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?

  • $\begingroup$ I liked the Q. However from an observational stand point we can at least go back to recombination epoch / cosmic background radiation. $\endgroup$
    – Alchimista
    Jan 3, 2019 at 8:41
  • $\begingroup$ The CMB was emitted 380 000 years only after the Big Bang, and is extremely homogeneous (one part in 100 000). This is a good argument for homogeneity in the early times of the universe. $\endgroup$
    – Cham
    Jan 3, 2019 at 14:56

1 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.


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.

  • $\begingroup$ What kind of symmetries would change due to this? Also, all of what we observe is after the surface of last scattering. Is there any reason why b(r) needs to be taken 0 before the recombination epoch? e^{-lt} can make this term 0, by the time recombination point is reached. $\endgroup$
    – Angela
    Jan 3, 2019 at 14:46
  • $\begingroup$ I edited my post. As I said, my ideas are just thoughts. I am not an expert on the field. I think the problem is; Is the additional term affects the spatial geometry such a way that cosmological principle is no longer valid ? $\endgroup$
    – seVenVo1d
    Jan 3, 2019 at 19:08
  • $\begingroup$ Now I am sure that its not homogeneous, hence its not valid for Cosmological principle and so we cannot use it to describe the universe (ıf we agree on that universe obeys Cosmological principle) $\endgroup$
    – seVenVo1d
    Jan 4, 2019 at 20:21
  • $\begingroup$ Just a question: Doesn't the 1/(1-kr^2) term make the universe non-homogenous? $\endgroup$
    – Angela
    Jan 6, 2019 at 7:06
  • $\begingroup$ You mean by alone ? or with the b(r) term ? . $\endgroup$
    – seVenVo1d
    Jan 6, 2019 at 8:01

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