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:(adsbygoogle = window.adsbygoogle || []).push({});

[itex] \gamma_{\mu \nu}(t,r)=\frac{2}{3R} \frac{d^2 q_{\mu \nu}}{dt^2} \bigg|_{t'=t-R/c} [/itex]

I have the following field from a rotating binary

[itex]

\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}

[/itex]

With [itex]r_o[/itex] being the orbital radius. I've computed (and double-checked) 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?if you have Mathematica available, here is a notebook with the detailed calculation, with explanations:

EDIT:

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

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# Einstein tensor of a gravitational source

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