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

Zanket · 12-09-2006 11:50am

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 · 12-09-2006 7:27pm

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 · 12-09-2006 7:46pm

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 · 12-09-2006 8:14pm

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 · 12-09-2006 8:53pm

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 · 12-09-2006 9:36pm

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 · 12-09-2006 10:21pm

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 · 13-09-2006 2:02am

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 · 14-09-2006 3:58am

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 · 14-09-2006 5:08pm

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.

planck2 · 15-09-2006 1:41am

oh i wouldn't say that at all. I am perfectly capable of finding the flaw in an arguement. Your's being a prime example

Zanket · 15-09-2006 1:49am

planck2 wrote:

I am perfectly capable of finding the flaw in an arguement. Your's being a prime example

What post # of yours contained a flaw of my paper that I did not subsequently refute? Empty/incomplete claims don't count.

planck2 · 15-09-2006 11:57am

take this for example, you claim your metric gives approximation to light bending effects and agrees with experimental measurements. I have not seen your calculation. You cannot do a single calculation yourself and rely on others to do them and you shoot them down when they say something which you claim you can't even verify. You talk nonsense. You are wasting my time. If I were the mod I would have finished this discussion ages ago.

breadmonkey · 15-09-2006 8:01pm

planck2 wrote:

take this for example, you claim your metric gives approximation to light bending effects and agrees with experimental measurements. I have not seen your calculation. You cannot do a single calculation yourself and rely on others to do them and you shoot them down when they say something which you claim you can't even verify. You talk nonsense. You are wasting my time. If I were the mod I would have finished this discussion ages ago.

Good job you aren't a mod then.

planck2 · 15-09-2006 11:59pm

Well I don't about that. I will listen to well reasoned arguments and people to are willing to discuss, but not those who consistently claim to be correct when they are not

Zanket · 16-09-2006 1:04am

planck2 wrote:

take this for example, you claim your metric gives approximation to light bending effects and agrees with experimental measurements. I have not seen your calculation.

There’s already a reader comment in section 6 that covers this:

Reader: No experiments are listed.

Author: For the experiment referenced and all those in weaker gravity (i.e. all other experimental tests of the Schwarzschild metric to date), this section shows that the new metric and the Schwarzschild metric make the same predictions. Then, for example, the new metric predicts that the relativistic orbital precession of Mercury is 42.98 arc seconds per century, the same as the Schwarzschild metric predicts.

Do you think it is impossible to show with logic alone that my metric will agree with the Schwarzschild metric to all significant digits for all experimental tests to date? Why are you insistent that predictions for specific experiments must be calculated? Only one experiment is needed: the one with the strongest gravity to date. That’s the one that section 6 references.

You cannot do a single calculation yourself and rely on others to do them ...

You’re the pot calling the kettle black. R_ab != 0 is your claim. If you want me to do the calculation for that, then you are relying on me to do it for you. Anyone can post an equation here. Unless you show your work you have not supported your claim. (I did the calculation for my metric and it shows that R_ab = 0. See, by your own logic I just proved you wrong. So what are you complaining about?)

... and you shoot them down when they say something which you claim you can't even verify. You talk nonsense.

I am not obligated to simply trust you when I can’t verify parts of your incomplete claim. For example, if there’s an experiment of natural phenomena that tests Ricci curvature, then list it here. How hard could that be, if you know about such an experiment? You should have had the experiment in mind before you ever made the claim. If you refuse to list it, and I can’t verify it despite an extensive search, then of course I’m going to be unconvinced. Any reasonable person would think the same.

If I were the mod I would have finished this discussion ages ago.

Censorship does seem to be back in favor these days.

I will listen to well reasoned arguments and people to are willing to discuss, but not those who consistently claim to be correct when they are not

Your insistence that logic alone cannot prove anything shows that you are not willing to listen to a well-reasoned argument.

If you think the paper is wrong, then by all means quote some part of it and make a supported claim that refutes it. I don’t just claim to be correct, I show it, whereas you do not support your claims. I am willing to discuss any claim you can make that has a basis. Every time someone has quotes the paper here or anywhere and says what they think is wrong with it, I give my reasons to refute that, unless I can’t, in which case I change the paper if I can save it. Whereas I give my bases and argue my points to the last side standing, you make empty claims and want to have the thread locked if I don’t take them on faith alone. Who’s being scientific here?

Professor_Fink · 16-09-2006 1:24am

Ok, this ends now. Here is a derivation for R_00 for both Zankets metric and the Schwarzschild metric. Note that it took 71 equations to derive this one element of the tensor. I do not plan on repeating this another 15 times.

Clearly R_00 is nonzero for Zankets metric, and R_00=0 for the Schwarzschild and Minkowski metrics.

