Varying an action with respect to a scalar field

[JD_PM]

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!

 

[etotheipi]

\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.

Hey JD, I don't know anything about this metric or topic, although from the original action I derived the same condition that $\partial^{\mu} \partial_{\mu} \phi = \partial V / \partial \phi$ but then instead considered that with the $(-+++)$ sig, and with $i \in \{1,2,3 \}$$$
\begin{align*}
\partial^{\mu} \partial_{\mu} \phi & = \partial^0 \partial_0 \phi + \partial^i \partial_i \phi \\ \\

\partial^{\mu} \partial_{\mu} \phi &= - \partial_0 \partial_0 \phi + \partial_i \partial_i \phi \\ \\

\partial^{\mu} \partial_{\mu} \phi &= - \ddot{\phi} + \nabla^2 \phi \overset{!}{=} \frac{\partial V}{\partial \phi} \implies \ddot{\phi} - \nabla^2 \phi = - \frac{\partial V}{\partial \phi}
\end{align*}$$I don't know how you might turn the $\nabla^2 \phi$ term into something involving scale factors, but wondered if perhaps you didn't take into account the summation?

[And also, the second time derivative $\partial^2 / \partial t^2 = \partial_0 \partial_0$ with indices down, right? ]

 

[JD_PM]

[...] perhaps you didn't take into account the summation?

[And also, the second time derivative $\partial^2 / \partial t^2 = \partial_0 \partial_0$ with indices down, right? ]

Absolutely right, thanks James I indeed missed the spatial term! We have

$$\ddot{\phi} - \nabla^2 \phi = - \frac{\partial V}{\partial \phi}$$

All what's left is to show that

\begin{equation}
\nabla^2 \phi = -3 \left( \frac{\dot a}{a}\right) \phi \tag{1}
\end{equation}

I am quite confident $(1)$ can be derived out of the fact that the real scalar field $\phi$ satisfies Poisson's equation $\nabla^2 \phi = 4 \pi G_N \rho$ (for instance, for some reference, check Tong's beautiful lecture notes EQ. (0.1)) AND the flat first Friedmann equation (let us not include the cosmological constant)

\begin{equation}
\left( \frac{\dot a}{a} \right)^2 = \frac{8 \pi G_N}{3} \rho \tag{F1}
\end{equation}

Remark: we can use Friedmann equations given that we are assuming an homogeneous and isotropic universe (which is implied in the metric we are using in this particular exercise). If you are interested in Cosmology, I strongly recommend probably the most complete, enlightening GR lecture notes I've ever come across: Matthias Blau's GR lecture notes. Friedmann equations are in EQ. (35.108).

Hence, using $\nabla^2 \phi = 4 \pi G_N \rho$ and $(F1)$ we find

\begin{equation*}
\nabla^2 \phi = -3 \left( \frac{\dot a}{a}\right) \phi \iff \phi := -\frac 1 2 \left( \frac{\dot a}{a}\right)
\end{equation*}

Argh... honestly I am not satisfied with my answer. I have been "forcing" the pieces to fit...

I will have a walk to refresh my mind!

 

[TSny]

\begin{align*}
& \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*}

The factor $\sqrt{-\det g} = a^3$ is part of the Lagrangian density. So, after doing the integration by parts in order to "shift" the derivative $\partial_{\nu}$,you get
\begin{align*}
& \int d^4 x \left[ a^3 \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[ \left( g^{\mu \nu} \partial_{\nu} \left( a^3 \partial_{\mu} \phi \right) - a^3 \frac{ \partial V(\phi)}{\partial \phi} \right)\right]( \delta \phi)+ {\text{Surface term}}
\end{align*}

If you want to take a shortcut, you can use the Euler-Lagrange equations as given by equation (10) here.
Be sure to include $\sqrt{-\det g}$ as part of $\mathcal L$.

 

[etotheipi]

[Ahh clever, @TSny! I'd forgotten that with this metric the Christoffel symbols (obviously!) don't vanish, so $\nabla_{\mu} \neq \partial_{\mu}$ and for this particular term from the IBP, the second and third terms are not equal:$$\int d^4 x \partial_{\mu} (\sqrt{-g} g^{\mu \nu} \partial_{\nu} \phi ) \delta \phi = \int d^4 x (\sqrt{-g} g^{\mu \nu} \nabla_{\mu} \nabla_{\nu} \phi) \delta \phi \, \, \neq \, \, \int d^4 x (\sqrt{-g} g^{\mu \nu} \partial_{\mu} \partial_{\nu} \phi) \delta \phi$$... ]

...Anyway, I will be quiet now

 

[JD_PM]

The factor $\sqrt{-\det g} = a^3$ is part of the Lagrangian density. So, after doing the integration by parts in order to "shift" the derivative $\partial_{\nu}$,you get
\begin{align*}
& \int d^4 x \left[ a^3 \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[ \left( g^{\mu \nu} \partial_{\nu} \left( a^3 \partial_{\mu} \phi \right) - a^3 \frac{ \partial V(\phi)}{\partial \phi} \right)\right]( \delta \phi)+ {\text{Surface term}}
\end{align*}

If you want to take a shortcut, you can use the Euler-Lagrange equations as given by equation (10) here.
Be sure to include $\sqrt{-\det g}$ as part of $\mathcal L$.

Ohhh I understand what my mistake was now! . We have

\begin{equation}
\delta S = 0 \iff \partial^{\mu}\left(a^3 \partial_{\mu} \phi\right) - a^3 \frac{ \partial V(\phi)}{\partial \phi} = 0 \tag{2}
\end{equation}

Oops I forgot to mention that we are dealing with a homogeneous scalar field (i.e. $\partial_i \phi = 0$) . Hence $(2)$ becomes

\begin{align}
&-\partial_{0}\left(a^3 \dot \phi \right) - a^3 \frac{ \partial V(\phi)}{\partial \phi} = 0 \Rightarrow \nonumber \\
&\Rightarrow 3a^2 \dot a \dot \phi + a^3 \ddot \phi + a^3 \frac{ \partial V(\phi)}{\partial \phi} = 0 \Rightarrow \tag{3} \\
&\Rightarrow
\ddot \phi + 3\left( \frac{\dot a}{a}\right) \dot \phi = -\frac{\partial V(\phi)}{\partial \phi} \nonumber
\end{align}

Where to get to the last line I simply multiplied both sides of $(3)$ by $1/a^3$

Once again @TSny, hats off to you!