- #1
gboff21
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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