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Q&A How are gravitational waves derived?

The typical 'most simple' derivation of the gravitational wave equation (GWE) starts by a perturbation of the 'background metric' $\bar{g}$ to get $g_{\mu\nu} = \bar{g}_{\mu\nu} + h_{\mu\nu}$, wher...

posted 1y ago by Mithrandir24601‭

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#1: Initial revision by user avatar Mithrandir24601‭ · 2023-06-13T23:12:35Z (over 1 year ago)
The typical 'most simple' derivation of the gravitational wave equation (GWE) starts by a perturbation of the 'background metric' $\bar{g}$ to get $g_{\mu\nu} = \bar{g}_{\mu\nu} + h_{\mu\nu}$, where $h$ is the perturbation that will be described by a wave equation.

'Far away' from the source of this perturbation, this derivation then considers the case where there is no (background) matter, gravity etc. to give that the stress-energy-momentum tensor far away from the source is approximated as $T_{\mu\nu} = 0$ and $\bar{g}_{\mu\nu} = \eta_{\mu\nu}$.

We then define $\bar{h}^{\mu\nu} = h^{\mu\nu} - \frac{1}{2}\eta^{\mu\nu}h^\alpha_\alpha$, which in the gauge $\partial_\mu \bar{h}^{\mu\nu} = 0$, **gives the wave equation** $$\Box\bar{h} = 0.$$

This is the equation of a wave travelling through the vacuum of free space, far away from the source, at the speed of light. As such, *because* this is a wave travelling at the speed of light, it is frame-independent in the sense that **any spatial and temporal co-ordinates are defined by the observer as for e.x. light/EM waves** and that the gravitons don't experience 'time' any more than photons do.

While the equations would be considerably more complicated, this could be extended to non-vacuum regions of space, although due to gravitational waves behaving so similarly to EM waves, I see no reason that the above idea of the observer defining the co-ordinate system wouldn't hold as above.