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If there is a spherically symmetric shell collapsing to form a black hole and there is an observer at the center of the shell, will he be instantly destroyed by the singularity (assuming no quantum gravity) when the radius of the shell reaches the Schwarzschild radius? If not, would the observer see the shell moving at the speed of light relative to him when it reaches its Schwarzschild radius?

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The shell would be invisible. The observer inside the collapsing shell observes no gravitational potential by Gauss' law. The shell interior to it experiences gravtational field and implodes. Our observer can only observe the shell by shining a light on it. The velocity of the shell would be $v~=~c(2m/r)$ that approaches the speed of light as $r~\rightarrow~2m$. As this reaches $r~<~2m$ the observer will never again receive a reflected photon from the imploding shell. An observer some radial distance from the imploding shell and free falling behind it also fails to see it after $r~<~2m$. The shell disappears.

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