Friedmann's equation for a^3 with Λ, ρm

[QuarkDecay]

Homework Statement

We have a flat Universe with a positive Λ, density matter ρ[SUB]m[/SUB]. Need to prove the formula for a[SUP]3[/SUP](t) below from the first Friedmann equation and changing the variable from the u=(...) given below

Relevant Equations

- Friedmann's first equation; H[SUP]2[/SUP]= 8πGρ[SUB]total[/SUB]/3 - k/a[SUP]2[/SUP]
- New variable; u=2Λa[SUP]3[/SUP]/ρ[SUB]o[/SUB]

We need to prove that a³(t)= ρ_o/2Λ [cosh(sqrt(24πGΛ)*t) -1] by changing into a variable of u, where
u=2Λa³/ρ_o

From Friedmann's second equation we know that Λ= ρ_m/ 2
Also ρ_m= ρ_o/ a³

[First attempt]
I begin from Friedmann's equation where (for here), ρ_total= ρ_m + Λ and k=0;

a'²/a² = 8πG(ρ_m + Λ)/3 , (a'=da/dt)

⇒ a'/a = sqrt ((8πG/3) * (3ρ_o/2a³) )

⇒∫ da/a = ∫sqrt((8πG/3) * (3ρ_o/2a³))dt

⇒ lna = sqrt((8πG) * (ρ_o/2a³)) t

⇒ a= exp [sqrt((8πG) * (ρ_o/2a³))t ]

which is not close with what we're trying to prove, and I didn't use the variable u I was supposed to. The solution is the correct for a flat universe, but when we're looking for the a(t). Seems like I have to change the a into u completely inside friedmann's equation.

So I tried this too; (second attempt)

a'²/a² = 8πG(ρ_m + Λ)/3

⇒ a'/a = sqrt ((8πG/3) * (3ρ_o/2a³)) = sqrt(4πGρ_o/a³) ⇒

⇒ da/dt = a*sqrt(4πGρ_o/a³) = sqrt(a²4πGρ_o/a³)=

= sqrt(4πGρ_o/a) (a)

and now changing the variable from a to u;

u=2Λa³/ρ_o ⇒ a³ = uρ_o/2Λ ⇒

a= [uρ_o/2Λ] ^3/2 (1)

So,
da= (ρ_o/2Λ) ^3/2 u^1/2du (2) (a),(1),(2) → ... du/u^1/2= sqrt(8πGΛ)dt = sqrt(8πGΛ)t

Where the cosh() is still missing...
In the first attempt I had the exact same problem too. The first integral was always da/sqrt(a) that doesn't give a cosh for a solution

 

[Orodruin]

From Friedmann's second equation we know that Λ= ρm/ 2

This is not true. You have not been given the relation between $\Lambda$ and $\rho_m$ and the proportion between them is arbitrary. Furthermore, $\rho_m$ is going to change as the universe expands, which $\Lambda$ will not.

8πG(ρm + Λ)/3 , (a'=da/dt)

⇒ a'/a = sqrt ((8πG/3) * (3ρo/2a3) )

Which means that this is wrong. As stated above $\Lambda$ will depend on $a$ in a different way from $\rho_m$.