Einstein tensor of a gravitational source

Question

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

Answer 1

The problem here is that in general case, a metric with some given coordinate dependence, will contain a priori both TT (Traceless-Traverse) fields that correspond to propagating fields, and "non-TT" fields that demand sources to be distributed on the spacetime

When one does the process of collapsing a metric to the TT "gauge" (notice the very intentional double quotes) one is essentially killing all matter sources from the metric coordinates, and just keeping the propagating transverse degrees of freedom.

Once one does this collapse operation, then it becomes easy to verify that the Ricci tensor (and by implication, the Einstein tensor as well) will vanish.

There are two ways to accomplish this. One is described in box 35.1 of Misner-Thorne-Wheeler, and it basically amounts to acknowledging that in the TT gauge, the following equation holds everywhere:

$$ R^i_{0j0}= - 2 \frac{ \partial^2 h^{TT}_{i j} }{\partial t^2}$$

If one simply integrates twice in time $- (1/2) R^i_{0j0}$, one is left with $h^{TT}_{i j}$ directly

In this case, trying this is pointless since Mathematica chokes on the time integrals.

The other alternative, is using a radial projector:

$$ P_{ij} = \delta_{ij} - n_i n_j$$

with $n_i = x_i / \sqrt{ x_j x^j }$, and constructing the TT projector:

$$ P^{TT}_{ijkl} = P_{ik} P_{jl} - (1/2) P_{ij} P_{jk} $$

and obtain $h^{TT}$:

$$ h^{TT}_{ij}= P^{TT}_{ijkl} h_{kl} $$

There are still some unanswered questions though: When one does this TT-projection operation, since the indices run from 1..3, basically the $h_{\mu 0}$ components of your original metric aren't just being projected, they are simply ignored!. This seems too harsh. Even if on the TT-gauge they were all to be zero, I imagine that one would want to do a more subtle operation that simply killing whatever values they had on the non-TT original coordinates...?