Expert: Philip A. Stahl - 1/11/2010

Question
 
I've read that the age of the universe can be calculated from first determining the rate of the its expansion and working backward.

My question is: working backward to what?  As an analogy, if we measured the speed of a marathon runner to determine T-zero when he started, wouldn't we also have to know the location of the starting line to make that calculation?
 
So even if we know the rate of expansion of the universe, what physical distance are we 're-winding' back to?  Is it the point where all the galaxies would crash in to each other (in reverse)?
 
Thanks,
John  

Answer
Hello,

Actually, it isn't an issue of 'working back' with respect to anything (in terms of the actual practical computation) but yes, most cosmologists do take the expansion to be from the putative Big Bang.

Anyway, the age of the universe can be computed once the Hubble constant is known. Then, the age of the universe (in seconds) is:

t = 1/ H_o

where H_o is the Hubble constant.

The beauty of this computation is that it renders redundant all distractive issues and questions such as "what distances are we rewinding back to", which truly aren't much use because in each new data generation cycle (almost!) the Hubble constant changes! Hence the distance and scaling must be changing!

Currently, we estimate H   ~ 70 km/ sec/Mpc*, where Mpc denotes ‘megaparsec’ – e.g. 3.26 light years is one parsec

(* See below)

The above is little use, however, without changing a lot of units and ensuring their consistency. I do the computations now simply to show you we don't need to address the questions you asked to find the age of the universe.

The key main initial step is to obtain the megaparsec equivalent for kilometers

MPC_km =  (c) (86400)(365.25)(3.26)x 10^6

where c = 300,000 km/sec, the velocity of light

Then: MPC_km = 3.08 x 10^19 km/Mpc

Now, what is usually colloquially called the "Hubble constant" is in reality the Hubble scale factor (a = 70 km/ sec/Mpc)

The REAL Hubble constant (H_o) is the scale factor divided by MPC_km:

H_o =  a/ MPC_km  =  2.26 x 10^-18 s^-1

then, t_o = 1/ H_o  =  1/ {2.26 x 10^-18 s^-1}  = 4.4 x 10^17 s



which when converted figures to be about 1.3 x 10^10 yrs. or 13 billion years in age

Again, the key point is obtaining the scale factor as the technical Hubble constant! One doesn't need to map a "distance back to" .....anything!

Hope this is of some assistance.