nonuniform gravitational field, where the spacetime is flat and where an inertial frame can exist. Section 2 implicitly applies the relativistic rocket equations to every such part to draw its conclusion. It does not apply SR to curved spacetime, as you suggest.

Zanket

planck2 wrote:

I disagree with your interpretation of what they are saying. The citation is Chapter 2 page 22, correct? I read it and I understand it.

Yes, the page is correct. You don’t specify your disagreement, so I have no further comment.

Any even you say that your paper does not deal with uniform gravitational fields, ...

I said it does not feature a globally uniform gravitational field.

... but R_ab is still nonzero. so how is the claim R_ab non zero refuted by your gravity field being not globally uniform.

It seemed that you were saying that if my paper features a globally uniform gravitational field, then R_ab is non zero. OK, so you’re not saying that.

Both the Professor and Son Goku claim that when R_ab is non zero, then my metric will not agree with some observations. But section 6 shows that the new metric is fully experimentally confirmed and nobody has refuted that, so I don’t find the argument convincing so far. I think it is likely that some misunderstanding leads to this claim, but it’s impossible for me to tell what that might be, because the claim has so far been made without support.

Professor_Fink

Zanket wrote:

Now this is a little better. I see some support for your claim here, but it’s incomplete. What is your basis for “your g_11 and g_44 don't satisfy this”? You don’t say. My paper does not mention g_11 or g_44, so you must have some other way to determine that. What is it?

Zanket, this shows a complete lack of knowledge about what a metric really is.

g is the metric tensor. If you give a line element (which is what you are actually giving) you can extract the metric tensor from it. As mentioned earlier, twice, g_11 and g_44 are the coefficients of dr^2 and dt^2.

So from your paper g_11=-(r / (r + R)), g_44 = 1/(r / (r + R)) = 1 + R/r since in your paper you give:

Zanket wrote:

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

As to how we get R_ab, well:

with


As you can see, knowing g_ab allows R_ab to be calculated, as Planck2 has done for you:

Planck2 wrote:

R_ab= 2*R^3/((r+R)^3*r^2), modulo a sign of course because I used -+++.

For this to be true, space must have R_ab!=0 when T_ab = 0, however the Schwarzchild geometry is not the only testable geometry with T_ab=0, and so section 6 isn't even remotely applicable.

And before you rant about the equations above being wrong, or not assumed in your solution, let me point something out to you. The come from differential geometry which is an area of mathematics, not physics. The are provable from first principals, and are beyond thought experiments, analogies or any other attack. It would be like trying to disprove dx^2/dx = 2x, impossible!

Nowhere do the field equations appear here, or in fact any general relativity.

Professor_Fink

Zanket wrote:

The equivalence principle is clear that SR, which includes the relativistic rocket equations, can be used in every locally uniform part of a nonuniform gravitational field, where the spacetime is flat and where an inertial frame can exist. Section 2 implicitly applies the relativistic rocket equations to every such part to draw its conclusion. It does not apply SR to curved spacetime, as you suggest.

Zanket, section 6 of your paper only says that experiments at large r should agree with the your metric too if the Schwarzchild metric holds. I'm not sure I accept that all experiments take place at sufficient distance, but it doesn't matter. Clearly they cannot agree for small r as the schwarzchild metric predicts an event horizon, which yours does not. But set all that aside for a minute, and think about the implications of your paper.

It implies that for T_ab=0, R_ab!=0. This must be true for all geometries, not just the Schwarzchild one, and it is in these other geometries that we can explicitly test the difference between your predictions and general relativity.

T_ab is always observed as implying R_ab=0, which disproves your metric.

It is important that you understand that this is all mathematical, there are no additional physical assumptions made, so there is no way you can say this isn't true because you only deal with the Schwarzchild geometry.

Your metric implies a relationship between matter and spacetime curvature which cannot be restricted to that particular arrangement of mass!

planck2

Zanket wrote:

I said it does not feature a globally uniform gravitational field.

What else could a uniform gravitational field be? It can only be globally uniform.

Son Goku

Zanket we have offered several rebukes. All you say is "That is baseless".

The reason I hold of on actually rebuking people like you is that calculations take a lot of pen and paper time for me because I have no access to any mathematical software.

However, I will rebuke you, however could you please not claim "That is baseless" without even reading or understanding what I have to say.

1. In your paper T_ab = 0 as you have a vacuum, ignoring your test particles which I assume are of negligible mass.

2. Also in your paper g_ab = ((r / (r + R)) * dt^2) - (dr^2 / (r / (r + R))) - (r^2 * dφ^2)

Computation:
First computing the Christoffel Symbols for a generic spherically symmetric metric:

The only nonvanishing components are:
T_000, T_011, T_001, T_100, T_111, T_101, T_122, T_133, T_233, T_212, T_332 and finally T_313.
In the above notation the Contravariant(Vector) component is listed first and the two Covariant(Form) components are listed second.

