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I'm dealing with some extension of GR, with action:

$S=\int d^4x\Big[\sqrt{-g} f(R,R_{\mu\nu}R^{\mu\nu})$

Varying this action gives:

$\delta S=\int d^4x\Big[\delta\sqrt{-g} f(R,R_{\mu\nu}R^{\mu\nu})+\sqrt{-g} (\delta R f_R+ \delta(R_{\mu\nu}R^{\mu\nu})f_Y)$ where $Y=R_{\mu\nu}R^{\mu\nu}$ and $f_Y=\frac{\partial f}{\partial{Y}}$ etc

$f(R)$ part is relatively simple; while variation of the second term is:

$\delta(R_{\mu\nu}R^{\mu\nu})f_Y=2(R^{\mu\nu}\delta R_{\mu\nu}+g^{\rho\sigma}R_{\mu\rho}R_{\nu\sigma}\delta g^{\mu\nu})f_Y$

Using relation:

$ \delta R_{\mu\nu}=\frac{1}{2}g^{\alpha\beta}[\nabla_{\alpha}\nabla_{\nu}\delta g_{\mu\beta}+\nabla_{\alpha}\nabla_{\mu}\delta g_{\nu\beta}-\nabla_{\mu}\nabla_{\nu}\delta g_{\alpha\beta}-\nabla_{\alpha} \nabla_{\beta}\delta g_{\mu\nu}] $

Now, integration by parts 2 times, gives:

$\int...g^{\alpha\beta}\nabla_{\alpha}\nabla_{\nu}(\delta g_{\mu\beta})R^{\mu\nu}f_Y= \int ...g^{\alpha\beta}\nabla_{\alpha}\nabla_{\nu}(R^{\mu\nu}f_Y)\delta g_{\mu\beta} $

and

$\int ...g^{\alpha\beta}\nabla_{\alpha}\nabla_{\mu}(\delta g_{\nu\beta})R^{\mu\nu}f_Y= \int ...g^{\alpha\beta}\nabla_{\alpha}\nabla_{\mu}(R^{\mu\nu}f_Y)\delta g_{\nu\beta}$

$\int...-g^{\alpha\beta}\nabla_{\mu}\nabla_{\nu}(\delta g_{\alpha\beta})R^{\mu\nu}f_Y=\int ...-g^{\alpha\beta}\nabla_{\mu}\nabla_{\nu}(R^{\mu\nu}f_Y) \delta g_{\alpha\beta}$

$\int ... -g^{\alpha\beta}\nabla_{\alpha} \nabla_{\beta}(\delta g_{\mu\nu})R^{\mu\nu}f_Y=\int ...-g^{\alpha\beta}\nabla_{\alpha} \nabla_{\beta}(R^{\mu\nu}f_Y)\delta g_{\mu\nu}$

Where boundary term is assumed to vanish. Apllying $\delta g_{\mu\nu}=-g_{\mu\rho} g_{\nu\sigma}\delta g^{\rho\sigma}$

leads to

$g^{\alpha\beta}\nabla_{\alpha}\nabla_{\nu}(R^{\mu\nu}f_Y)\delta g_{\mu\beta}=-g^{\alpha\beta}\delta g^{\rho\sigma}g_{\mu\rho} g_{\beta\sigma}\nabla_{\alpha}\nabla_{\nu}(R^{\mu\nu}f_Y)=-\delta g^{\rho\sigma}\nabla_{\sigma}\nabla_{\nu}(R_{\rho}^{\;\;\nu}f_Y)$

$ g^{\alpha\beta}\nabla_{\alpha}\nabla_{\mu}(R^{\mu\nu}f_Y)\delta g_{\nu\beta}=-g^{\alpha\beta}\delta g^{\rho\sigma}g_{\nu\rho}g_{\beta\sigma}\nabla_{\alpha}\nabla_{\mu}(R^{\mu\nu}f_Y)=-\delta g^{\rho\sigma}\nabla_{\sigma}\nabla_{\mu}(R^{\mu}_{\;\;\rho}f_Y ) $

$ -g^{\alpha\beta}\nabla_{\mu}\nabla_{\nu}(R^{\mu\nu}f_Y) \delta g_{\alpha\beta}=g^{\alpha\beta}\delta g^{\rho\sigma}g_{\alpha\rho}g_{\beta\sigma} \nabla_{\mu}\nabla_{\nu}(R^{\mu\nu}f_Y) =\delta g^{\rho\sigma} g_{\rho\sigma}\nabla_{\mu}\nabla_{\nu}(R^{\mu\nu}f_Y)$

$-g^{\alpha\beta}\nabla_{\alpha} \nabla_{\beta}(R^{\mu\nu}f_Y)\delta g_{\mu\nu}=\delta g^{\rho\sigma}\Box(R_{\sigma\rho}f_Y)$

while

$g^{\rho\sigma}R_{\mu\rho}R_{\nu\sigma}\delta g^{\mu\nu}f_Y=\delta g^{\rho\sigma}R^{\nu}_{\rho}R_{\nu\sigma}f_Y$

Then, variation will be:

$ \delta(R_{\mu\nu}R^{\mu\nu})f_Y=2(R^{\mu\nu}\delta R_{\mu\nu}+g^{\rho\sigma}R_{\mu\rho}R_{\nu\sigma}\delta g^{\mu\nu})f_Y=\delta g^{\rho\sigma}[\Box(R_{\sigma\rho}f_Y)+g_{\rho\sigma}\nabla_{\mu}\nabla_{\nu}(R^{\mu\nu}f_Y)+2R^{\nu}_{\rho}R_{\nu\sigma}f_Y-2\frac{1}{2}\nabla_{\sigma}\nabla_{\mu}(R^{\mu}_{\;\;\rho}f_Y )-2\frac{1}{2}\nabla_{\sigma}\nabla_{\nu}(R_{\rho}^{\;\;\nu}f_Y)]$.

Terms in bracket will appear in final field quations when other $...$ terms are present.

This part should be equal to:enter image description here

Where i have some problems with antysymetrization $C_{(ab)}=\frac{1}{2}(C_{ab}+C_{ba})$ - i know that since i am dealing with symmetric tensors i can switch indices:

$-\big(\frac{1}{2}\nabla_{\sigma}\nabla_{\mu}(R^{\mu}_{\;\;\rho}f_Y )+\frac{1}{2}\nabla_{\sigma}\nabla_{\nu}(R_{\rho}^{\;\;\nu}f_Y)\big)=-\frac{1}{2}(\nabla_{\sigma}\nabla_{\mu}(R^{\mu}_{\;\;\rho}f_Y )+\nabla_{\rho}\nabla_{\nu}(R_{\sigma}^{\;\;\nu}f_Y)\big))$

Hovewer, shouldn't antisymmetrization of the indices come from actual variational principle? I mean in paper https://arxiv.org/abs/1503.08751 they've could just use $-2\nabla_{\mu}\nabla_{\alpha}(R^{\alpha}_{\nu}f_Y)$ - am i missing something in my derivation of this part of the field equations?

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  • $\begingroup$ If this is homework, please add the homework-and-exercises tag. What is the screenshot an image of? Is it a textbook or handout? If so, then please credit the source appropriately. Please use standard punctuation and capitalization, e.g., "I," not "i." $\endgroup$ – Ben Crowell Oct 14 at 21:36
  • $\begingroup$ Sorry for this issues. It is not homework. I get correct results but method i've used looks ugly for me - im looking for more rigorous approach $\endgroup$ – Adam Kaczmarek Oct 15 at 8:00

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