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I've just answered Dipping a Dyson Ring below the event horizon, and while I'm confident my answer is correct I'm less certain about the exact consequences. To simplify the situation consider a diatomic molecule falling into a Schwarzschild black hole with its long axis in a radial direction.


The inner atom cannot interact with the outer atom because no radially outwards motion is possible, not even for electromagnetic fields travelling at the speed of light. However I find I'm uncertain exactly what the outer atom would experience.

We know that the outer atom feels the gravitational force of the black hole even though gravitational waves cannot propagate outwards from the singularity. That's because it's experiencing the curvature left behind as the black hole collapsed. Would the same be true for the interaction of the outer atom with the inner atom? Would it still feel an electromagnetic interaction with the inner atom because it's interacting with the field (or virtual photons if that's a better description) left behind by the inner atom? Or would the inner atom effectively have disappeared?

If the latter, presumably the fanciful accounts of observers falling into the black hole (large enough to avoid tidal destruction) are indeed fancy since it's hard to see how any large scale organisation could persist inside the event horizon.


I realise I didn't ask what I originally intended to. In the above question my molecule is freely falling and the question arose from a situation where the object within the event horizon is attempting to resist the inwards motion. I'll have to go away and re-think my question, but thanks to Dmitry for answering what I asked even if it wasn't what I meant :-)

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up vote 1 down vote accepted

I might be mistaken, but: Even though the molecule is inside the event horizon (relative to a distant observer), from the point of view of the molecule the event horizon is still ahead of it, and has not yet been reached. The inner atom would still be able to communicate with the outer atom well after they've both crossed the event horizon from our point of view.

It's only when the molecule is much closer to the singularity (i.e. about to be spaghettified), that the inner atom will disappear into the event horizon relative to the outer atom; any bond between them will be broken, and any charge of the inner atom will be added to the charge of the black hole.

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Thanks Dmitry. Since posting the question I've realised that I didn't ask quite what I intended to. –  John Rennie Oct 23 '13 at 14:06
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