I have read these questions:
Does a photon in vacuum have a rest frame?
Based on dmckee's answer, the answer is no to a photon's rest frame.
In the modern view each particle has one and only one mass defined by the square of it's energy--momentum four vector (which being a Lorentz invariant you can calculate in any inertial frame): $$ m^2 \equiv p^2 = (E, \vec{p})^2 = E^2 - \vec{p}^2 $$ For a photon this value is zero. In any frame, and that allows people to reasonably say that the photon has zero mass without needing to define a rest frame for it.
Now I understand that the graviton is hypothetical.
But it is meant to be massless like the photon. It is a gauge boson like the photon.
So we can write the same for a graviton: $$ m^2 \equiv p^2 = (E, \vec{p})^2 = E^2 - \vec{p}^2 $$
And this should be 0 in any inertial reference frame for the graviton too.
Now as per SR (second postulate):
- the speed of light in vacuum is the same for all observers, regardless of the light source
Now for GWs (like EM waves), it should be true that:
- the speed of GWs in vacuum is the same for all observers regardless of the motion of the GW's source
In this case, gravity could be thought of as the same thing as light, and we could basically build SR onto GWs (instead of light). We could build the postulates of SR onto GWs, derive the laws of SR from the behavior of GWs (instead of light).
Question:
Does a graviton (I understand it is theoretical, but what is it meant to be) in vacuum have a rest frame?
Do GWs obey the laws of SR, so is the speed of GWs in vacuum the same for all observers regardless of the motion of the GW's source?