How Is the Particle Horizon Distance Calculated in a Universe with Dark Energy?

[June_cosmo]

Homework Statement

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).

Homework Equations

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?

 

[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.

 

[June_cosmo]

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.

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

 

[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.

 

[June_cosmo]

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.

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. ? (a(x) means a(t) here). I think this isn't right?