I Observing a Collapsing Shell: Time Dilation Explained

blademan9999
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Observer inside collapsing shell problem.
What does and observer inside of a collapsing shell observe? Lets say we have a shell of matter collapsing to a black hole. What would observers near the center see? How would the rest of the universe appear when,

The shell is approaching the Schwarzschild radius?

After the shell passes the Schwarzschild radius?

Time Dilation inside a hollow shell According to here you get time dialation, so what happens as the shell contracts?

Say the observer has a clock, and the inside of the shell is like a mirror, theres a clock attached to the shell and finally there's a clock at a great distance. How fast would these clocks go relative to eachother?

Until the shell reaches the observer, the observer is free to move aorund, so the issue to me seems to be that whe have a shell observer inside of a black hole, which normally you're not supposed to be able to have.
 
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See https://www.physicsforums.com/threa...of-a-singularity-before-it-is-formed.1017274/. A lot of your question is answered by that. Basically, the event horizon forms inside the shell and expands outward at the speed of light. Once it passes you (you can't detect anything locally as it does) you're doomed. You just don't see any local curvature until the shell passes you in the other direction.
 
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I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

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