When was the matter density equal to the vacuum energy density?

Click For Summary
SUMMARY

In the ΛCDM model, the current Hubble parameter is H(t0) = 70 km/s/Mpc, with matter density Ωd(t0) = 0.3 and vacuum energy density ΩΛ(t0) = 0.7, confirming a spatially flat universe. To determine when matter density equals vacuum energy density, the equation Ωd(t) + ΩΛ(t) = 1 is utilized, leading to both densities being equal at 0.5. The relationship between matter density and redshift is expressed as $$\Omega_{m,0}(1+z_{eq})^3 = \Omega_{\Lambda,0}$$, allowing for the calculation of redshift (z_eq) and time (t) using a cosmological calculator.

PREREQUISITES
  • Understanding of the ΛCDM cosmological model
  • Familiarity with Hubble's Law and Hubble parameter
  • Knowledge of energy density parameters (Ωd, ΩΛ)
  • Basic grasp of redshift and its implications in cosmology
NEXT STEPS
  • Research the derivation of the equation $$\Omega_{m,0}(1+z_{eq})^3 = \Omega_{\Lambda,0}$$
  • Explore the use of cosmological calculators for time and redshift calculations
  • Study the implications of spatial flatness in cosmology
  • Investigate the relationship between matter density and scale factor (a) in cosmological models
USEFUL FOR

Astronomers, cosmologists, and physics students interested in the dynamics of the universe's expansion and the interplay between matter and vacuum energy density.

LeoChan
Messages
5
Reaction score
1
TL;DR
In ΛCDM, to find t and z when the matter density equal to the vacuum energy density.
In ΛCDM, H(t0) = 70km/s/Mpc,
Ωd(t0) = 0.3, Ωr(t0) = 0 and ΩΛ(t0) =0.7,
so that Ω(t0) = Ωd(t0) + Ωr(t0) + ΩΛ(t0) = 1and the universe is spatially flat.

I want to know the t and z when the matter density equal to the vacuum energy density. By total energy density equation, I think Ωd(t) + ΩΛ(t) = 1, so they are both equal to 0.5 .

Maybe 0.5 = Λ / ( 3 * H(t) ^ 2 ). As for the matter, I am not sure since I only know it is proportional to a^-3. Is it related to the redshift dependent Hubble parameter, H(z)?

Thank you for your attention.
 
Space news on Phys.org
Just use

$$\Omega_{m,0}(1+z_{eq})^3 = \Omega_{\Lambda,0}$$
to find ##z_{eq}## and then type it into some cosmological calculator to obtain ##t##
 
Arman777 said:
Just use

$$\Omega_{m,0}(1+z_{eq})^3 = \Omega_{\Lambda,0}$$
to find ##z_{eq}## and then type it into some cosmological calculator to obtain ##t##
Thank you.That is quite straight forward.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 134 ·
5
Replies
134
Views
11K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K