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

planck2

Zanket wrote:

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.


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.


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.

yes, i know, but your statement indicated that you believe a uniform gravitational field is not globally uniform.

Look T&W and GR itself says nothing can escape r=2M, you stated that the escape velocity v asymptotes to c at r=2M, but this is not true.


And your arguement about R_ab being non zero is incorrect, you simply show that you can't do the calculation yourself by arguing "logically..".

Also Section 6 is nothing more than a few lines long and does not contain one iota of evidence to back up your claims.

Zanket

Professor_Fink wrote:

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.

Show me the equation for R_ab for the Schwarzschild metric. There is logically no way that you can get a different equation than I gave, given the differences between the metrics. Make sure that the Schwarzschild metric you use is exactly the same as mine, except that eq. 9 is replaced with eq. 8.

In fact R_ab=0 is used to derive g_11 and g_44 for the scwarzchild metric.

This is a fatal problem; your argument dies here. You take R_ab = 0 as a given for the Schwarzschild metric, whereas you derive it for my metric. If you derive it for my metric, then you must derive it for the Schwarzschild metric as well. You must use the same method for both metrics. The Schwarzschild metric cannot be treated differently in an analysis comparing the metrics. You said:

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)

Nope, no R_ab here, like you “used to derive g_11 and g_44 for the scwarzchild metric”. By the basis you gave here (“since...”), and given the difference between eqs. 8 and 9, for the Schwarzschild metric the following must be true:

g_11=-(1 – (R / r)), g_44 = 1/(1 – (R / r)) = r / (r – R)

Now, what is R_ab for 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).

What observation? I’ll plug it in myself. Son Goku did not list any observation. You mean this one, ...

Son Goku wrote:

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.

... the “framework” which is nothing more than SR, with which my metric perfectly agrees? Show me something my metric does not agree with.

Professor_Fink wrote:

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.

You’ve said that if r_ab!=0 for my metric, then it won’t match observations. I’m saying that if no observation disagrees with my metric, then your conclusion is unproven. Nothing about my metric is silly like 1+1 != 2, which can be seen in fig. 3. More on this below.

We live in a universe governed by mathematics, the only question is which mathematics govern what phenomena.

This is contradictory. Yes, that is the question, because we live in a universe governed by nature, which the math follows. The tail does not wag the dog.

Oh how wrong you are. Gravitational redshift includes curvature effects from the photosphere all the way out.

Oops, I meant r / R < 5000. The current “record” is r / R = 104900. That fixes it. (Units are the same for both r and R.)

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!

Above you said that Son Goku showed “that this does not happen (at least to within experimental accuracy)”, which implies an observation (of natural phenomena). Now you’re talking about an “experiment” on paper only. Which is it?

There are only two ways to refute a theory of physics: show that it is internally inconsistent, or show that it conflicts with observations.

Unless you or anyone can show that my metric is internally inconsistent, the fact that it agrees with all observations trumps any problem you think it might have on paper. You’re basically saying that something you see in the math can preclude an observation that agrees with my metric while disagreeing with the Schwarzschild metric. But nature does what it wants, and it always wins.

So I’d like to see the equation for R_ab for the Schwarzschild metric if you please, but regardless, unless you or anyone can show that my metric is internally inconsistent, or can show an observation that agrees with the Schwarzschild metric while disagreeing with my metric, I’m not going to be convinced, and rightly so, for an “experiment” comparing R_ab and T_ab does not meet a scientific standard for invalidating a theory.

Son Goku

Zanket wrote:

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?

I can see why you would think that and at a first glance that is what you would think. However the slight differences are enough to give your metric Ricci Curvature.

Zanket wrote:

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

The observations of "G. B. Walker et al". As well as a lot of earlier observations.
Testing for the absence of Ricci Curvature in the absence of matter was one of the main tests of GR in the late 80s and the late 90s.
There are several other observations of other research groups, if you want their names listed.

Zanket

planck2 wrote:

yes, i know, but your statement indicated that you believe a uniform gravitational field is not globally uniform.

Not what I meant.

Look T&W and GR itself says nothing can escape r=2M, you stated that the escape velocity v asymptotes to c at r=2M, but this is not true.

Do you think it is possible for a theory to be inconsistent, to contradict itself? If yes, then see that it is possible that “T&W and GR itself says nothing can escape r=2M”, and it is inferable from GR that escape velocity is always less than c.

And your arguement about R_ab being non zero is incorrect, you simply show that you can't do the calculation yourself by arguing "logically..".

Actually I show that I can derive the correct equation for R_ab for the Schwarzschild metric (assuming that your equation for my metric is valid) using only logic.

Also Section 6 is nothing more than a few lines long and does not contain one iota of evidence to back up your claims.

The evidence is the logic which rigorously proves the case. You apparently didn’t read it carefully. If you did and disagreed, then you could quote the first statement with which you disagree. Scientifically, I must disregard claims that show no indication that the paper was read.

Zanket

Son Goku wrote:

I can see why you would think that and at a first glance that is what you would think. However the slight differences are enough to give your metric Ricci Curvature.

