Harmonic coordinates in General Relativity I was going through chapter 10 of Wald's book on GR. The part I am reading concerns the harmonic gauge:
\begin{equation}
H^\mu\equiv \Box x^\mu=0.
\end{equation}
In eq. (10.2.34) he gives the Ricci tensor in the harmonic gauge. The way I understand this (correct me if I'm wrong) is that since
$$H^\alpha=0,$$
you can add a term containing $H^\alpha$ in your expressions/equations and leave them unchainged. In particular, if you add the term $\partial_{\alpha(\mu}\partial_{\nu)}H^\alpha$ to the Ricci tensor you retrieve an expression quasilinear in the second derivatives of the metric, namely
\begin{equation}
R_{\mu\nu}\rightarrow R^H_{\mu\nu}\equiv R_{\mu\nu}+\partial_{\alpha(\mu}\partial_{\nu)}H^\alpha=-\frac{1}{2}\Box g_{\mu\nu}+\hat{F}(g,\partial g),
\end{equation}
where $\hat{F}$ is some nonlinear function and the index $H$ stands for Harmonic. Is this correct thus far?
Then, the thing that puzzled me is that he claims that the field equations reduce to $R^H_{\mu\nu}=0$, but $G_{\mu\nu}$ also contains the Ricci scalar. So what, exactly, happens with the Ricci invariant $R$. Is it zero in second derivatives?
 A: Your understanding is correct, I think. We can decompose the Ricci tensor into a part that is the d'Alembertian of the metric and a combination involving the harmonic functions $H^{\alpha}$.
$$R_{\alpha\beta}=R^{H}_{\alpha\beta} - H_{(\alpha,\beta)}$$
I think that the EFEs in the question are necessarily considered in the vacuum case. In harmonic coordinates, the EFEs reduce to $R^{H}_{\mu\nu}=0$, which involves the reasoning above, as well as the fact that in vacuum, $G_{\mu\nu}=0$ is equivalent to $R_{\mu\nu}=0$.
The relevant statement is the following:
In the absence of the matter fields (vacuum), the Einstein field equations reduce to $R_{\mu\nu}=0$.
We show that in vacuum, $R_{\mu\nu}$=0 and $G_{\mu\nu}=0$ are equivalent differential equations for the metric.
First:
$$R_{\mu\nu}=0 \implies G_{\mu\nu}=0$$
Proof:
Since $G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}$, we show that $R=0$ easily, by contracting $R_{\mu\nu}=0$ with the inverse metric $g^{\mu\nu}$, getting $R_{\mu\nu}g^{\mu\nu}=R=0$.
Therefore, it holds that $G_{\mu\nu}=R_{\mu\nu}$ and $G_{\mu\nu}=0$.
Second:
$$G_{\mu\nu}=0 \implies R_{\mu\nu}=0$$
Proof:
Contract $G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=0$ with $g^{\mu\nu}$, getting:
$$ R - \frac{1}{2}R \cdot 4 = 0$$
$$ R = 0$$.
This means $0=G_{\mu\nu}=R_{\mu\nu}$.
The Ricci scalar is identically zero in vacuum (assuming that the EFE's hold), and the equivalence follows.
Combining this equivalence and the argument with harmonic arguments, the desired reduction to $R^{H}_{\mu\nu}=0$ follows.
References:
1: https://www.math.uni-potsdam.de/fileadmin/user_upload/Prof-Part-Diff/GR_Workshop/, Lecture 2.
