Variation with respect to $R_{abcd}$? How to compute$\frac{\partial R}{\partial R_{abcd}}=\frac{1}{2}(g^{ac} g^{bd} - g^{ad} g^{bc})$?

Question

Recently, I have been interested in Wald's formula of entropy. In this formalism, there is a tensor $P^{abcd}=\frac{\partial L}{\partial R_{abcd}}$.

Among some papers they mention $P^{abcd}$, describing its symmetric anti-symmetric properties. But I want to know the explicit form of $P$ in certain cases.

There are known results from Einstein-Hilbert action (they mention, $P^{abcd}=\frac{1}{2}(g^{ac} g^{bd} - g^{ad} g^{bc})$ for Einstein Hilbert case), thus I assume that \begin{align} &\frac{\partial R}{\partial R_{abcd}}=\frac{1}{2}(g^{ac} g^{bd} - g^{ad} g^{bc})(?) \end{align} where $g$ is the usual symmetric metric, $R_{abcd}$ is Riemann curvature tensor, and $R$ is Ricci scalar.

It seems that they treat $R_{abcd}$ and $g_{ab}$ independently, so my first trial is decompose $R=g^{ac}g^{bd}R_{abcd}$, and try to compute $ \frac{\partial R_{pqrs}}{\partial R_{abcd}}$, from the symmetric arguments \begin{align} \frac{\partial R_{abcd}}{\partial R_{pqrs}}=\delta_{ab}^{pq}\delta^{rs}_{cd}+\delta_{cd}^{pq}\delta^{rs}_{ab} \end{align} But plugging this to $R$, I obtain a somewhat different answer shown above.

Am I doing something wrong?

If you have experienced this kind of derivative, do you have any hints or advice for this kind of algebraic computation?

Answer 1

First, note that $$R=R_{ac}g^{ac}=R_{abcd}g^{ac}g^{bd}.$$ The antisymmetries of $R$ imply $$R_{abcd}\left( g^{ac}g^{bd}+g^{ad}g^{bc} \right) = 0,\, R=R_{abcd}\left( \left( 1-k\right) g^{ac}g^{bd}-k g^{ad}g^{bc}\right)$$ for any constant $k$. Since $\dfrac{\partial R}{\partial R_{abcd}}$ must be antisymmetric when exchanging $a$ with $b$ or $c$ with $d$, it is the coefficient obtained by choosing $k$ so that $1-k=k$, i.e. $k=\frac{1}{2}$. Hence $$\frac{\partial R}{R_{abcd}}=\frac{1}{2}\left( g^{ac}g^{bd}-g^{ad}g^{bc}\right).$$ Note the $abcd\to cdab$ symmetry.