GR: 3-d star metric deriving from a general form

[binbagsss]

Homework Statement

attached:

I am stuck on question 2, and give my working to question 1 - the $B(r)$ part I am fine with the $A(r)$ part which clearly is the same in both regions seen by looking at $G_{rr}$ , and attempt, however I assume I have gone wrong in 1 please see below for details.

Homework Equations

$G_{uv}=R_{uv}-\frac{1}{2}Rg_{uv}=8\pi T_{uv}$

$T_{uv}=0 \implies R=0$ by trace of EFE which plugging this into EFE with RHS zero $\iff R_{uv}=0$

Therefore we have for $r>R$ : $G_{uv}=0$
for $r < R$ : $G_{tt}=8 \pi G \rho e^{2A(r)}$ and all other Einstein tensor components zero for this region.

The Attempt at a Solution

[/B]
I think my$e^{2 B(r) }$ must be wrong because, in order to get a metric expression for all $r$ we are asked to and need to impose some constraints that enable continuity at $r=R$ and $r=0$. I think my methodology for $r= R$ is okay, to relate the two integration constants via setting $B(r)$ obtained for $r<R$ equal to the constant obtained for $r>R$. However, the only way I can see that you could talk to $r=0$ is via perhaps setting some integration constant to zero, such that you are preventing there being a singularity at $r=0$ , for e.g as done in the derivation of the FRW metric see here:

However with my expression obtained in (1) there is no such singularity at $r=0$, I can't think of any other way we can impose some constraint from $r=0$?

Here's how I got my $B(r)$, looking at $G_{tt}$ with $A(r)=0$ gives, for $r<R$:

$e^{-2B(r)} d B = 8\pi G r dr$
$e^{-2B(r)}=-2\pi Gr^2 \rho + r_{01}$

$r>R$:
$e^{-2B(r)}/r dB/dr =0$
$r_{02}=e^{-2B(r)}$

and so looking at $r=R$ I have:
$r_{02}= -2\pi GR^2 \rho + r_{01}$and trying to look at $r=0$:
$e^{2B(r)}= \frac{1}{r_{01}-2\pi G \rho r^2}$

(all I can conclude is that $r_{01} \neq 0$ )

$r_{01}$ and $r_{02}$ both integration constants.

thanks...

 

[binbagsss]

bump. many thanks,

 

[binbagsss]

anyone? many thanks