The linearized Einstein field equations in the Lorenz gauge (with $g_{\mu \nu}=\eta_{\mu \nu}+h_{\mu \nu}$ and $\bar h_{\mu \nu}=h_{\mu \nu}-\frac{\eta_{\mu \nu}}{2}h$) are given by: $$ \Box \bar h_{\mu \nu}=-\frac{16\pi G}{c^4}T_{\mu \nu}$$ This has the general solution: $$\bar h_{\mu \nu}=\frac{4\pi G}{c^4}\int \frac{T_{\mu \nu}(\vec{r}_s,t-R/c)}{R} d^3\vec r_s+\phi_{\mu \nu}$$ Where $R=|\vec r-\vec r_s|$ and $\phi_{\mu \nu}$ is any tensor that satisfies $\Box \phi_{\mu \nu}=0$. I am guessing that any arbitrary choice of $\phi_{\mu \nu}$ will not lead to a solution $\bar h_{\mu \nu}$ that is in the Lorenz gauge, and may even change the curvature. My question is therefore: for a situation where no other sources are present (or ever have been) how do we go about choosing $\phi_{\mu \nu}$ such that we are still in the correct gauge. I have seen people take $\phi_{\mu \nu}=0$ is this always allowed?
Edit
After some research I am unsure we are allowed to simply add on $\phi_{\mu \nu}$ since this would change our boundary conditions and our boundary conditions have already been predetermined by our use of Green's function. If this the case then we appear to have lost all gauge freedom in our solution. I.e. $$\bar h_{\mu \nu}=\frac{4\pi G}{c^4}\int \frac{T_{\mu \nu}(\vec{r}_s,t-R/c)}{R} d^3\vec r_s$$ only holds for a specific gauge (since I can plug in a $T_{\mu \nu}$ and get a unique $\bar h_{\mu \nu}$). Yet I have seen people use this formula to derive the quadrapole formula in the $TT$-gauge. I my edit here is correct, where is the gauge freedom in this expression that allows us to produce a solution in the $TT$-gauge?