This means that Zankets metric is not conformal to flat space, which means he cannot use arguements based on special relativity with his metric and remain consistent. Additionally R_ab is always observed to be 0 when T_ab=0 otherwise special relativity would essentially be wrong. Muon decay experiments, etc., all verify special relativity, so Zanket's metric cannot be self consistent, or match experimental observations.

I think this should close the case!

Professor_Fink · 16-09-2006 1:30am

Zanket wrote:

You’re the pot calling the kettle black. R_ab != 0 is your claim. If you want me to do the calculation for that, then you are relying on me to do it for you. Anyone can post an equation here. Unless you show your work you have not supported your claim. (I did the calculation for my metric and it shows that R_ab = 0. See, by your own logic I just proved you wrong. So what are you complaining about?)

Actually, Zanket, I was the first to point out that R_ab was non-zero for your metric. Also, you 'proof' that R_ab is zero for your metric is a joke. It's not even close to being right.

You asked earlier if all functions of two metrics must be different if the metrics are different (I believe this was a large part of you 'proof'). The answer is no! d/dx (x + 4) and d/dx (x+100012) are both 1.

Sorry!

P.S. Logic has essentially 3 axioms:
1: The law of identity: A if and only if A
2: The law of the excluded middle: Either A or not-A
3: The law of non-contradiction: Not A and not-A
Your proof isn't logic, it's hand waving.

planck2 · 16-09-2006 3:42am

Zanket wrote:

You’re the pot calling the kettle black. R_ab != 0 is your claim. If you want me to do the calculation for that, then you are relying on me to do it for you. Anyone can post an equation here. Unless you show your work you have not supported your claim. (I did the calculation for my metric and it shows that R_ab = 0. See, by your own logic I just proved you wrong. So what are you complaining about?

Your insistence that logic alone cannot prove anything shows that you are not willing to listen to a well-reasoned argument.

The difference between you and me is this. I can calculate R_ab for and metric given to me, whether I have the time to do so is another matter.

I am willing to listen to well reasoned arguement, but your argument doesn't belong to that equivalence class.

Son Goku · 16-09-2006 9:07am

Professor_Fink wrote:

Ok, this ends now. Here is a derivation for R_00 for both Zankets metric and the Schwarzschild metric. Note that it took 71 equations to derive this one element of the tensor. I do not plan on repeating this another 15 times.

Clearly R_00 is nonzero for Zankets metric, and R_00=0 for the Schwarzschild and Minkowski metrics.

This means that Zankets metric is not conformal to flat space, which means he cannot use arguements based on special relativity with his metric and remain consistent. Additionally R_ab is always observed to be 0 when T_ab=0 otherwise special relativity would essentially be wrong. Muon decay experiments, etc., all verify special relativity, so Zanket's metric cannot be self consistent, or match experimental observations.

I think this should close the case!

I'm only on the fourth component, I think I'll wait to see what Zanket makes of this before I continue.

Now remember Zanket, there is nothing you can do to refute the fact that R_ab = 0 when T_ab = 0. If it wasn't true Special Relativity wouldn't work. There would be no regime where SR would be an approximation.
Your universe has constant curvature.

The other thing is you have never given the calculations that show certain things:

Then, for example, the new metric predicts that the relativistic orbital precession of Mercury is 42.98 arc seconds per century, the same as the Schwarzschild metric predicts.

Do it, calculate the precession of Mercury in your metric and show us that they agree.
You can't expect us to take it on faith.

Zanket · 16-09-2006 11:11am

Professor_Fink wrote:

Ok, this ends now. Here is a derivation for R_00 for both Zankets metric and the Schwarzschild metric. Note that it took 71 equations to derive this one element of the tensor. I do not plan on repeating this another 15 times.

Thanks! Now that is one fine example of supporting an argument.

I am in the process of analyzing this. I won't have a lot of free time until Monday, so it'll probably be a few days until I get an answer back to you.

In the meantime, do you confirm or deny that eq. 6 has a typo? There is no such expression in the Schwarzschild metric (at least not in the version in my paper’s reference, the version I posted above), whereas eq. 38, its complement for my metric, is an expression in my metric. It seems that the plus sign in eq. 6 should be a minus sign, in concordance with the minus sign in your usage of eq. 6 in eq. 12. If it’s not a typo, please explain the inconsistency between eqs. 6 and 38, and between eqs. 6 and 12.

Also, why is the fact that eqs. 6 and 38 are to the -1 power not reflected in the usage of those equations in eqs. 12 and 44 respectively?

Actually, Zanket, I was the first to point out that R_ab was non-zero for your metric.

It fully became planck2’s claim too when he argued it too.

You asked earlier if all functions of two metrics must be different if the metrics are different (I believe this was a large part of you 'proof'). The answer is no! d/dx (x + 4) and d/dx (x+100012) are both 1.

