Horizon size of a photon in a matter dominated universe

[EnSlavingBlair]

Homework Statement

The maximum proper distance a photon can travel in the interval (0,t) is given by the horizon size

h(t) = R(t) ∫₀^t dt' / R(t')

Show that, for a matter dominated universe

h(z) = H₀^-1(1+z)^-1(Ω-1)^-1/2cos^-1(1-2(Ω-1)/(Ω(1+z))) for Ω>1
= 2H₀^-1(1+z)^-3/2 for Ω=1
= H₀^-1(1+z)^-1(Ω-1)^-1/2cosh^-1(1+2(Ω-1)/(Ω(1+z))) for Ω<1

Also show that d_H ≈ 3H₀^-1Ω₀^-1/2(1+z)^-3/2 for (1+z)>>Ω^-1

Homework Equations

As is a matter dominated universe, can assume:
Ω_m = 1, Ω_r = 0, Ω_λ = 0

d_H(z) = c H₀^-1((1-Ω)(1+z)²+Ω_m(1+z)³)^-1/2

dt = -d_H(z)dz/(c(1+z))

cosh(ix) = cos(x)

cosh(x) = (e^x+e^-x)/2

The Attempt at a Solution

For the Ω=1 case, which seems to me as if it should be the simplest, my main problem seems to be with the R(t'). I don't really know how I'm supposed to integrate that! Anyway, this is what I've tried thus far:

d_H(z) = c H₀^-1(1+z)^-3/2

∫₀^t dt = -H₀^-1 ∫_∞^z (1+z)^-5/2
= 2(1+z)^-3/2/(3H₀)

problem is I haven't taken into consideration R(t') as I really don't know how. I know R(t₀)/R(t_e) = z+1 but I don't think that can be used here. And even if it was I end up with:

h(z) = 2H₀^-1 (1+z)^-1/2

which is wrong anyway.

At this point in time I'm only trying to get this part of the question out, as I believe it will make the other parts easier.

If you have any ideas on what I'm doing wrong or what I seem to be missing, that would be greatly appreciated.

Cheers,
nSlavingBlair

 

[clamtrox]

. And even if it was I end up with:

h(z) = 2H₀^-1 (1+z)^-1/2

which is wrong anyway.

Did you remember the extra factor of a in the front of the integral?