In section 4.4 of gravitational radiation chapter in Wald's general relativity, eq.4.4.49 shows the far-field generated by a variable mass quadrupole:
$$ \gamma_{\mu \nu}(t,r)=\frac{2}{3R} \frac{d^2 q_{\mu \nu}}{dt^2} \bigg|_{t'=t-R/c} $$
I have the following field from a rotating binary
$ \gamma^{\mu \nu} = \frac{2 M}{3 R} \begin{bmatrix} 2 & - r_o \omega(t \omega \cos{\omega t} + 2 \sin{\omega t} )& - r_0 ( \omega t \sin{\omega t} - 2 \cos{\omega t}) & 0 \\ - r_o \omega(t \omega \cos{\omega t} + 2 \sin{\omega t} ) & -2 r_o^2 \omega^2 \cos{2 \omega t} & - 2 r_o^2 \omega^2 \sin{2 \omega t} & 0 \\ - r_0 ( \omega t \sin{\omega t} - 2 \cos{\omega t}) & - 2 r_o^2 \omega^2 \sin{2 \omega t} & r_o^2 \omega^2 \cos{2 \omega t} & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} $
With $r_o$ being the orbital radius. I've computed the Einstein tensor of this radiation field, taking
$$g_{\mu \nu} = \eta_{\mu \nu} + \gamma_{\mu \nu}(t',r) \Bigg|_{t'=t-R/c}$$
And the resulting Einstein tensor has the following components:
$$ G^{\mu \nu} = \frac{M}{c^2} \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & \frac{2 x^2 - y^2 - z^2}{ R^5} & \frac{3 x y}{R^5} & \frac{3 x z}{R^5} \\ 0 & \frac{3 x y}{R^5} & \frac{2 y^2 - x^2 - z^2}{ R^5} & \frac{3 y z}{R^5} \\ 0 & \frac{3 x z}{R^5} & \frac{3 y z}{R^5} & \frac{2 z^2 - x^2 - y^2}{ R^5} \end{bmatrix} $$
Now, I'm trying to interpret the fact that the Einstein tensor is not zero outside the binary, as one would expect from the Einstein's equation
$$ G_{\mu \nu} = \frac{G}{c^4} T_{\mu \nu}$$
Question: why is the Einstein tensor non-zero for this metric?
Edit
if you have Mathematica available, here is a notebook with the detailed calculation, with explanations:
https://github.com/CharlesJQuarra/GravitationCalcs/blob/master/GravWaveAnalysis.nb
If someone wants to improve on it or propose changes, please, send me a pull request