I'm a grad student and I feel like this is an extremely elementary question that I should know, but I admit that I don't. Occasionally in science I find myself solving an equation for some answer where it feels like the units don't add up.

For instance, in a spatially flat matter-dominated universe with nonrelativistic matter, the accepted age of such a universe is:

$$t_0 = \frac{2}{3}H_0$$

where $H_0$ has units of ${\rm km} \cdot {\rm s}^{-1} \cdot {\rm Mpc}^{-1}$.

Why am I getting a time from this? I know it works, I know this is correct. This isn't the only example of this I've ever seen, but it is the freshest in my memory. It conceptually makes sense, but usually I'm in the habit of doing meticulous dimensional analysis and just wondered if there's some deep metaphysics or philosophy behind when that applies and when it doesn't.

  • $\begingroup$ Welcome to astronomy SE! I pretty-formated your formulas and units for you. $\endgroup$
    – B--rian
    Feb 12, 2021 at 21:47

1 Answer 1


The Hubble constant effectively has a unit of inverse time only: ${\rm s}^{-1}$:

It is usually given as ${\rm km}\cdot{\rm s}^{-1}{\rm Mpc}^{-1}$. Yet looking at it, km and Mpc are both units of length, so they cancel eachother out when converted to the same unit. Thus add the appropriate unit conversion from km to Mpc and it remains an inverse time only:

$$t = \frac{1}{H_0} = \frac{\rm Mpc}{60 {\rm km}\cdot{\rm s}^{-1}} = \frac{3\cdot 10^{22} {\rm m}{\rm s}}{60\cdot 1000 {\rm m}} = 5\cdot 10^{17}{\rm s} = 15{\rm Ga}$$

Add dimensionless factors to the equation to account for different cosmological models and inflation theories. This is anyway only a ballpark number - but an easy one to get for the age of the universe.

  • 2
    $\begingroup$ I fixed the LaTeX-typesetting for you, hope you don't mind. $\endgroup$
    – B--rian
    Feb 12, 2021 at 21:50
  • 1
    $\begingroup$ On the contrary, much appreciated $\endgroup$ Feb 13, 2021 at 7:57

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