# How does cosmological principle imply an expanding space?

My understanding of cosmological principle is this:
Space looks same in all directions and is homogeneous. But this principle was under heading of Observational Evidence for expansion of space in this article:

My question is, how does Cosmological Principle imply an expanding space?
If it's my understanding of cosmological principle which is flawed, I would highly appreciate if someone corrects me. Thanks in advance!!

• The cosmological principle states that at any given time the Universe is isotropic and homogeneous - which allows for constant curvature. In general relativity, there are typically 3 choices for curvature, $k=1,0,-1$. As far as an expanding universe is concerned, it's by fiat. It's consistent with Big Bang Theory - which is consistent with the Old Testament of the Bible. The Hubble data, however, looks like random sequence - it's not clear that red shifts imply velocity - and quasars which were believed to be on the "edge" of the Universe are actually in our Local Super Cluster. – Cinaed Simson Apr 4 '19 at 7:28
• It does if combined with other hypothesis or observations. – Alchimista Apr 4 '19 at 8:51

The Cosmological Principle doesn't actually imply that the universe is expanding, instead, what this principle implies that the spatial part of the space-time manifold is maximally symmetric. What this actually means that the spatial manifold has a constant curvature everywhere given by Ricci Scalar, $$R=-6\kappa$$ using this principle along with defining the cosmological standard time $$t$$ as the time co-ordinate for co-moving cosmological fluids, we can construct metric given by:
$$d s^{2}=- d t^{2}+R^{2}(t)\left[\frac{d r^{2}}{1-\kappa r^{2}}+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)\right]$$
where $$R(t)$$ is the scaling factor. Now, whether the universe is contracting/expanding depends on this scaling factor. In order to determine $$R(t)$$, we have to solve Einstein's field equation which is given by the Friedmann Equation for this metric: $$\frac{\dot{R}}{R}=-\frac{4 \pi}{3} G (\rho+3p)+\frac{1}{3} \Lambda c^{2}$$ and $$\left(\frac{\dot{R}}{R}\right)^{2}-\frac{8}{3} \pi G \rho=-\frac{\kappa c^{2}}{R^{2}}$$ $$\rho$$ being energy density and $$p$$ being the pressure for cosmological fluids.