The recent detection of gravitational waves made me wonder how the amplitude of the waves fell off with distance. My first naive thought was that it was probably by the cube of the distance. However, it seems the waves are usually generated by an orbiting pair of massive bodies, and that the radiation would tend to be in the plane of the orbit.

I did find this CalTech page on gravitational waves. It seems to suggest, as far as I can understand it, that the wave radiation is directed in the orbital plane and falls off only linearly with distance. But surely one need not be exactly in-line with the orbital plane in order to observe the gravitational waves, raising the question of how the wave amplitude falls off around the plane of the orbit, that is, what is the shape of the gravitational wave radiation?

  • 3
    $\begingroup$ The amplitude falls off as $1/r$, which means that the relevant quantity, the energy density, falls of as $1/r^2$, no different from an electromagnetic wave. The waves are not restricted to the orbital plane, but it is very hard to visualize the 3d pattern, so most drawings pretend that there is only a planar motion to simplify things. $\endgroup$
    – CuriousOne
    Feb 12, 2016 at 0:11

1 Answer 1


The gravitational wave amplitude depends on the time variation of the quadrupole moment. In the weak-field limit, expected to be a good approximation at reasonable distances from the binary, then the amplitude (strain) of the wave falls as $r^{-1}$ and is given by $$h_{jk} = \frac{2G}{c^2r} \frac{\partial^2}{\partial t^2} [D_{jk}(t-r/c)]^{TT}\ \ \ \ \ \ \ (j,k=1,2,3),$$ where $D_{jk}^{TT}$ is the "transverse traceless" projection of the quadrupole moment of the mass, evaluated at the retarded time $t-r/c$. $$D_{jk} = \int \rho(t)[x^i x^j - x^2 \delta^{jk}/3]$$ What this means is that it is motions of masses transverse to the line of sight which produce the radiation.

For a binary system, the strain amplitude when it is viewed perpendicular to the orbital plane ("face-on") is twice the amplitude of the binary seen "edge on". The "face-on" radiation consists of equal amounts of the two gravitational wave polarisations $h_{+}$ and $h_{\times}$, but out of phase, to produce a circularly polarised wave. The "edge on" wave contains only the $h_{+}$ polarisation state and hence is half the amplitude.

In between, the amplitude is in between these two limiting cases. Handy formulae, which you can find in Abbott et al. (2016) are $$ h_{+} = A_{GW}(1+\cos^2 i)\ \cos \phi_{GW}(t)$$ $$h_{\times} = -2A_{GW}\cos i\ \sin \phi_{GW}(t),$$ where $i$ is the binary inclination ($i=0$ face-on, $i=\pi/2$ edge-on) and $A_{GW}$ is a leading amplitude. For an equal mass binary of separation $R$ and with two components of mass $M$ and orbital angular velocity $\Omega$, then $A_{GW} \simeq 4M\Omega^2R^2/r$ and $\phi_{GW} =2\Omega t$.

You would add these orthogonal polarisations in quadrature to get the net amplitude.


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