- #1

- 1,131

- 158

- Homework Statement
- Given the action

\begin{equation*}

S = \int d^4 x \sqrt{-\det g} \left( -\frac 1 2 g^{\mu \nu} \partial_{\mu} \phi \partial_{\nu} \phi - V(\phi)\right)

\end{equation*}

Vary the action with respect to the scalar field ##\phi(x)## and obtain

\begin{equation*}

\ddot \phi + 3\left( \frac{\dot a}{a}\right) \dot \phi = -\frac{\partial V(\phi)}{\partial \phi}

\end{equation*}

- Relevant Equations
- Please check below

Let us work with ##(-+++)## signature

Where the metric ##g_{\mu \nu}## is the flat version (i.e. ##K=0##) of the Robertson–Walker metric (I personally liked how Weinberg derived it in his Cosmology book, section 1.1)

\begin{equation*}

(ds)^2 = -(dt)^2 + a^2(t) (d \vec x)^2

\end{equation*}

Hence ##\sqrt{-\det g} = a^3##

My computation is the following

\begin{align*}

\delta S &= \int d^4 x \delta \left[ \sqrt{-\det g} \left( -\frac{1}{2} g^{\mu \nu} \partial_{\mu} \phi \partial_{\nu} \phi - V(\phi) \right)\right]\\

&= \int d^4 x \left[ \sqrt{-\det g} \left( -\frac{1}{2} g^{\mu \nu} \delta \left(\partial_{\mu} \phi \partial_{\nu} \phi\right) - \underbrace{\delta V(\phi)}_{=\frac{\partial V(\phi)}{\partial \phi}\delta \phi} \right)\right]\\

&= \int d^4 x \left[ \sqrt{-\det g} \left( -\underbrace{g^{\mu \nu} \left(\partial_{\mu} \phi \ \delta( \partial_{\nu} \phi) \right)}_{=g^{\mu \nu} \left(\partial_{\mu} \phi \ \partial_{\nu}( \delta \phi) \right)} - \frac{\partial V(\phi)}{\partial \phi}\delta \phi \right)\right]\\

&= \int d^4 x \left[ \sqrt{-\det g} \left( g^{\mu \nu} \partial_{\mu} \partial_{\nu} \phi - \frac{\partial V(\phi)}{\partial \phi} \right)\right]( \delta \phi)\\

&+ \underbrace{\int d^4 x \left[ \sqrt{-\det g} \left( \partial_{\nu} \left( g^{\mu \nu} \partial_{\mu} \phi \delta \phi\right) \right)\right]( \delta \phi)}_{\text{Surface term vanishes}}

\end{align*}

Hence I get

\begin{equation*}

\delta S = 0 \iff \sqrt{-g}\left( \partial^{\mu}\partial_{\mu} \phi - \frac{\partial V(\phi)}{\partial \phi}\right) = 0

\end{equation*}

But the above equation leads to the following timelike component

\begin{equation*}

-\partial^{0}\partial_{0} \phi - \frac{\partial V(\phi)}{\partial \phi} = 0 \Rightarrow \ddot \phi = - \frac{\partial V(\phi)}{\partial \phi}

\end{equation*}

So I am missing the ##3(\dot a / a)\dot \phi## term.

First thought that my mistake was not including the following well-known variation

\begin{equation*}

\delta \sqrt{-g} = \frac{\sqrt{-g}}{2}g^{\mu \nu} \delta g_{\mu \nu} = -\frac{\sqrt{-g}}{2} g_{\mu \nu}\delta g^{\mu \nu}

\end{equation*}

But given that they explicitly state that we should vary wrt the scalar field I interpreted this as taking ##\delta g_{\mu \nu} \to 0##. Is such interpretation wrong? Is the solution to my issue simply including this variation in my computation?

Thank you!

Where the metric ##g_{\mu \nu}## is the flat version (i.e. ##K=0##) of the Robertson–Walker metric (I personally liked how Weinberg derived it in his Cosmology book, section 1.1)

\begin{equation*}

(ds)^2 = -(dt)^2 + a^2(t) (d \vec x)^2

\end{equation*}

Hence ##\sqrt{-\det g} = a^3##

My computation is the following

\begin{align*}

\delta S &= \int d^4 x \delta \left[ \sqrt{-\det g} \left( -\frac{1}{2} g^{\mu \nu} \partial_{\mu} \phi \partial_{\nu} \phi - V(\phi) \right)\right]\\

&= \int d^4 x \left[ \sqrt{-\det g} \left( -\frac{1}{2} g^{\mu \nu} \delta \left(\partial_{\mu} \phi \partial_{\nu} \phi\right) - \underbrace{\delta V(\phi)}_{=\frac{\partial V(\phi)}{\partial \phi}\delta \phi} \right)\right]\\

&= \int d^4 x \left[ \sqrt{-\det g} \left( -\underbrace{g^{\mu \nu} \left(\partial_{\mu} \phi \ \delta( \partial_{\nu} \phi) \right)}_{=g^{\mu \nu} \left(\partial_{\mu} \phi \ \partial_{\nu}( \delta \phi) \right)} - \frac{\partial V(\phi)}{\partial \phi}\delta \phi \right)\right]\\

&= \int d^4 x \left[ \sqrt{-\det g} \left( g^{\mu \nu} \partial_{\mu} \partial_{\nu} \phi - \frac{\partial V(\phi)}{\partial \phi} \right)\right]( \delta \phi)\\

&+ \underbrace{\int d^4 x \left[ \sqrt{-\det g} \left( \partial_{\nu} \left( g^{\mu \nu} \partial_{\mu} \phi \delta \phi\right) \right)\right]( \delta \phi)}_{\text{Surface term vanishes}}

\end{align*}

Hence I get

\begin{equation*}

\delta S = 0 \iff \sqrt{-g}\left( \partial^{\mu}\partial_{\mu} \phi - \frac{\partial V(\phi)}{\partial \phi}\right) = 0

\end{equation*}

But the above equation leads to the following timelike component

\begin{equation*}

-\partial^{0}\partial_{0} \phi - \frac{\partial V(\phi)}{\partial \phi} = 0 \Rightarrow \ddot \phi = - \frac{\partial V(\phi)}{\partial \phi}

\end{equation*}

So I am missing the ##3(\dot a / a)\dot \phi## term.

First thought that my mistake was not including the following well-known variation

\begin{equation*}

\delta \sqrt{-g} = \frac{\sqrt{-g}}{2}g^{\mu \nu} \delta g_{\mu \nu} = -\frac{\sqrt{-g}}{2} g_{\mu \nu}\delta g^{\mu \nu}

\end{equation*}

But given that they explicitly state that we should vary wrt the scalar field I interpreted this as taking ##\delta g_{\mu \nu} \to 0##. Is such interpretation wrong? Is the solution to my issue simply including this variation in my computation?

Thank you!