Particle horizon distance

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1. Nov 1, 2015

June_cosmo

1. The problem statement, all variables and given/known data
Numerically integrate and report the particle horizon distance today for the currently fa-
vored model $\Omega_M=1-\Omega_{DE}=0.25,\omega=-1$. Assume the scaled Hubble constant to be h = 0.72, and report the particle horizon in billions of lyr (Gyr).

2. Relevant equations

3. The attempt at a solution
Horizen distance $d=\int_0^{t0}\frac{dt}{a(t)}$, so in a flat universe $a(t)=(t/t0)^{2/(3(1+3\omega))}$,
so that we have $$d=\frac{2}{1+3\omega}H_0^{-1}$$,but this has nothing to do with $\Omega_M=1-\Omega_{DE}=0.25$?

Last edited: Nov 1, 2015
2. Nov 1, 2015

phyzguy

The $a(t) \propto t^{\frac{2}{3}}$ result is only true for a flat universe with $\Omega_{DM} =0$. You need to review the Friedmann equations for the case with $\Omega_{DM} \neq 0$.

3. Nov 1, 2015

June_cosmo

Oh that's right! So how do I derive a(t) from Friedmann equations?

4. Nov 1, 2015

phyzguy

Well, can you write down the first Friedmann equation for H in terms of the Ω parameters? Once you have done that, remember that $H = \frac {\dot a}{a}$. Then you should be able to write a differential equation for a that you can numerically integrate.

5. Nov 1, 2015

June_cosmo

Thank you! So that would be $\frac{H^2}{H_0^2}=\frac{0.25}{a^3}+0.75,$if we combine this with $H=\frac{\dot{a}}{a}$ and I solved this equation (online), it gave me http://www4f.wolframalpha.com/Calculate/MSP/MSP226920fg7hgi3c9658be000033830f2a7886e3e4?MSPStoreType=image/gif&s=20&w=550.&h=47. [Broken] ? (a(x) means a(t) here). I think this isn't right?

Last edited by a moderator: May 7, 2017