# How does calculating Hubble time work?

For calculating we take $$t=\frac dV$$, where $$V=Hd$$ . But velocity isn't uniform for an object as distance also increases with time. And if velocity isn't constant the equation $$t=\frac H d$$ also isn't applicable. So how is the Hubble time the age of the universe intuitively?

You're right, $$V$$ is not necessarily constant, and there are so many different ways to model the universe.

The fact that $$H$$ is related to the "Hubble time" comes from when we look at these models, we find that estimates of the age of the universe are usually (for simple models at least)

$$t_0\propto\frac{1}{H_0}$$

where $$t_0$$ is the amount of cosmic time that has elapsed between the beginning of the universe and now, and $$H_0$$ is the Hubble parameter evaluated today. Note that in general models, $$H$$ itself is not constant, but a function of time, but the relationship

$$V=Hd$$

is still true (just now it's not a simple linear relationship, since $$H$$ is not constant).

In a very simple universe which is flat and has no cosmological constant, (called an Einstein-de-Sitter universe), one can show that the age of the universe is

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

Note also that the value of the Hubble parameter is $$H_0\sim70\text{ km s}^{-1}\text{ Mpc}^{-1}$$ or $$\sim2\times10^{-18}\text{ s}^{-1}$$ in sensible units - i.e. inverse time. So it makes sense that $$1/H_0$$ is a time-scale for the universe, but it just so happens that using $$t_0=H_0^{-1}$$ is just a simple estimation.

To properly answer this question, you need to look at the details of cosmology and the cosmological field equations (Friedmann equations).

• Great answer. I would also add that in some cosmological models the Hubble time is the age of the universe, e.g. in the Milne model. Commented Jan 15, 2019 at 19:58

The unit of hubble parameter is $$1/s$$ so simply we can define Hubble time as $$\frac {1} {H}$$, which the unit becomes second. You dont have to think about the distances or etc. But if you want to you can write

$$t=\frac {r} {V}=\frac {r} {Hr}=\frac{1} {H}$$ In the derivation distances cancel out.

For current hubble parameter, $$H_0=70kms\pm 7^{-1}Mpc^{-1}$$ Hubble time is, $$H_0^{-1}=14.0\pm 1.4Gyr$$