general relativity - D'Alembertian for a scalar field

I have read that the D'Alembertian for a scalar field is $$ \Box = g^{\nu\mu}\nabla_\nu\nabla_\mu = \frac{1}{\sqrt{-g}}\partial_\mu (\sqrt{-g}\partial^\mu). $$

1. 
   

   Exactly when is this correct? Only for $\Box \phi$ where $\phi$ is a scalar-field?

   
   
   


2. 
   

   How exactly is it shown?

   
   


3. 
   

   Is it true only on-shell, from the Euler-Lagrange equations of a scalar-field?

   
   

Answer

This is based on the observation that, given some vector $V^\mu$,

$$\nabla_\mu V^\mu=\frac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}V^\mu)$$

We can show explicitly that this is true:

$$\nabla_\mu V^\mu=\partial_\mu V^\mu +\Gamma^\mu_{\mu\lambda}V^\lambda$$

Let's examine the last term:

$$\Gamma^\mu_{\mu\lambda}=\frac{1}{2}g^{\mu\rho}(\partial_\mu g_{\lambda\rho}+\partial_\lambda g_{\mu\rho} -\partial_\rho g_{\lambda\mu})$$ we can cancel the first and third term on the right hand side, yielding \begin{equation} \Gamma^\mu_{\mu\lambda}=\frac{1}{2}g^{\mu\rho}\partial_\lambda g_{\mu\rho} \end{equation}

The idea is to show that this equals $\frac{1}{\sqrt{-g}}\partial_\lambda\sqrt{-g}$:

\begin{align*} \partial_\lambda\sqrt{-g}&=-\frac{1}{2\sqrt{-g}}\partial_\lambda g=-\frac{1}{2\sqrt{-g}}\frac{|g_{\mu\nu}+\delta g_{\mu\nu}|-|g_{\mu\nu}|}{\delta x^\lambda}\\ &=-\frac{1}{2\sqrt{-g}}\frac{|g_{\mu\nu}||\mathbb{I}+(g_{\mu\nu})^{-1}\partial_\lambda g_{\mu\nu}\delta x^\lambda|-|g_{\mu\nu}|}{\delta x^\lambda}\\ &=-\frac{1}{2\sqrt{-g}}\frac{|g_{\mu\nu}|\bigr(1+\text{Tr}((g_{\mu\nu})^{-1}\partial_\lambda g_{\mu\nu}\delta x^\lambda)\bigr)-|g_{\mu\nu}|}{\delta x^\lambda}\\ &=\sqrt{-g}\frac{1}{2}(g^{\mu\nu}\partial_\lambda g_{\mu\nu}) \end{align*} Here, I used $|g_{\mu\nu}|$ to denote the determinant of $g_{\mu\nu}$. Multiplying by $\frac{1}{\sqrt{-g}}$ and comparing shows that $$\Gamma^\mu_{\mu\lambda}=\frac{1}{\sqrt{-g}}\partial_\lambda\sqrt{-g}=\partial_\lambda \ln\sqrt{-g}$$ It follows that $$\nabla_\mu V^\mu =\frac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}V^\mu) $$ Thus, you see that your formula of the d'Alembertian holds only for scalars: the first covariant derivative reduces to a partial derivative, and for $\partial^\mu\phi$ we appeal to the formula I derived. Note, however, that we never used the equations of motion. This is all just a much longer and more explicit version of what Qmechanic already posted while I was typing this up.