# Explanation: $H^{-1}$ is the time-scale over which the universe changes by $\mathcal{O}(1)$ [duplicate]

The Hubble parameter $H$ has dimensions equal to $[T]^{-1}$, and hence there is a natural time-scale for the Universe $H^{-1}$. This lecture by Neal Weiner says (he wrote at around 4:40)

$H^{-1}$ is the time-scale over which the universe changes by $\mathcal{O}(1)$.

He also said that unlike cosmologists this is how particle physicists think about the time scale $H^{-1}$.

Can some explain what does he mean by the statement above?

• Apr 28, 2018 at 17:35
• Possible duplicate of Hubble time, the age of the Universe and expansion rate Apr 28, 2018 at 18:44
– SRS
May 1, 2018 at 18:48

By definition, $H = \dot a/a$. In terms of $t_H = H^{-1}$, this reads

$$a = \dot a\cdot t_H$$

So if you assumed a fixed expansion rate $\dot a = \text{const}$, the universe would have needed a time $t_H$ to grow to scale $a$.

I haven't wached the video, but here's my guess what the lecturer was getting at:

If you do a Taylor-expansion of the scale factor, you end up with $$\Delta a = \dot a(t_0)\cdot\Delta t + \mathcal O(\Delta t^2)$$ If you want that change to be "$\mathcal O(1)$", ie $\Delta a \approx a(t_0)$, you end up with $$\Delta t \approx \frac{a(t_0)}{\dot a(t_0)} = H(t_0)^{-1}$$ This of course assumes the validity of our first order approximation, and I also might be completely wrong about the intended meaning of "changes by $\mathcal O(1)$".

• I have edited the question to clarify what exactly I'm looking for. @Christoph
– SRS
May 4, 2018 at 10:28
• @SRS: see edit... May 4, 2018 at 16:43