Cosmology: Horizon of the universe

rabbit44
Messages
28
Reaction score
0

Homework Statement


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 \OmegaM and sketch it. (You will need a computer or calculator to do this).

Homework Equations


Friedmann Equations

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!
 
Physics news on Phys.org
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.
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Back
Top