With which power law of distance is decreasing the amplitude of the distortion caused by a gravitational wave ? $r^{-1}$ ? $r^{-2}$ ? something else ?

-> The coalescence between 2 black holes about a billion light years away caused a distance variation of $10^{-18}$ m in the Advanced LiGO detector. How large was the distortion in the black holes nearborhood, say, at 1 AU ?

A $r^{-2}$ law would give $(\frac{10^9}{1.5 .10^{-5}})^2*10^{-18} = 4.10^9 m$, which seems a lot !

A $r^{-1}$ law would give $7.10^{-5} m$, i.e. a 10th of millimeter, which seems less "planet torning" :-)

  • $\begingroup$ Just like electromagnetic waves from an oscillating dipole, the scaling of amplitude is $r^{-1}$. $\endgroup$
    – ProfRob
    May 13, 2016 at 5:50

1 Answer 1


The "strain" (usually $h$) is a measure of the 'distortion' (in the metric) from a gravitational wave (GW). The energy carried by GW (its 'luminosity' of sorts) scales like, $L_{gw} \propto h^2$, and for energy to be conserved as it radiates away, we need $L_{gw} \propto r^{-2}$, so we find that,

$$h \propto r^{-1}$$

So your latter estimate is more accurate, and you can see that even nearby the BHs, the effect is fairly small!


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