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Varying the Ricci tensor with respect to the metric $g^{\mu\nu}$, one would get

$ \delta R_{\mu\nu} = \delta(g^{\rho\sigma} R_{\rho\mu\sigma\nu}) = g^{\rho\sigma}\delta R_{\rho\mu\sigma\nu} + \delta g^{\rho\sigma} R_{\rho\mu\sigma\nu}, $

but in all my references I found that $\delta R_{\mu\nu} = g^{\rho\sigma}\delta R_{\rho\mu\sigma\nu}$ (including Wikipedia), which implies that the last term of the above equation is identically zero. Why is that so?

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    $\begingroup$ Your first relation is correct. Check the conditions under which quantities are being calculated. Probably up to linear order and/or perturbing around flat spacetime (in which case, $R_{\rho \mu \sigma \nu}=0$) $\endgroup$
    – Avantgarde
    Jun 20, 2017 at 19:07
  • $\begingroup$ There's no perturbation around Minkowski here, this is only the standard procedure to get Einstein's equations from the Hilbert-Einstein action. $\endgroup$
    – Mr. K
    Jun 21, 2017 at 9:39
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    $\begingroup$ Why do you think this is even true? I don't see where they say it in the Wiki article you link. $\endgroup$
    – Ryan Unger
    Jun 21, 2017 at 12:31
  • $\begingroup$ @0celouvskyopoulo7, I realized it's not! In the Wiki page when they write $\delta R_{\mu\nu} = \delta R^\rho_{\;\mu\rho\nu}$ made me believe at first glance that $\delta R_{\mu\nu} = g^{\rho\sigma}\delta R_{\rho\mu\sigma\nu}$, but that's not true. Instead, $\delta R_{\mu\nu} = g^\sigma_\rho \delta R^\rho_{\mu\sigma\nu}$ holds as I proved below. $\endgroup$
    – Mr. K
    Jun 21, 2017 at 13:04
  • $\begingroup$ Wiki says $\delta R_{\mu\nu} = \delta R^\rho{}_{\mu\rho\nu}$ which is absolutely correct. I guess what you are assuming is $\delta R^\rho{}_{\mu\rho\nu} = g^{\rho\sigma} \delta R_{\rho\mu\sigma\nu}$ which is NOT true. $\endgroup$
    – Prahar
    Apr 19, 2021 at 8:30

3 Answers 3

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Consider a metric $\widetilde{g}_{\mu\nu}=g_{\mu\nu}+\delta g_{\mu\nu}$ and its inverse metric $\widetilde{g}^{\mu\nu}=g^{\mu\nu}-\delta g^{\mu\nu}$. (Fittingly I've just explained the negative sign here yesterday.) We have: \begin{align*} R_{\mu\nu}[\widetilde{g}] &=\widetilde{g}^{\kappa\lambda} R_{\kappa\mu\lambda\nu}[\widetilde{g}] =(g^{\kappa\lambda}-\delta g^{\kappa\lambda}) (R_{\kappa\mu\lambda\nu}[g]+\delta R_{\kappa\mu\lambda\nu}) \\ &=R_{\mu\nu}[g] +\underbrace{g^{\kappa\lambda}\delta R_{\kappa\mu\lambda\nu} -\delta g^{\kappa\lambda}R_{\kappa\mu\lambda\nu}[g]}_{=\delta R_{\mu\nu}}+\mathcal{O}(\delta g^2). \end{align*} I therefore don't think this formula is correct in a stricter sense, somebody probably approximated $\widetilde{g}^{\kappa\lambda}\approx g^{\kappa\lambda}$ directly in the first step and therefore left out a term in first order. Because of this I'd also use a different formula for the pertubation of the curvature tensor. Taking the formula for the Christoffel symbols and putting the perturbed metric in, you get: \begin{equation} \Gamma_{\mu\nu}^\kappa[\widetilde{g}] =\Gamma_{\mu\nu}^\kappa[g] +\underbrace{\frac{1}{2}g^{\kappa\lambda}\left( \nabla_\mu\delta g_{\lambda\nu} +\nabla_\nu\delta g_{\lambda\mu} -\nabla_\lambda\delta g_{\mu\nu}\right)}_{=\delta\Gamma_{\mu\nu}^\kappa} +\mathcal{O}(\delta g^2). \end{equation} This calculation is not short, but pretty straight forward. You immediately see, that $\delta\Gamma_{\mu\nu}^\kappa$ is a tensor, as it is a sum of covariant differentiations of a tensor. Taking the formula for the Riemann curvature tensor and putting this equation in, you get: \begin{equation} R_{\lambda\mu\nu}^\rho[\widetilde{g}] =R_{\lambda\mu\nu}^\rho[g] +\underbrace{\partial_\mu\delta\Gamma_{\lambda\nu}^\rho -\partial_\nu\delta\Gamma_{\lambda\mu}^\rho +\delta\Gamma_{\mu\sigma}^\rho\Gamma_{\lambda\nu}^\sigma[g] +\Gamma_{\mu\sigma}^\rho[g]\delta\Gamma_{\lambda\nu}^\sigma -\delta\Gamma_{\nu\sigma}^\rho\Gamma_{\lambda\mu}^\sigma[g] -\Gamma_{\nu\sigma}^\rho[g]\delta\Gamma_{\lambda\mu}^\sigma}_{=\delta R_{\lambda\mu\nu}^\rho} +\mathcal{O}(\delta g^2). \end{equation} Notice, that you can shorten this using the covariant differentiation: \begin{equation} \delta R_{\lambda\mu\nu}^\rho =\nabla_\mu\delta\Gamma_{\lambda\nu}^\rho -\nabla_\nu\delta\Gamma_{\lambda\mu}^\rho. \end{equation} You again immediately see, that $\delta R_{\lambda\mu\nu}^\rho$ is a tensor, as it is the sum of covariant differentiations of a tensor. Contracting to $\delta R_{\lambda\nu}$ gives you the simple Palatini identity.

