0

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.

  1. $$\delta(\sqrt{-g})hg^{\mu \nu}\nabla_{\mu}\phi\nabla_{\nu}\phi$$
  2. $$\sqrt{-g}\delta h g^{\mu \nu}\nabla_{\mu}\phi\nabla_{\nu}\phi$$
  3. $$\sqrt{-g}h \delta(g^{\mu \nu})\nabla_{\mu}\phi\nabla_{\nu}\phi$$
  4. $$\sqrt{-g}h g^{\mu \nu}\nabla_{\mu}\delta(\phi)\nabla_{\nu}\phi$$
  5. $$\sqrt{-g}h g^{\mu \nu}\nabla_{\mu}\phi\nabla_{\nu}\delta(\phi)$$

So I have found

  1. $$(-\frac{1}{2}\sqrt{-g}hg_{\mu\nu}\nabla_{\beta}\phi\nabla^{\beta}\phi)\delta g^{\mu\nu}$$
  2. $$(\sqrt{-g}\nabla_{\mu}\phi\nabla^{\mu}\phi\frac{\partial h}{\partial \phi})\delta \phi$$
  3. $$(\sqrt{-g}h\nabla_{\mu}\phi\nabla_{\nu}\phi)\delta(g^{\mu \nu})$$

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

$$\delta(\sqrt{-g}h(\phi)g^{\mu \nu}\nabla_{\mu}\phi\nabla_{\nu}\phi) = \sqrt{-g}\Big[\delta g^{\mu\nu}\big[-\frac{hg_{\mu\nu}}{2}\nabla_{\beta}\phi\nabla^{\beta}\phi+ h\nabla_{\mu}\phi\nabla_{\nu}\phi\big] + \delta \phi \big[\nabla_{\mu}\phi\nabla^{\mu}\phi\frac{\partial h}{\partial \phi} -2[\nabla_{\mu}h\nabla^{\mu}\phi + \square\phi h] \big]\Big]$$

camarman
  • 3,188

2 Answers2

1

In your variation of (1) you shouldn't use the same indices, so write $g_{\alpha \beta} \delta g^{\alpha \beta}$ instead. Then for (4) and (5) you'll have to integrate by parts in order to get $\delta \phi$ terms. I'll make a guess that the field variations in $\phi$ will vanish on your boundary, leaving terms in your integral of the form $\delta \phi \nabla (...)$. You should have seen something similar for a standard scalar field theory with partial derivatives.

This should be enough to solve your problem.

Eletie
  • 3,343
1

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.

Noone
  • 2,558