# Decomposition of variation of metric derivatives (and why is Gibbons Hawking York term not identically zero?)

So, I've been studying the Gibbons Hawking York (GHY) boundary term, which involves variations of the derivatives of the metric, $\delta(\partial_\lambda g_{\mu\nu})$. But, since the metric and its derivatives are functions only of spacetime coordinates, can they not just be written/decomposed as

$$\delta g_{\mu\nu} = \frac{\partial g_{\mu\nu}}{\partial x^\alpha}\delta x^\alpha\tag{1}$$

and

$$\delta(\partial_\eta g_{\mu\nu}) = \frac{\partial(\partial_\eta g_{\mu\nu})}{\partial x^\beta}\delta x^\beta~?\tag{2}$$

If so, then we can construct \begin{align*} \frac{\delta g_{\mu\nu}}{\delta g_{\mu\nu}}&=\frac{\partial g_{\mu\nu}}{\partial x^\alpha}\delta x^\alpha \left( \frac{\partial g_{\mu\nu}}{\partial x^\beta}\delta x^\beta \right)^{-1}\\ &=\frac{\partial g_{\mu\nu}}{\partial g_{\mu\nu}}\frac{\delta x^\alpha}{\delta x^\beta}\frac{\partial x^\alpha}{\partial x^\beta} \end{align*} Is this correct? If so, then we can compute this and I think it turns out to be the number 10, right? Since there are 10 independent components of the metric and we don't want to overcount contributions to the partial derivative that are not independent, that come from the sum over $\alpha$ and $\beta$. Then we can write $$1=\frac{1}{10}\frac{\delta g_{\mu\nu}}{\delta g_{\mu\nu}}$$

But then we can write \begin{align*} 1\times \delta(\partial_\eta g_{\rho\sigma})&= \frac{1}{10}\delta(\partial_\eta g_{\rho\sigma}) \frac{\delta g_{\mu\nu}}{\delta g_{\mu\nu}}\\ &= \frac{1}{10}\frac{\partial(\partial_\eta g_{\rho\sigma})}{\partial x^\beta}\delta x^\beta \frac{\delta g_{\mu\nu}}{\delta g_{\mu\nu}}\\ &= \frac{1}{10}\frac{\partial(\partial_\eta g_{\rho\sigma})}{\partial x^\beta}\delta x^\beta \left( \frac{\partial g_{\mu\nu}}{\partial x^\alpha}\delta x^\alpha \right)^{-1} \delta g_{\mu\nu}\\ &= \frac{1}{10}\frac{\partial(\partial_\eta g_{\rho\sigma})}{\partial g_{\mu\nu}} \delta g_{\mu\nu} \end{align*} so that we have converted this variation of the metric derivatives into a variation of the metric, multiplied by some partial derivative which should be identically zero - since the metric and its derivatives are considered independent. Even if they weren't independent, the GHY term is a boundary term, where the variation of the metric goes to zero, so the contribution would vanish.

So, can somebody explain what is wrong here?

That's not how variations work.

Assume you have a metric $g_{\mu\nu}(x)$. When you vary it, you take a one-parameter family of metrics, let's write them as $$g_{\mu\nu}(\epsilon,x),$$ where this is a separate metric for every value of $\epsilon$. We only demand two things, that for $\epsilon=0$, we have $g_{\mu\nu}(0,x)=g_{\mu\nu}(x)$, so you get back the original metric, and this one parameter family is smooth in $\epsilon$. We may also demand that if $\mathcal D$ is some smooth regular domain (on which we'll integrate), then at least for $|\epsilon|<<1$ we have $g_{\mu\nu}(\epsilon,x)=g_{\mu\nu}(x)+O(\epsilon^2)$ if $x\in\partial\mathcal D$, eg. the variation vanishes on the boundary.

Then what the $\delta$ variation is, is simply $$\delta g_{\mu\nu}\equiv\left.\frac{d g_{\mu\nu}(\epsilon)}{d\epsilon}\right|_{\epsilon=0}.$$ The same is true of any function of the metric. If $F(g)$ is some function of the metric, then we have $$\delta F\equiv\left.\frac{dF(g(\epsilon))}{d\epsilon}\right|_{\epsilon=0}.$$

Which means your "chain rule" doesn't work.

In some special cases, variations do come from variations of the coordinates, however the formula $$\delta g_{\mu\nu}=\partial_\sigma g_{\mu\nu}\delta x^\sigma$$ is still not correct.

Why? Because when you perform a coordinate transformation $x_\epsilon^\mu=x^\mu+\epsilon\xi^\mu(x)$, the metric actually transforms into $$g_{\mu\nu}(\epsilon,x)= \frac{\partial x^\rho_\epsilon}{\partial x^\mu}\frac{\partial x^\sigma_\epsilon}{\partial x^\nu}g_{\rho\sigma}(x+\epsilon\xi),$$ which, to first order in $\epsilon$ is $$(\delta^\rho_\mu+\epsilon\partial_\mu\xi^\rho)(\delta^\sigma_\nu+\epsilon\partial_\nu\xi^\sigma)(g_{\rho\sigma}+\epsilon\xi^\alpha\partial_\alpha g_{\rho\sigma})=g_{\mu\nu}+\epsilon\left(g_{\mu\rho}\partial_\nu\xi^\rho+g_{\rho\nu}\partial_\mu\xi^\rho+\xi^\rho\partial_\rho g_{\mu\nu}\right)+O(\epsilon^2),$$ so we have $$\delta g_{\mu\nu}=\xi^\rho\partial_\rho g_{\mu\nu}+g_{\mu\rho}\partial_\nu\xi^\rho+g_{\rho\nu}\partial_\mu\xi^\rho\equiv\mathcal L_\xi g_{\mu\nu}.$$

In your notation, $\delta x^\mu=\xi^\mu$, and the notation $\mathcal L_\xi$ denotes Lie derivative with respect to the vector field $\xi$, which is exactly what we have actually calculated.

• Hi Uldreth - thank you! That's a brilliant answer. I'll figure out how to add some kudos after this response. So then, if you want $\delta F / \delta g_{\mu\nu}$ is it correct to equate that with $dF \ dg_{\mu\nu}$? Mar 27 '18 at 12:56
• You say the chain rule doesn't work - but the derivatives of the metric can also be written as functions of $\epsilon$ can't they? So why can't we use the chain rule with the $d\epsilon$ instead of the coordinates, in the same fashion as the original post? Mar 30 '18 at 21:01
• @larrylonglegs You can, but that wont lead to anything involving $\delta x^\mu$. Mar 31 '18 at 19:48
• sorry I wasn't clear before. I meant, why can we not write: \begin{align*} \delta g_{\mu\nu,\lambda} &= \delta g_{\mu\nu,\lambda} \frac{\delta g_{\rho\sigma}}{\delta g_{\rho\sigma}} \\ &= \frac{d g_{\mu\nu,\lambda}}{d\epsilon} \left(\frac{d g_{\rho\sigma}}{d \epsilon}\right)^{-1}\delta g_{\rho\sigma} \\ &= \frac{d g_{\mu\nu,\lambda}}{d g_{\rho\sigma}}\delta g_{\rho\sigma} \end{align*} Apr 2 '18 at 11:27