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Is there any generalized Schwarzschild solution for an arbitrary number of dimensions? Is it necessary to calculate each individually, or is there a relationship between them?

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In $d>4$ dimensions, the analogue of the Schwarzschild solution is $$ ds^2 = - \left( 1 - \frac{2 M}{r^{d-3} } \right) dt^2 + \left( 1 - \frac{2M}{r^{d-3}} \right)^{-1} dr^2 + r^2 d \Omega_{d-2}^2 $$

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  • $\begingroup$ Most of the derivations of the Schwarzschild solution I've seen assume that as $c$ tends to infinity, it must reduce to Newtonian motion. That can't be assumed for higher dimensions, so how is this derived? $\endgroup$ – user1825464 Oct 5 '13 at 3:52
  • $\begingroup$ This is an asymptotically flat solution, though it may not be the only asymptotically flat static solution (as the Schwarzschild solution is for $d=4$). You derive the solution by solving Einstein's equations and imposing asymptotic flatness. $\endgroup$ – Prahar Oct 5 '13 at 3:55
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    $\begingroup$ @user1825464: Most of the derivations of the Schwarzschild solution I've seen assume that as c tends to infinity, it must reduce to Newtonian motion. No such assumption is needed. The Schwarzschild spacetime is the unique spherically symmetric solution of the Einstein field equations in a vacuum. The Newtonian limit might help with finding an interpretation of the constant $M$, but that's about it. $\endgroup$ – Ben Crowell Oct 5 '13 at 4:05
  • $\begingroup$ And you can similarly derive this solution by assuming a metric $ds^{2} = -A^{2} dt^{2} + B^{2} dr^{2} + r^{2}$(n-sphere metric), and then applying the condition that the metric should go flat as $r\rightarrow \infty$ along with the condition that $R_{ab}=0$ $\endgroup$ – Jerry Schirmer Oct 5 '13 at 4:49

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