I said (paraphrasing) that equations that compare metrics (or any other type of equation) must fully reflect the differences among the metrics.

Zanket · 16-09-2006 11:15am

planck2 wrote:

The difference between you and me is this. I can calculate R_ab for and metric given to me, whether I have the time to do so is another matter.

This is yet another empty claim. How do we know that you can calculate R_ab when you don’t show your work?

I am willing to listen to well reasoned arguement, but your argument doesn't belong to that equivalence class.

If you think your empty claims are well-reasoned arguments, then you’re right.

Zanket · 16-09-2006 11:34am

Son Goku wrote:

I'm only on the fourth component, I think I'll wait to see what Zanket makes of this before I continue.

If you show it, I suggest you lay it out like the Professor did, which is nicely done.

Now remember Zanket, there is nothing you can do to refute the fact that R_ab = 0 when T_ab = 0. If it wasn't true Special Relativity wouldn't work. There would be no regime where SR would be an approximation.
Your universe has constant curvature.

I’ll come back to that if I can’t refute the Professor.

Now this is how science is done! I’m glad to see that this claim has support.

The other thing is you have never given the calculations that show certain things:
Do it, calculate the precession of Mercury in your metric and show us that they agree.

Section 6 shows that results of the new metric match results of the Schwarzschild metric to all significant digits for all experimental tests of the latter to date, in which case such calculations are superfluous. Have you not read section 6?

You can't expect us to take it on faith.

I certainly don’t expect that, which is why I wrote a rigorous proof that covers every experimental test of the Schwarzschild metric to date. Read section 6, and then tell me the first statement with which you disagree, if any.

Professor_Fink · 16-09-2006 12:57pm

Indeed eq. 6 has a typo, but it is just in that equation. I used g_33 = (1-2GM/r)^-1 the whole way through. I've fixed the typo in the pdf.

g^33 is not the same as g_33. g_ii*g^ii =1. And so g^ii = g_ii^-1. It standard notation in differential geometry.

Professor_Fink · 16-09-2006 1:07pm

Zanket wrote:

This is yet another empty claim. How do we know that you can calculate R_ab when you don’t show your work?

I know for a fact that Planck2 has a first in theoretical physics and took both differential geometry and general relativity. So he can definitely calculate R_ab.

Just as it took me ages to write up the derivation for one component, I am sure that he probably does not have the time or the inclination.

Also a few days ago, Son Goku offered to write it out, and there is hardly any point in us all deriving the same thing.

planck2 · 16-09-2006 6:58pm

Further your metric is the Schwarzschild metric with M replaced with (-M). As I have said already this is gives and a naked singularity and moreover the spacetime is unstable. See the paper by Reinaldo J Gleiser and Gustavo Dotti in Classical and Quantum Gravity 23 (2006) 5063-5077.

Zanket · 16-09-2006 9:09pm

Professor_Fink wrote:

Indeed eq. 6 has a typo, but it is just in that equation. I used g_33 = (1-2GM/r)^-1 the whole way through. I've fixed the typo in the pdf.
g^33 is not the same as g_33. g_ii*g^ii =1. And so g^ii = g_ii^-1. It standard notation in differential geometry.

OK, thanks for the quick response on that.

I know for a fact that Planck2 has a first in theoretical physics and took both differential geometry and general relativity. So he can definitely calculate R_ab.

Be that as it may, he has to support his claims. We don’t simply trust even Hawking. He’s free to depend on your calculations, or Son Goku’s, but he didn’t do that. Instead he wants us to believe that my paper has a problem based on his word alone.

Zanket · 16-09-2006 9:11pm

planck2 wrote:

Further your metric is the Schwarzschild metric with M replaced with (-M).

No, my metric is clearly the Schwarzschild metric with sqrt(1 - (R / r)) replaced with sqrt(r / (r + R)), which is how my metric is derived in section 5. If my metric was the Schwarzschild metric with M replaced with –M, then sqrt(1 - (R / r)) in the latter would be replaced with sqrt(1 - (-R / r)) = sqrt(1 + (R / r)), which is the reciprocal of sqrt(r / (r + R)).

planck2 · 17-09-2006 12:21am

Zanket wrote:

OK, thanks for the quick response on that.

Be that as it may, he has to support his claims. We don’t simply trust even Hawking. He’s free to depend on your calculations, or Son Goku’s, but he didn’t do that. Instead he wants us to believe that my paper has a problem based on his word alone.

i don' think so. you expect us to believe you on your word alone.
You expected us to believe R_ab was non zero for Schwarschild based on your logical arguements. As for relying on Prof Fink and Son Goku they would probably agree that I need no help from them
"experimental" verifications my eye.

as for replacing with -M. well then it seems you might be correct. However to correctly get your metric you replace M with - M and swap g_00 and g_11.

