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I'm going to propose a thought experiment, based on two ideas.

One: A uniform spherical shell, by the Shell Theorem, does not exert any gravitational force on objects existing in the interior of the shell.

Two: A black hole, created by matter dense enough to lie within its Schwarzschild radius, is inescapable, even by light.

Suppose (don't ask how) we create a uniform spherical shell composed entirely of matter dense enough to form a black hole, such that we have a continuous event horizon that appears as two concentric spherical regions. One can think of this as a spherical shell with an infinite number of infinitesimally sized black holes (at least, from a macroscopic point of view, to avoid Pauli's Exclusion Principle), or simply with a very large mass density across the surface (which is allowed thickness in the radial direction, to remain a three-dimensional construct). The overlapping event horizons make this system appear as a single black hole from the outside. Obviously, this system is unstable, and will collapse into a messy crunch fairly quickly, but before it does, its properties seem contradictory.

So, what happens on the inside? This question should probably be addressed for completely overlapping interior event horizons (such that no region inside the shell sits outside of the collective event horizon, and all of the black hole "cores" sit inside of the event horizon of every other black hole), and for some space existing between the event horizons of black holes on opposite sides of the shell, so that an event horizon free region exists within the shell. Will objects on the inside feel the effects of the gravitational force, or will it be a happy island of no external gravity (that is, simply flat space)? Does the answer vary based on what major theory is used to address it?


marked as duplicate by Qmechanic Nov 25 '18 at 6:37

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The shell theorem holds in full general relativity, a result known as Birchoff's theorem.

Nothing special happens on the inside until the matter collapses to the radius of your observer, or unless your observer is trying to see through the shell of matter to the outside world. This, of course, won't work. But that isn't a local measurement.


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