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Imagine a black hole, shaped like a sphere, with a big hole through the centre, so that its centre of mass contains no matter.

If an object were dropped from outside the event horizon of the black hole, to fall through the hole, does the first law of thermodynamics, that energy can never be created or destroyed, imply that the object must pass through the hole and out of the event horizon on the other side of the black hole?

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closed as off-topic by David Hammen, Jon Custer, Qmechanic Feb 9 '17 at 21:39

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The situation you describe is not possible. We cannot simply make black holes in shapes of our choosing. A black hole is a solution to the equations of general relativity, and it is constrained by those equations. There is no way to make a black hole with a hole at its centre.

As it happens there is a solution to the equations of GR that is not too far from what you describe. This is the solution for a static charged black hole called the Reissner-Nordström metric. With this geometry objects can fall in through the event horizon, miss the singularity at the centre and pass back out of the event horizon again. Unfortunately this geometry takes an infinite time to form, so we are never going to see a Reissner-Nordström black hole in our universe.

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  • $\begingroup$ Are you saying that hyperdense material in the shape described couldn't happen in the universe, couldn't physically exist, or wouldn't be a black hole? $\endgroup$ – Issa Chanzi Feb 9 '17 at 7:31
  • $\begingroup$ @IssaChanzi: It couldn't physically exist. Any attempt to construct such an object would collapse into a regular Schwarzschild black hole. $\endgroup$ – John Rennie Feb 9 '17 at 8:26
  • $\begingroup$ Again it is not physically relevant, but in a five dimensional spacetime it is possible to create a black hole in the shape of a ring. $\endgroup$ – John Rennie Feb 9 '17 at 9:06

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