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Imagine a system where two massive objects are orbiting each other, something like a binary black hole or neutron star system. Such a system should emit gravitational waves. I'm curious on the distribution pattern of these waves. Can we only detect such system through gravitational waves if their plane of rotation happens to coincide with our location?

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  • $\begingroup$ "*It has been proven that such a system would emit gravitational waves. *" - Can you provide a reference? $\endgroup$ – safesphere Sep 1 '18 at 21:43
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    $\begingroup$ @safesphere Isn't that what LIGO had detected? $\endgroup$ – Ray Arifin Sep 1 '18 at 22:27
  • $\begingroup$ No, LIGO has detected mergers of of two black holes or neutron stars, but definitely not the waves emitted by "two massive objects orbiting each other": en.wikipedia.org/wiki/List_of_gravitational_wave_observations $\endgroup$ – safesphere Sep 2 '18 at 4:38
  • $\begingroup$ @safesphere I don't want to debate on this, so I just edited my question. $\endgroup$ – Ray Arifin Sep 2 '18 at 19:54
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Gravitational radiation is a type of radiation called quadrupole radiation. Quadrupole radiation has a certain intensity pattern as a function of the direction of emission -- or, more accurately, there is a family of such patterns. It's exactly analogous to the radiation patterns of certain kinds of radio antennas. Any such pattern has nonzero intensity in almost all directions, although it will be more intense in some directions than in others. It is not just confined to the equatorial plane.

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    $\begingroup$ In that case, would it be more intense in its equatorial plane? $\endgroup$ – Ray Arifin Sep 2 '18 at 19:52
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The gravitational waves are emitted in two polarisation states (plus and cross). For a binary orbiting with its plane perpendicular to the line of sight, both polarisations are seen with equal amplitude.

If viewing the orbit with the plane in the line of sight only one polarisation is present, and whether it is observed would depend on the detector orientation.

More generally, if the orbital inclination angle is $i$ (where $i=0$ is perpendicular to the line of sight), the relevant formula for the strain is $$ h = h_{+} (1 + \cos^2 i)/2 + h_{\times}\cos i\, $$ where $h_{+}$ and $h_{\times}$ have the same amplitude, but are out of phase by $\pi/2$.

To turn this into power you take the quadrature sum of the two orthogonal polarisation components. Thus there is 8 times as much power radiated along the orbital axis as a direction perpendicular to the orbital axis.

The above can be used to determine $i$ if you have two detectors with different sensitivities to the two polarisations (e.g. two laser interferometers with their arms rotated by 45 degrees with respect to each other).

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    $\begingroup$ It should be noted that this formula is an approximation rather than an exact result. In particular, it holds only for the $l=m=2$ mode. This mode is dominant in most cases, particularly so for equal mass quasi-circular systems. The other modes become more relevant if there is significant mass-ratio, spin or eccentricity. (The different $\iota$ dependence of the higher mode, will allow a more accurate determination of $\iota$ in those systems.) $\endgroup$ – mmeent Feb 25 at 7:09

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