Quadrupole formula in TT gauge In weak gravity limit, we can derive the celebrated quadrupole formula in Lorenz gauge $\partial^{\mu} \bar{h}_{\mu \nu} = 0$:
$$
\bar{h}_{ij} = \frac{2G}{r}\frac{d^2}{dt^2}I_{ij}(t_r)
$$
To get $h^{TT}_{ij}$ in transverse traceless gauge, one needs to find a gauge transformation $h_{\mu \nu} \rightarrow h_{\mu \nu} + 2\partial_{(\mu}\xi_{\nu)}$, such that
$$
h = 0\quad h_{0 \nu} = 0 \quad \partial^{\mu}\bar{h}_{\mu \nu} = 0.
$$
But Sean Carroll says that we can just project the $h_{\mu \nu}$ in Lorenz gauge by the transverse-traceless projector to get $h^{TT}_{\mu \nu}$.
The definition of transverse-traceless projector is:
$$
\mathcal{P}_{j k m n} \equiv P_{j m} P_{k n}-\frac{1}{2} P_{j k} P_{m n}
$$
where
$$
P_{j k} \equiv \delta_{j k}-n_{j} n_{k}
$$
and $n^i$ is the unit vector normal to the wavefront.
The quadrupole formula after projection becomes
$$
\bar{h}^{TT}_{ij} = \frac{2G}{r}\frac{d^2}{dt^2}I^{TT}_{ij}(t_r),
$$
but how to prove that such projection can be done by a gauge transformation? In other words, why we can get the same result by simply applying transverse-traceless projector instead of solving a bunch of differential equations?
 A: Since you edited your question after my initial comment, I think that I now know what you were trying to ask.
Any metric perturbation $h_{\mu\nu}$ (in any gauge) can be decomposed uniquely into several pieces, which correspond to the irreducible representations of $SO(3)$ as outlined here:
$$h_{00} = -2\phi \\ h_{0i} = \beta_i + \partial_i \lambda \\ h_{ij} = h^{\text{TT}}_{ij} +  \underbrace{\frac{1}{3} \psi \delta_{ij}}_{\text{trace}} + \underbrace{\partial_{(i} \omega_{j)}}_{\text{solenoidal}} + \underbrace{(\partial_i \partial_j - \frac{1}{3} \delta_{ij} \nabla^2) \theta}_{\text{longitudinal}}$$
with $$\psi = \delta^{ij}h_{ij} \\ \partial^i\beta_i = \partial^i\omega_i = 0 \\ \partial^i h^{\text{TT}}_{ij} = \delta^{ij}h^{\text{TT}}_{ij} = 0$$
and the boundary conditions that they must go to zero as $r \to \infty$. Under a gauge transformation $h_{\mu\nu} + \partial_\mu \xi_\nu + \partial_\nu \xi_\mu$, it can be shown that only $h^{\text{TT}}_{ij}$ remains invariant while all the other pieces change. This suggests that $h^{\text{TT}}_{ij}$ corresponds to something physical, as after all physical quantities do not depend on the choice of coordinates.
Using the pieces, three more gauge-invariant objects can be constructed: $$A = \psi - \nabla^2 \theta \\ B = \phi + \dot{\lambda} - \frac{1}{2} \ddot{\theta} \\ C_i = \beta_i - \frac{1}{2} \dot{\omega}_i$$
Decomposing the stress-energy tensor $T_{\mu\nu}$ in a similar way (which I will not bother repeating), and applying Einstein's equation, we get $$\nabla^2 A = \ldots \\ \nabla^2 B = \ldots \\ \nabla^2 C_i = \ldots \\ \Box^2 h^{\text{TT}}_{ij} = -16 \pi G T^{\text{TT}}_{ij}$$ where the dots correspond to the pieces of $T_{\mu\nu}$.
We have proven an important fact which is that only $h^{\text{TT}}_{ij}$ contains the propagating gravitational waves, as only $h^{\text{TT}}_{ij}$ obeys a wave equation. The rest of the pieces of $h_{\mu\nu}$ do not.
However, if we go by the traditional method which is to first impose the Lorentz gauge, we get $\Box^2 \bar{h}_{\mu\nu} = -16 \pi G T_{\mu\nu}$, which seems to make the entire $h_{\mu\nu}$ obey a wave equation. This obscures the important fact that the other pieces inside are only waving because of gauge choice. These are removed when we impose the TT gauge on top of the Lorentz gauge.
Therefore, we have two ways to get to the TT gauge from an arbitrary starting point. The first way (which always works) is to find $\xi_\mu$ explicitly by solving differential equations and using it to transform the gauge such that all the other pieces of $h_{\mu\nu}$ go to zero except $h^{\text{TT}}_{ij}$ (you can check that this requires $8$ constraints, which is consistent with $10-8=2$ degrees of freedom corresponding to the two polarizations of gravitational waves). This is shown here.
The second way (which applies to unidirectional waves) is to use the projection operator in the direction of propagation, which simply removes the rest of the pieces, leaving $h^{\text{TT}}_{ij}$ behind (remember that $h^{\text{TT}}_{ij}$ is gauge-invariant, anyway!)
Therefore, with either method, we arrive at the desired outcome which is to obtain $h^{\text{TT}}_{ij}$.
