# Acceleration universe

1. Jun 28, 2014

### Ledsnyder

Using the friedmann equations, is it possible to find the acceleration of the universe at a given time?

2. Jun 28, 2014

### Staff: Mentor

Hi, Ledsnyder, and welcome to PF!

If you define "the acceleration of the universe" as the second time derivative of the scale factor, then yes, the Friedmann equations (specifically the second one) can be used to solve for it. However, in order to find a solution, you have to know the density $\rho$ and the pressure $p$ as a function of time. If there is a cosmological constant (as we think there is in our actual universe), you have to know that as well.

3. Jun 28, 2014

### marcus

Not if you think of acceleration as something measured in Length/Time2

The scale factor is a dimensionless quantity a(t) normalized so that a(present) = 1.

It has no Length scale.

So the first derivative of the scale factor is a "PER UNIT TIME" quantity. It is not a speed. It is not a Length/time quantity. And the second derivative of the scale factor is not an acceleration (not a change in speed).
It is a "PER UNIT TIME PER UNIT TIME" quantity.

the "speed of expansion of the universe" at a given time is not defined,
and the "acceleration of the universe" at a given time is not defined.

Ben Crowell has a cosmology FAQ about this. Shouldn't we promote his FAQ items more?, some are pretty good. :^)

4. Jun 29, 2014

### Staff: Mentor

Doesn't this depend on which form of the FRW metric you adopt? I believe there is at least one form where all three spatial coordinates are "angular", so to speak, i.e., they don't have any "length scale" by themselves. In this form, the length scale is in the scale factor.

I agree, though, that since even if the above is true, it's coordinate-dependent, the scale factor is best not thought of as a "distance", so its second time derivative is not an "acceleration" in the ordinary sense of the term. I should have made that clearer in my post.

5. Jun 29, 2014

### Jorrie

The solution I use is from the handbook 'Principles of Physical Cosmology', PJE Peebles, 1993 (Princeton Press), Eq 5.54, pp. 100:

$$\ddot{a} = a H_0^2(\Omega_{\Lambda}-\Omega_m/(2a^3))$$

To find it at a given time, you need to find $a$ for your desired time, which is not so easy when a cosmological constant is present. With nifty use of cosmological calculators (like Lightcone 7 in my sig), one can however find it quite accurately.

Give it a try...

PS: if you want to work with a < ~1/3000 (which is t < ~65,000 yrs), you should include a radiation energy density term, because radiation was then dominant.

$$\ddot{a} = a H_0^2(\Omega_{\Lambda}-\Omega_m/(2a^3) -\Omega_r/a^4)$$

The 2013 density paramaters parameters are (in the order they appear in the brackets): 0.7, 0.3, 0.3/3400

Last edited: Jun 29, 2014
6. Jul 1, 2014

### Ledsnyder

T (Gy) R (Gly)
0.00037338 0.00062840
0.00249614 0.00395626
0.01530893 0.02347787
0.09015807 0.13632116
0.52234170 0.78510382
2.97769059 4.37361531
13.78720586 14.39993199
32.88494318 17.18490043
47.72506282 17.29112724
62.59805320 17.29930703
77.47372152 17.29980205
92.34940681 17.29990021

is there an equation that relates T and R?

7. Jul 1, 2014

### Jorrie

I am not aware of a direct equation, so I use numerical integration to independently find T and R (the latter being the Hubble radius in 'Lightcone 7'). It is then possible to read off (or plot) R against T. The equations used in Lightcone 7 are here: http://cosmocalc.wikidot.com/advanced-user.

The same goes for any of the D's, e.g. if you are interested in the radius of the observable universe over time, you can make a table including T and and the particle horizon (Dparticle in Lightcone 7).