# Cosmology: Horizon of the universe

1. Nov 29, 2009

### rabbit44

1. The problem statement, all variables and given/known data
This is what the question says exactly:

Assume the universe today is flat with both matter and a cosmological constant but no radiation. Compute the horizon of the Universe as a function of $$\Omega$$M and sketch it. (You will need a computer or calculator to do this).

2. Relevant equations
Friedmann Equations

3. The attempt at a solution
So I took the Friedmann Equation with k and the radiation density as 0 and solved it to find a(t). I got:

a = $$(\frac{\Omega_{\Lambda}}{\Omega_{M0}})^{-1/3}$$$$[sin(\frac{3H_{0}(\Omega_{M0})^{1/2}t}{2}$$]2/3

Latex takes ages so I don't really want to go through how I got there, but I'm pretty sure of it. Then I assumed the question is talking about the particle horizon rather than the event horizon. Either way I need to integrate 1 over this wrt t. Is this analytically possible, or is this the bit where I need a computer? The next part of the question asks me for the current horizon size if $$\Omega_{M0}$$ = 1/3 and h=1/$$\sqrt{2}$$, where I think h is something to do with H0. Just in case that next part is a clue to what I need to do.

Thanks for any help people!

2. Nov 30, 2009

### bombadil

Your on the right track. Though your scale factor seems to have one error in it: the $\Omega_M^{1/2}$ in the argument for sin should actually be $(\Omega_M-1)^{1/2}$. (check out equation 6.26 in "Introduction to Cosmology" by Ryden)

Yes, just integrate over 1/a times the speed of light to get the particle horizon. It does look like doing a numerical integral on a computer would be easiest.

On the second part of the problem, the parameter h is the hubble constant in units of 100 km/s/Mpc. So if h = 1/sqrt(2) then that means that H_0 ~ 70.7 km/s/Mpc.

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