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