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From the acceleration equation in cosmology the condition for acceleration comes out if $$\rho+3P\lt0$$,$$P\lt\frac{-1}{3}\rho$$,$$\omega\lt\frac{-1}{3}$$. For $\omega=-1$ it gives a constt. energy density & gives an exponential expansion which is quite suitable with the accelerated expansion. My question is why $\omega=-1$ is only taken to explain accelerated expansion and named dark energy?? Any $\omega\lt\frac{-1}{3}$ will work for accelerated expansion.

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Because there are three reasonably-compelling physical models which give $w=-1$!

The first is a cosmological-constant term in the Einstein-Hilbert action. If you expand the action in powers the curvature, is simply the zeroth-order term. We know there is a first-order term proportional to the Ricci scalar. Wouldn’t it be strange if there were no zeroth-order term? (And higher-order terms?)

The second is vacuum energy due to quantum fluctuations. This is an inescapable part of quantum field theory. Why should it be exactly zero? In fact, the real question is why it isn’t huge. Maybe it’s zero because of supersymmetry. But supersymmetry cannot be an unbroken symmetry or we would see selectrons that are the same mass as electrons, etc. With broken supersymmetry, the exact cancellation of bosonic and fermionic vacuum energy is broken.

The third is a uniform and static-or-slowly evolving scalar field filling space. We know such scalar fields exist in nature, because the Higgs field is one. The Higgs field does not seem to be suitable for explaining cosmological expansion, but we should not rule out the possibility that other scalar fields exist that do have cosmological effects.

By contrast, I am not aware of arguments as natural as these for another value of $w$ that is less than -1/3.

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