# Gravitational waves as dark energy?

Is the energy carried by gravitational radiation a viable candidate for $\Lambda$ / dark energy?

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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.
Hi! The same derivation holds for any particles or waves moving by the speed of light. Take a graviton of momentum $\vec p$ in a box $L^3$. It takes $L/v_x$ of time to go from the left boundary to the right one; in each collision, the momentum given to the walls is $2 p_x$. That's $2 p_x\cdot v_x/Lc$ of momentum per unit time. Sum over $x,y,z$ to get momentum per time $p\cdot v/Lc$. Divide by the area of the cube, $6L^2$, to get $pressure=Force/Area = p\cdot v/3L^3 = E/3L^3 = \rho/3$ for any particles/waves moving by the speed $c$. – Luboš Motl Nov 3 '11 at 11:05
Alternatively, you may argue that in 4 dimensions, the stress-energy tensor of radiation has to be traceless because the classical theory describing the radiation has no dimensionful constants (conformal symmetry). That means that $p_{xx}=p_{yy}=p_{zz}$ by rotational symmetry and all of them have to be $\rho/3$ to get zero for $\rho-3 \times \rho/3$. – Luboš Motl Nov 3 '11 at 11:10