I was reading about gravitational waves and about laser based detectors. I also read this. As mentioned in the answer, when ever there is a deformation in spacetime, doesn't it also create a minute time dilation? So in a laser based detector, even if the distance between the objects change when a gravitational wave passes, doesn't the time dilation affect the time taken by the laser to travel between the objects and make it a constant?


Later I read this What makes us think we can actually detect gravitational waves? as well. But I still think the time dilation argument makes sense.


1 Answer 1


Here is the metric of the gravitational waves in spherical coordinates (source):

$\eta_{\mu \nu} = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & r^2 & 0 \\ 0 & 0 & 0 & r^2 \sin^2\theta \end{bmatrix}$

As you can see, it changes only the space coordinates. And only the transversal ones. If there is a change also in the time coordinate, it is not a gravitational wave any more. So, the short answer in the literal sense is a clear no.

But, the gravitational waves are coming from the Einstein Field Eqations by calculating the metric far from the accelerating bodies. Somewhere I've read there are other wave-like gravitational effects with different metric, but they decay with higher powers with the distance from the source. Hopefully somebody with a higher understand in the topic will explain its details.

Extension: Fortunately it already happened in this wonderful answer.

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    $\begingroup$ That's the Minkowski metric. The full metric, including $h_{\mu\nu}$ (but not $\bar{h}_{\mu\nu}$), does modify time, though by a very small amount. $\endgroup$
    – Javier
    Aug 13, 2016 at 12:51
  • $\begingroup$ @Javier It is about the Petrov Classification? $\endgroup$
    – peterh
    Aug 13, 2016 at 14:08
  • $\begingroup$ Not at all. It's just that the metric you posted is the flat metric, which doesn't have any gravitational waves in it. Wikipedia tells you $\bar{h}_{\mu\nu}$; from that you calculate $h_{\mu\nu} = \bar{h}_{\mu\nu}-\frac12 \bar{h}\eta_{\mu\nu}$, and the full metric is $\eta_{\mu\nu}+h_{\mu\nu}$, which very much does affect time. $\endgroup$
    – Javier
    Aug 13, 2016 at 14:10

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