How to thoroughly distinguish a coordinate singularity and a physical singularity In a course on general relativity I am following at the moment, it was shown that the singularity $r=2M$ in the Schwarzschild solution is a consequence of the choice of coordinates. Introducing Kruskal-Szekeres coordinates $(u,v)$ resolves this problem: the singularity at $r=2M$ disappears, but if one draws a $(u,v)$ graphs with light cones and such, one still recognizes the event horizon at $r=2M$. The singularity at $r=0$ remains and is said to really be an essential singularity.
So in general: if you can find a coordinate transformation to get rid of a divergence in your metric, it is not a true singularity. However, it struck me that the Kruskal-Szekers coordinates were only discovered in 1960 (44 years after the Schwarzschild solution). 
This leaves me to wonder: is there a more systematic way of distinguishing physical vs. 'fake' singularities? In Carroll's book, I've read something about contractions of curvature quantities diverging at real singularities: E.g. $R^{\alpha \beta \gamma \delta}R_{\alpha\beta\gamma\delta}\propto r^{-6}$ such that $r=0$ is a real singularity (and $r=2M$ not). Could anyone make this ad-hoc rule more quantitative?
 A: I'm expanding on what @twistor59 says. Coordinatization of spacetime does not have physical consequences, it's is just some way you choose to parametrize the manifold (and some parametrizations might be singular). So whether there is actually something physically weird can be decided only by calculating the physical quantities ("observables"). Presumably, this qualitative behaviour must be independent of which coordinates you choose to describe the physics.
What are the physical observables which are independent of coordinatizations? They would have to be scalars, since otherwise they'd transform under a change of coordinates.
At the level of (torsion-free) GR with the Einstein-Hilbert action (no higher derivative terms), the only quantities/variables you can play with are: the metric, the connection (not really a tensor) and the curvature. So, nontrivial scalars must be built from the curvature (and the metric, when you contract indices). Wikipedia seems to have a handy list. :-)
If we include the Levi-Civita symbol $\varepsilon^{\mu \nu \rho \sigma}$ then we can also get $\det(g)$ ($\varepsilon^{\mu \nu \rho \sigma}$ is a pseudotensor but since the determinant has two of those, the quantity has a definite sign, independent of the handedness of your coordinates). I can't think of a good example illustrating how this quantity (~volume scale) tells us about singularities, or a reason for why it doesn't.
