# What is the difference between coordinate- and curvature-singularities?

I recently came across the following question:

The Schwarzschild metric in ingoing Eddington coordinates has the line element $$d s^{2}=-\left(1-\frac{2 M}{r}\right) d v^{2}+2 d v d r+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)$$ What are the coordinate singularities? What are the curvature singularities?

The first issue is that I don't really know what the difference between these two things is. If I had to guess, I would say that the coordinate singularities are singularities that are due to a poor choice of coordinate system (I have the Schwarzschild radius in mind, in the standard form of the line element) while curvature singularities are "true" singularities, i.e. they exist regardless of your choice of coordinate system.

If this is true, it seems quite straightforward to read these things of just from looking at the line element. In the above example I would have said that there are no coordinate singularities and only one true singularity ($$r=0$$). Are there examples where it is not so easy to see/guess these two things?

One additional thing that confuses me a bit about these two concepts is that people sometimes talk about "curvature invariants" $$R^{\mu\nu}R_{\mu\nu}$$ ($$R_{\mu\nu}$$ being the Ricci tensor) and $$R_{\alpha\beta\gamma\delta}R^{\alpha\beta\gamma\delta}$$ ($$R_{\alpha\beta\gamma\delta}$$ being the curvature tensor) and seem to make statements about singularities from these two quantities. How are these two things linked to curvature and coordinate singularities?

You are correct in your definitions of coordinate and curvature singularities.

Note that the scalar resulting from the full contraction of a tensor is a coordinate independent quantity.

If you can find a coordinate system in which a tensor is zero then it is necessarily zero in all coordinate systems (a nice example is that we can always find a coordinate system such that the Christoffel symbols vanish since it is not a tensor, but we can't always find a coordinate system such that the Riemann tensor vanishes, since this would imply zero curvature in all coordinate systems).

By the same reasoning if you can find a coordinate system in which a tensor is singular (blows up to infinity) then it is necessarily singular in all coordinate systems. Hence $$R_{\alpha\beta\gamma\delta}R^{\alpha\beta\gamma\delta}$$ being singular implies that we haven't just made a poor choice of coordinates, there is an unavoidable singularity at the point in spacetime at which we are evaluating the tensor.

• So just to complete the loop of reasoning, if you want to find curvature singularities, you should calculate the Ricci scalar and see where it blows up? Would there be different results if you calculated the square of the Ricci tensor versus the Ricci scalar? Aug 23, 2020 at 13:19
• Well, the Schwarzschild solution is a vacuum solution, $T_{\mu\nu}=0$ and $R_{\mu\nu}=0$, so the square of the Ricci tensor and the Ricci scalar are both identically zero. It's a good exercise to show that $T_{\mu\nu}=0$ implies $R_{\mu\nu}=0$. Aug 23, 2020 at 17:35

A curvature singularity is a real physical singularity. All observers should agree that at that point the curvature is infinite. A curvature invariant is of course co-ordinate invariant. If for example the Kretschmann scalar (the norm of the Riemann tensor, the Riemann tensor squared) blows at that point for a given metric then if we perform a coordinate redefinition we will not change the geometry of spacetime, we change the coordinates we measure with, so the Kretschmann scalar will still be infinite at that point.

In your case the Ricci tensor squared will be zero because you are in vaccum.