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.
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
1
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).
-
3$\begingroup$ This is much better than the accepted answer. $\endgroup$– user4552Commented Feb 22, 2018 at 1:08
$\begingroup$
$\endgroup$
8
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 $r\to0$, $K\to\infty$.
answered Dec 2, 2014 at 18:29
-
$\begingroup$ Is there a reasonable interpretation for other values of $K$? Such as when it vanishes, $K = 0$? $\endgroup$– XYZTCommented Dec 2, 2014 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. $\endgroup$ Commented Dec 2, 2014 at 19:20
-
-
1$\begingroup$ @KyleKanos For pseudo-Riemannian manifolds, you can have non-zero null curvature tensors, just like you can have nonzero, null vectors. $\endgroup$– asperanzCommented Dec 3, 2014 at 4:11
-
1$\begingroup$ This doesn't really answer the question. The same statements would be true for various other curvature scalars. $\endgroup$– user4552Commented Feb 22, 2018 at 1:08