Anamitra Palit 11 Posted January 20 Share Posted January 20 (edited) Link to file on the Google Drive https://drive.google.com/file/d/1C2-ru6uuDIw9u_e4HQ1bdwa01oKysQ-B/view?usp=sharing Material in Latex[It might be necessary to refresh the page for viewing the formulas and the equations] The paper establishes mathematically that the Riemann Tensor is a zero tensor Riemann Curvature |Tensor \[R^{\mu\nu}_{\quad\gamma\delta}=g^{\alpha\mu}g^{\beta\nu}R_{\alpha\beta\gamma\delta}\] (1) Interchanging the dummy indices alpha and beta we have, \[R^{\mu\nu}_{\quad\gamma\delta}=g^{\beta\mu}g^{\alpha\nu}R_{\beta\alpha\gamma\delta}\](2) Therefore, \[g^{\alpha\mu}g^{\beta\nu}R_{\alpha\beta\gamma\delta}=g^{\beta\mu}g^{\alpha\nu}R_{\beta\alpha\gamma\delta}\] (3) But \[R_{\beta\alpha\mu\nu}=-R_{\alpha\beta\mu\nu}\] (4) Thus we have \[g^{\alpha\mu}g^{\beta\nu}R_{\alpha\beta\gamma\delta}=-g^{\beta\mu}g^{\alpha\nu}R_{\alpha\beta\gamma\delta}\] (5) \[R_{\alpha\beta\gamma\delta}\left[g^{\alpha\mu}g^{\beta\nu}+g^{\beta\mu}g^{\alpha\nu}\right]=0\] (6) With (6) alpha and beta are dummy indices;others are free indices.Treating the Riemann tensor components as variables[unknowns], there are sixteen of them for the sixteen alpha beta combinations any given gamma , delta and mu, nu combinations.Each equation has gamma,delta,mu and nu as constant quantities. In total we have 256 equations[four values for each gamma,delta,mu and nu].Since these equations are of homogeneous nature, \[R_{\alpha\beta\gamma\delta}=0\] (7) In the orthogonal system of coordinates we obtain from (6) for distince mu and nu \[R_{\mu\nu\gamma\delta}\left[g^{\mu\mu}g^{\nu\nu}+g^{\nu\mu}g^{\mu \nu}\right]=0\] (8) In (8) we have considered different values of mu and nu as well as different values for gamma and delta. One should also take note of the fact that there is no summation on mu and on nu. \[R_{\mu\nu\gamma\delta}g^{\mu\mu}g^{\nu\nu}=0\] (9) In (8) also there is no summation on mu and on nu. Since g_mu mu not equal to 0 and g_nu nu not equal to zero we have for unequal mu ,nu and unequal gamma delta, \[R_{\mu\nu\gamma\delta}=0\] (10) If the Riemann tensor is a zero tensor ten the Ricci tensor is also a zero tensor.The Ricci scala becomes a zero valued scalar. Riemann Curvature.pdf Edited January 20 by Anamitra Palit Link to post Share on other sites

Markus Hanke 519 Posted January 21 Share Posted January 21 19 hours ago, Anamitra Palit said: If the Riemann tensor is a zero tensor ten the Ricci tensor is also a zero tensor.The Ricci scala becomes a zero valued scalar. Yes - this is trivially true, since a vanishing Riemann tensor means you are on a flat manifold. More generally, if a tensor vanishes then so do all of its contractions - this is true for any tensor. 19 hours ago, Anamitra Palit said: The paper establishes mathematically that the Riemann Tensor is a zero tensor Not really. It establishes only that its contractions vanish if Riemann itself vanishes. The reverse is not true, however - the vanishing of the Ricci tensor and/or scalar do not necessarily imply that Riemann is zero. Ricci flatness is a necessary but not a sufficient condition for the absence of Riemann curvature; to make it a sufficient condition, you need to demand the vanishing of Weyl curvature as well. The Ricci tensor is the trace of Riemann, whereas the Weyl tensor encodes the trace-free part of Riemann (the decomposition isn’t exactly trivial, though). Link to post Share on other sites

