Wednesday, 27 September 2017

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



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?



Answer



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 $\mathcal O(1)$".


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