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I know that, for example we have

$$\frac{\delta g^{jk}}{\delta g^{lm}}=\delta^{j}_{(l}\delta^{k}_{m)}.$$ This topic was discussed previously e.g. on Physicsforums.com and on Phys.SE.

So I was wondering, how can I compute $\frac{\delta R^{j}_{klm}}{\delta R^{a}_{bcd}}$?

And in general, how can I compute this quantity for a tensor with some symmetries?

Qmechanic
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Alabarda 980
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  • Why do you want to compute this? and will you want your final variation to be a functional differential in terms of the $\delta g_{ab}$ as you typically do when taking a variation? Is this a term in a larger variation you're doing? Are you just abstractly asking a question out of curiousity? – Zo the Relativist Jul 22 '20 at 14:36
  • I was wondering if I was able to compute the variation of the Hilbert Lagrangian taking as coordinates the inverse metric and the Riemann tensor (not the metric Riemann tensor) treated as independent coordinate. So I need to vary this quantity. I would expect to get product of deltas, but I'm stuck due to the symmetries of the tensor. – Alabarda 980 Jul 22 '20 at 14:46

1 Answers1

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The expression you are looking for is \begin{equation} \frac{\partial R_{\alpha\beta\mu\nu}}{\partial R_{\rho\sigma\lambda\gamma}} = \frac{1}{2}\left(\delta_{\alpha\beta}^{\rho\sigma}\delta_{\mu\nu}^{\lambda\gamma} + \delta_{\alpha\beta}^{\lambda\gamma}\delta_{\mu\nu}^{\rho\sigma}\right). \end{equation} Where the deltas in the equation are \begin{equation} \delta_{\alpha\beta}^{\rho\sigma} = \frac{1}{2}\left(\delta_{\alpha}^{\rho}\delta_{\beta}^{\sigma} - \delta_{\alpha}^{\sigma}\delta_{\beta}^{\rho}\right) \end{equation} With this, both the symmetric and antisymmetric properties of the Riemann tensor are taken into account. A more explicit expression is the following \begin{align} \frac{\partial R_{\alpha\beta\mu\nu}}{\partial R_{\rho\sigma\lambda\gamma}} &= \frac{1}{8}\left[(\delta_{\alpha}^{\rho}\delta_{\beta}^{\sigma}\delta_{\mu}^{\lambda}\delta_{\nu}^{\gamma} - \delta_{\alpha}^{\rho}\delta_{\beta}^{\sigma}\delta_{\mu}^{\gamma}\delta_{\nu}^{\lambda} - \delta_{\alpha}^{\sigma}\delta_{\beta}^{\rho}\delta_{\mu}^{\lambda}\delta_{\nu}^{\gamma} + \delta_{\alpha}^{\sigma}\delta_{\beta}^{\rho}\delta_{\mu}^{\gamma}\delta_{\nu}^{\lambda})\right. \nonumber \\[5pt] &\left.+ (\delta_{\alpha}^{\lambda}\delta_{\beta}^{\gamma}\delta_{\mu}^{\rho}\delta_{\nu}^{\sigma} - \delta_{\alpha}^{\lambda}\delta_{\beta}^{\gamma}\delta_{\mu}^{\sigma}\delta_{\nu}^{\rho} - \delta_{\alpha}^{\gamma}\delta_{\beta}^{\lambda}\delta_{\mu}^{\rho}\delta_{\nu}^{\sigma} + \delta_{\alpha}^{\gamma}\delta_{\beta}^{\lambda}\delta_{\mu}^{\sigma}\delta_{\nu}^{\rho})\right] \end{align} I've also been told by my doctoral advisor that the following equation also works \begin{equation} \frac{\partial R_{\alpha\beta\mu\nu}}{\partial R_{\rho\sigma\lambda\gamma}} = \frac{1}{2}\delta_{\alpha\beta}^{\rho\sigma}\delta_{\mu\nu}^{\lambda\gamma}, \end{equation} but I fail to see how is the symmetric property of the Riemann tensor is considered.

Víctor Zavala
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