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This is related to my previous question Variation with respect to $R_{abcd}$? How to compute$\frac{\partial R}{\partial R_{abcd}}=\frac{1}{2}(g^{ac} g^{bd} - g^{ad} g^{bc})$? In this case i'd like to compute Ricci tensor \begin{align} \frac{\partial R_{ab} R^{ab}}{\partial R_{abcd}}= \end{align}

In this case how do i compute the derivatives?

And further for \begin{align} \frac{\partial R_{abcd} R^{abcd}}{\partial R_{efgh}} \end{align} can i use former derivatives $(X^2)'= 2X$, and say above thing as $2 R_{efgh}$?

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We begin by noting that $$\frac{\partial R_{ab}}{\partial R_{cdef}}=\frac{\partial\left(R_{caeb}g^{ce}\right)}{\partial R_{cdef}}=\frac{\partial\left(R_{cdef}\delta_{a}^{d}\delta_{b}^{f}g^{ce}\right)}{\partial R_{cdef}},$$ suggesting an answer along the lines of $\delta_{a}^{d}\delta_{b}^{f}g^{ce}$. But by the antisymmetry properties of the Riemann tensor, there's more than one way to write $R_{ab}$ as a contradiction of $R_{cdef}$ with a tensor.

We need an antisymmetry when exchanging $c$ with $d$, suggesting an answer along the lines of $\frac{1}{2}\left(\delta_{a}^{d}\delta_{b}^{f}g^{ce}-\delta_{a}^{c}\delta_{b}^{f}g^{de}\right)$. But that can't be quite right either: we also need an antisymmetry when exchanging $e$ with $f$, suggesting an answer along the lines of $\frac{1}{4}\left(\delta_{a}^{d}\delta_{b}^{f}g^{ce}-\delta_{a}^{c}\delta_{b}^{f}g^{de}-\delta_{a}^{d}\delta_{b}^{e}g^{cf}+\delta_{a}^{c}\delta_{b}^{e}g^{df}\right)$. But we still need $cdef\to efcd$ to be a symmetry, giving the final result

$$\frac{\partial R_{ab}}{\partial R_{cdef}}=X_{ab}^{cdef}:=\frac{1}{8}\left(\left(\delta_{a}^{d}\delta_{b}^{f}+\delta_{a}^{f}\delta_{b}^{d}\right)g^{ce}-\left(\delta_{a}^{c}\delta_{b}^{f}+\delta_{a}^{f}\delta_{b}^{c}\right)g^{de}-\left(\delta_{a}^{d}\delta_{b}^{e}+\delta_{a}^{e}\delta_{b}^{d}\right)g^{cf}+\left(\delta_{a}^{c}\delta_{b}^{e}+\delta_{a}^{e}\delta_{b}^{c}\right)g^{df}\right).$$

Note that every term has $ab$ as lower indices and $cdef$ as upper indices.

By the product rule, $$\frac{\partial\left(R_{ab}R^{ab}\right)}{\partial R_{cdef}}=\frac{\partial R_{ab}}{\partial R_{cdef}}R^{ab}+R_{ab}\frac{\partial R^{ab}}{\partial R_{cdef}}.$$ We can change the heights of $a,\,b$ in the second term, viz. $$\frac{\partial\left(R_{ab}R^{ab}\right)}{\partial R_{cdef}}=2R^{ab}X_{ab}^{cdef}.$$ Expressions such as $R^{ab}\delta_{a}^{d}\delta_{b}^{f}g^{ce}=g^{ce}R^{df}$ give $$\frac{\partial\left(R_{ab}R^{ab}\right)}{\partial R_{cdef}}=\frac{1}{2}\left(g^{ce}R^{df}-g^{de}R^{cf}-g^{cf}R^{de}+g^{df}R^{ce}\right).$$ Note that every term has $cdef$ as upper indices and $ab$ don't exist on the right-hand side, since they're dummy indices contracted out on the left-hand side.

For the second derivative, imagine we instead wanted $\frac{\partial \left(V_aV^a\right)}{\partial V_b}$ for a vector; the answer would be $2V^b$, suggesting an answer like $2R^{efgh}$. This already has the right properties, so we're done.

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  • $\begingroup$ Thanks!. I made another post, physics.stackexchange.com/questions/301884/…. In this time i am trying to varying products of two contracted Riemann tensor. If you interested in please make some comment. $\endgroup$
    – phy_math
    Commented Dec 30, 2016 at 12:28
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From the helpful comment @J.G.,

\begin{align} \frac{\partial (R_{ab} R^{ab})}{\partial R_{cdef}} = 2 R^{ab} X^{cdef}_{ab}, \qquad X^{cdef}_{ab} = \frac{\partial R_{ab}}{\partial R_{cdef}} \end{align}

Try to compute \begin{align} R_{\mu\nu} &= R_{abcd} g^{bd} \delta_\mu^a \delta_\nu^c \\ & = \frac{1}{8} R_{abcd} (g^{bd} \delta_\mu^a \delta_\nu^c + g^{bd} \delta_\mu^c \delta_\nu^a - g^{ad} \delta_\mu^b \delta_\nu^c - g^{ad} \delta_\mu^c \delta_\nu^b - g^{bc} \delta_\mu^d \delta_\nu^a - g^{bc} \delta_\mu^a \delta_\nu^c + g^{ac} \delta_\mu^d \delta_\nu^b + g^{ac} \delta_\mu^b \delta_\nu^d) \end{align} where i antisymmetrized (a,b) and (c,d) and symmetrized the pairs (ab, cd), and symmetrized $\mu,\nu$ in side paranthesis

Thus i guess that \begin{align} X^{abcd}_{\mu\nu} =\frac{\partial R_{\mu\nu}}{\partial R_{abcd}} =\frac{1}{8} (g^{bd} \delta_\mu^a \delta_\nu^c + g^{bd} \delta_\mu^c \delta_\nu^a - g^{ad} \delta_\mu^b \delta_\nu^c - g^{ad} \delta_\mu^c \delta_\nu^b - g^{bc} \delta_\mu^d \delta_\nu^a - g^{bc} \delta_\mu^a \delta_\nu^c + g^{ac} \delta_\mu^d \delta_\nu^b + g^{ac} \delta_\mu^b \delta_\nu^d) \end{align}

\begin{align} \frac{\partial R_{\mu\nu} R^{\mu\nu}}{\partial R_{abcd}} & = R^{a[c} g^{b]d} + R^{b[d} g^{a]c} =\frac{1}{2} \left( R^{ac} g^{bd} - R^{bc} g^{ad} - R^{ad} g^{cb} + R^{bd} g^{ca}\right) \end{align}

Am i right?

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  • $\begingroup$ I'll write an answer soon. Your formula for $X$ makes no sense; it has two bonus indices, plus it places some of the indices at the wrong height. $\endgroup$
    – J.G.
    Commented Dec 30, 2016 at 8:38
  • $\begingroup$ @J.G, sorry there are some typo. I correct it $\endgroup$
    – phy_math
    Commented Dec 30, 2016 at 10:37
  • $\begingroup$ @J.G, After correcting indices properly i obtain the same answer in your post. Thanks. $\endgroup$
    – phy_math
    Commented Dec 30, 2016 at 10:43

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