Zanket · 17-09-2006 7:51am

Professor_Fink wrote:

Here is a derivation for R_00 for both Zankets metric and the Schwarzschild metric.

It looks like there is another typo, in eq. 22, copied to eq. 23. It seems that the plus sign should be a minus sign, in concordance with eqs. 14 and 24. Do you confirm?

Zanket · 17-09-2006 7:54am

planck2 wrote:

i don' think so. you expect us to believe you on your word alone.
You expected us to believe R_ab was non zero for Schwarschild based on your logical arguements.

A logical argument is obviously not my word alone.

As for relying on Prof Fink and Son Goku they would probably agree that I need no help from them

That’s beside the point.

"experimental" verifications my eye.

If only theories could be refuted that way.

However to correctly get your metric you replace M with - M and swap g_00 and g_11.

I don’t confirm or deny that (too lazy to verify in the absence of an identified problem). Do you see a problem with that?

Professor_Fink · 17-09-2006 1:12pm

Yes eqs 22 and 23 have typos. Essentially I made the mistake with g_33 the whole way through, and then noticed my mistake, so I went back through the paper fixing the equations before I posted the original, but it seems I have left a few typos uncorrected. As you can see, though, it is correct again from eq 24 onwards. I've fixed the typo in the PDF.

Zanket · 19-09-2006 9:07am

Professor_Fink wrote:

Here is a derivation for R_00 for both Zankets metric and the Schwarzschild metric.
...
Clearly R_00 is nonzero for Zankets metric, and R_00=0 for the Schwarzschild and Minkowski metrics.

I’ve analyzed your derivations. (Thanks again for writing them up.) First, based on the assumption that your derivation for the Schwarzschild metric is correct (I see no obvious mathematical mistakes), I give here a rigorous proof that R_00 for my metric too:

The only inputs to your derivation for the Schwarzschild metric are g_00, g_11, g_22, and g_33. You said:

g^33 is not the same as g_33. g_ii*g^ii =1. And so g^ii = g_ii^-1. It standard notation in differential geometry.

Then it is clear from eyeballing your derivation that g_11 and g_22 always drop out. That leaves only g_00 and g_33 as inputs, both of which refer to eq. 8 in my paper for the Schwarzschild metric:

Eq. 8 = 1 / gamma = sqrt(1 – (R / r))

My metric is derived by just replacing instances of eq. 8 with eq. 9:

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

You can see that the only difference between eqs. 8 and 9—hence the only difference between g_00 and g_33 for the metrics in your derivations—is that the sole r in eq. 8 is replaced with r + R in eq. 9. Then R_00 for my metric is calculable by plugging in the value (not the symbols) r + R for r into g_00 and g_33 for your derivation for the Schwarzschild metric, which always returns R_00 = 0. (For example, if R = 1 and r = 2 for the Schwarzschild metric, then plug in R = 1 and r = (2 + 1) = 3 to get R_00 for my metric for the inputs R = 1 and r = 2.) Then it is a mathematical certainty that R_00 = 0 for my metric too (again, assuming that your derivation for the Schwarzschild metric is correct). This completes my proof.

The proof tells me that there is at least one error in your analysis for my metric. To find them, I carefully entered your equations for each metric side by side in a spreadsheet, using sample inputs (e.g. R = 1 and r = 2). I verified the equations using the cross-reference equations (equivalent equations) in your analysis, and confirmed that R_00 = 0 for the Schwarzschild metric. Then I copied into a third column your equations for the Schwarzschild metric, except that I changed the value (the input value, not the symbol) for r, to r + R. (For example, if R = 1 and r = 2 for my metric, then in the third column I used R = 1 and r = (2 + 1) = 3.) Results in the column in my spreadsheet for my metric that do not match the corresponding result in the third column is either an equation in error, or based on an equation in error.

There are errors in eqs. 50, 55, and 63. I analyzed those equations further. This is what I found:

Eq. 50: The expression ∂/(∂x^3) should be equivalent to the expression in ∂/(∂x^3) in eq. 18 except that r is replaced with r + R, but it’s not. Instead it is equivalent to twice the expression ∂/(∂x^3) in eq. 18, for no apparent justifiable reason.

Eq. 55: The second line should be equivalent to the second line in eq. 23 except that r is replaced with r + R, but it’s not. Instead it is equivalent to the second line in eq. 23, for no apparent justifiable reason.

Eq. 63: The expression ∂/(∂x^3) is equivalent to R / (r^2 * g_00), the same as the expression ∂/(∂x^3) in eq. 31, for no apparent justifiable reason. The r not in g_00 in eq. 63 should be replaced with r + R.

When these errors are corrected and the corrections propagated as appropriate, R_00 = 0 for my metric.

Can you refute that?