As I recall, the indicator of failure originally stated by y’all is when R_ab != 0, not “slight differences”. Are you changing your story? What do you now say is the indicator of failure, precisely?

The observations of "G. B. Walker et al". As well as a lot of earlier observations.
Testing for the absence of Ricci Curvature in the absence of matter was one of the main tests of GR in the late 80s and the late 90s.
There are several other observations of other research groups, if you want their names listed.

In the absence of matter, r is effectively infinity, in which case the Schwarzschild metric and my metric return the same values for the same inputs. I don’t need to look at any observation to see that. Both metrics become equivalent to the metric for flat spacetime in the absence of matter. My basis for that is as simple as plugging a huge value for r into either eqs. 8 and 9 and seeing that both return approximately unity.

Section 6 shows that for any observation where r / R > 5000, which includes any observation in the absence of matter, the difference between the metrics is too slight to detect. The current “record” is r / R = 104900. So I don’t find this argument about R_ab != 0 convincing. The only argument that might be convincing is one that has an observation where r / R < 5000.

Note that the Professor showed immediately above that he takes R_ab = 0 as a given for the Schwarzschild metric (convenient when the indicator of failure is R_ab != 0), whereas he derives it for my metric, which is a scientifically invalid way to compare the metrics. So his argument about R_ab died. There must be a problem in your analysis too, since you think that my metric disagrees with observations in the absence of matter, when that is easily seen to be incorrect.

Professor_Fink

Zanket wrote:

Show me the equation for R_ab for the Schwarzschild metric. There is logically no way that you can get a different equation than I gave, given the differences between the metrics. Make sure that the Schwarzschild metric you use is exactly the same as mine, except that eq. 9 is replaced with eq. 8.

Your logic here is horribly flawed.

I'll use Q here instead of the Christoffel symbols (since I can't write the equations here).

g_11 =1-(GM/r)

g_44 = 1/(1-GM/r)

So only,

Q^1_14 = - (1/2) (1-GM/r)^-1 GM r^-2 = -(1/2) (r^2 - GMr)^-1
Q^1_41 = - (1/2) (1-GM/r)^-1 GM r^-2 = -(1/2) (r^2 - GMr)^-1

Q^4_11 = (1/2) (1-GM/r) GM/r^2
Q^4_44 = (1/2) (1-GM/r)^-1 GM/r^2

are non zero.

Clearly these are functions only of r, and not t, theta or phi.

So,

d(Q^L_ij)/dx^L !=0 only if i=j=1,4
d(Q^L_iL)/dx^j !=0 only if j=4 and i=1,4
etc.

You will see that the coefficients all vanish if either i or j is not either 1 or 4. If it is 1 or 4, add up all the coefficients, and you will find them equal to zero.

So

R_ab=0 for the Schwarzchild metric. R_ab!=0 for your metric.

Professor_Fink

You are just showing that you cannot even handle calculus. It really is something you should learn if you are interested in physics. You need it for everything.

Son Goku

Zanket wrote:

As I recall, the indicator of failure originally stated by y’all is when R_ab != 0, not “slight differences”. Are you changing your story? What do you now say is the indicator of failure, precisely?

Zanket, I have no idea what you're talking about.

The slight differences between your metric and the Schwarzchild metric lead to a non-vanishing Ricci Tensor for your metric, but not for the Schwarzchild metric. The fact that it doesn't vanish for your metric is a flaw.
I'm not changing my story.
I'm not joking here, I actually don't understand what you're talking about.

Section 6 shows that for any observation where r / R > 5000, which includes any observation in the absence of matter, the difference between the metrics is too slight to detect. The current “record” is r / R = 104900. So I don’t find this argument about R_ab != 0 convincing. The only argument that might be convincing is one that has an observation where r / R < 5000.

Whoa!, Zanket you have got things totally backwards.
Just because the two metrics themselves may agree as r grows says nothing about the Ricci Tensor. It is another observable. It can be formed from the metric, but that doesn't mean that agreements between your metric and the Schwarzchild metric can be extended to agreement between their Ricci Curvatures.
The metric isn't the only observable in a geometric theory of gravitation.
Observations in the late 90s show that R_ab = 0 when T_ab = 0.
You metric has R_ab != 0 for those conditions. This is the biggest difference between your metric and the Schwarschild metric.
Your metric leads to greater "Volume Reduction".

planck2

Zanket wrote:

Not what I meant.


Do you think it is possible for a theory to be inconsistent, to contradict itself? If yes, then see that it is possible that “T&W and GR itself says nothing can escape r=2M”, and it is inferable from GR that escape velocity is always less than c.


Actually I show that I can derive the correct equation for R_ab for the Schwarzschild metric (assuming that your equation for my metric is valid) using only logic.


The evidence is the logic which rigorously proves the case. You apparently didn’t read it carefully. If you did and disagreed, then you could quote the first statement with which you disagree. Scientifically, I must disregard claims that show no indication that the paper was read.

You are talking BS at this stage. You can't even compute the Ricci tensor.
Logic doesn't replace hard calculations. Your logic is flawed and obviously so.
I won't repeat myself again.

