What is the proof that the Schwarzschild metric is not static inside the horizon?

Question

In Lecture Notes on General Relativity, Sean M. Carroll shows that the Schwarzschild metric is not only stationary but also static (Chapter 7, page 169, Eq. 7.20 and following interpretation). On the other hand, in reply on Physics, Ben Crowell says that the Schwarzschild metric is not static inside the horizon. How is this to be understood?

Answer 1

As @John Rennie’s answer says, in order to have a static (region of) spacetime, you first need a timelike Killing field (in that region) which satisfies the extra condition of having an involutive orthogonal distribution (equivalently, the orthogonal distribution must be Frobenius-integrable).

The problem is that the black hole region doesn’t even have any timelike Killing fields, so it cannot be stationary, let alone static. One indication for this is that the coordinate vector field $\partial_t$ is now spacelike rather than timelike, however this does not imply that the black hole region is not stationary/static. There could be some other willy-wonka timelike Killing vector field in the black hole (but perhaps such that it becomes spacelike outside). Well, this actually does not happen, but we need to prove this (I won’t do it here though).

For notation sake, let $M=E\cup H\cup B$ be the smooth spacetime consisting of the Schwarzschild exterior, the horizon and the black hole region. The we have the following results the proof of which you can find in Barrett O’Neill’s Semi-Riemannian Geometry with Applications to Relativity (an excellent book with very clear treatment of the basics of Schwarzschild and FRW spacetimes):

1. $\xi$ is a Killing vector field on the exterior $E$ if and only if it equals $a\partial_t+\sigma$ for some constant $a$ and some vector field $\sigma$ which is the vertical lift of a Killing vector field on the round sphere $S^2$. In particular, the only timelikelike Killing vector fields on $E$ are non-zero constant multiples of $\partial_t$.
2. $\xi$ is a Killing vector field in the black hole $B$ if and only if it equals $a\partial_t+\sigma$ for some constant $a$ and some vector field $\sigma$ which is the vertical lift of a Killing vector field on the round sphere $S^2$. In particular, every Killing vector field on $B$ is spacelike.
3. $\xi$ is a Killing vector field in the black hole $B$ if and only if it equals $aT+\sigma$ for some constant $a$ and some vector field $\sigma$ which is the vertical lift of a Killing vector field on the round sphere $S^2$ (here $T=\frac{\partial}{\partial v}$ in Eddington-Finkelstein coordinates, which makes sense on all of $M$, and it restricts to the usual $\partial_t$ on $E$).

Note that the “if” direction is very easy and obvious; the “only if” direction is the more non-trivial one (it’s not entirely trivial because solving for $\xi$ such that $L_{\xi}g=0$ is actually a PDE, but a clever use of the various symmetries of Schwarzschild allows us the simplify the calculations). Also, you can’t really deduce statements 1 and 2 from each other unless you have some other fancy result about extending Killing vector fields from $E$ to $E\cup B$ or from $B$ to $E\cup B$ respectively; hence why I listed these two statements separately. Finally, statement 3 follows immediately from (1) or (2) and the fact that $(M,g)$ is a real-analytic Lorentzian manifold, so the result follows trivially by uniqueness of analytic continuation.

Note that for your question, it is only statement (2) that matters (I included the rest just for fun). All Killing vector fields on $B$ are spacelike, and so in particular none are timelike, hence $B$ is not stationary, and hence not static.

Answer 2

Formally, a spacetime is static if it admits a global, non-vanishing, timelike Killing vector field $K$, which is irrotational, i.e., whose orthogonal distribution is involutive.

(source Wikipedia)

Outside the horizon the $t$ coordinate gives the timelike Killing field $\partial_t$, but inside the horizon the $t$ coordinate is spacelike not timelike so there is no timelike Killing field. Hence the interior is not static.

Answer 3

Even in the classic Schwarzschild-Droste coordinates the metric is independent of the coordinate time t, but since that t is just the proper time of a stationary observer at infinity it might be more convincing to transform to coordinates that use the time coordinate of free-falling (Gullstrand-Painlevé) or proper-accelerating (Eddington-Finkelstein) local observers.

Then you will find that the metric is still independent of time, meaning it doesn't make a difference when you jump into the black hole. It will always take you the same amount of proper time to fall from one position to an other, you will encounter the same tidal forces at any given position on your way, and all the other invariants will also be the same.

It seems to me that the argument that the Schwarzschild metric wasn't static inside builds on either an overzealous definition of the word static, or even a misconception about coordinates.

The classic coordinates where the sign flip occurs use accelerated tachyons as local observers inside the horizon since those are the only ones capable of staying stationary with respect to the fixed stars outside and the singularity inside, so in that sense it is only natural that the time coordinate of the fixed stars is spacelike to them.

Before the proper coordinate transformations were discovered that used to be a common misconception, but nowadays we know the structure of spacetime doesn't change abruptly at the horizon, since as you cross it everything is smooth and you notice nothing special, at least from the relativistic perspective.

If we ain't talking about an eternal Schwarzschild but an astrophysical black hole formed by a collapsing star as described by the Oppenheimer-Snyder metric we have a different story though, in that case the interior is not the usual static vacuum solution but that of collapsing matter, which is locally dynamic, although still frozen due to time dilation from the outside perspective.