JesseM

Science Advisor

- 8,492

- 12

**Re: Redshifting**

Ah, so "a" stands for the "scale factor" and not the acceleration--does that mean the vertical axis of the graph is "a", and that the scale factor at a given time is proportional to the distance of some randomly-selected test particle at that time? But if so, why would they use the term "accelerating" to mean that the slope (proportional to theIt is really simple. If you look at the graph, ordinate represents average distance between galaxies, or scalefactor - a. Wherever slope of the curve is more then 45 deg relative to abcissa, universe is accelerating.

*velocity*of a test particle if the above is correct) is greater than 45 degrees? This section of a wikipedia article seems to say that the terminology of the expansion "accelerating/decelerating" is based on the sign of the deceleration parameter which can be defined either in terms of the Hubble constant and its first derivative or the scale factor "a" and its first and second derivatives...are you saying there's some mathematical argument that shows that a positive deceleration parameter (decelerating expansion) implies a slope closer to horizontal than 45 degrees, while a negative deceleration parameter (accelerating expansion) implies a slope closer to vertical than 45 degrees? i.e. that for any FLRW solution, if [tex]\dot{a}[/tex] is less than 1 then [tex]-\frac{\ddot{a} a }{\dot{a}^2}[/tex] will always be positive, while if [tex]\dot{a}[/tex] is greater than 1 then [tex]-\frac{\ddot{a} a }{\dot{a}^2}[/tex] will always be negative?

**edit:**never mind, I see you corrected yourself. I guess the "deceleration parameter" has no simple intuitive physical meaning in terms of the motion of a test particle though, it's just defined that way because it simplifies various cosmological equations?