Cosmology: Horizon of the universe

In summary, the conversation discusses computing the horizon of the universe as a function of the cosmological parameters \Omega_M and \Omega_\Lambda, and sketching it. The Friedmann Equations are used to solve for the scale factor a(t), and the particle horizon is found by integrating 1/a times the speed of light. This integration may be done numerically on a computer. The second part of the problem involves using the Hubble constant, denoted by h, to calculate the current horizon size if specific values are given for \Omega_{M0} and h.
  • #1
rabbit44
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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 [tex]\Omega[/tex]M 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 = [tex](\frac{\Omega_{\Lambda}}{\Omega_{M0}})^{-1/3}[/tex][tex][sin(\frac{3H_{0}(\Omega_{M0})^{1/2}t}{2}[/tex]]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 [tex]\Omega_{M0}[/tex] = 1/3 and h=1/[tex]\sqrt{2}[/tex], 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!
 
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  • #2
Your on the right track. Though your scale factor seems to have one error in it: the [itex]\Omega_M^{1/2}[/itex] in the argument for sin should actually be [itex](\Omega_M-1)^{1/2}[/itex]. (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.
 

1. What is the horizon of the universe?

The horizon of the universe refers to the distance beyond which we cannot see or detect any light or information. This is because the expansion of the universe has moved these objects beyond our observable limit.

2. How is the horizon of the universe related to the age of the universe?

The horizon of the universe is directly related to the age of the universe. As the universe expands, the horizon also expands, and the objects beyond the horizon become further away and older. This means that the farther we look, the farther back in time we are seeing.

3. What is the observable universe?

The observable universe is the portion of the entire universe that we are able to observe and gather information from. It is limited by the horizon of the universe and is estimated to be about 93 billion light-years in diameter.

4. Can the horizon of the universe change?

Yes, the horizon of the universe can change. As the universe continues to expand, the horizon also expands, allowing us to see more and more of the universe. However, due to the finite speed of light, there will always be a limit to how much of the universe we can observe.

5. How does the horizon of the universe affect our understanding of the universe?

The horizon of the universe plays a crucial role in our understanding of the universe. It allows us to see how the universe has evolved over time, and by studying objects at different distances, we can learn about the history and composition of the universe. It also helps us to understand the structure and size of the universe as a whole.

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