On Sunday, March 31, 2019 at 8:23:02 AM UTC-6, John Clark wrote: > > On Sun, Mar 31, 2019 at 8:43 AM Lawrence Crowell <goldenfield...@gmail.com > <javascript:>> wrote: > > >> > An antenna or any receiver of electromagnetic waves in effect measures >> the displacement of electrons or equivalently a current is produced. >> > > A radio receiver detects the power in a AC circuit, and that is the Root > Mean Square voltage times the Root Mean Square current. Unlike LIGO > radios don't detect peak to peak values. > > > A gravitational wave is measured according to strain, but a strain >> through distance has an energy content as well. >> > > Yes but LIGO detects the peak to peak displacement of a wave not its power > or energy as cameras and radios do. And that means LIGO's ability to detect > wave producing things is reduced with distance much more slowly than with > telescopes that deal with electromagnetic waves. Peak-to peak displacement > is proportional to the Root Mean Square of the wave and the RMS is > proportional to the square root of the power. So if there is 4 times less > power in the gravitational wave (because the source is twice as far away) > the peak to peak displacement is only reduced by a factor of 2. >

I guess you will have to give a reference on this. I looked at some references and I can't find anything on what you say here: https://labcit.ligo.caltech.edu/~dhs/Adv-LIGO/old/interferometric-gray-paper.pdf https://labcit.ligo.caltech.edu/~dhs/Adv-LIGO/old/interferometric-gray-paper.pdf I can see in one sense what you are saying about RMS, but I don't think your quite correct still. The interferometer measures a quadrupole displacemement. There is the quadrupole tensor Q = 3d_id_j - d^2δ_{ij}, where fields are Q_{ij}/r^4. The distance to the source is r and the distance between the two inspiralling black holes is d. The displacement is given by the metric g_{ab} = δ_{ab} + h_{ab} for the ++ and xx polarizations. The ++ polarization metric will be h_{++} = 2d_+^2/r^2, for r the distance to the source, and the curvature is R_++ = ½h_{++}R = d_{++}/r^4. The Einstein space criterion the metric proportional to the Ricci curvature and the metric giving the displacement means the displacement is ~ 1/r^2. LC > > So gravitational waves have intensities that drops with the square of >> the distance > > > I'm not disputing that, but that fact is not inconsistent with the fact > that LIGO's ability to detect gravitational wave sources only decreases > linearly with distance because with LIGO the key thing is peak to peak > displacement of the wave not its intensity. > > John K Clark > > > > >> -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at https://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/d/optout.