Kino 10 Posted January 23 Share Posted January 23 (edited) (7) does not follow from (6). The term in square brackets in (6) is symmetric in \( \alpha \) and \( \beta \), so (6) is satisfied by any arbitrary four-index tensor \( R_{\alpha\beta\gamma\delta} \) (not just the Riemann tensor) as long as it is antisymmetric in \( \alpha \) and \( \beta \). This is not surprising, since (6) is nothing more than a restatement of (4), which itself is nothing more than a statement that \( R \) is antisymmetric in its first two indices (true of the Riemann, but also of a great many other tensors). It certainly does not follow that all components of the Riemann tensor (or any other arbitrary four index tensor antisymmetric in its first two indices) are zero, as can be verified by (for example) calculating the Riemann tensor for the Schwarzschild metric and confirming that (6) is satisfied for at least one choice of \( \gamma \) and \( \delta \) for which at least some of the \( R_{\alpha\beta\gamma\delta} \) are non-zero. E.g: pick \( \gamma=r, \delta=t \). The non-zero \( R_{\alpha\beta r t} \) in (+---) signature Schwarzschild coordinates are \( R_{trrt}=2m/r^3 \) and \(R_{rtrt}=-2m/r^3 \). Thus (6) expands to \[ R_{trrt}(g^{t\mu}g^{r\nu}+g^{r\mu}g^{t\nu})+R_{rtrt}(g^{r\mu}g^{t\nu}+g^{t\mu}g^{r\nu})=\frac{2m}{r^3}(g^{t\mu}g^{r\nu}+g^{r\mu}g^{t\nu}-g^{r\mu}g^{t\nu}-g^{t\mu}g^{r\nu}) \]The term in brackets on the right hand side is clearly zero, so (6) is satisfied even though the Riemann tensor is clearly not zero. Edited January 23 by Kino 2 Link to post Share on other sites

joigus 492 Posted January 23 Share Posted January 23 (edited) 7 hours ago, Kino said: (7) does not follow from (6). The term in square brackets in (6) is symmetric in α and β , so (6) is satisfied by any arbitrary four-index tensor Rαβγδ (not just the Riemann tensor) as long as it is antisymmetric in α and β . Exactly. I can barely add anything significant. Contraction of pairs of symmetric indices with pairs of antisymmetric indices always gives zero, no matter what the value of the non-zero components of both. The OP clearly has problems with the tensor formalism. \[ A_{\alpha\beta}=-A_{\beta\alpha}\] \[ S_{\alpha\beta}=S_{\beta\alpha}\] \[ Q=S^{\alpha\beta}A_{\alpha\beta}=S^{\beta\alpha}A_{\alpha\beta}\] (swapping dummies) and, \[ S^{\beta\alpha}A_{\alpha\beta}=\left(S^{\alpha\beta}\right)\left(-A_{\alpha\beta}\right)=-Q\] (applying symmetry properties) As \( Q=-Q \), \(Q\) must be zero, even if \( A_{\alpha\beta} \) and \( S_{\alpha\beta} \) are not. The rest of the indices are along for the ride. Edited January 23 by joigus Link to post Share on other sites

Markus Hanke 519 Posted January 24 Share Posted January 24 7 hours ago, joigus said: The OP clearly has problems with the tensor formalism. I think the problem is one of physical meaning. It is not enough to just blindly do index gymnastics, if one does not understand the physical significance of the objects that one is manipulating, as well as of the manipulations themselves. All physical situations can be described mathematically, but not everything that is mathematically doable is automatically physically valid or meaningful. This has been a recurring issue across all of the threads he has opened here. It also isn’t clear what the actual claim really is, which has also been an issue on the other threads. Based on his last sentence, I understood the OP’s claim to be that if the Riemann tensor is zero, then all contractions derived from it are zero as well, which is trivially true. Only when reading Kino’s excellent (+1) post did I realise that what he actually seems to claim is that Riemann is always zero - which is of course both trivially wrong and physically meaningless. It is, in fact, so meaningless that it didn’t even occur to me that this might be what he trying to show! Link to post Share on other sites

