Interpreting the Kretschmann scalar

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.

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).

• This is much better than the accepted answer.
– user4552
Commented Feb 22, 2018 at 1:08

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 $r\to0$, $K\to\infty$.

• Is there a reasonable interpretation for other values of $K$? Such as when it vanishes, $K = 0$?
– XYZT
Commented Dec 2, 2014 at 19:15
• 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. Commented Dec 2, 2014 at 19:20
• Is (1) necessarily true?
– XYZT
Commented Dec 2, 2014 at 19:20
• @KyleKanos For pseudo-Riemannian manifolds, you can have non-zero null curvature tensors, just like you can have nonzero, null vectors. Commented Dec 3, 2014 at 4:11
• This doesn't really answer the question. The same statements would be true for various other curvature scalars.
– user4552
Commented Feb 22, 2018 at 1:08