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Okay, so gravitational wave is the "new" cool thing people are talking about. And my question is going to sound silly but it's completely logical.

I read the Wikipedia article and grasped a bit.

Now we know that waves (Mechanical and Electromagnetic) have amplitude and any object coming in contact is going to gain energy passed on by them.

If we look at this image from wikipedia article which shows earth and spacetime curvature we can think of earth as a plastic ball resting on water. And the gravitational waves are ripples in the curvature of spacetime which propagate as waves, travelling outward from the source.

So my question is did/will a gravitational wave displace earth ?

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    $\begingroup$ "And my question is going to sound silly but it's completely logical." (not a good way to start) $\endgroup$ – Danu Feb 15 '16 at 1:25
  • $\begingroup$ "If we look at this image from wikipedia article" - ah, the embedding diagram strikes again. Yes, to the extent of your knowledge, the question is logical. But, in fact, an embedding diagram employs an unphysical dimension to 'aid' in visualizing spatial curvature. But the Earth cannot move in in this fictional dimension. $\endgroup$ – Alfred Centauri Feb 15 '16 at 2:18
  • $\begingroup$ Scroll down to "Dimensionless Amplitude" in this link: tapir.caltech.edu/~teviet/Waves/gwave.html. If the Earth really rested on a rubber sheet, a passing wave would displace Earth in space by the amplitude of the wave. But space time is not a rubber sheet. A gravitational wave changes the curvature of space time. The shape of the Earth is distorted because the curvature of space time changes. But I don't think this is the same as displacement within space. $\endgroup$ – Ernie Feb 15 '16 at 2:43
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Gravitational waves were predicted 100 years ago and the LIGO observatory has been operational for over a decade... hardly "the new cool thing". They are also part of basically every physicist's standard education (or should be), so this is more of standard fare.

Now to your question... do gravitational waves deform objects? Yes. That does, of course, include the Earth. Do they deposit significant amounts of energy? No, they don't. The deformations that LIGO has to measure to detect them, at all, are smaller than the diameter of a proton. The deformation is roughly a strain in one direction and a stress in the perpendicular direction and vice versa at the frequency of the wave.

In the 1970s Braginsky and Ruffini and Wheeler calculated the flux of these waves at the Earth for different scenarios. Based on Box 36.3 in Misner, Thorne, Wheeler the only relevant objects that will produce a significant energy flux are

  1. Hypothetical neutron star binary of one solar mass each, 10,000 km distance between the stars, 12.2s orbital period, at a distance of 1000 parsec. The spiral time is 3.2 years, the emitted power is $3.25\times 10^{41}erg/s$ and the flux at Earth is $2.7\times 10^{-3}erg/m^2s$. This amounts to $2.7\times 10^{-6}W/m^2$.

  2. Hypothetical neutron star binary of one solar mass each, 1000 km distance between the stars, 0.39s orbital period, at a distance of 1000 parsec. The spiral time is 2.8 hours, the emitted power is $3.25\times 10^{46}erg/s$ and the flux at Earth is $2.7\times 10^{2}erg/m^2s$. This amounts to $2.7\times 10^{-1}W/m^2$.

  3. Hypothetical neutron star binary of one solar mass each, 100 km distance between the stars, 12.2ms orbital period, at a distance of 1000 parsec. The spiral time is 1 second, the emitted power is $3.25\times 10^{51}erg/s$ and the flux at Earth is $2.7\times 10^{7}erg/m^2s$. This amounts to $2.7\times 10^{4}W/m^2$.

  4. Hypothetical neutron star binary of one solar mass each, 10 km distance between the stars, 0.39ms orbital period, at a distance of 1000 parsec. The spiral time is 0.1ms, the emitted power is $3.25\times 10^{56}erg/s$ and the flux at Earth is $2.7\times 10^{12}erg/m^2s$. This amounts to $2.7\times 10^{9}W/m^2$.

We have nothing to fear in either case because only a tiny fraction of this gravitational wave energy can actually be absorbed by the planet. Should such a binary collapse ever happen in our next vicinity (like one parsec away), then the flux at Earth's surface would increase by another six orders of magnitude and then we are probably in the range where some geological effects could be seen. At that point, of course, these effects are irrelevant, since the radiation of the system would have literally fried everything on the planet, anyway.

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  • $\begingroup$ Thanks. Dumb questions: 1) what is spiral time? 2) how do they know it is 1.3b light years distant? $\endgroup$ – Mike Dunlavey Feb 15 '16 at 2:56
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    $\begingroup$ @MikeDunlavey: Spiral time is the time that is left before the two neutron stars (or black holes) merge. All of these estimates are based on general relativistic models of how the merger happens. As the two compact objects get closer, the emitted power goes up very quickly, i.e. they become ever "brighter" in terms of gravitational waves. Since we know the sensitivity of our detector and the attenuation of gravitational waves with distance, we can estimate the distance from the signal strength that was measured here on Earth. $\endgroup$ – CuriousOne Feb 15 '16 at 3:52

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