GR and T&W say that for an infalling body observed by an observer at radius r the velocity of the body is (2M/r)^(1/2). This happens to agree with Newtonian theory, but what they mean is different. In Newtonian theory the escape velocity at r is (2M/r)^(1/2). This is not so in GR. Using Newtonian theory as a guide we have at r=2M v_esc =c, but GR says not even light can escape this radius. Therefore v does not asymptote to c ever.

Zanket wrote:

Do you think it is possible for a theory to be inconsistent, to contradict itself? If yes, then see that it is possible that “T&W and GR itself says nothing can escape r=2M”, and it is inferable from GR that escape velocity is always less than c.

How does this imply a logical inconsistency?

Can we discontinue this thread Son Goku?, he doesn't even know what he is doing.

Zanket

Professor_Fink wrote:

g_11 =1-(GM/r)
g_44 = 1/(1-GM/r)

This is inconsistent with what you said above:

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)

The only difference between the metrics (between mine and the one from T&W; see below) is that r / (r + R) in my metric is replaced with 1 – (R / r) in the Schwarzschild metric. Then, and in agreement with the basis you gave here (“since...”), for the Schwarzschild metric the following must be true:

g_11=-(1 – (R / r)), g_44 = 1/(1 – (R / r)) = r / (r – R)

Which does not match yours above.

In my paper, R = 2M and G = 1, as noted in the conventions section. Perhaps that is the whole problem with your logic here. Did you think that R = M and/or that G != 1?

What Schwarzschild metric are you using? You should use the one by T&W here on pg. 2-19. That’s the base metric that I use to derive mine.

You are just showing that you cannot even handle calculus. It really is something you should learn if you are interested in physics. You need it for everything.

Actually I’m showing that you’re being inconsistent. I’m helping you to see where the problem is in your analysis. (I don’t know where it is because you haven’t completely supported your argument by, say, showing a complete derivation of R_ab for both metrics. It is not incumbent upon me to do such derivation for you.) There’s no need for me to calculate R_ab the hard way; I already showed above what the equation for R_ab for the Schwarzschild metric must be if planck2’s equation for R_ab for my metric is correct, and you agree with planck2’s equation. See also my reply to planck2 below.

Professor_Fink

Zanket wrote:

In my paper, R = 2M and G = 1, as noted in the conventions section. Perhaps that is the whole problem with your logic here. Did you think that R = M and/or that G != 1?

Zanket, I made a little bit of a mess typing up that reply. I meant to write (1 - 2GM/r), my bad. Plug it in, and go through the steps (I was a little lazy and missed my typo). You get R_ab=0.

Zanket wrote:

Actually I’m showing that you’re being inconsistent. I’m helping you to see where the problem is in your analysis. (I don’t know where it is because you haven’t completely supported your argument by, say, showing a complete derivation of R_ab for both metrics. It is not incumbent upon me to do such derivation for you.) There’s no need for me to calculate R_ab the hard way; I already showed above what the equation for R_ab for the Schwarzschild metric must be if planck2’s equation for R_ab for my metric is correct, and you agree with planck2’s equation. See also my reply to planck2 below.

Zanket, as mentioned above, R_ab=0 for the Schwarzchild metric, and not for yours. I made a mistake in typing up the derivation (I was a little hung over), and now I look a bit foolish. None of this changes the fact that R_ab=0 for the Schwarzchild metric and not for yours. I really can't see any point in continuing this conversation until you can adequately address the problem of non-zero Ricci curvature (R_ab!=0).

Frankly, anyone with even a basic idea of how to do calculus should be able to just pop g into the formula R_ab. I've given you the definitions for the Christoffel symbols (the capital gammas), so what's wrong? You didn't show anything about R_ab, you just showed that you don't know how to differentiate.

Professor_Fink

Zanket wrote:

Actually I’m showing that you’re being inconsistent. I’m helping you to see where the problem is in your analysis. (I don’t know where it is because you haven’t completely supported your argument by, say, showing a complete derivation of R_ab for both metrics. It is not incumbent upon me to do such derivation for you.) There’s no need for me to calculate R_ab the hard way; I already showed above what the equation for R_ab for the Schwarzschild metric must be if planck2’s equation for R_ab for my metric is correct, and you agree with planck2’s equation. See also my reply to planck2 below.

Actually, it is incumbant on you to support your claims. Anyone can work out the Ricci curvature, and indeed we all have, and see it is not equal to zero. We are not all independantly making the same mistake in determining that your metric has a non-zero Ricci tensor.

Also, what reply to Planck2?

Zanket

Son Goku wrote:

Zanket, I have no idea what you're talking about.

It seems that I misread your quote. I had noted that, given the equation for R_ab from planck2, R_ab must be != 0 for the Schwarzschild metric too; the R_ab it returns is only slightly different in weak gravity than the R_ab that planck2’s equation returns for my metric. I thought that is what you meant by “slight differences”. Apparently you ignored my point on that, and you meant the slight differences between the metrics, and you still think that R_ab = 0 for the Schwarzschild metric.

