Cosmological epoch of matter-radiation equality

  • Thread starter Thread starter gboff21
  • Start date Start date
  • Tags Tags
    Cosmological
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 4K views
gboff21
Messages
48
Reaction score
0

Homework Statement


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
 
Physics news on Phys.org
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?