Son Goku · 19-09-2006 10:01am

You can see that the only difference between eqs. 8 and 9—hence the only difference between g_00 and g_33 for the metrics in your derivations—is that the sole r in eq. 8 is replaced with r + R in eq. 9. Then R_00 for my metric is calculable by plugging in the value (not the symbols) r + R for r into g_00 and g_33 for your derivation for the Schwarzschild metric, which always returns R_00 = 0. (For example, if R = 1 and r = 2 for the Schwarzschild metric, then plug in R = 1 and r = (2 + 1) = 3 to get R_00 for my metric for the inputs R = 1 and r = 2.) Then it is a mathematical certainty that R_00 = 0 for my metric too (again, assuming that your derivation for the Schwarzschild metric is correct). This completes my proof.

No Zanket, you have to completely work out the Ricci Tensor again for a different metric. At no point are you allowed to convert r to r + R except at the very start, before the derivation. The whole process is nonlinear so you can't do this.

Eq. 50: The expression ∂/(∂x^3) should be equivalent to the expression in ∂/(∂x^3) in eq. 18 except that r is replaced with r + R, but it’s not. Instead it is equivalent to twice the expression ∂/(∂x^3) in eq. 18, for no apparent justifiable reason.

It shouldn't Zanket, work out the differential in both cases. From Eq. 15 in the first case and Eq. 47 in the second case. You'll see that they result in very different answers.
I can show you if you want.

Eq. 55: The second line should be equivalent to the second line in eq. 23 except that r is replaced with r + R, but it’s not. Instead it is equivalent to the second line in eq. 23, for no apparent justifiable reason.

Look at the two Christoffel symbols being multiplied. If you work out the derivation you'll see they can't come out the same.
I can show you if you want.

Eq. 63: The expression ∂/(∂x^3) is equivalent to R / (r^2 * g_00), the same as the expression ∂/(∂x^3) in eq. 31, for no apparent justifiable reason. The r not in g_00 in eq. 63 should be replaced with r + R.

Zanket do you understand these equations?
I just worked that out and got the answer Professor_Fink got.
What do you think ∂/(∂x^3) means?
Please answer this question because it is important in order to show you how the derivation works.

Remember the difference between raised and lowered components involves letting the metric or inverse metric act on one to give the other.

i.e., x^i = g^ik x_k

Professor_Fink · 19-09-2006 12:35pm

Zanket wrote:

You can see that the only difference between eqs. 8 and 9—hence the only difference between g_00 and g_33 for the metrics in your derivations—is that the sole r in eq. 8 is replaced with r + R in eq. 9. Then R_00 for my metric is calculable by plugging in the value (not the symbols) r + R for r into g_00 and g_33 for your derivation for the Schwarzschild metric, which always returns R_00 = 0. (For example, if R = 1 and r = 2 for the Schwarzschild metric, then plug in R = 1 and r = (2 + 1) = 3 to get R_00 for my metric for the inputs R = 1 and r = 2.) Then it is a mathematical certainty that R_00 = 0 for my metric too (again, assuming that your derivation for the Schwarzschild metric is correct). This completes my proof.

No, you are making a mistake here. The problem is that x^4 = r. If we replace r+R with r in you metric, then we also have to change x^4 to r+R. You're only doing the coordinate transform on the metric but not on x. This is why we are getting different answers. You cannot ignore x^4=r.

Surely you can see that my derivation is correct (up to a possible typo). There is no arguement you can put forth to show it's wrong, unless you spot a mistake in the mathematics.

Professor_Fink · 19-09-2006 12:42pm

Zanket wrote:

There are errors in eqs. 50, 55, and 63. I analyzed those equations further. This is what I found:

Eq. 50: The expression ∂/(∂x^3) should be equivalent to the expression in ∂/(∂x^3) in eq. 18 except that r is replaced with r + R, but it’s not. Instead it is equivalent to twice the expression ∂/(∂x^3) in eq. 18, for no apparent justifiable reason.

Eq. 55: The second line should be equivalent to the second line in eq. 23 except that r is replaced with r + R, but it’s not. Instead it is equivalent to the second line in eq. 23, for no apparent justifiable reason.

Eq. 63: The expression ∂/(∂x^3) is equivalent to R / (r^2 * g_00), the same as the expression ∂/(∂x^3) in eq. 31, for no apparent justifiable reason. The r not in g_00 in eq. 63 should be replaced with r + R.

When these errors are corrected and the corrections propagated as appropriate, R_00 = 0 for my metric.

Can you refute that?