joigus 492 Posted January 24 Share Posted January 24 3 hours ago, Markus Hanke said: I think the problem is one of physical meaning. Yes, this surfaced in the 1st post already. 3 hours ago, Markus Hanke said: It also isn’t clear what the actual claim really is, Yes, that's another problem. But also, the OP is not familiar with tensor algebra. They have a tendency to use repeated indices both for summation convention and for representing fixed diagonal elements, so no wonder the conclusions are wrong, already just at the mathematical level. Also, I've observed them being very cavalier in asserting other properties about tensors. Symmetric or antisymmetric only make sense for tensors twice covariant or twice contravariant, etc. The delta tensor, or tensor products of delta tensors; and the epsilon tensor, or tensor products of them--, are isotropic tensors only when they are pure-covariant or pure-contravariant. And so on. Seeing tensor algebra used like this brings tears to my eyes. That's why I'm taking certain distance from this particular OP's posts. I'm only too glad I have you and Kino to help with this. Link to post Share on other sites

joigus 492 Posted January 24 Share Posted January 24 (edited) 1 hour ago, joigus said: The delta tensor, or tensor products of delta tensors; and the epsilon tensor, or tensor products of them--, are isotropic tensors only when they are pure-covariant or pure-contravariant. And so on. Sorry. I made a mistake here. The delta tensor is an isotropic tensor only when it is a once-covariant, once-contravariant tensor --similarly for tensor products of them--. And the epsilon tensor is an isotropic tensor only when it's totally covariant or totally contravariant. I already pointed this out in a previous post. Same OP. Different thread. Edited January 24 by joigus Link to post Share on other sites

Kino 10 Posted January 24 Share Posted January 24 (edited) 10 hours ago, Markus Hanke said: I think the problem is one of physical meaning. In this thread, I think the problem is that OP did not realise that using the label \( R_{\alpha\beta\gamma\delta} \) for a tensor does not make it the Riemann tensor (a zero tensor satisfies the constraint placed on the tensor here but not all constraints on the actual Riemann). In the "Enigma of the tensors" thread I think it is incorrect index gymnastics. In "Norm Square of the Four Acceleration Vector" it is a misunderstanding of calculus. In "On Four Velocity and Four Momentum" it is that two vectors with a specified norm will not add to another vector with the same norm. I do not think OP has just one single problem. Edited January 24 by Kino Link to post Share on other sites

Anamitra Palit 11 Posted January 25 Author Share Posted January 25 (edited) [It might be necessary to refresh the page for proper viewing] Thanking Kino for his first comment.[posted on Saturday,3:43 pm] It is true that the product of a symmetric and antisymmetric tensor is always zero. Nevertheless we may consider the following: \[R_{\alpha\beta\gamma\delta}\left(g^{\alpha\mu}g^{\beta\nu}+ g^{\beta\mu}g^{\alpha\nu}\right)=0\](1) [A significant aspect of (1) is that it holds for any arbitrary(mu,nu) pair] Keeping gamma and delta constant[and we maintain so throughout this post] we vary alpha and beta for each equation in (1). In order to obtain new equations of the type (1) we vary mu and nu There are 16 unknowns in R_αβγδ for the four alpha and the four beta while we have 16 equations[linear homogeneous equations ] in that mu and nu take on four values of each. Unless the determinant of the coefficient matrix is zero the solutions to R_αβγδ are trivial in their nature. This cannot be violated subject to a coefficient matrix with non zero determinant] We have \[R_{\alpha\beta\gamma\delta}=0\] (2) [The null tensor is always antisymmetric[ asides being symmetric]] We may write (1) in the form \[\left(R_{\alpha\beta\gamma\delta}+R_{\beta\alpha\gamma\delta}\right)\left(g^{\alpha\mu}g^{\beta\nu}+ g^{\beta\mu}g^{\alpha\nu}\right)=0\](3) We have eight variables of the form \[R_{\alpha\beta\gamma\delta}+R_{\beta\alpha\gamma\delta}\] while the number of equations continue to remain sixteen. Gamma and delta have been maintained constant throughout the writing as mentioned earlier. Edited January 25 by Anamitra Palit Link to post Share on other sites