Look at the post from the Professor above. He uses different reasoning to derive R_ab for the Schwarzschild metric than for my metric. I’m not convinced that R_ab is being calculated correctly for my metric. More on that below.

Just because the two metrics themselves may agree as r grows says nothing about the Ricci Tensor. It is another observable.

I’d like to understand this. Are you talking about an observation of natural phenomena, or an “experiment” on paper only, like a comparison of R_ab for the metrics, like the Professor mentioned? How could there be an observation of natural phenomena that invalidates my metric, but is not an experimental test of Schwarzschild geometry, when my metric applies to only Schwarzschild geometry (including experimental tests of SR in the flat spacetime at r = a limit of infinity)?

The metric isn't the only observable in a geometric theory of gravitation.

A metric is not an observable; it’s an equation that is used to predict an observable. You’re not making sense here. Perhaps, like the Professor did, you’re talking here about an “observation” on paper, which can refute a theory of physics only where it shows that it is internally inconsistent, which is not the case here.

Observations in the late 90s show that R_ab = 0 when T_ab = 0.
You metric has R_ab != 0 for those conditions. This is the biggest difference between your metric and the Schwarschild metric.

Nobody here has shown full derivations of R_ab for both metrics, or even what the equation for R_ab is for both metrics, and what has been shown for R_ab for the metrics is consistent with each other. I’ve shown that R_ab must be != 0 for the Schwarzschild metric, given planck2’s equation for R_ab for my metric; more on that below. You’ve ignored that reasoning. Y’all are not adequately supporting your claims. All that is being done re R_ab so far amounts to simply insisting that I am wrong without showing the full basis for that or addressing my counterpoints. Then y’all are not engaging in a scientific discussion. Anyone can claim that they’re right without showing how and ignoring simple logic that shows otherwise.

Professor_Fink

Oh, and if you still doubt us about the Schwarzchild metric having zero Ricci curvature check out http://scienceworld.wolfram.com/physics/SchwarzschildBlackHole.html, specifically the second equation on the page.

Professor_Fink

Zanket wrote:

Nobody here has shown full derivations of R_ab for both metrics, or even what the equation for R_ab is for both metrics, and what has been shown for R_ab for the metrics is consistent with each other. I’ve shown that R_ab must be != 0 for the Schwarzschild metric, given planck2’s equation for R_ab for my metric; more on that below. You’ve ignored that reasoning. Y’all are not adequately supporting your claims. All that is being done re R_ab so far amounts to simply insisting that I am wrong without showing the full basis for that or addressing my counterpoints. Then y’all are not engaging in a scientific discussion. Anyone can claim that they’re right without showing how and ignoring simple logic that shows otherwise.

The reason we have not typed up a full derivation for the Ricci tensor for both metrics (bar my somewhat botched attempt) is that it is painfully slow. That's it. No hidden agenda. It's just a long derivation because there are 16 entries in the tensor, and each has to be calculated seperately.

Also, you cannot just multiply the Ricci tensor for one metric by some scaling factor to get the Ricci curvature for another. It is a function of the metric that includes partial derivatives. Why on Earth would you think you can just scale it?

Zanket

planck2 wrote:

Logic doesn't replace hard calculations.

Hard calculations depend on logic. Can you prove that no mathematical proof contains the word “then”? Logic alone is valid. Can you refute Einstein’s “relativity of simultaneity” thought experiment? You cannot refute it by making your quote above. You’re not being scientific here. You’re just showing bias for hard calculations.

You said:

planck2 wrote:

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

The only difference between the metrics (between mine and the one from T&W) is that r / (r + R) in my metric is replaced with 1 – (R / r) = (r - R) / r in the Schwarzschild metric.

The difference between the metrics must be fully reflected in anything that compares them. (If you disagree, then why?) Then the r + R in your equation must be r in the equation for R_ab for the Schwarzschild metric, and the r in your equation must be r – R in the equation for R_ab for the Schwarzschild metric.

Then, assuming that your equation is correct, 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 > 0. There is no logical way that the same method can be used to derive R_ab for both metrics, and not derive the equation above for the Schwarzschild metric, when it derives your equation for my metric.

What equation do you get for R_ab for the Schwarzschild metric? If you don’t respond, then I’ll assume that you cannot support your claim that R_ab = 0 for the Schwarzschild metric.

GR and T&W say that for an infalling body observed by an observer at radius r the velocity of the body is (2M/r)^(1/2). This happens to agree with Newtonian theory, but what they mean is different. In Newtonian theory the escape velocity at r is (2M/r)^(1/2). This is not so in GR.

It is so in GR. Why do you think T&W plug r=2M into that equation to find the maximum escape velocity for GR? Do you think they’re plugging it into an equation that does not apply to GR?

Since they aren’t explicit, I’ve tried to find a better reference for that. In the meantime, see here.

Using Newtonian theory as a guide we have at r=2M v_esc =c, but GR says not even light can escape this radius. Therefore v does not asymptote to c ever.

