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

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?

• Hi @phy_math: Is this from Wald's GR book? Which precise reference? Which page? – Qmechanic Dec 29 '16 at 19:25
• @Qmechanic, Its comes from Wald's paper related on BlackHoles. I see this kinds of equations a lot in like modified gravity, $f(R)$, and Gauss Bonnet gravity, higher derivatives theories, etc. – phy_math Dec 29 '16 at 23:59

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.
• thanks!, I noticed that symmetric argument works for variation of $R$, but how about $R_{ab} R^{ab}$ case?, I will make it as another post. – phy_math Dec 29 '16 at 23:54
• @phy_math Hint: if $X_{ab}^{cdef}:=\dfrac{\partial R_{ab}}{\partial R_{cdef}}$ then $\dfrac{\partial \left( R_{ab}R^{ab}\right)}{\partial R_{cdef}}=2R^{ab}X_{ab}^{cdef}$. – J.G. Dec 30 '16 at 0:08
• Symmetries: $R_{xy}=-R_{yx}$,$\;\;\langle R_{xy}v,w\rangle=-\langle R_{xy}w,v\rangle$,$\;\;R_{xy}z+R_{yz}x+R_{zx}y$,$\;\;\langle R_{xy}v,w\rangle=\langle R_{vw}x,y\rangle\;\;$ where all vectors are an elements of $T_{p}M$. – Cinaed Simson Nov 20 '19 at 21:54