# 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 $$r_0$$?

• Compute a few scalars like the Ricci and Kretschmann scalars and see if they take infinite values anywhere. – John Rennie May 22 at 15:17
• Your answer most likely depends on the metric function $f(r)$. So what is its behavior at origin ($r \to 0$)? And note that, all information about curvature singularities is stored in Kretschmann scalar. – SG8 May 22 at 15:51
• @SG8 all information about curvature singularities is stored in Kretschmann scalar Do you have a reference for this claim? I’m skeptical of it, since there are other curvature scalars. – G. Smith May 22 at 16:54
• @G.Smith - The best way to determine the singularities of a spacetime is to examine all components of the Riemann tensor, $R_{\mu \nu \rho \lambda}$. But, the components of the Riemann tensor depend on the coordinate representation of a spacetime. Since the Kretschmann 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 singularities of a spacetime (by definition it preserves all the information about the singularities). – SG8 May 22 at 17:51
• @G.Smith - This may be more evident from this relation $${R_{\mu \nu \delta \lambda }}{R^{\mu \nu \delta \lambda }} = {C_{\mu \nu \delta \lambda }}{C^{\mu \nu \delta \lambda }} + \frac{4}{{D - 2}}{R_{\mu \nu }}{R^{\mu \nu }} - \frac{2}{{(D - 1)(D - 2)}}{R^2}$$ – SG8 May 22 at 17:52

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.
$$ds^2 = f(r)(dt^2-dr^2-dz^2) - \frac{1}{f(r)} d\phi^2,$$
$${R_{\mu \nu \delta \lambda }}{R^{\mu \nu \delta \lambda }} = \frac{{12f{{(r)}^2}{{\left( {\frac{{{d^2}f(r)}}{{d{r^2}}}} \right)}^2} - 32f(r){{\left( {\frac{{df(r)}}{{dr}}} \right)}^2}\left( {\frac{{{d^2}f(r)}}{{d{r^2}}}} \right) + 27{{\left( {\frac{{df(r)}}{{dr}}} \right)}^4}}}{4{f{{(r)}^6}}}.$$
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.