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Main Question or Discussion Point
I'm trying to find Schwarzschild solution for 3dimensional spacetime (i.e. [tex]time\otimes space^2[/tex]). The problem is, I can't take the 4dimensional solution
Starting with metric in the general form (see http://en.wikipedia.org/wiki/Deriving_the_Schwarzschild_solution" [Broken] for defining analogous 4dimensional metric) [tex]\[ds^2=e^{\nu(r)} dt^2e^{\lambda(r)} dr^2r^2d\phi^2, \] [/tex]
I calculate the Ricci tensor (prime denoting differentiation with respect to [tex]r[/tex])
Note that the Ricci tensor is completely determined by metric tensor. Thus, once the metric in general form is defined  with [tex]e^\nu[/tex] and [tex]e^\lambda[/tex], one inevitably finds himself at differential equations resulting in flat spacetime (or so it seems); and I don't see any way to define metric differently.
I've created a text with a little more details (in attached file), comparing the 4 and 3dimensional derivations in parallel. Hope it would somehow help.
I'm sort of stuck. Any ideas are greatly appreciated.
[tex]\[ds^2=\left(1\frac{r_g}{r}\right) dt^2\left(1\frac{r_g}{r}\right)^{1} dr^2r^2\left(d\theta^2+sin^2\theta d\phi^2\right)\][/tex]
and substitute [tex]\theta = \frac{\pi}{2}[/tex], which would yield a tidy result. I have to solve this explicitly, starting with 2 spatial dimensions instead of 3, calculating Ricci tensor and equating its components to zero: [tex]R_{\mu\nu}=0[/tex]. Now, that does not seem to be very hard, but the results obtained are not satisfactory.Starting with metric in the general form (see http://en.wikipedia.org/wiki/Deriving_the_Schwarzschild_solution" [Broken] for defining analogous 4dimensional metric) [tex]\[ds^2=e^{\nu(r)} dt^2e^{\lambda(r)} dr^2r^2d\phi^2, \] [/tex]
I calculate the Ricci tensor (prime denoting differentiation with respect to [tex]r[/tex])
[tex]R_{00}=e^{\nu\lambda} \left(\frac{1}{2}\nu''+\frac{1}{4}\nu'^2\frac{1}{4}\nu'\lambda'+\frac{1}{2}\frac{\nu'}{r} \right)=0 [/tex]
[tex]
R_{11}=\frac{1}{2}\nu''\frac{1}{4} \nu'^2 +\frac{1}{4}\nu'\lambda'+\frac{1}{2}\frac{\lambda'}{r}=0 \nonumber [/tex]
[tex] R_{22}=\frac{1}{2}re^{\lambda}(\lambda'\nu')=0 [/tex]
Solving this gives[tex]
R_{11}=\frac{1}{2}\nu''\frac{1}{4} \nu'^2 +\frac{1}{4}\nu'\lambda'+\frac{1}{2}\frac{\lambda'}{r}=0 \nonumber [/tex]
[tex] R_{22}=\frac{1}{2}re^{\lambda}(\lambda'\nu')=0 [/tex]
[tex]\lambda'+\nu'=0 \nonumber [/tex]
[tex]\lambda'\nu'=0 \nonumber [/tex]
or [tex]\lambda'=\nu'=0[/tex]. Applying the condition for [tex]r \to \infty: \; e^\nu \to 1, \ e^\lambda \to 1[/tex], the metric turns out to be Euclidean:[tex]\lambda'\nu'=0 \nonumber [/tex]
[tex]\[ds^2= dt^2 dr^2r^2d\phi^2\][/tex]
which does not seem right at all.Note that the Ricci tensor is completely determined by metric tensor. Thus, once the metric in general form is defined  with [tex]e^\nu[/tex] and [tex]e^\lambda[/tex], one inevitably finds himself at differential equations resulting in flat spacetime (or so it seems); and I don't see any way to define metric differently.
I've created a text with a little more details (in attached file), comparing the 4 and 3dimensional derivations in parallel. Hope it would somehow help.
I'm sort of stuck. Any ideas are greatly appreciated.
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