Talk:Deceleration parameter

Hubble factor business

This business of the Hubble factor still decreasing needs some explanation. What I came to realize is that a, as the scale factor, determines whether there is an acceleration or not. Beynod that, the Hubble factor will decrease even with q=0 since as the universe gets bigger the galaxies moving away at a given rate get farther away. Hence the decrease. That the Hubble factor is still decreasing indicates that this is not a rapid expansion. In any case, some explanation of this is needed in the article. --EMS | Talk 04:04, 12 August 2006 (UTC)

Okay, I've rewritten q as:

$q={\frac {4\pi G(\rho +{\frac {3p}{c^{2}}})-\Lambda }{3H^{2}}}$

Which shows independence of the sign of q from the sign of H. --Michael C. Price ^talk 09:34, 12 August 2006 (UTC)