# What basic cosmo stuff can you calculate from two numbers and two equations?

1. Nov 14, 2008

### marcus

Standard cosmology is a numerical science (math models instead of verbal concepts and analogies). So a way to get understanding and test your grasp is to try and see what you can calculate on your own.

We can get a lot of basic cosmo stuff from just two numbers (71 and 0.73) and two short equations. The simple beginnings of this were introduced in post#39 of the basic cosmo Sticky thread:

In that post we saw how easily to calculate that distances are increasing at a rate which is 7.3 percent per billion years.
Also that that presentday total energy density is about 0.85 nJ per cubic meter and the pressure (due to dark energy) is - 0.62 in the same units. I am going to try sometimes using purple-colored words to name physical quantities, instead of conventional symbols such as Greek letters, which may put some folks off.

To proceed then a bit further along the same lines, let's bring in the two Friedmann equations.
In barebones form (space being nearly flat, we restrict to the case of zero spatial curvature) the two Friedmanns are:

First Friedmann
(a'(t)/a(t))^2 = (8 pi G/(3c^2)) density
Second Friedmann
a"(t)/a(t) = - (4 pi G/(3c^2)) ( density + 3 pressure)

Note that a(t) is usually normalized to equal one at t=now, the present era. So when calculating presentday numbers we can ignore the a(t) in the denominators. Recalling that in the Sticky an easy way was shown to calculate that presentday density= 0.85 nJ per m3 and presentday pressure= - 0.62 nJ per m3 (or equivalently, if you like, 0.62 nanopascal), the reason why the pressure's quoted here in nJ per m3 is that to add density-and-pressure we need them both expressed in the same units.

Obviously density + 3 pressure = 0.85 - 1.86 = -1.01 nJ per m^3
So Second Friedmann says that for t=now:
a"(t)/a(t) = (4 pi G/(3c^2)) ( 1.01 nJ per m^3)

That is something we can type into the Google calculator pretty much verbatim,
and it will give us the current acceleration of expansion. The only downside is it gives it in terms of per second per second. And then the rate is a tiny tiny number. To get it on a per billion years per billion years basis, so it is more reasonable sized, multiply the above thing by (10^9 year)^2, put this verbatim into Google, and press return.
(10^9 year)^2 (4 pi G/(3c^2)) ( 1.01 nJ per m^3)

When I put the blue thing into Google searchbox, what it calculates is 0.00313

That means 0.00313 (or if you like, 0.313 percentage points) per billion years per billion years.
To illustrate. the present rate that distances are increasing is 7.3 percent per billion years (see calculation in Sticky thread) and we now know that percentage rate of increase is itself growing by 0.313 per billion years. Heading toward 7.613 in other words. The rough linear approximation is meant to give a feel for what the second time derivative a"(t) of the scale factor signifies and how it affects things.

Last edited: Nov 14, 2008
2. Nov 14, 2008

### marcus

Simply doing some calculation with the standard cosmo model can help get one comfortable and familiar with it. The main concern here is that we share an understanding of the mainstream consensus picture as a common ground, however one choses to deviate in various ways.

For example we know the Hubble parameter H(t) is changing with time. Would you say that at the present moment it is increasing or decreasing? According to the standard cosmo model, it is decreasing and will continue to decrease forever, but more and more slowly.

Part of having a mathematical (rather than verbal) understanding of basic cosmo is understanding why, since H(t) tells us the rate distances are expanding, and since we believe the expansion is accelerating, that H(t) should be decreasing.

The explanation is that H(t) = a'(t)/a(t), the time derivative of the scalefactor divided by the scalefactor itself. All acceleration means is that a'(t) is increasing, it doesn't mean that H is increasing.

In fact H(t) is decreasing and is expected to decrease forever, though it levels out asymptotically and approaches a limit. Can you calculate this limit? What is the final asymptotic value of H(t), that it will be nearing a few billion years from now? In other words, what is the value of Hubble parameter in the late universe?

That is a simple application of First Friedmann. As distances and volume increase, the density of matter goes to zero and the total density, now 0.85 nJ per m^3, will approach the constant density of dark energy, namely 0.62 nJ per m^3. That is less by a factor of 0.73---this was how we calculated it in the Sticky thread.