General Relativity - FRW Metric - FRW Equations show that ...

[binbagsss]

Homework Statement

Homework Equations

see above

The Attempt at a Solution

Using the conservation equation for $p=0$

I find: $\rho =\frac{ \rho_0}{a^3}$; (I am told this is $\geq0$ , is $a\geq0$ so here I can conclude that $\rho_0 \geq =0$ or not?)

Plugging this and $p=0$ into the first Einstein equation I get:

$\dot{a^2}+k-\Lambda a^2=\frac{8\pi G \rho_0}{a}$

So a stationary solution is to solve for $a$ and get no time independence, so don't we need something of the form:

$\frac{da}{dt} a^k =0$ or can I find a more general expression to this?

This is ofc not possible to get since $\Lambda$ and $k$ are constants and can not depend on $a$?

Many thanks

 

[phyzguy]

In a time independent solution, a is constant, so Λ and k can depend on a and still be constant.

 

[binbagsss]

In a time independent solution, a is constant, so Λ and k can depend on a and still be constant.

mmm

so $k=8 \pi G \rho_0 / a$ and $\Lambda =0$ ?

 

[phyzguy]

so $k=8 \pi G \rho_0 / a$ and $\Lambda =0$ ?

How did you conclude Λ=0? Remember k = 0, +/-1.

 

[binbagsss]

How did you conclude Λ=0? Remember k = 0, +/-1.

$k=0$ and $\Lambda = \frac{-8\pi G \rho_0}{a}$ ?

Basically need to solve for $8\pi G \rho_0/a +\Lambda a^2 -k = 0$ subject to the above values of $k$ only allowed ?

 

[phyzguy]

$k=0$ and $\Lambda = \frac{-8\pi G \rho_0}{a}$ ?

Basically need to solve for $8\pi G \rho_0/a +\Lambda a^2 -k = 0$ subject to the above values of $k$ only allowed ?

Aren't there two equations?

 

[binbagsss]

Aren't there two equations?

equating $a$ coefficients, but baring in mind that $\Lambda$ can depend on $a$ and $k$ can not, so $k$ for the 0th order equation?

 

[binbagsss]

I find: $\rho =\frac{ \rho_0}{a^3}$; (I am told this is $\geq0$ , is $a\geq0$ so here I can conclude that $\rho_0 \geq =0$ or not?)

Plugging this and $p=0$ into the first Einstein equation I get:

$\dot{a^2}+k-\Lambda a^2=\frac{8\pi G \rho_0}{a}$

apologies to re-bump but
re-looking at this , $\rho =\frac{ \rho_0}{a^3} \geq 0$
$\rho=\frac{\rho_0}{a} \frac{1}{a^2} \geq 0$ and so since $\frac{1}{a^2}>0$ it must be that $\rho=\frac{\rho_0}{a} \geq 0$

so from the other equation $0+k-\Lambda a^2 \geq 0$ is the required constraint ?

 

[binbagsss]

apologies to re-bump but
re-looking at this , $\rho =\frac{ \rho_0}{a^3} \geq 0$
$\rho=\frac{\rho_0}{a} \frac{1}{a^2} \geq 0$ and so since $\frac{1}{a^2}>0$ it must be that $\rho=\frac{\rho_0}{a} \geq 0$

so from the other equation $0+k-\Lambda a^2 \geq 0$ is the required constraint ?

I'm getting a bit confused since it says $k=-1,0,1$ , but since the question doesn't specify whether $\Lamda$ is $>0$ or $<0$ there is no need to/no way to narrow down this further