Euler-Lagrange equations of a scalar field Given the Robertson-Walker metric

for a scalar field $\phi(t)$, how can we obtain the equation of motion for this scalar field?


I took the contravariant derivative of the scalar field is which is nothing more thant the gradient of that same field and then I applied the covariant derivative to that quantity, since the gradient of the scalar field is a vector, I took the covariant derivate of his components, which are by themselfs contravariant, although the spatial components of the 4-vector are zero the Christoffel are not, therefore having a component that is not zero and is $H\dot{\phi}$ ,where H is $H = \frac{\dot{a(t)}}{a(t)}$.


Finally, is this approach right? If not, how would you approach it?
 A: The following hints might be useful:

*

*The Euler-Lagrange equations can also be written as follows:

\begin{equation}
\frac{1}{\sqrt{-g}}\partial_{\alpha}\left[\sqrt{-g}\frac{\delta L}{\delta (\partial_{\alpha}\phi)}\right] - \frac{\delta\mathcal{L}}{\delta\phi} = 0\ ,
\end{equation}
due to the fact that the divergence can be written in terms of the determinant of the metric tensor (see for instance Tensors - Computing the Divergence formula for a given metric tensor or https://en.wikipedia.org/wiki/Divergence). This form could be useful for diagional metrics, reducing the amount of calculations to do.


*Although the results must be the same, there could be terms that you included and should not be there, due to the fact that $\phi$ depends on $t$ only. From the expression I wrote,

\begin{align}
\frac{1}{\sqrt{-g}}\partial_{\mu}\left[\sqrt{-g}g^{\mu\nu}\partial_{\nu}\phi\right]& = \frac{1}{\sqrt{-g}}\partial_{\mu}\left[\sqrt{-g}g^{\mu\ t}\partial_{t}\phi\right]\\
& = \frac{1}{\sqrt{-g}}\partial_{t}\left[\sqrt{-g}g^{tt}\partial_{t}\phi\right]\\
& = \frac{1}{\sqrt{-g}}\partial_{t}\left[\sqrt{-g}\partial_{t}\phi\right]\\
& = \frac{\dot{a}}{a}\partial_{t}\phi+\partial^{2}_{t}\phi\\
& = H(t)\dot{\phi}+\ddot{\phi}\ .
\end{align}
