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Cosmological epoch of matter-radiation equality

  1. Nov 12, 2012 #1
    1. The problem statement, all variables and given/known data
    If Ω_0m=0.25 and Ω_0R=7.4*10^-5 calculate the redshift when the two densities Ω_m and Ω_R are equal.

    Relevant Equations
    1+z=1/a
    [itex]\Omega = \frac{rho}{rho_{crit}}[/itex]
    [itex]\rho_{0,crit} = \frac{3 H_{0}^{2}}{8 \pi G}[/itex]

    The attempt at a solution

    convert matter density: [itex]\epsilon_{0,m} = \rho_{0,m} c^{2} = \Omega_{m,0} \rho_{crit,0} c^{2}[/itex]

    sub in for critical density: [itex]\epsilon_{0,m} = \Omega_{m,0} \frac{3 H_{0}^{2}}{8 \pi G} c^{2}[/itex]

    calculate ratio of matter to radiation: [itex]\frac{\epsilon_{R}}{\epsilon{M}} = \frac{\Omega_{R,0}}{\Omega{M,0}} \frac{8 \pi G c^{2}}{3 H_{0}^{2}}[/itex]

    and as [itex]\epsilon_{R} \propto 1/a^{4}[/itex] and [itex]\epsilon_{M} \propto 1/a^{3}[/itex] and ρ0/a^3 = ρ

    [itex]\frac{\epsilon_{R}}{\epsilon{M}} = \frac{\epsilon_{0,R}}{\epsilon{0,M}} 1/a[/itex]

    put [itex]\frac{\epsilon_{R}}{\epsilon{M}} = 1[/itex] so

    [itex]1 = \frac{\Omega_{M,0}}{\Omega{R,0}} \frac{3 H_{0}^{2}}{8 \pi G c^{2}} (1+z)[/itex]

    This comes out as 1+z = 3.215*10^-6
    and so gives a negative redshift!

    Now I have either done something terribly wrong or the Omegas given are for an arbitrary universe in which the equality epoch has yet to occur!


    Thanks
     
  2. jcsd
  3. Nov 12, 2012 #2
    Sudden realisation (maybe):

    Is it because
    (ϵR)(ϵM) = (ϵ_0,R) (ϵ_0,M) a
    not

    (ϵR)(ϵM) = (ϵ_0,R) (ϵ_0,M) 1/a??

    1+z≈311000?
     
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