How does this imply a logical inconsistency?

That is not the basis for the inconsistency shown in section 2, which shows that GR also says that light can escape r=2M. Why should I repeat that basis here, when you already ignore it?

Can we discontinue this thread Son Goku?, he doesn't even know what he is doing.

Like you knew what you were doing when you mistakenly insisted that my metric is not asymptotically flat? Don’t be so quick to assume that you are right. Do you request censorship because you can’t support your claims? Obviously I am giving bases for mine here. By locking the thread Son Goku would just be proving that this forum is not for scientific discussion.

Professor_Fink

Zanket wrote:

Hard calculations depend on logic. Can you prove that no mathematical proof contains the word “then”? Logic alone is valid. You can prove otherwise only by refuting them directly. Can you refute Einstein’s “relativity of simultaneity” thought experiment? You cannot refute it by making your quote above. You’re not being scientific here. You’re just showing bias for hard calculations.

I believe he means solid/rigorous when he uses hard, and does not mean complicated.

Professor_Fink

Zanket wrote:

The only difference between the metrics (between mine and the one from T&W) is that r / (r + R) in my metric is replaced with 1 – (R / r) = (r - R) / r in the Schwarzschild metric.

The difference between the metrics must be fully reflected in anything that compares them. (If you disagree, then why?) Then the r + R in your equation must be r in the equation for R_ab for the Schwarzschild metric, and the r in your equation must be r – R in the equation for R_ab for the Schwarzschild metric.

This is where you are making a mistake Zanket. You cannot just replace r with r-R, because this screws up the solid angle bit. Also r appears explicitly in the formula for R_ab.

What you're doing is not even remotely consistent.

planck2

I am talking about solid,logical, rigorous thought which is what differential geometry is built on. This gives us our answers. There are a number of ways to obtain your metric, but I am afraid the Ricci tensor is still non zero in each case. By the way you still haven't shown us that you can calculate R_ab. Goodnight.

Zanket

Professor_Fink wrote:

Zanket, I made a little bit of a mess typing up that reply. I meant to write (1 - 2GM/r), my bad. Plug it in, and go through the steps (I was a little lazy and missed my typo). You get R_ab=0.

OK, I’ve taken this about as far as I can in a spreadsheet. Immediately I ran into a snag. You said:

Q^1_14 = - (1/2) (1-GM/r)^-1 GM r^-2 = -(1/2) (r^2 - GMr)^-1
Q^1_41 = - (1/2) (1-GM/r)^-1 GM r^-2 = -(1/2) (r^2 - GMr)^-1

Q^4_11 = (1/2) (1-GM/r) GM/r^2
Q^4_44 = (1/2) (1-GM/r)^-1 GM/r^2

Given the above correction of yours, I correct and simplify to:

Q^1_14 = - (1/2) (1-2GM/r)^-1 GM r^-2 = -(1/2) (r^2 - GMr)^-1
Q^1_41 = Q^1_14
Q^4_11 = (1/2) (1-2GM/r) GM/r^2
Q^4_44 = -Q^1_14

The snag is that the red equal sign is incorrect. These equations are not equivalent. Can you correct it please, so I don’t go down the wrong path?

Assuming that the equation to the left of the red equal sign is the correct one, I entered some sample values:

R = 1
r = 2

I get:

Q^1_14 = -0.12500
Q^1_41 = -0.12500
Q^4_11 = 0.03125
Q^4_44 = 0.12500

How am I supposed to get zero from that? I didn’t get your explanation of that.

Frankly, anyone with even a basic idea of how to do calculus should be able to just pop g into the formula R_ab. I've given you the definitions for the Christoffel symbols (the capital gammas), so what's wrong?

What’s wrong is that I have already known for a while in this discussion that—using a method that is a few orders of magnitude easier than what you’re presenting—whatever R_ab is for the Schwarzschild metric, it is identical for my metric (more on that below), so I’m not too interested in doing the calculus. Instead I’m focusing on figuring out what your fundamental error is, at a lower level than the ones mentioned below.

You didn't show anything about R_ab, you just showed that you don't know how to differentiate.

Actually I can show that you are still being inconsistent about R_ab, in two ways.

First, you said:

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)

Using your same implied logic to get g_11 for the Schwarzschild metric, which (from the online T&W reference listed in my paper) is:

ds^2 = (-(1 - (R / r)) * dt^2) + (dr^2 / (1 - (R / r))) + (r^2 * dφ^2)

I get:

g_11 = -(1 – (R / r))

Which does not match what you said above, with your correction added in bold:

g_11 =1-(2GM/r)

Which simplifies to:

g_11 = 1 - (R / r)

Can you show that I’m using different logic than you did to derive g_11 for my metric? Your logic looks simple.

The second way you are being inconsistent about R_ab is because you agree with planck2’s equation for R_ab for my metric, but not mine for the Schwarzschild metric. More on this below.

Actually, it is incumbant on you to support your claims.

