Solving Schwarzschild Equation in 5D: Ricci Tensors

[binbagsss]

Homework Statement

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 $B$?

Homework 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
$R_{\theta_1 \theta_1}=0$

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

 

[mighty2000]

for any help!Thank you for posting your question. Solving for the Ricci tensors in this case can be a bit tricky, but I believe I have found a way to do it.

First, I would recommend rewriting the equations in a slightly different form:

$R_{tt} = e^{2A}(A'' + 2A'^2 - 2B'A')$
$R_{rr} = e^{2A}(-A'' + 2A'^2 - 2B'A')$
$R_{\theta_1 \theta_1} = 2 + e^{2A}(-2B' + 2A'^2)$
$R_{\theta_2 \theta_2} = sin^2 \theta_1 R_{\theta_1 \theta_1}$
$R_{\phi \phi} = sin^2 \theta_1 sin^2 \theta_2 R_{\theta_1 \theta_1}$

Now, we can see that all of the equations have the same term $A'' + 2A'^2 - 2B'A'$, so we can set this term equal to some constant, say $C$. This gives us the following equations:

$R_{tt} = e^{2A}C$
$R_{rr} = -e^{2A}C$
$R_{\theta_1 \theta_1} = 2 + e^{2A}(-2B' + C)$
$R_{\theta_2 \theta_2} = sin^2 \theta_1 R_{\theta_1 \theta_1}$
$R_{\phi \phi} = sin^2 \theta_1 sin^2 \theta_2 R_{\theta_1 \theta_1}$

Next, we can use the equation for $R_{\theta_1 \theta_1}$ to solve for $B'$ in terms of $C$ and $A'$:

$B' = \frac{1}{2} + \frac{1}{2}e^{-2A}C - \frac{1}{2}A'$

Now, we can substitute this into the equations for $R_{tt}$ and $R_{rr}$ to get:

##R_{tt} = \frac{1