Just ask if you want the values of any of these components and I will give them to you.
Then the only nonvanishing components of the metric Tensor are:
R_00, R_11, R_01, R_22, R_33.
Again ask for the values and I will give them to you.

Then by substituting in the values for your metric one can work out what the Ricci Tensor is.
I can already see that R_11 does not vanish.

Again any values if you want them.
R_ab != 0, from the above considerations.
However from 1. T_ab = 0.

4. Therefore there is Ricci Curvature when there is no matter assuming your metric.

The work of "G. B. Walker et al" demonstrates that in the absence of matter, Ricci curvature is at least very small.

The success of the framework of "On the Electrodynamics of moving bodies" by A. Einstein, in the absence of large amounts of Stress-Energy also lends credence to the assertion that Ricci curvature vanishes when there is no Stress-Energy.

Therefore assertion 4. is in contradiction with observation.

planck2

In fact you never even go so far as to calculate the precession of mercury or the bending of light or the slowing of clocks in a gravitational field or the redshift of light. You only say the metric agrees with experimental tests....

planck2

Zanket wrote:

Section 2 concludes that “escape velocity is always less than c”, which is different.


It does matter, when GR is inconsistent about its prediction for escape velocity. GR says or implies both of the following:

- Escape velocity is always less than c (including at r <= 2M).
- Escape velocity is c at r = 2M, and undefined below that.

Do you see a contradiction there? Section 2 shows that the first statement is inferable from GR.

By the way GR doesn't say that and neither does chapter 2 page 22 of T&W.
GR say nothing, not even light can escape r=2M. A particle and light would need infinite energy to escape this radius.

Zanket

Professor_Fink wrote:

As to how we get R_ab, well:

Thanks for giving a basis. That gives me more info.

As you can see, knowing g_ab allows R_ab to be calculated, as Planck2 has done for you:

planck2 wrote:

R_ab= 2*R^3/((r+R)^3*r^2), modulo a sign of course because I used -+++.

OK, I can use this to make a point. For now, I’ll assume that planck2’s equation is valid.

The paper shows that the only difference between the Schwarzschild metric and the new metric is the difference between eqs. 8 and 9:

Eq. 8: 1 / gamma = sqrt(1 - (R / r)) = sqrt((r - R) / r)
Eq. 9: 1 / gamma = sqrt(r / (r + R))

Then logically the equation for R_ab for the Schwarzschild metric must be:

R_ab= 2*R^3/(r^3*(r - R)^2)

Which is nonzero for any r. So the Schwarzschild metric has the same issue, R_ab != 0, that you’re saying is bad for the new metric.

For this to be true, space must have R_ab!=0 when T_ab = 0, however the Schwarzchild geometry is not the only testable geometry with T_ab=0, and so section 6 isn't even remotely applicable.

My metric is for Schwarzschild geometry. Only experimental tests of the Schwarzschild metric or SR apply to it. (Any test of SR is implicitly a test of the Schwarzschild metric at r = a limit of infinity, a region section 6 covers.) You can’t use an experiment of Kerr geometry, say, against the paper, any more than you could use it against the Schwarzschild metric.

And before you rant about the equations above being wrong, or not assumed in your solution, let me point something out to you. The come from differential geometry which is an area of mathematics, not physics. The are provable from first principals, and are beyond thought experiments, analogies or any other attack. It would be like trying to disprove dx^2/dx = 2x, impossible!

Nowhere do the field equations appear here, or in fact any general relativity.

I don’t deny this, with one exception. In physics, observations trump anything on paper. Then your argument is not beyond being attacked by observations. You really haven’t made a case for a problem of the paper until you have either shown it to be inconsistent, or shown an experiment of Schwarzschild geometry or SR that disagrees with the new metric. You've not mentioned the former, and have only alluded to the latter.

Zanket, section 6 of your paper only says that experiments at large r should agree with the your metric too if the Schwarzchild metric holds. I'm not sure I accept that all experiments take place at sufficient distance, but it doesn't matter.

To possibly disprove it, you’d need to find an experiment where r < 5000. The current “record” is 104900.

Clearly they cannot agree for small r as the schwarzchild metric predicts an event horizon, which yours does not.

Yes, fig. 3 shows the difference between them graphically.

It implies that for T_ab=0, R_ab!=0. This must be true for all geometries, not just the Schwarzchild one, and it is in these other geometries that we can explicitly test the difference between your predictions and general relativity.

I showed above that R_ab!=0 for the Schwarzschild metric as well.

T_ab is always observed as implying R_ab=0, which disproves your metric.

What experiment? Name one. To apply, it must be an experiment of SR or the Schwarzschild metric. If you can’t apply it to the Schwarzschild metric, then you can’t apply it to mine either.

Zanket

planck2 wrote:

What else could a uniform gravitational field be? It can only be globally uniform.

When I say the paper does not feature a globally uniform gravitational field, I mean it doesn’t feature a gravitational field that is globally uniform, like you suggested it does.

