# Specific gravitational wave polarisation direction for "edge-on" binary system

The gravitational wave signal from a binary system approaching merger consists of a mixture of the two $$h_+$$ and $$h_\times$$ orthogonal polarisation states.

If the binary is "face-on" (with an orbital inclination $$i=0$$) then the two states have equal amplitudes. The situation has rotational symmetry around the line of sight and so it doesn't matter in which directions we define the oscillations associated with the two polarisations.

If the binary is "edge-on" however, with $$i=\pi/2$$, then only $$h_+$$ polarisation is seen. There is now no rotational symmetry - the binary's orbital plane has a definite orientation when viewed on the sky.

In which directions are the axes of the $$h_+$$ polarisation in this case? Is one of the axes aligned with the binary orbital plane? If so, how is a symmetric $$h_+$$ metric perturbation produced from an intrinsically asymmetric situation?

There is a relatively simple symmetry argument that explains why there can be no "cross" polarization when a system is viewed "edge-on". The argument revolves around the fact that a binary with aligned spins and angular momentum (i.e. a binary with a well-defined, non-precessing orbital plane) satisfies a relection symmetry in the orbital plane (i.e. in standard spherical coordinates the system is invariant under the transformation $$\theta\mapsto \pi-\theta$$).

Consequently, the gravitational field (including gravitational waves) generated by the binary needs to satisfy this same symmetry.

Now consider what this symmetry does with a gravitational wave travelling in the orbital plane:

• A "plus" polarized wave (i.e. one with one of its axes aligned with the orbital plane, and the other perpendicular to it) is mapped to itself under this transformation.
• A "cross" polarized wave (i.e. one with both axes at 45 degree angles with the orbital plane) on the other hand ends up phase shifted by half a period.

That is, we have, \begin{align} h_{+} &\mapsto h_{+}\\ h_{\times} &\mapsto -h_{\times}\\ \end{align}

Consequently, requiring that the generated gravitational wave strain is invariant under reflection in the orbital plane, implies that $$h_{\times}$$ has to vanish in the orbital plane, i.e. when the system is view "edge-on".

This resolves the first part of the question.

The second part wonders how it can be that motion that is constrained to a plane can lead to displacement in a direction that is perpendicular to that plane. This due to the quadrupolar nature of gravitational waves. Stretching spacetime in one direction, contracts it in a perpendicular direction (and vice versa).

This behaviour should be intuitive to anybody that has ever played with play-dough. If you stretch a piece of dough, it will become thinner causing a displacement in a direction perpendicular to the direction you are pulling.

• It just seems to lack an intuitive sense like the emission of say linear polarised waves from a rotating dipole (when viewed "edge on") does. The displacement of the masses in a binary is only in one plane, but the GWs produce displacements at right angles to that plane? Your explanation certainly convinces me why we should get plus polarisation, given a choice between the two. Mar 24, 2021 at 15:47
• @ProfRob, That is just the quadrupolar nature of the gravitational wave. Stretching spacetime in one direction, contracts it in a perpendicular direction (and vice versa). Intuitively, this is not really that different than stretch a piece of dough. If you stretch it, it will become thinner causing a displacement in a direction perpendicular to the direction you are pulling. Mar 24, 2021 at 15:53
• Please add something like that - or even that - to your answer and I will accept. That is the intuition I was missing. Mar 24, 2021 at 16:17
• @ProfRob, updated Mar 24, 2021 at 16:57

I think your intuition that the symmetry of the wave differs from the symmetry of the source may be because you have forgotten to take into account that the wave is owing to the dynamic not the static part of the source. First take the time-averaged quadrupole moment of the source, then add to this the dynamic part to get the total quadrupole moment as a function of time. The dynamic part has the symmetry you were asking about.