# Why can't we use the line element to distinguish coordinate from gravitational singularities?

I am a bit confused as to why we can't use the line element to identify coordinate from gravitational singularities. My question stems from learning about the Schwarzschild Metric and the singularity present at the Schwarzschild Radius $$R_s$$ where we have $$ds^2 = (1-\frac{R_s}{r})dt^2 -(1-\frac{R_s}{r})^{-1}dr^2 +r^2d\Omega^2$$ so then $$ds^2$$ goes to negative infinity as $$r$$ approaches $$R_s$$.

Because $$ds^2$$ is a scalar I would assume it is invariant under coordinate transformations so we should find that even if we switch out of Schwarzschild coordinates we should still get that $$ds^2$$ goes to negative infinity at $$R_s$$ although I have of course seen this is not the case, though I'm not sure exactly why. More fundamentally, I assume that $$(1-\frac{R_s}{r})^{-\frac{1}{2}}dr$$ can be used to find the radial distance between two events and in this case if one event is outside of the Schwarzschild Radius and the other is at the Schwarzschild Radius this tells us that the radial distance between them is infinite. Because this is a physical measurement, presumably it should be unaffected by what our choice of coordinates are, otherwise the metric tensor and therefore the line element would not really serve as the metric of our space and would not measure distances on our manifold which also seems like the wrong conclusion.

Note: I am specifically talking about a case where the mass producing this curvature is within the Schwarzschild Radius so the Schwarzschild solution still holds at the radius. Also, I know a bit about manifolds so I can somewhat follow along with the formalism of differential geometry but I am by no means an expert so I would greatly appreciate it if you can explain a bit of the differential geometry you use if that is the best way to answer the question. Thanks!

There is a delicate interplay of factors going on here, but first: $$ds^2$$ is not a scalar, it is a tensor. Specifically, $$ds^2$$ is just bad notation for the coordinate-invariant metric $$g$$ which may be written in coordinates/components as $$g=g_{\mu\nu}dx^\mu\otimes dx^\nu$$.

The singularity at the Schwarzschild radius you have pointed out is, in fact, a coordinate singularity. This should perhaps not be surprising when using the notation I have written above for the metric tensor -- what you have identified is a singularity in one of the components of $$g_{\mu\nu}$$, which is itself a coordinate-dependent quantity.

To identify intrinsic singularities, we need to use scalars, as you point out. The most common tool for doing this is the Riemann curvature and its derived invariants. For example, the Ricci scalar is, as its name would imply, a scalar. Hence if we were to compute the Ricci scalar and found it to have a singularity, there would be a true intrinsic singularity in the spacetime.

But the Ricci scalar is only one quantity and so finding a singularity in the Ricci scalar is good enough to establish the existence of a singularity, the converse does not hold: the lack of a singularity in Ricci scalar does not mean there isn't a singularity in the spacetime. Generally, we need all the curvature invariants. For example, we could build $$R^\mu_{\mu\alpha\beta}R^{\alpha\beta}$$ just to make something up. This is clearly a scalar quantity, and so we would need to check this also for more singularities to see if any were missed when we checked the Ricci scalar. There are infinitely many such invariants, but for the most common examples usually just checking the Ricci scalar is good enough.

I recommend looking in Carroll's GR book (or the nearly identical online notes). It is a generally good book on GR and a discussion of coordinate vs. intrinsic singularities can be found within it.

• I will just note here that Schwarzschild is ricci flat, so you will not detect the $r=0$ singularity by checking the ricci scalar alone (or at least without noting that there is a $\nabla^{2}\frac{1}{r}$ term in the ricci scalar). Nov 28, 2020 at 6:46
• Thanks for your reference to the Carroll, it does provide some nice formalism on what the spacetime interval actually is. However, I am still a bit confused since I thought the spacetime interval was invariant under coordinate transformations so although formally being a tensor, we should still not be able to get rid of coordinate singularities using a coordinate transform. Additionally, I still don't understand how the spacetime interval can be a coordinate dependent quantity and yet still measure distances in spacetime which are physical measurements and thus should be coordinate invariant. Nov 28, 2020 at 20:52
• @user446153 The spacetime interval is the intrinsic metric I have called $g$. This object is coordinate independent in the sense that for any pair of vectors $v=v^\mu\partial_\mu$ and $w=w^\mu\partial_\mu$ the scalar given by $g(w,v)=g_{\mu\nu}v^\mu w^\nu$ is coordinate independent. The singularities you are pointing to in the question are singularities in the components $g_{\mu\nu}$, not singularities in $g$. Nov 28, 2020 at 21:06