Is the energy carried by gravitational radiation a viable candidate for $\Lambda$ / dark energy?
Nope. Gravitational radiation is a kind of radiation and it has a completely different equation of state than the cosmological constant.
The cosmological constant has pressure equal to the energy density with a minus sign, $p=-\rho$: the stress-energy tensor is proportional to the metric tensor so the spatial and temporal diagonal components only differ by the sign. Radiation has $p=+\rho/3$, much like for photons. Most of the energy density of the Universe has $p/\rho = -1$; that's what we know from observations because the expansion accelerates. A radiation-dominated Universe wouldn't accelerate (and didn't accelerate: our Universe was indeed radiation-dominated when it was much younger than today).
The ratio $p/\rho$ must be between $-1$ and $+1$ because of the energy conditions (or because the speed of sound can't exceed the speed of light). The $-1$ bound is saturated by the cosmological constant, the canonical realization of "dark energy"; $-2/3$ and $-1/3$ comes from hypothetical cosmic domain walls and cosmic strings, respectively; $0$ is the dust, i.e. static particles; $+1/3$ is radiation; and higher ratios may be obtained for "somewhat unrealistic" types of matter such as the dense black hole gas for which it is $+1$. This ratio determines the acceleration rate as a function of the Hubble constant.
protected by ACuriousMind♦ Apr 28 '17 at 10:54
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