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Suppose I'm hovering 1 meter away from the horizon of a Schwarzschild black hole and I'm holding a 2 meter long metal rod. Now I thrust the rod towards the horizon? What do I observe?

A satisfactory answer to this question needs to include a discussion of the following:

  1. When the Schwarzschild radius is much larger than 1 meter.

  2. When the Schwarzschild radius is much smaller than 1 meter.

  3. What happens to the rod? Probably some time dilation happens and I can't actually see it going through the event horizon.

  4. People often say that in a big enough black hole, the horizon isn't anything special because gravity isn't strong. But if I can't even move my stick by a meter, something really special must be happening?

  5. If I can indeed observe the stick going through the horizon, what happens when I pull back on it?

Please ignore quantum effects. The question is labeled as a GR question, nothing else.

Qmechanic
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Ryan Unger
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  • Possible duplicates: https://physics.stackexchange.com/q/126929/2451 and links therein. – Qmechanic Aug 01 '19 at 17:31
  • See also https://physics.stackexchange.com/q/158195/ – Ryan Unger Aug 01 '19 at 17:31
  • Re: "in a big enough black hole...gravity isn't strong." Don't conflate the gradient of the gravity field with its absolute strength. If the BH is sufficiently large, then you will not be spaghettified, if you are in free fall, because the force pulling your nose toward the center won't be hugely different from the force pulling your toes, but in the scene that you describe, both of those forces will be astronomically, mind-bogglingly, incomprehensibly huge. – Solomon Slow Aug 01 '19 at 18:32
  • You'd need a stick of an infinite length to reach the event horizon from a hovering position. Space around a black hole is curved in such a way that it infinitely expands in the radial direction toward the event horizon according to the Schwarzschild metric. – safesphere Aug 01 '19 at 20:26
  • @safesphere no it doesn't. The proper radial distance $s(r) = \int_{r_s}^r (1-r_s/r')^{-1/2}dr'$ for the Schwarzschild metric is finite. See length contraction in a gravitational field. – John Rennie Aug 02 '19 at 09:35
  • @JohnRennie Thanks John! I appreciate the clarification. So space does infinitely expand in the sense of $ds/dr\to\infty$, but the distance doesn't diverge. Interesting. I'll need to ponder what Kevin Brown meant here: "a hovering observer [...] is actually a great distance (perhaps even light-years) from the $r = 2m$ boundary, even though he is just $1$ inch above" - https://www.mathpages.com/rr/s7-03/7-03.htm – safesphere Aug 02 '19 at 20:29

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