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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 $r<r_{s}$ 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?

JanG
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1 Answers1

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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:

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.

Yukterez
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