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Particle horizon distance

  1. Nov 1, 2015 #1
    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 [itex]\Omega_M=1-\Omega_{DE}=0.25,\omega=-1[/itex]. 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 [itex]d=\int_0^{t0}\frac{dt}{a(t)}[/itex], so in a flat universe [itex]a(t)=(t/t0)^{2/(3(1+3\omega))}[/itex],
    so that we have $$d=\frac{2}{1+3\omega}H_0^{-1}$$,but this has nothing to do with [itex]\Omega_M=1-\Omega_{DE}=0.25[/itex]?
     
    Last edited: Nov 1, 2015
  2. jcsd
  3. Nov 1, 2015 #2

    phyzguy

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    The [itex] a(t) \propto t^{\frac{2}{3}}[/itex] result is only true for a flat universe with [itex]\Omega_{DM} =0[/itex]. You need to review the Friedmann equations for the case with [itex]\Omega_{DM} \neq 0[/itex].
     
  4. Nov 1, 2015 #3
    Oh that's right! So how do I derive a(t) from Friedmann equations?
     
  5. Nov 1, 2015 #4

    phyzguy

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    Well, can you write down the first Friedmann equation for H in terms of the Ω parameters? Once you have done that, remember that [itex] H = \frac {\dot a}{a}[/itex]. Then you should be able to write a differential equation for a that you can numerically integrate.
     
  6. Nov 1, 2015 #5
    Thank you! So that would be [itex]\frac{H^2}{H_0^2}=\frac{0.25}{a^3}+0.75,[/itex]if we combine this with [itex]H=\frac{\dot{a}}{a}[/itex] 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
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