Is $\left(-\frac{\partial^2}{\partial t^2}+\nabla^2\right)\phi=0$ same as $\partial_\mu\left( g^{\mu\nu}\sqrt{-g} \partial_\nu\phi\right)=0$? I'm a beginner when we talk about General Relativity, right now I reading an article about gravitational waves but I need help here. Until I know the equation that describes a gravitational wave is,
\begin{equation}
\left(-\frac{\partial^2}{\partial t^2}+\nabla^2\right)\phi=0,
\end{equation}
where $\phi$ is a scalar field.
It is the wave equation in 3+1D, but in the article says the equation that describes a gravitational wave is,
\begin{equation}
\partial_\mu\left( g^{\mu\nu}\sqrt{-g} \partial_\nu\phi\right)=0.
\end{equation}
My question is, they are both the same equations, aren't they? Is the first one a special case of the second equation? If that is the case, then how can we obtain the first from the second one?
 A: [EDITED]
Almost. The "correct" form of the second equation is
$$
\frac{1}{\sqrt{-g}}\ \partial_{\mu}(\sqrt{-g}\ g^{\mu\nu}\partial_{\nu}\ \phi)=0
$$
which is to say
$$
\text{div}\ \text{grad}\ \ \phi=0
$$
if you are familiar with the gradient and divergence operators in curved spacetimes, or
$$
\Delta\ \phi=0
$$
if you are familiar with the Laplace-Beltrami operator on curved spacetimes. Of course the former is verified if and only if $\partial_{\mu}(\sqrt{-g}\ g^{\mu\nu}\partial_{\nu}\ \phi)=0$, so the statement is indeed correct. The wave equation in flat spacetime is indeed the particular form of the above equation in flat coordinates, which are such that, with $x^{0}=t$, $x^{i}=\vec{x}$, 
$$
g_{00}=-1\qquad \qquad g_{0i}=0\qquad g_{ij}=\delta_{ij}
$$
with $i=1,2,3$. Then you have
$$
g^{00}=-1\qquad \qquad g^{0i}=0\qquad g^{ij}=\delta^{ij}
$$
and
$$
\det g=-1
$$
and you can see for yourself that the equation reduces to the wave equation.
$$
$$
Now as for gravitational waves, they don't actually obey the wave equation. It is possible to use the so called "weak field approximation" (you have a metric of the form $g=\eta+h$, with $\eta$ the Minkowski metric and $h$ a small perturbation of $\eta$), together with a special choice of coordinates, in order to get an equation which has the same form as the wave equation. In this case, the unknown function(s) would be the $h_{\mu\nu}$'s, so that you have
$$
\eta^{\sigma\lambda}\ \partial_{\sigma}\partial_{\lambda}\ h_{\mu\nu}=\bigg(-\frac{\partial^{2}}{\partial t^{2}}+\nabla^{2}\bigg)\ h_{\mu\nu}=0
$$
But the said special choice of coordinates brings with it some additional equations:
$$
\eta^{\mu\sigma}\ \partial_{\mu}h_{\sigma\nu}=0\qquad\qquad \eta^{\mu\nu}h_{\mu\nu}=0
$$
Which is to say, $h_{\mu\nu}$ is divergence-free (the first one) and trace-free (the second one). Notice that, as I said, the former only are an approximation to the full equation, which is nonlinear.
