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So I read here that the non-zero Riemann tensors for the Schwarzschild metric are:

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However, I am very curious on what these non-zero Riemann tensors would look like in a 5-dimensional spacetime(4 spatial + 1 time). Could it be possible for someone to tell me what values 5-dimensional nonzero Riemann tensors for the Schwarzschild metric would hold?

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  • $\begingroup$ By definition, the Schwarzild metric is (1,3)! You should clarify what do you mean by a "5 dimensional spacetime for Schwarschild metric". $\endgroup$
    – LSS
    Commented Aug 19, 2022 at 0:13
  • $\begingroup$ You might want to check out (4+1)D Kaluza-Klein theory: en.wikipedia.org/wiki/Kaluza%E2%80%93Klein_theory $\endgroup$ Commented Aug 19, 2022 at 0:22
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    $\begingroup$ @LSS I meant a spherically symmetric non rotating chargeless mass in 4 spatial dimensions and 1 time dimension. I hope this clarified what I meant. $\endgroup$ Commented Aug 19, 2022 at 0:42
  • $\begingroup$ @JungwoonSong In this case, you should check out the Schwarzschild-Tangherlini metric, which is the generalization of the Schwarzschild metric to other dimensions. $\endgroup$ Commented Aug 19, 2022 at 2:13

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As @Samuel Adrian Antz pointed out in his comment: the $d$-dimensional generalisation of the Schwarzschild metric is the so called Schwarzschild-Tangherlini metric with the line element $$ ds^2=-\Big[1-\Big(\frac{a}{r}\Big)^{d-3}\Big] dt^2 + \Big[1-\Big(\frac{a}{r}\Big)^{d-3}\Big]^{-1} dr^2 + r^2 d\Omega^2_{d-2}, $$ and $d\Omega_1^2=d\varphi^2,\quad d\Omega^2_{i+1}=d\theta_i^2+\sin^2\theta_i d\Omega_i^2\;(i\ge 1)$ [de.wikipedia.org/wiki/Schwarzschild-Tangherlini-Metrik]. From this line element -- the corresponding metric -- one can compute the Riemann tensor as usual.

In $d=4$ with coordinates $(q_i)=(t,r,\theta,\phi)$ we recover the know result for the six independent, non-vanishing components of the Riemann tensor for the Schwarschild metric: $$ \begin{array}{cc} \text{R}_{1212}=-\text{R}_{2112}=-R_{1221}=R_{2121}\text{(=}R_{1212}\text{=...)}= & -\frac{a}{r^3} \\ R_{1313}=-R_{3113}=-R_{1331}=R_{3131}\text{(=}R_{1313}\text{=...)}= &-\frac{a (a-r)}{2 r^2} \\ R_{1414}=-R_{4114}=-R_{1441}=R_{4141}\text{(=}R_{1414}\text{=...)}= & -\frac{a (a-r) \sin ^2(\theta)}{2 r^2} \\ R_{2323}=-R_{3223}=-R_{2332}=R_{3232}\text{(=}R_{2323}\text{=...)}= & \frac{a}{2 a-2 r} \\ R_{2424}=-R_{4224}=-R_{2442}=R_{4242}\text{(=}R_{2424}\text{=...)}= & \frac{a \sin ^2(\theta)}{2 a-2 r} \\ R_{3434}=-R_{4334}=-R_{3443}=R_{4343}\text{(=}R_{3434}\text{=...)}= & a r \sin ^2(\theta) \\ \end{array} $$

In $d=5$ with coordinates $(q_i)=(t,r,\theta_1,\theta_2,\phi)$ we find: $$ \begin{array}{cc} R_{1212}=-R_{2112}=-R_{1221}=R_{2121}\text{(=}R_{1212}\text{=...)}= & -\frac{3 a^2}{r^4} \\ R_{1313}=-R_{3113}=-R_{1331}=R_{3131}\text{(=}R_{1313}\text{=...)}= & -\frac{a^2 \left(a^2-r^2\right) \sin ^2(\theta (2))}{r^4} \\ R_{1414}=-R_{4114}=-R_{1441}=R_{4141}\text{(=}R_{1414}\text{=...)}= & \frac{a^2 r^2-a^4}{r^4} \\ R_{1515}=-R_{5115}=-R_{1551}=R_{5151}\text{(=}R_{1515}\text{=...)}= & -\frac{a^2 \left(a^2-r^2\right) \sin ^2(\theta_1) \sin ^2(\theta_2)}{r^4} \\ R_{2323}=-R_{3223}=-R_{2332}=R_{3232}\text{(=}R_{2323}\text{=...)}= & \frac{a^2 \sin ^2(\theta_2)}{a^2-r^2} \\ R_{2424}=-R_{4224}=-R_{2442}=R_{4242}\text{(=}R_{2424}\text{=...)}= & \frac{a^2}{a^2-r^2} \\ R_{2525}=-R_{5225}=-R_{2552}=R_{5252}\text{(=}R_{2525}\text{=...)}= & \frac{a^2 \sin ^2(\theta_1) \sin ^2(\theta_2)}{a^2-r^2} \\ R_{3434}=-R_{4334}=-R_{3443}=R_{4343}\text{(=}R_{3434}\text{=...)}= & a^2 \sin ^2(\theta_2) \\ R_{3535}=-R_{5335}=-R_{3553}=R_{5353}\text{(=}R_{3535}\text{=...)}= & a^2 \sin ^2(\theta_1) \sin ^4(\theta_2) \\ R_{4545}=-R_{5445}=-R_{4554}=R_{5454}\text{(=}R_{4545}\text{=...)}= & a^2 \sin ^2(\theta_1) \sin ^2(\theta_2) \\ \end{array} $$

So when moving from $d=4$ to $d=5$ we go from six independent, non-vanishing components to ten.

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  • $\begingroup$ Nice answer, +1. I searched for a bit, but just didn't find a complete list of all components of the Riemann curvature tensor anywhere. $\endgroup$ Commented Aug 19, 2022 at 12:03
  • $\begingroup$ Thanks so much! $\endgroup$ Commented Aug 19, 2022 at 12:29

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