Editing Hubble's law (section)

=== Hubble time ===
The Hubble constant {{math|''H''{{sub|0}}}} has units of inverse time; the '''Hubble time''' {{mvar|t{{sub|H}}}} is simply defined as the inverse of the Hubble constant,<ref>{{cite book |last1=Hawley |first1=John F. |title=Foundations of modern cosmology |last2=Holcomb |first2=Katherine A. |date=2005 |publisher=Oxford University Press |isbn=978-0-19-853096-1 |edition=2nd |location=Oxford [u.a.] |pages=304 |language=en-uk}}</ref> i.e.

<math display="block">t_H \equiv \frac{1}{H_0} = \frac{1}{67.8 \mathrm{~(km/s)/Mpc}} = 4.55\times 10^{17} \mathrm{~s} = 14.4 \text{ billion years}.</math>

This is slightly different from the [[age of the universe]], which is approximately 13.8&nbsp;billion years. The Hubble time is the age it would have had if the expansion had been linear,<ref>{{Cite book |last=Ridpath |first=Ian |title=A Dictionary of Astronomy |publisher=Oxford University Press |year=2012 |isbn=9780199609055 |edition=2nd |page=225 |language=en |doi=10.1093/acref/9780199609055.001.0001}}</ref> and it is different from the real age of the universe because the expansion is not linear; it depends on the energy content of the universe (see {{slink||Derivation of the Hubble parameter}}).

We currently appear to be approaching a period where the expansion of the universe is exponential due to the increasing dominance of [[vacuum energy]]. In this regime, the Hubble parameter is constant, and the universe grows by a factor [[E (mathematical constant)|{{mvar|e}}]] each Hubble time:

<math display="block">H \equiv \frac{\dot a}{a} = \textrm{constant} \quad \Longrightarrow \quad a \propto e^{Ht} = e^{\frac{t}{t_H}}</math>

Likewise, the generally accepted value of 2.27&nbsp;[[Exa-|Es]]<sup>−1</sup> means that (at the current rate) the universe would grow by a factor of {{mvar|e}}{{sup|2.27}} in one [[exasecond]].

Over long periods of time, the dynamics are complicated by general relativity, dark energy, [[Inflation (cosmology)|inflation]], etc., as explained above.