# How can the distance to the event horizon, as measured by a tape attached to a falling mass, be reconciled with the mass passing through it?

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?

• Greg Egan gives a nice analysis of the related problem for a Rindler observer: 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. Commented Sep 18, 2023 at 9:48
• 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. ;) Commented Sep 18, 2023 at 9:50

• 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. Commented Sep 18, 2023 at 6:31
• @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 $+1$. Commented Sep 19, 2023 at 4:57
• @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. Commented Sep 19, 2023 at 5:30