What are the equations of motion for the scalar field in the tetrad formalism? The action of a massless scalar field in curved spacetime is given by:
\begin{equation}
  S(\phi)=\int d^{4}x \sqrt{-g}\left(g^{\mu\nu}\phi_{,\mu}\phi_{,\nu}\right)
\end{equation}
Now the action can be rewritten using the tetrad formalism as:
\begin{equation}
  S(\phi)=\int d^{4}x e\left(\eta^{ab}\phi_{,a}\phi_{,b}\right)
\end{equation}
where $e$ is the determinant of the veilbein $e^{a}_{\mu}$ and we have used the identity $\phi_{,a}=e^{\mu}_{a}\phi_{,\mu}$.
Is it correct to assume that the equation of motion can be given by:
$$\partial_{a}\left(\frac{\partial\mathcal{L}}{\partial\left(\partial_{a}\phi\right)}\right) - \frac{\partial\mathcal{L}}{\partial\phi} = 0
$$
where $\mathcal{L}$ is the Lagrangian density?
In which case we have explicitly:
$$e \eta^{ab}\phi_{,ab}+\eta^{ab}\phi_{,b}e_{,a}=0$$ 
 A: The action of  a massless scalar field is given by:
\begin{eqnarray}
  S(\phi)&=&\int{\cal{L}}dt\\
  &=&\int d^{4}x \sqrt{-g}\left(g^{\mu\nu}\phi_{,\mu}\phi_{,\nu}\right)
  \end{eqnarray}
Now choosing a tetrad, i.e., a basis of one form at each spacetime point $\{e^{a}=e^{a}_{\mu}dx^{\mu}\}$ we can rewrite the action as:
\begin{eqnarray}
  S(\phi)&=&\int{\cal{L}}dt\\
  &=&\int d^{4}x e\left(e^{\mu}_{a}e^{\nu}_{b}\eta^{ab}\phi_{,\mu}\phi_{,\nu}\right)
  \end{eqnarray}
where $e$ is the determinant of $e^{a}_{\mu}$ which is equal to $\sqrt{-g}$ and we have use the identity $e_{a}^{\mu}e_{b}^{\nu}\eta^{ab}=g^{\mu\nu}$.
Now the equations of motion are given by the Euler-Lagrange equations:
\begin{equation}
     \frac{\delta \mathcal{S}}{\delta\phi}=\frac{\partial\mathcal{L}}{\partial\phi} -\partial_\mu \left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)=0. 
\end{equation}
So we have
\begin{eqnarray}
    \frac{\partial\mathcal{L}}{\partial\phi}&=&0\\
     \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}&=&  e e_{a}^{\mu}e_{b}^{\nu}\eta^{ab}\phi_{,\nu}
\end{eqnarray}
where the equation the motion takes the form:
\begin{equation}
  \partial_{\mu}\left(e e_{a}^{\mu}e_{b}^{\nu}\eta^{ab}\phi_{,\nu}\right)=0
\end{equation}
Now taking into account that $\partial_{b}=e^{\nu}_{b}\partial_{\nu} $  gives:
\begin{equation}
  \partial_{\mu}\left(e e_{a}^{\mu}\eta^{ab}\phi_{,b}\right)=0
\end{equation}
Then the equation of motion takes the form:
\begin{eqnarray}
e\eta^{ab}\partial_{a}\phi_{,b}+e\eta^{ab}\phi_{,b}\partial_{\mu}(e_{a}^{\mu})+\eta^{ab}\phi_{,b}\partial_{a}e=0\\
\end{eqnarray}
