# B In what direction do gravitational waves get emitted?

1. Oct 19, 2016

### Meerio

2. Oct 19, 2016

### Jonathan Scott

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.

3. Oct 19, 2016

### 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".