Deceleration parameter

Remarks

With the Hubble parameter $H = \dot a / a$, one has $q = -1 - \dot H / H^2$. Example: for $a(t) = t^{2/3}$, $\dot a = \tfrac{2}{3} t^{-1/3}$ and $\ddot a = -\tfrac{2}{9} t^{-4/3}$, so $q = \tfrac{2}{9} / \tfrac{4}{9} = \tfrac{1}{2}$. For $a(t) = e^{Ht}$ with $H$ constant, $q = -1$. Non-example: for a static model, $\dot a = 0$, so $q$ is not defined. The sign of $q$ is opposite to the sign of $\ddot a$: $q < 0$ exactly when the expansion accelerates. For a mixture of perfect fluids with constant equation of state parameters $w_i$ and density parameters $\Omega_i$, the second Friedmann equation gives $q = \tfrac{1}{2} \sum_i \Omega_i (1 + 3 w_i)$.