Obtaining the KG equation from Action

Question

After solving the field equation for

$$S = \int \sqrt{-g}dx^4[f(\phi)R + h(\phi)g^{\mu \nu}\nabla_{\mu}\phi\nabla_{\nu}\phi - V(\phi)]$$

I have obtained

$$2h\square \phi + \frac{\partial h}{\partial \phi}\nabla^{\mu}\phi\nabla_{\mu}\phi - R\frac{\partial f}{\partial \phi} + \frac{\partial V}{\partial \phi} = 0~~(*)$$

It seems that the action for the inflation is

$$S = \int \sqrt{-g}dx^4[\frac{1}{2}g^{\mu \nu}\nabla_{\mu}\phi\nabla_{\nu}\phi - V(\phi)]$$

Thus for $f=0$ and $h=1/2$ $(*)$ gives,

$$\square \phi +\frac{\partial V}{\partial \phi}=0$$

so I am getting

$$\ddot{\phi} + \frac{\partial V}{\partial \phi}=0$$ from here but the inflation field equation is

$$\ddot{\phi} + 3H\dot{\phi} + \frac{\partial V}{\partial \phi} = 0$$

So is there terms missing in $(*)$, or am I am doing something wrong..?

Or can we obtain KG equation from the given $(*)$..maybe there should be something at the right side of the $(*)$ ?

Answer 1

$\square \phi = V'(\phi)$ does not imply $\ddot{\phi} = V'(\phi)$ as you have implied. The covariant derivative term $\square \phi$ contains the metric tensor, and consequent connection terms, so that is where the Hubble factor $H(t)$ comes from.

Another hint, consider that $$\nabla_{\mu} \nabla^{\mu}\phi = \partial_{\mu} \nabla^{\mu} \phi + \Gamma^{\mu}_{\mu \nu} \nabla^{\nu} \phi$$