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How do you interpret the Kretschmann scalar (in general relativity)? What can you tell from it?


The Kretschmann scalar is defined as

$$K = R_{abcd} R^{abcd} $$

where $R_{abcd}$ is the Riemann curvature tensor.

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The Kretschmann scalar can be used as an indicator of curvature singularities in the manifold. For instance, in the Schwarzschild black hole (given in the Wikipedia link in your post), $$ K\propto\frac1{r^6} $$ so as $r\to0$, $K\to\infty$.

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  • $\begingroup$ Is there a reasonable interpretation for other values of $K$? Such as when it vanishes, $K = 0$? $\endgroup$ – XYZT Dec 2 '14 at 19:15
  • $\begingroup$ Sure: there are two other cases: (1) $K=0$, then there's no curvature (2) if $K>0$ (but finite), then the manifold is not flat. $\endgroup$ – Kyle Kanos Dec 2 '14 at 19:20
  • $\begingroup$ Is (1) necessarily true? $\endgroup$ – XYZT Dec 2 '14 at 19:20
  • $\begingroup$ Actually, I'm not 100% sure on that note. I presume yes, but I'm willing to be corrected. $\endgroup$ – Kyle Kanos Dec 2 '14 at 19:26
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    $\begingroup$ @KyleKanos For pseudo-Riemannian manifolds, you can have non-zero null curvature tensors, just like you can have nonzero, null vectors. $\endgroup$ – asperanz Dec 3 '14 at 4:11
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For vacuum solutions, since the Ricci tensor $R_{ab}$ vanishes, the Kretschmann scalar is equal to the norm of the Weyl tensor, $K = C_{abcd}C^{abcd}$. This means it is telling you something about the tidal forces at a given point. I might use $K^{1/2}$ to characterize the strength of the tidal forces. This can be used in Schwarzschild or Kerr spacetimes to see that the tidal forces go like $M/r^3$ (at least in the equatorial plane for Kerr).

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    $\begingroup$ This is much better than the accepted answer. $\endgroup$ – Ben Crowell Feb 22 '18 at 1:08

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