Yes, of course I can refute this. Those quantities are christoffel symbols (the capital gammas) and so already contain partial derivatives with respect to r, so of course the there is differences between the two sections, d/dr g_00 and d/dr g_33 are not the same for both metrics!

planck2 · 19-09-2006 6:58pm

Didn't I say he could not calculate

Zanket · 19-09-2006 11:39pm

Son Goku wrote:

No Zanket, you have to completely work out the Ricci Tensor again for a different metric. At no point are you allowed to convert r to r + R except at the very start, before the derivation. The whole process is nonlinear so you can't do this.

In my proof, all I do is convert r to r + R at the very start, before the derivation. You agree that I can do that, so the proof holds.

Now I elaborate on my proof. Keep in mind that it involves only the Professor’s derivation for the Schwarzschild metric; it does not involve the derivation for my metric. In the Professor’s derivation for the Schwarzschild metric, consider:

g_00 = -1 + (R / r)
g_33 = (1 – (R / r))^1

Let y = r. (Nothing amiss there; it’s just algebraic substitution, which is always allowed.) Then for the Schwarzschild metric:

g_00 = -1 + (R / y)
g_33 = (1 – (R / y))^1

In the derivation, g_11 and g_22 drop out, and there are no other inputs. So had the Professor done the derivation using the versions of g_00 and g_33 having r replaced with y, the result would have been the same, namely R_00 = 0. And nothing prevents him from being able to do that. Do you agree? (Please answer that.)

For my metric let y = r + R. (Nothing amiss there; it’s just algebraic substitution, which is always allowed.) Then for my metric:

g_00 = -1 + (R / y)
g_33 = (1 – (R / y))^1

Notice that these inputs are identical to those for the Schwarzschild metric immediately above. Then all the inputs are the same for both metrics, which means that the derivation for R_00 for the Schwarzschild metric can substitute for the derivation for my metric. Nothing prevents me, when the derivation is done, from plugging in a value of r + R for y, instead of r, to calculate a value for R_00. Then it is a mathematical certainty that R_00 = 0 for my metric too, assuming that the Professor’s derivation for the Schwarzschild metric is valid. Do you agree? (Please answer that.)

Zanket wrote:

Eq. 50: The expression ∂/(∂x^3) should be equivalent to the expression in ∂/(∂x^3) in eq. 18 except that r is replaced with r + R, but it’s not. Instead it is equivalent to twice the expression ∂/(∂x^3) in eq. 18, for no apparent justifiable reason.

It shouldn't Zanket, work out the differential in both cases. From Eq. 15 in the first case and Eq. 47 in the second case. You'll see that they result in very different answers.
I can show you if you want.

I agree that they should differ; in what you quoted of mine, I said they should differ (“should be equivalent except that ...”). What I disagree with is how they differ. There’s no justification for one being twice the other. But yes, please show me the work. I’d like to see it.

Zanket wrote:

Eq. 55: The second line should be equivalent to the second line in eq. 23 except that r is replaced with r + R, but it’s not. Instead it is equivalent to the second line in eq. 23, for no apparent justifiable reason.

Look at the two Christoffel symbols being multiplied. If you work out the derivation you'll see they can't come out the same.
I can show you if you want.

You misread my quote. I said what you’re saying; namely, the results should be different (“should be equivalent except that ...”). But the expressions for the second line in eqs. 23 and 55 do not differ in the Professor's work; both are equivalent to 4 / r. Then you agree with me, in which case you implicitly agree that the Professor's derivation is invalid. But yes, please show me the work.

What do you think ∂/(∂x^3) means?
Please answer this question because it is important in order to show you how the derivation works.

I don’t fully know, but it became clear in my analysis that fully knowing the meaning of ∂/(∂x^3) is unnecessary to make my points. What I do know is that they are equivalent to another expression containing r, and it seems that this is all I need to know.

For example, it is easy to see that, given eqs. 25 and 57, the second line in both eqs. 23 and 55 is equivalent to 4 / r. You agreed with me that these should differ, so you implicitly agree that the Professor’s derivation is invalid.

Son Goku · 20-09-2006 12:20am

Zanket wrote:

You agreed with me that these should differ, so you implicitly agree that the Professor’s derivation is invalid.

Hilarious Zanket, you know that's not what I'm saying.

I don’t fully know, but it became clear in my analysis that fully knowing the meaning of ∂/(∂x^3) is unnecessary to make my points.

It is completely necessary. Saying that it isn't doesn't make any sense.
Every calculation in that pdf involves something of the form ∂/(∂x^i), you need to know how it works.

To be more simplistic look at the derivative with respect to x_3 of your g_00 and the Schwarschild g_00 and you will see how the answers are very different.
You're using linear logic from elementary algebra on a nonlinear tensor calculation.

I'll respond to your other points later unless somebody else does first.

As Professor_Fink said ∂/(∂x^i) is built into most of the Tensorial quantities.

Professor_Fink · 20-09-2006 12:28am

Zanket wrote:

In my proof, all I do is convert r to r + R at the very start, before the derivation. You agree that I can do that, so the proof holds.