Sure, but “R_ab != 0” is your claim. If someone says to me, “R_ab != 0”, it not incumbent on me to show that R_ab = 0. Otherwise, by the same logic, someone could refute any theory whose author is dead by making any refutation at all, no matter how ridiculous. Given that you haven’t shown the full work for your claim, I’m going beyond the call of duty here to figure out where your mistake is.

We are not all independantly making the same mistake in determining that your metric has a non-zero Ricci tensor.

That is exactly what y’all are doing, as y’all will learn. I know this for rock-solid reasons I have already given, which have been ignored. But still I’ll elaborate below. I suspect that the mistake stems from a misunderstanding, not a math error.

Oh, and if you still doubt us about the Schwarzchild metric having zero Ricci curvature check out http://scienceworld.wolfram.com/physics/SchwarzschildBlackHole.html, specifically the second equation on the page.

Thanks for the link. I don’t deny that R_ab = 0 for the Schwarzschild metric when correctly calculated. What I dispute is that R_ab != 0 for my metric in that case. I say that when the same method is used to calculate R_ab for both metrics, it must be zero for both metrics or nonzero for both metrics.

The reason we have not typed up a full derivation for the Ricci tensor for both metrics (bar my somewhat botched attempt) is that it is painfully slow. That's it. No hidden agenda. It's just a long derivation because there are 16 entries in the tensor, and each has to be calculated seperately.

I believe you. I haven’t doubted that it’s slow. I am curious though, as to why y’all are so sure that R_ab != 0 for my metric, given all the work to calculate it, but you can’t show much of your work here. How hard can it be to just show the equation for R_ab for the Schwarzschild metric, which I requested, if you’ve already formulated it for my metric to make the initial claim?

Suppose three people tell you what you told me above, but about X. They claim that X != 0, therefore GR is invalid, but they can’t show their full work because it’s too tedious. Meanwhile you show that X = 0 using seemingly irrefutable simple logic, which is ignored by them even as they insist that you back up their claim or else they can't see any point in continuing the conversation. Would you be convinced that X != 0? Or would you think they all made a mistake, if only due to some misunderstanding?

Using the link you gave, I accept that R_ab = 0 for the Schwarzschild metric. I show again (but now I’ll elaborate) that it must be the same for my metric. The link says, “In empty space, the Einstein field equations become R_ab = 0”. Here are the spacelike versions of metrics in question:

Schwarzschild = ds^2 = (-(1 - (R / r)) * dt^2) + (dr^2 / (1 - (R / r))) + (r^2 * dφ^2)
Mine = ds^2 = (-(r / (r + R)) * dt^2) + (dr^2 / (r / (r + R))) + (r^2 * dφ^2)

I bolded all the differences, which are the squares of eqs. 8 and 9 respectively. Here they are extracted:

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

It doesn’t take long to confirm, if only by eyeballing them, that at a limit of R = 0, which is empty space, these equations both return unity. Then in empty space, the metrics are identical. I emphasized that in the hope that for once it will not be ignored. (One can hope!)

I elaborate even further. When the metrics are identical in empty space, then whatever R_ab is for the Schwarzschild metric in empty space, so it is for my metric there. Then if R_ab = 0 for the Schwarzschild metric in empty space, so it is for my metric there.

I can already predict the response: “You’re just using logic. Only math can prove something!” Nevertheless I ask: if the metrics are identical in empty space, then how can R_ab differ between them there? Surely the same method applied to the same metric does not yield two different results.

Also, you cannot just multiply the Ricci tensor for one metric by some scaling factor to get the Ricci curvature for another. It is a function of the metric that includes partial derivatives. Why on Earth would you think you can just scale it?

You missed the point. I was giving a clue as to where y’all went astray. My equation for R_ab for the Schwarzschild metric need not make sense to make my point, which was that I used a valid method to derive my equation, given the assumption that planck2’s equation for R_ab for my metric is valid. If my equation doesn’t make sense to you, then you should reevaluate your acceptance of planck2’s equation. More on that below.

I believe he means solid/rigorous when he uses hard, and does not mean complicated.

Yes, I thought the same when I replied. I say that logic alone can be solid/rigorous. It must be, since hard math depends on logic.

This is where you are making a mistake Zanket. You cannot just replace r with r-R, because this screws up the solid angle bit. Also r appears explicitly in the formula for R_ab.

What you're doing is not even remotely consistent.

What I did is valid, given the assumption that planck2’s equation is valid. (Now that I accept that R_ab = 0 for the Schwarzschild metric, I know that planck2’s equation is invalid. The valid equation should return R_ab = 0 too.) It is your acceptance of planck2’s equation that is inconsistent. I said:

Zanket wrote:

The difference between the metrics must be fully reflected in anything that compares them. (If you disagree, then why?)

By disputing my equation for R_ab, you are implying that the whole difference between the metrics need not be reflected in a comparison between them. Can you justify that? There is only one r by itself in planck2’s equation. If it is not replaced, then the whole difference between the metrics (the difference between eqs. 8 and 9 above) is not reflected. (Once r + R from eq. 9 is replaced, the only change that can complete the conversion—to fully reflect the difference—is to replace the remaining r.) Will you just ignore this logic and repeat yourself? If so, then I emphasize: I don’t confirm or deny the validity of my equation—that is irrelevant. The point is, if it’s invalid, then so is planck2’s equation, which you accept.

