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The polarization depends on the relative orientation between the line of sight and the orbital plane of the binary (the inclination angle $\iota$). To leading order in $v/c$, where $v$ is the orbital velocity, the two polarizations $h_+$ and $h_\times$ are related to the inclination $\iota$ by
\begin{eqnarray}
h_+ &\propto& \frac{1 + \cos^2 \iota}{4} ...
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Therefore, for the exact same reason our "observable" universe is limited by the travel distance of photons since the CMB, our local gravitational potential shifts over time for the same reason.
It's true that the gravitational field at a spacetime point is due to the matter on the past light cone of that point, but that's automatically taken into ...
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The problem is not the particles, it is that one cannot control and focus a gravitational wave. The gravitational waves detected by LIGO, the first evidence of gravitational waves, were coming from the merging of two black holes .
There is no way to detect or predict such occurrences, and the LIGO "screen" is continents long.
Gravitons are another ...
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You are describing LIGO: a beam of photons is sent through an empty region and allowed to interfere with a reference beam. When gravitational waves arrive they cause a shift of the interference fringes.
You may want the wave to be fully inside the test chamber rather than affecting it as a whole, but note that the effect of the wave - periodic changes in the ...
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Yes, it is the divergence theorem. The term is simply $$\int\partial_k(y^iT^{kj})d^3y=\int y^iT^{kj}n_k d^2y$$
where the integral is over the surface at infinity and $\boldsymbol{n}$ is the unit vector away from the surface. We presume that $T^{kj}$ vanishes at infinity (since the source is isolated) and hence the integral is zero.
You are right that the ...
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