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)[reply]

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)[reply]

All that you have done is to confuse things even more. For a reader,

What is the Hubble parameter?
How can it be decreasing when the universe's expansion is accelerating?
What is the significance of its decreasing while the expansion is accelerating?

These issues need to be answered with a well thought-out write-up, not by throwing another equation at the reader. I took that business out of the general relativity article because it was more confusing than informative given the issues that it raised. I can accept it here, but only if the issues are dealt with. --EMS | Talk 18:12, 12 August 2006 (UTC)[reply]

Your first point completely baffles me. Have you looked the article itself, where the Hubble parameter is defined and wikilinked? What more of an explanation are you after? --Michael C. Price ^talk 18:21, 12 August 2006 (UTC)[reply]

Regarding the second point, that can only be explained by presenting the equations, which show that the sign of q and the sign of ${\dot {H}}$ are independent.

Regarding the third point, I have no idea of its significance.

If you find my explanation confused them please expand it. But statements in the literature that the cosmic expansion is accelerating do beg the question of how this can be compatible with H still decreasing. This is what I have tried to address in this article.--Michael C. Price ^talk 18:45, 12 August 2006 (UTC)[reply]