Zanket

planck2 wrote:

I am talking about solid,logical, rigorous thought which is what differential geometry is built on. This gives us our answers.

Given that it is built on that and can be rigorous, then—logically!—so can solid, logical, rigorous thought by itself. But you reject that notion. Logically, you can’t be consistent and accept something while rejecting a standalone part of its foundation. That’s like accepting GR while rejecting SR.

There are a number of ways to obtain your metric, but I am afraid the Ricci tensor is still non zero in each case.

I’ve shown that your equation for R_ab for my metric is invalid, in a number of ways.

By the way you still haven't shown us that you can calculate R_ab.

Actually I did, but you can’t see it because you erroneously reject logic without math.

Goodnight.

If you’re leaving us, then thanks again for your comments.

Son Goku

Right Zanket, I will in a day or two post up the complete derivation of R_ab for your metric and the Schwarzschild metric.

Post up what your metric is in the next post, with the coordinates in bold, so that I make no mistakes.

Also you can't get the Ricci Tensor from one metric by looking at the Ricci Tensor of another metric. Two metrics can be nearly similar and have almost totally different Ricci Tensors.

Zanket

Professor_Fink wrote:

You didn't show anything about R_ab, you just showed that you don't know how to differentiate.

Another problem with your argument that I just noticed. You agreed with planck2 here, for his equation for R_ab for my metric:

planck2 wrote:

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

The problem claimed is that R_ab != 0. But the link that you gave says “In empty space, the Einstein field equations become R_ab = 0”. Note "become". Then presumably R_ab should be zero for the Schwarzschild metric only in empty space, and is nonzero in nonempty space.

In empty space, R = 0 or r = a limit of infinity, take your pick. Either way, planck2’s equation returns 0. So what is the problem? Not even planck2’s equation identifies a problem with my metric.

Now I have nothing from y’all but your word that R_ab != 0 for my metric. Just exactly what is your basis for this claim? Have you calculated R_ab for my metric, using R = 0 or r = a limit of infinity, and seen that it is nonzero? Or did you use nonempty space (R > 0 or r < a limit of infinity), in which case R_ab is presumably nonzero even for the Schwarzschild metric?

Here is more support for my position:

From T&W from the online link in my references, or google for it:

Here we check off the ways in which [the Schwarzschild metric] makes sense.
...
Second, as the r-coordinate increases without limit, the curvature factor (1 - 2M/r) approaches the value unity, as it must. Why must it? Because an observer far from the center of attraction can carry out experiments in her vicinity without noticing the presence of the distant object at all. For her spacetime is locally flat. In other words, for large r the Schwarzschild metric [10] must go smoothly into the metric for flat spacetime [9].

Third, as the mass M goes to zero, the curvature factor (1 - 2M/r) approaches the value unity, as it must. Why must it? Because a center of attraction with zero mass is the same as the absence of a massive body at that center, in which case equation [10] becomes equation [9], the expression for the interval in flat spacetime.

Both of those ways in which the Schwarzschild makes sense apply to my metric as well, as is easy to see looking at eq. 9 in my paper (versus eq. 8 for the Schwarzschild metric). My metric also becomes the metric for flat spacetime when r = a limit of infinity or R = 0. Then R_ab must be zero for both metrics in empty space.

Son Goku

Zanket wrote:

The problem claimed is that R_ab != 0. But the link that you gave says “In empty space, the Einstein field equations become R_ab = 0”. Note "become". Then presumably R_ab should be zero for the Schwarzschild metric only in empty space, and is nonzero in nonempty space.

Schwarschild spacetime is empty everywhere.

Have you calculated R_ab for my metric, using R = 0 or r = a limit of infinity, and seen that it is nonzero? Or did you use nonempty space (R > 0 or r < a limit of infinity), in which case R_ab is presumably nonzero even for the Schwarzschild metric?

Zanket, this entire paragraph makes no sense. If you don't know what the Ricci Tensor is, or what Tensors in general are, then don't talk about them or you will say foolish things.
R_ab is a Tensor Field, you can't calculate it at one point. You calculate and then you can check its value at any point.
R_ab = 0 over the whole of the Schwarzschild spacetime, but not over the whole of yours.
Do you know what R_ab measures?

planck2

By the way. I have just noticed the following, his final reference for his paper is MTW, but he still wants us to derive R_ab for Schwarzschild for him. It is shown how to do so in the book.

Zanket

Son Goku wrote:

If you don't know what the Ricci Tensor is, or what Tensors in general are, then don't talk about them or you will say foolish things.

Presumably this is a scientific discussion, not an ego trip, so I’ll talk about whatever I think I understand enough to discuss. I won’t always have the understanding.

R_ab is a Tensor Field, you can't calculate it at one point. You calculate and then you can check its value at any point.

That is what I meant: derive the equation for R_ab, then check its value for a point (that is, calculate a value for the point by plugging values into the equation).

