Does a Closed Universe Satisfy the FRW Equations?

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unscientific
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Homework Statement



(a) Show that the equations satisfy FRW equations.
(b) Show the metric when ##\eta## is taken as time

frw1.png

Homework Equations

The Attempt at a Solution



[/B]
The FRW equation is:
[tex]3 \left( \frac{\dot a}{a} \right)^2 = 8\pi G \rho[/tex]

Using ##\frac{da}{dt} = \frac{da}{d\eta} \frac{d\eta}{dt}##:

[tex]\dot a = \frac{\frac{1}{\eta_*} sin \left( \frac{\eta}{\eta_*}\right)}{1 - cos \left( \frac{\eta}{\eta_*} \right)}[/tex]
[tex]\dot a = \frac{C sin \left( \frac{\eta}{\eta_*} \right)}{\eta_* a}[/tex]

The LHS is then

[tex]3 \left( \frac{\dot a}{a} \right)^2 = 3 \left[ \frac{C^2 sin^2 (\frac{\eta}{\eta_*})}{\eta_*^2 a^4} \right][/tex]Not sure how to show this equals RHS..
 
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unscientific said:

Homework Statement



(a) Show that the equations satisfy FRW equations.
(b) Show the metric when ##\eta## is taken as time

frw1.png

Homework Equations

The Attempt at a Solution



[/B]
The FRW equation is:
[tex]3 \left( \frac{\dot a}{a} \right)^2 = 8\pi G \rho[/tex]

Using ##\frac{da}{dt} = \frac{da}{d\eta} \frac{d\eta}{dt}##:

[tex]\dot a = \frac{\frac{1}{\eta_*} sin \left( \frac{\eta}{\eta_*}\right)}{1 - cos \left( \frac{\eta}{\eta_*} \right)}[/tex]
[tex]\dot a = \frac{C sin \left( \frac{\eta}{\eta_*} \right)}{\eta_* a}[/tex]

The LHS is then

[tex]3 \left( \frac{\dot a}{a} \right)^2 = 3 \left[ \frac{C^2 sin^2 (\frac{\eta}{\eta_*})}{\eta_*^2 a^4} \right][/tex]Not sure how to show this equals RHS..

In a closed universe the FRW equation has a spatial curvature term with a 'k' in it. You omitted that.
 
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Dick said:
In a closed universe the FRW equation has a spatial curvature term with a 'k' in it. You omitted that.

FRW equation is given by:
[tex]3 \left( \frac{\dot a}{a} \right)^2 + \frac{kc^2}{a^2} = 8\pi G \rho[/tex]

Curvature parameter is given by ##\Omega = \frac{8\pi \rho G}{3H_0^2}##.

[tex]\frac{C^2 sin^2 (\frac{\eta}{\eta_*})}{\eta_*^2 a^4} + \frac{kc^2}{a^2} = \frac{\Omega}{H_0^2}[/tex]

At ##t=0##, ##\frac{\eta_0}{\eta_*} = sin (\frac{\eta_0}{\eta_*})##, so ##a_0 = C\left[ 1 - \sqrt{1 - (\frac{\eta_0}{\eta_*})^2} \right]##.

Rearranging, ##(\frac{\eta_0}{\eta_*})^2 = (\frac{a_0}{c})(1 - \frac{a_0}{c})##.

How am I to relate ##\eta## to ##\eta_0##?
 
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I'm just confused at this stage, shouldn't ##a_0 =1##? I have a feeling this problem is much simpler than it seems..Would appreciate some help
 
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5th Attempt

[tex]\left( \frac{\dot a}{a}\right)^2 = \frac{8\pi G \rho}{3} - \frac{kc^2}{a^2}[/tex]

Using ##\rho = \Omega \rho_c = \Omega \cdot \frac{3H^2}{8\pi G}##:

[tex]\left( \frac{\dot a}{a}\right)^2 = \Omega H^2 - \frac{kc^2}{a^2}[/tex]

[tex]\dot {a}^2 = \Omega \dot {a}^2 - kc^2[/tex]

[tex]\frac{da}{dt} = c \sqrt{\frac{k}{\Omega -1}}[/tex]

But there is no factor of ##C## on the LHS, as ##\frac{da}{dt} = \frac{sin(\frac{\eta}{\eta_*})}{\eta_* \left[1-cos(\frac{\eta}{\eta_*})\right]}##