Robertson-Walker metric and cosmic homogeneity

Question

The Robertson-Walker metric is of the form

$$\tag{1} ds^2 = dt^2 - a(t)^2 \Big(\frac{dr^2}{1 - kr^2} + r^2 d\theta^2 + r^2 \sin^2\theta \, d\phi^2 \Big).$$

My question is related to the $a^2(t)$ term. Since the cosmos is found to be homogeneous, one argues that it cannot be a function of $r$. It will only be a function of $t$. Thus, its of the form $a^2(t)$.

However, what prevents it from being something like

$$\tag{2} a(t)^2 + b(r) \, e^{-lt},$$

where $l$ is some constant. More specifically,

$$\tag{3} ds^2 = dt^2 - \big(\, a(t)^2 + b(r)e^{-lt} \big) \Big(\frac{dr^2}{1 - kr^2} + r^2 d\theta^2 + r^2 \sin^2 \theta \, d\phi^2 \Big).$$

After some 14 billion years, the $b(r) \, e^{-lt}$ term can become vanishingly small, and become negligible. But it may not have been negligible once upon a time. The universe may not have been homogeneous once upon a time. What is the evidence we have to say that the universe was homogeneous even in the early begining?

Answer 1

I am not sure but here some of my thoughts.

The early universe also must be homogeneous and isotropic. We can see that from the CMBR. Hence r dependence on the new scale factor $S(r,t)$, might affect the homogeneity in the early universe. If we assume universe must hold the cosmological principle, we are allowed to choose 3 different spatial geometries. r dependence on the metric will change the topology hence the symmetry in large that we observe now.

Edit: Also, when you add a new scale that depends on r, then

$$S(r,t)=a^2(t)+b(r)e^{-lt}$$ then for $$t\rightarrow -\infty $$ $$S(r,t)\rightarrow \infty $$

Which means that when we go back in time expansion becomes faster.

Further Note: Let us take the simplest case where $\kappa=0$ then the matric becomes (for $S(t,r)$)

$$ds^2=-c^2dt^2+[(a^2(t)+b(r)e^{-lt})(dr^2+r^2d\Omega^2)]$$ Or we can write it as, $$ds^2=-c^2dt^2+[(a^2(t)(dr^2+r^2d^2\Omega)]+[b(r)e^{-lt}(dr^2+r^2d\Omega^2)]$$

But we can just try to look at the spetial metric for r dependence to understand the geometry.

$$ds^2=[b^2(r)e^{-lt}(dr^2+r^2d^2\Omega)]$$

Lets take $d^2\Omega=0$ for simplicity then we have

$$ds^2=b(r)e^{-lt}dr^2$$ then lets say $b(r)=r^{2n}$ $$ds^2=r^ne^{-lt}dr^2$$ this is actually something like, $$ds^2=q(t)r^ndr^2$$ In normal FLRW metric at this point we would have, $$ds^2=a^2(t)dr^2$$

So the distance from the object does not just depend on time but also depends on some power of the radial distance ?

$$ds=e^{-lt/2}\int r^{n}dr$$ $$s=(r^{n+1}/n+1) e^{-lt/2}$$

Which I think its not one of the Spatial metric that gives the cosmological principle ? I am not really expert but maybe someone can help to clarify and expand the idea.