In what direction do gravitational waves get emitted?

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Jonathan Scott
Gold Member
Gravitational waves are emitted in all directions but the amplitude and polarization depends on the angle between the observer's line of sight and the plane of the orbit. The visual depiction is a representation of what happens in the 2D plane of the orbit using the third dimension to represent the wave.

pervect
Staff Emeritus
The visual depiction of the gravitational wave is just a rather loose metaphor for the mathematics. The mathematical description involves the two polarizations of the gravitational wave, symbolized by ##h_+##, the "plus" polarization, and ## h_x##, the "times" polarization.

A real-world GR detector like Ligo will respond to only one of the two polarization components. Multiple detectors, with different orientations, could in principle be arranged such to measure both components.

We will set up a coordinate system based on a celstrial sphere, where we place the origin of the celestial sphere at the source of the GW's, oriented so that the equator corresponds to the orbital plane of the binary source of the GW's. Note that this is different from the usual "earth-centric" defintion of the celestial sphere one will see if one looks up the term in wiki, but the basic idea is the same. Given these coordinates, one can write the amplitudes of the two components of the GW: https://en.wikipedia.org/w/index.php?title=Gravitational_wave&oldid=744529583

$$h_+ \propto (1 + \cos^2 \theta) \quad h_x \propto \cos \theta$$

Here ##\theta## represents what one might loosely call the "lattitude" (the convention is different because the lattiutde of the equator is taken as 90 degrees and not zero). The detector response is independent of "longitude".

One can see that the ##h_+## component's amplitude varies from one to two, and the ##h_x## component varies from zero to one as ##\theta##, the inclination away from the celestial equator, varies. The amplitude of the waves is independent of ##\phi##, the "longitude".

The overall amplitude can be described by the formula ##h = \sqrt{(h_+)^2 +(h_x)^2}.

One can see from these formula that if the detector is oriented correctly, to detect the plus (+) components of the GW's, that one will always get a signal. If one happens to have the detector oriented to capture the times (x) component, one could miss GW's if the detector was as the celestial north or south pole. Since our current setups only mesaure one component, they aren't guaranteed to detect all sources. A complicating factor here is the transverse nature of the GW's, the detector might not and probably will not be transverse to the direction of propagation of the GW's. I believe that this won't kill the signal but only cut it's amplitude in half in the worst case - I could be wrong.

This is a mathematical description of what can actually be measured, which hopefully answers the question.

Unfortunately, the visual metaphors used to "explain" GW's are imperfect - but they give some idea of the flavor of the idea, which is probably all that can be expected without going into all the rather advanced math.

Meerio
Is it in a sphere or on a flat(?) surface?
The amplitude of the waves is independent of ##\phi##, the "longitude".
However, the phase of the waves depends on ##2 \phi## hence the waves are emitted in the double spiral pattern shown in the animation. It makes some sense to me to take the illustrated deflection out of the disc as representative of the amplitude of the strain ##h## that pervect described, don't think of it as a displacement.