In radiation gauge ($\gamma=0$), the Einstein field equation in vacuum for a perturbation $\gamma_{\mu\nu}:=g_{\mu\nu}-\eta_{\mu\nu}$ is given by $$ \boxed{ \partial^\alpha\partial_\alpha \gamma_{\mu\nu}=0 }$$ where we use the Lorenz gauge condition $ \partial^\mu\gamma_{\mu\nu}=0$.
This equation is equal to the Einstein field equation to linear order in $\gamma_{\mu\nu}$. However, it is no longer equal to the Einstein field equation to second order in $\gamma_{\mu\nu}$.
To fix this, we re-define our correction to the metric: $$ \eta_{\mu\nu} + \gamma_{\mu\nu} \mapsto \eta_{\mu\nu}+\gamma_{\mu\nu}+\tilde{\gamma}_{\mu\nu} $$
Where $\tilde{\gamma}_{\mu\nu}$ is a an additional dynamical field, in addition to $\gamma_{\mu\nu}$, which, in order to satisfy the Einstein field equations to second order in $\gamma_{\mu\nu}$ (we assume here that $\tilde{\gamma}_{\mu\nu}$ is of the same order of magnitude as $(\gamma_{\mu\nu})^2$ and so mixed terms proportional to $\gamma_{\mu\nu}\tilde{\gamma}_{\mu\nu}$ are proportional to third order in $\gamma_{\mu\nu}$ or higher.), $\tilde{\gamma}_{\mu\nu}$ must obey the following equation: $$ ^{(1)}\tilde{R}_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}\eta^{\alpha\beta}\,^{(1)}\tilde{R}_{\alpha\beta} =8\pi\underbrace{\left[\frac{-1}{8\pi}\left(\,^{(2)}R_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}\eta^{\alpha\beta}\,^{(2)}R_{\alpha\beta}\right)\right]}_{t_{\mu\nu}} $$
where $\tilde{R}_{\mu\nu}$ is the Ricci tensor containing only terms with the correction $\tilde{\gamma}_{\mu\nu}$ and $R_{\mu\nu}$ is the Ricci tensor containing only terms with the original perturbation $\gamma_{\mu\nu}$, and the left-hand superscript $^{(j)}R_{\mu\nu}$ means exclusively only terms proportional to the $j^{th}$ power of the corresponding variable ($\gamma$ or $\tilde{\gamma}$) are contained in that term.
Then we say even though we have vacuum, $t_{\mu\nu}$ represents the energy-momentum tensor of a gravitational wave travelling through that vacuum.
I have a few questions:
Is my description so far accurate?
If a gravitational wave is passing throughout some part of spacetime, why do we ascribe this segment of spacetime as vacuum to begin with? My logic would say this part of spacetime should have $T_{\mu\nu}$ codify the energy-momentum density of said wave, hence, the vacuum Einstein equations should not hold, even in linear order.
What happens if we don't use the linear approximation for the Einstein equation? How do we see that there is some self-energy contained in the metric of the vacuum without in the exact equation?
What happens if we go to third order?
How do you prove that $t_{\mu\nu}$ is conserved, that is, that $$ \partial^{\mu}t_{\mu\nu}=0$$ I tried to just differentiate the term from here but things went south. Should I be using the radiation gauge condition? Should I use any other trick?
At any particular point in spacetime, should I use $\boxed{\eta_{\mu\nu} + \gamma_{\mu\nu}}$, $\boxed{\eta_{\mu\nu} + \tilde{\gamma}_{\mu\nu}}$, or $\boxed{\eta_{\mu\nu}+\gamma_{\mu\nu}+\tilde{\gamma}_{\mu\nu}}$ as the metric? It seems like it must be the second option, since $\gamma_{\mu\nu}$ has already been "used up" in the other side of the equation.
If I have a radiating source, for example, a spinning rod, then as far as I understand, the procedure is to first see what kind of perturbation to flat space $\gamma_{\mu\nu}$ it generates, which is proportional to the second time derivative of its mass quadrupole tensor divided by the distance from it. What happens if I then plug that into $t_{\mu\nu}$? How is that different from the actual $T_{\mu\nu}$ of the spinning rod?
