Curvature of Schwarzschild spacetime In Eq. (1.1) of a paper (R Balbinot & E Poisson, Phys. Rev. D41, 395 (1990)) it is mentioned that the curvature of the Schwarzschild spacetime is $$
M / r^{3}
$$
but I don't get what do they mean by curvature (and it is not mentioned in the paper).
What could it be?
 A: This is essentially the Weyl Scalar $\Psi_2$ (the only non vanishing Weyl scalar for Schwarzschild metric) . In NP formalism, if we consider an orthonormal null tetrad ($l, n, m, \bar{m}$) , then the Weyl scalar $\Psi_2$ is defined as$$ \Psi_2=-C_{abcd}l^am^bn^c\bar{m}^d$$ For Schwarzschild metric, we can consider the null tetrad:
$l=\frac{1}{\Delta}(r^2, \Delta, 0,0) $, $n=\frac{1}{2r^2}(r^2, -\Delta, 0,0) $  (where $\Delta=r^2-2Mr$) and $m=\frac{1}{r\sqrt{2}}(0, 0,1,i\csc{\theta}) $ and after substituting in the above expression we get $\Psi_2=M/r^3$.
EDIT: This is the closest scalar quantity I could find and it agrees well with the functional form of $M/r^3$ (except there is an ambiguity of (-1) factor depending on definition of $\Psi_2$). Physically, $\Psi_2$ corresponds to "coulombic potential energy", however, I'm not sure how this quantity is relevant to the discussion in attached paper. We can try to construct curvature invariants for Schwarzschild spacetime and lets say we have a non-zero curvature invariant where there are $n$ number of Weyl or Riemann curvatures contracted to one another, then the functional form of this  curvature invariant will be $(M/r^3)^n$ (upto some numerical factor). Overall, we can assume $M/r^3$ to be a rough estimate for invariant curvature quantity for Schwarzschild spacetime.
