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.

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    $\begingroup$ 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? $\endgroup$ – Marton Trencseni Nov 3 '11 at 10:22
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    $\begingroup$ 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 $c$. $\endgroup$ – Luboš Motl Nov 3 '11 at 11:05
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    $\begingroup$ 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 $\rho-3 \times \rho/3$. $\endgroup$ – Luboš Motl Nov 3 '11 at 11:10
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    $\begingroup$ 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). $\endgroup$ – Marton Trencseni Nov 3 '11 at 11:23
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    $\begingroup$ Dear @mtrencseni, as WIMP says, your comment that the waves don't really get reflected is valid. A material that would be able to reflect them nearly perfectly cannot exist, in fact. Still, I need some well-defined action of them to calculate the pressure using naive mechanics. Of course, if you don't need such visual aid, you may just use the formulae for general relativity where the pressure is given by the stress-energy tensor. Of course, the (matter) stress-energy tensor for pure gravity waves is really zero and one deals with many issues about the "triviality" of energy in GR. $\endgroup$ – Luboš Motl Nov 3 '11 at 19:39

protected by ACuriousMind Apr 28 '17 at 10:54

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