Computing the longitudinal and traceless part of the left hand side of Einstein's equation I am reading a textbook on cosmology. Consider $G^i_j$, the left hand side of Einstein's equation. If $\Psi$ and $\Phi$ are first order perturbations to the time and spatial components respectively of the metric, $G^i_j$ can be written as
$$G^i_j = F(\Phi,\Psi)\delta^i_j + k^ik_j\frac{\Phi+\Psi}{a^2}$$
where $F$ is a function of $\Phi$ and $\Psi$
The textbook then tries to consider only the longitudinal and traceless part of $G^i_j$ by contracting $G^i_j$ with the operator $\hat{k}_i\hat{k}^j -\frac{1}{3}\delta^j_i$. I have a few questions here. What does "longitudinal" mean? Also, how does one concoct the operator $\hat{k}_i\hat{k}^j -\frac{1}{3}\delta^j_i$ and why does it pick out the longitudinal and traceless part of $G^i_j$?
 A: We are working in Newton gauge which is also called longitudinal gauge. The perturbed metric can be written as $g_{\mu\nu} = g_{\mu\nu}^{(0)} + g_{\mu\nu}^{(1)}$ with,
\begin{equation}
g_{\mu\nu}^{(1)} = \left(\begin{matrix}
2\psi & v_i \\
v_i & 2\phi \delta_{ij}+h_{ij} 
\end{matrix}\right)
\end{equation}
Unlike in synchronous gauge, here, due to presence of the potential, the observers experience the velocity. The velocity components can be separated into longitudinal ($v_L$) and transverse components ($v_T$) such that, the former is curl free ($\nabla \times v_L =0 $) and the later is divergence free. Since $v_L$ is curl free it can be written as gradient of the scalar; and the scalar here would be the potential. We knew this beforehand as we are in Newtonian gauge there has to be velocity associated with the potential.
We can carry the same analogy to the spatial perturbations $h_{ij}$ which can be decomposed into transverse, longitudinal and trace part. These modes are independent of each other (-you can take the diveregnce of the metric to get only the transverse part, similarly for the longitudinal part you can take the curl) - this is the decomposition theorem. We take vectors $\partial^ih_{ij}^L, \partial^ih_{ij}^T$, which are longitudinal and transverse respectively (analogous to our arguments for a vector).
\begin{equation}
\epsilon_{ijk}\partial_i\partial_kh^L_{ij} = 0 \implies h^L_{ij} = \left( \partial_i\partial_j - \frac{1}{3}\delta_{ij}\nabla^2\right)S
\end{equation}
where $S$ is a scalar. The last term in momentum space turns out to be a projection operator you are looking for (of course in your case you are acting it on the Ricci tensor).
Sorry for the late reply, I completely forgot. Let me know if you need more.
