What exactly goes wrong with Schwarzschild coordinates at the event horizon?

It is well known that if one uses the Schwarzschild coordinates (t, r, $\theta$, $\phi$) to solve Einstein's equations, the components of the metric tensor blow up at the "event horizon", r = 2M (in units where c = G = 1).
It is also well known that this is an "artifact of using bad coordinates", a so-called coordinate singularity.
My questions is: What exactly goes wrong with the coordinate system at r = 2M ?

The standard examples of coordinate singularities that I have seen are something like polar coordinates at the origin of $R^2$ or longitude and latitude at the poles of a sphere and the usual problem is that multiple coordinate lines (of the form $\theta$ = constant) intersect at the same point.
Is that what happens in this case ? Or something else altogether ?

Intuitively, if a manifold is smooth, the only way I can imagine an inner product misbehaving when expressed as a metric tensor, is if the tangent vectors to some coordinate lines are not well-defined. But I can't quite see why exactly this is happening with the Schwarzschild coordinates.

Recall that a coordinate system is for $M$ is given by an open set $\Omega\in \mathbb{R}^n$ and a map $\psi: \Omega\to M$ that is a diffeomorphism between $\Omega$ and its image. Then the inverse $\psi^{-1}$ gives a local coordinate system $\psi^{-1}: \psi(\Omega) \to \mathbb{R}^4$.

In the Schwarzschild case it is useful to think in terms of the corresponding map $\psi$ and not the actual coordinate functions $\psi^{-1} = (r,t,\theta,\psi)$. In other words, what you want to do is look at the map that sends points in $\mathbb{R}^4$ with the coordinate value $(r,t,\theta,\psi)$ to points on the Schwarzschild manifold.

Looking at it this way, the first problem with the Schwarzschild coordinate is that when $r = 2M$, $$\psi(2M,t_1,\theta,\psi) = \psi(2M,t_2,\theta,\psi)$$ for any pair of $t_1, t_2$. This automatically implies that $\partial_t\psi =0$ and hence $\psi$ is not a diffeomorphism.

Another way to say this is that when $r = 2M$, the mapping $\psi$ corresponding to the inverse of the coordinate functions is independent of $t$; for every $t$, the image of $\psi(2M,t,\cdot,\cdot)$ is a fixed 2-sphere in the Schwarzschild manifold (the bifurcate sphere for the past and future event horizons).

This is analogous to the situation of the function $$u: \mathbb{R}^2 \to \mathbb{S}^2\subset \mathbb{R}^3$$ where $$(\theta,\phi) \mapsto (\cos\theta, \sin\theta \cos\phi, \sin\theta\sin\phi)$$ This is a smooth map and is (up to some notational changes) the common way of parametrizing the sphere (with $\theta\in [0,\pi]$ and $\phi\in [0,2\pi)$). When $\theta = 0$ or $\pi$, you get the north/south pole independent of $\phi$.

Wait, but you ask: the metric for the sphere in the $(\theta,\phi)$ coordinates degenerates, but it doesn't blow up! Why is it that the metric in the Schwarzschild case blows up?

The answer is that the problem with the $t$ coordinate is not the only problem with the coordinate system. The $r$ function can be geometrically defined as the area-radius function on the spherically symmetric manifold that is the Schwarzschild solution. It turns out that the function $r$ has a critical point exactly on the bifurcate sphere. The critical point is a "saddle point". Away from the bifurcate sphere $r$ is non-critical, and so we can define the smooth function $\psi$. But the criticality of $r$ (which is part of the inverse of $\psi$) implies that $\psi$ cannot be extended to a differentiable map near where the image is the bifurcate sphere.

Putting everything together: there are two things that go wrong with the Schwarzschild coordinates at $r = 2M$.

