I ran into something peculiar while attempting to carefully derive the Newtonian limit of general relativity, specifically for the geodesic equation.
To set it up, we assume that the curve $q:[a,b]\rightarrow M^{1,3}$ is a timelike geodesic. Then its components satisfy the geodesic equation
$$\frac{d^{2}q^{k}(\tau)}{d\tau^{2}} =-\Gamma^{k}_{ij}\frac{dq^{i}(\tau)}{d\tau}\frac{dq^{i}(\tau)}{d\tau} $$ where the $\Gamma^{k}_{ij}$ are the unique metric-compatible connection coefficients and $\tau$ is the arc length parameter (i.e. the proper time). To obtain a Newtonian limit, we'll first change the parametrization variable to $x^{0}=t$ using the chain rule to obtain
$$\frac{d^{2}q^{k}(t)}{dt^{2}} = -\Gamma^{k}_{ij}\frac{dq^{i}(t)}{dt}\frac{dq^{i}(t)}{dt}+\Gamma^{0}_{ij}\frac{dq^{k}(t)}{dt}\frac{dq^{i}(t)}{dt}\frac{dq^{i}(t)}{dt}.$$
This expression vanishes identically for $k=0$ as expected, so we now only look at terms with $k\neq 0.$ We write this in powers of the velocity $v^{k}(t)=\frac{dq^{k}(t)}{dt}$ to obtain the famous result that
$$\frac{d^{2}q^{k}(t)}{dt^{2}} = -\Gamma^{k}_{00}-2\sum\limits_{j\neq 0}\Gamma^{k}_{0j}v^{j}(t)-\sum\limits_{i,j\neq 0}\Gamma^{k}_{ij}v^{i}(t)v^{j}(t)+\Gamma^{0}_{ij}\frac{dq^{i}(t)}{dt}\frac{dq^{i}(t)}{dt}v^{k}(t)=-\Gamma^{k}_{00}+\mathcal{O}(v).$$
So we have that the acceleration is $a^{k}(t)\sim-\Gamma^{k}_{00}$ as $v\searrow 0$. We move on to estimating the effect of a small perturbation to the Minkowski metric. This is where I get something weird. For $\epsilon>0,$ suppose the metric tensor takes the form $$\textbf{g}=\boldsymbol{\eta}+\epsilon \textbf{h}=\boldsymbol{\eta}(\textbf{1}+\epsilon\boldsymbol{\eta}\textbf{h})$$ where $\boldsymbol{\eta}$ is the ordinary Minkowski metric and $\textbf{h}$ is a matrix depending on the coordinates. We expand the inverse metric in powers of $\epsilon$ using the Neumann series to obtain $$ \textbf{g}^{-1}=(\textbf{1}+\epsilon\boldsymbol{\eta}\textbf{h})^{-1}\boldsymbol{\eta}=\sum\limits_{n=0}^{\infty}(-\epsilon\boldsymbol{\eta}\textbf{h})^{n}\boldsymbol{\eta}=\boldsymbol{\eta}-\epsilon\boldsymbol{\eta}\textbf{h}\boldsymbol{\eta}+\mathcal{O}(\epsilon^{2}).$$
So the inverse metric's components are $g^{ij}\sim \eta^{ij}-\epsilon \eta^{im}\eta^{jn}h_{mn}$ as $\epsilon\searrow 0.$ We now express the connection coefficients in terms of the metric. For $k\neq 0$, this gives us $$a^{k}(t)\sim-\Gamma^{k}_{00}=\frac{1}{2}g^{k\ell}(\partial_{\ell}g_{00}-2\partial_{0}g_{0\ell})\sim\frac{1}{2}(\eta^{k\ell}-\epsilon \eta^{km}\eta^{\ell n}h_{mn})(\epsilon\partial_{\ell}h_{00}-2\epsilon\partial_{0}h_{0\ell})\\ \sim \frac{\epsilon}{2}\eta^{k\ell}\partial_{\ell}h_{00}-\epsilon\eta^{k\ell}\partial_{0}h_{0\ell}+\mathcal{O}(\epsilon^{2}).$$ So the components of the acceleration in this limit are $\boxed{a_{k}(t) \sim -\frac{\epsilon}{2}\partial_{k}h_{00}+\epsilon\partial_{0}h_{0k}}$
The 1st term is correct, as it ends up containing the Newtonian potential. But the 2nd term $\epsilon\partial_{0}h_{0k}$ isn't supposed to be included in the final answer. Is there a reason it should vanish?
After obtaining this, I found https://einsteinrelativelyeasy.com/ where it states that a 3rd assumption in the Newtonian limit is that the metric is static. This would, of course, kill the 2nd anomalous term. But this is strange to me, since these two terms seem to have the same relative size. Can't Newtonian gravity handle time-varying gravitational fields? Why is this static assumption made in the Newtonian limit, and does this 2nd term have any physical meaning?