Schwarschild spacetime is empty everywhere.
...
R_ab = 0 over the whole of the Schwarzschild spacetime, but not over the whole of yours.

I see what you’re getting at here. I’ll look more into this. I understand it better from here.

Then you suggest that the equation for R_ab for the Schwarzschild metric is R_ab = 0, period, no inputs (unlike planck2’s equation for R_ab for my metric that accepts inputs for r and R).

Right Zanket, I will in a day or two post up the complete derivation of R_ab for your metric and the Schwarzschild metric.

Yes, I’d like to see that, thanks in advance. I’m curious as to how the equation you derive for my metric will have inputs, but the equation for the Schwarzschild metric will not. Hopefully your derivation is amenable to being put into a spreadsheet, so I can verify it (like what I did for the Professor’s work above).

I’d also like to know why you thought that R_ab != 0 for my metric, when you have not yet done the derivation of that. What was your original basis?

Post up what your metric is in the next post, with the coordinates in bold, so that I make no mistakes.

Here are the spacelike versions, with the differences in bold, from post #201:

Schwarzschild = ds^2 = (-(1 - (R / r)) * dt^2) + (dr^2 / (1 - (R / r))) + (r^2 * dφ^2)
Mine = ds^2 = (-(r / (r + R)) * dt^2) + (dr^2 / (r / (r + R))) + (r^2 * dφ^2)

The definitions at the top of my paper apply. The Schwarzschild metric comes from T&W, pg. 2-19 in the online link in my references.

Also you can't get the Ricci Tensor from one metric by looking at the Ricci Tensor of another metric. Two metrics can be nearly similar and have almost totally different Ricci Tensors.

Your conclusion, the first statement, seems to be a non sequitur. I used a valid method to convert planck2’s equation to an equation for R_ab for the Schwarzschild metric. (That does not mean that my equation is correct, for I assumed that planck2’s equation is valid.)

Let two metrics be identical except that x/y in one is y/x in the other. Let there be only one instance of x and y in either equation. Then in any equation that is derived from one of the metrics, the x’s and y’s can simply be swapped to derive that equation for the other metric. Can you refute that? Can you show or make up one example of any such derivation where this logic does not hold valid? Maybe your derivation of R_ab for the metrics will show that.

Son Goku

Zanket wrote:

I’d also like to know why you thought that R_ab != 0 for my metric, when you have not yet done the derivation of that. What was your original basis?

Calculating something half on paper and half in your head is a lot simpler than posting it in a forum that doesn't support TEX.

Zanket

Son Goku wrote:

Calculating something half on paper and half in your head is a lot simpler than posting it in a forum that doesn't support TEX.

OK. Straightforward is better than fancy. I’ll be comparing your derivations side-by-side, as far as I am able to.

The observations of "G. B. Walker et al". As well as a lot of earlier observations.
Testing for the absence of Ricci Curvature in the absence of matter was one of the main tests of GR in the late 80s and the late 90s.
There are several other observations of other research groups, if you want their names listed.

Yes, I’d like more names please. A search has not turned up anything yet. I’d still like confirmation from you that these are observations of natural phenomena, and not just on paper.

planck2

This is laughable. Your talking about a subject when you don't even understand the math.
Needless to point out that you will reply "I don't need to know the math to know that there is a problem."

The point is this you can't prove that R_ab =0 for your metric (because it isn't) and you can't do so for the Schwarzschild metric either, you rely on others to do so.

Zanket

planck2 wrote:

This is laughable. Your talking about a subject when you don't even understand the math.
Needless to point out that you will reply "I don't need to know the math to know that there is a problem."

The point is this you can't prove that R_ab =0 for your metric (because it isn't) and you can't do so for the Schwarzschild metric either, you rely on others to do so.

You misunderstand how theories are refuted. It is not up to the author to know everything you know. It is up to you to support your claims. R_ab != 0 is your claim, not mine, and you haven’t shown your work. I showed with basis that if your equation for R_ab for my metric is valid, then R_ab != 0 for the Schwarzschild metric too, and you have not refuted that. (Which invalidates your equation, your sole basis for your claim here, since we know from reliable sources that R_ab = 0 for the Schwarzschild metric.) I already have a paper showing in two ways that GR is internally inconsistent, in which case your argument about R_ab is for naught, and you haven’t refuted that either. So the real point is that you cannot refute the paper. I’ve shown my work, whereas you have not. Instead of showing your work, you requested that the thread be locked.

As far as I can tell in an extensive search, there is no experimental test of Ricci curvature involving natural phenomena. There is loads of information online about the experimental confirmation of GR, but there is no mention of a test of Ricci curvature therein. The loose references given so far by y’all have not panned out. If there is no such test (and y’all need to show your work on that too—I’ve already helped you enough by searching for it) then yes, "I don't need to know the math to know that there is a problem" with this R_ab argument, because theories cannot be refuted that way. Nature governs physics, not the other way around. The only purely mathematical argument that can refute a theory is one showing that it is internally inconsistent, which is not the case here since you are comparing my theory to another theory.