8
$\begingroup$

So the FLRW metric takes the following form in reduced-circumference polar coordinates.

$$\mathrm ds^2 = -c^2 \mathrm dt^2 + a^2(t) \left(\frac{\mathrm dr^2}{1 - k\, r^2} + r^2 (\mathrm d\theta^2+\sin^2\theta\,\mathrm d\phi^2)\right)$$

It's clear to me how this is derived from the restrictions of the cosmological principle applied to the most general possible metric but what is not clear to me is the reason behind assuming a constant curvature term $k$. It cannot be position dependent and be compatible with the cosmological principle but it seems like it should have the freedom to be time dependent much like the scale factor as long as it varies everywhere the same. Is there some coordinate redefinition possible such that the time dependence can be removed or am I missing something more?

$\endgroup$
0
5
$\begingroup$

Definition 1. A spacetime is said to be spatially homogeneous if there is a one-parameter family of spacelike hypersurfaces $\Sigma_t$ foliating the spacetime such that for each $t$ and for any points $p,q\in\Sigma_t$ there is an isometry of the spacetime metric $g$ which takes $p$ to $q$.

Definition 2. A spacetime is said to be isotropic if at each point there is a congruence of timelike curves, with tangents denoted $u$, satisfying: Given any point p and two unit spacelike vectors in $T_pM$, there is an isometry of $g$ which leaves $p$ and $u$ fixed but rotates one of these spacelike vectors into the other.

Restrict $g$ to a Riemannian metric $h$ on $\Sigma_t$. The geometry of each "leaf" of the foliation must inherit homogeneity and isotropy.

Let ${}^{(3)}\operatorname{Riem}$ be the Riemann tensor on $\Sigma_t$, $R_\Sigma$ be the scalar curvature and $T$ be the tensor field $$T(X,Y)Z=6\left[h(Z,Y)X-h(Z,X)Y\right]$$ for vector fields $X,Y,Z$.

Theorem. Homogeneity and isotropy of $\Sigma_t$ $\Leftrightarrow$ ${}^{(3)}\operatorname{Riem}=R_\Sigma T$, $R_\Sigma=\text{const.}$

Proof. Construct the Riemann tensor of $\Sigma_t$ using $h$. One may view this as an endomorphism $L$ of the space of $2$-forms $W$. By the symmetry properties of the Riemann tensor, $L$ is symmetric, and by a theorem in linear algebra, $W$ has an orthonormal basis of eigenvectors of $L$. If the eigenvalues were distinct, one could pick out a preferred $2$-form on $\Sigma_t$. Using the Hodge star on $\Sigma_t$, one could then construct a preferred vector. Since this would violate isotropy, the eigenvalues must be equal. We call this value $K$: $$L=K\operatorname{id}_W$$ In other words, $${}^{(3)}R_{ab}{}^{cd}=K\delta^c{}_{[a}\delta^d{}_{b]}$$ where ${}^{(3)}R_{ab}{}^{cd}$ are the components of ${}^{(3)}\operatorname{Riem}$. Contracting everything appropriately gives $$R_\Sigma=3K$$ Homogeneity automatically fixes $K$ to be a constant. $\quad\Box$

This proof very closely follows the one given in Wald, R. M. 1984, General Relativity (Chicago University Press).

$\endgroup$
0
0
$\begingroup$

The radius of curvature for the standard FLRW metric does vary with time.

The radius of curvature, RoC, equals a(t)/k^(1/2), which is time varying.

k equals 1/RoC^2 for a(t) equal to 1.

The 1/(1-k R^2) factor can be viewed as being derived from the derivative of the ARCSIN(a(t)R/RoC) function from the metric below. The R is the full scaled R, a(t)R. The k is the full reciprocal of square of RoC, the full value with a(t). The two a(t)'s cancel.

The equivalent formulation of the FLRW metric using angles w, u & v, using w instead of R, is:

ds^2 = -c^2 dt^2 + RoC(t)^2 (dw^2 + sin(w)^2 (du^2 + sin(u)^2 dv^2))

The conventional a(t)R equals RoC(t)sin(w) in this metric.

For a(t) equal to 1 in the current Universe and R and RoC equal to the inferred co-moving radius of the observable Universe, k would be very small.

$\endgroup$
0
-11
$\begingroup$

The FLRW metric starts with "the assumption of homogeneity and isotropy of space". But Einstein described a gravitational field as space that is "neither homogeneous nor isotropic" :

enter image description here

Hence setting expansion aside, for the universe as a whole there's no overall gravitational field, and light goes straight. Because of this we say the universe is flat, as per the WMAP findings.

There is no reason to assume a constant curvature term $k$. Assuming a constant curvature term is the wrong assumption. And let's face it, two out of three "shapes of the universe" were always going to be wrong:

enter image description here Public domain image courtesy of NASA

IMHO the universe is flat, it was flat a billion years ago, and a billion years before that. It's always been flat, and always will be.

$\endgroup$
0

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