I am trying to find the Field equation for
$$S = \int \sqrt{-g}dx^4[f(\phi)R + h(\phi)g^{\mu \nu}\nabla_{\mu}\phi\nabla_{\nu}\phi – V(\phi)$$
but I could not take the variation of $$\delta(\sqrt{-g}h(\phi)g^{\mu \nu}\nabla_{\mu}\phi\nabla_{\nu}\phi)$$
I know that $\delta(\sqrt{-g}) = -\frac{1}{2}\sqrt{-g}g_{\mu\nu}\delta g^{\mu\nu}$. We can also take $\delta h = \frac{\partial h}{\partial \phi}\delta \phi$.
We can separate the expression into five terms.
- $$\delta(\sqrt{-g})hg^{\mu \nu}\nabla_{\mu}\phi\nabla_{\nu}\phi$$
- $$\sqrt{-g}\delta h g^{\mu \nu}\nabla_{\mu}\phi\nabla_{\nu}\phi$$
- $$\sqrt{-g}h \delta(g^{\mu \nu})\nabla_{\mu}\phi\nabla_{\nu}\phi$$
- $$\sqrt{-g}h g^{\mu \nu}\nabla_{\mu}\delta(\phi)\nabla_{\nu}\phi$$
- $$\sqrt{-g}h g^{\mu \nu}\nabla_{\mu}\phi\nabla_{\nu}\delta(\phi)$$
So I have found
- $$(-\frac{1}{2}\sqrt{-g}hg_{\mu\nu}\nabla_{\beta}\phi\nabla^{\beta}\phi)\delta g^{\mu\nu}$$
- $$(\sqrt{-g}\nabla_{\mu}\phi\nabla^{\mu}\phi\frac{\partial h}{\partial \phi})\delta \phi$$
- $$(\sqrt{-g}h\nabla_{\mu}\phi\nabla_{\nu}\phi)\delta(g^{\mu \nu})$$
From $(4)$ and $(5)$ I obtain,
$$(4)+(5) = 2\sqrt{-g}h\nabla^{\mu}\phi \nabla_{\mu}\delta(\phi)$$
and by taking,
$$\nabla_{\mu}(h\delta\phi\nabla^{\mu}\phi)=\nabla_{\mu}h\nabla^{\mu}\phi\delta \phi+h\nabla_{\mu}(\delta\phi)\nabla^{\mu}+h\square\phi \delta\phi$$
$(4)+(5)$ becomes,
$$(4)+(5)=-2\sqrt{-g}[\nabla_{\mu}h\nabla^{\mu}\phi + \square\phi h]\delta\phi$$
So In summary I have found
$$\delta(\sqrt{-g}h(\phi)g^{\mu \nu}\nabla_{\mu}\phi\nabla_{\nu}\phi) = \sqrt{-g}\Big[\delta g^{\mu\nu}\big[-\frac{hg_{\mu\nu}}{2}\nabla_{\beta}\phi\nabla^{\beta}\phi+ h\nabla_{\mu}\phi\nabla_{\nu}\phi\big] + \delta \phi \big[\nabla_{\mu}\phi\nabla^{\mu}\phi\frac{\partial h}{\partial \phi} -2[\nabla_{\mu}h\nabla^{\mu}\phi + \square\phi h] \big]\Big]$$
Best Answer
For 1 do not repeat the same index more than two times. As Eletie says, you may rename the indices in one of the terms, then your expression will be meaningful. 4,5 are the same. The whole term is a summation. You can rename $\mu \to \nu$ in 4 to obtain 5 since the metric is symmetric. In order to procced from there the answers to these question will be useful.
Derivation of Klein-Gordon equation in General Relativity
Variation of a scalar field
EDIT
Ok. So the term is
$$h(\phi)\nabla^{\mu}\delta\phi\nabla_{\mu}\phi = \nabla^{\mu}(h(\phi)\delta\phi\nabla_{\mu}\phi) - \frac{dh}{d\phi}\nabla^{\mu}\phi\nabla_{\mu}\phi \delta\phi - h(\phi)\Box\phi\delta\phi$$
The total derivative vanishes due to the boundary conditions.