Sign of Lambda cosmology homework

[latentcorpse]

In question 4 of this paper, I am trying to find the sign of [itex]\Lambda[/itex]

Since we have the vacuum einstein eqns [itex]G_{\mu \nu}=-\Lambda g_{\mu \nu}[/itex]

we can just contract to get [itex]R-\frac{4}{2} R=-4\Lambda \Rightarrow \Lambda=\frac{R}{4}[/itex]

So [itex]\Lambda[/itex] will have the same sign as [itex]R[/itex]

Now given the components of [itex]R[/itex] in the question, we compute [itex]R=3 \frac{\ddot{a}}{a} + ( a \ddot{a} + 2 \dot{a}^2 +2 ) ( 1 + \sin^2{\chi} + \sin^2{\chi} \sin^2{\theta})[/itex]

Now the second bracket is clearly positive.

I'm not sure about the rest of the expression though because, how do we know the signs of [itex]a , \dot{a} , \ddot{a}[/itex]? We aren't told whether the universe is accelerating or not?



Secondly, in the next bit of the question, we have to obtain the scale factor.

I have solved the continuity equation [itex]\dot{\rho} + 3\gamma \frac{\dot{a}}{a} \rho=0[/itex]
However, [itex]\gamma=0[/itex] when we have a cosmological constant fluid.
This means [itex]\dot{\rho}=0 \Rightarrow \rho = k[/itex] where [itex]k[/itex] is constant.

This would meant that the Friedmann equation will read [itex]\left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi G k}{3} + \frac{\Lambda}{3}[/itex]

But then I don't see how to solve for [itex]a[/itex]? Also I think I may have made a mistake or missed something as I don't see how that time symmetry and rescaling stuff is going to help.

I think we are trying to get to a de Sitter spacetime ultimately!

Cheers.

 

[NateD]

The Friedmann equation can be rewritten to solve for the scale factor a: \frac{\dot{a}^2}{a^2} = \frac{8 \pi G k}{3} + \frac{\Lambda}{3} \Rightarrow \dot{a}^2 = a^2 \left( \frac{8 \pi G k}{3} + \frac{\Lambda}{3} \right)From the expression for \Lambda, we can see that \Lambda will be positive. Therefore, the right hand side of this equation is always positive. This means that \dot{a}^2 is also always positive, so \dot{a} must be positive. Now, in order to solve for a, we need to use an integrating factor. We can use the time symmetry of the universe and the rescaling of the scale factor to arrive at the relation a(t)=a_0 e^{\lambda t} where \lambda is a constant. By substituting this into the Friedmann equation, we get \lambda^2=\frac{8 \pi G k}{3} + \frac{\Lambda}{3} \Rightarrow \lambda = \sqrt{\frac{8 \pi G k}{3} + \frac{\Lambda}{3}}Therefore, the scale factor of the universe is a(t)=a_0 e^{\sqrt{\frac{8 \pi G k}{3} + \frac{\Lambda}{3}} t} This implies that the universe is expanding exponentially, thus it is a de Sitter spacetime.