# Birkhoff's theorem and Schwarzschild vacuum solution [duplicate]

Birkhoff's theorem states that any spherically symmetric solution of the vacuum field equations must be static and asymptotically flat, but the well-known Schwarzschild solution satisfies these conditions only for $$r\ge r_{S}$$.

What about the rest of spacetime?

The usual explanation, that behind the event horizon time and space exchange their role, leads to paradoxes. For example we cannot define that part of spacetime by the inequality $$r because the $$r$$ means time and the $$r_S$$ means space, clearly two different things. Even the term "behind the horizon" seems to be wrong. We should rather use the term "after the horizon has happened".

Another problem is the solution itself. As a time-dependent vacuum metric, it should be derived from the non-static Einstein equations.

But, for me, the biggest contradiction in the story of an astronaut crossing the horizon, or better, to whom the horizon has happened is that he follows further his timelike geodesic although the time and space directions have exchanged their places. He is not moving in space, in respect to the event horizon, but in time.

Moreover, the plausibility of this picture is severely challenged by the fact that we are trying to describe the properties of vacuum spacetime in terms of the motion of matter there. What happens to the astronaut after a finite period of his proper time? My conclusion is that he remains spatially on the event horizon, which has expanded in area to accommodate his mass.

Which of these conclusions is incorrect?

• Using the Schwarzschild coordinates at and inside the event horizon is not helpful. Use any of the coordinate systems that do not misbehave (Kruskal-Szekeres, Eddington, Gullstrand–Painlevé) and you avoid the problems. Commented Jun 4 at 15:37
• Your question seems a little unclear. For example you say he since then follows a space-like geodesic and no longer a time-like geodesic but the nature of a geodesic is coordinate independent. Since the geodesic is timelike in the astronaut's rest frame it is timelike, and timelike everywhere. Remember that coordinates are just labels we apply to points in spacetime and do not necessarily have the physical significance that we think they do in everyday life. The misbehaviour of the $r$ coordinate inside the horizon doesn't matter - it's just a coordinate. Commented Jun 4 at 15:47
• @JohnRennie The Schwarzschild coordinates are perfectly valid inside the horizon. No coordinate system is valid at the horizon, because the coordinate transformation there is singular and thus mathematically meaningless. “Removing” the coordinate singularity (by an invalid singular transformation) is a widespread misconception, if not a deception. Commented Jun 4 at 16:19
• @JanG You are correct on the horizon happening, something hardly anyone here understands. However, you really need to edit the question to fix references to geodesics, as everyone already has mentioned. For example: “After the event horizon has happened, there are no time-like circular geodesics in that part of spacetime.” - It should be “space-like”, not “time-like” here. And, as Yuktetez mentioned, this is incorrect. The angular coordinates $\phi$ and $\psi$ don’t change through (or after) the hirizon. (Cont.) Commented Jun 4 at 16:54
• Does this answer your question? Does Birkhoff's theorem hold inside the event horizon? Commented Jun 4 at 17:07

JanG wrote: »Schwarzschild solution satisfies these conditions only for $$r>r_S$$«

That's not true, the Schwarzschild metric is spherically symmetric all the way down to the singularity. Even if you replace the singularity with a collapsing star, the inner shells don't feel the outer shells as long as everything is spherically symmetric.

JanG wrote: »he since then follows a space-like geodesic and no longer a time-like geodesic«

Also wrong, see MTW exercise 13.5:

You can't go from timelike to spacelike. You will never overtake a photon, not even behind the horizon. The general consensus is that if you cross the horizon you notice nothing special, so you neither become a photon at the horizon nor a tachyon behind it.

JanG wrote: »We should rather use the term "after the horizon has happened"«

You seem to confuse the horizon at $$r=2$$ with the singularity at $$r=0$$. The horizon doesn't happen to everyone who hasn't crossed it yet since you're still free to fly away if you haven't crossed it yet. If you decide to cross it you're not free to escape the singularity anymore, though.

• While your answer is correct, you overlook the fact that the OP simply confused the geodesics terminology. He already admitted and explained it in a comment above and should really edit the question to correct it, but clearly there is no need to base your answer in his typo. Therefore the only tangible point in your answer is that the metric is spherically symmetric. While true, this does not answer the question and is really just a comment. Commented Jun 4 at 16:30
• @safesphere - if his question is a typo he needs to fix it, otherwise i'm gonna assume he asked what he asked. if you have a better answer feel free to write it down, this is all i got for now. Commented Jun 4 at 16:32
• Also, the OP is correct on the horizon “happening” to falling objects. You blame him for being confused while it is actually you who is being confused here. Commented Jun 4 at 16:34
• @JanG - If you fall from infinity so that your local velocity is simply the negative escape velocity v=-c√(rs/r) the spatial part of the metric becomes euclidean in your frame, see Gullstrand Painlevé coordinates where the local observers are such free fallers. In the classic Schwarzschild Droste coordinates the local observers are stationary, which in english means they'd have to be photons at the horizon or accelerated tachyons behind the horizon, that gives you weird sign flips which are unphysical. In the frame of any physical observer the metric is regular all the way down to the center Commented Jun 4 at 18:52
• When you finally hit the center you do so with an infinite relative velocity though, which is kind of like zero velocity with a flipped space time axis, but until you hit the center everything is regular, just strong tidal forces and the fact you can't escape. Commented Jun 4 at 18:58