Now I elaborate on my proof. Keep in mind that it involves only the Professor’s derivation for the Schwarzschild metric; it does not involve the derivation for my metric. In the Professor’s derivation for the Schwarzschild metric, consider:

g_00 = -1 + (R / r)
g_33 = (1 – (R / r))^1

Let y = r. (Nothing amiss there; it’s just algebraic substitution, which is always allowed.) Then for the Schwarzschild metric:

g_00 = -1 + (R / y)
g_33 = (1 – (R / y))^1

In the derivation, g_11 and g_22 drop out, and there are no other inputs. So had the Professor done the derivation using the versions of g_00 and g_33 having r replaced with y, the result would have been the same, namely R_00 = 0. And nothing prevents him from being able to do that. Do you agree? (Please answer that.)

I'm afraid you are wrong. x^3 = r, as I have mentioned in my derivation. You need to change x^3 to y-R. You're just using y. That is incorrect. You're not using the substitution consistently. So, no, I do not agree with you, and I have proved, rigorously, that R_00!=0. My derivation is accurate, and you seem not to understand how calculus works.

Zanket wrote:

I don’t fully know, but it became clear in my analysis that fully knowing the meaning of ∂/(∂x^3) is unnecessary to make my points. What I do know is that they are equivalent to another expression containing r, and it seems that this is all I need to know.

For example, it is easy to see that, given eqs. 25 and 57, the second line in both eqs. 23 and 55 is equivalent to 4 / r. You agreed with me that these should differ, so you implicitly agree that the Professor’s derivation is invalid.

Aaaaaaaaarrrrrrrrggghhh! ∂/(∂x^3) = ∂/∂r which is the partial derivative with respect to r. Your claim that R_00=0 is completely wrong, as you are not changing this r when you make the substitution. You NEED to do this.

Zanket · 20-09-2006 1:07am

Professor_Fink wrote:

Yes, of course I can refute this. Those quantities are christoffel symbols (the capital gammas) and so already contain partial derivatives with respect to r, so of course the there is differences between the two sections, d/dr g_00 and d/dr g_33 are not the same for both metrics!

Before I reply to the rest of your posts, let’s focus on this. Look closely at what you quoted of mine. I agree that there should be differences between the two sections for the expressions I mentioned in that quote; that’s what I’m saying there.

Let’s focus just on the second lines in eqs. 23 and 55. You say that these expressions should be different between the sections, but clearly they are equivalent. Both of the second lines are equivalent to 4 / r. Why is that not a problem, when you agree they should differ?

Zanket · 20-09-2006 1:25am

Son Goku wrote:

Hilarious Zanket, you know that's not what I'm saying.

Seriously, I don’t. Let’s focus on this:

Zanket wrote:

Eq. 55: The second line should be equivalent to the second line in eq. 23 except that r is replaced with r + R, but it’s not. Instead it is equivalent to the second line in eq. 23, for no apparent justifiable reason.

Look at the two Christoffel symbols being multiplied. If you work out the derivation you'll see they can't come out the same.

By “they”, you mean the second lines in eqs. 23 and 55, right? But they are the same. They are both equivalent to 4 / r. We both think that is a problem. So how is it that you are not contradicting yourself? I really don’t get it.

Zanket · 20-09-2006 2:03am

Professor_Fink wrote:

Yes, of course I can refute this.

Son Goku wrote:

Hilarious Zanket, you know that's not what I'm saying.

There seems to be some miscommunication here. To be clear, when I say above that the second lines in eqs. 23 and 55 are the same, both equivalent to 4 / r, I’m not saying that I derived them and they are the same. I’m saying that they are the same right in the Professor’s derivation. In other words, it is the Professor who says that they are the same, not me.

Professor_Fink · 20-09-2006 2:12am

Ok, here is the derivation using your substitution. As you see, R_00 is still non-zero. You are not making the correct coordinate substitutions.

Professor_Fink · 20-09-2006 2:18am

Zanket wrote:

There seems to be some miscommunication here. To be clear, when I say above that the second lines in eqs. 23 and 55 are the same, both equivalent to 4 / r, I’m not saying that I derived them and they are the same.

Why should they be different? It doesn't matter that the metrics are different, this particular function of both metrics is the same for both.

The g_00 and g_33 terms cancel in both, only the g_11 and g_22 terms contribute here, and g_11 and g_22 are the same for both!

What is your problem with this?

Zanket · 20-09-2006 8:11am

Professor_Fink wrote:

Ok, here is the derivation using your substitution. As you see, R_00 is still non-zero. You are not making the correct coordinate substitutions.

Thanks, I’ll come back to this if needed.

Why should they be different? It doesn't matter that the metrics are different, this particular function of both metrics is the same for both.

