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Consider a gravitational wave in linearized gravity $d_{\mu \nu}(X_{\eta}) = D_{\mu \nu} e^{i X_{\eta} K^{\eta}}$ with $K^{\eta} = (-\omega t, \textbf{k})$. Let $d=| \textbf{D}|$ the scalar maximum amplitude of the wave, measured in distance units.

A gravitational wave makes in a single oscillation of period:

$$T= \frac{ 2\pi}{ \omega}$$

an orthogonal displacement between points separated by a distance $2d$. It would seem to me that if an amplitude of a gravitational wave exceeds the distance that rays could travel at the speed of light, then you could outrun a Light-cone, at least in principle.

If the above analysis is correct and it allows you to outrun a light-cone, maybe there is some solution to the breakdown of causality:

  1. We ignore causality, or develop a quantum theory of causality that accounts for FTL tinkering
  2. Certain wavelengths in gravitational waves are forbidden

If we apply the Occam's razor, it would seem that the second option should be the first option to study. Is it possible that this expression holds for gravitational waves?

$$ \frac{d \omega}{\pi} < c $$

Basically disallowing wavelengths that would allow wavefronts to communicate spacelike-separated points of space faster than a light-cone could in the same time.

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  • $\begingroup$ I seem to recall reading an article on wave mechanics regarding wave group velocity that appears to be breaking the light speed limit, however being a false interpretation. But I've forgotten the source, details. I'll be interested in seeing answers, other comments. $\endgroup$ – docscience Nov 18 '15 at 15:50
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    $\begingroup$ the gravitational waves are transversal, so the points that are becoming reachable via FTL are separated along the direction that is orthogonal to propagation. Because of that the group velocity of the waveform itself is irrelevant. Even more, the waveforms can be entirely periodic $\endgroup$ – diffeomorphism Nov 18 '15 at 15:59
  • $\begingroup$ I don't follow your argument at all. What has the amplitude to do with a "movement between points"? The (group) velocity of the wave is given by $\mathrm{d}\omega/\mathrm{d}k$, not something related to the amplitude. $\endgroup$ – ACuriousMind Nov 18 '15 at 16:28
  • $\begingroup$ Ok, let me put it in these terms: if the gravitational wave is travelling along the z-axis, imagine that the wave is confined inside a gaussian beam shape, and that the wave is polarized along the x-axis. So points near the beam placed along the x-axis at opposing sides of the beam, will be able to send messages back and forth with less delay that a light ray would be able to, if the wave was not present $\endgroup$ – diffeomorphism Nov 18 '15 at 16:40
  • $\begingroup$ Correct. The point is that the waveform of the gravitational wave must be shaped before any FTL communication can occur. $\endgroup$ – lurscher Nov 18 '15 at 17:07
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The physical effects of gravitational wave (GW) is best understood in the transverse, traceless gauge. So, if a linearly polarized GW is propagating in the z-direction, it can have only $h_{11}= - h_{22}$ and $h_{12}= h_{21}$ as non-zero components which are orthogonal to the direction of propagation, along with the condition that the magnitudes of these components are much much less than unity.

In a time interval $T= 2 \pi/\omega$ (where $\omega$ is the angular frequency of a monochromatic GW), the separation $\Delta L$ between two test particles intercepting the GW can change by an amount of the order of $h L$, where $h$ and $L$ are the magnitude of GW amplitude and initial separation between the particles (provided $L$ is much much less than the wavelength of the GW, which is $2\pi c/\omega$).

Hence, $\Delta L/T= h L/T = h L \omega/2 \pi$, which is much much less than $h c$.

Since, $h$ is much much less than unity, $\Delta L/T$ is much much less than $c$. So, the rate of change of separation between the test particles can never exceed the speed of light.

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  • $\begingroup$ you said: "$hL \omega / 2 \pi$, which is much much less than $hc$". Actually you are missing the key point that you are free to choose $\omega$ such that $\omega > 2 \pi c / hL$, which means that $\nabla L/T > c$ $\endgroup$ – lurscher Nov 19 '15 at 13:47
  • $\begingroup$ you said: "(provided $L$ is much much less than the wavelength of the GW, which is $2\pi c/\omega$)". Why are you making this assumption? $\endgroup$ – lurscher Nov 19 '15 at 14:06
  • $\begingroup$ The separation L between test particles has to be much less than radius of curvature of space-time geometry in order that it changes by virtue of geodesic deviation equation. For a GW, the radius of curvature of space-time geometry is of the order of its wavelength. So, hLω/2π is necessarily less than h c. $\endgroup$ – Patrick Das Gupta Nov 19 '15 at 17:57

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