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Simple quadrupole field not yet in Lorenz gauge?

  1. Dec 29, 2015 #1
    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} =##
    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

    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. jcsd
  3. Jan 3, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
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