Anamitra Palit Posted January 23 Share Posted January 23 (edited) [One my have to refresh the page for viewing the formulas and the equations] We consider the transformation of the rank two contravariant tensor: \[\bar A^{\mu \nu}=\frac{\partial \bar x^{\mu}}{\partial x^\alpha}\frac{\partial \bar x^{\nu}}{\partial x^\beta}A^{\alpha\beta}\](1) Inverse transformation[for non singular transformations] \[A^{\alpha \beta}=\frac{\partial x^\alpha}{\partial \bar x^\mu}\frac{\partial x^{\beta}}{\partial \bar x^\nu}\bar A^{\mu\nu}\](2) For the diagonal components[alpha=beta] \[ A^{\alpha \alpha}=\frac{\partial x^\alpha}{\partial \bar x^\mu}\frac{\partial x^{\alpha}}{\partial \bar x^\nu}\bar A^{\mu\nu}\] (3) We consider the situation where the off diagonal components of A[alpha not equal to beta] are all zero The diagonal elements of A-bar are given by \[\bar A^{\mu\mu}=\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)^2 A^{\alpha\alpha}\] (4) For off diagonal elements of A-bar are given by ' \[\bar A^{\mu\nu}=\frac{\partial \bar x^\mu}{\partial x^\alpha} \frac{\partial \bar x^\mu}{\partial x^\alpha} A^{\alpha\alpha}\] (4’) Subject to the situation that the off diagonal elements of A are all zero we consider the following three cases: Case 1. Assume A-bar^ mu nu=0for all μ≠ν for all barred reference frames[the general condition that off diagonal elements of A are zero continues to hold] we have \[\frac{\partial \bar x^\mu}{\partial x^\alpha} \frac{\partial \bar x^\mu}{\partial x^\alpha} A^{\alpha\alpha}=0\] (4’’) We have (4'') irrespective of the transformation elements[all reference frames being considered]. From (4’’) A-bar^mu mu=0.Again from (4’) and (4’’) |A-bar ^mu nu becomes zero for μ≠ν. Then the tensor becomes null Case 2 Off diagonal elements are non zero for all A-bar[that the off diagonal elements of A are all zero continues to hold] \[\bar A^{\mu\mu}=\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)\frac{\partial x^\alpha}{\partial \bar x^\rho} \frac{\partial x^\alpha}{\partial \bar x^\sigma} A^{\rho\sigma}\] \[\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)\frac{\partial x^\alpha}{\partial \bar x^\rho} \frac{\partial x^\alpha}{\partial \bar x^\sigma} A^{\rho\sigma}=\delta^{\mu}_{\rho} \delta^{\mu}_{\sigma}\] \[\frac{\partial \bar x^\rho}{\partial x^\beta}\frac{\partial \bar x^\sigma}{\partial x^\gamma}\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)\frac{\partial x^\alpha}{\partial \bar x^\rho} \frac{\partial x^\alpha}{\partial \bar x^\sigma} A^{\rho\sigma}=\frac{\partial \bar x^\rho}{\partial x^\beta}\frac{\partial \bar x^\sigma}{\partial x^\gamma}\delta^{\mu}_{\rho} \delta^{\mu}_{\sigma}\] \[\Rightarrow\left(\frac{\partial \bar x^\mu}{\partial x^\alpha}\right)^2\delta^\alpha_\beta \delta^\alpha_\gamma=\frac{\partial \bar x^\mu} {\partial x^\beta}\frac{\partial \bar x^\mu}{\partial x ^\gamma}\] For β≠γ we have from (5) \[\frac{\partial \bar x^\mu}{\partial x ^\beta}\frac{\partial \bar x^\mu}{\partial x ^\gamma}=0\](6) \[\Rightarrow\frac{\partial \bar x^\mu}{\partial x^\beta}=0\]or \[\frac{\partial \bar x^\mu}{\partial x ^\beta}=0\] (7) \[\Rightarrow \bar A^{\mu\nu}=0\Rightarrow A^{\alpha\beta}=0\] (8) Equations represented by (7) and (8)are not true! We do have an enigma of a persistent nature. [In fact (8) implies that the metric tensor is the null tensor;the Riemann tensor and the Ricci tensor are null tensors; the Ricci scalar is zero valued.] Case 3 For A as per our initial postulation the off diagonal elements of A are all zero. For A-bar there we assume the existence of at least one mu ,nu pair in each barred frame such that A-bar^mu nu=0.These mu nu pairs may not be identical for all barred frames. Since we may have an infinitude of reference frames catering to our assumption it follows that for some specific (μ,ν)pair also A-bar mu mu is zero in an infinitely many frames of reference \[\bar A^{\mu\nu}=\frac{\partial \bar x^\mu}{\partial x^\alpha} \frac{\partial \bar x^\nu}{\partial x^\alpha}A^{\alpha\alpha}\] Since A^m mu=0 for a specific pair in infinitely many frames,we have, \[\frac{\partial \bar x^\mu}{\partial x^\alpha} \frac{\partial \bar x^\nu}{\partial x^\alpha}A^{\alpha\alpha}=0\](7) For the same mu nu pair (7) we have an infinitude of equations like (7) whee the transformation elements are distinct. We have included infinitely many barred frames in equation (7) \[\Rightarrow A^{\alpha\alpha}=0\Rightarrow A=0\] (8) Case 1 is an instance of case 3 Points to Observe: 1. Anti symmetric tensors transform to anti symmetric tensors in all frames of reference. Noe the null tensor is an antisymmetic tensor asides being a symmetric one.An arbitrary non zero[non null] tensor may be expressed as the sum of a symmetric an an antisymmetric tensor. Our analysis might be considered for an arbitrary tensor which is neither symmetric or antisymmetric.It reduces the null tensor making both the symmetric and the antisymmetric parts zero. 2.If a tensor has a component of identical value in all reference frames it has to be the null tensor. 3. Let us represent (1) in a standard matrix form \[\bar A=MAM^T\](9) where M is the transformation matrix For an arbitrary M even if M is non singular in nature, we will not necessarily have both A and A-bar as diagonal[both with the off diagonal elements as zero].If one of A or A-bar is diagonal with the other non diagonal then A ,A-bar both become null tensors as we have already seen! One may test the situation using a diagonal tensor with A and an arbitrary square matrix for M For an arbitrary transformation matrix ,a diagonal tensor A does not necessarily produce a diagonal A-bar. Only by specific valid choice of the space time transformation can we have both sides of (10) or (1) for that matter, as diagonal[components zero for alpha not equal to beta]. An arbitrary[non singular ] transformation will produce from a diagonal tensor a non diagonal. It is always possible to have infinitely many transformation for which the off diagonal elements are non zero[diagonal elements of the tensor in some reference frame are assumed to be non zero] Edited January 23 by Anamitra Palit Link to comment Share on other sites More sharing options...

