Gravitational waves as dark energy?

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

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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.

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Thanks for the answer. How can it be shown that gravitational waves (grav. radiation) ie. ripples in space-time have the same equation of state as photons? –  mtrencseni Nov 3 '11 at 10:22
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
Again, thanks for the answers. Replying to your first answer: thinking of gravitational radiation as ripples in spacetime (and not gravitons), why would they bounce off the wall? I'd think the wave would go straight through it, the same way the gravitational force goes right through it (ie. no shielding). –  mtrencseni Nov 3 '11 at 11:23
Gravitational waves do in fact not bounce off any wall, but the argument is just about scaling. Lubos' is right that the equation of state is the same for all relativistic stuff, but you don't need that to see that gravitational waves make no dark energy candidate: Their density decreases as does the density of all normal stuff (relativistic or not). You can't make anything constant that way. That's another way of saying it doesn't have the right ratio p/\rho. –  WIMP Nov 3 '11 at 14:33