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When hovering 2km. above the horizon of a black hole with a mass of the sun, at r=5km., the distance you measure with a measuring tape attached to a mass you throw in the hole will tell you the distance to the hole's horizon is about 4km. See this answer.

Now when we look at the situation in the frame to which the mass is attached, you will fall through the horizon and hit the singularity in a finite time. All masses that fell in before you are in front of you, and all that will be thrown in after you, will end up behind you.

How can this be reconciled with the 4km. that the observer hovering above the horizon measures? Won't the measuring tape break or show a lot more than 4km.?

And suppose I throw in two masses from a hovering platform. Say one minute apart. They will both show 4km. But when they pass the horizon they will get pulled apart by the tidal force, so the measuring tape of the first seems to measure a greater distance than the second mass. How can this be resolved?

Il Guercio
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1 Answers1

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The answer you linked is about dangling a measuring tape above a black hole, not letting it fall freely. The end of the tape never crosses the event horizon, and its speed, relative to static coordinates, is zero (except while you're lowering or raising it, which you can do arbitrarily slowly). If you imagine that the tape is perfectly unstretchable, then the greatest length that you can dangle and then recover is 4 km; any more than that and you'll lose the end to the black hole, no matter how strong it is.

If you just let it fall, it will keep unspooling indefinitely. You never see any of it cross the horizon, but it appears to length-contract near the horizon, so an unlimited amount can fit outside – see Would something falling into a black hole appear to be flattened to an outside observer? and Nic Christopher's answer.

benrg
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