Bufofrog Posted January 23 Share Posted January 23 Have you ever considered finishing some of your open threads before starting new ones? Link to comment Share on other sites More sharing options...

Kino Posted January 23 Share Posted January 23 1 hour ago, Anamitra Palit said: For off diagonal elements of A-bar are given by ' A¯μν=∂x¯μ∂xα∂x¯μ∂xαAαα This expression is clearly invalid because the indices do not match. 1 hour ago, Anamitra Palit said: Off diagonal elements are non zero for all A-bar[that the off diagonal elements of A are all zero continues to hold] A¯μμ=(∂x¯μ∂xα)∂xα∂x¯ρ∂xα∂x¯σAρσ This expression is clearly invalid because the indices do not match. Is \( \alpha \) a dummy index? Is \( \mu \) a dummy index? 1 hour ago, Anamitra Palit said: If one of A or A-bar is diagonal with the other non diagonal then A ,A-bar both become null tensors as we have already seen! Obviously not true, and counterexamples are common. See, for example, Eddington-Finkelstein coordinates (non-diagonal) and Schwarzschild coordinates (diagonal) on Schwarzschild spacetime. Link to comment Share on other sites More sharing options...

joigus Posted January 23 Share Posted January 23 13 minutes ago, Kino said: This expression is clearly invalid because the indices do not match. The OP wants to fix the value of the alpha index and at the same time keep Einstein's summation convention. Very dangerous practice. No wonder they get inconsistent results. Link to comment Share on other sites More sharing options...

