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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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  • Greg Egan gives a nice analysis of the related problem for a Rindler observer: https://www.gregegan.net/SCIENCE/Rindler/RindlerHorizon.html This is a good approximation for a black hole so massive that the spacetime curvature at its horizon is negligible. The big advantage is that the analysis doesn't use GR mathematics, only SR. – PM 2Ring Sep 18 '23 at 09:48
  • BTW, a $1 M_\odot$ BH has a Schwarzschild radius ~2.953 km, and the tidal force will rupture a 1 m long steel beam in freefall at 77 km. Of course, shell observers in the vicinity of the EH would be crushed to an atomic paste. ;) – PM 2Ring Sep 18 '23 at 09:50

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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  • Imagine we had $10^6$ km of luminous tape attached to the mass, and the mass can transmit back it's $r$ coordinate. If the tape can run out indefinitely, where does it go? We would be able to see the line of tape stretching from the platform to the mass and the distance to its $r$ coordinate would be less than 4 km. Is there a length-contraction effect I am not considering? Either way, your answer could do with more exposition of the last sentence. – ProfRob Sep 18 '23 at 06:31
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    @ProfRob There is a length-contraction effect. I added a link to another answer about it. (Incidentally, your earlier comment never pinged me for some reason.) – benrg Sep 18 '23 at 06:56
  • @ProfRob “Is there a length-contraction effect I am not considering?” - The gravitational length contraction is in the Schwarzschild metric formula (e.g. radial): $$d\tau^2=(Adt)^2-(dr/B)^2$$ The zero Ricci tensor conserves the infinitesimal spacetime volume. In physical terms this means that in vacuum the time dilation always equals the length contraction: $A=B$. If near the horizon your time is dilated $10$ times, then your length is radially contracted also $10$ times. Your misconception may be caused by overusing the raindrop coordinates. This answer is correct $+1$. – safesphere Sep 19 '23 at 04:57
  • @ProfRob “Imagine we had $10^6$ km of luminous tape” - Starting roughly at $1,km/s^2$, its free fall will take only several seconds before the whole million-kilometer-long tape is contracted to under one Planck length, as observed remotely from outside. – safesphere Sep 19 '23 at 05:30
  • @benrg While this answer is correct and I upvoted it, the existing answer to the question you link is wrong, as I explain in the comments. Your comment there also is incorrect, which is puzzling, because the gravitational length contraction is elementary (see above). While the velocity contraction adds to the effect, the objects hypothetically hovering close to the horizon are still length contracted exactly to the same degree as they are redshifted. Just look at the Schwarzschild metric. – safesphere Sep 19 '23 at 05:48