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For a universe with a constant expansion rate, i.e. the Hubble constant $H_0$ is a constant, the Hubble law is $$v=H_0d$$ where $v$ is relative velocity between two galaxies and $d$ is the separation between them.

I read that the age of the universe is given by $$\frac{d}{v}={1\over H_0}$$ which is the Hubble time $t_H$.

This calculation implies that at all times, the relative velocity between two galaxies is constant. But Hubble's law say that galaxies that are nearer to each other have a smaller relative velocity. So an earlier instant of time, when the two galaxies are nearer to each other, the relative velocity should be smaller.

How then can we use a constant relative velocity between two galaxies to calculate the age of the universe?

Qmechanic
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TaeNyFan
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1 Answers1

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Age of the universe is $$t = H_0^{-1}\int_0^{\infty}\frac{dz}{(1+z)E(z)}~~(1)$$

where $E(z) = \sqrt{\Omega_{\rm m,0}(1+z)^3 + \Omega_{\rm r,0} (1+z)^4 + \Omega_{\rm \Lambda,0}+ \Omega_{\rm k,0}(1+z)^2}$

For instance if you assume a flat-matter dominated universe ($\Omega_{\rm m,0}, \Omega_{\rm \Lambda,0}, \Omega_{\rm k,0}$), the age of the universe becomes $$t = \frac{2}{3H_0}$$

or for radiation dominated universe

$$t = \frac{1}{2H_0}$$

so as you can see from (1) and for two cases above, the age of the universe is inversely proportional to $H_0$ (i.e., $t \propto H_0^{-1}$)

camarman
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