In fact you never even go so far as to calculate the precession of mercury or the bending of light or the slowing of clocks in a gravitational field or the redshift of light. You only say the metric agrees with experimental tests....

I don’t just say it, I prove it. Your implied assumption that I must calculate the prediction for a particular experiment is invalid, as section 6 shows. Read it carefully to see that.

By the way GR doesn't say that and neither does chapter 2 page 22 of T&W.
GR say nothing, not even light can escape r=2M. A particle and light would need infinite energy to escape this radius.

If GR outright said it, or if T&W said it, then there’d be no point in me writing a paper to show it. It is inferable from GR, as section 2 shows. The reference from T&W is just for the sentence it follows, and not for the conclusion it precedes.

Zanket

Son Goku wrote:

Zanket we have offered several rebukes. All you say is "That is baseless".

If what I say that about is baseless, then no rebuke to me is warranted.

The reason I hold of on actually rebuking people like you is that calculations take a lot of pen and paper time for me because I have no access to any mathematical software.

There’s no reason to rebuke someone because it takes too much time for you. In that case you just make the best case you can in the time you have, but you give some basis. I don’t expect you to spend even half an hour on anything. If it takes too much time, then just say so, and I can offer suggestions as to how you give me something to work with quicker.

However, I will rebuke you, however could you please not claim "That is baseless" without even reading or understanding what I have to say.

Up to this point, you have done no more than assert that you can prove it to me. That is not a basis.

1. In your paper T_ab = 0 as you have a vacuum, ignoring your test particles which I assume are of negligible mass.

Now this is good stuff, thanks. I assume that this is the case for the Schwarzschild metric as well.

Again any values if you want them.
R_ab != 0, from the above considerations.
However from 1. T_ab = 0.

Alas, I don’t have the skill at that level to do the calculation. So I take your word for it for now. I assume that you think for the Schwarzschild metric that R_ab = 0. But I show above in my post to the Professor that R_ab must be nonzero for the Schwarzschild metric as well, given the differences between the metrics. So what’s the problem?

Therefore assertion 4. is in contradiction with observation.

What observation? Section 6 shows that no such observation exists.

Professor_Fink

Zanket wrote:

The paper shows that the only difference between the Schwarzschild metric and the new metric is the difference between eqs. 8 and 9:

Eq. 8: 1 / gamma = sqrt(1 - (R / r)) = sqrt((r - R) / r)
Eq. 9: 1 / gamma = sqrt(r / (r + R))

Then logically the equation for R_ab for the Schwarzschild metric must be:

R_ab= 2*R^3/(r^3*(r - R)^2)

Which is nonzero for any r. So the Schwarzschild metric has the same issue, R_ab != 0, that you’re saying is bad for the new metric.

That's complete crap. You can't just multiply one Ricci tensor by some factor. You have to put the new g_11 and g_44 into the equations I gave in my last post. The result is R_ab=0. In fact R_ab=0 is used to derive g_11 and g_44 for the scwarzchild metric.

Zanket wrote:

My metric is for Schwarzschild geometry. Only experimental tests of the Schwarzschild metric or SR apply to it. (Any test of SR is implicitly a test of the Schwarzschild metric at r = a limit of infinity, a region section 6 covers.) You can’t use an experiment of Kerr geometry, say, against the paper, any more than you could use it against the Schwarzschild metric.

It is easy to show, as I have and as Planck2 has, that your metric implies R_ab!=0 for T_ab=0, independant of the configuration of the system. Many observations, such as those quoted by Son Goku, show that this does not happen (at least to within experimental accuracy).

Zanket wrote:

I don’t deny this, with one exception. In physics, observations trump anything on paper. Then your argument is not beyond being attacked by observations.

Eh, that's rubbish. First of all, they are mathematical results, not physics. Can an observation shows 1+1!=2? No. If it could, then Physics could not exist. We live in a universe governed by mathematics, the only question is which mathematics govern what phenomena.

Zanket wrote:

You really haven’t made a case for a problem of the paper until you have either shown it to be inconsistent, or shown an experiment of Schwarzschild geometry or SR that disagrees with the new metric. You've not mentioned the former, and have only alluded to the latter.

To possibly disprove it, you’d need to find an experiment where r < 5000. The current “record” is 104900.

Oh how wrong you are. Gravitational redshift includes curvature effects from the photosphere all the way out. All redshifts measured from neutron stars must be substantialls less than that. Also, units would be nice. I'm assuming you either mean km or m.

Zanket wrote:

I showed above that R_ab!=0 for the Schwarzschild metric as well.

No you didn't.

Zanket wrote:

What experiment? Name one. To apply, it must be an experiment of SR or the Schwarzschild metric. If you can’t apply it to the Schwarzschild metric, then you can’t apply it to mine either.

Son Goku has already answered this. The schwarzchild metric implies R_ab-0 for T_ab=0, and yours implies that R_ab!=0 for T_ab=0. Anything which distinguishes between the two cases is a valid experiment. It does not have to be conducted in a schwarzchild setup!