Does the isotropy of the CMB imply a homogeneous universe? Some generalized EGS theorems

Abstract

We demonstrate that the high isotropy of the Cosmic Microwave Background (CMB), combined with the Copernican principle, is not sufficient to prove homogeneity of the universe -in contrast to previous results on this subject. The crucial additional factor not included in earlier work is the acceleration of the fundamental observers. We find the complete class of irrotational perfect fluid spacetimes admitting an exactly isotropic radiation field for every fundamental observer and show that they are Friedman-Lemaître-Robertson-Walker (FLRW) if and only if the acceleration is zero. While inhomogeneous in general, these spacetimes all possess three-dimensional symmetry groups, from which it follows that they also admit a thermodynamic interpretation.In addition to perfect fluid models we also consider multi-component fluids containing non-interacting radiation, dust and a quintessential scalar field or cosmological constant in which the radiation is isotropic for the geodesic (dust) observers. It is shown that the non-acceleration of the fundamental observers forces these spacetimes to be FLRW.While it is plausible that fundamental observers (galaxies) in the real universe follow geodesics, it is strictly necessary to determine this from local observations for the cosmological principle to be more than an assumption. We discuss how observations may be used to test this.

Metric-affine f(R) theories of gravity

“…This applies to a universe filled with any perfect fluid that is geodesic and barotropic (Clarkson and Barrett, 1999;Ellis et al, 1983a,b). Moreover, an "almost-EGS theorem" holds: spacetimes that are close to satisfying the EGS conditions are close to FLRW universes in an appropriate sense (Stoeger et al, 1995).…”

Cosmic microwave background and scalar-tensor theories of gravity

“…We note that the preceding calculation did lead to additional restrictions on φ, namely equations (18) and (20) must be satisfied and ∇ a F b = 0. However, if we take the field equations (2) and 'contract' them using u a u b and g ab , respectively, using the Brans-Dicke field equation (3) to substitute for ∇ a ∇ a φ, we obtain two expressions which can be written, using the equations above, as a sum of terms each of which has a particular (separable) dependence on time, space or a functional dependence on φ. Setting these two expressions equal to zero, it is then straightforward to show that these two equations can only be satisfied in general for φ = φ(t).…”

An Anisotropic Cosmological Model with Isotropic Background Radiation

“…This is important in modern cosmology because, in order to explain the observed acceleration detected with type Ia supernovae, scalar fields are often introduced to generate a quintessence component. It was recently shown that the conclusions of the EGS theorem remain valid in the presence of a quitessence field unless the gradient of the field is orthogonal to the dust congruence [4]. This alternative to the FRW geometry has been neglected so far because it is believed to lead to unphysical situations.…”

“…The EGS theorem ensures that a spacetime region satisfying Einstein equations and containing only dust and radiation has a Friedmann-Robertson-Walker (FRW) geometry provided that the dust velocity field u µ is (geodesic and) expanding, and that the distribution function of the photons is a solution to the Liouville equation which is isotropic with respect to u µ . In fact, the result that the geometry is FRW depends critically on the hypotheses of the theorem, and there exist counterexamples in which all but one of the assumptions are satisfied [2][3][4]. The theorem admits generalizations for matter consisting in a generic (barotropic) perfect fluid [2], and for an almost isotropic CMB [5], case in which the geometry is approximately of the FRW form.…”

“…We assume that the remaining matter consists of dust and radiation, which form a two-component perfect fluid with four-velocity u µ . The energy density and pressure are thus given by = A 2 /a 4 + D/a 3 and p = A 2 /(3a 4 ), where A and D are two positive constants. One then arrives at the following solution of (3):…”