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

  • #1
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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)= ρo/2Λ [cosh(sqrt(24πGΛ)*t) -1] by changing into a variable of u, where
u=2Λa3o

From Friedmann's second equation we know that Λ= ρm/ 2
Also ρm= ρo/ a3

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

a'2/a2 = 8πG(ρm + Λ)/3 , (a'=da/dt)

⇒ a'/a = sqrt ((8πG/3) * (3ρo/2a3) )

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

⇒ lna = sqrt((8πG) * (ρo/2a3)) t

⇒ a= exp [sqrt((8πG) * (ρo/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(ρm + Λ)/3

⇒ a'/a = sqrt ((8πG/3) * (3ρo/2a3)) = sqrt(4πGρo/a3) ⇒

⇒ da/dt = a*sqrt(4πGρo/a3) = sqrt(a24πGρo/a3)=

= sqrt(4πGρo/a) (a)

and now changing the variable from a to u;

u=2Λa3o ⇒ a3 = uρo/2Λ ⇒

a= [uρo/2Λ] 3/2 (1)

So,
da= (ρo/2Λ) 3/2 u1/2du (2)


(a),(1),(2) → .... du/u1/2= 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
 

Answers and Replies

  • #2
Orodruin
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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##.
 

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