# Horizon size of a photon in a matter dominated universe

1. Jun 11, 2013

### nSlavingBlair

1. The problem statement, all variables and given/known data

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

h(t) = R(t) ∫0t 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 dH ≈ 3H0-1Ω0-1/2(1+z)-3/2 for (1+z)>>Ω-1

2. Relevant equations

As is a matter dominated universe, can assume:
Ωm = 1, Ωr = 0, Ωλ = 0

dH(z) = c H0-1((1-Ω)(1+z)2m(1+z)3)-1/2

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

cosh(ix) = cos(x)

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

3. 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:

dH(z) = c H0-1(1+z)-3/2

0t dt = -H0-1z (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(te) = 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

2. Jun 11, 2013

### clamtrox

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

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