How are gravitational waves derived?

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

We then define ${\overline{h}}^{\mu \nu }=h^{\mu \nu }-\frac{1}{2}{\eta }^{\mu \nu }{h}_{\alpha }^{\alpha }$, which in the gauge ${\partial }_{\mu }{\overline{h}}^{\mu \nu }=0$, gives the wave equation $$◻\overline{h}=0.$$

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.

posted almost 2 years ago

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Mithrandir24601‭ staff

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