- #1

QuarkDecay

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- 2

- 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 a

u=2Λa

From Friedmann's second equation we know that Λ= ρ

Also ρ

[First attempt]

I begin from Friedmann's equation where (for here), ρ

a'

⇒ a'/a = sqrt ((8πG/3) * (3ρ

⇒∫ da/a = ∫sqrt((8πG/3) * (3ρ

⇒ lna = sqrt((8πG) * (ρ

⇒ a= exp [sqrt((8πG) * (ρ

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'

⇒ a'/a = sqrt ((8πG/3) * (3ρ

⇒ da/dt = a*sqrt(4πGρ

= sqrt(4πGρ

and now changing the variable from a to u;

u=2Λa

a= [uρ

So,

da= (ρ

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

^{3}(t)= ρ_{o}/2Λ [cosh(sqrt(24πGΛ)*t) -1] by changing into a variable of u, whereu=2Λa

^{3}/ρ_{o}From Friedmann's second equation we know that Λ= ρ

_{m}/ 2Also ρ

_{m}= ρ_{o}/ a^{3}[First attempt]

I begin from Friedmann's equation where (for here), ρ

_{total}= ρ_{m}+ Λ and k=0;a'

^{2}/a^{2}= 8πG(ρ_{m}+ Λ)/3 , (a'=da/dt)⇒ a'/a = sqrt ((8πG/3) * (3ρ

_{o}/2a^{3}) )⇒∫ da/a = ∫sqrt((8πG/3) * (3ρ

_{o}/2a^{3}))dt⇒ lna = sqrt((8πG) * (ρ

_{o}/2a^{3})) t⇒ a= exp [sqrt((8πG) * (ρ

_{o}/2a^{3}))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}/a^{2}= 8πG(ρ_{m}+ Λ)/3⇒ a'/a = sqrt ((8πG/3) * (3ρ

_{o}/2a^{3})) = sqrt(4πGρ_{o}/a^{3}) ⇒⇒ da/dt = a*sqrt(4πGρ

_{o}/a^{3}) = sqrt(a^{2}4πGρ_{o}/a^{3})== sqrt(4πGρ

_{o}/a)**(a)**and now changing the variable from a to u;

u=2Λa

^{3}/ρ_{o}⇒ a^{3}= uρ_{o}/2Λ ⇒a= [uρ

_{o}/2Λ]^{3/2}**(1)**So,

da= (ρ

_{o}/2Λ)^{3/2}u^{1/2}du**(2)**(a),(1),(2) → ... du/u^{1/2}= sqrt(8πGΛ)dt = sqrt(8πGΛ)tWhere 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