Transverse traceless gauge in linearized GR I'm reading about gravitational waves and I'm wondering how we know we can always go to the transverse and traceless gauge? Going to the de Donder gauge is fine, I can follow that, but then showing that we can always go to traceless part and also have $\bar{h}_{0\mu}=0$ I don't follow (where I have defined $\bar{h}_{\mu\nu}=h_{\mu\nu}-\frac{1}{2}g_{\mu\nu}h$ as usual). I get these coupled PDEs and I don't know why we would assume they always have a solution? For the trace to be $0$ I get
$\bar{h} + 2 \partial^{\mu}\xi_{\mu}=0$ where $\bar{h}$ is the old trace.
Then to get the $\bar{h}_{0i}$ components to be 0, I get
$\bar{h}_{0i}-\partial_0\xi_i-\partial_i\xi_0=0$
The textbook I'm reading (maggiore's one on GW) says use the $0^{th}$ component of $\xi$ to get rid of the trace and the the i components to get rid of the ($0$i) components of h.
I'm also trying to get to the transverse traceless gauge in FRW. I know in Maggiore's second textbook he shows how to do it using Bardeen variables, but my professor said it's possible without them and just using the gauge symmetry. I've tried it and I just get more coupled PDEs that are way more complicated that the ones for flat spacetime. Again, how would I know that these always have a solution?
 A: Let's consider the same question in a simpler context: Free Maxwell theory.
One can fix the Lorenz gauge (the analog of de Donder gauge in GR) in which $\partial_\mu A^\mu=0$. In this gauge the fields equations take the form $\Box A_\mu=0$.
Now in this gauge one is still free to perform a gauge transformation $A\to A+ d\lambda$ for gauge parameters satisfying $\Box \lambda=0$ (so called residual gauge symmetries). Now one can use the residual gauge symmetries to fix the gauge further. For example the analog of transverse gauge in this example is to set $A_0=0$:
\begin{align}
A_0\to A_0'=A_0+\partial_0\Lambda=0
\end{align}
There exist a solution $\Lambda=-\int dt A_0$ using which the transformed gauge field has $A_0=0$. Note that
\begin{align}
\Box \Lambda=-\int dt \,\Box A_0=0
\end{align}
This implies that the above gauge parameter is a residual symmetry of the Lorenz gauge.
The same argument works for GR. In de Donder gauge the linearized field equations take the form $\Box h_{\mu\nu}=0$ and the residual symmetries take the form $\Box \xi^\mu=0$ (where $\Box$ is defined w.r.t the background). To go to the TT gauge, you have 4 ordinary differential equations to solve for 4 functions $\xi^\mu$, so there always exist a solution. Explicitly from the second equation we get
\begin{equation}
\bar{h}_{0 i}-\partial_{0} \xi_{i}-\partial_{i} \xi_{0}=0\implies \xi_{i}=\int dt (\bar{h}_{0 i}-\partial_{i} \xi_{0})
\end{equation}
Using this in the first equation we get
\begin{align}
0&=\bar{h}+2(\dot \xi^0+\partial_i \xi^i)=\bar{h}+2\dot \xi^0+2\int dt (\partial^i \bar{h}_{0 i}-\partial_{i}\partial^i \xi_{0})
\end{align}
De Donder gauge implies $\partial_{i}\partial^i \xi^0=\partial_{0}\partial^0 \xi^0$ and $\partial^i \bar{h}_{0 i}=-\partial_0 \bar{h}_{0 0}$ we get
$$\bar h-2 \bar h_{00}+4\dot \xi^0=0\implies \xi^0=-\frac{1}{4}\int dt (\bar h-2 \bar h^{00} )$$
Note that the fact that both dynamical variables and residual symmetries solving the same (wave) equation is a necessary condition.
However, note that my argument was for linearized GR. I am not sure if the same is true for full GR. Maybe in full GR you can do it only asymptotically.
