Is there anything beyond the event horizon of a black hole?

Question

About 10 years ago, I read Schwarzschild's paper on black holes, and I noticed, in particular, that the roles of the $r$ and $t$ coordinates reverse in their meanings when some mass goes from outside the event horizon (where $r$ is the spatial, radial variable, and $t$ is the time variable) to inside the event horizon (where $r$ is time and $t$ is spatial). My reaction to that was to conclude that there is nothing inside the event horizon. Shortly after that, I discovered that Dirac had published a paper in 1962 concluding the same thing.

Answer 1

I turns out that nothing particularly significant happens to a local observer as they cross the event horizon of a typical black hole. The structure of spacetime is actually smooth across the horizon, with equally well behaved exterior and interior regions.

The event horizon (at $r_{S}=\frac{2GM}{c^{2}}$ for a Schwarzschild black hole) marks a change in the global geodesic structure; once something is inside, it can never subsequently escape back out to infinity. However, locally there is no physical singularity at $r_{S}$, in spite of the coordinate artifact in the Schwarzschild metric $$ds^{2}=-\left(1-\frac{r_{S}}{r}\right)\,dt^{2}+\left(1-\frac{r_{S}}{r}\right)^{-1}\,dr^{2}+r^{2}\left(d\theta^{2}+\sin^{2}\theta\,d\phi^{2}\right).$$ (Henceforth, I set $c=1$.)

The fact that the metric components can diverge and change sign is actually an artifact of a somewhat poor choice of coordinates. The variables $r$ and $t$ are very useful at large distances away from the black hole singularity, but close in they are inopportune. Written in terms of $r$ and $t$, the metric behaves badly around $r=r_{S}$, but there is no actual singularity in the curvature tensor $R^{\mu}{}_{\nu\rho\sigma}$. The Riemann tensor $R^{\mu}{}_{\nu\rho\sigma}$ is finite even for $0<r\leq r_{S}$; and, in fact, its Ricci part vanishes, $R_{\mu\nu}=0$, everywhere except at $r=0$, since the Schwarzschild metric is a vacuum solution of the Einstein field equations.

We can make the fact that the spacetime continues smoothly down past the event horizon manifest by choosing different coordinates—the Kruskal–Szekeres coordinates, $$T=\left(\frac{r}{r_{S}}-1\right)^{1/2}\exp\left(\frac{r}{2r_{S}}\right)\sinh\left(\frac{t}{2r_{S}}\right)\\ X=\left(\frac{r}{r_{S}}-1\right)^{1/2}\exp\left(\frac{r}{2r_{S}}\right)\cosh\left(\frac{t}{2r_{S}}\right)$$ outside the horizon ($r\geq r_{S}$); and $$T=\left(1-\frac{r}{r_{S}}\right)^{1/2}\exp\left(\frac{r}{2r_{S}}\right)\cosh\left(\frac{t}{2r_{S}}\right)\\ X=\left(1-\frac{r}{r_{S}}\right)^{1/2}\exp\left(\frac{r}{2r_{S}}\right)\sinh\left(\frac{t}{2r_{S}}\right)$$ inside ($r\leq r_{S}$). Note that $T$ and $X$ are continuous and nonsingular through $r=r_{S}$. In these coordinates, the metric takes a form that is free of singularities except at $r=0$, $$ds^{2}=\frac{4r_{S}^{3}}{r}\exp\left(-\frac{r}{r_{S}}\right)\left(-dT^{2}+dX^{2}\right)+r^{2}\left(d\theta^{2}+\sin^{2}\theta\,d\phi^{2}\right).$$ This form is slightly awkward, since it still involves the radial coordinate $r$ (which is a function of $X$ and $T$). Moreover, not all values of $X$ and $T$ are permitted; $T^{2}-X^{2}$ must be less than 1. (Note that $T^{2}-X^{2}=1$ corresponds to the singularity at $r=0$.) However, the event horizon lies at $T^{2}-X^{2}=0$, and in these coordinates, there is nothing singular there at all; not only is the curvature tensor finite in all its components, but so is the metric tensor $g_{\mu\nu}$! We must conclude that the there is nothing locally special about the event horizon, and the singularities that occur there are just an artifact of choosing poor coordinates.