Computing curvature singularities from a metric Suppose I have the metric
$$ds^2 = f(r)(dt^2-dr^2-dz^2) - \frac{1}{f(r)} d\phi^2. $$
How would you calculate the curvature singularities of this metric if we assume that $f(r)$ takes value $0$ for $r_0$?
 A: You have to compute some curvature scalars such as $R$, ${R_{\mu \nu }}{R^{\mu \nu }}$ etc. for your spacetime in order to find where those curvature invariants diverge. In order to understand whether there is singularity at all or not, it is enough to show that one of those curvature scalars diverges. But, the information related to the spacetime singularity may be lost for some curvature scalars such as $R$ and ${R_{\mu \nu }}{R^{\mu \nu }}$ due to the contraction of Riemann tensor (For example, in the case of the Schwarzschild metric, one obtains $R=0$ while the Kretschmann invariant diverges at the origin). Since the Kretschmann (scalar) invariant, ${R_{\mu \nu \delta \lambda }}{R^{\mu \nu \delta \lambda}}$, is a sum of squares of Riemann tensor components, it can be used to find the true (essential) singularities of a spacetime since, by definition, it preserves all the information about the singularities. So the Kretschmann invariant is preferred.
Considering the metric in your question, i.e.,
$$ds^2 = f(r)(dt^2-dr^2-dz^2) - \frac{1}{f(r)} d\phi^2,$$
the Kretschmann invariant is computed as (using the Maple's GRtensor package)
$${R_{\mu \nu \delta \lambda }}{R^{\mu \nu \delta \lambda }} = \frac{{12f{{(r)}^2}{{\left( {\frac{{{d^2}f(r)}}{{d{r^2}}}} \right)}^2} - 32f(r){{\left( {\frac{{df(r)}}{{dr}}} \right)}^2}\left( {\frac{{{d^2}f(r)}}{{d{r^2}}}} \right) + 27{{\left( {\frac{{df(r)}}{{dr}}} \right)}^4}}}{4{f{{(r)}^6}}}.$$
Next, more information about the metric function $f(r)$ is needed. At first glance, it seems that the Kretschmann invariant diverges at $r=r_0$ since $f(r_0)=0$. But, this is a naïve guess. To be sure, you have to put the explicit form of the metric function $f(r)$ in it and examine different limits such as $r \to 0$, $r \to r_0$ etc.