1. From the point of view of the mapping $\psi:\mathbb{R}^4\supset\Omega \to M_S$: when $r = 2M$ the function becomes constant in $t$, and hence is not a diffeomorphism.
2. Furthermore, since the "inverse" function $r:M_S\to\mathbb{R}$ has a critical point on the bifurcate sphere [which as we recall corresponds to exactly when $r = 2M$ in the Schwarschild coordinates], this means $\psi$ is also not differentiable at $r = 2M$, and so is not a diffeomorphism.

The first gives rise to the fact that one of the metric coefficients in the Schwarzschild metric being 0, the second gives rise to the fact that one of the metric coefficients become infinite.

• In "mathematician notation," one thinks of the metric $g$ as a rank (0,2) tensor, and then, as you describe, one component blows up and one goes to zero. It may be more natural to think about this in index notation, where we would consider $g_{\mu\nu}$ or $g^{\mu\nu}$ on more of an equal footing. The components of the upper-index metric that blow up are the ones that vanish in the lower-index metric, and vice versa. – Ben Crowell Feb 16 '17 at 15:54
• In other words, what you want to do is look at the map that sends points in $\mathbb{R}^4$ with the coordinate value $(r,t,\theta,\psi)$ to points on the Schwarzschild manifold. A difficulty here is that we don't know a priori what "the" manifold is. You can chop out $r \le 2M$, and then Schwarzschild coordinates are perfectly well behaved. Actually the maximal extension of the S. metric includes more stuff: a white hole and a 2nd copy of Minkowski space. As a simpler example, polar coords in the plane are well behaved if we remove a point, but the maximal extension restores the point. – Ben Crowell Feb 16 '17 at 16:05
• @BenCrowell: I read the question to mean "why is Schwarzschild coordinates on the Kruskal-extension bad". I admit I have not considered any other interpretation. – Willie Wong Feb 16 '17 at 20:35
• Thanks a lot, Willie. As Ben points out, I should have perhaps specified "Schwarzschild coordinates on a suitably extended manifold", but you addressed precisely the difficulty I was having. – Anindya Feb 17 '17 at 1:36

The Schwarzschild coordinates are defined such that the square root of the t-t component of the metric gives a time dilation factor relative to a distant observer. Such a definition stops working at the event horizon, where the time dilation diverges.

Intuitively, if a manifold is smooth, the only way I can imagine an inner product misbehaving when expressed as a metric tensor, is if the tangent vectors to some coordinate lines are not well-defined. But I can't quite see why exactly this is happening with the Schwarzschild coordinates.

To build intuition, you may want to look at Rindler coordinates, which are coordinates for flat spacetime that in a certain sense can be interpreted as the coordinates of an accelerated observer. (GR doesn't have any coordinate system that really has all the properties we would like for an accelerated frame.) Rindler coordinates have an event horizon, which is a misbehavior of the coordinate system very much like the misbehavior of Schwarzschild coordinates at the event horizon. The t-t component of the metric vanishes at the horizon, just as it does for the Schwarzschild metric. In both cases this is because the horizon is lightlike, and the metric measure of a lightlike vector is zero. (Although they are similar coordinate singularities, the event horizons are different in the sense that the existence of the Schwarzschild event horizon can be described in coordinate-independent terms as a boundary of a region from which a causal curve can never escape to null infinity.)

If you want to mentally classify coordinate singularities, you are unlikely to get a really satisfying answer. A coordinate singularity is a singularity that isn't a curvature singularity, but there is no known way to define a curvature singularity that everyone agrees makes sense in all cases. The classic paper on this, which is fairly readable, is Geroch, "What is a singularity in general relativity?," Ann Phys 48 (1968) 526. For instance, you can define a curvature singularity as one where a curvature polynomial blows up, but in 2+1 or more dimensions, curved spaces exist such that all their curvature invariants vanish: H-J Schmidt, http://arxiv.org/pdf/gr-qc/9404037v1.pdf . We want to rule out silly examples like Minkowski space with one point removed, and so on. The most widely accepted approach is to use geodesic incompleteness as a definition, but that still leaves a lot of ambiguity.