Critical density and total observable mass

[jimjohnson]

Assuming a Hubble constant of 74.3 km/sec/Mpc, the critical density is about E-29 gm/cm3. To calculate observable matter based on 5% of this density (the other 95% is dark matter and dark energy), a volume has to be used. But which volume, the one with a 13.7 billion light year radius or the approximate 45 billion light year radius of the expanded universe?
The two results vary by 39 times, 4.6 x E54 gm or 1.8 x E56 gm.

 

[jimjohnson]

After pondering this, I think the answer should be to use the volume based on the 13.7 billion year radius. Since matter density dilutes with expansion, it is more logical to assume the amount of mass is based on the 13.7 billion year radius.Then, matter density would be 39 times less at the larger radius. Sound correct?

 

[ibysaiyan]

After pondering this, I think the answer should be to use the volume based on the 13.7 billion year radius. Since matter density dilutes with expansion, it is more logical to assume the amount of mass is based on the 13.7 billion year radius.Then, matter density would be 39 times less at the larger radius. Sound correct?

Here's a guess.. since Hubble's constant can be defined : (dR/dt) /R_{0} so it'd make sense to take into account the volume of the universe corresponding to hubble's time ( although Hubble's time is a misleading way of finding the age of the universe) . Just a thought . I don't see anything wrong with your logic . I maybe wrong wait for other ( more experienced users) to post.

 

[jimjohnson]

I have two conflicting sources.
The first says use the larger volume: http://en.wikipedia.org/wiki/Observable_universe
The second says use the "present event horizon" which is the smaller volume: NASA, Ask the Astrophysicst, Feb 11 ,1998 by Jim Lochner
Other input? Thanks.