The order of the time-space components of the metric tensor in post-Newtonian expansion

Question

In the book Gravitational waves Vol.1: theory and experiment by M.Maggiore, in chapter 5, page 239, the author announces that when we expand the metric tensor by $v \over c$, the components $g_{0i}$ start from order $({v \over c})^3$. But why? Why does the order $({v \over c})^1$ vanish?

I have a thought that this can be guessed from something like the Schwarz solution, where $g_{00}$ and $g_{ij}$ has order $({v \over c})^2$ (the $GM \over r$) while $g_{0i}$ is simply $0$. But in Schwarz solution this is because we did some gauge transformations. So, now in the post-Newtonian expansion, if we first have non-zero $g_{0i} ^{(1)}$ (the $({v \over c})^1$ order of $g_{0i}$), is it also possible that we do a gauge transformation so $g_{0i} ^{(1)}$ become $0$ ? If not, why can we assume $g_{0i} ^{(1)}$ to be simply $0$ ?

Answer 1

In the weak-field (Newtonian) limit, if you were to calculate the Einstein tensor and the Einstein equation, you would get:

\begin{equation} \Delta ^{(2)}g_{00} = \Delta \phi = 4\pi G\rho \end{equation} with $\phi$ the Newtonian potential. Using the Virial theorem, this is indeed of order $(v/c)^2$.

Now the main idea of the Post-Newtonian expansion is that you expand up to a specific order in $(v/c)$ in the Einstein equation. The question is then what order of the metric's components corresponds to that order in the Einstein tensor. Suppose you had a term $(v/c)^1$ in the metric, then grouping all spatial derivatives of this component in the Einstein tensor gives a first order term in the Einstein tensor, while we have just proven that the lowest order term should be of order two. Therefore, the lowest order term we can have for the $g_{oi}$ components should indeed be third order in $(v/c)$.