Markus Hanke 519 Posted January 25 Share Posted January 25 16 hours ago, Kino said: In this thread, I think the problem is that OP did not realise that using the label Rαβγδ for a tensor does not make it the Riemann tensor (a zero tensor satisfies the constraint placed on the tensor here but not all constraints on the actual Riemann). Good point! Link to post Share on other sites

Anamitra Palit 11 Posted January 25 Author Share Posted January 25 40 minutes ago, Markus Hanke said: Good point! It is not a correct point.General perspectives cannot be violated in the special cases.Why don't you go through my last post in this discussion? -2 Link to post Share on other sites

Markus Hanke 519 Posted January 25 Share Posted January 25 5 minutes ago, Anamitra Palit said: It is not a correct point.General perspectives cannot be violated in the special cases.Why don't you go through my last post in this discussion? All you have done is repeat what you had already stated, you did not actually address any of the points raised. 1 Link to post Share on other sites

Anamitra Palit 11 Posted January 25 Author Share Posted January 25 (edited) 17 hours ago, Kino said: In this thread, I think the problem is that OP did not realise that using the label Rαβγδ for a tensor does not make it the Riemann tensor (a zero tensor satisfies the constraint placed on the tensor here but not all constraints on the actual Riemann). In the "Enigma of the tensors" thread I think it is incorrect index gymnastics. In "Norm Square of the Four Acceleration Vector" it is a misunderstanding of calculus. In "On Four Velocity and Four Momentum" it is that two vectors with a specified norm will not add to another vector with the same norm. I do not think OP has just one single problem. 1)General perspective always pertain to the special cases.Whatever ensues from: \[X_{\alpha\beta\gamma\delta}\left(g^{\alpha\mu}g^{\beta\nu}+g^{\alpha\nu}g^{\beta\mu}\right)=0\] will also apply to the Riemann tensor. For constant delta and gamma there are 16 variables of the type X_{alpha beta gamma delta} for four values of each alpha and beta..These are treated as the unknowns. There are 16 equations for the four values of each of mu and nu. We have sixteen linear homogeneous equations with sixteen variables[unknowns].If the determinant of the coefficient matrix is non zero then each X_{alpha beta gamma delta}=0. If you consider the Riemann tensor in place of X the component values will be zero[subject to the determinant of the coefficient matrix being non zero] . This idea is quite evident in my last post in the discussion. [gamma and delta have been held constant everywhere in the comment] 2)Two proper velocity vectors will not add up to a proper velocity since the norm square of the sum in general will be different from c^2: gre ater than c^2 3)The comment made by Kino on the "Norm of the Four Velocity Vector" is quite misleading. He is not specific with the error. He is searching for it as I understand.He will definitely report it at his earliest convenience.I hope so! Edited January 25 by Anamitra Palit Link to post Share on other sites

Anamitra Palit 11 Posted January 25 Author Share Posted January 25 (edited) 7 hours ago, Anamitra Palit said: 1)General perspective always pertain to the special cases.Whatever ensues from: Xαβγδ(gαμgβν+gανgβμ)=0 will also apply to the Riemann tensor. For constant delta and gamma there are 16 variables of the type X_{alpha beta gamma delta} for four values of each alpha and beta..These are treated as the unknowns. There are 16 equations for the four values of each of mu and nu. We have sixteen linear homogeneous equations with sixteen variables[unknowns].If the determinant of the coefficient matrix is non zero then each X_{alpha beta gamma delta}=0. If you consider the Riemann tensor in place of X the component values will be zero[subject to the determinant of the coefficient matrix being non zero] . This idea is quite evident in my last post in the discussion. [gamma and delta have been held constant everywhere in the comment] \[X_{\alpha\beta\gamma\delta}\left(g^{\alpha\mu}g^{\beta\nu}+ g^{\beta\mu}g^{\alpha\nu}\right)=0\] The coefficient matrix is independent of gamma and delta. For the same alpha and beta various gamma and delta should produce the same value for the corresponding component ,that is, X_{alpha beta gamma delta}=X_{alpha beta gamma’ delta’} If the coefficient matrix is non zero(or zero) for some (gamma,delta ) pair it should remain non zero (or zero) for other (gamma, delta) pairs. Edited January 25 by Anamitra Palit Link to post Share on other sites

