I'm trying to understand the empty universe model. My first textbook (which doesn't cover the empty universe) gives the Friedmann equation as

$$\left[\frac{1}{R}\frac{dR}{dt}\right]^{2}=\frac{8\pi G}{3}\rho-\frac{kc^{2}}{R^{2}}$$ Am I right in thinking all I need do is put $\rho=0$ so $$\frac{dR}{dt}=\pm c$$

for $k=-1$ and then say $$R=\pm ct$$ for an expanding or contracting empty universe?

Does that also mean an empty universe is expanding or contracting at the speed of light?

Many thanks (apologies if this is blindingly obvious)


1 Answer 1


The curvature radius is increasing at a rate of the speed of light. To see how fast a comoving observer is receding from you (assuming you are a comoving observer as well), you need the Hubble parameter and the observer's distance.

$$ v=H\cdot d $$

$$ H = \dot R/R $$

You can see that everything past the Hubble radius $d_H = c/H$ recedes faster than the speed of light, independent of the content of the Universe. Why don't you read up on it, there is an excellent paper by Davis and Lineweaver about common misconceptions.

  • $\begingroup$ Thanks for that. I looked at the paper - superluminal galaxies are not travelling faster than light in our or any other observer's inertial frame so don't contradict STR. That is amazing. $\endgroup$
    – Peter4075
    Apr 3, 2012 at 18:34

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