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A Flaw of General Relativity, a New Metric and Cosmological Implications [Technical]
Comments
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planck2 wrote: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.0
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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.0 -
Professor_Fink wrote:You didn't show anything about R_ab, you just showed that you don't know how to differentiate.planck2 wrote:R_ab= 2*R^3/((r+R)^3*r^2), modulo a sign of course because I used -+++.
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.
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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.0 -
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.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?
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?0 -
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.0
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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.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.Schwarschild spacetime is empty everywhere.
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R_ab = 0 over the whole of the Schwarzschild spacetime, but not over the whole of yours.
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.
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.
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.
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.0 -
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?0
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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.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.0 -
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.0 -
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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.
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.0 -
oh i wouldn't say that at all. I am perfectly capable of finding the flaw in an arguement. Your's being a prime example0
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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.0
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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.0
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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 not0
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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.
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 ...... and you shoot them down when they say something which you claim you can't even verify. You talk nonsense.If I were the mod I would have finished this discussion ages ago.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
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?0 -
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!0 -
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.0 -
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.0 -
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!
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.
You can't expect us to take it on faith.0 -
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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.
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.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 1.0 -
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.I am willing to listen to well reasoned arguement, but your argument doesn't belong to that equivalence class.0
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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.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.
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.You can't expect us to take it on faith.0 -
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.0 -
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.0 -
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) 5063-5077.0
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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.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.0 -
planck2 wrote:Further your metric is the Schwarzschild metric with M replaced with (-M).0
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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_11.0 -
Professor_Fink wrote:Here is a derivation for R_00 for both Zankets metric and the Schwarzschild metric.0
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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.As for relying on Prof Fink and Son Goku they would probably agree that I need no help from them"experimental" verifications my eye.However to correctly get your metric you replace M with - M and swap g_00 and g_11.0 -
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.0
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Professor_Fink wrote:Here is a derivation for R_00 for both Zankets metric and the Schwarzschild metric.
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Clearly R_00 is nonzero for Zankets metric, and R_00=0 for the Schwarzschild and Minkowski metrics.
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.
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?0 -
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.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.
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.
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.
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_k0 -
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.0 -
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!0 -
Didn't I say he could not calculate0
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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.
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.
I can show you if you want.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.
I can show you if you want.What do you think ∂/(∂x^3) means?
Please answer this question because it is important in order to show you how the derivation works.
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.0 -
Zanket wrote:You agreed with me that these should differ, so you implicitly agree that the Professor’s derivation is invalid.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.
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.0 -
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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.0 -
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!
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?0 -
Son Goku wrote:Hilarious Zanket, you know that's not what I'm saying.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.0
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Professor_Fink wrote:Yes, of course I can refute this.Son Goku wrote:Hilarious Zanket, you know that's not what I'm saying.0
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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.0
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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?0 -
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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.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!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.
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.0 -
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.0 -
Ok, I'll wait.0
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Actually, Zanket does point out mistakes in my calculation. I'm just correcting it now. I'll post it in about 10 mins.0
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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.0 -
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