Anisotropic perturbations due to dark energy

Abstract

A variety of observational tests seem to suggest that the universe is anisotropic. This is incompatible with the standard dogma based on adiabatic, rotationally invariant perturbations. We point out that this is a consequence of the standard decomposition of the stress-energy tensor for the cosmological fluids, and that rotational invariance need not be assumed, if there is elastic rigidity in the dark energy. The dark energy required to achieve this might be provided by point symmetric domain wall network with $P/ρ=-2/3$, although the concept is more general. We illustrate this with reference to a model with cubic symmetry and discuss various aspects of the model.

Figures (3)

• Figure 1: Perturbation evolution for $k=10^{-3}{\rm Mpc}^{-1}$ for a component with $w=-2/3$, $\hat{\mu}_{L}=0.18$ and $\Delta\hat{\mu}=0.01$ in the direction $\theta=\pi/2$ and $\phi=\pi/8$. On the left is the scalar velocity of the dark energy component $|V_{\rm DE}^{\rm S}|$ (solid line) and the vector velocity component $(|V^{V1}_{\rm DE}|^2+|V^{V2}_{\rm DE}|^2)^{1/2}$ and on the right are the vector (dotted line) and tensor metric components. Note that the vector and tensor perturbation are non-zero even though the initial conditions were pure scalar.
• Figure 2: $P_{\rm T}$ (left) and $|V^{V1}_{\rm DE}|^2+|V^{V2}_{\rm DE}|^2$ (right) at $k=10^{-3}{\rm Mpc}^{-1}$ as function with $\theta$ and $\phi$ plotted in the Hammer-Aitoff projection when $w=-2/3$, $\hat{\mu}_{L}=0.18$ and $\Delta \hat{\mu}=0.01$. Note the anisotropy of the $P_{\rm T}$ and that $(|V_{\rm DE}^{\rm V1}|^2+|V_{\rm DE}^{\rm V2}|^2)^{1/2}$ is non-zero. Both have obvious cubic symmetry.
• Figure 3: The normalized variance of the power spectrum $K$ (left) and the ratio of the vector and scalar velocities $R$ (right) for $w=-2/3$ and $\hat{\mu}_{L}=0.18$. Curves show varying levels of the anisotropy $\Delta \hat{\mu}=10^{-1} {\rm (solid)}$, $10^{-2} {\rm(dotted)}$, $10^{-3} {\rm(short-dash)}$, $10^{-4} {\rm(long-dash)}$.