Markus Hanke Posted January 24 Share Posted January 24 8 hours ago, Anamitra Palit said: We do have an enigma of a persistent nature. To me it is an enigma why you keep trying to propose things that are just trivially wrong, and easily contradicted by simple examples. 8 hours ago, Anamitra Palit said: In fact (8) implies that the metric tensor is the null tensor;the Riemann tensor and the Ricci tensor are null tensors; the Ricci scalar is zero valued. Even in ordinary Euclidean space (the kind you deal with in high school), the metric tensor is evidently not null. Have you considered what it would mean for all tensors to be null? It would mean (among other things) that there would be no notion of distance, area, volume, or angles, and hence that the length of any arbitrary curve in that space is identically null, irrespective of your choice of parametrisation. It would be a world without any concept of separation in space, or duration in time. This is evidently nothing like the world we actually live in, so why would you make such a claim? 9 hours ago, Anamitra Palit said: An arbitrary non zero[non null] tensor may be expressed as the sum of a symmetric an an antisymmetric tensor. This is only true across two indices though: \[A_{\alpha \beta \gamma ...} =A_{( \alpha \beta ) \gamma ...} +A_{[ \alpha \beta ] \gamma ...}\] 9 hours ago, Anamitra Palit said: It reduces the null tensor making both the symmetric and the antisymmetric parts zero. See above - the vanishing of symmetric and antisymmetric parts does not necessarily yield a null tensor. 9 hours ago, Anamitra Palit said: If a tensor has a component of identical value in all reference frames it has to be the null tensor. Not true. For example, the Minkowski metric is locally the same in all reference frames, yet it is most certainly not a null tensor. 9 hours ago, Anamitra Palit said: For an arbitrary transformation matrix ,a diagonal tensor A does not necessarily produce a diagonal A-bar. It doesn’t need to. For example, you can cover a patch of Schwarzschild spacetime with a Gullstrand-Painleve chart, or with a Schwarzschild coordinate chart; they are related by a simple coordinate transformation, but the former contains off-diagonal elements in the metric, whereas the latter doesn’t. They both describe the same spacetime. Link to comment Share on other sites More sharing options...

Kino Posted January 24 Share Posted January 24 15 hours ago, joigus said: The OP wants to fix the value of the alpha index and at the same time keep Einstein's summation convention. Very dangerous practice. No wonder they get inconsistent results. It's worse than that. (4') has two free indices on the left and either zero or one on the right depending on how you interpret the repeated \( \mu \). And look at the indices in the first equation in the un-numbered block after (4''). Indices \( \rho \) and \( \sigma \) are used on the unbarred tensor and the barred coordinate differentials. Compare with the correctly stated transformation, (1). So the two derivatives not in brackets are (probably...) the barred-to-unbarred transform applied to the already unbarred tensor. Then we get to your point about whatever it is OP thinks they are doing with the triply repeated \( \alpha \). 8 hours ago, Markus Hanke said: To me it is an enigma why you keep trying to propose things that are just trivially wrong, and easily contradicted by simple examples. Having looked at a few more of the OP's postings I think they believe that there is something fundamentally wrong with maths. Therefore they expect illogical and contradictory results and, when they get them, do not bother to check for their own errors. Link to comment Share on other sites More sharing options...

Markus Hanke Posted January 25 Share Posted January 25 17 hours ago, Kino said: Having looked at a few more of the OP's postings I think they believe that there is something fundamentally wrong with maths. Therefore they expect illogical and contradictory results and, when they get them, do not bother to check for their own errors. I would tend to agree. Link to comment Share on other sites More sharing options...

joigus Posted January 25 Share Posted January 25 18 hours ago, Kino said: It's worse than that. (4') has two free indices on the left and either zero or one on the right depending on how you interpret the repeated [...] Oh, but that's not because it's much worse than I pointed out. It's because it's bound to get worse if you make a notational blunder of that magnitude. If you want to discuss anything in terms of a 2-index tensor being diagonal in a certain point --o perhaps everywhere?, the OP didn't tell us--, you could arrange to distinguish this by using Latin capital letters, e.g., \[A^{BB}=\frac{\partial x^{B}}{\partial\bar{x}^{\mu}}\frac{\partial x^{B}}{\partial\bar{x}^{\nu}}\bar{A}^{\mu\nu}\] Meaning, \[A^{00}=\frac{\partial x^{0}}{\partial\bar{x}^{\mu}}\frac{\partial x^{0}}{\partial\bar{x}^{\nu}}\bar{A}^{\mu\nu}\] \[A^{11}=\frac{\partial x^{1}}{\partial\bar{x}^{\mu}}\frac{\partial x^{1}}{\partial\bar{x}^{\nu}}\bar{A}^{\mu\nu}\] etc. So it can be done, but not the way the OP is doing it. Not that it's very useful to consider tensors as objects that are or aren't diagonal in any invariant geometrical sense, as they are objects referred to two different bases. 18 hours ago, Kino said: Having looked at a few more of the OP's postings I think they believe that there is something fundamentally wrong with maths. Therefore they expect illogical and contradictory results and, when they get them, do not bother to check for their own errors. Absolutely. When I'm doing maths and I get to such surprising results as "the whole of tensor algebra/calculus is bonkers, because all tensors are null" --or something like that, I'm not completely sure if that's the point--, I try to retrace my steps and, sure enough, I can spot a silly mistake. The last thing that would cross my mind is to highlight the "result" and announce to the world, "hey, I've found an enigma". 2 Link to comment Share on other sites More sharing options...

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