Kino 10 Posted January 25 Share Posted January 25 12 hours ago, Anamitra Palit said: 1)General perspective always pertain to the special cases.Whatever ensues from: Xαβγδ(gαμgβν+gανgβμ)=0 will also apply to the Riemann tensor. Since that is merely a disguised form of your (4), all you are saying is that a tensor that is antisymmetric is antisymmetric. Taking six equations to get to a tautology seems excessive, but isn't wrong. (7) still does not follow from (6). I already gave an example of a non-zero Riemann tensor that satisfies (6). All you've done is shown that a zero tensor is not inconsistent with some of the properties of the Riemann. 12 hours ago, Anamitra Palit said: The comment made by Kino on the "Norm of the Four Velocity Vector" is quite misleading I have no idea which comment you are referring to. 2 Link to post Share on other sites

Markus Hanke 519 Posted January 26 Share Posted January 26 (edited) Let’s look very briefly at what the symmetries of Riemann actually constrain. We have two sets of symmetries - first, those that are present even in the absence of a metric: \[R{^\alpha}{_{\beta \gamma \delta}}=R{^\alpha}{_{\beta [\gamma \delta]}}\] \[R{^\alpha}{_{[\beta \gamma \delta]}=0}\] \[R{^\alpha}{_{\beta [\gamma \delta ||\mu]}}=0\] Second, we have metric-induced symmetries: \[R_{\alpha \beta \gamma \delta}=R_{[\alpha \beta]\gamma \delta}\] \[R_{\alpha \beta \gamma \delta}=R_{\gamma \delta \alpha \beta}\] \[R_{[\alpha \beta \gamma \delta]}=0\] What this means: in dimension n, every index pair can take on n(n-1)/2 independent values (due to their anti-symmetry), which initially leaves us with a symmetric matrix with \[\frac{1}{2}\left(\frac{n( n-1)}{2}\right)\left(\frac{n( n-1)}{2} +1\right) =\frac{n( n-1)\left( n^{2} -n+2\right)}{8}\] independent components. The Bianchi identities (last relation in non-metric symmetries above) constrain a further n!/(n-4)!/4! components. This finally leaves us with \[c_{n} =\frac{n^{2}\left( n^{2} -1\right)}{12}\] functionally independent components of the Riemann tensor. In n=4 dimensions, this evaluates to 20. So on a spacetime manifold with 4 dimensions, the symmetries of Riemann leave 20 tensor components unconstrained and functionally independent, meaning those components are not identically zero in the general case. Hopefully this clears things up, since it is trivially obvious that geodesic deviation does in fact exist in the real world (contrary to the OP’s claim), just like the theory says it does. Edited January 26 by Markus Hanke 2 Link to post Share on other sites

joigus 492 Posted January 26 Share Posted January 26 (edited) An alternative proof from direct Taylor expansion in the metric coefficients and counting how many parameters are left that I cannot set to zero by changing the coordinate system: https://www.youtube.com/watch?v=gf-G4QiAHLY&list=PLaNkJORnlhZnwjIXnOHrX50FEyoyiTh4o&index=5 Those must coincide with the number of independent components of the Riemann. \( \frac{1}{12}n^2\left(n^2-1\right) \) Uses Young tableaux, which allows you to count free parameters very easily. Edited January 26 by joigus 2 Link to post Share on other sites

Kino 10 Posted January 28 Share Posted January 28 I think the point (and what the OP proved) is that zero is consistent with the symmetries of the Riemann, naturally so because zero is the Riemann tensor of flat spacetime. But it isn't the only tensor that satisfies those symmetries. Link to post Share on other sites

Markus Hanke 519 Posted January 29 Share Posted January 29 13 hours ago, Kino said: I think the point (and what the OP proved) is that zero is consistent with the symmetries of the Riemann, naturally so because zero is the Riemann tensor of flat spacetime. But it isn't the only tensor that satisfies those symmetries. Yes, and it also doesn’t imply that Riemann is always identically zero for all manifolds. Which it evidently isn’t. 1 Link to post Share on other sites

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