- #1

QuarkDecay

- 47

- 2

- TL;DR Summary
- Radiation and Matter equilibrium point and expansion rate calculation

We want to calculate the a

My book solves it this way;

ρ

⇒ (ρ

= (ρ

I don't understand the a(t

ρ

After that it continues like;

ρ

and it goes a

=(Ω

what does the z

_{o}/a(t_{eq}) of the equilibrium point between ρ_{m}and ρ_{r}(ρ_{m}= ρ_{r})My book solves it this way;

ρ

_{m}(t) / ρ_{r}(t)= a(t) ⇒⇒ (ρ

_{m}/ ρ_{r})_{teq}=1 == (ρ

_{m}/ ρ_{r})_{o}* a(t_{eq})/ a_{o}I don't understand the a(t

_{eq})/ a_{o}part. If ρ_{m}(t)= ρ_{ο}/α_{o}^{3}and ρ_{r}(t)= ρ_{ο}/α_{o}^{4}then it should beρ

_{m}(t)/ ρ_{r}(t) = ρ_{o}^{m}/ ρ_{o}^{r}* a_{o}After that it continues like;

ρ

_{m}/ρ_{r}= Ω_{m}/Ω_{r}and it goes a

_{o}/ a(_{teq})= 1 + z_{eq}==(Ω

_{m}/Ω_{r})_{o}what does the z

_{eq}mean and how did that come up from the equation?**(2)**Also, there's an equilibrium point for dark energy and radiation density as well and a similar problem like above. Does dark energy have a known ω value? Like the ω=1/3 for radiation. I was looking for its value and couldn't find it, so I thought we maybe don't know it because it's dark energy?