1
$\begingroup$

I know that the derivative of the contraction of two Ricci tensors with respect to the Riemann tensor must be \begin{equation} \left(\frac{\partial \left(R^{\alpha\beta}R_{\alpha\beta}\right)}{\partial R_{\mu\nu\rho\sigma}}\right)_{g^{\gamma\delta}}=2g^{\alpha[\nu}g^{\mu][\rho}g^{\sigma]\beta}R_{\alpha\beta}, \end{equation} but when I compute it, I keep getting $2g^{\nu\sigma}R^{\mu\rho}$. I would like to know how to proceed in order to obtain the correct result.

Edit:

It seems that my problem is actually more fundamental than I thought, I shall derive in two ways the derivative of the Ricci scalar:

  1. The first one, which leads to the correct result is

\begin{equation} \left(\frac{\partial R}{\partial R_{\mu\nu\rho\sigma}}\right)_{g^{\omega\psi}}=\frac{1}{2}g^{\lambda\gamma}g^{\tau\delta}\delta^{\alpha\beta}_{\gamma\delta}\left(\frac{\partial R_{\alpha\beta\lambda\tau}}{\partial R_{\mu\nu\rho\sigma}}\right)_{g^{\alpha\beta}}=\frac{1}{2}g^{\rho\gamma}g^{\sigma\delta}\delta_{\gamma\delta}^{\mu\nu}=\frac{1}{2}\left(g^{\rho\mu}g^{\sigma\nu}-g^{\rho\nu}g^{\sigma\mu}\right)=g^{\rho[\mu}g^{\nu]\sigma}, \end{equation} where I used \begin{equation} R=\frac{1}{2}\delta^{\alpha\beta}_{\gamma\delta}R_{\alpha\beta}^{\quad \gamma\delta}. \end{equation} 2. The second one, is a more direct approach, recalls that $R=g^{\alpha\gamma}g^{\beta\delta}R_{\alpha\beta\gamma\delta}$ \begin{equation} \left(\frac{\partial R}{\partial R_{\mu\nu\rho\sigma}}\right)_{g^{\omega\psi}}=g^{\alpha\gamma}g^{\beta\delta}\left(\frac{\partial R_{\alpha\beta\gamma\delta}}{\partial R_{\mu\nu\rho\sigma}}\right)_{g^{\alpha\beta}}=g^{\rho\mu}g^{\nu\sigma}. \end{equation} Derivation 2 does not yield the correct result. What is the mistake?

$\endgroup$
  • $\begingroup$ How is that derivative even defined? What does it mean to vary $R_2^2$ with respect to $R_4$? $\endgroup$ – AccidentalFourierTransform May 25 '18 at 23:28
  • $\begingroup$ If you want to vary a Lagrangian that is a function of Riemann tensors (and its contraction) you can realise that this derivative will yield automatically the variation. It is actually a useful definition for greater curvature orders than just the Einstein-Hilbert term. In Padmanabhan's Gravitation book this topic is covered. I actually obtain correctly the quadratic invariant terms consisting in $R^2$ and $R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}$ but I don't find the expression for this one. $\endgroup$ – Janstillerion May 25 '18 at 23:37
  • $\begingroup$ You should edit the question to provide some steps to your work so we have a better understanding of where you are having an issue. $\endgroup$ – Triatticus May 27 '18 at 18:34
  • $\begingroup$ It also looks like you are holding the metric constant during the variation. I'd be curious to know what this is in reference to. Having studied higher derivative theory long ago taking the variation of a quadratic action w/r to the metric is well understood. I agree with the comment made by Triatticus, that you should provide steps in your derivation. We can more easily offer help, rather than go off and take the derivatives and posting more results. $\endgroup$ – ggcg Jun 3 '18 at 20:53
  • $\begingroup$ Edited and added the required info. $\endgroup$ – Janstillerion Jun 4 '18 at 17:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.