Deriving Klein-Gordon equation in curved spacetime [closed]

Question

I try to drive The Klein-Gordon equation for a massless scalar field in case of FRW metric:

$$ ds^2= a^2(t) [-dt^2 + dx^2] $$

So I start by:

$$\left(\frac{1}{g^{1/2}}\partial_{\mu}(g^{1/2}g^{\mu\nu}\partial_{\nu}) \right)\phi = 0$$

The determinant of the metric is $(\sqrt{-g}=a^4)$. So that the equation leads to:

\begin{eqnarray} \nonumber &&\frac{1}{a^4} \left( \partial_0 a^4 g^{0\nu} \partial_\nu + \partial_i a^4 g^{i\nu} \partial_\nu \right) \phi=0, \\ \nonumber && \partial_0 a^4 g^{00} \partial_0 \phi + a^4 g^{ij} \partial_i \partial_j \phi=0,\\ \nonumber && - \partial_0 a^2 \partial_0 \phi + a^4 \nabla^2 \phi=0, \\ \nonumber && - 2 a \dot{a} \dot{\phi} - a^2 \ddot{\phi} + a^4 \nabla^2 \phi =0,\\ && 2 \frac{ \dot{a}}{a^3} \dot{\phi} + \frac{1}{a^2} \ddot{\phi} - \nabla^2 \phi=0. \end{eqnarray}

Please let me know if this derivation is correct.

Answer 1

I soppuse that the answer should be

\begin{eqnarray} \nonumber &&\frac{1}{a^4} \left( \partial_0 a^4 g^{0\nu} \partial_\nu + \partial_i a^4 g^{i\nu} \partial_\nu \right) \phi=0, \\ \nonumber && \partial_0 a^4 g^{00} \partial_0 \phi + a^4 g^{ij} \partial_i \partial_j \phi=0,\\ \nonumber && (\partial_0 a^4) g^{00} \partial_0 \phi+ a^4(\partial_0 g^{00}) \partial_0 \phi +a^4 g^{00} \partial^2_0 \phi+ a^4 \nabla^2 \phi=0, \\ \nonumber && 4 a^3 \dot{a} (-a^{-2}) \partial_0 \phi+ a^4 (2 a^{-3} \dot{a}) \partial_0 \phi +a^4 (-a^{-2}) \ddot{\phi} + a^4 \nabla^2 \phi=0, \\ \nonumber && -4 a \dot{a} \dot {\phi}+2 a \dot{a} \dot{\phi} -a^2 \ddot{\phi} +a^4 \nabla^2 \phi=0, \\ \nonumber && 2 \frac{ \dot{a}}{a^3} \dot{\phi} + \frac{1}{a^2} \ddot{\phi} - \nabla^2 \phi=0. \end{eqnarray} so I believe it is fine