The g_00 and g_33 terms cancel in both, only the g_11 and g_22 terms contribute here, and g_11 and g_22 are the same for both!

Then why does Son Goku disagree with you?:

Son Goku wrote:

Look at the two Christoffel symbols being multiplied. If you work out the derivation [for the second lines in eqs. 23 and 55] you'll see they can't come out the same.

Please show your work here, for the second line in eq. 55. The g_00 term cannot cancel. You said that g_00 * g^00 = 1. Then the g_00 term in the second line must = R / (r * (r + R)), to be consistent with eqs. 43 to 45. I believe the 1 / r in eq. 56 should be 1 / (r + R).

Also please show your work between eqs. 50 and 51. There is a mistake there too; more on that below.

Eqs. 64 and 65 are invalid. Between eqs. 63 and 64 you have:

(∂ / ∂x^3) * g_00 = -(R / r^2)

But between eqs. 43 and 45 you have:

(-∂ / ∂x^3) * g_00 = (R / (r + R)^2)

Eq. 43 to 45 are consistent with eqs. 11 to 13, but eqs. 63 to 65 are inconsistent with eqs. 31 to 33. Eq. 65 should be = R^2 / (2 * (r + R)^4).

Also, a minor problem in eq. 32: the expression in parentheses should have a negative sign, to be consistent with eq. 64. This doesn’t affect your result.

Some background on the problem between eqs. 50 and 51:

Eq. 19 is equivalent to: (R * ((3 * R) - (2 * r))) / (2 * r^4)

Eq. 19 depends on only the inputs g_00 and g_33, so if I replace every r with r + R, I should get an equation that is equivalent to eq. 51. When I replace every r with r + R and simplify, I get:

(R * (R - (2 * r))) / (2 * (r + R)^4)

But eq. 51 is equivalent to:

(R * (R - (2 * r))) / (r + R)^4

There’s no justification for the difference in red. Eq. 51 is twice what it should be, if it was consistent with eq. 19. You must have made a mistake.

Son Goku · 20-09-2006 11:09am

Zanket, this is a very simple derivation of a standard quantity in differential geometry. You're asking such confusing questions that I've probably made a slip up along the way.

To reset everything and lose the confusion, I ask that you list by number your current contentions with Professor_Fink's derivation.

i.e.:
1. Eq. 64 has........
2. Eq. 45 should be the same as.....
e.t.c.

Then the thread will continue from there. I ask for Professor_Fink and planck2 to wait until Zanket has done this and only respond based on Zanket's next post.

Professor_Fink · 20-09-2006 11:11am

Ok, I'll wait.

Professor_Fink · 20-09-2006 11:37am

Actually, Zanket does point out mistakes in my calculation. I'm just correcting it now. I'll post it in about 10 mins.

Professor_Fink · 20-09-2006 12:06pm

Zanket wrote:

Please show your work here, for the second line in eq. 55. The g_00 term cannot cancel. You said that g_00 * g^00 = 1. Then the g_00 term in the second line must = R / (r * (r + R)), to be consistent with eqs. 43 to 45. I believe the 1 / r in eq. 56 should be 1 / (r + R).

No, you're wrong. I have added the derivation as eq 69/70.

Zanket wrote:

Also please show your work between eqs. 50 and 51. There is a mistake there too; more on that below.

Eqs. 64 and 65 are invalid. Between eqs. 63 and 64 you have:

(∂ / ∂x^3) * g_00 = -(R / r^2)

But between eqs. 43 and 45 you have:

(-∂ / ∂x^3) * g_00 = (R / (r + R)^2)

Eq. 43 to 45 are consistent with eqs. 11 to 13, but eqs. 63 to 65 are inconsistent with eqs. 31 to 33. Eq. 65 should be = R^2 / (2 * (r + R)^4).

Also, a minor problem in eq. 32: the expression in parentheses should have a negative sign, to be consistent with eq. 64. This doesn’t affect your result.

Some background on the problem between eqs. 50 and 51:

Eq. 19 is equivalent to: (R * ((3 * R) - (2 * r))) / (2 * r^4)

Eq. 19 depends on only the inputs g_00 and g_33, so if I replace every r with r + R, I should get an equation that is equivalent to eq. 51. When I replace every r with r + R and simplify, I get:

(R * (R - (2 * r))) / (2 * (r + R)^4)

But eq. 51 is equivalent to:

(R * (R - (2 * r))) / (r + R)^4

There’s no justification for the difference in red. Eq. 51 is twice what it should be, if it was consistent with eq. 19. You must have made a mistake.

Yes, I made a few slips. All are fixed in the new version here. Note that this is a new link!

R_ab is still non-zero though. It is absolutely impossible to make R_ab = 0 for your metric.

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

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