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This is for whoever is interested in the solution. The second term in the equation of my original post is not zero as it is. But, from $g_{\mu\alpha}g^{\alpha\nu}=\delta_\mu^\nu$, one can prove that $\delta g_\rho^\sigma = - g_\alpha^\sigma g_\rho^\beta \delta g_\beta^\alpha$. Then the variation of the Ricci tensor gives

$ \delta R_{\mu\nu} = \delta(g_\rho^\sigma R^\rho_{\;\mu\sigma\nu})= g^\sigma_\rho\delta R^\rho_{\;\mu\sigma\nu} + \delta g_\rho^\sigma R^\rho_{\;\mu\sigma\nu}, $

where

$ \delta g_\rho^\sigma R^\rho_{\;\mu\sigma\nu} = - g_\alpha^\sigma g_\rho^\beta \delta g_\beta^\alpha R^\rho_{\;\mu\sigma\nu} = -\delta g_\beta^\alpha R^\beta_{\;\mu\alpha\nu}, $

which vanishes as $\rho,\sigma,\alpha,\beta$ are all dummy indices. Hence,

$ \delta R_{\mu\nu} = g^\sigma_\rho\delta R^\rho_{\;\mu\sigma\nu}. $

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    $\begingroup$ This is not correct. One should be particularly careful when raising and lowering indices on varied quantities. For instance, $\delta g^{\mu \nu}$ is usually taken to mean the variation of the inverse metric, not the raised index version of the variation of the metric. The two possible interpretations differ by a sign. Furthermore, note that the final equation you've arrived at is not the equation in your question – the index placement is crucial. $\endgroup$
    – gj255
    Jun 21, 2017 at 12:01
  • $\begingroup$ @gj255 nowhere I contradicted your statement that '$\delta g^{\mu\nu}$ is usually taken to mean the variation of the inverse metric, not the raised index version of the variation of the metric'. I completely agree with that and that was the main reason why I posted my question in the first place as we can't simply commute the metric with the variation sign. $\endgroup$
    – Mr. K
    Jun 21, 2017 at 13:14
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    $\begingroup$ Of course the index placement is crucial. What I proved is that the equation in my question is wrong. But in any case, the equation $\delta R_{\mu\nu} = \delta R^\rho_{\;\mu\rho\nu}$ still holds, and that was the main cause of my misunderstanding. $\endgroup$
    – Mr. K
    Jun 21, 2017 at 13:15
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    $\begingroup$ Note that $g^\sigma_\rho = \delta^\sigma_\rho$ $\endgroup$
    – doetoe
    Jun 21, 2017 at 13:53
  • $\begingroup$ @doetoe that's true! I guess your comment explains everything now. $\endgroup$
    – Mr. K
    Jun 21, 2017 at 14:26
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The variation of the Ricci tensor $$ \delta R_{\mu\nu}~=~\delta(g^{\rho\sigma} R_{\rho\mu\sigma\nu}) = g^{\rho\sigma}\delta R_{\rho\mu\sigma\nu} + \delta g^{\rho\sigma} R_{\rho\mu\sigma\nu}~=~g^{\rho\sigma}\delta R_{\rho\mu\sigma\nu}~+~\delta s\left(\frac{D}{ds} g^{\rho\sigma}\right)R_{\rho\mu\sigma\nu}. $$ The last term is zero by the covariant constancy of the metric. This is seen with $$ \frac{D}{ds} g^{\rho\sigma}~=~\frac{d}{ds} g^{\rho\sigma}~+~\left(\Gamma^\rho_{\alpha\beta}g^{\alpha\sigma}~+~\Gamma^\sigma_{\alpha\beta}g^{\alpha\rho}\right)\frac{dx^\beta}{ds}. $$ This is why the $\delta g^{\rho\sigma} R_{\rho\mu\sigma\nu}~=~0$

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    $\begingroup$ This hardly makes sense. If $\delta g=\delta s\frac{D}{ds}g$, then how could $\delta g$ ever be nonzero? $\endgroup$
    – Ryan Unger
    Jun 21, 2017 at 2:25
  • $\begingroup$ I don't find any valid reason why this answer is being downvoted. Indeed this is the correct answer. The covariant derivative of the metric tensor is always zero and this makes the term inside the braces of the last term in the first equation zero. This is just basic general relativity. $\endgroup$
    – Richard
    Dec 13, 2019 at 5:56
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    $\begingroup$ @Richard Several years late to the party, but the reason this doesn't make sense is that $\delta g$ is not the infinitesimal change in the metric obtained by parallel transporting along some curve in spacetime. Indeed, if it were, then $\delta g$ would always be zero as per Ryan's comment, and the entire concept of variation with respect to the metric would fall apart. Instead, the smooth variation $\delta g$ is arbitrary. $\endgroup$
    – J. Murray
    Aug 17, 2021 at 13:04

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