Understanding the Density Parameter in Cosmology: Matts Roos' Text Explanation

In summary: Making use of the relation (1.19), ##\dot{a}(t_0) = H_0##, we find that the final +1 comes from the very constant of integration that you were asking about before!
  • #1
robertjford80
388
0
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

Screenshot2012-06-04at45600AM.png
 
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  • #2
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
Or you can view it as a definite integral with limits:
[tex]\int^t_{t_0}d(\dot a^2) = -{H_0}^2 \Omega_0\int^t_{t_0}\frac{da}{a^2}[/tex]

[tex]\dot a^2(t) - \dot a^2(t_0)={H_0}^2 \Omega_0(\frac{1}{a(t)}-\frac{1}{a(t_0)})[/tex]
[tex]\dot a^2(t) - {H_0}^2={H_0}^2 \Omega_0(\frac{1}{a(t)}-1)[/tex]
[tex]\dot a^2(t) ={H_0}^2( \frac{\Omega_0}{a(t)}-\Omega_0+1)[/tex]

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.
 
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  • #4
i still don't understand the progression from 1.35 to 1.36
 
  • #5
robertjford80 said:
i still don't understand the progression from 1.35 to 1.36

I edited my earlier post to add more detail. Does this clear it up?
 
  • #6
i'm falling asleep so i'll look at it tomorrow
 
  • #7
robertjford80 said:
i still don't understand the progression from 1.35 to 1.36

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
cepheid said:
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##.

If omega H20 = 8piGrho/3 then I don't see where the final +1 comes from in equation 1.36
 
  • #9
robertjford80 said:
If omega H20 = 8piGrho/3 then I don't see where the final +1 comes from in equation 1.36

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

robertjford80's cosmology textbook said:
making use of the relation (1.19), ##\dot{a}(t_0) = H_0##, we find
 
  • #10
ok, thanks
 

Related to Understanding the Density Parameter in Cosmology: Matts Roos' Text Explanation

1. What is the density parameter in cosmology?

The density parameter is a measure of the overall density of the universe, and is denoted by the symbol Ω. It compares the actual density of matter in the universe to the critical density, which is the density required for the universe to be flat.

2. How is the density parameter calculated?

The density parameter is calculated by dividing the actual density of matter in the universe by the critical density. The actual density is determined by measuring the amount of matter in the universe, including both visible matter (such as stars and galaxies) and dark matter. The critical density is calculated using the Hubble constant and the speed of light.

3. What does the density parameter tell us about the universe?

The density parameter is an important factor in understanding the overall composition and evolution of the universe. If Ω is less than 1, it indicates that the universe is open and will continue to expand forever. If Ω is equal to 1, the universe is flat and will continue to expand at a decreasing rate. If Ω is greater than 1, the universe is closed and will eventually collapse in a "Big Crunch."

4. Why is the density parameter important in cosmology?

The density parameter is important because it helps us understand the overall structure and fate of the universe. It also plays a crucial role in theories of dark matter and dark energy, as well as the study of the expansion rate of the universe.

5. How does the density parameter relate to other cosmological parameters?

The density parameter is closely related to other important cosmological parameters, such as the Hubble constant, the age of the universe, and the amount of dark matter and dark energy. It is also used in conjunction with other parameters to create models and theories about the origin and evolution of the universe.

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