Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

If dark energy contributes mass-energy density $\rho$ and pressure $p$ to the stress-energy tensor, then you can define $w=p/\rho$, where $w=-1$ gives a cosmological constant, $w<-1$ gives a big rip, and $w<-1/3$ if we want to use dark energy to explain cosmological acceleration. The WP "Big Rip" article cites a paper that dates back to 2003 http://arxiv.org/abs/astro-ph/0302506 , which states that the empirical evidence was at that time only good enough to give $-2 \lesssim w \lesssim -.5$.

Have observations constrained $w$ any more tightly since 2003?

I've heard many people express the opinion that $w<-1$ is silly or poorly motivated, or that "no one believes it." What are the reasons for this? Other than mathematical simplicity, I don't know of any reason to prefer $w=-1$ over $w\ne -1$. Considering that attempts to calculate $\Lambda$ using QFT are off by 120 orders of magnitude, is it reasonable to depend on theory to give any input into what values of $w$ are believable?

share|improve this question
add comment

1 Answer 1

I do not know if this answer will address fully your Question, anyway:

Clustering of Photometric Luminous Red Galaxies II: Cosmological Implications from the Baryon Acoustic Scale (2011/Apr)

Combining with previous measurements of the acoustic scale, we obtain a value of $w_0$ = -1.03 +/- 0.16 for the equation of state parameter of the dark energy

cited here : Cosmology today–A brief review (25 pages on Theory and data, 2007/Jul)

You can find a new model of the Universe without Dark Energy here:
A self-similar model of the Universe unveils the nature of dark energy (21 pages Jul/2011, not peer-reviewed, it uses only Newton and Coulomb laws). I do not know how this novel viewpoint can be discarded.
Argument:
From $$F=m\cdot a,\, F=G\cdot\frac{m_{1}\cdot m_{2}}{d^{2}},\, F=\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{q_{1}\cdot q_{2}}{d^{2}},\, c=\frac{1}{\sqrt{\varepsilon_{0}\mu_{0}}}. $$ obtain the dimensional equations $$\left[G\right]=M^{-1}L^{3}T^{-2},\left[\varepsilon\right]=M^{-1}Q^{2}L^{-3}T^{2},\left[c\right]=LT^{-1}$$ because the the sum of the exponents is zero, a cute coincidence ;), we know that $[M],[Q],[L],[T]$ can scale in the same way and $[c],[G],[\varepsilon]$ are constants. If the universe can scale thru time, keeping always the same basic physical laws, it will scale.
The scaling law $\alpha(t_{S})=e^{-H_{0}\cdot t_{S}}$ is derived from the observational data, where $t_S$ is considered from the viewpoint of a comoving invariant referential, and Dark Energy is absent from it.

The accelerated expansion is an artifact of the standard model
The statement that space expansion is accelerating is not the result of some direct measurement more or less independent of the cosmological model but, on the contrary, it is a consequence of the theoretical framework of the standard model. The deceleration parameter at the present moment, $q_{0}$, in the $\Lambda$CDM model, for flat space and $\Omega_{R}=0$ , is given by $q_{0}=\frac{1}{2}\left(\Omega_{M}-2\Omega_{\Lambda}\right)$ therefore, for $\Omega_{M}+\Omega_{\Lambda}=1$ , the value of $q_{0}$ is negative for $\Omega_{\Lambda}>1/3$ ; a value of $\Omega_{\Lambda}$ lower than 1/3 leads to a comoving distance largely in disagreement with observations, hence, in the framework of $\Lambda$CDM model it has to be $\Omega_{\Lambda}>1/3$ and, so, $q_{0}<0$ .

(Needless to say: this is my preferred viewpoint because it has no free lunches: growing space, growing dark energy, and apply to all scales, even to Solar system)

share|improve this answer
2  
Thanks for the reference to the Carnero paper -- that's exactly what I was looking for! If your answer had stopped there, I would have accepted and upvoted. But the Oliveira paper is pure crackpottery. –  Ben Crowell Aug 8 '11 at 16:09
    
I would still be interested in posts as to why Big Rip scenarios seem not to be taken seriously by so many people. –  Ben Crowell Aug 8 '11 at 16:29
    
@Ben Crowell I used a few lines to present the 'argument' that a scaling model is inscribed in the gravitation and electrostatic laws. Somewhere you argued that $M_e/M_p$ is constant. D'accord it is as constant as the ratio of the mass of my hand versus the one of my body. But my hand is not invariant. There is no evidence that the mass of the proton is an 'absolut' invariant (irt itself). From you I expect more that nasty word : Be the first to present evidence of the invariance or to advance a physical counter-argument to the scaling model. To you it should be a piece of cake. –  Helder Velez Aug 8 '11 at 18:05
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.