The density parameter

1. Jun 4, 2012

robertjford80

This is from Matts Roos' text on cosmology. I don't follow the final step and when he integrates I understand the first part of the left side, and the right side. But I don't see where -a'^2(t0) comes from

2. Jun 4, 2012

twofish-quant

It's a constant of integration. Since you can add an arbitrary constant and have the differential equation still work, instead of writing C, he writes the constant so that the numbers end up nice later.

3. Jun 4, 2012

phyzguy

Or you can view it as a definite integral with limits:
$$\int^t_{t_0}d(\dot a^2) = -{H_0}^2 \Omega_0\int^t_{t_0}\frac{da}{a^2}$$

$$\dot a^2(t) - \dot a^2(t_0)={H_0}^2 \Omega_0(\frac{1}{a(t)}-\frac{1}{a(t_0)})$$
$$\dot a^2(t) - {H_0}^2={H_0}^2 \Omega_0(\frac{1}{a(t)}-1)$$
$$\dot a^2(t) ={H_0}^2( \frac{\Omega_0}{a(t)}-\Omega_0+1)$$

Note that we define the current scale parameter a(t0) to be 1, and since the Hubble constant H is defined to be adot/a, that H(t0) = adot(t0)/a(t0) = adot(t0), since a(t0)=1.

Last edited: Jun 4, 2012
4. Jun 4, 2012

robertjford80

i still don't understand the progression from 1.35 to 1.36

5. Jun 4, 2012

phyzguy

I edited my earlier post to add more detail. Does this clear it up?

6. Jun 4, 2012

robertjford80

i'm falling asleep so i'll look at it tomorrow

7. Jun 4, 2012

cepheid

Staff Emeritus
The progression from 1.35 to 1.36 is just algebra. Notice that if you take the leftmost side of 1.35, and the rightmost side of it, and you multiply them both by H02, then you end up with$$\Omega_0 H_0^2 = \frac{8\pi G}{3}\rho_0$$So, anywhere that you see $8\pi G \rho_0 / 3$ in 1.34, just replace it with $\Omega_0 H_0^2$.

8. Jun 4, 2012

robertjford80

If omega H20 = 8piGrho/3 then I don't see where the final +1 comes from in equation 1.36

9. Jun 4, 2012

cepheid

Staff Emeritus
It comes from the very constant of integration that you were asking about before!

10. Jun 4, 2012

ok, thanks