Reference: archive.org/details/GeneralRelativity/page/n250/mode/1up

I am studying the phenomenon of gravitational waves as one of the conseguences of the linearized Einstein field equation.
In order to do this I am starting from the fact that the linearized Einstein field equation in vacuum is given by: $$\square \gamma_{\mu\nu}=0$$ where $$\gamma_{\mu\nu}=h_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}h,$$ with $h_{\mu\nu}$ the perturbation to add to the background metric, given by Minkowski metric, in order to obtain the perturbed metric.

In order to obtain the linearized equation above we have to require that the Hilbert gauge is satisfied $$\gamma^{\mu\nu}_{,\nu}=0$$ and it is possible thanks to an infinitesimal coordinates transformation $x'^{\mu}=x^{\mu}-\xi^{\mu}$ with $\xi^\mu$ such that $\square \xi^{\mu}=\gamma^{\mu\nu}_{,\nu}$.
Now I want to find out another infinitesimal coordinate transformation in order to ensure that the trace of $\gamma$ in these new coordinates is null, i.e. $\gamma'=0\iff \gamma'=\gamma-2\xi^{\lambda}_{,\lambda}=0$.

Now in order to obtain this new coordinate transformation that does not destroy the Hilbert gauge ($\gamma^{\mu\nu}_{,\nu}=0$) ensured with the previous coordinate transformation I want that the function $\xi$ to add to the old $\xi$ (of the old coordinate transformation) is such that $$\square \xi^\mu=0.$$

$\textbf{What I have not understood is:}$ why $\square \xi^\mu=0$ has the consistency condition that $$\square \gamma=0~?$$ If $\square \gamma\neq 0$ does not continue to hold $\gamma'=\gamma-2\xi^{\lambda}_{,\lambda}$?

I have thought that the reason is: since I want $2\xi^{\lambda}_{,\lambda}=\gamma$, if $\square \xi^{\lambda}=0$ this implies that $\square \xi^{\lambda}_{,\lambda}=0$ and so $2\square \xi^{\lambda}_{,\lambda}=\square \gamma=0$, do you think it is right?

  • $\begingroup$ Could you show us how the "consistency condition" $\square\gamma^\mu{}_\mu=0$ is derived? (I personally haven't seen it before) $\endgroup$ Apr 2 at 9:59
  • $\begingroup$ This is in as matter of fact my question! I don't know how understand why $\square \gamma=0$ is a consistency condition required by $\square \xi^\mu=0$ $\endgroup$
    – pawel
    Apr 2 at 10:10
  • $\begingroup$ I have tried to argue as I have shown in the very last part of the question but I am not so confident that this is right! $\endgroup$
    – pawel
    Apr 2 at 10:12
  • $\begingroup$ Oh I see. (Also, I think you're missing some comma's in $\xi_\lambda^\lambda$) $\endgroup$ Apr 2 at 10:23
  • $\begingroup$ Oh yes! Thanks it is a typo! $\endgroup$
    – pawel
    Apr 2 at 10:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.