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What does this equal to, and how do I actually calculate this correctly? $$ \frac{\delta G_{ab}}{\delta g_{cd}}=? $$

BinaryBurst
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  • I am not familiar with the notation. What are you trying to get? The divergence? – mfl Jul 10 '14 at 14:07
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    I'm guessing he's talking about the "functional" (Gateaux) derivative of $G(g)$. – Anthony Carapetis Jul 10 '14 at 14:15
  • Can you add a bit more detail about what you are looking for. Normally one wants to take the variation of a functional integral (least action type calculation). Is this the case here (then you need to have some scalar quantity in the action s.a. $G_{ab}A^{ab}$ for some tensor $A^{ab}$ - this is relevant for when one does integration by parts)? Or are you just looking for the function derivative itself? Note that in the last case the answer will have to contain terms such as $\frac{\delta g_{ab,c}}{\delta g_{cd}}$ or equivalently $\delta \Gamma$ where $\Gamma$ is the Christoffel symbols. – Winther Sep 19 '14 at 16:35

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The Einstein tensor is defined as $$G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R$$

so by taking the variation we find

$$\frac{\delta G_{\mu\nu}}{\delta g_{\alpha\beta}} = \frac{\delta R_{\mu\nu}}{\delta g_{\alpha\beta}} - \frac{1}{2}\delta_{\mu}^{\alpha}\delta_{\nu}^{\beta}R - \frac{1}{2}g_{\mu\nu}\frac{\delta R}{\delta g_{\alpha\beta}}$$

Now $R = g_{\gamma\sigma}R^{\gamma\sigma}$ so $\frac{\delta R}{\delta g_{\alpha\beta}} = R^{\alpha\beta} + g_{\gamma\sigma}\frac{\delta R^{\gamma\sigma}}{\delta g_{\alpha\beta}}$ giving us

$$\frac{\delta G_{\mu\nu}}{\delta g_{\alpha\beta}} = \frac{\delta R_{\mu\nu}}{\delta g_{\alpha\beta}} - \frac{1}{2}\delta_{\mu}^{\alpha}\delta_{\nu}^{\beta}R - \frac{1}{2}g_{\mu\nu}R^{\alpha\beta} - \frac{1}{2}g_{\mu\nu}g_{\gamma\sigma}\frac{\delta R^{\gamma\sigma}}{\delta g_{\alpha\beta}}$$

I will now assume that the variation is to be taken under a integral $\int \sqrt{-g} dx^4$ (we will use the notation $\dot{=}$ for this), as this is usually where we want to take the variation and the expression becomes extremely complicated otherwise. The variation of $R_{\mu\nu}$ can now be written as a divergence and can be integrated using Stokes theorem to yield a surface term (which we will assume vanishes as usual). Then we obtain

$$\frac{\delta G_{\mu\nu}}{\delta g_{\alpha\beta}} \dot{=} -\frac{1}{2}\left( \delta_{\mu}^{\alpha}\delta_{\nu}^{\beta}R + g_{\mu\nu}R^{\alpha\beta}\right)$$

Winther
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Hint: Write down $G_{ab}=Rc_{ab}-\frac{R}{2}g_{ab}$ in terms of the metric (you can express both the Ricci tensor and the scalar curvature in terms of the Christoffel symbols, and those can be written as functions of the metric), then differentiate. Probably it's better if you subdivide the problem into smaller pieces, though.

Daniel Robert-Nicoud
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