I have a question about the linearized Einstein Field Equations, and in particular about the Newtonian limit. It goes as follows. If one uses the trace-reversed form of the EFE for the 00-component and uses the linearized Ricci tensor, one gets:
$$-\frac{1}{2}\square h_{00} + \partial_{0}V_{0} = \frac{8\pi G}{c^{4}}T_{00}$$
with $V_{0}:= \partial_{\rho}h^{\rho}_{0} - \frac{1}{2}\partial_{0}h$. From here, if one assumes a static spacetime ($\partial_{0}$ annihilates everything) and models the matter field as dust comoving with the reference frame ($T^{\mu\nu} = \delta^{\mu}_{0}\delta^{\nu}_{0}\rho c^{2}$) then approximating to lowest nontrivial order in the metric perturbation one gets the classical Poisson equation for the newtonian potential:
$$-\frac{1}{2}\nabla^{2}h_{00} = \frac{4\pi G}{c^{2}}\rho.$$
Simple enough... but what I don't understand is the normal form of the linearized EFE in this setting (not the trace-reversed form). In that case, it seems that one really needs to choose a gauge in order to obtain the Newtonian limit, since the linearized Einstein tensor contains a lot of undesired terms that complicate everything, yet the approximation of the right hand side remains almost unchanged...
So: trace-reversed form requires no gauge fixing, but normal form seems to require it. Why is this the case? Is the Newtonian limit dependent on some specific gauge condition, and if so, what is the relevance of this fact?
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