- The problem statement, all variables and given/known data The most general form:
$ds^2=e^{2A(r)}dt^2-e^{2B(r)}dr^2-r^2(d\theta_1^2 +sin^2\theta_1(d\theta^2_2+sin^2\theta_2d\phi^2))$
Ricci tensors:
$R_{tt}=e^{2(A-B)}(\frac{3A'}{r}+A'^2-B'A'+A'')$ $R_{rr}=-A''+\frac{3B'}{r}+A'(B'-A')$ $R_{\theta_1 \theta_1}=2+e^{-2B}(-2+r(B'-A'))$ $R_{\theta_2 \theta_2} = sin^2 \theta_1 R_{\theta_1 \theta_1}$ $R_{\phi \phi} = sin^2 \theta_1 sin^2 \theta_2 R_{\theta_1 \theta_1}$
To solve the Ricci tensors for $A$ and $B$?
Relevant equations
The attempt at a solution
I have tried preceeding as you do in the 4-d case. This is to do $e^{-2(A-B)} R_{tt} + R_{rr} =0 $ which gives $A=-B$
and then to solve $R_{\theta_1 \theta_1}=0$ and then $R_{rr}$
However doing this i get:
$1+e^{2A}(1-A') = 0$ multiply though by $e^{-3A}$ to get $d/dr(e^{-A}r)=-e^{-A}$
$$e^{-A}r=\int -e^{-A} dr $$
and I can't solve for $A(r)$ explicitly
I can't see a better way to proceed as with the other components there are terms including $A'' , A'^2$ things.
Many thanks