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:


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.


> > 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

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