# Simple quadrupole field not yet in Lorenz gauge?

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1. Dec 29, 2015

### CharlesJQuarra

I'm having trouble reproducing some of the results regarding gravitational waves in the Wald's General Relativity

In section 4.4 of gravitational radiation, eq.4.4.49 shows the far-field generated by a variable mass quadrupole:

$$\gamma_{i j}(t,r)=\frac{2}{3R} \frac{d^2 q_{i j}}{dt^2} \bigg|_{t'=t-R/c}$$

I want to verify that this simple field satisfies the Lorenz gauge (eq.4.4.25)

$$\partial_{i} \gamma_{i j}=0$$

I wrote the $q_{i j}$ for a simple rotating binary

$\ddot{q}_{i j} =$
\begin{bmatrix}
2 \omega^2 \cos{2\omega(t-R/c)} & - 2 \omega^2 \sin{2\omega(t-R/c)} & 0 \\
- 2 \omega^2 \sin{2\omega(t-R/c)} & - 2 \omega^2 \cos{2\omega(t-R/c)} & 0 \\
0 & 0 & 0
\end{bmatrix}

then, I wrote $R=\big|(x,y,z)\big|$

$\partial_{i} \gamma_{i 3}$ trivially cancels, but when I compute the other components of the divergence I get

$$\partial_{i} \gamma_{i 1} = \frac{2 \omega^2([-c x+2 y R \omega]\cos{2\omega(t-R/c)}+[c y+2 x R \omega]\sin{2\omega(t-R/c)})}{c R^3}$$

$$\partial_{i} \gamma_{i 2} = \frac{2 \omega^2([c y+2 x R \omega]\cos{2\omega(t-R/c)}+[c x-2 y R \omega]\sin{2\omega(t-R/c)})}{c R^3}$$

Which as you might have noticed, are not zero in general

Given that the Lorenz gauge is used everywhere one wants to study gravitational wave propagation, it seems unexpected that the far-field of a simple binary quadrupole is not automatically in the Lorenz gauge

Question: I want to understand what is wrong here, if there is anything wrong. Am I wrong/naive in expecting this simple physical system field to be in the Lorenz gauge? Is there a simple transformation that can be applied to this field in order to be manifestly in harmonic coordinates?

2. Jan 3, 2016